问:
Precise definition of the support of a random variable
$\newcommand{\F}{\mathcal{F}} \newcommand{\powset}[1]{\mathcal{P}(#1)}$
I am reading lecture notes which contradict my understanding of random variables. Suppose we have a probability space $(\Omega, \mathcal{F}, Pr)$, where

$\Omega$ is the set of outcomes

$\F \subseteq \powset{\Omega}$ is the collection of events, a $\sigma$-algebra

$\Pr:\Omega\to[0,1]$ is the mapping outcomes to their probabilities.

If we take the standard definition of a random variable $X$, it is actually a function from the sample space to real values, i.e. $X:\Omega \to \mathbb{R}$.

What now confuses me is the precise definition of the term support.

According to Wikipedia:

the support of a function is the set of points where the function is
  not zero valued.

Now, applying this definition to our random variable $X$, these lectures notes say:

Random Variables – A random variable is a real valued function defined
  on the sample space of an experiment. Associated with each random
  variable is a probability density function (pdf) for the random
  variable. The  sample space is also called the support of a random
  variable.

I am not entirely convinced with the line the sample space is also callled the support of a random variable.

Why would $\Omega$ be the support of $X$? What if the random variable $X$ so happened to map some element $\omega \in \Omega$ to the real number $0$, then that element would not be in the support?

What is even more confusing is, when we talk about support, do we mean that of $X$ or that of the distribution function $\Pr$?

This answer says that:

It is more accurate to speak of the support of the distribution than
  that of the support of the random variable.

Do we interpret the support to be

the set of outcomes in $\Omega$ which have a non-zero probability,

the set of values that $X$ can take with non-zero probability?

I think being precise is important, although my literature does not seem very rigorous.

答:
You appear to have misquoted these lecture notes.  What they actually say is

Random Variables – A random variable is a process, which when followed, will result in a numeric output. The set of possible outputs is called the support, or sample space, of the random variable. Associated with each random variable is a probability density function (pdf) for the random variable.

It appears to me that the author of the notes is simply using a very old notion of random variable that predates Kolmogorov's definition of it as being a function on the set of outcomes of a probability space, as has now been almost universally adopted by modern mathematicians.
The notes appear to have been written for the probability unit of a course in actuarial science. They also seem to me to depart from standard modern mathematical terminology in at least one other respect—namely, in using the term "probability density function" not only in its usual sense, but also to refer to the probability mass function of a discrete distribution.  Wikipedia does say, however, that the probability mass function is sometimes also referred to as a "discrete density function" (a term I can't recall ever having previously come across), so perhaps this apparent departure from standard terminology is not quite as egregious as I had originally thought.

What these notes are calling the "support, or sample space" of a random variable is not, as your misquotation has characterised it, the domain of a function constituting the random variable, but the set of possible numerical values which the random variable can have. It is essentially the same thing as what would now be called the support of the random variable's distribution in standard modern mathematical terminology.

If the set of outcomes $\ \Omega\ $ of a probability space is a topological space, with topology $\ \mathscr{T}\ $, say, and the probability $\ \text{Pr}\ $ is a Borel probability measure, then the support $\ supp_{\,\text{Pr}}\ $ of $\ \text{Pr}\ $, as Stefan Hansen and Did have pointed out in the comments, is the closed set equivalently defined as either
$$
supp_{\,\text{Pr}}=\left\{x\in\Omega\,|\,\text{Pr}(\mathscr{O})>0\ \text{ for all }\ \mathscr{O}\in\mathscr{T}\ \text{ with }\ x\in\mathscr{O}\right\}\ ,
$$
or
$$
supp_{\,\text{Pr}}=\bigcap\left\{\mathscr{C}\,|\,\text{Pr}(\mathscr{C})=1\ \text{ and }\ \mathscr{C}^{\,c}\in\mathscr{T}\right\}\ .
$$
It is easy to show that these definitions are equivalent, and that $\ supp_{\,\text{Pr}}\ $ is closed. While it is not necessarily true that $\ \text{Pr}\big(supp_{\,\text{Pr}}\big)=1\ $ in the most general case, it is true in most cases of practical interest.  It is true, for instance, if $\ \Omega\ $ is a $\sigma$-compact metric space.

If $\ X:\Omega\rightarrow\mathbb{R}\ $ is a real-valued random variable, then the distribution $\ \mu_X\ $ of $\ X\ $ is the probability measure  $
\mu_X:\left\{S\subseteq\mathbb{R}\,|\,X^{-1}(S)\in\mathcal{F}\,\right\}\rightarrow$$[0,1]\ $ defined by
$$
\mu_X(S)=\text{Pr}\left(X^{-1}(S)\right)\ .
$$
If $\ X\ $ is Borel measurable, then $\ supp_{\,\mu_X}\ $ is a closed subset of $\ \mathbb{R}\ $ with $\ \mu_X\big(supp_{\,\mu_X}\big)=1\ $, and it is the smallest such closed subset.  If $\ X\ $ has a probability density function $\ f\ $, then $\ supp_{\,\mu_X}\ $ will also be the support of $\ f\ $, as that term is normally defined—namely the closure of the set of points $\ x\ $ for which $\ f(x)\ne0\ $.  Although the lecture notes you cite don't define random variables as functions on sample spaces, their concept of the "support of a random variable" is essentially the same thing as the set $\ supp_{\mu_X}\ $ I've just described.

I also have my doubts about the extent to which the general definition for the support of a function which you quote from Wikipedia is used in practice.  I've previously only ever seen this notion defined as it is in Wikipedia's second sentence:

If the domain of $\ f\ $ is a topological space, the support of $\ f\ $ is ... defined as the smallest closed set containing all points not mapped to zero.

where I have replaced the possibly redundant "instead" with ellipses.  Here, for instance, is the online Encyclopedia of Mathematics's definition, here is Wolfram  MathWorld's definition, and here is planetmath's definition.  As it happens, the support of a function with respect to the discrete topology is exactly the set of points on which the function does not vanish, so Wikipedia's general definition is equivalent to taking the discrete topology as being the default when no other topology has been specified.  However, I've never come across an instance in technical mathematical literature where the support of a function has been defined in this way.

As far as I can recall, I have never seen an instance in modern mathematical literature where the support of any probability measure, even when it is the distribution of a random variable, has been called the support of a random variable, and, in my opinion, it would be a poor choice of terminology to use the term in that sense. It's quite conceivable to me, however, that if the sample space on which a random variable $\ X\ $ is defined happens to be a topological space, then studying the support of $\ X\ $, considered as a function on that sample space, in the sense in which the term "support of a function" is defined by the online Encyclopedia of Mathematics, Wolfram  MathWorld and planet math, is something one may well wish to do. It seems to me, therefore, that use of the term "support of $\ X\ $" in that sense would be completely unobjectionable.


问:
If X is a topological space, U is open in X, A is dense in X, then cl(A∩U)=cl(U)
One direction is trivial so I just need to show any element in closure of U is in the other set.

If x is in cl(U) then any open neighborhood excluding x intersects U non trivially. Since A is dense the open neighborhood intersects A non trivially. If I can show that x is a limit point of A∩U, I would be done, however how do I show that this open neighborhood intersects U and A at the same points? I can imagine situations where the open neighborhood of x intersect U and A but not the points that U and A share in common.

答:
Let $x\in\overline U$. Then for every open set $V$ such that $x\in V$ we have $V\cap U\ne\varnothing$.

Now, $V\cap U$ is open and nonvoid, so the denseness of $A$ guarantees that $A\cap (U\cap V)\ne\varnothing$. So, given the set $A\cap U$ for every set $V$ with $x\in V$ we have that $(A\cap U)\cap V=A\cap (U\cap V)\ne\varnothing$, wich is the definition of $x\in \overline{A\cap U}$.

ContentLicense: CC BY-SA 4.0


答:
You haven't used that $U$ is open yet...

If $x \in \operatorname{cl}(U)$, let $V$ be any open neighbourhood of $x$; we have to show $V$ intersects $A \cap U$.

Firstly, $V$ intersects $U$ as $x \in \operatorname{cl}(U)$, so pick $y \in V \cap U$, and $V \cap U$ is an open neighbourhood of $y$ so it must intersect $A$ as $A$ is dense and so $y \in \operatorname{cl}(A)=X$, ergo:

$$\emptyset \neq (U \cap V) \cap A  = V \cap (U \cap A)$$

and so we're done, $x \in \operatorname{cl}(A \cap U)$ as $V$ was arbitrary.

Title: If X is a topological space, U is open in X, A is dense in X, then cl(A∩U)=cl(U)

Body:
One direction is trivial so I just need to show any element in closure of U is in the other set.

If x is in cl(U) then any open neighborhood excluding x intersects U non trivially. Since A is dense the open neighborhood intersects A non trivially. If I can show that x is a limit point of A∩U, I would be done, however how do I show that this open neighborhood intersects U and A at the same points? I can imagine situations where the open neighborhood of x intersect U and A but not the points that U and A share in common.


问:
How to compute the indefinite integral $\int \frac u{u+1}\,\mathrm du$?
How do you compute $$ \int \frac{u}{u+1}\,\mathrm du$$</p>

答: integrate by parts using : $f = u$ and $g' = 1/ (1+u)$. this solves the problem</p>

ContentLicense: CC BY-SA 3.0


问:
Binomial Theorem Inequality
How do I proof the right side of the following inequality:

$(x^p+y^p)\leq(x+y)^p\leq2^{p-1}(x^p+y^p)$ for $0\leq x,y$ and $p > 1$

The left side can be easily proved with the binomial theorem and I assume that the right side can be proved with $\sum_{n=0}^{p}\binom{p}{n}=2^p$, but I can't get it working.

答:
Hint

Neither inequality is true unless you place some restrictions on the values of $\ x\ $, $\ y\ $ and $\ p\ $.  However, the inequality
$$
\left(x+y\right)^p\le2^{p-1}\left(x^p+y^p\right)
$$
is equivalent to
$$
\left(\frac{x}{2}+\frac{y}{2}\right)^p\le\frac{x^p+y^p}{2}\ ,
$$
and this will hold if the function $\ x\mapsto x^p\ $ is convex on its domain of definition.  I'll leave it to you to determine the values of $\ p\ $ for which this is the case.

Title: Binomial Theorem Inequality


问:
if $2a_n+b_n$ and $a_n+3b_n$, both converge to a finite limit, then do $a_n,b_n$ also converge to a final limit?
What I did: if if $lim_{n\to \infty}2a_n+b_n=L_1$ and $lim_{n\to \infty}a_n+3b_n=L_2$.

Then $lim_{n\to \infty}(2a_n+b_n-2(a_n+3b_n))=L_1-2L_2=lim_{n\to\infty}(-2b_n)$.

But here I'm wondering, I know that If I know that $c_n$,$d_n$, converge then $lim_{n\to\infty}c_n*d_n$ converges.

Here I know that $lim(-2)=-2$, but I don't know that $lim_{n\to\infty}b_n$ converges, it's what I'm trying to prove, so my question is:
 Is it valid to say that if $lim_{n\to\infty}(-2b_n)=L_1-2L_2$, then $lim_{n\to\infty}b_n=\frac {L_1-2L_2}{-2}$?

答:
If you know that $c_n$ converges to $L$ then $Ac_n$ converges to $AL$ for any constant $A$

If $A=0$ this is trivial, and otherwise I can choose $N$ so that with $n\gt N$ we have $|c_n-L|\lt \frac {\epsilon}{|A|}$ (using the freedom I have to beat any small number I choose) and $|Ac_n-AL|\lt \epsilon$.

You are right to ask questions about this - generally limits work when you add and multiply (you should see how much you can prove yourself), but you have to be very careful about dividing by zero (in cases where you are dividing by $a_n$ and $a_n$ has a non-zero limit, it is prudent to choose $n$ large enough that you can bound $a_n$ away from zero).


答:
RHS is mean power inequality:If $p>1$, then $$M_p>M_1$$

$$\implies \left(\frac{x^p+y^p}{2}\right)^{1/p} >\frac{x+y}{2}~~~(1)$$
$$\implies \left(\frac{x^p+y^p}{2}\right) >\left(\frac{x+y}{2}\right)^p~~~(2)$$
Also (2) is Jensen's inequality for $f(x)=x^p \implies f'(x)=px^{p-1}, f''(x)= p(p-1)x^{p-2}>0, ~if~ p>1.$$

Title: Binomial Theorem Inequality

Body:
How do I proof the right side of the following inequality:

$(x^p+y^p)\leq(x+y)^p\leq2^{p-1}(x^p+y^p)$ for $0\leq x,y$ and $p > 1$

The left side can be easily proved with the binomial theorem and I assume that the right side can be proved with $\sum_{n=0}^{p}\binom{p}{n}=2^p$, but I can't get it working.


问:
Is there an open mapping theorem for affine morphisms?
Let $A$ and $B$ be rings. If $\varphi : A \longrightarrow B$ is such that $^a\varphi : Spec(B) \longrightarrow Spec(A)$ is bijective, then in what conditions $^a\varphi$ is a homeomorphism? Or, more general, in what conditions $^a\varphi$ is open if one assume only surjectiveness?

If I pick the constructible topology on the spectrum (applying the functor $(\bullet)^{Bool}$ that makes the space Boolean), then it will be a homeomorphism. So I was thinking about creating some way of returning back any information from the Boolean space to the original space when I restrict $A$ and $B$ to be some kind of rings.

Thanks in advance.

答: If $A$ is normal, and $^a\varphi$ bijective, then $^a\varphi$ is an isomorphism of affine varieties. This is Corollary 17.4.6 in Tauvel & Yu (Lie Algebras and Algebraic Groups), I am sure it can be found in other places as well.</p>


问:
Min/max dice problem

Suppose Anne throws two fair dice and the faces show $4$ and $3$. It is your turn and Anne states that if both dice show a number greater than $4$ and $3$, you will win the game. What is the probability of you winning the game? (i.e. if your outcome is $4$ and $5$ you will win, as the biggest gets paired with the biggest and smallest with the smallest).

By listing the favorable possibilities, we obtain:

$$\begin{aligned}
&(4,5)&\qquad (4,6) &\qquad (5,6) &\qquad (5,5)&\\
&(5,4)&\qquad (6,4) &\qquad (6,5) &\qquad (6,6)&
\end{aligned}$$

so the proability must be $8/36$ or $2/9$. However, I am interested in knowing why the following way does not work.
Let $X_1\sim\text{Unif}\,\{1,6\}$ and $X_2\sim\text{Unif}\,\{1,6\}$. I need

$$\begin{aligned}
P(\max\{X_1,X_2\}>4,\min\{X_1,X_2\}>3)&=P(\max\{X_1,X_2\}>4)P(\min\{X_1,X_2\}>3)\\
&=\left[1-P(\max\{X_1,X_2\}\leq4)\right]P(X_1>3,X_2>3)\\
&=\left[1-P(X_1\leq4,X_2\leq4)\right]P(X_1>3)P(X_2>3)\\
&=\left[1-P(X_1\leq4)P(X_2\leq4)\right]\left[1-P(X_1\leq3)\right]\left[1-P(X_2\leq3)\right]\\
&=\left(1-\frac{4}{6}\cdot\frac{4}{6}\right)\left(1-\frac36\right)\left(1-\frac36\right)\\
&=\left[1-\left(\frac{4}{6}\right)^2\right]\left(1-\frac36\right)^2=\frac5{36}.
\end{aligned}
$$

What is wrong with this approach?

Another approach I considered is,
$$\begin{aligned}
P(\max\{X_1,X_2\}>4,\min\{X_1,X_2\}>3)&=P(\{X_1>4,X_2>3\}\cup\{X_1>3\,X_2>4\})\\
&=P(\{X_1>4,X_2>3\})+P(\{X_1>3\,X_2>4\})\\ &=P(X_1>4)P(X_2>3)+P(X_1>3)P(X_2>4)\\
&=\frac26\cdot\frac36+\frac36\cdot\frac26=\frac13.
\end{aligned}$$

Why are the above different? Is it not true that $\max\{X_1,X_2\}>4,\min\{X_1,X_2\}>3 \implies \{X_1>4,X_2>3\}\cup\{X_1>3\,X_2>4\}$?

答:
You can use:

$$P(A\cap B) = P(A|B)P(B)$$

Wikipedia : Conditional_probability

So, $P(\max|\min)=\frac89$, and $P(\min)=\frac14$, or, alternatively, $P(\min|\max)=\frac{8}{20}$, and $P(\max)=\frac{20}{36}$.

Title: Min/max dice problem


问:
"Isolator polynomials" don't work with complex roots?
I have just implemented the algorithm explained at https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.133.2233&rep=rep1&type=pdf to isolate the real roots of a quintic polynomial. My algorithm works perfectly when my polynomial has all real roots, but when it has two complex roots, for example, it is failing.

My input polynomial is:
$$p(x) = x^5 -0.375x^3 - 0.1625x^2 -2.7x - 1$$
which has the real roots:
$$x_0 = -1.224$$
$$x_1 = -0.374$$
$$x_2 = 1.449$$
which means that (according to the example isolator polynomial formulae for quintics in the paper at the bottom of page 3) my isolator polynomials and their roots are:
$$a(x) = -39.94453125x^2 + 7.3921875x + 10.66875, (x_0 = 0.61756, x_1 = -0.4325)$$
$$b(x) = -3.75x^2 + 2.4375x + 0.5625, (x_0 = 0.8306, x_1 = -0.1806)$$

The idea of the isolator polynomials, as the paper explains, is that the roots of these "IPs" can be used as the interval bounds to isolate the roots of the original polynomial.

I have graphed these equations using Desmos (link: https://www.desmos.com/calculator/bwkm7yx7pt), and it is clear that the two quadratics do not isolate the roots (at least, the two right roots of the quadratics don't - it appears that the two left roots still contain the middle root of the quintic, however I'm not sure if that is a coincedence or not.)

Can anyone explain why this is happening? Is there a flaw or my calculation? Or perhaps the algorithm? My goal is to isolate at least two real roots so that I can solve them using Newton's method, deflate the quintic to a cubic, and then solve using the cubic formula.

Help would be much appreciated!

Edit: After checking my calculations again, I cannot find an arithmetic error

答:
I saw your comment on my post on SO. I am using IPs for solving quintics, and for your example I seem to get the appropriate real roots.

roots.rkt> (roots-quint 1.0 0.0 -0.375 -0.1625 -2.7 -1.0)

'(-1.22423523014232 -0.3742386961297533 1.4494051911699204)`

(From Mathematica: x == -1.22424 || x == -0.374239 || x == 0.0745344 - 1.22489 I ||
x == 0.0745344 + 1.22489 I || x == 1.44941)

My isolator polynomial roots are (sorted):
-0.43249415130797386 -0.1805937104039171 0.6175554668040721 0.8305937104039172

So your IP roots seems right, of course that's evident from the plot. These roots separate real roots of the original quintic if any. So if there are roots in any of the following intervals, there must exactly be one root.

[-∞, -0.43249] -1.22423523014232

[-0.43249, -0.18059] -0.3742386961297533

[-0.18059, 0.61756]

[0.61756, 0.83059]

[0.83059, +∞] 1.4494051911699204

This is just like your plot shows.

Note that I used Lagrange's root bounds instead of +∞. and -1 * upper-bound of P(-x) instead -∞. Then used https://www.researchgate.net/publication/228712261_Bisected_Direct_Quadratic_Regula_Falsi. My Scheme/Racket implementation of Brent's is slightly slower.

5 roots cannot be bounded with 3 intervals, which is what 4 roots of the isolator polynomial can give you. But the do isolate the roots like the name says. You would need an upper and lower bound for the real roots of the original poly along with the isolator polynomial roots. There are several bounds available for you to pick. Since my strategy is to only use Brent's or BDQRF to find and deflate the left-most root, then solve the quartic, I didn't think it was worth to find a tighter upper/lower root bound.

I hope this answers your question.


问:
Revisit: If $A$ is full column rank, then $A^TA$ is always invertible
This is possibly a stupid question. I've read the question and answer in this link If $A$ is full column rank, then $A^TA$ is always invertible already.

My question is from $x=0$ how can we conclude $A^TA$ is invertible (or nonsingular)? I hope to get a simple (uncomplicated) explanation. Many thanks!

答:
If the columns $c_1, \ldots, c_n$ of an $n \times n$ matrix $B$ do not generate the whole space, then they must be linearly dependent. Thus, there exist $x_1, \ldots,x_n$ not all zero for which

$$
x_1c_1 + \ldots + x_n c_n = 0.
$$

In other words, if $x := (x_1, \ldots, x_n)$ then $x \neq 0$ and

$$
Bx = 0.
$$

This way, we see that if $B$ does not have full rank, then there exits $x \neq 0$ for which $Bx = 0$.

The answer you cite shows that if $A^T Ax = 0$, then $x = 0$, hence $A^T A$ must have full rank by our previous remark.

Now, if a matrix $B$ has full rank, it is invertible: for each $e_i = (0,\ldots, \overbrace{1}^i,\ldots 0)$ there exists $(a_{ij})_i$ for which

$$
d_{1i}c_1 + \ldots + d_{ni}c_n = e_i,
$$

since $c_1, \ldots, c_n$ generate the whole space. If $C$ is the matrix defined by $D_{ij} = d_{ij}$, we see that

$$
BD = I.
$$

This in turn shows that $B$ is invertible.


问:
How to find the Fourier Coefficient to this discrete time signal?
I'm having trouble finding the fourier coefficients to this discrete time signal:

My approach:

I start by first finding the discrete period, $N$, and discrete frequency, $\Omega_0$

From the problem, we can easily identify that $N = 10$ and so $\Omega_{0} = \frac{2\pi}{10} = \frac{\pi}{5}$

$$x[n] = \sum_{k = 0}^{9}F[K]e^{j\left( \frac{\pi}{5}\right)kn} $$

where $F[K]$ are the Fourier coefficients we want to find.
$$F[K] = \frac{1}{N}\sum_{n = 0}^{9}x[n]e^{-j\left( \frac{\pi}{5}\right)kn}$$

$$F[K] = \frac{1}{10}\left( \sum_{n = 0}^{4}\frac{7}{2}e^{-j\left( \frac{\pi}{5}\right)kn} + \sum_{n = 4}^{9}\frac{-7}{2}e^{-j\left( \frac{\pi}{5}\right)kn}
\right)$$

$$F[K] = \frac{7}{20}\left( \sum_{n = 0}^{4}e^{-j\left( \frac{\pi}{5}\right)kn} - \sum_{n = 4}^{9}e^{-j\left( \frac{\pi}{5}\right)kn}
\right)$$

$$F[K] = \frac{7}{20} \left( 1 + \cos(\frac{\pi}{5}k) + j\sin(\frac{\pi}{5}k) + \cos(\frac{2\pi}{5}k) + j\sin(2\frac{\pi}{5}k) + \cos(\frac{3\pi}{5}k) + j\sin(3\frac{\pi}{5}k) + \cos(\frac{4\pi}{5}k) + j\sin(\frac{4\pi}{5}k) - \cos(\pi k) - j\sin(\pi k) - \cos(\frac{6\pi}{5}k) - j\sin(\frac{6\pi}{5}k) - \cos(\frac{7\pi}{5}k) - j\sin(\frac{7\pi}{5}k) - \cos(\frac{8\pi}{5}k) - j\sin(\frac{8\pi}{5}k) - \cos(\frac{9\pi}{5}k) - j\sin(\frac{9\pi}{5}k)
\right)$$

This looks like such a mess to reduce... or is there something elegant that I'm missing?

According to the solutions

$
X[K]=
\begin{cases}
 \frac{7}{10} + \frac{7j \pi}{5}\left( \sin(\frac{\pi}{5}) + \sin( \frac{2\pi}{5}) \right) , &\text{if}\, k = - 1\\
 \frac{7}{10} + \frac{7j \pi}{5}\left( \sin(\frac{\pi}{5}) + \sin( \frac{2\pi}{5}) \right) , &\text{if}\ k = + 1\\
 \frac{7}{10} - \frac{7j \pi}{5}\left( \sin(\frac{2 \pi}{5}) - \sin( \frac{\pi}{5}) \right) , &\text{if}\ k = + 3\\
 \frac{7}{10} + \frac{7j \pi}{5}\left( \sin(\frac{2 \pi}{5}) - \sin( \frac{\pi}{5}) \right) , &\text{if}\ k = - 3\\
      0 &\text{otherwise}
\end{cases}
$

HOW DO YOU FIGURE THAT OUT!!? Is there some super clever technique I'm not employing? Or is all my math wrong?

答:
Your second sum should index from $n=5$.  Then here is a small simplification:
\begin{align}
F[k] &= \frac{7}{20} \Big( \sum_{n=0}^4 e^{-ink\pi/5}-\sum_{n=5}^9e^{-ink\pi/5} \Big) \\
&=\frac{7}{20} \Big( \sum_{n=0}^4 e^{-ink\pi/5}-\sum_{n=0}^4   e^{-5ik \pi/5} e^{-ink\pi/5} \Big) \\
&= \frac{7(1 - (-1)^k)}{20} \sum_{n=0}^4 e^{-ink\pi/5} \\
&=\left\{ \begin{array} ~0 & \text{when } k \text{ is even} \\
\frac{7}{10} \sum_{n=0}^4 e^{-ink\pi/5} & \text{otherwise} \\
\end{array}\right .
\end{align}
Thus for even $k$, $F[k] = 0$.  Otherwise, pair the terms in the sum using the fact that $k$ is odd,
\begin{align}
e^{-ik\pi/5}+e^{-i4k\pi/5}&=e^{-ik\pi/5}-e^{ik\pi/5}=-2i\sin(k\pi/5), \quad\text{ pairing } n = 1 \text{ and } n=4\\
e^{-i2k\pi/5}+e^{-i3k\pi/5}&=e^{-2ik\pi/5}-e^{i2k\pi/5}=-2i\sin(2k\pi/5), \quad\text{ pairing } n = 2 \text{ and } n = 3 .
\end{align}
So taking the real and imaginary, the $\cos$ terms in the sum cancel for all values of $k$ and we are left only with two $\sin$ terms in the imaginary part,
\begin{align}
F[k]=\left\{
\begin{array}
~0 & \text{when } k \text{ is even} \\
~\frac{7}{10} -\frac{7}{5}i \Big(\sin\big(\frac{1}{5}k\pi\big)+\sin\big(\frac{2}{5}k\pi\big) \Big)&\text{when } k \text{ is odd},
\end{array} \right.
\end{align}
which is close to the answer you wanted;  the stated answer has a $\pi$ factor in the imaginary part which I think must be an error, and we differ in some signs.

PS.  I have used $i$ instead of $j$ out of habit.


答:
$$
\int \frac{u}{u+1} du = \int \frac{u+1-1}{u+1} du = \int 1 \, du - \int \frac{1}{u+1} du.
$$

Now it should be easy to solve.

Title: How to compute the indefinite integral $\int \frac u{u+1}\,\mathrm du$?

Body: How do you compute $$ \int \frac{u}{u+1}\,\mathrm du$$</p>


问:
Can the sum of all rationals under a number be finite?
Does there exist any non zero number x such that the sum of all rationals between 0 and x is finite?</p>

答: Hint: use the density of $\mathbb Q$ in $\mathbb R$ to show that for any $\varepsilon > 0$, there exists infinitely many rationals in $[\frac \varepsilon 2, \varepsilon]$, and conclude.


答:
WLOG, assume $x > 0$. Note that there are infinitely many $n \in \Bbb N$ such that $\frac1n \in [0, x).$ Note that $\sum_{n \ge 1} \frac1n = \infty$. Conclude.

Edit: Technically all I've stated above is not sufficient. You need to use the fact that there exists $N_0 \in \Bbb N$ such that $\frac1n \in [0, x)$ for all $n > N_0$.

Title: Can the sum of all rationals under a number be finite?

Body: Does there exist any non zero number x such that the sum of all rationals between 0 and x is finite?</p>


问:
What does Frobenius reciprocity say in the case of $G = S_3$ and $H = C_2$?
Using character theory the statement of Frobenius reciprocity would be:
$$ \langle \text{Ind}_H^G \psi , \phi \rangle_G = \langle \psi , \text{Res}_H^G \phi\rangle_H $$
Let's try to evaluate this expression with $G = S_3$ (the permutation group of three objects) and $H = C_2$ is a 2-cycle.

The character table of $S_3$ looks like this:
$$ \begin{array}{c|crr}
 & (1) & (12) & (123) \\ \hline
\phi_1 & 1 & 1 & 1 \\
\phi_2 & 1 & -1 & 1 \\
\phi_3 & 2 & 0 & -1 \\
\end{array} $$

The character table of $C_2$ is even simpler:
$$ \begin{array}{c|cr}  & (1) & (12) \\ \hline
\psi_1 & 1 & 1\\
\psi_2 & 1 & -1 \end{array} $$

How do we compute the induced characters $\text{Ind}_{H}^{G} \psi$ (from $H = C_2$ to $G = S_3$) and in the other direction, the restricted characters $\text{Res}_H^G \phi$.

We also might need the Schur orthogonality relations to compute the inner product.  So what does the equation look like in the end?

Related:

Find character table for symmetric group $S_3$

Character theory questions on Chemistry.SE
 Why does $(2z^2-x^2-y^2,x^2 - y^2)$ have -1 trace in the character table of $T_d$ for $C_3$ rotation?

https://physics.stackexchange.com/questions/306995/decoding-the-character-table

答:
Prof. Holt is right, in the character table of $S_3$ the last two columns under the line should be switched: I corrected this for you. Now let us do some calculations. First let us find out what $\psi_1^G=(1_H)^G$ must be (I am using a less involved notation for the characters). Using Frobenius' Reciprocity: $[(1_H)^G,1_G]=[1_H,1_H]=1$, hence $(1_H)^G=1_G + \delta$, where $\delta$ is a character with $[\delta,1_G]=0$, so $\delta$ is a certain sum of the non-principal characters of the table of $S_3$. Note that $3=(1_H)^G(1)=1_G(1) + \delta(1)=1+\delta(1)$, so $\delta(1)=2$. Observe from the character table that $(\phi_3)_H=\psi_1+\psi_2=1_H+\psi_2$, This implies that $\delta = \phi_3$ (again use Frobenius' Reciprocity to calculate $[(1_H)^G, \phi_3]$). We conclude that $\psi_1^G=\phi_1+\phi_3$. In a similar vein now try to calculate $\psi_2^G$. Your answer should be $\psi_2^G=\phi_2 + \phi_3$.

The example you mentioned is just a small group example. For larger groups calculations get more involved and one needs to apply some efficiency, taking into account how conjugacy classes of $G$ and $H$ "behave". See for example I.M. Isaacs, Character Theory of Finite Groups, p. 64, where the character table of $G=A_5$ is constructed from $H=A_4$.


问:
If $f$ is uniformly continuous on the dense set $E \subset X$ then a continuous extension of $f$ exists on $X$.

Let $E \subset X$ which is a dense subset of $X$.The function $f$ has its range in $\mathbb{R}$.We need to show that there exists a continuous extension from $E$ to $X$.

My attempt :

let $x \in X$ and we choose a cauchy sequence {$e_n$} $\to x$.Now, since $\{e_n\}$ is a convergent sequence so it is a cauchy sequence. As $f$ is a uniformly continuous function, then $\{f(e_n)\}$ is also a cauchy sequence and so it is convergent to a point,we define that to be $f(x)$[As $\mathbb{R}$ is a complete metric space].

Now, we choose another sequence $\{(e_n)'\} \to x$.We see that
$d(e_n,x)<\frac{\delta}{2}$ for all $n \ge K_1$ and $d((e_n)',x)) < \frac{\delta}{2}$ for all $n \ge K_2$
Then, we see that $d_y(f((e_n)'),f(e_n)) \le \frac{\epsilon }{2}$ when $d(e_n,(e_n)')\le \delta$ for all $n \ge max\{K_1,K_2\}$.Using this equation we can show that $f((e_n)') \to f(x)$.

To prove:$f$ is uniformly continuous. We choose $d(x_m,x_n) < \frac{\delta}{3}$.

Let $\{(e_n^m)\} \to x_m$ and $\{(e_n^k)\} \to x_k$.Now,  $\{f(e_n^m)\} \to f(x_m)$ and $\{f(e_n^k)\} \to f(x_k)$.So,

$d_y(f(e_n^k),f(x_k)) < \frac{\epsilon}{3}$ for all $n \ge K_1$ and

$d_y(f(e_n^m),f(x_m)) < \frac{\epsilon}{3}$ for all $n \ge K_2$.

Now by some manipulations we can find a $N$ such that when $n_1,n_2 \ge N$ we have $d((e_{n_1}^m), (e_{n_2}^k))  < \delta $ and by choosing such a delta we can have $d_y(f(e_{n_1}^k),f(e_{n_2}^m) )< \frac{\epsilon}{3}$ for all $n_1,n_2\ge N$.Also we see that,

$d_y(f(x_m),f(x_k)) \le d_y(f(e_n^k),f(x_k)) + d_y(f(e_n^m),f(x_m)) + d_y(f(e_n^k),f(e_n^m))< \epsilon$ for all $n \ge max\{K_1,K_2,N\}$

This has been an attempt. Can someone go through my proof?I am not sure about the part where I am proving uniform continuity.

答:
Your proof is mainly correct, although the notation might seem a bit confusing at times. I will try to write a cleaner version of what you did in the last paragraph, where you proved the uniform continuity of the extension of $f$, which is denoted exactly the same. Of course, I shall assume that $(X, d)$ is a complete metric space and by abuse of notation, I shall also denote the standard euclidian distance on $\mathbb{R}$ by $d.$

Let $\varepsilon > 0.$ Since $f$ is uniformly continuous on the dense set $E,$ there is $\delta_\varepsilon > 0$ such that $d(f(e_1), f(e_2)) < \frac{\varepsilon}{3}$ whenever $d(e_1, e_2) < \delta_\varepsilon$ and $e_1, e_2 \in E.$ Take now $x, y \in X$ such that $d(x, y) < \frac{\delta_\varepsilon}{3}.$ From the definition of the extension (thus implicitly the density of $E$), we infer that there are $e_x, e_y \in E$ such that $d(e_x, x) < \frac{\delta_\varepsilon}{3} > d(y, e_y)$ and $d(f(x), f(e_x)) < \frac{\varepsilon}{3} > d(f(e_y), f(y)).$ In particular, it follows from the triangle inequality that we also have that $d(e_x, e_y) < \delta_\varepsilon.$ Hence we have that $d(f(e_x), f(e_y)) < \frac{\varepsilon}{3}.$ Finally, applying once again the triangle inequality, we deduce that $d(f(x), f(y)) \leq d(f(x), f(e_x)) + d(f(e_x), f(e_y)) + d(f(e_y), f(y)) < \varepsilon.$ This proves the claim that the extension of $f$ is uniformly continuous.


答:
Integration by parts needlessly complicates this problem.

A simpler approach uses partial fraction decomposition.  Since
$$\frac{u}{u+1} = 1-\frac{1}{u+1},$$
we have
$$\int \frac{u}{u+1} du = \int 1du - \int \frac{1}{u+1}du = u - \log \vert u+1 \vert + c.$$


问:
An order ideal in of a poset $P$ is join-irreducible in $J(P)$ if and only if it is generated by a single element?
An order ideal of $P$ is a subposet $Q \subset P$ with the property that if $x,y \in P$, $x \in Q$ and $y \leq x$ then $y \in Q$.

Given a lattice $L$, an element $x \in L$ is called join-irreducible if it can't be written as the join of two other elements. That is, if $x = y \vee z$ then either $x = y$ or $x=z$.

For a finite poset $P$, let $J(P)$ denote the set of order ideals of $P$. It is a basic result in combinatorics that $J(P)$ is a distributive lattice.

I'm trying to show that an order ideal in $P$ is join-irreducible in $J(P)$ if and only if it is generated by a single element.

I'd appreciate some guidance.. thank you!!

答:
The join of order ideals is simply their union.

The order ideal generated by a single element $x\in P$ is the set $A_x:=\{y\in P:y\le x\}$.

If $A_x=B\cup C$ for order ideals $B,C$, then $x\in B$ or $x\in C$, but then $A_x\subseteq B\subseteq A_x$ in the first case, and similarly $A_x=C$ in the second case.

For the converse, we use finiteness of $P$ (otherwise it could fail, for example consider $(-\infty,0)\subseteq\Bbb R$).

Let $B$ be a join irreducible order ideal, and consider the sets $A_x$ for all $x\in B$, these must all be included in $B$, and thus we get
$$B=\bigcup_{x\in B} A_x\,.$$
Since $B$ is join irreducible, and this is a finite join, we must have $A_x=B$ for some $x\in B$.


问:
Localization of the minimum of a convex function
I'd like to prove a result but I do not know where to start. Any help or little idea is more than welcome (:

If I had to ask my proof issue in therm of an exercise question, It would look like that :

Let $x \in \mathbb{R}^2$, $f(x): \mathbb{R}^2 \to \mathbb{R}$ a convex function with $f(x^*)$ as minimum and $\chi := \{x:||x||_F = r \in \mathbb{R}\}$. If we know that $f(x^*)\leq 0$ and $\text{inf}\{f(x):x\in\chi\} > 0$, then, is it possible to prove that $||x^*||_F \leq r$ ?

Here $||x||_F$ is the Frobenius, i.e., the usual Euclidean, norm of $x$ in $\mathbb R^2$.

Thanks in advance,

Deramake

答:
I would start by proving that the sublevel sets $C_t:=\{f\leq t\}$ for all real $t$ form a nested family of convex closed sets.  What you're after is $C_{t^*}=:C_*$ where $t^*:=\inf_D f$ where $D$ is the domain of $f$.  Now there are two cases $C_*$ is a single point, or not.

$C_*$ single point case (which you called $x_*$): in this case $x_*\in C_*\subseteq C_{t_\flat}=:C_\flat\subseteq B\subseteq C_{t_\sharp}=:C_\sharp$ where $t_\flat:=\int_S f$, $t_\sharp:=\sup_\chi f$ and $\chi=\partial B$ is the sphere or radius $r$, boundary of the ball of radius $r$, $B$. In other words $|x_*|\leq r$, but if $t_*<t_\flat$ then $C_*\subsetneq C_\flat$ and (because $C_*$ has only one point) then $x_*\in B\smallsetminus \chi$; so in fact you have more $|x_*|<r$.

$C_*$ is not a single point: it has to be convex and the same argument as above applies for any point $x_*$, but you need to be careful when the two sublevel sets are tangent: this cannot occur because it would make $f$ discontinuous, which is false because it's convex.

If the function is strictly convex, only the first case applies.


问:
Can we show $\sigma_p(T)\setminus\sigma_p(T')=\sigma_r(T')$?
Let $X$ be a complex Banach space and $T\in\mathfrak L(X)$.

Can we show that $\sigma_p(T)\setminus\sigma_p(T')=\sigma_r(T')$?

I think the claim is true: Let $\lambda\in\mathbb C$. Then$^1$, $$\mathcal N(\lambda-T)=\mathcal R(\lambda-T')_\perp\tag1.$$ If $\lambda\in\sigma_p(T)$, then $$\overline{\mathcal R(\lambda-T')}\ne X'\tag2$$ by $(1)$. And if additionally $\lambda\not\in\sigma_p(T')$, then $\lambda\in\sigma_r(T')$ by $(2)$.

On the other hand, if $\lambda\in\sigma_r(T')$, then $\mathcal N(\lambda-T')=\{0\}$ (hence $\lambda\not\in\sigma_p(T')$) and it holds $(2)$, which should yield $$\mathcal N(\lambda-T)\ne\{0\}\tag3$$ by $(1)$ (hence $\lambda\in\sigma_p(T)$).

Right?

$^1$ If $V\subseteq X'$, then $$V_\perp:=\{x\in X:\langle x,\varphi\rangle=0\text{ for all }\varphi\in V\}.$$

答:
No, (2) does not imply (3) in general.

The implication ``$\overline{ R(\lambda-T')} \ne X'$ $\Rightarrow$ $N(\lambda-T)\ne\{0\}$'' is true for reflexive Banach space $X$ (invoke Hahn-Banach in $X'$, represent the so-obtained functional in $X''$ by an element in $X$). It is not true non-reflexive Banach spaces:

Take $X=c_0$, $X'=l_1$, $T=I-L$, $T'=I-R$, where $L$ and $R$ are left and right shift. Then
$$
R(T') \subset \{ x\in l^1: \ \sum_{n=1}^\infty x_n=0\},
$$
where the set on the right is a closed, proper subspace of $X'$, but $N(T)=\{0\}$.


问:
prove if f is continuous and $\lim\limits _{x\to\infty}f\left(x\right)=L\neq0$ then $\int\limits _{1}^{\infty}f\left(x\right)\sin x\,dx$ diverges

The Question

let $f:\left[1,\infty\right)\to\mathbb{R}$ be a continuous function with $\lim\limits _{x\to\infty}f\left(x\right)=L\neq0$. then $\int\limits _{1}^{\infty}f\left(x\right)\sin x\,dx$ diverges.

I came across this question while preparing to my exam in calculus. I wasn't able to solve it: couldn't prove it nor give a counterexample.

What have I tried?

I tried using tests such as the comparsion test, but I realised $f\left(x\right)\sin x$ isn't positive. Tried looking for counter examples but failed.

Here is my best attempt so far:
since $L\neq0$ we can assume without loss of generality $L>0$, therefore $f>0$ almost always.
for some $\mathbb{N}\ni k,\, M=2k$ we have $$\int\limits _{M\pi}^{k\pi}f\left(x\right)\sin x\,dx=\sum_{i=M}^{k}\left(-1\right)^{i}\int\limits _{0}^{\pi}f\left(x\right)\sin x=Q_{k}$$ and since $$2\left(L+\epsilon\right)\leq\int\limits _{0}^{\pi}f\left(x\right)\sin x\leq2\left(L+\epsilon\right)$$ for some $\epsilon$, we can write $$\sum_{i=M}^{k}\left(-1\right)^{i}\cdot2\cdot\left(L+\left(-1\right)^{i+1}\epsilon\right)\leq Q_{k}\leq\sum_{i=M}^{k}\left(-1\right)^{i}\cdot2\cdot\left(L+\left(-1\right)^{i}\epsilon\right)$$

But here I got stuck, I'm not sure how to continue and I'm not sure if this is the right direction.

答: Without loss of genrality we may suppose $L >0$. If the intergal converges then $\int_a^{b} f(x)\sin x dx$ tends to $0$ as $b >a \to \infty$. Let us pove that this is not true. Consider $\int_{2n\pi}^{2n\pi +\pi}  f(x)\sin x dx$. Note that $\sin x$ is positve in this interval Also, $f(x) >\frac L 2$ in this interval if $n$ is large enough. This gives $\int_{2n\pi}^{2n\pi +\pi}  f(x)\sin x dx >\frac L 2 \int_{2n\pi}^{2n\pi +\pi}  \sin x dx=L$. This is a contradiction.


问:
Probability of rolling three dice without getting a 6 in 8 rolls
I know the propability of rolling three dice without getting a $6$ in $1$ roll is $91/216$, but what about rolling it in $8$ rolls consecutively with three dices?</p>

答:
The probability of rolling one die and not getting a $6$ is $\frac56$, so the probability of rolling three dice without getting a $6$ is $\left(\frac56\right)^3=\frac{125}{216}$, not $\frac{91}{216}$.

$\frac{91}{216}=1-\frac{125}{216}$ is therefore the probability of getting at least one $6$ when rolling three dice.

The probability of getting no sixes when you roll three dice $8$ times is the same as the probability of getting no sixes when you roll $3\cdot8=24$ dice, which is $\left(\frac56\right)^{24}$; this is about $0.01258$.


问:
UMVUE OF $\sum_{k=4}^{\infty}\frac{e^{-\theta}\theta^k}{k!}$

Let $x_1=1, x_2=4$ be the data on a random sample of size $2$ from a
Poisson($\theta$) distribution, where $\theta \in (0,\infty)$. Let T
be the uniformly minimum variance unbiased estimate of T($\theta$)=
$\sum_{k=4}^{\infty}\frac{e^{-\theta}\theta^k}{k!}$ based on the given
data. Then T equals____?

I am using $x$ instead of $k$.

$P(X\ge 4)=\sum_{x=4}^{\infty}\frac{e^{-\theta}\theta^x}{x!}$

$I(X_1)=1$ if $X\ge4$,$0$ otherwise

So using Rao–Blackwell theorem

$E(I(X_1)|T=\sum_{i}X_i)$
=$1\cdot P\left( \dfrac{X_1\ge4,X_1+X_2+...+X_n=T}{T=\sum_{i}X_i} \right)$

How do I solve this? if I was given $X_1=1$ I would've utilized it and then used the fact of independence but I am not sure how to proceed here. Also, tell me if there is any other way possible way to solve this problem.

答:
Let $g(T)$ be the unbiased estimator of $\tau(\theta)=\sum\limits_{k=4}^{\infty}\frac{e^{-\theta}\theta^k}{k!}$ based on $T=X_1+X_2\sim \text{Poisson}(2\theta)$.

That is, $$\operatorname E_{\theta}\left[g(T)\right]=\tau(\theta)\quad,\forall\,\theta>0$$

Or, $$\sum_{t=0}^\infty g(t) \frac{e^{-2\theta}(2\theta)^t}{t!}=\sum_{k=4}^{\infty}\frac{e^{-\theta}\theta^k}{k!}\quad,\forall\,\theta$$

So for every $\theta$,

\begin{align}
\sum_{t=0}^\infty g(t)\frac{2^t}{t!}\cdot\theta^t &=e^{\theta}\sum_{k=4}^{\infty}\frac{\theta^k}{k!}
\\&=\sum_{\ell=0}^\infty \frac{\theta^\ell}{\ell!}\sum_{k=4}^{\infty}\frac{\theta^k}{k!}
\\&=\sum_{t=0}^\infty \left(\sum_{k+\ell=t,k\ge 4,\ell\ge 0}\frac1{k! \ell!}\right)\theta^t
\end{align}

Comparing coefficients of $\theta^t$,

\begin{align}
g(t)&=\frac{t!}{2^t} \sum_{k+\ell=t,k\ge 4,\ell\ge 0}\frac1{k! \ell!}
\\&=\frac1{2^t} \sum_{k=4}^t \frac{t!}{k!(t-k)!}\mathbf1_{\{t\ge 4\}}
\\&=\begin{cases} \frac1{2^t} \sum\limits_{k=4}^t \binom{t}{k} &,\text{ if }t\ge 4 \\ 0 &,\text{ if }0\le t\le 3\end{cases}
\end{align}

This suggests that UMVU estimate of $\tau(\theta)$ is $0$ when observed value of $T$ is less than $4$ (which is of course not the case here). As pointed out by @Henry, $g(T)$ is probably not a useful estimator in that situation.


答:
This is a solution my friend came up with (credit doesn't go to me)

we can assume without loss of generality $L>0$. let $m:=\frac{L}{2}$. since $\lim\limits _{x\to\infty}f\left(x\right)=L\neq0$ there exists $x_0$ such that $\forall x\geq x_{0},\;f\left(x\right)\geq m$. let $\epsilon:=2m$. for all $N\in\mathbb{N}$ exists $N\in\mathbb{N}$ such that $2\pi n\geq N$, then $$\left|\int\limits _{2\pi n}^{2\pi n+\pi}\sin xf\left(x\right)\,dx\right|=\int\limits _{2\pi n}^{2\pi n+\pi}\sin xf\left(x\right)\,dx\geq m\int\limits _{2\pi n}^{2\pi n+\pi}\sin x\,dx=2m=\epsilon$$ so be Cauchy's theorem the integral diverges.

Title: prove if f is continuous and $\lim\limits _{x\to\infty}f\left(x\right)=L\neq0$ then $\int\limits _{1}^{\infty}f\left(x\right)\sin x\,dx$ diverges

Body:
The Question

let $f:\left[1,\infty\right)\to\mathbb{R}$ be a continuous function with $\lim\limits _{x\to\infty}f\left(x\right)=L\neq0$. then $\int\limits _{1}^{\infty}f\left(x\right)\sin x\,dx$ diverges.

I came across this question while preparing to my exam in calculus. I wasn't able to solve it: couldn't prove it nor give a counterexample.

What have I tried?

I tried using tests such as the comparsion test, but I realised $f\left(x\right)\sin x$ isn't positive. Tried looking for counter examples but failed.

Here is my best attempt so far:
since $L\neq0$ we can assume without loss of generality $L>0$, therefore $f>0$ almost always.
for some $\mathbb{N}\ni k,\, M=2k$ we have $$\int\limits _{M\pi}^{k\pi}f\left(x\right)\sin x\,dx=\sum_{i=M}^{k}\left(-1\right)^{i}\int\limits _{0}^{\pi}f\left(x\right)\sin x=Q_{k}$$ and since $$2\left(L+\epsilon\right)\leq\int\limits _{0}^{\pi}f\left(x\right)\sin x\leq2\left(L+\epsilon\right)$$ for some $\epsilon$, we can write $$\sum_{i=M}^{k}\left(-1\right)^{i}\cdot2\cdot\left(L+\left(-1\right)^{i+1}\epsilon\right)\leq Q_{k}\leq\sum_{i=M}^{k}\left(-1\right)^{i}\cdot2\cdot\left(L+\left(-1\right)^{i}\epsilon\right)$$

But here I got stuck, I'm not sure how to continue and I'm not sure if this is the right direction.


问:
Double counting $n2^{n-1} = \sum_{k=0}^{n} k\binom{n}{k}$
I am trying to prove by double-counting that

$$k2^{k-1} = \sum_{i=0}^{k} i\binom{k}{i}$$

where $n$ is a positive integer.

I did it algebraically. However, the professor told me that I have to do this also by double-counting but I am not sure how to do this. Can someone help me?

答:
Consider the problem of forming a committee with a president from a group of $n$ people.

On one hand, you can first fix a president ($n$ choices) and then choose a subset of the remaining people to form the committee ($2^{n-1}$ choices). This gives you LHS.

On the other, you can fix a size $k$ (between $1$ and $n$) first and then pick a committee with $k$ people ($\binom{n}{k}$ many choices) and then pick a president from that ($k$ choices). Sum over all possible $k$ to get the RHS.


问:
How much should I know about Category theory for A&M introduction to commutative algebra?
I'm recently reading A&M introduction to commutative algebra, and I found that in the text, they implicitly assume I know some basic notions about category theory (for example, in the exercise, they assume I know about Tor functor). Should I know some of Category theory to read this text? If so, could you suggest some textbook or something for that?</p>

答: You'll be fine with not knowing category theory. As guidoar also says, you may want to be acquainted with basic notions like exact sequences and diagrams. It is true that some exercises in section 2 require knowledge of Tor-groups, but if you don't have that knowledge, you can probably safely skip these exercises.</p>


问:
The evaluation of Complex integration
$K>0$

$C_K$ : $z=(K-t)+it \,\,(0\leqq t \leqq K)$ (Straight Line on $\mathbb{C}$)

Prove that

(i) $|z|\geqq \dfrac{K}{\sqrt{2}}$ for $z\in C_K$

(ii) $\bigg|\dfrac{e^{iz}}{z}\bigg|\leqq \dfrac{\sqrt{2}}{K} e^{-t}$ for $z\in C_K$

(iii) $\lim_{K\to \infty} \displaystyle\int_{C_K} \dfrac{e^{iz}}{z} dz=0$

I could prove (i) and (ii), but I cannot prove (iii) .

\begin{align}
\Bigg| \displaystyle\int_{C_K} \dfrac{e^{iz}}{z} dz \Bigg|
&\leqq \sup_{z\in C_K} \Bigg| \dfrac{e^{iz}}{z} \Bigg| \cdot L(C_K) \\
&\leqq \dfrac{\sqrt{2}}{K}e^{-t}\cdot L(C_K) \\
&=\dfrac{\sqrt{2}}{K}e^{-t}\cdot \sqrt{K^2+K^2} \\
&=2e^{-t}.
\end{align}

The term $K$ vanishes so it didn't work.

I would like you to give me some ideas.

答:
Simply start to evaluate the integral,
\begin{align}
\left\lvert \int_{C_K} \frac{e^{iz}}{z} dz \right \rvert &=\left\lvert
\int_0^K \frac{e^{(K-t)i-t}}{(K-t)+it} \cdot(-1+i) dt \right\rvert\\
&\leqslant\sqrt{2}\int_0^K \frac{e^{-t}}{\lvert (K-t) + it\rvert} dt
\end{align}
The denominator is bounded below by $\frac{1}{\sqrt 2} K$, so you obtain,
\begin{align}
\left\lvert \int_{C_K} \frac{e^{iz}}{z} dz \right \rvert &\leqslant\frac{2}{K} \int_0^K e^{-t} dt \\
&=\frac{2(1-e^{-K})}{K}
\end{align}
and this last term has limit zero as $K \to \infty$.

Comment on approach taken:  the approximation using the length of $C_K$ and the maximum value of the integrand doesn't work because it cannot exploit the fact that $e^{iz}$ decreases rapidly along $C_K$.  Moreover, in the answer you gave you still have a term $e^{-t}$, but the $t$ variable is only relevant inside the integral and therefore should never appear as part of the final estimate.

Title: The evaluation of Complex integration


问:
Number of roots of the equation $f(xf(x)) = \sqrt{9 - x^2f^2(x)}$

Given the following graph of a $5$th degree polynomial $y = f(x)$:

Find the number of solutions of the equation: $$f(xf(x)) = \sqrt{9 - x^2f^2(x)}$$

My attempt was to figure out what $f(x)$ is, by noticing the roots of its derivative based on the graph to know $f'(x)$, then find $\int f'(x)$, but the function I found did not match the original graph. (Even if I founded, I think it would be very messy to plug it in the equation)

One other idea is that let $t = xf(x)$, then the equation become:

$$f(t) = \sqrt{9-t^2}$$

But I can't process from here.

Is my idea correct? If yes, then how do I continue? Or is there a better way to solve this?

答:
According to the given question, you need to find the solutions of $$f(xf(x)) = \sqrt{9 - x^2f^2(x)}$$

Substituting $x(f(x))$ by $t; t\in (-\infty , \infty)$, like you did, we obtain

$$f(t) = \sqrt{9-t^2} \\
\Rightarrow    f(x) = \sqrt{9-x^2}$$

ie., We need to find the number of points where the graph of $f(x)$ and $ \sqrt{9-x^2} =g(x) \space \text{(say)}$ intersect.

Analysing $g(x)$, we can say that
$$ x^2 + \left(g(x)\right)^2 =9 ; g(x) \geq 0$$
ie., $g(x)$ traces a semicircular region as shown below

Superimposing the graphs of $f(x)$ and $g(x)$ we obtain,

(Zoomed in view)

As we can notice four points of intersection, there are four solutions to the given expression $f(t) = \sqrt{9 - t^2}$

Now, the values of $t$ that satisfy are

$t$ approximately $-1.5$ or something close to that, which the product $xf(x)$ can give in 2 different ways. First when $x<0$, second when $f(x)<0$.

$t = 3$ which can be obtained in 4 ways, once for $x=1, f(x)=3$ then, for $x=3, f(x)=1$ and twice for $x<1$.

once more check it for $t$ between $0$ and $1$

and lastly for $t$ between $1$ and $2$.

Edit #1: The last two calculations would get messier the more we try to analyse it raw-handedly. Holding on to the suggestion by @windows prime in the comments, we can draw an approximate graph of $t= x f(x) =t_{\small 0}$ having $t_{\small 0}$ as a solution of $t$

These graphs would be rectangular hyperbolas if you observe them clearly with $f(x)$ at $y-axis$.

For both, $t_1$ whose value lies between $0$ and $1$ and $t_2$ whose value lies between $1$ and $2$, the graphs will have 4 intersections each with that of $y= f(x)$.

So, we have $2+4+4+4 = 14$ solutions in total.

Edit #2: The intersection of hyperbolic graphs and $f(x)$ would look approximately like the one below:

$\color{#Ff3537}{xy \approx -1.5}\\
\color{#4466ff}{xy = 3}\\
\color{#008800}{xy = t_1;\space t_1\in (0,1)}\\
\color{#8800ff}{xy = t_2;\space t_2\in(1,2)}\\$

Here, we have 14 intersections pointing towards 14 distinct solution.


问:
What are the automorphisms acting on $Q\sqrt[3]{2}$ that fixes Q?
For the field $Q(\sqrt[3]{2})$, the elements in the field are in the form of $a + b\sqrt[3]{2}+c\sqrt[3]{4}$, where a,b,c, are elements of Q.

Is the automorphism group that fixes Q  $C_{3}$?

multiplying all elements in $Q(\sqrt[3]{2})$ by $\sqrt[3]{2}$, will result a cyclic cycle of period 3, but what about mapping $\sqrt[3]{2}$  to  $-\sqrt[3]{2}$? For extension fields like $Q(\sqrt{2},\sqrt{3})$, the mapping towards a radical's negative counterpart is part of the automorphism group. Should this also be part of the automorphism group that fixes Q?

答:
Hint: note that such an automorphism is determined by the image of $\sqrt[3]{2}$. The latter has to get mapped to a root of the minimal polynomial $m(\sqrt[3]{2}, \Bbb Q)$ that additionally lies in $\Bbb Q(\sqrt[3]{2})$.

 The only such root is $\sqrt[3]{2}$ itself, hence the group is trivial.


问:
Energy estimates (Evans chapter 7 , theorem 2)
I meet some problem in theorem 2(energy estimates)  Evans's book  chapter 7.

Step 4. Fix any  $v\in H_0^1(U)$, with $||v||_{H_0^{U}}\leq 1$, and write $v=v^1+v^2$, where $v^1\in span \{w_k \}$ and $(v^2,w_k)=0 (k=1,\cdots)$. Since the functions $\{w_k\}_{k=0}^{\infty}$ are orthogonal in $H_0^1(U)$, $||v^1||_{H_0^1(U)}\leq ||v||_{H_0^1(U)}\leq 1$. Utilizing (16), we deduce for  a.e. $0\leq t\leq T$ that $$(u_m',v^1)+B[u_m,v^1,t]=(f,v^1)$$. Then (14) implies
\begin{equation}
 \left\langle u_m',v\right \rangle=(u_m',v)=(u_m',v^1)=(f,v^1)-B[u_m,v^1;t]
\end{equation}
$ \cdots $

Why $\left\langle u_m',v\right \rangle=(u_m',v) $ (the last sentence in step 4), the left side is the pairing of $H^{-1}(U)$ and $H_0^1(U)$ , the right side is inner product in $L^2(U)$.
Could someone give me some advice , thank you!
(14),(16) are equation under below

(14)$$u_m(t):=\Sigma_{k=1}^{m}d_m^k(t)w_k,  $$
(16)$$(u_m',w_k)+B[u_m,w_k,t]=(f,w_k) (0\leq t \leq T, k=1, \cdots ,m) $$

Whole theorem is under below.

答:
My guess is the following :

let $\alpha \in H^1_0(Q)$. We can say that $\alpha \in H^{-1}(U)$ with the following construction (using your notation for $()$ and $<>$) :

$$<\alpha|u> := (\alpha  |u), \quad \forall u \in H^1_0(U).$$

Indeed, we have

$$|<\alpha|u>| \leq ||\alpha||_{L^2(U)} ||u||_{L^2(U)} \leq  \underbrace{||\alpha||_{L^2(U)}}_{:=C} ||u||_{H^1_0(U)}$$
and therefore $\alpha$ can be seen as en element of the dual of $H^1_0(U)$.

Now use that with

$$\alpha=u_m'(t)=\sum_{k=1}^m d_m^{k}{'}(t) w_k \in H^1_0(U)$$
and $v \in H^1_0(U)$.


问:
Describe all functions which are defined as $f:\mathbb R \to \mathbb R $ and satisfying $f(x)=\frac1{f(x)}$
Describe all functions which are defined as $f:\mathbb R \to \mathbb R $ and satisfying $$f(x)=\dfrac1{f(x)}$$

My Attempt:

If $f$ is constant function then.

$$(f(x))^2=1$$

Then $f(x)=1$ or $f(x)=-1$

I cannot think non-constant function satisfying this

What about $f$ being continuous?

答:
Hint:
$$\begin{align}
&\qquad \quad f(x)=\dfrac1{f(x)} \\
&\implies \frac{f(x)^2-1}{f(x)}=0 \\
&\implies (f(x)-1)(f(x)+1)=0 \end{align}\tag{for all $x$}$$


答:
$f$ satisfies this equation iff there exists a set $A$ such that $f =2\chi_A-1$. If $f$ is continuous then it is identically $1$ or identically $-1$.

[For the proof define $A$ as $\{x: f(x)=1\}$; $ \chi_A$ is defined by $ \chi_A(x)=1$ if $ x \in A$ and $0$ if $x \notin A$].

Title: Describe all functions which are defined as $f:\mathbb R \to \mathbb R $ and satisfying $f(x)=\frac1{f(x)}$

Body:
Describe all functions which are defined as $f:\mathbb R \to \mathbb R $ and satisfying $$f(x)=\dfrac1{f(x)}$$

My Attempt:

If $f$ is constant function then.

$$(f(x))^2=1$$

Then $f(x)=1$ or $f(x)=-1$

I cannot think non-constant function satisfying this

What about $f$ being continuous?


问:
Is it true that $ \int_{[1,\infty)} f_n\to \int_{[1,\infty)} f$?
Can you please help me solve this on measure theory? My TA did not go over this. He said we are not going over this but you can do this if you want. Can someone please explain to me? Thanks.

Suppose $f_n\to f$ on $[1,\infty)$ and $|f_n(x)| \le 1 / x^2$ where $1\le x$.

Is it true that $ \int_{[1,\infty)}  f_n\to \int_{[1,\infty)} f$ ?

答:
you have $|f_n (x)|\leq \frac{1}{x^2}$ , for $x \geq 1$.

By the dominated convergence theorem you have

$$\lim_{n \rightarrow \infty } \int f_n =  \int \lim_{n \rightarrow \infty } f_n = \int f .$$
In the first equality was use the dominated convergence theorem


答:
If you allow discontinuous functions, then the general solution is that you pick some subset $A$ of $\Bbb{R}$ and define a function $f_A : \Bbb{R} \to \Bbb{R}$ by:

$$
f_A(x) = \left\{\begin{array}{l@{\quad}l} 1 & \mbox{if $x \in A$} \\
-1 & \mbox{if $x \not\in A$}\end{array}\right.
$$

Any solution is $f_A$ for some unique $A$.
The two continuous solutions correspond to $A = \Bbb{R}$ and $A = \emptyset$.


问:
What are the (real) coefficients $a (t), b (t)$ of the representation $r (t)$?
A spaceship flies in $\mathbb{R}^2=\mathbb{C}$ with constant speed in the direction of the x-axis.

Physicist P1 uses the canonical unit vectors $e_1 = (1, 0) = 1$ and $e_2 = (0, 1) = i$ as the coordinate system; in this the location of the spaceship at time $t$ is given by $r (t) = (t, 0) = t$.

Physicist P2 uses a coordinate system rotating around $0$, whose (time-dependent) Axes he holds for the guidelines of the universe, and that are given by
$\tilde{e}_1 (t) = (\cos (t), \sin (t)) = \cos (t) + i\sin (t)$, $\tilde{e}_2 (t) = (- \sin (t), \cos (t)) = - \sin (t) + i \cos (t)$

What are the (real) coefficients $a (t), b (t)$ of the representation $r (t) =
a (t) \tilde{e}_1 (t) + b (t) \tilde{e}_2 (t)$ in the system of P2 explicit?

$$$$

From $r (t) =a (t) \tilde{e}_1 (t) + b (t) \tilde{e}_2 (t)$ we get $(t,0)=a (t) (\cos (t), \sin (t)) + b (t) (- \sin (t), \cos (t)) $.

So we have to solve the system $$t=a(t)\cos t-b(t)\sin t \\ 0 =a(t)\sin t+b(t)\cos t$$ or not? Do we have to differentiate these equations?

Do we use also that the speed is constant?

答:
Hint:

Fix $t$.  For a given orthonormal basis $\{\tilde{e}_1 (t),\tilde{e}_2 (t)\}$, find the coordinates for a given vector $r(t)$ with respect to that basis using inner products:

$$
a(t) = r(t) \cdot \tilde{e}_1 (t),\\
b(t) = r(t) \cdot \tilde{e}_2 (t) .
$$

Yes, we are solving the system you wrote.  But the inverse of an orthogonal matrix is easy!


问:
Moment of inertia of sphere of changing density
I need to calculate moment of inertia of a sphere. The density of sphere changing $\rho=cz^2$ All the three components $I_x,~I_y,~I_z~.$

I did the assumption $dm=\rho dV$ then I found $c$ as $=\frac{15m}{4\pi r^5}~.$
 Then called
$$I=\int\rho dv z^2=c\int r^4 \cos^4\theta r^2 \sin\theta dr d\theta d\phi=c\int r^6 dr \int \cos^4\theta \sin\theta d\theta \int d\phi $$
Finally I found the $I=\frac{3}{7}\frac{R^7}{r^5}$

But I guess it's for $I_z$. How can I calculate $I_x=I_y$

答:
$I_x = \int dm \ (y^2+z^2), I_y = \int dm \ (x^2+z^2), I_z = \int dm \ (x^2+y^2)$

As $\rho = cz^2, dm = \rho dV = cz^2 dV$

Using cylindrical coordinates,

$x^2+y^2 = r^2, r^2 + z^2 = R^2 \ $ where $R$ is the radius of the sphere.

$dm = \rho dV = c z^2 \ r \ dr \ dz \ d\theta \ $

$\displaystyle I_z = c \int_0^{2\pi} \int_{-R}^{R} \int_0^{\sqrt{R^2-z^2}} z^2 r^3 \ dr \ dz \ d\theta = \frac{8 \pi c R^7}{105}$

Please note that density is only a function of $z$, so $I_x$ and $I_y$ will be equal due to symmetry of sphere.

That leads to $\displaystyle I_x = I_y = \frac{I_x + I_y}{2} = \frac{c}{2} \int_0^{2\pi} \int_{-R}^{R} \int_0^{\sqrt{R^2-z^2}} z^2 (r^2+2z^2) \ r \ dr \ dz \ d\theta$

$\displaystyle  = \frac{I_z}{2} + c \int_0^{2\pi} \int_{-R}^{R} \int_0^{\sqrt{R^2-z^2}} z^4 \ r \ dr \ dz \ d\theta$

I will leave the integral for you to complete.

Lastly,

$m = c \displaystyle \int_0^{2\pi} \int_{-R}^{R} \int_0^{\sqrt{R^2-z^2}} z^2 \ r \ dr \ dz \ d\theta$

Once you find $m$, you can also write moment of inertia in terms of $m, R$.


问:
How to solve this system of set equalities?
I have been trying to solve this for over an hour but I am still stuck on how to solve it.

Given the sets $A_1, A_2, \ldots, A_{2014}$ and a system of $2014$ equations with variable X:

\begin{align}
A_1 \cup X &= A_2 \cap A_3 \\
A_2 \cup X &= A_3 \cap A_4 \\
&\vdots \\
A_{2012} \cup X &= A_{2013} \cap A_{2014} \\
A_{2013} \cup X &= A_{2014} \cap A_1 \\
A_{2014} \cup X &= A_1 \cap A_2
\end{align}

Determine when a solution exists and when it does, find it.

I usually solve such equations by rearranging them into $(P \cap X) \cup (Q \cap X)=\emptyset$ and then write the condition $Q \subseteq P^C$ and the solutions are all $X$ such that $Q \subseteq X \subseteq P^C$, but I have no idea how to do that with this system. Any help would be appreciated!

答:
If I am not mistaken, your equations imply that for all $i$, we have that $A_i \subseteq A_{i+1}$.

Hence, $(A_i)_{1 \leq i \leq 2014}$ is an increasing sequence. But the last equation tells us that $A_{2014} \subseteq A_1$. Hence, all the sets $A_i$ are equal to the same set, denoted by $A$.

Finally, all your equations can be rewritten as $A \cup X = A$ which is equivalent to $X \subseteq A$.

Conclusion: the system has a solution if and only all the sets $A_i$ are identical, in which case the solutions are exactly the subsets of $A_1$.

(I hope I have not overlooked anything because this solution seems to completely miss the point of "system of set equations" ...)


问:
Expectations of the Wiener Process
In the Wiener process, by definition, we have that $E(W(t))=0$.
What would be the $E(W(e^{2t})$?

My best guess would be that this would equal to 0 again because the expectation with the Wiener process of any t would be equal to 0.

Furthermore, what would be the variance of this Wiener Process?

答:
Since for any $s\ge0$ we have that $W(s) = W(s)-W(0)\sim\mathcal{N}(0,s)$, we have $W(e^{2t})\sim\mathcal{N}(0,e^{2t})$, with mean $0$ and variance $e^{2t}$.

It is not a Wiener process as it has the wrong variance.

Title: Expectations of the Wiener Process


问:
Is this group a Kahler group?
Let $S$ be a non-orientable closed surface with $b_{1}(S)=2$, is $\pi_{1}(S)$ a Kahler group?

答:
No: https://mathscinet.ams.org/mathscinet-getitem?mr=3033516

It follows from the Biswas-Mj result that a one-relator Kahler group has no torsion in $G^{ab}$. But your group is a one-relator group with torsion in $G^{ab}$.

Title: Is this group a Kahler group?


问:
Cesaro means $\frac{1}{N} \sum_1^{N} a_n$ of a positive sequence converging to zero while the sequence itself does not
In this post: Cesaro means of a positive sequence, OP is referring to a "famous example" of a sequence $a_n \geq 0$ whose Cesaro means $\frac{1}{N} \sum_1^{N} a_n$ converge to zero while the sequence itself does not. What "famous example" is OP referring to?

答:
Take
$$a_k = 1 \quad \text{ if } k=2^n \text{ for a }n \in \mathbb{N} \quad \quad \text{and} \quad a_k=0 \text{ otherwise}$$


问:
What does Frobenius reciprocity say in the case of $G = S_3$ and $H = C_2$?
Using character theory the statement of Frobenius reciprocity would be:
$$ \langle \text{Ind}_H^G \psi , \phi \rangle_G = \langle \psi , \text{Res}_H^G \phi\rangle_H $$
Let's try to evaluate this expression with $G = S_3$ (the permutation group of three objects) and $H = C_2$ is a 2-cycle.

The character table of $S_3$ looks like this:
$$ \begin{array}{c|crr}
 & (1) & (12) & (123) \\ \hline
\phi_1 & 1 & 1 & 1 \\
\phi_2 & 1 & -1 & 1 \\
\phi_3 & 2 & 0 & -1 \\
\end{array} $$

The character table of $C_2$ is even simpler:
$$ \begin{array}{c|cr}  & (1) & (12) \\ \hline
\psi_1 & 1 & 1\\
\psi_2 & 1 & -1 \end{array} $$

How do we compute the induced characters $\text{Ind}_{H}^{G} \psi$ (from $H = C_2$ to $G = S_3$) and in the other direction, the restricted characters $\text{Res}_H^G \phi$.

We also might need the Schur orthogonality relations to compute the inner product.  So what does the equation look like in the end?

Related:

Find character table for symmetric group $S_3$

Character theory questions on Chemistry.SE
 Why does $(2z^2-x^2-y^2,x^2 - y^2)$ have -1 trace in the character table of $T_d$ for $C_3$ rotation?

https://physics.stackexchange.com/questions/306995/decoding-the-character-table

答:
As an alternative to Nicky Hekster's answer, if you wanted to compute the induced characters directly, rather than by using Frobenius Reciprocity, you could do it using the formula

$$\psi_H^G(g) = \frac{1}{|H|} \sum_{x \in G,\,x^{-1}gx \in H}\psi(x^{-1}gx).$$
So, in your example $G=S_3$, $H=C_2$, for $g = (1,2,3)$, since $x^{-1}gx \not\in H$ for all $x \in G$, we have $\psi_H^G(g)=0$.

For $g = (1)$, we get $\psi_H^G(g) = 6/2 = 3$ for $\psi=\psi_1$ or $\psi_2$.

Finally, for $g=(1,2)$, we have $x^{-1}gx \in H$ iff $x=(1)$ or $(1,2)$, and we find $(\psi_1)_H^G(g) = 1$ and $(\psi_2)_H^G(g) = -1$.

Now you can see directly that $(\psi_1)_H^G = \phi_1+\phi_3$ and $(\psi_2)_H^G = \phi_2+\phi_3$


问:
Locally finite Borel measure on $\mathbb{R}^2$ that is not a product measure.
In the construction of Lebesgue-Stieltjes measures on $\mathbb{R}$, I have learned that a Borel measure that is finite on bounded intervals corresponds to a right-continuous increasing real-valued function up to constants. Then we can construct a product measure in $\mathbb{R}^2$ from two such locally finite Borel measures.

My question is whether every locally finite Borel measure on $\mathbb{R}^{d}$ can be expressed as a product measure from those on $\mathbb{R}$ in this way.
The question arose from a separate construction of Lebesgue-Stieltjes measures on $\mathbb{R}^2$ by generating it from function $F:\mathbb{R}^2\rightarrow \mathbb{R}$ satisfying the following properties:

F is increasing in the sense that $a_0\leq a_1$, $b_0\leq b_1$ then $F((a_1,b_1))-F((a_0,b_0))\geq 0$.

$F((a_1,b_1))-F((a_0,b_1))-F((a_1,b_0))+F((a_0,b_0))\geq 0$ for $a_0\leq a_1$, $b_0\leq b_1$.

$F$ is right-continuous for each coordinate.

Then $\mu_F((a_0,a_1] \times (b_0,b_1])=F((a_1,b_1))-F((a_0,b_1))-F((a_1,b_0))+F((a_0,b_0))$ can be extended to a Borel measure using a similar proof in $\mathbb{R}$, and similar correspondence to locally finite Borel measures hold as well.
It is easy to see that a product measure from $f$ and $g$ satisfies these conditions by defining $F((a,b))=f(a)g(b)$.
But I was not able to show the converse, that is, this construction is just a product measure construction.

I tried to come up with a locally finite Borel measure on $\mathbb{R}^2$ that cannot be constructed as a product measure, but I was not able to do so.
The only thing I could see was that if $F((x,y))=f(x)+g(y)$ where each function depends on each variable only, then the resulting measure generated by $F$ is just zero measure.
I have also found that there is a theory of disintegration of measures, but I am not familiar with the topic.

Any help would be appreciated.

答:
Let $\mu$ be the image measure of the Lebesgue measure, $\lambda$, on $\Bbb R$ by the map $d: \Bbb R \rightarrow \Bbb R ^2$ such that $d(x) = (x,x)$. Then we have $\mu((a_0,a_1] \times (b_0,b_1])=\lambda((a_0,a_1] \cap (b_0,b_1])$.

Let us prove that there are no functions $f$ and $g$ such that $\mu((a_0,a_1] \times (b_0,b_1])= (f(a_1)-f(a_0))(g(b_1)-g(b_0))$. So $\mu$ is not a product measure of Borel measures on $\Bbb R$.

Proof:
Suppose such functions $f$ and $g$ exist. It is clear that $f$ can not be constant. So there are $a_1$ and $a_0$ such that $f(a_1)-f(a_0) \neq 0$. So given any $b_0$ and $b_1$ such that $(a_0,a_1] \cap (b_0,b_1] = \emptyset$, we have
$$ (f(a_1)-f(a_0))(g(b_1)-g(b_0)) = \mu((a_0,a_1] \times (b_0,b_1]) = \lambda(\emptyset)=0$$
Since  $f(a_1)-f(a_0) \neq 0$, we must have $g(b_1)-g(b_0)=0$.

But then for any $b_0$ and $b_1$ such that $(a_0,a_1] \cap (b_0,b_1] = \emptyset$, we have
$$\lambda((b_0,b_1])= \mu((b_0,b_1] \times (b_0,b_1])= (f(b_1)-f(b_0))(g(b_1)-g(b_0))= 0 $$
Contradiction.

So, there are no functions $f$ and $g$ such that $\mu((a_0,a_1] \times (b_0,b_1])= (f(a_1)-f(a_0))(g(b_1)-g(b_0))$. So $\mu$ is not a product measure of Borel measures on $\Bbb R$.


问:
Does this property hold for fundamental vector fields?
Let G be a lie group and let M be a G-manifold. Let $\mathfrak{g}$ be the lie algebra of G. If $X \in \mathfrak{g}$, we denote by $X_M$ the vector field on M such that

$(X_M.f)(m) = \frac{d}{dt} f (exp(-tX)m)|_{t=0}$

for $f \in C^\infty(M), m \in M.$

Suppose that $X = Y + Y'$ , where  $Y,Y' \in \mathfrak{g}$, $\lambda \in \mathbb{R}$. Does this imply that $X_M = Y_M + Y'_M$ ?

答:
Let $\phi:G\rightarrow M$ defined by $\phi(g)=g.m$, $f(exp(-tY).m)=f(\phi(exp(-tY))$,

${d\over{dt}}f(exp(-tY)m)_{\bar t=0}={d\over {dt}}_{t=0}f(\phi(exp(-tY))=df_m.d\phi_{Id}.{d\over {dt}}_{t=0}exp(-tY))={d\over {dt}}_{t=0}f(\phi(exp(-tY))=df_m.d\phi_{Id}(-Y)$.

This implies that $(Y_M+Y'_M)(m)=df_m.d\phi_{Id}(-Y)+df_m.d\phi_{Id}(-Y')=df_m.d\phi_{Id}(-(Y+Y'))=X_M(m)$.


问:
What would happen if in the BigM method we don’t set the penalty for the artificial variables?
The usual line of argument is that since the artificial variables are introduced solely for the purpose of obtaining a basic feasible solution, and have no meaning in the context of the LP, they must be forced to be zero eventually. But what would happen if we let them be strictly positive?

As far as I can see, them being positive would not affect the feasibility of the other variables, and not including the penalty terms in the objective cannot make it worse. Would this result in infeasibility of the program or sub-optimal result somehow?

答:
As you mentioned, artifical variables are used to find a feasible solution.

If you don't set the penalty, then they might stay in the basis, resulting in an infeasible solution.  There is equivalence between the two formulations only when the artificial variables equal $0$.

For example, if your constraint is
$
x \ge 10
$
you need to add a slack variable $e$ and an artificial variable $a$:
$$
x-e +a = 10
$$
Now, if $a=10$, then $x=0\not \ge 10$, which is not feasible.


问:
Is there an expression for the ith term of this sequence $1,1,2,1,2,3,1,2,3,4,1,2,3,4,5...?$
I'm trying to do some work in Excel and if I found a formula for this sequence it would help a lot. I don't particularly need to know why the formula works.

I have found the sequence here .

But there in no nice form for it.

Thanks!

答:
The first expression given under FORMULA is quite straightforward:

$$a_n=n-\frac{m(m+1)}2+1,\quad\text{where}\quad m=\left\lfloor\frac{\sqrt{8n+1}}2\right\rfloor\;.$$

Everything here can be done in Excel.


问:
There is a number divisible by all integers from 1 to 200, except for two consecutive numbers. What are the two?
To reiterate the question, basically there is some number, $n$ that exists that is divisible all the integers $1, \dots, 200$, except for two consecutive numbers in that range. The goal is to find what those two consecutive integers are. The answer isn't trivial though, since $n$ needs to be divisible by all those numbers, it is difficult to find two numbers next to each other such the multiples of those numbers aren't less than $200$ and such that those cannot be prime factorized into numbers that are in the prime factorization of $n$.

I have tried doing this computationally, but the LCM of all the numbers in the range (less two of them) is ginormous and checking the divisibility condition doesn't seem to work on my computer. The problem would be simple if the two numbers didn't have to be consecutive, since we could just select two prime numbers, but since one must be even, this is not possible.

I am trying to think of properties of divisibility that could help, but haven't found anything that worked yet. For example, I was looking for numbers that a prime such that a number before or after it is the square of a prime number. This way, we could say that the prime number itself is omitted from $n$ and that there is only one factor of the square root of the other number in $n$. I am not sure if that would definitely work, but regardless I couldn't find those numbers. I tried another perfect square and a prime number, $196$ and $197$, but there must be enough factors to make two $14$s in $n$, so that doesn't work either.

I am not experienced at all in number theory or discrete math, this is just a brainteaser I have heard. (Also for reference, I do not know the answer to reverse engineer something from). Any help would be appreciated!

Thanks!

答:
First of all, one of those consecutive numbers must be a prime number. Because;

The divisibility of an arbitrary number to a prime number is not affected by its divisibility to numbers smaller than that prime number.

But, that prime number should be chosen in such a way that it must be greater than $200$ when multiplied by $2$. Let $p$ be a prime number in the $[1,200]$ closed interval. We can state that: $$0 \equiv n\pmod {2p} \space \Rightarrow \space 0 \equiv n\pmod p$$ $$2p \gt 200 \space \Rightarrow \space p \gt 100$$

Although this reasoning is valid, it is not enough for us to assign the value of the second integer. Let us take a look at the maximum possible number of a prime factor making up the integer $n$. For example, the number of $2$ as a factor in $n$ is $7$ at most. This boils down to the fact that the number of 2 as a prime factor in each divisor except for 128 in the [1,200] closed interval is 6 at most. If we exclude $128$ as one of the divisors of $n$, this does not affect the divisibility of $n$ to other integers in the $[1,200]$ closed interval. When the same thought process is applied to other prime factors making up the integer $n$, we see that predefined conditions do not meet. Therefore, one of the consecutive numbers is $128$.

Now, look at the numbers $127, 128$, and $129$. Among them $128$ is the one assigned but what is the other one? We have two options: $127$ and $129$. Since one of the consecutive numbers must be prime, we can easily deduce that the other one is $127$. So, $127$ and $128$ are the integers that do not divide $n$. Hope it helps!


问:
How does multiplication and power influence on domain and range?
given $f(x) = x^2 + 4x + 2, f(x)\rightarrow [2, 3]$

$ran_{f} = [14, 23]$

let $g(x) = (f(x))^2 = (x^2 + 4x + 2)^2$

let $h(x) = 3 \cdot (f(x)) = 3 \cdot (x^2 + 4x + 2) $

given $m(x) = 2x, f(x) \rightarrow [0, 10]$

let $n(x) = f(x) \cdot m(x) = (x^2 + 4x + 2) \cdot (2x)$

what is the domain and range of $g(x)$, $h(x)$ and $n(x)$?

I am sorry if the question seems stupid, I am trying to build a Segment Common Algebraic Expression object, I have achieved multiplication and power by recursive function successfully, but I have two variables xlim and ylim I don't know what to do with it. I have forgotten all I learned in high school. I am truely sorry if you were my math teacher who is looking at this.

答:
I'd say it works like this.

For multiplication by $c$ we just get a range of $[ca, cb]$ for $af(x)$ if $f(x)$ has range $[a,b]$ and $c$ is positive. If $c$ is negative the range is $[cb, ca]$.

Powers are a little more complicated since for even powers the sign dissappears and negative powers are also a thing.
Suppose $f(x)$ has range $[a, b]$:
For odd powers we get the range of $f^c$ is $[min(a^c, b^c), max(a^c, b^c)]$.
For even powers we get the range of $f^c$ is $[min(|a|,|b|)^c, max(a^c,b^c)]$ if $c$ is even and either $a$ and $b$ are both odd or both even.
If they have opposite sign and even power, then we get as range $[0, max(a^c,b^c)]$.
We get this min max thing because the even power of a negative number is negative.


问:
A 2nd countable metric space is separable

A 2nd countable metric space is separable

Can anyone give an idea on how to prove that, I could prove the converse quite easily but I am not sure which set to consider in order to show that it is dense.

答: HINT: Start with a countable base $\mathscr{B}$ for the topology and pick one point from each member of $\mathscr{B}$.</p>


问:
A surjective ring homomorphism which is not an isomorphism

Give an example of a ring $R$ and a surjective homomorphism $R \to R$ that is not an isomorphism.

Here is my approach. I know a lemma that says that if $R$ is a Noetherian ring, and $f: R \to R$ is a surjective homomorphism, then $f$ is an isomorphism. So the ring $R$ above must be non-Noetherian, which is hard(?) to find, as 'most' rings are Noetherian (or so I've been told in lectures). One example of a non-Noetherian ring that I know is $K[X,Y]/\langle X,XY, XY^2, XY^3, \dots \rangle$, and from this, I was trying to construct such a homomorphism. But I'm stuck; I can see no intuition, and no systematic way to do this except for guessing. Does anyone have a nice solution to this?

Note: I've done some searching, and it seems that upon learning higher mathematics, this is a standard problem. For example, When is a ring homomorphism $A \to A$ surjective but not injective. has a solution which uses a lot of mathematics that I am unfamiliar with. This problem is from a basic undergraduate algebra course, and I'm trying to work out the answer with only what I know.

Edit: The example of a 'non-noetherian' ring I give above is wrong.

答:
Analogous to @Arturo's answer, consider a "left-shift" on the infinite product of $\Bbb Z_n$'s say.  That's $(x_1,x_2,\dots)\mapsto(x_2,x_3,\dots)$.

As a corollary, invoking the theorem you mentioned, we have that $\prod_{k=1}^\infty\Bbb Z_n$ is not Noetherian.


问:
If the value of this improper integral is positive, then can one say that the integrand is positive almost everywhere?
I have a real valued continuous function $f(t)$ over the real line such that,
$$\int\limits_0^\infty{f(t)\,ds}\geq 0.$$
Then can it be concluded that the,
$$\int\limits_0^t{f(s)\,ds}\geq 0, \quad \forall t>0?$$

My attempt at an answer.

Since $$\int\limits_0^\infty{f(t)\,ds}\geq 0,$$ and $f(t)$ is continuous, I conjecture that, $f(t)$ is positive a.e. on $[0,\infty).$ Then clearly, $f(t)$ is positive a.e. on $[0,t]$ for every $t\in\mathbb{R}.$ This implies that, $$\int\limits_0^t{f(s)\,ds}\geq 0, \quad \forall t>0.$$

答:
A simple counterexample:
$$f(x) =
\begin{cases}
-\cos x  & \text{if $x\le\frac{3\pi}{2}$} \\
0 & \text{if $x>\frac{3\pi}{2}$}
\end{cases}$$


问:
Let $G$ be a group and $H = \{ g \in G \mid gx = xg \ \forall x \in G \}$. Show that $H \leqslant G$ and that $H$ is commutative.

Let $G$ be a group and $H = \{ g \in G \mid gx = xg \ \forall x \in G \}$. Show that $H \leqslant G$ and that $H$ is commutative.

How can I show that $H \leqslant G$? This seems to me that I would want to show that $H$ is a subgroup of $G$? That would seem to imply the statement $H \leqslant G$. So I would need to show that $H$ is non empty and closed under products and inverses?

Let $h\in H$, then $hh^{-1} = e_G$ right? So the inverse is there. How can I show the closure under products here?

答:
If $gx=xg$ for all $x\in G$ i.e. $g\in H$ and $hx=xh$ for all $x\in G$ i.e. $h\in H$ then
$$(gh)x=g(hx)=g(xh)=(xh)g=x(hg)=x(gh)$$for all $x\in G$ and thus $gh\in H$ i.e. $H$ is closed under multiplication.


答:
Recall that for $H$ to be a subgroup of $G$, it suffices to prove that for all $a, b \in H$, we have $ab^{-1} \in H$.

Suppose $a, b \in H$. We have $x^{-1} b = b x^{-1}$ for all $x \in G$, which is equivalent to $b^{-1}x = xb^{-1}$. We now multiply with $a$ from the left and get
$ab^{-1}x=axb^{-1}$. Now use that $a$ commutes and associativity, and arrive at

$$(ab^{-1})x = x(ab^{-1}),$$

which finishes the proof.

EDIT: In order to use this subgroup test, it is also necessary to show that $H$ is non-empty. This is trivial since the identity element of $G$ will always be in $H$.

Body:
Let $G$ be a group and $H = \{ g \in G \mid gx = xg \ \forall x \in G \}$. Show that $H \leqslant G$ and that $H$ is commutative.

How can I show that $H \leqslant G$? This seems to me that I would want to show that $H$ is a subgroup of $G$? That would seem to imply the statement $H \leqslant G$. So I would need to show that $H$ is non empty and closed under products and inverses?

Let $h\in H$, then $hh^{-1} = e_G$ right? So the inverse is there. How can I show the closure under products here?


答:
You need to use the axiom of choice, you can't always write an explicit countable dense subset.

Hint: Use the fact that you have a countable basis to the topology, and those sets are all non-empty.

Body:
A 2nd countable metric space is separable

Can anyone give an idea on how to prove that, I could prove the converse quite easily but I am not sure which set to consider in order to show that it is dense.


问:
The formulas of prostapheresis: memorization technique
This question is related purely for my students of an high school and indirectly for me. The formulas below are the formulas of prostapheresis,

\begin{cases}
\sin\alpha+\sin\beta=2\,\sin \dfrac {\alpha+\beta}{2}\, \cos \dfrac {\alpha-\beta}{2} \\
\sin\alpha-\sin\beta=2\sin \dfrac {\alpha-\beta}{2} \,\cos \dfrac {\alpha+\beta}{2}\\
\cos\alpha+\cos\beta=2\cos \dfrac {\alpha+\beta}{2}\,\cos \dfrac {\alpha-\beta}{2}\\
\cos\alpha-\cos\beta=-2 \,\sin \dfrac {\alpha+\beta}{2} \,\sin \dfrac {\alpha-\beta}{2}
\end{cases}

and while I am able to find them, I am not able to find a technique to memorize them.

Is there a technique to be able to memorize them?

答:
If the students can recall

the first formula
\begin{equation}
\sin\alpha+\sin\beta=2\,\sin \dfrac {\alpha+\beta}{2}\, \cos \dfrac {\alpha-\beta}{2},
\end{equation}

the odd symmetry of $\sin(x)$ and basic differentiation rules,

then they can promptly derive the other three formulas. The second one follows by $(2)$, as already mentioned. Taking the derivative with respect to $\alpha$ in the first formula we obtain
\begin{align}
\cos(\alpha)=\cos\bigg(\frac{\alpha+\beta}{2}\bigg)\cos\bigg(\frac{\alpha-\beta}{2}\bigg)+\sin\bigg(\frac{\alpha+\beta}{2}\bigg)\sin\bigg(\frac{\alpha-\beta}{2}\bigg),
\end{align}
and switching $\alpha$ and $\beta$ the same formula for $\cos(\beta)$ follows. Now sum and subtract (using $(2)$ again) to obtain the two identities for the cosine.


答:
To check closure under inverses you have to show that $h^{-1}\in H$. You only know that $h$ is invertible as element of $G$, that is there is $h^{-1}\in G$ such that $hh^{-1}=e_G$ but it might be the case that $h^{-1}\in G\setminus H$, i.e. that the inverse is not in the subset $H\subseteq G$. Similarily, you have to show that $e_G\in H$ and that if $h_1,h_2\in H$ implies that $h_1h_2\in H$.

Let's take a closer look at the inverses. We are given $h\in H$ and want to show that $h^{-1}\in H$ which is to say that $h^{-1}x=xh^{-1}$ for all $x\in G$. We are given that $xh=hx$ since by assumption $h\in H$. Now, first multiply from the left by $h^{-1}$ and then from the right by $h^{-1}$ to get:

$$xh=hx\quad\implies\quad h^{-1}xh=x\quad\implies\quad h^{-1}x=xh^{-1}$$

Now, show that $xe_G=e_Gx$ for all $x\in G$ and that if $h_1,h_2\in H$ then $x(h_1h_2)=(h_1h_2)x$ for all $x\in G$ to complete the proof.

Aside: the subgroup you are dealing with is called the center of $G$, denoted by $Z(G)$ and measures "how abelian" the group is. We have, for instances $G=Z(G)$ if and only if $G$ is abelian.

Title: Let $G$ be a group and $H = \{ g \in G \mid gx = xg \ \forall x \in G \}$. Show that $H \leqslant G$ and that $H$ is commutative.


问:
Straightening an arc OR curving a line segment to compare their respective lengths.
With no available means for their ready measurement, how can it be geometrically determined whether an arc and a line segment, both of unknown lengths, are equal or not lengthwise?

In a similar vein, how can an arc be mapped such that it is straightened and it's true length preserved? OR, vice versa, how to map a line segment to a curve such that its length is maintained?

I have a project that is stalled because I cannot find a solution myself despite repeated experiments and googling without success until I am near blue in the face.
I found two supposed methods that were heavily mathematical and beyond my level of comprehension, and one was also coordinate based and I am not well versed in coordinate geometry.

I have opted not to include my geometric 'scratchings' because they are particularly crude and messy and are far from being a working solution. I have included a figure of a simpler example from which I can likely extrapolate the methods from any solutions you may provide here. I want to map or transform (straighten) the red arc and compare its length with (or against) that of the blue line segment. (NB:  The red arc and the distance between the 2 large red dots are of equal length, but assume that this is not known). I have added some auxiliary lines and marked the half- and quarter-points of the arc to show my partial efforts and thinking.

答:
Let the side length of the square be $s$, so its diagonal is equal to $s\sqrt{2}$.

Then, notice that the line is equal to $2$ of the square's diagonals, so its length is $2s\sqrt{2}$.

Now, note that the arc has a radius equal to the square's diagonal, or $s\sqrt{2}$. Then, if the arc were equal to a full circle, its length would be $2\pi r = 2\pi s\sqrt{2}$.

However, since the arc is only a quarter of a full circle, we have to divide by $4$. Thus, the length of the arc is $\frac{\pi s\sqrt{2}}{2}$.

Thus, $\boxed{\text{The line has length }2s\sqrt{2}\approx 2.828s\text{, and the arc has length }\frac{\pi s\sqrt{2}}{2}\approx 2.221s.\ \ }$

For the general case with an arbitrary line and arc, you can first determine the length of the line by using the Distance Formula on its endpoints.

Next, given $3$ points on the arc (two endpoints and another random point), find the center of the arc by finding the intersection of the perpendicular bisectors of $2$ pairs of the points. Once given the center, find the radius of the arc using the Distance Formula.

Now, find the angle between the $2$ radii intersecting the endpoints. You can do this by converting the radii to vectors, taking their dot product, dividing by the radius squared, and taking the inverse cosine. (You might have to add $180^{\circ}$ or so.)

Finally, given the angle subtended by the arc (in radians) and its radius, multiply the two to get the length of the arc.

Thus, you can determine the length of the line given its endpoints, and you can determine the length of the arc given the endpoints and $1$ other point on the arc.


问:
Sequent calculus: What does it mean to say “choose the sequents from which we can derive some other sequent via the right conjunction rule”?
My question is probably quite silly really, but I am puzzled what this exercise is asking:

Excercise. Choose all sequents from which we can derive some other sequent via the right-side conjunction rule.

$\color{green}{\lnot (A\land B), \lnot A \vdash A\land B}$ (correct)

$\lnot (A\land B), A \vdash A\land B$

$\color{green}{\lnot (A\land B), \lnot A \vdash \lnot A \land \lnot B}$ (correct)

$\lnot (A\land B), B \vdash \lnot A \land \lnot B$

$\lnot (A\land B), \lnot B \vdash A\land B$

(Source: homework excerise in since-discontinued course, translated)

The correct answers are marked green.

Here is the referenced

Right-side conjunction rule:
$$\cfrac{\Gamma \vdash \mathcal{F}\quad \Gamma \vdash \mathcal{G}}{\Gamma\vdash \mathcal{F}\land \mathcal{G}}$$
where $\mathcal{F}, \mathcal{G}$ are zeroth order formulae, and $\Gamma$ is a list of zeroth order formulae or is empty.

I believe I do understand the basics of Gentzen's sequent calculus. I am aware of the axioms and rules for allowed derivations, including the right-side conjunction rule; the intuition behind the formalisation is also clear. I have derived dozens upon dozens of formulae in sequent calculus (as trees), and proved, for example, the correctness and completeness of sequent calculus wrt tautologies in propositional calculus.

With this in mind,

the problem is not how to solve the excercise. Rather, my problem is in understanding what is even being asked in the first place. So: What exactly is the excercise after?

Even with the “correct” answers known, I am admittedly confused. Firstly, the multiple choices are all of the form $\Omega \vdash \mathcal{F} \land \mathcal{G}$ (with varying $\Omega, \mathcal{F}, \mathcal{G}$). One usually uses the right-side conjunction rule to derive sequents of the form $\Omega \vdash \mathcal{F} \land \mathcal{G}$, not use them as “sequents from which” to derive other sequents. Secondly, what are the “some other sequents” which I should check? Others from the multiple-choice list? Other sequents in general?

Perhaps the phrasing of the exercise is a bit off from what is common... Is this a common type of exercise, maybe you can recognize the intended idea behind it? To reiterate, what is the exercise asking for?

EDIT

Here is another exercise of the same type. I also know the “correct” answer(s) to the test excercise but I will omit this information at first. That way I have an elementary check if an answer has the intended interpretation.

Test excercise.
Choose all sequents from which we can derive some other sequent via the left-side conjunction rule.

$\vdash (X \lor \lnot X) \land (Y \lor \lnot Y)$

$Z, X \land Y \vdash \lnot X \lor \lnot Y \lor \lnot Z$

$X \land (Y \lor Z), Y \land (X \lor Z) \vdash X \land Y \land Z$

$\lnot Z \land \lnot X \vdash (Y \lor \lnot Z) \land (\lnot Y \lor \lnot X)$

$X\land Y \vdash X\lor Y$

(Source: homework excerise in since-discontinued course, translated)

Left-side conjunction rule:
$$\cfrac{\Gamma, \mathcal{F}, \mathcal{G} \vdash \mathcal{H}}{\Gamma, \mathcal{F} \land \mathcal{G} \vdash \mathcal{H}}$$
where $\mathcal{F}, \mathcal{G}$ are zeroth order formulae, and $\Gamma$ is a list of zeroth order formulae or is empty.

The correct answers were indeed (spoiler)

 2 and 3.

答:
"Firstly, the multiple choices are all of the form Ω⊢F∧G. "

By that sort of reasoning, also, all of the multiple choices are of the form Ω⊢F and Ω⊢G.  Note that every well-formed formula in formal logic has the form of a variable.

But, they are not all of the same form Ω⊢F∧G where Ω indicates a defined list, and they don't all have the same Ω.

"One usually uses the right-side conjunction rule to derive sequents of the form Ω⊢F∧G, not use them as “sequents from which” to derive other sequents."

Consider proving that $\land$ associates .  I believe that if you do so, you'll find using sequents with conjuncitons to derive other conjunctions.

"Secondly, what are the “some other sequents” which I should check?"

What other sequents have the same premises in the same order?

The idea behind the exercise lies in recognizing when premise lists are equivalent as lists and also recognizing that the right-side conjunction rule requires that the assumed sequents have equivalent premise lists, otherwise the rule can't get used properly.  Thus, 1. can't get used with 2., 3., nor 5. since the antecedents consist of different lists, nor 4. get used with 5., and so on.


问:
Question about the percentages of sets in other sets
I am puzzling over the question:
If there are three sets, $A, B, $ and $C$, where:

$A \cap B \ne \emptyset$

$A \cap C \ne \emptyset$

$B \cap C \ne \emptyset$

i.e. $A, B$ and $C$ have to intersect in some way.

From there:

a) what % of set $D$ is contained inside set $A$ if $D=A \cap B$ ?

b) what % of set $E$ is contained inside set $A$ if $E=D \cap B$  (i.e. $A \cap B \cap C)$ ?

c) what % of set $F$ is contained inside set $A$ if $F$ is the triangle in the diagram below ?

State any variables that will have to be included to solve these problems.

How should I start??

EDIT:
I meant what percentage of all items in $D, E$ or $F$ are contained in all of the items in $A$

答:
Either I don't understand what you mean or your question has too few restrictions.

Based on what you said in the comments and the interpretation of your question, my answer is, it could be any percentage. Assume, e.g. that $A=\{1,2\}$, $B=\{1,3\}$ and $C=\{1,4\}$ then $D$ contains $50$% of A's elements. Now in comparison, if $A=\{1,2,3,4,5\}$, then D only contains $20$% of A's elements.


问:
The spectrum of self adjoint bounded operator on a Hilbert space - Why not use $\sigma(T^*)=\bar{\sigma(T)}$?
Let $T$ be a bounded self-adjoint operator on a Hilbert space $\cal H$. Denote by $\sigma(T)$ the spectrum of $T$.

I wish to prove that the spectrum of a self-adjoint operator is contained in the real line.

I sense it is a dumb question, but is it not enough to use the fact that $\lambda\in \sigma(T)$ if and only if $\bar{\lambda}\in \sigma(T^*)$?

Since $T=T^*$, we'll have $\lambda=\bar{\lambda}$, thus $\lambda$ is real.

Note that I've encounter the several questions asked here regarding this issue (such as the ones I attached below), but none answered my question.

Spectrum of self-adjoint operator on Hilbert space real

On the spectrum of a self-adjoint operator

答: You can't deduce that $\lambda = \overline{\lambda}$ immediately. All you know is that if $\lambda \in \sigma(T)$, then there is $\mu \in \sigma(T)$ with $\lambda = \overline{\mu}$.  How do you conclude $\lambda = \overline{\lambda}$ from this?


问:
Does this characterisation of homeomorphism hold?
Given two topological spaces $X_1, X_2$, I am looking for a characterisation of homeomorphism that works something like this:

$$ X_1 \simeq X_2 \iff \forall Y \text{ topological space } \mathcal{C}(X_1,Y) \simeq \mathcal{C}(X_2,Y) $$

where the first $\simeq$ is of course a homeomorphism, while the second $\simeq$ is some sort of isomophism yet to be (if possible) defined.

I was wondering if this statement could be also taken:

$$ X_1 \simeq X_2 \iff \mathcal{C}(X_1,X_2) \simeq \mathcal{C}(X_2,X_1) $$

答:
Your intuition is spot on: this is Yoneda's lemma. If $\mathsf{C}$ is some (locally small) category, the lemma states that $x \simeq y$ precisely if $\hom(x,-)$ and $\hom(y,-)$ are naturally isomorphic functors. Moreover, all such natural isomorphisms arise as postcomposition of some arrow.

There is a lot of terminology to unpack, but applying this to spaces gives that if $\mathcal{C}(X_1,Y) \simeq \mathcal{C}(X_2,Y)$ naturally (in the category-theoretical sense) in $Y$, then $X_1$ and $X_2$ are homeomorphic.


Title: If the value of this improper integral is positive, then can one say that the integrand is positive almost everywhere?

Body:
I have a real valued continuous function $f(t)$ over the real line such that,
$$\int\limits_0^\infty{f(t)\,ds}\geq 0.$$
Then can it be concluded that the,
$$\int\limits_0^t{f(s)\,ds}\geq 0, \quad \forall t>0?$$

My attempt at an answer.

Since $$\int\limits_0^\infty{f(t)\,ds}\geq 0,$$ and $f(t)$ is continuous, I conjecture that, $f(t)$ is positive a.e. on $[0,\infty).$ Then clearly, $f(t)$ is positive a.e. on $[0,t]$ for every $t\in\mathbb{R}.$ This implies that, $$\int\limits_0^t{f(s)\,ds}\geq 0, \quad \forall t>0.$$

Body:
$$
f(x) = \begin{cases} -1 & \text{if } 0<x<1, \\ 2e^{-(x-1)} & \text{if } x>1. \end{cases}
$$
Then
$$
\int_0^1 f(x) \,dx = -1 \text{ and } \int_0^\infty f(x)\,dx = +1.
$$


问:
Prove: every strongly connected graph has a single stationary distribution
Let $G= \langle V,E ~\rangle$ be a finite graph. We define $A_g$ to be the matrix with $[A_g]_{i,j}=1$ if there is an edge $e=(i,j)$ and $0$ otherwise.

Let $P$ be the transition matrix for a random walk on the graph $G$. So $P_{i,j}=P(x_t=j|x_{t-1}=i)$

$P$ is a normalize version of $A_g$ such that every row sums to $1$.

We define $\pi$ to be a stationary vector if it holds  that $\pi_i \geq 0$ and $\lVert \pi \rVert_1=1$ and satisfy $\pi^T=\pi^TP$

For ergodic graphs, I know there is a single stationary vector $\pi$ and for every vector $x$ we have $\lim_{t\to\infty}x^TP^t=\pi^T$

What I want to prove, is that for not ergodic graphs, but still strongly connected (for every vertex $i$ and vertex $j$ there is a way to get from $i$ to $j$, or in other word there is some $k$ that $[P^k]_{i,j} > 0$), there is only a single stationary vector $\pi$.

I know I won't be promised that for every $x$ we get $\lim_{t\to\infty}x^TP^t=\pi^T$, but still there is only 1 $\pi$

I can find a stationary vector with $\pi_i=\frac{deg(i)}{2\cdot |E|}$. One can multiply it by $P$ and see it works. But I am struggling to show only this $\pi$ can work with strongly connected graph.

Thanks

答:
This is false for an infinity graph.

Take $G=(\mathbb{Z},E)$ with $(i,j) \in E $ if $|i-j|=1$

Then by an argument of symmetry, the probability measure should be the same on each integer but then can not sum to $1$.


问:
Confusion about domain of definition for rational functions
Many times I get confused about the domain of function when we deal with limit or continuity or differentiation at particular point. I want to get clear about this confusion.

From baby rudin.

In definition $5.1$, to define derivative of real valued $f$, it is first assured that $f$ is defined on $[a,b]$ and then derivative is defined for $x\in [a,b]$.

Theorem $5.3$ part (c) says, Suppose $f$ and $g$ are defined on $[a,b]$ and are differentiable at point $x\in [a,b]$, then $(\frac{f}{g})'(x)=\frac{g(x)f'(x)-g'(x)f(x)}{g(x)^2}$, provided $g(x)\neq 0$.

Here's my doubt : When we define $\frac{f}{g}$, we alter the domain by removing those points at which $g$ takes value $0$. Then, it is possible that the alterted domain for $\frac{f}{g}$ may not be even interval. Then how to define derivative? I think along with assumption $g(x)\neq 0$, we should put one more assumption that in some neighbourhood of $x$, $g$ doesn't take value $0$.

Such kind of confusion pops up many times in my mind. For simplicity, can I always make the assumption that given function(even if rational) is nicely defined in some neighbourhood of point $p$, (i.e. when domain is real, function makes sense for $(a,b)$ and $p\in (a,b)$), at which we wish to determine continuity/differentiability/limit.

答:
It seems that you missed the sentence that appears right before theorem 5.2:

If $f$ is defined on a segment $(a,b)$ and if $a<x<b$, then $f'(x)$ is defined […] as above. But $f'(a)$ and $f'(b)$ are not defined in this case.

So, consider, say, $f,g\colon[0,1]\longrightarrow\Bbb R$ defined by $f(x)=x$ and by $g(x)=2x-1$ respectively. Then the domain of $\frac fg$ is $\left[0,\frac12\right)\cup\left(\frac12,1\right]$. In particular, the domain of $\frac fg$ contains both intervals $\left(0,\frac12\right)$ and $\left(\frac12,1\right)$, and therefore it makes sense to talk about $\left(\frac fg\right)'$ when $x$ belongs to any of these intervals.


问:
Is there still any hope that the GCH could be equivalent to some large cardinal axiom?
Is there still any hope that the GCH could be equivalent to some large cardinal axiom?

Even a simple yes or not answer will be fine. Thanks!!

答: I don't think there was ever any hope that the GCH would be equivalent to a large cardinal axiom.  There was once a hope that CH or its negation would be implied by a large cardinal axiom, but this was dashed by the Levy–Solovay theorem and its generalizations, which say that large cardinals are preserved by "small" forcings such as the usual ones to force CH and not CH.


问:
Radon-Nikodym derivative $\frac{d\mathbb{P}_{X}}{d\mathbb{Q}_{X}}$ from $\frac{d \mathbb{P}}{d\mathbb{Q}}$?
Assume we have two probability measures $\mathbb{Q}$,$\mathbb{P}$ and assume we know the radon-nikodyn-derivative $\frac{d \mathbb{P}}{d\mathbb{Q}}$.

For some random variable $X$, $\mathbb{P}_{X}(A):= \mathbb{P}(X^{-1}(A))$

Can we say anything about the radon-nikodyn derivative

$$\frac{d\mathbb{P}_{X}}{d\mathbb{Q}_{X}} ?$$

Can we compute it from the given one?

答:
I am not entirely sure if something can be said in general about this Radon-Nikodym derivative, but I will provide a positive answer to your question in the case when $X$ is actually invertible. To this end assume $\mathbb{P}, \mathbb{Q}$ are probability measures on the measurable space $(\Omega, \mathcal{B})$ and denote the aforementioned Radon-Nikodym derivative by $f := \frac{d\mathbb{P}}{d\mathbb{Q}}.$ Also let $X : \Omega \rightarrow \mathbb{R}$ be an invertible random variable. I will denote the pushforwards of the probabilities in the classical manner, i.e. $X_\ast \mathbb{P}$ for example.

By the very definition of the Radon-Nikodym derivative $f,$ we have that for every measurable set $A \subseteq \Omega,$ $\mathbb{P}(A) = \int_A f d \mathbb{Q}.$ Let now $B \in \mathcal{B}(\mathbb{R}).$ We have that $X_\ast \mathbb{P}(B) = \int_B d X_\ast \mathbb{P} = \int_\mathbb{R} \chi_B d X_\ast \mathbb{P} = \int_\Omega \chi_B \circ X d \mathbb{P} = \int_\Omega \chi_B \circ X \cdot f \,\, d\mathbb{Q} = \int_\Omega (\chi_B \cdot f\circ X^{-1}) \circ X \,\, d\mathbb{Q} = \int_\mathbb{R} \chi_B \cdot f \circ X^{-1} \,\, d X_\ast \mathbb{Q} = \int_B f \circ X^{-1} \,\, d X_\ast \mathbb{Q}.$ Since $B$ was arbitrary, it follows that the Radon-Nikodym derivative of $X_\ast \mathbb{P}$ with respect to $X_\ast \mathbb{Q}$ is $\frac{d \, X_\ast \mathbb{P}}{d \, X_\ast \mathbb{Q}} = \frac{d \mathbb{P}}{d \mathbb{Q}} \circ X^{-1},$ which is quite a natural result to expect.


问:
Expected power of random vector by matrix
I need to find the expected power of Y = A X where X~CN(0,I) and A is a complex matrix.
Can anybody help me?
Thank you

答:
The definition of power of a random vector is the expected value of the squared norm of the vector: $E(||y||^2)$.

Moreover,  $||y||^2=y^H y$.

Now we calculate $E(|(|y|)|^2 )=E(x^H A^H Ax)$. Then since

$Tr(ABCD) = Tr(DABC) = Tr(CDAB)=...$

and because expectation and trace can be interchanged we can proceed as follows. Also note that $a = Tr(a)$ if $a$ is a scalar (and $x^H A^H Ax$ is a scalar). Then we get

$E(x^H A^H Ax) = E(Tr(x^H A^H Ax))= E(Tr(A^H Axx^H)) = Tr(E(A^H Axx^H))= Tr(A^H AE(xx^H))= Tr(A^H A)$

which gives the solution.