问:
What is an example and a non-example of a left-invariant vector field?
I am going through a few different books reviewing Lie Groups and Lie Algebras and came across the notion of a left-invariant vector field. Given a diffeomorphism $L_ag : G \to G, L_a(g) = ag.$ We then have a map from $T_gG$ to $T_{ag}G$ such that $L_{a*} X |_g = X|_{ag}.$ Additionally, we have in coordinates that $$X^u(g) \frac{\partial x^v(ag)}{\partial x^u(g)}\frac{\partial}{\partial x^v} |_{ag} = X^v(ag)\frac{\partial}{\partial x^v} |_{ag}.$$ However I have two issues. The fist is this definition seems so intuitively obvious I cannot think of a non-example (Note: I've checked previous Stackexchange questions on this matter and the solutions amounted to short comments that didn't fully answer my question) and the second is I am a bit shaky as to how use the coordinate definition to check if a vector field is or isn't left invariant.

Therefore, I am hoping someone could provide two examples. The first being an example of a left invariant vector field and the second a non-example. If you could show why these examples satisfy and fail to satisfy the requirements respectively I would be greatly apppreciative!

答:
Here are two fundamental examples.

Take the vector field $X=\frac{d}{d\theta}$ on $S^1$, which we think of as laying in the complex plane $\Bbb{C}\cong \Bbb{R}^2$. Viewed geometrically, this is the vector field which sends $(x,y)\in S^1$ to the tangent vector $(y,-x)\in T_{(x,y)}S^1.$ You should draw a picture of this and it should become clear why it is left invariant. It is because $e^{i\theta}\in S^1$ acts by rotation by $\theta$ radians counterclockwise for $\theta \in [0,2\pi)$ (say), and it is a good exercise to check that the rotation by $\psi$, $R_\psi$, acts on $X_\theta$ by $R_\psi(X_\theta)=X_{\theta+\psi}.$

It is easy to construct non-invariant vector fields on $\Bbb{R}^2$. Note that left-invariance is equivalent to translation invariance under all $v\in \Bbb{R}^2$. Pictorally, this is a vector field that does not "vary" from point to point, like $X:=\frac{\partial}{\partial x}+\frac{\partial}{\partial y}$ or really any vector field of the form $Y=a\frac{\partial }{\partial x}+b\frac{\partial }{\partial y}$. To construct non-invariant vector fields, just introduce non-constant $C^\infty$ coefficient functions like $Z=x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}$ for instance. At least for this type of example it is easy to see non-invariance. Left multiplication with $v\in \Bbb{R}^2$ is $\mu_v(x,y)=(x,y)+v$ and taking the differential of this affine transformation gives $(\mu_v)_*=\text{Id}.$ So, a general vector field
$$
X_{(x,y)}=f(x,y)\partial_x+g(x,y)\partial_y
$$  is invariant if and only if for all $v$, $(\mu_v)_*X_{(x,y)}=X_{(x,y)+v}$. Writing out the equations and comparing coefficients of $\partial_x$ and $\partial_y$ we see that this is in turn equivalent to $f(x,y)$ and $g(x,y)$ being translation invariant for all $v$. I.e. for all $v$ $f((x,y)+v)=f(x,y)$. This is equivalent to $f$ and $g$ being constant and hence equivalent to $X=a\partial_x+b\partial_y$.

This idea generalizes to $\Bbb{R}^n$ quite easily.


问:
Sequences and Cauchy proofs
I am solving the following problem:

a.) Let $(a_n)$ be a sequence such that $|a_{n+1} −a_n| < C^n$ for all $n$ for some constant $0 < C < 1$ . Prove that $(a_n)$ is a Cauchy sequence, therefore is convergent.

This is what I have:

$C^n +  C^n+1+…+C^m, S=C^n+C^{n+1}+…+C^m, CS=C^{n+1}+C^{n+2}+…+C^{m+1}(1-0)=C^n-C^m+1, S=C^n-C^m+1$

b.) Let $(a_n)$ be a sequence such that $|a_{n+1} −a_n| <\frac{1}{n}$ for all $n$. Is it true that such $(a_n)$ is always convergent?

Im not sure about what I have for part a, and I haven’t begun part b.)

Thanks.

答:
Consider the series $$\sum_{i=1}^\infty  |a_{i+i} - a_i|$$

Since the terms in this series satisfy $|a_{n+1} - a_n| <C^n$ for each $n$, we can say that $$\sum_{n=1}^\infty |a_{n+1} - a_n| < \sum_{n=1}^\infty C^n$$

The series on the right is a geometric series with common ratio $C \in (0,1)$. Hence, the geometric series converges. Furthermore, since the series on the left is bounded below by $0$ and consists of terms that are non-negative real numbers, the series on the left must converge. This implies that the "tail" of the series on the left gets arbitrarily small. In other words, if we fix $\epsilon > 0$, there exists $M > 0$ such that

$$\sum_{n=M}^\infty |a_{n+1} - a_n| < \epsilon$$

Then, we note that if we pick two numbers $n>m > M$, we have $$|a_n - a_m| = |(a_n - a_{n-1}) + (a_{n-1} - a_{n-2}) + \ldots + (a_{m+1} - a_m)| \leq \sum_{i=m}^n |a_{i+1}-a_i|\leq \sum_{i=M}^\infty|a_{i+1} - a_i|$$

where the first inequality that appears follows by repeated applications of the triangle inequality. The sum on the very right we know to be smaller than $\epsilon$, so it follows $|a_n - a_m| < \epsilon$. Hence, our sequence is Cauchy.

For part (b), we need to notice how crucial the fact was that we had a geometric series in the proof of part (a). If we simply let $|a_{n+1} - a_n| < \frac{1}{n}$, we may not necessarily have had a convergent series since $$\sum_{n=1}^\infty \frac{1}{n} = \infty$$

If you can construct a sequence $(a_n)$ satisfying $|a_{n+1} - a_n| < \frac{1}{n}$ but for which $$\sum_{n=1}^\infty |a_{n+1} - a_n|$$ does not converge, you may very well be in business for a counterexample to part (b). In fact, you could let $$a_n = \sum_{i=1}^n \frac{1}{i}$$

You would just have to show this satisfies the desired inequality and it does not converge.

ContentLicense: CC BY-SA 4.0

Title: Sequences and Cauchy proofs


问:
Approximation of Lebesgue zero sets in Euclidean metric space.
Let $O \subset \mathbb{R}^N$, $N \geq 3$, be a bounded set and $\mu$ the normalized probability Lebesgue measure on the induced $\sigma$-algebra of $O$. Therefore, $\mu(O) = 1$. Due to the regularity of $\mu$, we have

\begin{align}
\forall\ \varepsilon > 0,\ \exists\ K = K(\varepsilon) \subset O,\ K \text{ compact}:\ \mu(K) \geq 1 - \varepsilon.
\end{align}

Define $A := O \setminus K$ and let $d(\cdot,\cdot)$ be the Euclidean metric. Is the following statement true in general?

\begin{align}
\forall\ \epsilon > 0,\ \forall\ x \in A,\ \exists\ y = y(\epsilon, x) \in K:\ d(x,y) < \epsilon.
\end{align}

Remark: My intention with the above question is to get a better feeling of how metrically/topologically awkward a zero measure set, within a bounded set, can be.

答:
The set $A$ is open and so your statement is always false (because any point in $A$ is an interior point).

The answer is unrelated to the Lebesgue measure.


问:
Preparation for Putnam?
Currently, my math training includes Calc 1-3, linear algebra, and some introduction to set theory/discrete math.
What would you recommend that I study over summer in preparation for the Putnam? Real analysis, topology, abstract algebra (all of the above)? What would be the most pertinent? Thanks!</p>

答: Putnam does not require any knowledge of analysis/algebra/topology, just do past exams like suggested above by user6312.</p>

ContentLicense: CC BY-SA 3.0

Title: Preparation for Putnam?


问:
Applying Möbius' theorem

Let $\displaystyle T(Z)=\frac{z+1}{z+i}$. Find a straight line in $\mathbb{C}$ that is mapped by $T$ to a circle.

True or False: Every holomorphic function $f:\mathbb{C}\to\mathbb{C}$ maps any straight line in $\mathbb{C}$ to either a circle or a straight line.

My Answer:

For  $z= 1+i,z= 1-i, z=-1-i$ substituted in $T$, I get $3$ distinct outputs: $z=1+i$ gives $\frac{4}{5}-\frac{3i}{5}$, $z=1-i$ gives $2-i$, and $z=-1-i$ gives $i$. Since these point are not in straight line then by Möbius' theorem it should be a circle.

答:
Note that $T(z)\to\infty$ as $z\to -i$. Therefore to have a bounded (i.e. circular) image, the line must stay away from $-i$. Thus for example the real axis is such a line.

2 is trivially false, Try almost any function and almost any line (e.g. $z\mapsto (z+i)^2$ and the real axis).

Title: Applying Möbius' theorem


问:
Find the sum of all numbers greater than $10000$ formed by using all digits $0,1,2,4,5$
Find the sum of all numbers greater than $10,000$ formed by using all digits $0,1,2,4,5$ and no digit being repeated in any number.

I could find that the number of such numbers are $96$ but I could not find their sum.

答:
Let's keep placeholders for where we will keep the digits :

[] [] [] [] []

Now, in the first box i.e: ten thousands place, we can put any number in the following set $ \{1,2,4,5 \}$ and after that have 4! ways of filling in the remaining numbers. So, there are 4! numbers with 1 as first number, 4! numbers with 2 as first number.. till 5. So, the sum of digits in ten thousand place is given as:

$$ 4!(1+2+4+5) \cdot 10^4$$

Similarly, suppose we fix some digit in the thousand place, then we have $ 3 \cdot 3!$ way of arranging the rest of numbers (remember we can't keep zero in first place). Hence the sum of no.s in thousand place is given as:

$$ 3 \cdot 3! ( 1 + 2 + 4 +5) \cdot 10^3$$

And, similar results for hundreds , tens and ones place. Now, to get the whole number, we just need to add up the parts of it's expanded form. Hence,

$$ \text{number}  = 4!(1+2+4+5) \cdot 10^4 + 3 \cdot 3! (1+2+4+5) \cdot 10^3 + 3 \cdot 3! (1+2+4+5) \cdot 10^2 + 3 \cdot 3! (1+2+4+5) \cdot 10^1 + 3 \cdot 3! (1+2+4+5) \cdot 10^0  = 3119976$$


问:
How can I calculate distance from two pixels HSV?
I want to look for a better way to calculate the distance between two pixels in the HSV color space. In my program I used Euclidean distance.
$$
distance(p_1, p_2)  = \sqrt{(h_1 - h_2)^2 + (s_1 - s_2)^2 +(v_1 - v_2)^2}
$$
$where \,\, h_1, h_2, s_1, s_2, v_1, v_2 \,\, are\,\, the\,\, components\,\, of\,\, the\,\, two\,\, pixels\,\, p_1\,\, and \,\,p_2 \,\,from\,\, HSV\,\, color \,\,space. \quad
 p_1 = (h_1, s_1, v_1)\,\,
p_2 =(h_2, s_2,v_2)
 $

Using Euclidean distance is not a good approach because the hue value is expressed in degrees, $ hue \in [0^{\circ}, 360 ^{\circ}]$. How can I calculate this distance in another way? My proposal is to use polar coordinates to calculate the distance between two HSV pixels. But I don't know how to express hue, saturation, value in cylindrical space. I think that the x axis is saturation, the value represents the y-axis and $\theta$ is represented by hue value in degrees.

答:
You can use
$$\left\lbrace \begin{aligned}
x &= s \cos h \\
y &= s \sin h \\
z &= v
\end{aligned} \right.$$
with
$$d(x_1, y_1, z_1, x_2, y_2, z_2) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$
and therefore, after simplification,
$$d(h_1, s_1, v_1, h_2, s_2, v_2) = \sqrt{(v_2 - v_1)^2 + s_1^2 + s_2^2 - 2 s_1 s_2 \cos(h_2 - h_1)} \tag{Cylinder}\label{Cylinder}$$

This cylinder model has the problem that it does not take into account that all hues and saturations at (or very near) zero value are black.  The HSV cone (or hexagonal pyramid) described in the Wikipedia HSL and HSV article might work better.  Instead of saturation, we use chroma ($C = S V$) instead of the saturation for the radius:
$$\left\lbrace \begin{aligned}
x &= s v \cos h \\
y &= s v \sin h \\
z &= v \\
\end{aligned} \right .$$
giving a distance function of
$$d(h_1, s_1, v_1, h_2, s_2, v_2) = \sqrt{ (v_2 - v_1)^2 + s_1^2 v_1^2 + s_2^2 v_2^2 - 2 s_1 s_2 v_1 v_2 \cos(h_2 - h_1) } \tag{Cone}\label{Cone}$$

Do note that all above assumes $h$ has the same range as your $\sin$ and $\cos$ functions; typically radians.  If your $\cos$ function takes radians, and $h_1$ and $h_2$ are in degrees, use cos((h2 - h1)*M_PI/180.0; if your $\cos$ function takes radians, and $h_1$ and $h_2$ are within $[0, 1]$ or $[-0.5, +0.5]$, use cos((h2 - h1)*M_PI); and if your $\cos$ functon takes radians, and $h_1$ and $h_2$ are within $[0, 255]$ or $[-128, 127]$, then use cos((h2 - h1)*M_PI/128.0.


问:
Outer Measure on $\mathbb R$ with trivial measurable sets
This question arises from wondering how bad an outer measure can be. Some authors prefer to work with general outer measures (e.g. Evans and Gariepy), and it would be nice to be able to think of outer measures as just extensions of measures. As in every outer measure induces a measure and every measure induces an outer measure, and it would be nice if these were inverse to each other, though in general they are not. There are of course conditions (e.g. regularity) that allow you to know that this is the case, but one can wonder how nasty a general outer measure can be.

One way an outer measure can be bad is not having enough measurable sets. Hence the question is whether there is an example of the most extreme version of this (on $\mathbb R$ and maybe $\mathbb R^n$ if it's interesting): does there exist an outer measure on $\mathbb R$ such that the only measurable sets (under the Caratheodory criterion) are $\mathbb R, \emptyset$?

答:
Yes there is! Consider the function $\mu^{*}:P(\mathbb{R})\to [0,\infty)$ given by

$$\mu^{*}(A)= \begin{cases} 1, & \text{if} \hspace{2mm} A\neq \emptyset, \\
0, & \text{if} \hspace{2mm} A=\emptyset .
  \end{cases}$$
It is easy to see that $\mu^{*}$ is an outer measure on $\mathbb{R}$. Now, I claim that the only two subsets of $\mathbb{R}$ satisfying the Carathéodory criterion are $\mathbb{R}$ and $\emptyset$. Clearly, if we let $E=\mathbb{R}$ or $E=\emptyset$ then, for every $A\subseteq \mathbb{R}$, the following equality is true $$\mu^{*}(A)= \mu^{*}(A\cap E) + \mu^{*}(A\setminus E).$$

Conversely, suppose that $\emptyset\neq E \subset \mathbb{R}$ (here the symbol $\subset$ means proper inclusion). Let $a\in E$, $b\in \mathbb{R}\setminus E$ and put $A=\{a,b\}$. Then, $$\mu^{*}(A)= 1 \neq 2 = 1+1 =\mu^{*}(A\cap E) + \mu^{*}(A\setminus E).$$

As a final comment, if we consider any set $X$ with at least two elements, then the corresponding $\mu^{*}$ produces an outer measure on $X$ such that the only $\mu^{*}$-measurable sets are the empty set and $X$ itself.


问:
Algebraic manipulation in Uniform convergence
i´m trying to prove that a sequence defined as $f_{n}(x):=\sqrt{x^{2}+\frac{|x|}{n}}$ uniformly converges in $\mathbb{R}$. i have that the $\lim_{n \to \infty} f_{n}(x)=|x|$ so i get that i need to solve the next expression $|\sqrt{x^{2}+\frac{|x|}{n}}-|x||$ but i dont know how to manipulate that expression, any help would be of great value.

Body:
Hint:

$$x^2 + \frac{|x|}{n} < x^2 + 2 \frac{|x|}{n} + \frac{1}{n^2} = \left(|x| + \frac{1}{n} \right)^2$$


问:
Why are the prime ideals of $R_f$ exactly the prime ideals of $R$ not containing $f$?
Let $R$ be a ring and $D_f=\{\text{primes ideals in } R \text{ not containing } f\}$ be a basic open set. Let $R_f$ be the localization of $R$ at $f$, I am trying to show that $D_f=\operatorname{Spec}R_f$. I can see every prime in $D_f$ certainly gives a prime ideal of $R_f$, but how can I show the other direction?

答: For any prime ideal $P$ in $R_f$, its inverse image under the canonical map $R \to R_f$ is again a prime ideal. Show that this can not contain $f$.


问:
sum of the eigenvalues = trace($A$)?
Is it true that for a square matrix $A$, all of whose eigenvalues exist in the base field, sum of the eigenvalues = trace($A$)?

The result holds in all the matrices I've studied.

答: Since all eigenvalues exist in the base field, $A$ will be similar to a matrix $J$ in Jordan Canonical Form, with eigenvalues of $A$ along the diagonal, and assertion easily follows. (Hint: Use $tr(AB) = tr(BA)$ for $A, B$ square matrices)</p>

Title: sum of the eigenvalues = trace($A$)?


问:
Consecutive primitive roots of congruence $x^n\equiv \bmod (m) $
Consecutive primitive roots of congruence $x^n\equiv \mod (m) $

I saw in this site that: There is no algorithm for finding primitive roots of congruence $x^n\equiv \mod (m) $. Here I discuss a certain case:

Consider following identity:

$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+ \cdot\cdot\cdot+b^{n-1})$

suppose a and b are two consecutive numbers and $a=b+1$  we have:

$(b+1)^n-b^n=(1)[(b+1)^{n-1}+(b+1)^{n-2}b+ \cdot\cdot\cdot+b^{n-1}]$

Let $(b+1)^{n-1}+(b+1)^{n-2}b+ \cdot\cdot\cdot+b^{n-1}=m$

suppose $b+1$ and $b$ have a remainder $r$ when divided by m i.e we can write:

$(b+1)^n \equiv r \mod(m)$

$b\equiv r \mod (m)$

Subtracting these relations gives:

$(b+1)^n-b^n \equiv 0 \mod(m)$

rewriting this as:

$(b+1)^n\equiv b^n\equiv r \mod(m)$

For example :

$5^2\equiv 3 \mod (11)$

$6^2\equiv 3 \mod (11)$

5 and 6 are consecutive roots of congruence $x^2\equiv 3 \mod (11)$

Hence we can conclude  that any number of the form $(b+1)^n-b^n=(b+1)^{n-1}+(b+1)^{n-2}b+ \cdot\cdot\cdot+b^{n-1}=m$ has two consecutive roots in congruence $x^n\equiv r \mod(m)$ and these roots may be primitive for certain value of r which can be found by following relation:

$b^n \equiv r \mod(m)$

Clearly b can have any value, so the number of these types of congruence can be infinite.

My question is : can this method be correct for large m, in other words can b and b+1 always two of  primitive roots? can someone give a counter example?

Update:I checked number$13^3-12^3=469$:

$12^3\equiv 321\bmod{469}$

$13^3\equiv 321\bmod{469}$

I found some more solutions; 80, 146, 213, 243, 348 for remainder 321 but none of them are primitive.I do not think these type of number have primitive roots for any n.

答:
Your method is a way of finding two consecutive integers which have the same $n$th power for some modulus.

However, this is not what 'primitive root' means. See, for example, https://en.wikipedia.org/wiki/Primitive_root_modulo_n

In your own example, $5^5\equiv 1\pmod {11}$ and so $5$ is not a primitive root modulo $11$.


问:
Find the infimum and supremum of the set $G = \{x \in \Bbb R \mid x^2+x+1>0\}$.
Find the infimum and supremum of the set $G = \{x \in \Bbb R \mid x^2+x+1>0\}$.

Attempt:
For all real $x$, the discriminant is $1^2-4(1)(1) =-3<0$. So, for all real $x$ has $x^2+x+1>0$ i.e. $G = \Bbb R$.

Hence, $G = (-\infty,\infty)$ and hence, $\inf G = -\infty$ and $\sup G = \infty$.

But, for another consideration, for all real $x$, we have $x^2+x+1 = (x+\frac{1}{2})^2 + \frac{3}{4} \ge \frac{3}{4}$, which equality attained when $x = - \frac{1}{2}$. Therefore, $\inf G = \frac{3}{4}$ and $\sup G = \infty$.

What should I do?

答:
Seems you are mixing two different things here whose notations seem similar

(and maybe that's what confused you) but they are not the same thing.

$G_1 = \{x \in \Bbb R \mid x^2+x+1>0\}$

$G_2 = \{x^2+x+1 \mid x \in \Bbb R \}$

Note that $$G_1 = \Bbb R$$ while

$$G_2 = [3/4; \infty)$$

So $$\inf G_2 = 3/4$$

and so there is no contradiction here or any logical issue.


问:
Finding the determinant of $2A+A^{-1}-I$ given the eigenvalues of $A$

Let $A$ be a $2\times 2$ matrix whose eigenvalues are $1$ and $-1$. Find the determinant of $S=2A+A^{-1}-I$.

Here I don't know how to find $A$ if eigenvectors are not given. If eigenvectors are given, then I can find using $A=PDP^{-1}$. Please solve for $A$; the rest I can do.

答:
\begin{pmatrix}
0 &-1 \\
 -1&0
\end{pmatrix}

the Characteristic polynomial is $(x-1)(x+1)$.


问:
Suppose $(E,m)$ and $(E'm)$ are both equalizers of $f$ and $g$ show that there is an isomorphism $k\colon E\to E'$

Suppose $f,g\colon X\to Y$ are functions. Then there is a set $E$ and a function $m\colon E\to X$ with the following property: $f\circ m=g\circ m$, and for any other function $h \colon F \to X$ with $f \circ h = g \circ h$ there is a unique function $k\colon F\to E$ such that $m\circ k=h$. We call $(E,m)$ an equalizer of $f$ and $g$.

What I have so far,

Suppose there is a function $i\colon E\to E'$

I am trying to see what composition of function, from those given in the definition, will be such that

$$i \circ \text{that function}=I_E\quad\text{and}\quad\text{that function} \circ i=I_{E'}$$

The problem is, there is no function(s) or composition thereof that arrows to $E$ and its counterpart to $E'$.

Any ideas?

答:
As mentioned in the comments by azif00 it is not exactly clear if you want to show that $(E,m)$ exists or that it is unique up to (unique) isomorphism. As the existence is only a (simple) set-theoretic matter I will talk about uniqueness. What follows is the usual argument for showing that universal objects (i.e. objects defined by universal properties) are unique up to unique isomorphism.

Let $f,g\colon X\to Y$ be two set-functions between set $X,Y$ (so we are working in $\sf Set$). The equalizer $\operatorname{eq}(f,g)$ of $f$ and $g$ is characterized as a pair $(\operatorname{eq}(f,g),m)=:(E,m)$ such that $f\circ m=g\circ m$ (I fixed a small typo there in your questions) and such that for every other pair $(E',m')$ with $f\circ m'=g\circ m'$ there is a unique set-function $k\colon E'\to E$ with $m\circ k=m'$. I decided to state the universal property again as your (original) version is slightly confusing and hides the crucial fact.

So, let us take two pairs $(E,m),(E',m')$ which both satisfy the aforegoing universal property. Using this property both ways around (so one time with $(E,m)$ as the set-function-inducing-pair and one time with $(E',m')$, but both times with the other as second object) we obtain unique set-functions:

$$k\colon E'\to E,\quad k'\colon E\to E'$$

We note, however, that we can use the universal property on the pair $(E,m)$ in both places! This gives us a unique set-function $i\colon E\to E$ such that $m\circ i=m$. But there already is such a set-function, namely the identity $\operatorname{id}_E\colon E\to E$, and by uniqueness $i=\operatorname{id}_E$. A similar argument may be applied to the pair $(E',m')$.

Now note that the compositions $k\circ k'\colon E\to E$ and $k'\circ k\colon E'\to E'$ both satisfy the very same properties as the corresponding identites on $E$ and $E'$, respectively, do. Hence, by uniqueness again, we have $k\circ k'=\operatorname{id}_E$ and $k'\circ k=\operatorname{id}_{E'}$. So, $k\colon E'\to E$, in particular, is a uniquely defined bijective set-function; that is an isomorphism in $\sf Set$.


答:
Consider the category $\mathscr C$ that has

as objects, pairs $(F,h)$ where $F$ is a set and $h$ is a function from $F$ to $X$ such that $f \circ h = g \circ h$; and

a morphism $k$ from $(F_1,h_1)$ to $(F_2,h_2)$ is a function $k \colon F_1 \to F_2$ such that $h_1 = h_2 \circ k$.

Then note that an equalizer of $f$ and $g$ is an object $(E,m)$ in $\mathscr C$ such that for any other object $(F,h)$ in $\mathscr C$ there is a unique morphism $k$ from $(F,h)$ to $(E,m)$. In other words, $(E,m)$ is a terminal object in $\mathscr C$. Can you conclude from here?

Title: Suppose $(E,m)$ and $(E'm)$ are both equalizers of $f$ and $g$ show that there is an isomorphism $k\colon E\to E'$

Body:
Suppose $f,g\colon X\to Y$ are functions. Then there is a set $E$ and a function $m\colon E\to X$ with the following property: $f\circ m=g\circ m$, and for any other function $h \colon F \to X$ with $f \circ h = g \circ h$ there is a unique function $k\colon F\to E$ such that $m\circ k=h$. We call $(E,m)$ an equalizer of $f$ and $g$.

What I have so far,

Suppose there is a function $i\colon E\to E'$

I am trying to see what composition of function, from those given in the definition, will be such that

$$i \circ \text{that function}=I_E\quad\text{and}\quad\text{that function} \circ i=I_{E'}$$

The problem is, there is no function(s) or composition thereof that arrows to $E$ and its counterpart to $E'$.

Any ideas?


问:
Dropping continuity condition on functional equation
I have the following equation that is to be satisfied by some function $f:[0,1)\rightarrow\mathbb R$:

$$f\left(\frac{2x}{1+x^2}\right)=(1+x^2)f(x\sqrt{2-x^2}).$$

I managed to prove that, given continuity at $0$, the only possible solution is

$$f(x)=\frac{f(0)}{\sqrt{1-x^2}}.$$

I wonder, however, if this still applies if we drop the continuity condition. If it's possible to prove the continuity at $0$ given the functional equation alone, for example. I thought about it for a while but haven't gotten to anything useful.

答:
Idea:
Let's  denote $g(x) = \sqrt{1-x^2}f(x)$. From your equation, we have
$$g(\frac{2x}{1+x^2}) = g(x\sqrt{2-x^2})$$
Make the change of variables $x = \frac{1 - \sqrt{1 - t^2}}{t}$ with $0<t<1$ and denote
$$u(t)=u(x(t))=\frac{1 - \sqrt{1 - t^2}}{t}\sqrt{2-(\frac{1 - \sqrt{1 - t^2}}{t})^2}$$

We have then for all $0<t<1$
$$g(t)=g(u(t))$$

The function $u(t)$ have this property: $u(t)<t$ (here below is the function $t-u(t)$)

and the sequence $u^{(n)}(t)$ defined by $\{ u^{(n)}(t) = u(u^{(n-1)}(t)), u^{(0)}(t)=t \}$ converge to $0$.

Because the function $f(x)$ (and so $g(x)$) continues at $0$, we can conclude that $g(t) = g(0)$ for all $t\in (0,1)$.

PS: If we suppose the assumption that $f(x)$ is continue at $x=0$, the solution of $g(x)$ would be as follows:

We define a set $A$ in $\mathbb{R}\cap(0,1)$ such that for each $x\in A$, $\not\exists n\ge 1$ and $y\in A$ with $x = u^{(n)}(y)$ or $y = u^{(n)}(x)$

We define a function $k(x): A\rightarrow \mathbb{R}$, $k(x)$ is not necessary continue

The function $g(x)$ is defined as: $g(x)=g(k(x^*))$ with $x^*$ is the unique value such that there exists $n$:  $x = u^{(n)}(x^*)$
And the function $f(x)$ will be
$$f(x)=\frac{g(x)}{\sqrt{1-x^2}}$$


问:
Volume of solid of revolution about a line other than the axis - using Cylindrical Shells method
I am struggling with a concept of finding the radius in the shell method of finding the volume of the solid when revolved around a line other than the axis.  Maybe the issue here is that I am looking for a "rule" instead of actually visualizing the problem.  It was my understanding that when revolving around a line other than the axis a formula could be used as follows: (this example uses revolution about a vertical line other than the $y$ axis)
the radius would be $|verticalvalue|$ $+/-$ $x$.  Use $+$ if the region of revolution is to the right of the line and use $-$ if the region is to the left.

I thought I had worked many problems using this concept and gotten the correct answer.  However, doesn't work in this problem:

Use the method of cylindrical shells to find the volume generated by rotating $y=x^2$, $y=0$, $x=1$, $x=2$ about the line $x=1$.

My integral, using my above "concept" was to form the following integral:

$V=2\pi\int(1+x)(x^2) dx$ evaluated from lower limit of 1 to upper limit of 2.  I used the absolute value of $1$ PLUS (since the region of revolution is to the right of the vertical line of revolution.

This does not produce the book answer and the book used $x-1$ for the radius instead of $x+1$, obviously producing very different results.

My question:  is there a rule for writing the radius?  Why didn't my rule work?

答:
You would only "use the absolute value of 1 plus the x" if the region of rotation was to the right of the vertical line of revolution AND the left boundary of the region of rotation were to the left of the y-axis. Perhaps most of the problems you did in the past were this case.

There are rules for these types of problems. If you are rotating about a vertical line, and the axis of rotation is the left boundary, you use $x-h$. But if you're rotating about the right boundary, you use $h-x$.

As for your problem, it is usually better to draw a picture. I don't know anyone who doesn't at least visualize these types of find-the-volume-of-rotation problems.

Since $1\le 1< 2$, use $\int_1^2 (x-1)f(x)dx$. You get

$$2\pi\int_1^2(x-1)x^2dx$$

However if you were to rotate it around the line h=2, not shown, you get

$$2\pi\int_1^2(2-x)x^2dx$$

Note that you would get a larger volume for the first rotation (h=1) because the taller part of it has to rotate more, covering more volume, but for the second rotation (h=2), the larger part is in the "center" and creates less volume.


问:
A question on parallel transport of a vector field along a smooth curve
I was given the following exercise:

Consider the connection on $\mathbb{R}^2$ defined by $\Gamma_{1 2}^1 =1$ and all the other $\Gamma_{jk}^i$ equal to zero wrt $(x^1, x^2) = (x,y)$. Determine the equations for parallel transport of a vector field $Y = Y^i \frac{\partial}{\partial x^i}$.

I found that the equations of motion would be given by:

$$\frac{dY^1}{dt} + \frac{dy}{dt} Y^1 = 0$$

and $$\frac{dY^2}{dt} = 0$$

I can see that the second equations implies that $Y^2$ will be constant, but I can't figure how to solve the first ODE since it contains 2 variables. The solution given to me mentions that the first equation implies that $Y^1 e^y$ is constant, but I don't see why. Furthermore, it states the following:

The components of the vector field $Y$ along a curve from $C(t_0)$ to $C(t)$ are given by :
$$  \left[ \begin {array}{c} Y^1 \left( t \right) \\ Y^2
 \left( t \right) \end {array} \right]
 =  \left[ \begin {array}{cc} {{\rm e}^{y \left( t \right) -y \left( {
\it t_0} \right) }}&0\\ 0&1\end {array} \right] \left[ \begin {array}{c} Y^1 \left( t_0 \right) \\ Y^2
 \left( t_0 \right) \end {array} \right]
$$

Which apparently follows from the first equation. Can some help clarify my confusion about this?

答:
$$\frac{dY^1}{dt} + \frac{dy}{dt} Y^1 = 0$$
$$\frac{dY^1}{dt} =- \frac{dy}{dt} Y^{1}$$
$$\int \frac{dY^1}{Y^1} =- \int{dy}$$
$$\ln Y^1=-y+C$$
This gives the final result:
$$Y^1=Ke^{-y}$$
Are you sure it is $\frac{dY^1}{dt} +\frac{dy}{dt} Y^{1}=0$ and not $\frac{dY^1}{dt} - \frac{dy}{dt} Y^{1}=0$ ?

Note that this DE:
$$\frac{dY^1}{dt} + \frac{dy}{dt} Y^1 = 0$$
Is just:
$$\frac{Y'^1}{Y^1} + y' = 0$$
That you can rewrite as:
$$(\ln Y^1)'+y'=0$$
$$(\ln Y^1+y)'=0$$
Integrate both side.
$$\ln Y^1+y=C$$


问:
Is there a formula for finding binary numbers in a binary string?

One bit

Let's suppose that I have a short binary string: 01. This string contains the binary digits 0 and 1 representing the numbers from zero to one. So, a string 2 digits long is required to represent 2 values (1 bit).

Two bits

That wasn't very interesting. However if I want to represent all the numbers that are up to two bits long in a single string, it's:

```<code>00110
</code>```

This contains:

```<code>00 01 10 11
</code>```

In other words, 5 digits are required to represent 4 values (2 bits).

Three bits

I can do three bits:

```<code>0001110100
</code>```

containing:

```<code>000 001 010 011 100 101 110 111
</code>```

10 digits are required to represent 8 values (3 bits).

Four bits

```<code>0000111101100101000
</code>```

contains all of:

```<code>0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111
</code>```

19 digits are required to represent 16 values (4 bits).

Additional discovery

Finally I discovered an interesting thing.

Having found my first string for 4 bits 1001010000111101100, I realised I could move a digit from the beginning to the end or vice-versa, and the string would still always contain all the numbers, though sometimes (apparently, half the time) the bit needs to be flipped. So I could do it with:

```<code>0010100001111011001
0101000011110110010
1010000111101100101  # this one required flipping the moved digit
0100001111011001010  # this one required flipping the moved digit
1000011110110010100
0000111101100101000  # this one required flipping the moved digit
0001111011001010000
0011110110010100001  # this one required flipping the moved digit
0111101100101000011  # this one required flipping the moved digit
1111011001010000111  # this one required flipping the moved digit
1110110010100001111
1101100101000011110  # this one required flipping the moved digit
1011001010000111101
0110010100001111011
1100101000011110110
1001010000111101100  # this one required flipping the moved digit
</code>```

I also realised that this means that if the sequence of digits is looped, I can get away with a shorter sequence. i.e. a repeating 0000111101100101... - 16 digits, still containing all my numbers.

I can lop off two digits from the 3-bit sequence to make 00011101... - 8 digits.

Or remove one digit from the 2-bit sequence to make 0011... - 4 digits.

So if I loop the sequence, it looks like I need as many digits as numbers I can represent in the sequence, while unlooped, it's that number plus the bit-depth minus 1.

But, I am not certain if this holds for 5-bit and higher number rangers, I haven't tried to work that out yet, because so far I haven't yet worked out a sequence for more than 4 bits.

My questions

What is the smallest string that would contain all the 5-bit numbers?

Is there a formula or process for finding the string, other than intuition and trial-and-error (which is what I used)?

Will it always be possible, for any bit-depth, to find a string that contains all the numbers of that bit-depth, without repeating any of them?

1 bit requires 2 digits in the (unlooped) string; 2 bits requires 5; 3 bits requires 10; 4 bits requires 19. Is there a pattern to this and can we know how many digits in the string will be required to contain all the binary numbers of a particular bit-depth?

Is there a name for any of this, that I could look up to read more about it?

答:
If you allow wraparound, you need only $2^n$ bits to get all $n$-bit sequences; see De Bruijn sequences. If you don’t allow wraparound, you’ll need to repeat the first $n-1$ bits at the end of the string. For example, $00010111$ does the job with wraparound, while without wraparound you need to extend it to $0001011100$, which yields in turn $000$, $001$, $010$, $101$, $011$, $111$, $110$, and $100$.

In particular, to get all $5$-bit numbers you need $2^5=32$ bits with and $2^5+4=36$ bits without wraparound. One possible sequence is

$$00000100011001010011101011011111(0000)\,.$$


问:
Ask book to deeply understand partially ordered sets
I learn little about ordering and poset before, but I think it's not enough and want to learn more about ordering and Poset. Can anyone please recommend some best books to learn about this topic. I really appreciate it. Thanks so much</p>

答: E.Harzheim, Ordered Sets, Springer Verlag, 2005.</p>


问:
Circular seating with no restrictions
I was seeing this video by Eddie woo on circular seating's, in it , he discusses the following question:

Find the number of ways to arrange 5 boys and 5 girls in a circular seating

At 0:40, he places the first person and then keeps the rest and hence total ways to do it is $ 1 \cdot 9!$ , however I can't precisely frame why we only have one way to keep the first person. So, I'm looking for the understanding on why he said the first step like this... (he laters say it is related to the fact we can rotate the circle but I didn't get that either)

答:
Imagine you have placed your 10 persons around the circle. Imagine there are K distinct ways to do this placement (i.e. K such circular configurations).

Think about the circumference as if it's a circular rope e.g.

For each such configuration (circular placement) you can cut the circumference in between any two persons. You can do this in 10 different ways for each configuration (because you have 10 gaps between the 10 people). So each such circular configuration gives rise to 10 normal (linear) permutations (you can prove this is a bijection) of the 10 people.

But we know all the distinct linear permutations are $M = 10!$

So the number K should be such that $$K \cdot 10 = M$$ i.e. $$K = \frac{10!}{10}$$

Btw, it doesn't matter if you have 5 boys and 5 girls or say 8 boys and 2 girls. When the word is about people, it is always assumed that any two people are distinguishable/distinct.

Title: Circular seating with no restrictions


问:
Find the area of a regular polygon
I found a quiz on advanced trigonometry and geometry online and I was doing it for fun and it asks

Prove that for any regular polygon with side length $S$ and number of sides $N$ that $A =  S N \cot(\pi/N)/4$

I have been stumped for 30 minutes now. Can somebody help?

答:
HINT Here is one approach. Since it's a regular polygon, call its center $C$ and draw edges from $C$ to every vertex of the polygon. Note you end up with $N$ identical triangles, where the central angle is $2\pi/n$ and the length of the opposite side is $S$. You also know that all such triangles are isosceles by symmetry.

It suffices to prove the area of one such triangle is $S\cot(\pi/N)/4$, so now you can compute the length of the height from $C$ onto the side of length $S$ and compute the area. Can you finish it?

Title: Find the area of a regular polygon


问:
The definition of the complementary of a subgroup is symmetrical
Let $G$ be a group and $H,K$ be its subgroups. Neither of them has to be normal.

If

$$G = HK = \{ hk : h\in H, k\in K\} \text{ and } H\cap K = \{e\},$$

then we say $K$ is complementary to $H$. Wikipedia says this definition is symmetrical, namely we can also write $G=KH$.

I understand that $HK=G=KH$ when one of them is normal. But why is it true in general here?

答:
Given any subgroups $H$ and $K$ of a group $G,$ we have that $HK$ is a subgroup of $G$ if and only if $HK = KH.$ Considering that $G = HK$ is a group (and hence a subgroup), it follows that $HK = KH.$

We will establish that if $HK$ is a subgroup of $G,$ then $HK = KH$ (as we do not need the converse immediately). Given any element $hk \in HK,$ we have that $h^{-1} k^{-1}$ is in $HK$ (because $H$ and $K$ are both subgroups of $G$). Considering that $hkh^{-1} k^{-1}$ is in $HK$ by hypothesis that $HK$ is a subgroup of $G,$ we find that $hkh^{-1} k^{-1} = h_1 k_1$ so that $h_1^{-1} hk = k_1 kh.$ We claim that the map $\ell_{h_1^{-1}} : H \to H$ defined by $\ell_{h_1^{-1}}(h) = h_1^{-1} h$ is surjective, from which it follows that $HK \subseteq KH.$ Conversely, given any element $kh \in KH,$ we have that $h = h e_G \in HK$ and $k = e_G k \in HK$ so that $kh = (e_G k)(h e_G) \in HK$ by assumption that $HK$ is a subgroup of $G.$ We conclude that $HK = KH.$


问:
Is this Perfect Binary Tree Countably Infinite or Uncountably Infinite?
Given a perfect binary tree which at a certain fixed level has infinitely many leaves, hence each leaf at that fixed level has infinitely many ancestors. From that fixed level the tree also extends to infinite height.

Are the nodes of this tree countably infinite or uncountably infinite?

I know that a rooted binary tree, even if extended to a height of infinity is countably infinite; just start at the root and enumerate level by level. However, I cannot imagine a way to count the nodes of this tree or to apply the idea that the union of countably infinite subtrees is countable. Thus, I’m leaning towards it being uncountably infinite but have serious doubts.

How can a tree look like that?

Imagine that instead of the tree simply growing in height it is also expanding sideways as shown below, so it is infinitely high, infinitely wide and infinitely deep. Getting infinitely wide is illustrated below.

答:
Start with arbitrarily chosen node and map it to 1. There're 3 nodes at the distance 1 from it, map them to the numbers from 2 to 4. Then there're 6 nodes at the distance 3, so let's map them to the numbers from 5 to 10. Continuing in this manner you can count all the nodes.

Notice that each vertex has a degree of 3, so from the first node we went to 3 new nodes, but each time after that we can only go to 2 new nodes from each old node, which means that if we consider an arbitrarily chosen node a root each of its three children will be an infinite rooted binary tree.


问:
Improper integral $\int^{\pi/2}_0 (\operatorname{csc} x - \frac{1}{x})\,\mathrm dx$

$\def\cosec{\operatorname{csc}}$
  Calculate the value of the improper integral

$$\int^{\pi/2}_0\left (\cosec x - \dfrac{1}{x}\right)\,\mathrm dx.$$

You may use the standard integral $\int \cosec x \,\mathrm dx = -\ln|\cosec x + \cot x| + c $.

Please note this is a 3 point question. Here is my solution:

This definite integral is improper at $x=0$, so we have:
\begin{align*}
\lim_{b\to0}\int^{\pi/2}_b\cosec x - \dfrac{1}{x}\,\mathrm dx &= -\ln|\cosec x + \cot x|- \ln|x| \biggr|^{\pi/2}_b\\
&= \lim_{b\to0} -\left(\ln\left|\cosec \frac{\pi}{2} + \cot \frac{\pi}{2}\right|- \ln\left|\frac{\pi}{2}\right| \right) - (-\ln|\cosec b + \cot b|- \ln|b| )
\end{align*}

For $\lim\limits_{b\to0}-(\ln|\cosec b + \cot b|+ \ln|b|)$, using the properties of logarithms, I have $-\lim\limits_{b\to0} \ln|b\cosec b + b\cot b|$.

The limits are both solved similarly by L'Hôpital's rule, giving:

$$-\lim_{b\to0}\ln|b\cosec b + b\cot b| = -\lim_{b\to0}\ln|1 + 1| = -\ln|2|$$

Therefore, I have:

$$\lim_{b\to0} \left(-\ln|1+0|- \ln\left|\frac{\pi}{2}\right| \right) - (-\ln|2| ) = \ln\left|\frac{2}{\pi/2}\right| = \ln\left|\dfrac{4}{\pi}\right|$$

Is this correct and answered sufficiently? Have I made any errors?

Thanks so much!

答:
This CW answer is intended to remove this question from the Unanswered queue.

As Mhenni Benghorbal remarks, your answer is entirely correct. Also, the derivation is nicely written out.

Cheers!


问:
Roll $5$ dice. Record the product. How many rolls are required so that the recorded products generate a square number?
$5$ fair dice are rolled. The product of the displayed numbers is recorded (lets call it M1). Dice rolled again and product is recorded again (M2). How many rolls are required so that the products (M1 * M2) generate a square number?

Its been a long time since I did probabilities so I started going about this the long way. Generated an Excel spreadsheet with all possible $7776$ outcomes and came to the conclusion that there are $126$ distinct values for the product after the first roll ($7776$ being the largest).

And that's where I became stuck and lost. Any help would be greatly appreciated.

Thanks

答:
Here are some of the details for the procedure outlined in lulu's second comment. If the current value of the product is $\ 5^a3^b2^c\ $ then the state of the Markov chain is $\ \gamma\beta\alpha\ $, where $\ \gamma=c\hspace{-0.2em}\pmod{2}\ $, $\ \beta=b\hspace{-0.2em}\pmod{2},$ and $\ \alpha=a\hspace{-0.2em}\pmod{2}\ $.  For the throw of a single die, the probabilities of each of the eight states are given by the following table:

$$
\begin{array}{c|c|c|c|c|c|c|c|c|}
\text{throw}&1\text{ or }4&2&3&6&5\\
\text{state, }s&000&001&010&011&100&101&110&111\\
\hline
p_1(s)&\frac{1}{3}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&0&0&0\\
\hline
\end{array}
$$
The probability $\ p_n(s)\ $ of obtaining state $\ s\ $ when $\ n\ $ dice are thrown can be calculated recursively from the following convolution:
$$
p_n(s)=\sum_{t=000}^{111}p_1(t)p_{n-1}(t\oplus s)\ ,
$$
where $\ \oplus\ $ denotes the bitwise xor operation (i.e. componentwise mod $2$ addition).

The values of $\ p_5\ $ are given in the following table:
$$
\begin{array}{c|c|c|c|c|c|c|c|c|}
s&000&001&010&011&100&101&110&111\\
\hline
p_5(s)&\frac{139}{972}&\frac{137}{972}&\frac{137}{972}&\frac{137}{972}&\frac{107}{972}&\frac{35}{324}&\frac{35}{324}&\frac{35}{324}\\
\hline
\end{array}
$$
Thus, the initial state of the Markov chain, after the first throw, is the probability row vector
$$
\pi(1)=\pmatrix{\frac{139}{972}&\frac{137}{972}&\frac{137}{972}&\frac{137}{972}&\frac{107}{972}&\frac{35}{324}&\frac{35}{324}&\frac{35}{324}}\ .
$$
Whenever the chain enters state $\ 000\ $, the product at that time will be a square, and the goal has been reached.  This state is therefore absorbing:
$$
\mathbb{P}\big(S_{n+1}=s\left|
 S_n=000\right.\big)=\cases{1&if $\ s=000$\\
 0&otherwise.}
$$
Otherwise the transition probabilities are given by
$$
\mathbb{P}\big(S_{n+1}=s\left|\right.S_n=t\big)=p_5(s\oplus t)\ ,
$$
for $\ t\ne000\ $.  Therefore the transition matrix of the chain is
$$
\Pi=\pmatrix{&000&001&010&011&100&101&110&111\\
000&1&0&0&0&0&0&0&0\\
001&\frac{137}{972}&\frac{139}{972}&\frac{137}{972}&\frac{137}{972}&\frac{35}{324}&\frac{107}{972}&\frac{35}{324}&\frac{35}{324}\\
010&\frac{137}{972}&\frac{137}{972}&\frac{139}{972}&\frac{137}{972}&\frac{35}{324}&\frac{35}{324}&\frac{107}{972}&\frac{35}{324}\\
011&\frac{137}{972}&\frac{137}{972}&\frac{137}{972}&\frac{139}{972}&\frac{35}{324}&\frac{35}{324}&\frac{35}{324}&\frac{107}{972}\\
100&\frac{107}{972}&\frac{35}{324}&\frac{35}{324}&\frac{35}{324}&\frac{139}{972}&\frac{137}{972}&\frac{137}{972}&\frac{137}{972}\\
101&\frac{35}{324}&\frac{107}{972}&\frac{35}{324}&\frac{35}{324}&\frac{137}{972}&\frac{139}{972}&\frac{137}{972}&\frac{137}{972}\\
110&\frac{35}{324}&\frac{35}{324}&\frac{107}{972}&\frac{35}{324}&\frac{137}{972}&\frac{137}{972}&\frac{139}{972}&\frac{137}{972}\\
111&\frac{35}{324}&\frac{35}{324}&\frac{35}{324}&\frac{107}{972}&\frac{137}{972}&\frac{137}{972}&\frac{137}{972}&\frac{139}{972}}\ .
$$
After $\ n\ $ throws the state of the chain will be the probability row vector $\ \pi(n)=\pi(1)\Pi^{n-1}\ $, and the probability that a square has been reached at or before that time is the first entry of that vector.

If $\ t_s\ $ is the expected number of throws needed to reach state $\ 000\ $ from state $\ s\ $, then $\ t\ $ must satisfy the equations
\begin{align}
t_{000}&=0\\
t_s&=1+\sum_{k=000}^{111}\Pi_{sk}t_k\\
&=1+\sum_{k=001}^{111}\Pi_{sk}t_k\\
&=\Pi_{s\,000}+\sum_{k=001}^{111}\Pi_{sk}\big(t_k+1\big)\ \ \text{ for }\ s\ne000\ .
\end{align}
The solution of these equations is given in the following table
$$
\begin{array}{c|c|c|c|c|c|c|c|c|}
s&000&001&010&011&100&101&110&111\\
\hline
t_s&0&\frac{970}{121}&\frac{970}{121}&\frac{970}{121}&\frac{1752}{211}&\frac{212414}{25531}&\frac{212414}{25531}&\frac{212414}{25531}\\
\hline
\end{array}\ ,
$$
and the expected number of throws for the chain to reach the absorbing state $\ 000\ $ is
\begin{align}
1+\sum_{s=000}^{111}\pi(1)_st_s&=\sum_{s=000}^{111}\pi(1)_s\big(t_s+1\big)\\
&=8\ .
\end{align}


答:
It follows from the following standard exercise:

Theorem. Let $G$ be a group, and let $H$ and $K$ be subgroups. Then $HK=\{hk\mid h\in H, k\in K\}$ is a subgroup of $G$ if and only if $HK=KH$ as sets.

Proof. If $HK$ is a subgroup, then since $H,K\subseteq HK$, if $k\in K$ and $h\in H$, then $kh\in HK$; hence $KH\subseteq HK$. And if $hk\in HK$, then $(hk)^{-1}=k^{-1}h^{-1}\in HK$, so there exist $h’\in H$ and $k’\in K$ such that $k^{-1}h^{-1}=h’k’$. Taking inverses on both sides we get $hk = (k’)^{-1}(h’)^{-1}\in KH$, so $HK\subseteq KH$, giving equality.

Conversely, if $HK=KH$ as sets, since $HK$ is clearly nonempty it is enough to show that if $hk,h’k’\in HK$, then $(hk)(h’k’)^{-1}\in HK$. Now, $\Bigl(k{k’}^{-1}\Bigr)h’^{-1}\in KH=HK$, so there exist $h’’\in H$, $k’’\in K$ such that $k(k’)^{-1}(h’)^{-1} = h’’k’’$. Then
$$(hk)(h’k’)^{-1} =hk(k’)^{-1}(h’)^{-1} = hh’’k’’ = (hh’’)k’’\in HK,$$
proving that $HK$ is a subgroup. $\Box$

Note that normality is not required. If either $H$ or $K$ are normal (or more generally, if $H\subseteq N_G(K)$ or $K\subseteq N_G(H)$), then we will get $HK=KH$, and therefore that $HK$ is a subgroup. But you can have $HK=KH$ without either subgroup normalizing the other.

Example where $HK=KH$ but neither subgroup normalizes the other. In $G=S_4$, take $H=\langle (1234)\rangle$ and $K$ the subgroup of all permutations that fix $4$. It is clear that $H$ does not normalize $K$, while $K$ does not normalize $H$ since for example $(12)(1234)(12)=(2134)\notin H$. However, $|HK|=|H||K|/|H\cap K| = (4)(6)=24=|S_4|$, so $HK=S_4=KH$. More generally, if $G=S_n$, $n\geq 4$, $H=\langle(12\cdots n)\rangle$, and $K$ is the subgroup of permutations that fix $n$ (so $K\cong S_{n-1}$), then $|HK|=n!$, so $HK=S_n$, but neither normalizes the other (since we are requiring $n\geq 4$).

Title: The definition of the complementary of a subgroup is symmetrical

Body:
Let $G$ be a group and $H,K$ be its subgroups. Neither of them has to be normal.

If

$$G = HK = \{ hk : h\in H, k\in K\} \text{ and } H\cap K = \{e\},$$

then we say $K$ is complementary to $H$. Wikipedia says this definition is symmetrical, namely we can also write $G=KH$.

I understand that $HK=G=KH$ when one of them is normal. But why is it true in general here?


问:
Quadratic form of Hilbert space operator is equal when operators are equal

Show that $\omega(A) \leq ||A|| \leq 2 \omega(A)$ and deduce that $\langle A \xi , \xi \rangle = \langle B \xi, \xi \rangle$ when $A = B$.

I already showed the inequalities but I am stuck on showing $\langle A \xi , \xi \rangle = \langle B \xi, \xi \rangle$ when $A = B$.. Note that $A, B$ are Hilbert space operators and $\omega(A) = \sup \{ |\langle A \xi , \xi \rangle| : ||\xi|| = 1 \}$ is the numerical radius. Also note that we consider complex Hilbert spaces.

答:
Obviously when $A=B$ we have that $A\xi=B\xi$ so $\langle A\xi,\xi\rangle=\langle B\xi,\xi\rangle$, so I assume you mean the converse.

Note that if $\langle A\xi,\xi\rangle=\langle B\xi,\xi\rangle$ then $\langle (A-B)\xi,\xi\rangle=0$, so it suffices to show that if $T$ is a bounded operator with $\langle Tx,x\rangle=0$ for all $x$, then $T=0$. But if this is the case then $\omega(T)=0$, and from the inequalities we get that $\|T\|=0$, thus $T=0$.


问:
Cardinality of set of all straight lines in $\mathbb R^2$, that pass through at least two different rational coordinates.
Definition: A coordinate $(x,y)\in \mathbb R^2$ is rational if $x,y$ are both rational nos.

Let $S$ be the set of all straight lines in $\mathbb R^2$ that pass through at least two distinct rational points.

Let $C$ be the set of all rational points in $\mathbb R^2$. Since $\mathbb Q \times \mathbb Q$ is countable so are its subsets. Therefore, it is possible to write  $C=\{c_1,c_2,...,c_k,c_{k+1},...\}$. Of course $C=\mathbb Q \times \mathbb
 Q$.

Let $S_i$ be the set of all straight lines that pass through $c_i\in C$. For every $s\in S_i$, $\exists$ a rational point $c_{k(s)}\ne c_i$ thus we can have a bijection. Therefore, $S_i$ is countable.

Hence, $S=\cup_{i=1}^{\infty} S_i$ is countable collection of countable sets and therefore countable.

Is my proof correct? Thanks.

答:
Your primary question "Is my proof correct?" has been answered by
user $7\cdot 9011\cdot 2\,$.

The ensuing comment-discussion hints at some detour in your proof.

You may directly aim at a bijection by observing that an element in $S$ is either

parallel to the $y$-axis, hence a vertical line having a rational $\,x_0\,$ as $x$-intercept,

or both its slope $\,m\,$ and its $y$-intercept $\,y_0\,$ are rational.

Taken together, these correspondences define a bijection
$$S\:\longrightarrow\:\big(\{\perp\}\times\mathbb Q\big)\cup\big(\mathbb Q\times\mathbb Q\big)
\:=\: \big(\{\perp\}\cup\mathbb Q\big)\times\mathbb Q\,$$
whose range set is countably infinite.

With the symbol $\perp\,$ standing for "vertical line", and the variable $\,m\in\mathbb Q\,$ as slope we have
$$\big\{(x,y)\in\mathbb R^2 \mid x=x_0\big\}\;\longmapsto\;(\perp, x_0)\\[3ex]
\big\{(x,y)\in\mathbb R^2 \mid y=mx+y_0\big\}\;\longmapsto\;(m, y_0)\,.$$


问:
Folland real analysis 6.28

The Question

I'm trying to solve this exercise from Folland's real analysis book in chapter 6:

Let $I_\alpha$ be the $\alpha th\ $  fractional integral operator $I_\alpha f\left(x\right)=\frac{1}{\Gamma\left(\alpha\right)}\int_{0}^{x}{\left(x-t\right)^{\alpha-1}f(t)dt}$ for $\alpha>0 $

 And let $J_\alpha f\left(x\right)=x^{-\alpha}I_\alpha f\left(x\right)$

 a) prove: $J_\alpha$ is bounded on $L^P\left(0,\infty\right)$ for $1<p\le\infty$ more precisely $||J_\alpha f\left(x\right){||}_p\le\frac{\Gamma(1-p^{-1})}{\Gamma(\alpha+1-p^{-1})}||f{||}_p $

 b) there exist $f\in L^1(0,\infty)$ such that $J_1f\notin L^1(0,\infty)$

My Understanding

My attempt was to define $$ k(x,y)= \frac{1}{\Gamma\left(\alpha\right)}x^{-\alpha}\left(x-y\right)^{\alpha-1}\chi_{\left\{\left(x,y\right):x<y\right\}}\left(x,y\right) $$
and to use

but when i tried to calculate i get:$$ \int_{0}^{\infty}{\left|k\left(x,1\right)\right|x^{-1/p}dx}=\frac{1}{\mathrm{\Gamma}\left(\alpha\right)}\int_{0}^{1}{x^{-\left(\alpha+1/p\right)}\left(1-x\right)^{\alpha-1}dx} $$

which is a bit different from what i need, what did i do wrong?

答:
Notice that
\begin{align}
J_\alpha f (x) =&\ \frac{x^{-\alpha}}{\Gamma(\alpha)} \int^x_0(x-y)^{\alpha-1} f(y)\ dy\\
=&\ \int^\infty_0 K(x, y) f(y)\ dy
\end{align}
where
\begin{align}
K(x, y) = \frac{1}{\Gamma(\alpha)}x^{-\alpha}(x-y)^{\alpha-1}\chi_{\{(x, y)\in (0, \infty)\times (0, \infty)\mid y<x\}}.
\end{align}

Then, we see that
\begin{align}
\int^\infty_0 |K(x, 1)|x^{-1/p}\ dx =&\ \int^\infty_0\left| \frac{1}{\Gamma(\alpha)}x^{-\alpha}(x-1)^{\alpha-1}\chi_{\{ x\in  (0, \infty)\mid 1<x\}}\right| x^{-1/p}\\
=&\ \frac{1}{\Gamma(\alpha)} \int^\infty_1 x^{-\alpha-1/p}(x-1)^{\alpha-1}\ dx\\
=&\ \frac{-1}{\Gamma(\alpha)}\int^\infty_1 \frac{1}{x^{1/p-1}}\left(1-\frac{1}{x}\right)^{\alpha-1}\ d\left(\frac{1}{x}\right)\\
=&\ \frac{1}{\Gamma(\alpha)}\int^1_0 x^{\frac{1}{p}-1}(1-x)^{\alpha-1}\ dx = \frac{1}{\Gamma(\alpha)}\frac{\Gamma(\frac{1}{p})\Gamma(\alpha)}{\Gamma(\frac{1}{p}+\alpha)}
\end{align}

However, it seems that my answer is the Holder conjugate of the expected answer. I don't know whether I have made any mistakes or it is a typo in the textbook.

Title: Folland real analysis 6.28


问:
Doubt - Optimal Control Theory Problem with transversality condition
I'm trying to solve this optimal control theory problem, but i'm having some problem finding the constants $k_1$ and $k_2$. I will show what I have developed so far.

The problem:

$$V(y)=\int_{0}^{1}u^2+2yu+2y^2\,dt$$ with $$y'=y+u;\quad y(0)=0;\quad y(1)=y_{1}$$ where $y(1)$ can vary and $T=1$ is given.

I will skip all the calculations, but after finding the Hamiltonian and applying the canonic equations and solving the system of differential equations I have:

\begin{align}
u&=-y-\frac{1}{2}\lambda\\
y&=-\frac{1}{2}k_1e^t+\frac{1}{2}k_2e^{-t}\\
\lambda&=k_1e^t+k_2e^{-t}
\end{align}

I should use the initial condition $y(0)=0$ and the traversality condition $\lambda(T=1)=0$ to find $k_1$ and $k_2$. But I don't know why, I can't solve these simple system:

\begin{align}
y(0)&=-\frac{1}{2}k_1+\frac{1}{2}k_2=0\\
\lambda(T=1)&=k_1e+k_2e^{-1}=0
\end{align}

I find $k_1=k_2=0$ which seems wrong to me.

Did I miss something in the process?

Thank you very much.

答:
The problem is that the condition on the Lagrange multiplier $\lambda(1)=0$ doesn't apply here (recall this condition arises from the derivation of the Euler-Lagrange equations with the boundary term coming from the integration by parts step). Since the point $y_1$ is presumably fixed, assuming anything about $\lambda(1)$ overdetermines the resulting system of necessary optimality conditions.

As far as solving your system is concerned for completeness of this answer, note that, fortuitously,
$$u=\frac{\lambda}{2}-y
$$
implies
$$\dot{y}=u+y=\frac{\lambda}{2}.
$$
Now since
$$\frac{\delta V}{\delta\lambda}=0\to2u+4y-\dot{\lambda}-\lambda=2y-\dot{\lambda}=0,
$$
after substituting in for u, we find after differentiating the equation for $\dot{y}$, the following boundary value problem
$$\ddot{y}-y=0,\quad y(0)=0,\ y(1)=y_1
$$
whose simple solution is given by
$$y(t)=\frac{y_1}{\sinh1}\sinh t.
$$
For completeness, the lagrange multiplier is
$$\lambda=2\dot{y}=2\frac{y_1}{\sinh1}\cosh t,
$$
therefore the optimal control is given by
$$u=\frac{\lambda}{2}-y=\frac{y_1}{\sinh1}\cosh t-\frac{y_1}{\sinh1}\sinh t=\frac{y_1}{\sinh1}e^{-t}.
$$


答:
It sounds like you're considering an iterative construction which keeps embedding a binary tree inside a larger binary tree, always growing a bit in each direction. If so, this actually does stay countable: picking an arbitrary node and "rotating" everything around that node, what you have is just three copies of the usual (one-way-)infinite binary tree connected at that chosen node.

Of course it's a bit difficult to prove this, given that the definition you have is still rather vague. But I can't see a way of interpreting what you have in mind which doesn't lead to a countable graph. Certainly any time I define a sequence of finite graphs $G_1\subseteq G_2\subseteq G_3\subseteq ...$, their union $$\bigcup_{i\in\mathbb{N}}G_i$$ is countable, so if you want to wind up with something uncountable your construction has to be substantially more complicated.

Title: Is this Perfect Binary Tree Countably Infinite or Uncountably Infinite?

Body:
Given a perfect binary tree which at a certain fixed level has infinitely many leaves, hence each leaf at that fixed level has infinitely many ancestors. From that fixed level the tree also extends to infinite height.

Are the nodes of this tree countably infinite or uncountably infinite?

I know that a rooted binary tree, even if extended to a height of infinity is countably infinite; just start at the root and enumerate level by level. However, I cannot imagine a way to count the nodes of this tree or to apply the idea that the union of countably infinite subtrees is countable. Thus, I’m leaning towards it being uncountably infinite but have serious doubts.

How can a tree look like that?

Imagine that instead of the tree simply growing in height it is also expanding sideways as shown below, so it is infinitely high, infinitely wide and infinitely deep. Getting infinitely wide is illustrated below.


问:
Missing Step Showing That $M_t = \exp(\alpha B_t - \alpha^2 t/2)$ doesn't converge to $0$ in $L^1$ norm.
My plan is to confirm is that $M_t$ is a martingale. If true, then $E[M_{t+1}] = E[M_{t}]$, and since $E[M_0]=1$, then $E[M_t]=1$ for all $t$. We can then conclude that
$$||M_t||_1=E[|M_t|]=1$$
$$\lim_{t \to \infty}|M_t - 0|=1$$
So $M_t$ doesn't converge to zero in $L^1$ norm.

From my book I have the following direction for the proof that $M_t$ is in fact a Martingale.

For $t > s$ we have
$$E[M_t|\mathcal{F}_s] = E[\exp(\alpha (B_t - B_s + B_s) - \alpha^2 (t-s+s)/2)|\mathcal{F}_s]$$
$$= E[\exp(\alpha (B_s) - \alpha^2 (s)/2)\exp(\alpha (B_t - B_s) - \alpha^2 (t-s)/2)|\mathcal{F}_s]$$
$$= M_sE[\exp(\alpha (B_t - B_s) - \alpha^2 (t-s)/2)|\mathcal{F}_s]$$

The last step, which is not explained is that, somehow $E[\exp(\alpha (B_t - B_s) - \alpha^2 (t-s)/2)|\mathcal{F}_s]=1$, so we end up with $E[M_t|\mathcal{F}_s]=M_s$.

I think the idea is that $B_t - B_s = B_{t-s}$, so $\exp(\alpha B_{t-s})$ must equal $\exp(\alpha^2(t-s)/2)$, but I'm not quite sure how?

答:
Note that $B_t-B_s\sim N(0,t-s)$. The moment generating function of a random variable $X$ with normal distribution $N(\mu,\sigma^2)$ is given by
\begin{align}
\mathbb{E}\left[e^{tX}\right]= e^{\mu t+\frac{1}{2}\sigma^2t^2}
\end{align}
Using these two observations (and the fact that Brownian motion has independent increments) we conclude that
\begin{align}
\mathbb{E}\left[e^{\alpha (B_t-B_s)}|\mathcal{F}_s\right]=e^{\frac{1}{2}(t-s)\alpha^2}
\end{align}


答:
If the "to infinity" parts are only out to countable infinity, i.e. $\omega$ levels deep, then the number of nodes themselves are also countably infinite. To see this, just consider performing a breadth-first traversal, laying down numbers as you go. Because each level individually is at finite depth and has finitely many members, it will be possible to reach it eventually after a suitably large finite number of steps.

Now if you want to go beyond $\omega$ depth, then the tree can become uncountably large, but then you run into the question of how to "make that leap" since there will be no immediate predecessors for the nodes that you can connect that first transfinite level to. That is, you have to now define just what the transfinite binary tree means, and how you do that may affect the answer to what the cardinal number of its node set is.

That said, one perhaps good way to do that might be to note how that any node of a binary tree at a given depth $d$ can be "addressed" by a string of binary bits $d$ long, hence we could take the $\omega$-th level to have all its nodes for each countably infinite binary string, i.e. of length $\omega$, or equivalently, one node for each possible endless path starting at the root. Then your $\omega$-th level (so $\omega+1$ height of the tree all total) will be uncountably infinite (cardinal $\beth_1$) and so the whole tree also uncountable. But this may not be the only sensible definition to make.


问:
question on Cauchy-Hadamard formula
I know that the radius of convergence $R$ of a power series $\sum\limits_{k=0}^{\infty}a_kx^k$ is defined by $R:=\frac{1}{\limsup\limits_{k\to\infty}\sqrt[k]{|a_k|}}$ and it can be derived from the root test of infinite series.

However, I am wondering why we don't define the radius of convergence by $R:=\frac{1}{\limsup\limits_{k\to\infty}|a_k|}$?

If we assume that $a_k$ is bounded then for all $x$ with $|x|<\frac{1}{\limsup\limits_{k\to\infty}|a_k|}$ it follows that:
$
|a_kx|<\frac{|a_k|}{\limsup\limits_{k\to\infty}|a_k|}\leq 1$ such that $\sum\limits_{k=0}^{\infty}(|a_kx|)^k$ is a geomtric series. Further, $\sum\limits_{k=0}^{\infty}\sqrt[k]{|a_k|}(|a_kx|)^k$ converges and hence $\sum\limits_{k=0}^{\infty}a_kx^k$ converges as well.

Or is my reasoning flawed?

答:
Simple counterexample: $a_k=2$ has radius of convergence $1$, not $1/2$.

A more extreme one: $a_k=2^k$ has radius of convergence $1/2$, not $0$.

The flaw in your reasoning is in the algebra: in order to make the connection to geometric series, you should write $\sum |a_k| |x|^k = \sum \left ( |a_k|^{1/k} |x| \right )^k$. Then if $x<\frac{1}{\limsup |a_k|^{1/k}}$ then $|a_k|^{1/k} |x|<1$ for all but finitely many $k$. On the other hand if $x>\frac{1}{\limsup |a_k|^{1/k}}$ then $|a_k|^{1/k} |x|>1$ for infinitely many $k$.

Title: question on Cauchy-Hadamard formula


问:
Group Algebra of Direct Product is Tensor Product of Group Algebras
Given groups $G$ and $H$, I want to show that $R[G \times H] \cong R[G] \otimes R[H]$.  I'm not sure why, but I'm struggling with it.  I'd like to map the basis elements from one to the other in the obvious way, that is $$ F:R[G] \otimes R[H] \rightarrow R[G \times H]$$ $$ g \otimes h \mapsto (g,h) $$ but I'm having trouble with the details.  Clearly this is invertible, so I just need to show it respects the algebra structure.  I've got that $$ F((g\otimes h) \cdot (g'\otimes h')) = F((gg') \otimes (hh')) = (gg',hh') = (g,h)(g',h') = F(g \otimes h) F(g' \otimes h) $$ and that $$ F(r(g \otimes h)) = r(g,h) = rF(g \otimes h), $$ so how do I go about showing that $$ F( (g\otimes h) + (g' \otimes h')) = F( g\otimes h) + F(g' \otimes h'). $$  Clearly, $g \otimes h + g' \otimes h' \neq (g+g') \otimes (h + h')$ in general so I feel like I have to be missing something silly or I'm going about this completely wrong?  Any help is appreciated!

答:
It might be easier to appeal to the universal property of the tensor product and the fact that universal objects are defined uniquely up to (unique) isomorphism.

So, we want to show that a given algebra homomorphism $\varphi\colon R[G]\times R[H]\to A$ uniquely induces an algebra homomorphism $\overline\varphi\colon R[G\times H]\to A$ such that $\varphi=\overline\varphi\circ\iota$ where $\iota\colon R[G]\times R[H]\to R[G\times H]$ is the obvious embedding. But this is clear as $\varphi$ is uniquely determined by its values on $g\in G$ and $h\in H$ and hence we can easily define $\overline\varphi$ (uniquely) on the pairs $(g,h)$ which form a basis for $R[G\times H]$.


问:
Proving that $f_n$ doesn't converge uniformly
Given the succesion $f_{n}(x)=x^{2}+\frac{|x|}{n}$. I decided to do the next thing

Let $\epsilon=1$ . Then there is a N $\in\mathbb{N}$ such that for all $n\geq{N}$ we have $|x^{2}+\frac{|x|}{n}-x^{2}|$ < 1. Let us choose $n=N$, and x=n. Then we have $|x^{2}+\frac{|x|}{n}-x^{2}|$=$\frac{n}{n}=1\geq{1}$ so it doesn't converges uniformly in $\mathbb{R}$

答:
One way you might clarify your reasoning is to more explicitly identify why if the sequence converges uniformly in $\mathbb{R}$, it must be to the pointwise limit, i.e., to the function $f(x) = x^2$. You likely know this from a theorem, but if it's stated as a theorem, it wouldn't hurt to cite it.

It's true that the fact that $|f_n(n) - f(n)| = 1$ holds for all $n$ prevents uniform convergence. For if $(f_n)$ converged uniformly, there would be some $N$ such that for all $n \geq N$, the inequality $|f_n(x) - f(x)| < 1$ would hold for all $n \geq N$ and all $x$, and yet it doesn't (as you have noted by putting $n=x=N$).


问:
Find $\lim_{n \rightarrow \infty} \int_0 ^{\infty} \left(1+ \frac{x}{n}\right)^n e^{-ex} dx$
Find $$\lim_{n \rightarrow \infty} \int_0 ^{\infty} \left(1+ \frac{x}{n}\right)^n e^{-ex} dx.$$

I know that $\lim_{n \rightarrow \infty}  \left(1+ \frac{x}{n}\right)^n = e^x.$

Also, I showed that $ \left(1+ \frac{x}{n}\right)^n \leq e^x$ for all $x \geq -n$.

So I'm thinking about the Lebesgue Dominated Convergence Theorem. But I'm not sure how to deal with the $e^{-ex}$ part.

答:
Let
$$ f_n(x)=\left(1 + \frac{x}{n}\right)^ne^{-ex} \chi_{[0,n]}(x).$$
Then
$$ |f_n(x)|\le e^{-(e-1)x}, \lim_{n\to\infty}f_n(x)=f(x)=e^{-(e-1)x}. $$
Since
$$ \int_0^\infty e^{-(e-1)x}dx<\infty $$
the DCT, one has
$$ \lim_{n\to\infty}\int_0^\infty f_n(x)dx=\int_0^\infty\lim_{n\to\infty} f_n(x)dx. $$


问:
Convergent subsequence implies continuous function attains minimum
Let $(X,T)$ be compact and non-empty, then every sequence in $(x_n)_{n=0}^{\infty}$ in $(X,T)$ has a convergent subsequence.

Using this I want to show the following:

If $f:X \rightarrow \mathbb{R}$ is $(X,T)$-continuous, then $f$ attains its minimum on $X$. That is, there exists $\tilde{x} \in X$ such that $f(\tilde{x}) \leq f(x)$ for all $x \in X$.

My attempt:

Let $(x_n)_{n=0}^{\infty}$ be a sequence in $(X,T)$ such that $|f(x_n) - \inf_{x\in X} f(x)| \rightarrow 0$ as $n \rightarrow \infty$. How do I justify the existence of $(x_n)$?

Since $(X,T)$ is compact, $(x_n)_{n=0}^{\infty}$ has a convergent subsequence in $(X,T)$, say $(x_{n_k})_{k=0}^{\infty}$, denote the limit by $\tilde{x} \in X$. Then since $f$ is a continuous function $f(x_{n_k}) \rightarrow f(\tilde{x})$ in $(\mathbb{R},T_{\text{usual}})$.

Moreover, $\tilde{x}$ has the property we desire,  $f(\tilde{x}) \leq f(x)$ for all $x \in X$. How do I justify this?

答:
The usual way to prove this is by first showing that $f$ continuous and $X$ compact, non-empty implies $f(X)$ is compact. Once you know $f(X)$ is a compact subset of $\mathbb{R}$, you will know by the Heine-Borel Theorem that $f(X)$ is closed and bounded. Since $f(X)$ is bounded, $\inf f(X)$ is a real number, and it will be a limit point of $f(X)$. Since $\inf f(X)$ is a limit point and $f(X)$ is closed (i.e. $f(X)$ contains its limit points), you will know that $\inf f(X)$ must be an element of $f(X)$. This proves the desired result since $\inf f(X) \in f(X) \iff \exists x_0 \in X$ such that $f(x_0) = \inf f(X)$.

So, let's show that $f$ continuous and $X$ compact implies $f(X)$ is compact. Let $b_n$ be a sequence in $f(X)$. This means that there is a corresponding sequence $a_n$ in $X$ with $f(a_n) = b_n$. Since $X$ is compact, $(a_n)$ has a convergent subsequence, say $(a_{n_i})_{i=1}^\infty$ converging to $a$. Then, I claim $(b_{n_i})$ is a convergent subsequence of $(b_n)$ converging to $f(a)$. We see that $$\lim_{i \to \infty} b_{n_i} = \lim_{i \to \infty} f(a_{n_i}) = f\left(\lim_{i \to \infty} (a_{n_i})\right) = f(a) $$

where the middle equality holds by the continuity of $f$. Hence, $f(X)$ is compact which implies the result by the discussion in the first paragraph.


问:
How to solve this $3\times3$ system of inequalities
Suppose that $x$, $y$, $z$, $\nu$, $N$, $g$ are some postitive parameters and $A$, $B$ and $C$ are variables that belong to $(0,+\infty)$. I want to identify one or all of the feasible points taht solve the following system of inequalities with respect to $A$, $B$ and $C$

\begin{equation}xA-\nu gB-Ng C>0 \\
-\nu gA+yB-Ng C>0\\
-\nu gA-\nu gB +zC>0\end{equation}

答:
We can rewrite the system of inequalities, including the restrictions on $A,B,C$ to be positive, as follows:

$$  \begin{align*}
  A &\gt \frac{\nu g}{x} B + \frac{N g}{x} C \\
  B &\gt \frac{\nu g}{y} A + \frac{N g}{y} C \\
  C &\gt \frac{\nu g}{z} A + \frac{\nu g}{z} B \\ \\
A,B,C &\gt 0 \end{align*} $$

[Note that I've taken verbatim the terms as originally posted, but I asked for clarification about the third inequality in a Comment on the Question.]

The feasible region is convex, an intersection of six open half-spaces, each delineated by a plane passing through the origin.  It is possible for this intersection to be empty, unlike the relaxed "weak inequalites" system where the origin $(0,0,0)$ will always be a solution.

Note that if $(A,B,C)$ is a feasible point, then so too is any positive scalar multiple $r\gt 0$ of it, $(rA,rB,rC)$.  So geometrically, if the feasible region is not empty, it consists of a collection of open rays proceeding from the origin.

If the coefficients were known, we could represent the feasible region by taking its intersection with the plane:

$$ A + B + C = 1 $$

That intersection (if not empty) is an open polygon (the polygon excluding its perimeter).  Such a polygon would be convex and have at most six sides, and its vertices can be readily computed by many linear programming packages.  The feasible region of your problem is then all the open rays proceeding from the origin through a point in that open polygon.  The points in the open polygon can be parameterized by barycentric coordinates.

If actual values of the coefficients can be given, I'd be happy to use them to illustrate this representation of solutions.  Otherwise I could choose a few sets of coefficients to illustrate both the existence and non-existence of solutions.


答: One other way to see it is to consider the map $\varphi : R[G \times H] \to R[G] \otimes_R R[H]$ that sends $$\sum_{(g, h) \in G \times H} a_{(g, h)} (g, h) \mapsto \sum_{(g, h) \in G \times H} a_{(g, h)} (g \otimes_R h).$$ Understanding the sum on the right-hand side as a double sum $$\sum_{h \in H} \sum_{g \in G} b_g c_h (g \otimes_R h) = \biggl(\sum_{g \in G} b_g g \biggr) \otimes_R \biggl(\sum_{h \in H} c_h h \biggr)$$ with $b_g c_h = a_{(g, h)},$ it follows that $\varphi$ is surjective, as every element of $R[G] \otimes_R R[H]$ is a finite $R$-linear combination of elementary tensors. Of course, $\varphi$ is injective, so it is an isomorphism.

Title: Group Algebra of Direct Product is Tensor Product of Group Algebras

Body: Given groups $G$ and $H$, I want to show that $R[G \times H] \cong R[G] \otimes R[H]$.  I'm not sure why, but I'm struggling with it.  I'd like to map the basis elements from one to the other in the obvious way, that is $$ F:R[G] \otimes R[H] \rightarrow R[G \times H]$$ $$ g \otimes h \mapsto (g,h) $$ but I'm having trouble with the details.  Clearly this is invertible, so I just need to show it respects the algebra structure.  I've got that $$ F((g\otimes h) \cdot (g'\otimes h')) = F((gg') \otimes (hh')) = (gg',hh') = (g,h)(g',h') = F(g \otimes h) F(g' \otimes h) $$ and that $$ F(r(g \otimes h)) = r(g,h) = rF(g \otimes h), $$ so how do I go about showing that $$ F( (g\otimes h) + (g' \otimes h')) = F( g\otimes h) + F(g' \otimes h'). $$  Clearly, $g \otimes h + g' \otimes h' \neq (g+g') \otimes (h + h')$ in general so I feel like I have to be missing something silly or I'm going about this completely wrong?  Any help is appreciated!


问:
Equality involving cosine and binomial coefficients - check if true for every $n$
While doing some computations with complex numbers I think I proved that

$$(\cos{ \frac{n \cdot \pi}{4} )\cdot 2^{n/2}} = \binom{n}{0} - \binom{n}{2} + \binom{n}{4} - \binom{n}{6} + ... $$

How? Well, these are two different forms of

$$\frac{1}{2} \cdot ((1+i)^n + (1-i)^n)$$

But because of the late hour, I am not sure if I didn't make some mistake.

Is this equality really true for every natural n?

Also, I wonder how can one prove it without using complex numbers. I mean is these some sort of elementary (high-school) proof?

答:
Without complex numbers, by induction, assume that for some integer $n\ge 0$,

$$\begin{align*}
2^{n/2}\cos\frac{n\pi}{4} = \binom{n}{0}-\binom n2+\binom{n}{4}-\binom n6+\cdots\\
2^{n/2}\sin\frac{n\pi}{4} = \binom{n}{1}-\binom n3+\binom{n}{5}-\binom n7+\cdots\\
\end{align*}$$

Then by compound angle formulae,

$$\begin{align*}
\cos\frac{(n+1)\pi}{4} &= \cos\frac{n\pi}{4}\cos\frac\pi 4 - \sin\frac{n\pi}{4}\sin\frac\pi 4\\
&= \frac1{\sqrt2} \left(\cos\frac{n\pi}{4} - \sin\frac{n\pi}{4}\right)\\
2^{(n+1)/2}\cos\frac{(n+1)\pi}{4} &= 2^{n/2} \left(\cos\frac{n\pi}{4} - \sin\frac{n\pi}{4}\right)\\
&= \left[\binom{n}{0}-\binom n2+\cdots\right] - \left[\binom{n}{1}-\binom n3+\cdots\right]\\
&= \binom n0 + \left[-\binom n2+\binom n4 -\cdots\right] + \left[-\binom{n}{1}+\binom n3-\cdots\right]\\
&= \binom{n+1}0 - \binom{n+1}2 + \binom{n+1}4-\cdots
\end{align*}$$

Similarly

$$\begin{align*}
\sin\frac{(n+1)\pi}{4} &= \frac1{\sqrt2} \left(\cos\frac{n\pi}{4} + \sin\frac{n\pi}{4}\right)\\
2^{(n+1)/2}\sin\frac{(n+1)\pi}{4} &= \left[\binom{n}{0}-\binom n2+\cdots\right] + \left[\binom{n}{1}-\binom n3+\cdots\right]\\
&= \binom{n+1}{1}-\binom {n+1}3+\binom{n+1}{5}-\cdots
\end{align*}$$

What remains is to prove that the base case when $n=0$ is true.


问:
What are the random variables that constitute i.i.d. samples?
Consider the following passage from Wooldridge (2010):

"For much of this book we adopt a random sampling assumption. More precisely, we assume that (1) a population model has been specified and (2) an independent, identically distributed (i.i.d.) sample can be drawn from the population."

When we refer to a sample as i.i.d., what random variables are we referring to? It seems that there are two possibilities:

Are we saying that each observation is a random variable? For example, do we have a random variable for observation 1, another for observation 2, etc.? In this case, it seems each random variable would be a random vector since our population model has, at the very least, one dependent variable and one covariate.

Are the random variables that constitute the i.i.d. sample the dependent variable and each covariate in our population model? In this case, are the observations for each variable in our population model possible realizations for each random variable?

Source: Wooldridge, J. M. 2010. Econometric Analysis of Cross Section and Panel Data.
Second Edition.

答:
Yes, each observation is a random variable. Yes, if the observation has an outcome variable and covariates, then it would be a random vector. And you would have a random matrix for your entire sample.

Yes, that's also correct. Each random variable consists of a dependent part and the covariates that predict it, the independent part(s). For example, one observation could have the characteristics wage, education, experience, married. Wage is a dependent variable and education, experience, and married are the independent variables. Of course, you could define your model another way and choose different dependent variables, but generally you'd be interested in a particular dependent variable of interest that you're trying to predict.


答:
I will present two possible scenarios. The first one corresponds to the i.i.d vectors $Z_i = (Y_i, X_i)$. Assume that $X_i$ is diet type of the $i$th subject, and $Y_i$ is its weight diff after a certain time period. If you allocate the subject randomly to the diet type (assume that there are two types; $0$ - placebo, and $1$ - some diet), then you have $n$ independent random vectors $Z_i$, that have the same distribution.
Alternatively, assume that the allocation is not random and is determined by some rule (e.g., function of the initial weight), then you have $n_1$ i.i.d random variables of the type $Y_i|X_i=0$, and $n_2$ i.i.d random variables of the type $Y_i|X_i=1$, such that $n= n_1 + n_2$. Consequently, the data generating model for the first case is usually assumed to be something like  $Y_i = \beta_0 + \beta_1X_i + \epsilon_i$, where $X_i \sim Ber(p)$ for $i=1,...n$ while, for the second case $Y_i = \alpha_0 + \alpha_1x_i + \xi_i$, for $i=1,...,n$, where $x_i \in \{0, 1\}$.

Title: What are the random variables that constitute i.i.d. samples?

Body:
Consider the following passage from Wooldridge (2010):

"For much of this book we adopt a random sampling assumption. More precisely, we assume that (1) a population model has been specified and (2) an independent, identically distributed (i.i.d.) sample can be drawn from the population."

When we refer to a sample as i.i.d., what random variables are we referring to? It seems that there are two possibilities:

Are we saying that each observation is a random variable? For example, do we have a random variable for observation 1, another for observation 2, etc.? In this case, it seems each random variable would be a random vector since our population model has, at the very least, one dependent variable and one covariate.

Are the random variables that constitute the i.i.d. sample the dependent variable and each covariate in our population model? In this case, are the observations for each variable in our population model possible realizations for each random variable?

Source: Wooldridge, J. M. 2010. Econometric Analysis of Cross Section and Panel Data.
Second Edition.


问:
Finding the Root of a Polynomial over $\mathbb F_{16}$
Suppose we have the polynomial $p(x) = x^4 + x + 1$ over $\mathbb F_{16}.$  How does one find a root $\alpha$ of $p(x)$?

I have done this for $\mathbb F_2$ by listing and trying every possible element. That approach doesn't seem practical here.

答:
Considering that $16 = 2^4,$ the field $\mathbb F_{16}$ is isomorphic to a quotient of the polynomial ring $(\mathbb Z / 2 \mathbb Z)[x]$ by the ideal generated by an irreducible polynomial $f(x)$ of degree 4. Consequently, for any root $\beta$ of $f(x),$ the evaluation homomorphism $\varphi : (\mathbb Z / 2 \mathbb Z)[x] \to (\mathbb Z / 2 \mathbb Z)(\beta)$ that sends $g(x) \mapsto g(\beta)$ induces an isomorphism $\mathbb F_{16} \cong (\mathbb Z / 2 \mathbb Z)(\beta).$ One can show that $f(x) = x^4 + x^3 + x^2 + x + 1$ is irreducible in $\mathbb Z / 2 \mathbb Z,$ so we may identify $\mathbb F_{16}$ with the set $\{a \beta^3 + b \beta^2 + c \beta + d \,|\, a, b, c, d \in \mathbb Z / 2 \mathbb Z \}$ with $\beta^4 + \beta^3 + \beta^2 + \beta + 1 = 0.$

Observe that $\mathbb F_{16}$ has characteristic $2,$ hence we have the following identities. \begin{align*} \beta^5 &= \beta (\beta^4) = \beta(\beta^3 + \beta^2 + \beta
+ 1) = \beta^4 + \beta^3 + \beta^2 + \beta = 1 \\ \\ \beta^6 &= \beta(\beta^5) = \beta \\ \\ \beta^8 &= (\beta^4)^2 = (\beta^3 + \beta^2 + \beta + 1)^2 = \beta^6 + \beta^4 + \beta^2 + 1  = \beta^4 + \beta^2 + \beta + 1 \end{align*} From here, you can show that $\beta^3 + \beta$ is a root of $x^4 + x + 1$ over $\mathbb F_{16}.$

 \begin{align} (\beta^3 + \beta)^4 + (\beta^3 + \beta) + 1 &= \beta^{12} + \beta^4 + \beta^3 + \beta + 1 \\ &= (\beta^6)^2 + \beta^4 + \beta^3 + \beta + 1 \\ &= \beta^4 + \beta^3 + \beta^2 + \beta + 1 = 0 \end{align}

Of course, I do not claim that this is the simplest computation, but it works.


问:
Show that for all $a,b,c$ positive real numbers that $(a+b+c)(a^{-1}+b^{-1}+c^{-1})\geq 9$
Show that for all $a ,b , c $ positive real numbers that

$$(a + b + c)(a^{-1} + b^{-1} + c^{-1})⩾ 9 $$

答:
HINT

Are you aware of the Cauchy-Schwarz inequality?

If you take $x = (\sqrt{a},\sqrt{b},\sqrt{c})$ and $y = \left(\dfrac{1}{\sqrt{a}},\dfrac{1}{\sqrt{b}},\dfrac{1}{\sqrt{c}}\right)$, we have that
\begin{align*}
\|x\|\|y\|\geq|\langle x,y\rangle|
\end{align*}

Can you take it from here?


问:
$\int_{\Omega} f \psi = 0$ for every $\psi$ which is zero at $\partial \Omega$ imples $f=0$
This is a generalization of my previous question. Let $I=[t_{0},t_{1}]\subset \mathbb{R}$ be fixed and  $\Omega \subset \mathbb{R}^{n}$. Let $\psi$ and $f$ be sufficiently differentiable (we can assume it to be smooth) and let us assume that $\psi(t_{0},x)= \psi(t_{1},x) = 0$ for every $x \in \mathbb{R}^{n}$ and $\psi(t,x)|_{\partial \Omega} = 0$ for every $t \in I$. If:
$$\int_{I}\int_{\Omega}f(t,x)\psi(t,x)dxdt = 0$$
for every $\psi$ satisfying the above conditions, is it true that $f(t,x) \equiv 0$?

答:
Consider just those $\psi(t,x) = \phi(t)g(x)$.
$$
0=\int_I \int_\Omega f(t,x)\psi(t,x)\,dx\,dt = \int_I \phi(t) \left(\int_\Omega f(t,x)g(x)\,dx\right)\,dt = \int_I \phi(t) F_g(t)\,dt.
$$
where $F_g$ is the inner integral.
The assumptions imply (e.g. by your previous question) that $F_g(t) = 0$ for all $t \in I$.
Fix $t \in I$ and let $f_t(x) := f(t,x)$.
Since $g$ (smooth enough etc.)  was unrestrained, we have
$$
0 = \int_\Omega f_t(x) g(x)\,dx
$$
for all smooth $g$.
Assume for sake of contradiction that $f_t(x_0) \neq 0$ for some $x_0$ on the interior of $\Omega$.
Take $g$ to be a positive smooth approximation of $1_{B(x_0)}$ where $B(x_0)$ is a ball around an arbitrary point $x_0$ such that $B(x_0) \subseteq \Omega$ and $f$ does not change sign in $B(x_0)$. Then we find that $\int_{B(x_0)}f_t(x) g(x) \, dx = 0$ but the integral of a smooth function of constant sign being zero means the function is zero, so we arrive at a contradiction, $f_t(x_0)$ must be $0$.
Hence $f_t(x_0) = 0$ for all $t \in I$ and all $x$ on the interior of $\Omega$.
Since for each $t$, $f_t$ is smooth, its value on the boundary of $\Omega$ is the limit of its values on the interior (here is where we assume $\Omega$ is not a pathological set), and hence $f_t(x) =0$ for all $t \in I, x \in \Omega$.


答:
$f(x):=x^2+x+1$ is the only irreducible quadratic $\Bbb F_2$ polynomial. $p(x)$ is irreducible over $\Bbb F_2$ as it has no roots and $f(x)^2=x^4+x^2+1\neq p(x)$. $$\therefore\;\Bbb F_{16}\approx\frac{\Bbb F_2[x]}{\langle p(x)\rangle}$$

Now, the element $x+\langle p(x)\rangle$ is a root of $p(x)$ after identifying $\Bbb F_2$ with $\{\mathbf 0+\langle p(x\rangle),\mathbf 1+\langle p(x)\rangle\}$ where $$\mathbf 0:=0+2\Bbb Z\;\;\;\;\;\mathbf 1:=1+2\Bbb Z$$

Title: Finding the Root of a Polynomial over $\mathbb F_{16}$

Body:
Suppose we have the polynomial $p(x) = x^4 + x + 1$ over $\mathbb F_{16}.$  How does one find a root $\alpha$ of $p(x)$?

I have done this for $\mathbb F_2$ by listing and trying every possible element. That approach doesn't seem practical here.


问:
Difference between Factorization theorem and Fischer-Neymann theorem for t to be sufficient estimator of $\theta$

Difference between Factorization theorem and Fischer-Neymann theorem for t to be sufficient estimator of $\theta$

Factorization theorem says for parameter t to be sufficient statistic of $\theta$, joint density must be equal to product of 2 functions - one function is only function of random sample and other function is function of t and $\theta$

While Fischer-Neyman theorem just says the same but instead of that factor function to be that of t and $\theta$, it adds that it must be pdf(or pmf) of function of t and $\theta$

Can you please explain the difference?

答:
It seems you are quite confused. The Fisher-Neymann factorization theorem says that if the density f can be factored into nonnegative functions g, h such that $f(x)=h(x)g(T(x),\theta)$, then T(x) is a sufficient statistic for theta. Usually people say factorization theorem for short.

By contrast, the definition of sufficiency is that T(x) is a sufficient statistic for theta if f(x|T(x)) does not depend on theta.


问:
First-order logic: How to determine if a formula has a finite or an infinite model

Let $\sigma = \{ E \}$ be a signature with a binary relational symbol $E$. Does the following $[\sigma]$-formula have a finite model? Does it have an infinite model?
$$\varphi := \forall x \exists y (E(x,y) \land \forall z ( z \neq y \rightarrow \neg E(x,z))) \land \forall x \forall y \forall z (E(x,z) \land E(y,z) \rightarrow x = y) \land \exists x \forall y \neg E(y,x)$$

I'm not sure how to approach this question and don't know where to start, can someone give me a hint or point me in the right direction?

答:
Welcome to MSE!

An $E$ structure is just a graph where $E(x,y)$ says there is a (directed!) edge from $x$ to $y$.

So what does $\varphi$ say? If we can translate it into english, then we can do some combinatorics to see if it has finite or infinite models.

$\varphi$ breaks up into a conjunction of $3$ pieces. Let's look at each of them in turn.

$\forall x . \exists y . E(x,y) \land \forall z (z \neq y \to \lnot E(x,z))$

This says every $x$ sees a vertex $y$, and moreover if $z \neq y$, then $x$ doesn't see $z$.

So this says that every $x$ is $E$ related to exactly one vertex $y$ (do you see why?). If you like, we can phrase this as "each vertex has exactly one outgoing edge".

$\forall x,y,z . E(x,z) \land E(y,z) \to x=y$

This says if $x$ and $y$ both see $z$, then actually $x=y$. Graph theoretically, this says "every vertex has at most one incoming edge" (do you see why?).

$\exists x . \forall y . \lnot E(y,x)$

This says that some $x$ has no incoming edges (again, do you see why?).

So now we know what we're looking for! A model of $\varphi$ is a (directed) graph such that

Every vertex has exactly $1$ outgoing edge

Every vertex has at most $1$ incoming edge

At least one vertex has no incoming edges

Now I turn the question back to you:

Does this have a finite model? What about an infinite model?

I hope this helps ^_^


问:
Verify the following arguments are valid using Valid Argument Forms
I am given the following and need to break this down into a valid argument form chart

q -> r

~q -> (u ^ s)

s -> (w v r)

~r

----------

therefore
~w -> r

I so far have down in statement reason format (please enter answers in this format)

~r            given

q->r          given

~q            Modus Tollens

s -> (w v r)  given

w             elimination? (this is where I get confused should this be s -> w?)

thank you

答:
You have derived ${\sim}q$ by modus tollens, so look to the third premise ${\sim}q\to(u\land s)$.

Having thereby derived $u\land s$ by modus ponens, you may derive $s$ by whatisitcalled rule.

Now you may invoke the premise $s\to(w\vee r)$.

Having now derive $w\lor r$ by modus ponens, look again to the premise ${\sim}r$ so you may derive $w$ by whichisitnamed rule.

Having done that, you now need to derive ${\sim}w\to r$.

So what methods do you know that allow you to deduce such a conditional statement?

$$\begin{array}{|lrr}~~1.~q \to r\\~~2.~{\sim}q \to (u \wedge s)\\~~3.~s \to (w \vee r)\\~~4.~{\sim}r\\\hline ~~5.~{\sim}q &\text{modus tollens }&4, 1\\~~6.~u\wedge s&\text{modus ponens }&2, 5\\~~7.~s&\text{what is it called }&6\\~~8.~w\vee r&\text{modus ponens }&3, 7\\~~9.~w &\text{what is it named }&8, 4\\~~~~\vdots\\13.~{\sim}w\to r&\text{some reason}&\mathsf{???}\end{array}$$


问:
Let $a_1,\dots,a_n$ be real numbers, and set $a_{ij} = a_ia_j$.
Let $a_1,\dots,a_n$ be real numbers, and set $a_{ij} = a_ia_j$. Consider the $n \times n$ matrix $A=(a_{ij})$.
Then

It is possible to choose $a_1.\dots,a_n$ such that $A$ is non-singular

matrix $A$ is positive definite if $(a_1,\dots,a_n)$ is nonzero vector

matrix $A$ is positive semi definite for all $(a_1,\dots,a_n)$

for all $(a_1,\dots,a_n)$, $0$ is an eigen value of $A$

I have calculated upto $3\times 3$ that determimant is $0$ but I have no idea how to conclude rigoriously. please help?

答:
It depends.
With $a=(a_1,\ldots,a_n)$, we can more conveniently write $v\mapsto \langle a,v\rangle a$ instead of $v\mapsto Av$, which simplifies computations.

If $n=1$, then $a_1=1$ is a valid choice and makes $A$ regular (the identity). For $n>1$, there exists nonzero $v$ with $\langle a,v\rangle = 0$, hence $A$ is singular. (To be explicit, either some $a_i$ is $=0$ and then the $i$th standard base vector $e_i$ is in the kernel; or $a_1,a_2$ are nonzero and $a_1e_2-a_2e_1$ is in the kernel. That is: Such a choice is possible if and only if $n=1$.

$\langle Av,v\rangle = \langle \langle a,v\rangle a,v\rangle = \langle a,v\rangle^2\ge 0$, so we see that $A$ is positive semi-definite. But if $n>1$ or $a=0$, it is defnitely not positive definite, as follows from 1. Hence the claim is true only for $n=1$

As shown in 2, this is correct

As shown in 1, this is correct only if $n>1$.

So the only claim that is unrestrictedly correct, is 3.


问:
discrete math: n pennies among k children with each child having atleast 2 pennies
This problem is posed in Lovasz's Discrete math book chapter 3 and I understand the correct answer which is $$ \binom{n-k-1}{k-1} $$

However, why is my approach not right ? Here it goes.
We have n pennies. We group them into 2 each  and hence have
$$
\lfloor{\frac{n}{2}}\rfloor
$$
units. Lets call them U.
Now we have to distribute U units among k-1 children which will lead to

$$ \binom{\lfloor{\frac{n-2}{2}}\rfloor-2}{k-1} $$
ways. I subtracted 2 because 2 pennies will goto the first child.

答:
Assume you have n pennies, and you want to distribute them among k children, with each child having at least 2 pennies. You can go as follow:

Distribute 2 pennies to k child.

Since the pennies are identical, the order you distribute the pennies to the children does not matter.

Then there is only one way to do that.

Now you left with n-2K pennies.

And the number of distributing the remaining pennies is:

$(n-2k)+k-1 \choose k-1 $ = $ n-k-1 \choose k-1$.
[By the previous theorem in the textbook].

So the total number is:

$ (1) $ $ n-k-1 \choose k-1$ = $ n-k-1 \choose k-1$.


答:
It's quite possible that this is an exercise given to students in introductory courses such as number theory or discrete mathematics that have taken non-proof based calculus. If that's the case, I will give a proof which doesn't rely on Cauchy-Schwarz or the AM-GM inequality.

Lemma: For every positive real number $x$, we have $x + \frac{1}{x} \geq 2$.

Proof: Using Calculus, we take the derivative of the function $f(x) = x + \frac{1}{x}$ to obtain $f'(x) = 1 - \frac{1}{x^2}$. The critical values are thus $1, 0,$ and $-1$, and it is easy to show that $f(x)$ has a local minimum at $x  = 1$ $(f(1) = 2)$. Since we need not consider negative values of $x$, we can conclude the result.

So now we are ready to tackle the main question in this post: showing $(a+b+c)(a^{-1} + b^{-1} + c^{-1}) \geq 9$.

Proof: We observe that if we simply expand $(a+b+c)(a^{-1} + b^{-1} + c^{-1})$ we obtain $$(a+b+c)(a^{-1} + b^{-1} + c^{-1}) = 3 + (ab^{-1} + a^{-1}b) + (a^{-1}c + ac^{-1}) + (cb^{-1} + c^{-1}b)$$

Then, we note that if we let $x = ab^{-1}$, we have $(ab^{-1} + a^{-1}b) = x + \frac{1}{x} \geq 2$ by our lemma. We can argue in exactly the same way to show that $(a^{-1}c + ac^{-1}) \geq 2$ and $(cb^{-1} + c^{-1}b) \geq 2$. Thus, we obtain $$3 + (ab^{-1} + a^{-1}b) + (a^{-1}c + ac^{-1}) + (cb^{-1} + c^{-1}b) \geq 3 + 2 + 2 + 2 = 9$$ as desired.

Title: Show that for all $a,b,c$ positive real numbers that $(a+b+c)(a^{-1}+b^{-1}+c^{-1})\geq 9$

Body:
Show that for all $a ,b , c $ positive real numbers that

$$(a + b + c)(a^{-1} + b^{-1} + c^{-1})⩾ 9 $$


答:
Okay, here's another approach. First, multiply the left side all out, to get
3 + a/b + b/a + a/c + c/a + b/c + c/b.  Now consider a/b + b/a. Making a common demonimator, we get $\frac{a^2 + b^2}{ab} $. Next we note that $(a-b)^2 = a^2 + b^2 -2ab$, so $a^2 + b^2 = (a-b)^2 + 2ab.$  Putting that over ab, we get $\frac{(a-b)^2 + 2ab}{ab} > 2$.  Repeat this for the other pairs, and we get the result.


问:
$M^n-I$ invertible implies $M$ nilpotent over finite field
Let $M$ be a matrix over the finite field $\mathbb F_p$ ($p$ prime) such that $M^n-I$ is invertible for all positive integers $n$. Prove that there exists $k$ such that $M^k=0$ (i.e. $M$ is nilpotent).

The condition is equivalent to $\det(M^n-I)\neq0$ for all $n$. So using $$M^{2n}-I=(M^n-I)(M^n+I),$$ I have shown that $M^n+I$ must always be invertible too.

It would be enough to show that the only eigenvalue of $M$ is $0$, but I'm not sure how to do that.

答: There are only finitely many matrices in $\mathbb F_p$ with the same size as $M$, so the list $M, M^2, M^3,\dots$ repeats eventually, say $M^k=M^\ell$ for $k<\ell$. Then $M^k\left(M^{\ell-k}-I\right)=0\implies M^k=0$, as $M^{\ell-k}-I$ is invertible.


问:
Example of series of the form $\sum \frac{1}{n^{1+a_n}}$
How one can find a sequence $(a_n)$ of positive numbers such that $\lim a_n=0$ and

$$\sum^{\infty}_{n=i} \frac{1}{n^{1+a_n}}$$

converges.

Thank you!

答:
Let

$$ a_{n} = \frac{\log \log n}{\log n} + \frac{2 \log \log \log n}{\log n} $$

for large $n$. Then

$$\frac{1}{n^{1+ a_{n}}} = \frac{1}{n \log n (\log \log n)^{2}} $$

for large $n$ and by the integral test, the corresponding summation converges.


问:
P-values and statistical significance
The mean for the result of a regional test is $525$ points with a deviation of $80$.

A $90$ students class did the test and got a mean score of $535$.

Is the mean score of the class different from the regional mean score for a statistical significance of $0.05$?

I'm trying to understand the role of statistical significance and just can't wrap my head around how could these means not be different, would love to get some help understanding this problem.

答:
@Semoi's Answer (+1) in terms of a 95% confidence interval is a correct
approach for this two-sided test at the 5% level (but it should be centered at the class average 535). Here I show the test in
what I hope is a familiar format.

You need to think of $\mu = 525$ as a population mean and $\sigma = 80$ as the population standard deviation (for the 'region').  You need to think of
$\bar X = 535$ as the mean of a sample of size $n = 90$ (this one class) as
explained in @NuclearHoagie's Comment.

"Different from" implies a 2-sided alterative for your z-test. The test statistic is
$z = \frac{535 - 525}{80/\sqrt{90}}= 1.18585.$ The two-sided P-value is
$$P(|Z| >  1.18585) =  0.2357 > 0.05 = 5\%,$$ where $Z$ is standard normal.
So do not reject
$H_0:\mu = 525$ against the two-sided alternative at the 5% level.

In R:

```<code>z = (535-525)/(80/sqrt(90));  z
[1] 1.185854
2*pnorm(-z)
[1] 0.2356799
</code>```

From a recent release of Minitab statistical software (which has a one-sample z-test procedure for summarized data):

```<code>One-Sample Z

Test of μ = 525 vs ≠ 525
The assumed standard deviation = 80

 N    Mean  SE Mean       95% CI          Z      P
90  535.00     8.43  (518.47, 551.53)  1.19  0.236
</code>```

Below is a simulation of class averages a from a thousand
classes of size $n = 90.$ The summary shows class averages were
as small as 479 and as large as 557. Half of the 1000 classes
had scores between 518 and 532. So an average score of 535 in a class of
90 students is not surprising. [Such a class with an average score of 555 would be surprising.]

```<code>set.seed(2021)
a = replicate(1000, mean(rnorm(80, 525, 90)))
summary(a)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  479.1   518.0   525.3   525.4   532.3   557.7
</code>```

A histogram of the 1000 simulated class averages is shown below. The vertical dashed line is at 535.

Title: P-values and statistical significance

Body:
@Semoi's Answer (+1) in terms of a 95% confidence interval is a correct
approach for this two-sided test at the 5% level (but it should be centered at the class average 535). Here I show the test in
what I hope is a familiar format.

You need to think of $\mu = 525$ as a population mean and $\sigma = 80$ as the population standard deviation (for the 'region').  You need to think of
$\bar X = 535$ as the mean of a sample of size $n = 90$ (this one class) as
explained in @NuclearHoagie's Comment.

"Different from" implies a 2-sided alterative for your z-test. The test statistic is
$z = \frac{535 - 525}{80/\sqrt{90}}= 1.18585.$ The two-sided P-value is
$$P(|Z| >  1.18585) =  0.2357 > 0.05 = 5\%,$$ where $Z$ is standard normal.
So do not reject
$H_0:\mu = 525$ against the two-sided alternative at the 5% level.

In R:

z = (535-525)/(80/sqrt(90));  z
[1] 1.185854
2*pnorm(-z)
[1] 0.2356799

From a recent release of Minitab statistical software (which has a one-sample z-test procedure for summarized data):

One-Sample Z

Test of μ = 525 vs ≠ 525
The assumed standard deviation = 80

 N    Mean  SE Mean       95% CI          Z      P
90  535.00     8.43  (518.47, 551.53)  1.19  0.236

Below is a simulation of class averages a from a thousand
classes of size $n = 90.$ The summary shows class averages were
as small as 479 and as large as 557. Half of the 1000 classes
had scores between 518 and 532. So an average score of 535 in a class of
90 students is not surprising. [Such a class with an average score of 555 would be surprising.]

set.seed(2021)
a = replicate(1000, mean(rnorm(80, 525, 90)))
summary(a)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  479.1   518.0   525.3   525.4   532.3   557.7

A histogram of the 1000 simulated class averages is shown below. The vertical dashed line is at 535.


问:
How to find a lower bound on a smallest solution of $\ln( \frac{x}{a} ) +\frac{1}{x}=0$.
The equaiton
\begin{align}
 \ln( \frac{x}{a} ) +\frac{1}{x}=0
\end{align}
has two solutions for a sufficiently large $a$ (e.g., $a>3$).
I want a lower bound on the smaller of the two solutions.

I have found an upper bound on the largest of the two solutions as follows:
\begin{align}
0= \ln( \frac{x}{a} ) +\frac{1}{x}\le  1- \frac{a}{x}  +\frac{1}{x}
\end{align}
where we have used a lower boung $\ln(x) \ge 1-\frac{1}{x}$.  This implies that $x_2<a-1$.

However, I have no idea how to find a lower bound on the smallest solution.

答:
Set $u = \ln(x/a)$. Then this gives

$$
u + \frac{1}{ae^u} = 0
$$

which we rewrite as

$$
u e^u = -\frac{1}{a}
$$

Then we see

$$
u = W(-1/a)
$$

Where $W$ is the Lambert W function.

Now it's known (it's actually on the wikipedia page linked above) that

$$
W(x) = \sum_{n = 1}^\infty \frac{(-n)^{n-1}}{n!} x^n
$$
whenever $|x| < 1/e$.

Thankfully you added the requirement that $a > 3$ or so, and so we can use this formula to derive progressively better lower bounds, since we know the exact solution will be

$$
\ln(x/a) = u = W(-1/a) =
\sum_{n = 1}^\infty \frac{(-n)^{n-1}}{n!} \left ( \frac{-1}{a} \right )^n
$$

I hope this helps ^_^


问:
Two equal circles inscribed in a triangle
Two equal circles are inscribed in a triangle as shown, with AC = 20 cm, AB = 13 cm and BC = 21 cm. Find the radius of the circle. I have no clue how it can be done except the drawing made in the diagram as shown. [Only Pythagoras theorem  and algebraic manipulations are allowed to be used]

答:
With trig:

Hint: The cosine rule gives $$\cos A = \dfrac{b^2+c^2-a^2}{2bc} \ \text{ where }a=BC,b=CA,c=AB$$ so you know all trigonometric values like $f(z)$ or $f(z/2)$ where $z$ is the angle $A,B$ or $C$, $f$ is one of the six trigonometric functions.

 Further hint: Note that $$BC = r\cot \frac{B}2 + 2r + r\cot \frac{C}2$$

Without trig:

Hint: Call the centers of the circles $B',C'$, with $B'$ closer to $B$.

 $BB' \cap CC' = I =$ incenter of $\triangle ABC$

Lengths of $BI,CI$ are known from applications of internal angle bisector theorem. Use similarity, and you might even be able to use the fact that $ID$ (where $D$ is the midpoint of $BC$) passes through the mutual touchpoint of the two circles.


答:
Notice htat $\sum \frac{1}{n\log^2 n}$ converges.
So, make this happen by setting
$$n^{a_n}=\log^2 n,$$
Then $a_n\log n = 2\log\log n.$ </p>

Title: Example of series of the form $\sum \frac{1}{n^{1+a_n}}$

Body:
How one can find a sequence $(a_n)$ of positive numbers such that $\lim a_n=0$ and

$$\sum^{\infty}_{n=i} \frac{1}{n^{1+a_n}}$$

converges.

Thank you!