问:
Is the null space of a matrix the same after you multiply it by more matrices?
Suppose I know that the nullspace of matrix $P$ is $\mathbf{c}$. Now suppose I multiple $P$ by some matrix $C$, ie. $CP$. Is the nullspace of $CP$ also just $\mathbf{c}$? Or can there be other vectors in the nullspace not equal to $\mathbf{c}$?

答: $Px=0\Rightarrow CPx=0$ for any $P,~C$. So null space of $P$ is a subset of the null space of $CP$. They are not always equal, for example if $P$ is non-singular and $C$ is singular then since $CP$ is singular, the null space of $CP$ is strictly bigger.


问:
Where is my error in this combinations word problem?

A student has to answer $10$ out of $13$ questions in an examination. The number of ways in which he can answer if he must answer at least $3$ of the first five questions is:

The answer is not $^5C_3 \times {^{10}C_7}$. Why?????

$C$ here is combination.

答:
Your solution implies the exam taker answers $3$ questions from the first $5$, and then $7$ questions from the remaining $10$.

This counts some of the combinations more than once.

For example, label the first $5$ question $A,B,C,D,E$, and the last $8$ as $a,b,c,d,e,f,g,h$.

Then you count say $ABCDEabcde$ several times, e.g.:

(ABC)DEabcde, (ABD)CEabcde, (CDE)ABabcde, etc...

In fact it counts each one $\binom{5}{3}=10$ times.

A different solution is to notice that if you answer $10$ questions from $13$, then, by the Pigeonhole Principle, you must have answered at least $2$ from the first $5$, and there are $\binom{5}{2}=10$ ways to do that. So the answer is $\binom{13}{10}-\binom{5}{2}=286-10=276$.

ContentLicense: CC BY-SA 4.0


问:
Corollary 2.9 from Eisenbud's Commutative Algebra

Corollary 2.9. If $\varphi:M\longrightarrow N$ is a map of $R$-modules, then $\varphi$ is a monomorphism (or epimorphism, or isomorphism) if for every maximal ideal $\mathfrak{m}$ of $R$ the localized map
$$\varphi_\mathfrak{m}:M_\mathfrak{m}\longrightarrow N_\mathfrak{m}$$
is a monomorphism (or epimorphism, or isomorphism).

So I understand that this is a statement about $R$-modules. As I understand it, this should translate easily into an equivalent statement about rings. But I'm having a little trouble making this little mental gymnastics.

If $\varphi:R\to S$ is a map of rings. We can regard $R$ as an $R$-module, and $S$ as an $R$-module as well via the action $r\cdot s=\varphi(r)s$ for all $r\in R$ and $s\in S$. Then, say I want to localize at a maximal ideal $\mathfrak{m}$ of $R$. Then treating $R$ and $S$ both as modules:
$$ \varphi_{\mathfrak{m}} : R_{\mathfrak{m}} \to S_{\mathfrak{m}} $$
is the localized map. But I do not only want to view this as module map, I also want to view this as a ring map. I'm having a little trouble seeing what exactly $S_{\mathfrak{m}}$ means in terms of rings. At which multiplicative set is $S$ being localized at as a ring? My mental block comes from the fact that $\mathfrak{m}$ is not, at least a priori, a maximal ideal (or even ideal) of $S$.

Thanks.

答: In this case $S_m$ is a localization of $R$-algebra. There are at least two ways to work with this. You can repeat localization construction from scratch for algebras and check that there is a well-defined product of fractions if you localize an algebra. Or, you can notice that $S_m \cong R_m \otimes_R S$, since $S$ and $R_m$ are algebras over $R$ the tensor product also has an algebra structure.


问:
Calculate tangent of $f(x) = x^4 -2x^2 - x$
Given is the following function:

$f(x) = x^4 - 2x^2 - x$

Question

Give 2 points on the graph of this function that have the same tangent.

I am in a hurry and have no idea how to approach this, I've calculated the derivative $f'(x) = 4x^3 - 4x - 1$ but from this I don't know how to calculate the two points..

Can anybody please help me?

答:
This question appeared in the 2016 FAMAT Calculus January Regional,

As the question does not assume $2$ distinct points of tangency, you could theoretically give the same two points. For example, $(0,0)$ and $(0,0)$ have the same tangent line, i.e. $y = -x$.

ContentLicense: CC BY-SA 3.0


问:
Find the minimum of $\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$
Find the minimum of $$f(x,y)=\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$$ It seems that $f_x=f_y=0$ is very hard to compute. Is there any easier idea?

答:
$f(x,y)$ can be written as
$$f(x,y)=3\sqrt 2\bigg(\sqrt{\bigg(x-\frac 12\bigg)^2+\bigg(0-\frac{1}{6}\bigg)^2}+\sqrt{\bigg(x-\frac{2y-1}{6}\bigg)^2+\bigg(0-\frac{2y-1}{6}\bigg)^2}\bigg)+\sqrt{4y^2-12y+10}$$

Here, let
$$g(x):=\sqrt{\bigg(x-\frac 12\bigg)^2+\bigg(0-\frac{1}{6}\bigg)^2}+\sqrt{\bigg(x-\frac{2y-1}{6}\bigg)^2+\bigg(0-\frac{2y-1}{6}\bigg)^2}$$
which can be seen as the sum of the distance from $(x,0)$ to $\bigg(\dfrac 12,\dfrac 16\bigg)$ and the distance from $(x,0)$ to $\bigg(\dfrac{2y-1}{6},\dfrac{2y-1}{6}\bigg)$.

Case 1 : If $\dfrac{2y-1}{6}\le 0$, i.e. $y\le\dfrac 12$, then then $g(x)$ is minimized when $(x,0)$ is on the line passing through $\bigg(\dfrac 12,\dfrac 16\bigg)$ and $\bigg(\dfrac{2y-1}{6},\dfrac{2y-1}{6}\bigg)$. So, solving $$\bigg(0-\frac 16\bigg)\bigg(\frac 12-\frac{2y-1}{6}\bigg)=\bigg(x-\frac 12\bigg)\bigg(\frac 16-\frac{2y-1}{6}\bigg)$$ for $x$ gives $x=\dfrac{1-2y}{6(1-y)}$. Therefore, we have $$\begin{align}f(x,y)&\ge 3\sqrt 2\ g\bigg(\dfrac{1-2y}{6(1-y)}\bigg)+\sqrt{4y^2-12y+10}
\\\\&=\frac{1}{\sqrt 2}\sqrt{\frac{2y^2-6y+5}{(1-y)^2}} + \frac{1}{\sqrt 2}\sqrt{\frac{(1-2 y)^2 (2y^2 - 6 y + 5)}{(1-y)^2}} + \sqrt{4 y^2 -12 y + 10}
\\\\&=\frac{1}{\sqrt 2}\cdot\frac{\sqrt{2y^2-6y+5}}{1-y}+\frac{1}{\sqrt 2}\cdot\frac{(1-2y)\sqrt{2y^2-6y+5}}{1-y}+ \sqrt{4 y^2 -12 y + 10}
\\\\&=\frac{(2-2y)\sqrt{2y^2-6y+5}}{\sqrt 2\ (1-y)}+ \sqrt{4 y^2 -12 y + 10}
\\\\&=2\sqrt{4 y^2 -12 y + 10}
\\\\&=4\sqrt{\bigg(y-\frac 32\bigg)^2+\frac 14}\end{align}$$Therefore, we see that if  $y\le \dfrac 12$, then the minimum of $f(x,y)$ is $4\sqrt{\bigg(\dfrac 12-\dfrac 32\bigg)^2+\dfrac 14}=2\sqrt 5$ which is attained when $(x,y)=\bigg(0,\dfrac 12\bigg)$.

Case 2 : If $\dfrac{2y-1}{6}\gt 0$, i.e. $y\gt\dfrac 12$, then $g(x)$ is minimized when $(x,0)$ is on the line passing through $\bigg(\dfrac 12,\dfrac 16\bigg)$ and $\bigg(\dfrac{2y-1}{6},\color{red}-\dfrac{2y-1}{6}\bigg)$. So, solving $$\bigg(0-\frac 16\bigg)\bigg(\frac 12-\frac{2y-1}{6}\bigg)=\bigg(x-\frac 12\bigg)\bigg(\frac 16+\frac{2y-1}{6}\bigg)$$ for $x$ gives $x=\dfrac{2y-1}{3y}$. Therefore, we have $$\begin{align}f(x,y)&\ge 3\sqrt 2\ g\bigg(\dfrac{2y-1}{3y}\bigg)+\sqrt{4y^2-12y+10}
\\\\&=\sqrt{\frac{y^2 - 2 y + 2}{y^2}} + \sqrt{\frac{(2 y - 1)^2 (y^2 - 2 y + 2)}{y^2}} + \sqrt{4 y^2 -12 y + 10}
\\\\&=\frac{\sqrt{y^2-2y+2}}{y}+\frac{(2y-1)\sqrt{y^2-2y+2}}{y}+\sqrt{4y^2-12y+10}
\\\\&=2\sqrt{y^2-2y+2}+2\sqrt{y^2-3y+\frac 52}
\\\\&=2\bigg(\sqrt{(y-1)^2+(0-1)^2}+\sqrt{\bigg(y-\frac 32\bigg)^2+\bigg(0-\frac 12\bigg)^2}\bigg)\end{align}$$Here, let $$h(y):=\sqrt{(y-1)^2+(0-1)^2}+\sqrt{\bigg(y-\frac 32\bigg)^2+\bigg(0-\frac 12\bigg)^2}$$ which can be seen as the sum of the distance from $(y,0)$ to $(1,1)$ and the distance from $(y,0)$ to $\bigg(\dfrac 32,\dfrac 12\bigg)$. So, $h(y)$ is minimized when $(y,0)$ is on the line passing through $(1,1)$ and $\bigg(\dfrac 32,\color{red}-\dfrac 12\bigg)$. So, solving $$(0-1)\bigg(1-\frac 32\bigg)=(y-1)\bigg(1+\frac 12\bigg)$$ gives $y=\dfrac 43$. Therefore, we see that if $y\gt \dfrac 12$, then the minimum of $f(x,y)$ is $2h\bigg(\dfrac 43\bigg)=\sqrt{10}$ which is attained when $(x,y)=\bigg(\dfrac{5}{12},\dfrac 43\bigg)$.

In conclusion, it follows from the two cases that the minimum of $f(x,y)$ is $\color{red}{\sqrt{10}}$ which is attained when $(x,y)=\bigg(\dfrac{5}{12},\dfrac 43\bigg)$.


问:
If $x^{T}A^{T}Ax = x^Tx$ holds for every $x$, then $A^{T} A = I_n$

Given $A \in \mathbb R^{n \times n}$, if $$\left( \forall x \in\mathbb R^n \right) \left(x^{T} A^{T} A x = x^T x \right)$$ how to conclude that $A^{T}A = I_n$?

I appreciate any help!

答: Presumably the underlying field is real. It suffices to prove the stronger statement that if $S$ is a symmetric matrix and $x^TSx$ is identically zero, then $S=0$. To prove this, put $x=v+Sv$ for some arbitrary vector $v$.


问:
Solution of the ordinary differential equation $y'(t)=-y^3+y^2+2y$
Consider the solution of the ordinary differential equation $y'(t)=-y^3+y^2+2y$ subject to $y(0)=y_0 \in (0,2). $ then $\lim \limits_{t \to \infty}y(t)$ belongs to

{-1,0}

{-1,2}

{0,2}

{0, $+\infty $}

My Attempt:  $y'(t)=-y^3+y^2+2y \\ \Rightarrow y'(t)=-y(y^2-y-2) \\ \Rightarrow y'(t)=-y(y-2)(y+1) \\ \Rightarrow \frac{dy}{dt}=-y(y-2)(y+1) \\ \Rightarrow \frac{dy}{y(y-2)(y+1)}=-dt \\ \Rightarrow \int\frac{dy}{y(y-2)(y+1)}=\int-dt \\by \;\;partial \;fraction \; we \; get \\ \int [\frac{-1}{2y}+\frac{1}{6(y-2)}+\frac{1}{3(y+1)}]dy =-\int dt \\ \Rightarrow -\frac{1}{2} log|y| +\frac{1}{6} log|y-2| + \frac{1}{3} log|y+1|=-t + logc_1 \\ \Rightarrow \frac{-3log|y|+log|y-2|+2log|y+1|}{6}=-t + logc_1 \\ \Rightarrow log\frac{(y-2)(y+1)^2}{y^3}=-6t+6logc_1 \\ \Rightarrow log\frac{(y-2)(y+1)^2}{y^3}-logc_1^6=-6t \\ \Rightarrow log\frac{(y-2)(y+1)^2}{cy^3}=-6t \;\; where \;c_1^6=c \\ \Rightarrow \frac{(y-2)(y+1)^2}{cy^3}=e^{-6t}..........(1) \\ when \;\;t=0 \;\; \frac{(y_0-2)(y_0+1)^2}{y_0^3}=c \\ (1) \Rightarrow \frac{(y-2)(y+1)^2}{y^3}=\frac{(y_0-2)(y_0+1)^2}{y_0^3} e^{-6t}$

Now I am not getting how to apply the limit for this and how to find $\lim \limits_{t \to \infty} y(t)$

答:
Given that $$y'(t)=-y^3+y^2+2y=-y\left(y^2-y-2\right)=-y(y-2)(y+1)$$
Clearly at $~y=-1,~0,~2~$ we have $~y'=0~.$

So the direction field will be something like:

When $~y<-1~,$ slope of $~y~$ is increasing.

When $~-1<y<0~,$ slope of $~y~$ is decreasing.

When $~0<y<2~,$ slope of $~y~$ is increasing.

When $~y>2~,$ slope of $~y~$ is decreasing.

Given that $~y(0)=y_0\in(0,~2)~,$ which implies that $~\lim_{t\to\infty}y(t)=2~.$

Hence option $\bf{(2)}$ and option $\bf{(3)}$ are the only correct options.

Note:  I know that there is an accepted answer of this old question. But One can check that this is an totally new idea.  So I don't think the community should have any objection about it. Thank you.


问:
Example of a map from the natural numbers to the positive rational numbers less than 1
I want an example of a bijective function $f:\mathbb{N} \to \{n \in \mathbb{Q}:0<n<1\}$

There is already a question about an example of a map between $\mathbb{N}$ and $\mathbb{Q}$.

I am not really a layperson but you can approximate me as a layperson. So I cannot derive an example for this.

MOTIVATION:

a few days I was thinking about some crazy functions and roots of those function. The weierstrass factorization theorem also says that any continuous function can be written as the product involving its roots.

Yesterday I saw a video on the product formula of $\sin(x)$.

And I started wandering that can a function have roots which are all rational numbers or maybe roots which all positive rational numbers less than 1.

To get started at all I have to start here

$$\prod_{n=1}^\infty\left(x-a_nQ_n \right)$$

Where $Q_n$ is an example of a bijective function $f:\mathbb{N} \to \{n \in \mathbb{Q}:0<n<1\}$

答:
Let $S$ be the set of all $(x,y)\in \Bbb N^2$ such that $x<y$ and $\gcd(x,y)=1.$ For $q\in \Bbb Q$ with $0<q<1$ let $f(q) =(x,y)\in S$ such that $q=x/y.$  Then $f:\Bbb Q\cap (0,1)\to S$ is a bijection.

Consider the reverse-lexicographic order (i.e. backwards-dictionary order) $<^*$ on $S.$ That is, if $(x_1,y_1)$ and $(x_2,y_2)$ are in S then $(x_1,y_1)<^* (x_2,y_2)$ iff

(i) $\; y_1<y_2$ or (ii) $\; y_1=y_2 \land x_1<x_2.$

Observe that $<^*$ is a linear ("total") order.

Now for any $n\in\Bbb N$ there is exactly one $z\in S$ such that the set $$T(z)=\{w\in S: w=z\lor w<^*z\}$$ has exactly $n$ members (See Appendix below).

So for $q\in \Bbb Q$ with $0<q<1,$  let $g(q)=n$ iff $T(f(q))$ has exactly $n$ members. Then $g: \Bbb Q\cap (0,1)\to\Bbb N$ is a bijection. (Recall $f$ from the 1st paragraph).

Appendix. For $y'\in \Bbb N$ let $S(y')=\{(x,y')\in S\}.$ Then $S(y')$ is finite, with at most $y'-1$ members. So if $z=(x,y)\in S$ then $T(z)\subseteq \cup \{S(y'): y'\le y\}$ so $T(z)$ is finite. Now we have

(a). If $z$ and $z'$ are in $S$ with $z<^*z'$ then the finite set $T(z)$ is a proper subset of the finite set $T(z')$ so $T(z)$ has fewer members than $T(z').$

(b). $\;(1,2)$ is the $<^*$-least member of $S,$ so $T((1,2))$ has just one member.

(c). If  $(x,y)\in S$   then there is a $<^*$-least $w\in  S$ such that $(x,y)<^*w,$ namely

(c-i). $\; w=(x',y)$ where $x'$ is the least $x''>x$ such that $(x'',y)\in S$ IF such an $x''$ exists, or

(c-ii). $\; w=(1,y+1)$ if there is no such $x''$ as in (c-i).

And if $T((x,y))$ has $n$ members then $ T(w)$ has  $n+1$ members.

So by induction on $n\in \Bbb N,$ by (b) and (c), for every $n$ there exists some  $z\in S$ such that $T(z)$ has  $n$ members;  By (a) there is only one such $z.$

Remark: You may have noticed that $<^*$ is a well-order. That is, if $\emptyset\ne U\subseteq  S$ then $U$ has a $<^*$-least member. If $<^*$ is a well-order on an infinite set $S$ and if $T(z)$ (as defined above) is finite for each $z\in S$ then $(S,<^*)$ is order-isomorphic to $(\Bbb N,<).$ The first few members of $S$ in $<^*$-increasing order are $(1,2),(1,3),(2,3),(1,4),(3,4),(1,5),...,$ corresponding to the rationals $1/2,1/3,2/3,1/4,3/4,1/5,...$


问:
Proving that Ring of Complex Entire functions is neither Artinian nor Noetherian
Question: Prove that the Ring of Complex Entire functions is neither Artinian nor noetherian.

Proof: Clearly $R$ is not Artinian because it is a commutative integral domain which is not a field, and $R$ is not noetherian because it is  not  a factorisation domain.

Is there a proof of this theorem using the Ascending / Descending Chain condition for Artinian / Noetherian rings?

答: Let $J_n$ be the ideal of entire functions vanishing on the first $n$ positive integers, and let $I_n$ be the ideal of entire functions vanishing on positive integers greater than $n$.  Then $$J_1\supset J_2\supset J_3\supset\cdots,$$ $$I_1\subset I_2\subset I_3\subset\cdots,$$ and all of these containments are proper.  One way to see that the containments are proper is to use the fact that given any sequence of complex numbers $a_1,a_2,\ldots$, there is an entire function $f$ such that $f(n)=a_n$ for each positive integer $n$.  For more on this, see these MathOverflow questions.</p>

OwnerDisplayName: anonymous

ContentLicense: CC BY-SA 2.5


问:
Subset sum problem is NP-complete?
If I know correctly, subset sum problem is NP-complete. Here you have an array of n integers and you are given a target sum t, you have to return the numbers from the array which can sum up to the target (if possible).

But can't this problem be solved in polynomial time by dynamic programming method where we construct a table n X t and take cases like say last number is surely included in output and then the target becomes t- a[n]. In other case, the last number is not included, then the target remains same t but array becomes of size n-1. Hence this way we keep reducing size of problem.

If this approach is correct, isn't the complexity of this n * t, which is polynomial? and so if this belongs to P and also NP-complete (from what I hear) then P=NP.

Surely, I am missing something here. Where is the loophole in this reasoning?

Thanks,

答:
This is one of those fine points sometimes neglected when learning about the subject. The efficiency of an algorithm is always measured in regards to the size of representation of the input - how much bits you need to encode it. In the case of numbers this distinction is crucial since the number $n$ is usually represented by $\lg n$ (log in the base 2) of bits. Hence, a solution that is $O(n)$ is exponential in the input size, hence extremely inefficient.

The classic example for this distinction is primality checking; even the most naive algorithm is $O(n)$, but we can't think of something like this as truly efficient even if we adopt a real-life approach - we can (and do) work with numbers with hundereds of digits on a daily basis, and usual arithmetic with those numbers is quite fast (being polynomial in the number of digits), but naive primality testing methods will never finish in real life even for numbers with a hundred digits or so.

The same danger lurks in any problem involving numbers, in particular subset sum.

Title: Subset sum problem is NP-complete?


问:
how can $\sqrt{2}/4=1/2^{3/2}$
$\sqrt{2}/4=1/2^{3/2}$ this is a subpart of a work example question I have but I don't understand how I can convert the first part into the second.

Title: how can $\sqrt{2}/4=1/2^{3/2}$

Body:
$$\require{cancel}\frac{\sqrt{2}}{4}=\frac{\cancel{\sqrt{2}}}{\cancel{\sqrt{2}}\sqrt{2}\sqrt{2}\sqrt{2}}=\frac{1}{\sqrt{2}^3}=\frac{1}{(2^{1/2})^3}=\frac{1}{2^{3/2}}.$$

Or,

$$\frac{\sqrt{2}}{4}=\frac{2^{1/2}}{2^2} = 2^{(1/2)-2} = 2^{-3/2}=\frac{1}{2^{3/2}}.$$


问:
Representation of a complex number in the Argand Plane
Consider the complex number $\omega = \frac{1+i}{\sqrt{2}} $. Find the value of $(\sum_{k=0}^n z^{k^2}) (\sum_{j=0}^n z^{-j^2})$

I started by collecting the terms with similar but opposite sign indices and writing the sum in polar form, but I am unable to proceed with the other terms

答:
Denote $g_n=\displaystyle\sum_{k=0}^n \omega^{k^2}$ where $\omega=e^{\pi i/4}=\frac{1}{\sqrt{2}}(1+i)$ is a primitive $8$th root of unity.

Note that $\omega^{-1}=\overline{\omega}$ so you're interested in the value of $g_n\overline{g_n}=|g_n|^2$.

Observe that if $n=8q+r$ with $r<8$, then $g_n=qg_8+g_r$ because the squares $k^2$ in the exponent of $\omega$ witihin the $g_n$ sum are periodic modulo $8$.

Therefore, it suffices to compute $g_n$ for $n=0,1,\cdots,7$ and $n=8$.


Title: how can $\sqrt{2}/4=1/2^{3/2}$

Body: $\sqrt{2}/4=1/2^{3/2}$ this is a subpart of a work example question I have but I don't understand how I can convert the first part into the second.

Body:
$$\frac{\sqrt2}{2^2} = \frac{2^\frac{1}{2}}{2^2} =2^{-\frac{3}{2}} =
\frac{1}{2^\frac{3}{2}}$$


问:
A problem on number of injections from a set to the same set
Let a set be defined as $A = \{1,2,3,4,5\}$ and let $f: A \to A$ such that for all $i$ in $A$, at least one $f(i) = i$. How many such injective functions exist?

My approach:

As at least one $f(i) = i$, total number of injective functions is the same as sum of all injective functions where number of fixed elements lies between $1$ and $5$.

If one element fixed for which $f(i) = i$ and no others:

Let $(1,1)$ be fixed, number of functions is $(3 \times 2  \times 1)$.

There are $5$ such functions, where one of the five elements is fixed, total is $6 \times 5 = 30$.

I am unable to apply this logic where two or more pairs are fixed, for example $(1,1)$ and $(2,2)$ and no others such that $f(i) = i$. How can this be solved?

答:
Let $A_i$ be the event that $f(i) = i$

We want to find $\bigcup\limits_{i=1}^{5}A_i$

By Inclusion-Exclusion and symmetry of events, this equals
$${5 \choose 1}A_1  - {5 \choose 2}(A_1\cap A_2) + {5\choose3}(A_1 \cap A_2 \cap A_3) - {5 \choose 4}(A_1 \cap A_2 \cap A_3 \cap A_4) + 1$$

Counting the events is straightforward since now we don't need to consider the cases in which f(1) = 1, f(2) = 2 and no other f(i) = i


问:
Classify all groups of order 182
In studying for an upcoming prelim, I came across this problem:

Classify all groups of order $182 = 2*7*13$.

Now, the standard tricks here are to look at Sylow's theorems or semi-direct products. Let $n_p$ denote the number of Sylow $p$-subgroups. We can conclude that $n_7 =1$ since $n_7 |26$ and $n_7 \equiv 1 (\mod 7)$. Now, the standard element counting tricks seem to fail me here. Frequently, supposing that one of the other Sylow subgroups is not normal will lead to some sort of contradiction, but I think even assuming $n_{13} = 14$ (which is the next smallest thing it can be), there is just barely enough room in the group for $n_2$ to be $7$. Any ideas on where to proceed from here?

In a related question, does anyone have on hand a good resource for qualifying/prelim exam questions of the flavor `` Classify all groups of order...'' where the order given is in some regards manageable? I've been working through the examples on Stack exchange, but this is such a standard type of problem I'm inclined to think there might be a good resource for these types of questions.

答:
Here is my comment with some added detail:

Let $G$ be a group of order $182$. You already concluded that you have a normal $7$-Sylow subgroup, so let us call this subgroup $P$ and look at $G/P$.

$G/P$ has order $26$ so the $13$-Sylow subgroup has index $2$. This gives us a subgroup of $G$ of index $2$, which is normal due to the index. We will call this subgroup  $N$. So now we have that $G$ is a semidirect product of $N$ and a $2$-Sylow subgroup which we will call $Q$.

Let us consider $N$ a bit more carefully. It is a group of order $91 = 7\cdot 13$ and we easily see from the Sylow theorems that both its Sylow subgroups are normal, and hence $N$ is cyclic.

So we now know that our group is a semidirect product of a cyclic group of order $91$ with one of order $2$, so we need to look for subgroups of $\rm{Aut}(C_{91})$ of order $1$ or $2$ (the one of order $1$ just gives us a direct product, which will give the cyclic group of order $182$ which is the only abelian possibility).

Now, $\rm{Aut}(C_{91})$ is isomorphic to $C_6\times C_{12}$ so it has $3$ distinct subgroups of order $2$ and we need to determine which of these yield distinct semidirect products.

To do this, it is probably a good idea to see precisely how these automorphisms of order $2$ act on $C_{91} \simeq C_7\times C_{13}$. The three possible automorphisms are then $(g,h)\mapsto (g^{-1},h)$, $(g,h)\mapsto (g,h^{-1})$ and $(g,h)\mapsto (g^{-1},h^{-1})$.

To see that these give three distinct semidirect products, we can count the number of elements of order $2$ in each.

An arbitrary element can be written as $(g,h)\varphi$ where $\varphi$ is either the identity or the given automorphism of order $2$. Clearly if $\varphi$ is the identity, the element cannot have order $2$, so let us now compute the square of such an element in each of the three cases (remember that $\varphi$ is its own inverse and that conjugation by $\varphi$ corresponds to applying $\varphi$).

In the first case, we get $(g,h)\varphi(g,h)\varphi = (g,h)(g^{-1},h) = (1,h^2)$ which is the identity if and only if $h^2 = 1$ which means if and only if $h = 1$, so we get precisely $7$ elements of order $2$ in this group.

Similarly, in the second case, we get that the element has order $2$ if and only if $g = 1$ so we get $13$ element of order $2$ in this case.

In the final case, we see that all elements of the given form have order $2$, so we have $91$ such elements (this case gives us the dihedral group).

Title: Classify all groups of order 182


问:
Reference for Wold-Decomposition Theorem (Operator Theory)
I'm looking for full proves for the following theorems:

(Wold-decomposition theorem 1954.) Every isometry is a direct sum of unitary and unilateral shifts.

Every isometry is unitary equivalent to a direct sum of unitary and unilateral shifts.

Here is a sketched proof from Wikipedia for (1).
https://en.wikipedia.org/wiki/Wold%27s_decomposition#A_sequence_of_isometries

Can I get some suggestions or references for the full proofs?

$\bf{EDIT}$: From the comments, (2) implies (1), because of the orthogonal decomposition of the space.. I still don't really see why. Could someone help me on this part?

Thank you!

答:
Suppose $T : H\rightarrow H$ is an isometric linear map on a Hilbert space $H$, meaning that $\|Tx\|=\|x\|$ for all $x\in H$. Then $T$ is bounded, with an adjoint $T^*$ such that $T^*T=I$ because the polarization identity gives
\begin{align}
         \langle T^*Tx,y\rangle &= \langle Tx,Ty\rangle \\
       &=\frac{1}{4}\sum_{n=0}^{3}i^n\langle T(x+i^ny),T(x+i^ny)\rangle= \\
       &= \frac{1}{4}\sum_{n=0}^{3}i^n\langle x+i^ny,x+i^ny\rangle
       =\langle x,y\rangle, \;\;\; x,y\in H.
\end{align}
$TH=H$ iff $T$ is unitary. If $TH\ne H$, there exists a unit vector $x\in H$ such that $x\perp TH$. Then $\{ x,Tx,T^2x,\cdots \}$ is an orthonormal subset of $H$ on which $T$ is a simple shift operator. If $x\perp y$ and $x,y\perp TH$, then $[\{ x,Tx,T^2x,\cdots\}]$ and $[\{y,Ty,T^2y,\cdots\}]$ are mutually orthogonal sets on which $T$ is a unilateral shift. There are as many copies of such shifts as there are in an orthonormal basis of $(TH)^{\perp}$. What's left over is $Y=\bigcap_{n=1}^{\infty}T^n H$. This is a closed invariant subspace of $T$, and $T$ is unitary on this subspace because $TY=Y$, though $Y=\{0\}$ may occur. Together this gives a decomposition of the space into a subspace where $T$ is unitary, and a collection of mutually orthogonal invariant subspaces of $T$ on which $T$ is a unilateral shift.


问:
Ratio of segments in equally divided triangle

In a $\Delta PQR$, $M$ and $N$ are points on $PQ$ and $PR$, respectively such that $MN || QR$ and $MN$ divides $\Delta PQR$ into two parts equal in area, then find the ratio of $PM:PQ$.

Let '$[...]$' denote the area of a polygon. Then $[PNM]=[NMQR]=x$.

Joining $RM$, we see that $$\frac{PM}{PQ-PM}=\frac{[PNM]+[MNR]}{[NMQR]-[MNR]}=\frac{x+[MNR]}{x-[MNR]}$$

$$\frac{PM}{PQ-PM}=\frac{PN}{PR-PN}=\frac{[PNM]}{[MNR]}=\frac{x}{[MNR]}$$

Let $\frac{PM}{PQ-PM}=y$, then $$y=\frac{x+\frac1y}{x-\frac1y}$$

I am not able to proceed further from here except for simplifying the above equation. Help?

答:
We can use similar triangles. Let $\triangle PMN \sim \triangle PQR$ with similarity ratio $r$. Then we know
$$\frac{[PMN]}{[PQR]} = r^2 = \frac{1}{2} $$
$$\Rightarrow r=\frac{PM}{PQ}=\ldots$$


问:
Finiteness of the volume $Z(\mathbb A) G(\mathbb Q) \backslash G(\mathbb A)$ for $G = \operatorname{GL}_2$
I am reading Gelbart's lectures on the trace formula and am confused on how the Siegel domain is used to prove the finiteness of the volume of $Z(\mathbb A) G(\mathbb Q) \backslash G(\mathbb A)$ for $G = \operatorname{GL}_2$.  Here compact sets $C_1 \subset N(\mathbb A)$ and $C_2 \subset (\mathbb A^{\ast})^1$ are given, together with a real number $c$, and the Siegel set $\mathfrak S_c$ is defined to be the set of
$$g = znh_tmk \in G(\mathbb A)$$
with $z \in Z(\mathbb A), n \in N(\mathbb A), h_t = \begin{pmatrix} e^t \\ & e^{-t} \end{pmatrix}$, $m = \begin{pmatrix} a \\ & 1 \end{pmatrix}$ with $a \in \mathbb A^{\ast}$, and $k \in K$, such that

$$n \in C_1, a \in C_2, t > c/2.$$

From the Iwasawa decompostion, we can say that
$$\int\limits_{Z(\mathbb A)\backslash G(\mathbb A)} f(g)dg = \int\limits_K \int\limits_{N(\mathbb A)} \int\limits_{(\mathbb A^{\ast})^1} \int\limits_{\mathbb R} f(nh_tmk) dt dm dn dk.$$

Here is what Gelbart says about how we can prove the finiteness of the volume of $Z(\mathbb A) G(\mathbb Q) \backslash G(\mathbb A)$:

I don't understand the computation

$$\int\limits_{Z(\mathbb A) G(\mathbb Q) \backslash G(\mathbb A)} dg \leq \int\limits_{Z(\mathbb A) \backslash \mathfrak S_c} dg.$$

What is going on here?  Are we using the fact that

$$\int\limits_{Z(\mathbb A) \backslash G(\mathbb A)} dg = \int\limits_{\big(Z(\mathbb A) \backslash Z(\mathbb A)G(F)\big)\backslash \big(Z(\mathbb A) \backslash G(\mathbb A)\big)} \sum\limits_{\gamma \in Z(\mathbb A) \backslash Z(\mathbb A)G(F)} dg $$
with $\big(Z(\mathbb A) \backslash Z(\mathbb A)G(F)\big)\backslash \big(Z(\mathbb A) \backslash G(\mathbb A)\big) = Z(\mathbb A)G(F) \backslash G(\mathbb A)$?

答:
A special case of this was addressed in my later question here.

The main point is that if $G$ is a unimodular second countable, locally compact Hausdorff group, and $H$ a discrete subgroup of $G$, then for any Borel subset $B$ of $H \backslash G$, and any Borel set $A$ of $G$ whose image in $H \backslash G$ is $B$, there exists a Borel subset $X \subseteq A$ which projects bijectively onto $B$.  It follows that $\operatorname{meas}(X) = \operatorname{meas}(B)$ from the equation

$$\int\limits_G f(g) dg = \int\limits_{H \backslash G} \Bigg( \sum\limits_{h \in H} f(hg) \Bigg) dg$$

which holds for any positive measurable function $f$ (taking $f$ to be the characteristic function of $X$, the map $Hg \mapsto \sum\limits_{h \in H} f(hg)$ will be the characteristic function of $B$).

Now, let $G$ be a connected, reductive group over a number field $k$, and $\mathfrak S_c$ be a Siegel set for $G$, so that $G(\mathbb A) = G(k) \mathfrak S_c$.

If we consider the image $\overline{\mathfrak S_c}$ of $\mathfrak S_c$ in $Z(\mathbb A) \backslash G(\mathbb A)$, then this is easily seen to remain a Borel set, and it surjects to the quotient $(\mathbb Z(A) \backslash Z(\mathbb A)G(k) ) \backslash (Z(\mathbb A) \backslash G(\mathbb A)) = Z(\mathbb A)G(k) \backslash G(\mathbb A)$.

Letting $X$ be a Borel subset of $\overline{\mathfrak S_c}$ which maps bijectively onto $Z(\mathbb A)G(k) \backslash G(\mathbb A)$, we get

$$\operatorname{meas}(Z(\mathbb A)G(k) \backslash G(\mathbb A)) = \operatorname{meas}(X) \leq \operatorname{meas}(\overline{\mathfrak S_c}) < \infty.$$


问:
How to give a rigorous proof to the divergence of $\sin nx$
It's easy to know that sequence $\{\sin (nx)\} \ (x\neq k \pi)$ does not converge , but when I write the proof, I find it hard to give a rigorous one. Could someone please help me out here? A hint may help as well. Thanks in advance.

答:
Claim: If $\sin (nx)$ converges as $n\to\infty$, then $\sin x=0$.

Proof: Use the sum and difference formulas $\sin(A+B)=\sin A\cos B + \cos A\sin B$ and $\sin(A-B)=\sin A\cos B-\cos A\sin B$ to write
$$
\sin ((n+1)x) = \sin (nx)\cos x + \cos (nx)\sin x\tag1
$$
and
$$
\sin( (n-1)x) = \sin (nx)\cos x - \cos (nx)\sin x.\tag2
$$
If $\sin (nx)$ converges to a limit $L$ as $n\to\infty$, then $\sin((n+1)x)$ and $\sin((n-1)x)$ converge to the same $L$. Add formulas (1) and (2), let $n\to\infty$, and deduce that
$$2L=2L\cos x.$$

It follows that $\cos x=1$, or $L=0$. The case $\cos x=1$ immediately leads to $\sin x=0$.

So suppose $L=0$. Subtract (2) from (1), square both sides of the result, and take the limit as $n\to\infty$. Conclude that $\cos^2(nx)\cdot\sin^2 x$ converges to $0$ as $n\to\infty$.
Since $\cos^2(nx)=1-\sin^2(nx)$ tends to $1$, conclude  $\sin^2 x=0$ and therefore $\sin x=0$.


问:
Finding zeros of a function
I need to find when $f(x)=0$, where:
\begin{equation}
f(x)=1-\dfrac{k}{x}-\dfrac{k}{3}\dfrac{e^{-ax}}{x}+\dfrac{4k}{3}\dfrac{e^{-bx}}{x}
\end{equation}
Here, $k$, $a$ and $b$ are positive constants. I think this problem analytically is impossible, but I really don't know. I've tried to use Mathematica but it doesn't give any answer and I'm pretty desperate.

Thank you so much.

答:
Another point.

We have
$$g(x) >x-\frac{4 }{3}k+\frac{4}{3} k e^{-b x}$$ which means that the solution $x$ we are looking for has, as an upper bound,
$$x_{\text{max}} = \frac{4}{3} k+\frac 1 b W\left(-\frac{4}{3} b k e^{-\frac{4 b k}{3}}\right)$$
For the worked example in answers $(a=1,b=2,k=1)$, the iterations would be
$$\left(
\begin{array}{cccc}
 n & \text{Newton} & \text{Halley} & \text{Householder} \\
 0 & 1.21624210 & 1.21624210 & 1.21624210 \\
 1 & 0.94495684 & 0.92825892 & 0.92322342 \\
 2 & 0.92178566 & 0.92152597 & 0.92152579 \\
 3 & 0.92152582 & 0.92152579 &  \\
 4 & 0.92152579 & &
\end{array}
\right)$$

For the worked example in comments $(a=2,b=\frac 32,k=3)$, the iterations would be
$$\left(
\begin{array}{cccc}
 n & \text{Newton} & \text{Halley} & \text{Householder} \\
 0 & 3.9899342 & 3.9899342 & 3.9899342 \\
 1 & 2.9756564 & 2.9644279 & 2.9585726 \\
 2 & 2.9552065 & 2.9551855 & 2.9551854 \\
 3 & 2.9551854 & 2.9551854 &
\end{array}
\right)$$

From a purely numerical point of view, this could be a quite efficient solution using even much higher order methods (the successive derivatives of $g(x)$ being so unexpensive to compute.

For example, using a method of order $6$ gives as a first iterate $x=2.9552614$.

Edit

What we can also do is to use one iteration of Newton, Halley or Householder (even higher order method) around $x=k$. This does not make any problem with a CAS.

For the worked cases, this will give with Newton
$$x=\frac{128-36 e-117 e^2+18 e^3+18 e^4}{64-12 e-95 e^2+12 e^3+18 e^4}\sim 0.921758$$
$$x=\frac{16-97 e^{3/2}+156 e^3+26 e^6-80 e^{15/2}+6 e^{12}}{4-23 e^{3/2}+36 e^3+8
   e^6-24 e^{15/2}+2 e^{12}}\sim 2.955188$$

and this will be better and better when $\Delta=(4 e^{-b k}-e^{-a k})$ will close to $0$. For the first case $\Delta=\frac{4-e}{e^2}\sim 0.173462$; for the second $\Delta=\frac{4 e^{3/2}-1}{e^6}\sim 0.041957$. This seems to explain the results.

Title: Finding zeros of a function


问:
Great Circles and Inscribed Cube
A cube is inscribed within a sphere. How many distinct great circles are there that contain at least 2 vertices of the cube along its perimeter?

Intuition tells me any great circle that coincides with two or more vertices of the cube must intersect with 2 vertices that are diagonal from each other. Hence, there are 4 pairs of such diagonals.

However, this was incorrect. I'm not sure where I'm going wrong and any help would greatly be appreciated.

答:
There are infinitely many such great circles.  This is because:

Any great circle on the sphere must share its center with the center of the sphere.

Any two vertices of the cube that are diametrically opposite along a long diagonal of the cube will be collinear with the center; i.e., the long diagonal is a diameter of the sphere.

An infinite number of great circles can be drawn through these opposing vertices.

Title: Great Circles and Inscribed Cube


问:
Transformation of the MGF
Let X be a continuous random variable with the probabilty density function given by

$f_X(x) = \frac{24}{x^4} , x>2$.

Let $Y = \frac{1}{X^3}$. Compute the mgf and pdf of Y.

How do I find the mgf of Y? I know that I can find the pdf of Y using the formula $$f_Y(y)=f_X(g^{-1}(y)) \vert\frac{\partial}{\partial y}g^{-1}(y)\vert.$$

Update: I got the pdf of Y as Y~Uniform$[0, \frac{1}{8}]$

答: Once you have the density of $Y$, simply compute $$M_Y(t) = \operatorname{E}[e^{tY}] = \int_{y=0}^{1/8} 8e^{ty} \, dy.$$

Title: Transformation of the MGF


答:
Correct, more or less! $Y\sim U\left(0;\frac{1}{8}\right)$

(zero and 8 have to be excluded from the support)

Now the MGF is known but if you want to calculate it just use the definition

$$MGF_Y=\mathbb{E}\left(e^{tY}\right)$$

Title: Transformation of the MGF

Body:
Let X be a continuous random variable with the probabilty density function given by

$f_X(x) = \frac{24}{x^4} , x>2$.

Let $Y = \frac{1}{X^3}$. Compute the mgf and pdf of Y.

How do I find the mgf of Y? I know that I can find the pdf of Y using the formula $$f_Y(y)=f_X(g^{-1}(y)) \vert\frac{\partial}{\partial y}g^{-1}(y)\vert.$$

Update: I got the pdf of Y as Y~Uniform$[0, \frac{1}{8}]$


问:
Averaging over Mahalanobis distance vectors of different clusters
Given vectors from two different clusters (in particular in my example from two different experimental conditions, called "CS" and "US") where the Mahalanobis Distance is calculated according to:

$M^2 = (x-\mu)^T \Sigma^{-1}(x-\mu)$

the explanation reads as follows: "where x and $\mu$ are individual and mean population vectors for CS and US ensemble responses [that is n-dim vectors; my addition], respectively, and $x^T$ and $\mu^T$ are their transposes. We averaged the M over all points x in both sets". The source is here in the methods, under "Population vector analysis"

And the illustration looks like this:

My questions:

As I understand the take one cluster as the reference in turn and then average over vector-by-vector calculations of the mahalanobis distance? Is this a common and correct approach?

Does this lead to a distance measure which is no longer symmetric?

答:
Preamble

Whenever all the variables under consideration are expressed in the same unit of measurement and therefore they appear to be directly comparable, the Euclidean scalar product or the associated Euclidean distance is usually used,

$ d^2(\mu,x)\equiv \langle \mu - x,\mu - x \rangle_{eucl}:= \sum_{i=1}^n (\mu_i - x_i)^2 $

Often, the collected data do not have homogeneous units of measurement, for example in the pair of weight-height variables, the unit of measurement of the first variable is mass, while the second variable expresses the measurement of a length. Weight and height cannot be added together, so it is necessary to express the data in dimensionless numbers. If the variables refer to different units of measurement, the Weighted Euclidean Distance is used in order to transforms the observable variables into dimensionless and therefore comparable numbers.The transformation is operated by a suitable multiplier or weight, $\lambda_i$, which makes the observed difference $(\mu_i - x_i)$ of a character directly comparable with the observed difference in one of the other characters $(\mu_j - x_j)$ where $i\neq j$ and $i,j=1,\cdots, n$.

If the measurements of the single variables have non-univocal values, i.e. in probabilistic terms they present non-zero variance, it is common to choose the reciprocal of the standard deviation as the multiplier: $\lambda = 1 / \sigma $ (Mahalanobis transformation), where $\sigma $ designates the standard deviation.
The Mahalanobis transformation that operates on vectors by multiplying them by the reciprocal of the corresponding standard deviation $ x \mapsto f (x):= \sigma^{-1}x$ is an isometry with respect to the usual Euclidean scalar product:

$\langle\mu - x,\mu - x\rangle_{mahal} = \langle f(\mu - x),f(\mu - x)\rangle_{eucl}$

If the n variables are statistically independent, therefore uncorrelated, the square of the Mahalanobis distance is written as:

$d^2(\mu - x,\mu - x)_{mahal}:= \sum_{i=1}^n \tfrac{(\mu_i - x_i)^2} {\sigma^2_{ii}} $

where $\sigma^2_{ii}$ is the variance referred to the i-th variable.

Otherwise, if the n variables are correlated, the square of the Mahalanobis distance is written as

$d^2(\mu - x,\mu - x)_{mahal}:= \sum_{i=1}^n \sum_{j=1}^n (\mu_i - x_i) \cdot \sigma^{-2}_{ij} \cdot (\mu_j - x_j) $,

where $\sigma^{2}_{ij} $ is the element of the inverse matrix of the variance-covariance matrix $\Sigma$.

In short

$M=(\mu - x)^t \Sigma^{-1} (\mu - x) $

Question 1

Once a distance $d$ able to express the notion of proximity for the observations has been adopted, it is possible to introduce a classification rule for the observations with respect to a representative of each class. The statistical character of the discriminant analysis derives from the fact that the average or the sample mean of the populations under examination is chosen to be the representative of each class.

In the example of the two classes (sets "US" and "CS"), indicated with $ \mu_1$ the average of the first set (population) and with $ \mu_2$ the average of the second population, the observation $x$ belongs to class 1 if and only if:

$d(\mu_1,x) < d(\mu_2,x)$

that is if $x$ is closer to $\mu_1$.

Question 2

Just consider $d^2(\mu,x)=\|\mu-x\|^2=\langle \mu-x , \mu-x \rangle$ to express the rule of belonging to set 1 in terms of the dot product as
$\langle \mu_1-x, \mu_1-x\rangle < \langle \mu_2-x, \mu_2-x\rangle$

The relationship just written, exploiting the bilinearity property of the scalar product, can be rewritten after simple steps such as

$\langle \mu_1-\mu_2, x-1/2(\mu_1+\mu_2)\rangle > 0$.

Indicated with $x_0=1/2(\mu_1+\mu_2)$ the classification rule for set 1 elements becomes:

$Set_1: \langle \mu_1-\mu_2, x-x_0 \rangle > 0 $


问:
Radius of Convergence of $(1-x)^{1/4}$
This is my first post so I am very new to MathJax formatting - i apologize in advance for the messy formatting

This equation becomes $$\sum_{n=2}^\infty \frac{(4n-5) x^n}{(4^n)n!}$$

The textbook says the ratio test would be:

$$\lim_{n\to \infty} \left\lvert\frac{(4n-1)(4n-5)(x^{n+1})(4^n)n!}{(4n-5)(x^n)(4^{n+1})(n+1)n!}\right\rvert$$

But I thought it would be:
$$\lim_{n\to \infty} \left\lvert\frac{(4n-1)(x^{n+1})(4^n)n!}{(4n-5)(x^n)(4^{n+1})(n+1)!}\right\rvert$$

So no extra $(4n-5)$ - but that would result in an addition $(n+1)$ in denominator (because of the expanded factorial) and the radius of convergence would now be infinite (and not $|x| < 1$ as above fraction suggests).

Thanks

答:
The generalized binomial series is

$$ (1 \ + \ x)^p \ \ = \ \ \sum_{n=0}^\infty \   \left( \begin{array}{c} p \\ n \end{array}   \right) \ x^n  \ \ , $$

which becomes $$ (1 \ - \ x)^p \ \ = \ \ \sum_{n=0}^\infty \  (-1)^n \ \left( \begin{array}{c} p \\ n \end{array}   \right) \ x^n  \ \  $$

for the base $ \ (1 \ - \ x) \ $ .  Writing out the first few terms of the series with the fractional binomial coefficients for $ \ p \ = \ \frac14 \ $ gives us

$$ (1 \ - \ x)^{1/4} \ \ = \ \  1 \ ·  \left( \begin{array}{c} \frac14 \\ 0 \end{array}   \right) \ x^0 \ \ + \ \  -1 \ ·  \left( \begin{array}{c} \frac14 \\ 1 \end{array}   \right) \ x^1 \ \ + \ \  1 \ ·  \left( \begin{array}{c} \frac14 \\ 2 \end{array}   \right) \ x^2 \ \ + \ \  -1 \ ·  \left( \begin{array}{c} \frac14 \\ 3 \end{array}   \right) \ x^3 \ \ + \  \ldots        $$

$$ = \ \ 1 \ · \frac{  1}{0!} \ x^0 \ \ - \ \ 1 \ · \frac{  \frac14   }{1!} \ x^1 \ \ + \ \ 1 \ · \frac{ \frac14 · \left(  \frac{-3}{4}   \right)}{2!} \ x^2 \ \ - \ \ 1 \ ·\frac{   \frac14 · \left(  \frac{-3}{4}   \right) ·  \left(  \frac{-7}{4}   \right)  }{3!} \ x^3 \ \ + \  \ldots        $$

$$ = \ \ 1  \ \ - \ \  \frac{  1   }{4 \ · \ 1!} \ x^1 \ \ + \ \  \frac{ 1 · (  -3)}{4^2 \ · \ 2!} \ x^2 \ \ - \ \ \frac{ 1 · (-3) · (-7) }{4^3 \ · \ 3!} \ x^3 \ \ + \  \ldots        $$

There are various ways that people choose for writing the general term; my guess is that this author(s) for your textbook decided to leave the first two terms outside of the general summation to start it at $ \ n \ = \ 2 \ $ (this has no effect on the interval of convergence):

$$ (1 \ - \ x)^{1/4} \ \ = \ \ 1  \ \ - \ \  \frac14 \ x \ \ - \ \
\sum_{n=2}^\infty \ \frac{3 \ · \ 7 \ · \ · \ldots \ · \ (4n-5) }{4^n \ · \ n!} \ x^n  \ \ .  $$

This is a product of factors in arithmetic progression in the numerator of the general term; that is what must be used in the Ratio Test.  Almost all of them will cancel in the calculation, but what will remain is

$$ \lim_{n\to \infty} \  \left\lvert \ \frac{(4[n+1]-5)(4n-5)(x^{n+1})}{4^{n+1}  \ · \ (n+1)!} \ · \ \frac{4^n \ · \ n!}{(4n-5)(x^n) } \ \right\rvert \ \ = \ \ \lim_{n\to \infty} \  \left\lvert \ \frac{(4n-1)(x)}{4  \ · \ (n+1)}  \ \right\rvert \ \ , $$

which will yield the radius of convergence $ \ R \ = \ 1 \ $ that obtains for the series representing $ \ (1 \ + \ x)^p \ $ .  [The generalized binomial series is exactly the same as you would get by deriving a Maclaurin series for these functions.]

Title: Radius of Convergence of $(1-x)^{1/4}$


问:
In a Hilbert Space: Minimal $\Leftrightarrow$ Biorthogonal
Perhaps obvious, but I'm struggling a bit with (i) $\Rightarrow$ (ii) only.

Proposition. Let $\{x_n\}_{n \in \mathbb{N}}$ be a sequence in a Hilbert space $H$. Then the following are equivalent:

i) Let $S_n := \overline{\text{span}\{x_n : n \neq m\}}$. For each $m \in \mathbb{N}$, $x_m \notin S_n$.

ii) Exists $\{y_n\}_{n \in \mathbb{N}}$ in $H$ that is biorthogonal to $x_m$, i.e. the inner product is $1$ when $n = m$ and $0$ otherwise.

My idea: I suppose if we let $S_n^{\perp}$ be the orthogonal complement of $S_n$, then for any $y_n \in S_n^{\perp}$, we have $\langle x_m, y_n \rangle = 0$. But when $n = m$, how can I be sure that $\langle x_m, y_n \rangle = 1$?

答:
i) implies ii): There exists a continuous linear functional which has the value $1$ at $x_n$ and the value $0$ on $span \{x_i: i \neq n\}$. Continuous linear functionals are given by inner product with some element so we get $y_n$ such that $\langle  y_n, x_m \rangle=0$ for $ m \neq n$ and $1$ for $m=n$.

ii) implies i) is obvious: If $x_n $ is in the closure of the span of $\{x_i: i \neq n\}$ then the fact that $y_n$ is orthogonal to eaxh $x_i, i \neq n$ implies that $y_n$ is orthogonal to $x_n$, a contradiction.


问:
What class of topological spaces is this?
It is well-known but apparently I have forgotten the answer to the below question. Which topological spaces $V$ have this property:

For every open subset $U\subset V$ there exists a non-zero continuous function $f\colon V\to \mathbb{R}$ such that $\overline{\{x\in V\colon f(x)\neq 0\}}\subset U$?

Is it normality by chance?

答:
Every completely regular space has your property, whether or not it is $T_1$. For each $x\in X$ and open nbhd $U$ of $x$, complete regularity gives you a continuous $f_{x,U}:X\to[0,1]$ such that $f_{x,U}(x)=1$ and $f_{x,U}(y)=0$ for all $y\in X\setminus U$. Now let $U$ be any non-empty open set in the completely regular space $X$. Fix $x\in U$ arbitrarily, and let $$V=\left\{y\in U:f_{x,U}(y)>\frac12\right\}\;;$$ then $V$ is an open nbhd of $x$, and $f_{x,V}$ is a continuous real-valued function on $X$ whose support is contained in $U$.

However, there are also spaces with your property that are not completely regular. Let $X$ be the space constructed by John Thomas in ‘A Regular Space, Not Completely Regular’, The American Mathematical Monthly, Vol. $76$, No. $2$ (Feb., $1969$), pp. $181$-$182$, and described in detail in this answer. $X$ has a dense set of isolated points, so if $U$ is a non-empty open set in $X$, $x\in U$ be an isolated point, and define

$$f:X\to\Bbb R:y\mapsto\begin{cases}
1,&\text{if }y=x\\
0,&\text{otherwise}\;;
\end{cases}$$

clearly $f$ is continuous and has support $\{x\}\subseteq U$.


问:
Generically global generated vector bundles of rank $2$ on surfaces
Let $X$ be a projective surface over $\mathbb{C}$.

Let $E$ be a rank $2$ vector bundle whose general global section vanishes on finite points. (i.e. Codim $Z(s) = 2$ for general $s \in H^0(X, E)$)

In addition, we assume that $\text{dim}_\mathbb{C} H^0(X,E) \geq 2$.

Question

$E$ is generically global generated ?

(We say $E$ is generically global generated if the evaluation map $ ev: H^0(E) \otimes \mathcal{O}_X \rightarrow E$ is surjective on an open set $U \subseteq X $)

答:
A general section of $E$, say $s_1$, gives an exact sequence
$$
0 \to \mathcal{O}_X \to E \to \det(E) \to \mathcal{O}_Z \to 0,
$$
where $Z = Z(s_1)$. The cohomology exact sequence
$$
0 \to H^0(X,\mathcal{O}_X) \to H^0(X,E) \to H^0(X,\det(E))
$$
implies that the second section of $E$, say $s_2$, projects to a nontrivial global section of the line bundle $\det(E)$. This latter section is nonzero on an open subset $U \subset X$, and then a diagram chase shows that $E$ is generated by $s_1$ and $s_2$ on $U$.


问:
Inequality question involving a cyclic expression and moduli
Consider three real numbers a,b,c such that $a^2+b^2+c^2 =1$. What will be the maximum value of the expression $|a-b| + |b-c| + |c-a|$ ?

I tried two approaches. One I used the fact that $a^2+b^2+c^2 =1 = \frac{1}{2} (|a-b|^2 + |b-c|^2 + |c-a|^2) + ab + bc + ca$. Original plan was to somehow eliminate ab + bc + ca , but unable to do so - got stuck in a series of repeating expressions

Another variant was to try and assert that $a \geq b \geq c$ without loss of generality. Then
$(|a-b| + |b-c| + |c-a|)^2 \leq 4(a-c)^2$ , but I am unable to use the given fact about $a^2 + b^2 + c^2$.

I realize that I am missing something very trivial here as this seems a fairly simple problem - so any help would be greatly appreciated.

答:
Let's assume that $a\ge b\ge c$, then $|a-b|+|b-c|+|c-a|=2(a-c)$

And $(a-c)^2=a^2+c^2-2ac$

Because $a^2+c^2\le a^2+c^2+b^2\le1$ and $-2ac\le a^2+c^2\le a^2+c^2+b^2\le1$

We got $(a-c)^2=a^2+c^2-2ac\le 2$

So $(a-c)\le \sqrt 2$

So the answer is $2\sqrt 2$

Equality holds at $(a,b,c)=(\sqrt \frac12,-\sqrt \frac12,0)$


问:
If unit quaternions $q_1,q_2$ satisfy $q_1^2=q_2^2=-1$, then the map $S^3\to S^3$, $x\mapsto q_1xq_2^{-1}$ has a fixed point
Let $\Bbb H$ denote the quaternion algebra. We can identify $S^3$ as the subset of unit quaternions.
For fixed $q_1,q_2\in S^3$, consider the map $\phi:S^3\to S^3$ defined by $\phi(x)=q_1xq_2^{-1}$. How can we show that if $q_1^2=q_2^2=-1$ (so that $\phi^2=\text{id}$) then $\phi$ has a fixed point?

(This is claimed in a lecture note that I was reading, contained in the sketch of proof of the statement that if $G$ is a subgroup of $SO(4)$ of order $2$ acting freely on $S^3$, then $G=\{\pm \text{id}\}$.)

答:
In fact, we can characterize the set of fixed points geometrically.

The quaternion square roots of $-1$ are the unit vectors, I'll call them $\mathbf{u}$ and $\mathbf{v}$. The solutions to the equation $\mathbf{u}x\overline{\mathbf{v}}=x$ are the solutions to $x\mathbf{v}x^{-1}=\mathbf{u}$. If $x=\exp(\theta\mathbf{w})$ then this says $\mathbf{v}$ is obtained from $\mathbf{u}$ by rotating it around $\mathbf{w}$ by $2\theta$. Let's consider the case when $\mathbf{u},\mathbf{v}$ are parallel. If $\mathbf{v}=\mathbf{u}$ then $\mathbf{w}=\mathbf{u}$ too and $\theta$ is arbitrary, in other words $x\in\mathbb{R}[\mathbf{u}]=\mathrm{span}\{1,\mathbf{u}\}$. If $\mathbf{v}=-\mathbf{u}$ then $x$ anticommutes with $\mathbf{u}$, which means $x$ has no scalar part and must be a vector perpendicular to $\mathbf{u}$.

In general, if $S^3$ acts on $S^2$ by conjugation, the "transporter" from $\mathbf{v}$ to $\mathbf{u}$ (the set of solutions $x$) is a coset of $\mathbf{v}$'s stabilizer, so is a circle. To parametrize that circle in $S^3$ it suffices to find orthogonal elements of it.

If $\mathbf{u},\mathbf{v}$ are not parallel we can pick two very nice rotation axes: the one $\mathbf{w}_1$ which is between them, and $\mathbf{w}_2$ which is perpendicular to them. That is, $\mathbf{w}_1$ is $\mathbf{u}+\mathbf{v}$ and $\mathbf{w}_2$ is $\mathbf{u}\times\mathbf{v}$, but normalized. The former axis requires an angle of $\theta=\pi$ and the latter requires the angle $\theta=\angle\mathbf{uv}$. Then

$$ x_1=\frac{\mathbf{u}+\mathbf{v}}{\|\mathbf{u}+\mathbf{v}\|}, \quad x_2=\exp\left(\frac{\theta}{2}\frac{\mathbf{u}\times\mathbf{v}}{\|\mathbf{u}\times\mathbf{v}\|}\right) $$

can be used to parametrize all $x=(\cos\phi)x_1+(\sin\phi)x_2$. Note $x_2$ can be written purely in terms of $\mathbf{u},\mathbf{v}$ using Euler's formula, half-angle formulas, the dot product $\cos\theta=\mathbf{u}\cdot\mathbf{v}$, and Pythagorean identity.


问:
Show that $x^{2}\frac{\partial^{2}u}{\partial x^{2}}+2xy\frac{\partial^{2}u}{\partial x\partial y}+y^{2}\frac{\partial^{2}u}{\partial y^{2}}=0$
I want to prove that

$x^{2}\frac{\partial^{2}u}{\partial x^{2}}+2xy\frac{\partial^{2}u}{\partial x\partial y}+y^{2}\frac{\partial^{2}u}{\partial y^{2}}=0$

where $u(x,y)=\frac{xy}{x+y}$. I know that it could be done by calculating each partial derivative but I realized that

$x^{2}\frac{\partial^{2}u}{\partial x^{2}}+2xy\frac{\partial^{2}u}{\partial x\partial y}+y^{2}\frac{\partial^{2}u}{\partial y^{2}}=[(x,y)\cdot \nabla]^{2}u$.
So my question is if there exists some result that I can use to prove in an easy way that the equality holds.

答:
I get
$$x^{2}\frac{\partial^{2}u}{\partial x^{2}}+2xy\frac{\partial^{2}u}{\partial x\partial y}+y^{2}\frac{\partial^{2}u}{\partial y^{2}}=[(x,y)\cdot \nabla]^{2}u-[(x,y)\cdot \nabla]u,$$

$$[(x,y)\cdot \nabla]\left(\frac{xy}{x+y}\right)=\frac{xy}{x+y}$$
Then
$$x^{2}\frac{\partial^{2}u}{\partial x^{2}}+2xy\frac{\partial^{2}u}{\partial x\partial y}+y^{2}\frac{\partial^{2}u}{\partial y^{2}}=\frac{xy}{x+y}-\frac{xy}{x+y}=0$$


问:
Help solving this PDE
I'm starting to study PDEs and i'm having problem solving this one:

$$x^2u_x-xyu_y=2x^3+x^2y+x^2+\frac{x^3y}{x+1}$$
$$u(t,t+1)=t^2+t, t>0$$

What i've done so far:

Using the method of characteristics, we can say (defining $x(s)=\alpha(s)$ and $y(s)=\beta(s)$: $$\alpha'(s)=\alpha^2(s)$$ $$\beta'(s)=-\alpha(s)\beta(s)$$

from this, we have that: $$\alpha(s)=\frac{1}{c_1-s}$$ $$\beta(s)=e^{-s\frac{1}{c_1+s}+c_2}$$
for us to have $\alpha(0)=t$ and $\beta(0)=t+1$ we need to have $c_1=\frac{1}{t}$ and $c_2=ln(t+1)$, then, we have: $$\alpha(s)=\frac{t}{1-ts}$$ $$\beta(s)=e^{-s\frac{t}{1-ts}}(t+1)$$

then (and here is where i'm stucked):

$$u(\alpha(s),\beta(s))=\int_{0}^{s}\frac{d}{ds}u(\alpha(s),\beta(s))ds + u(\alpha(0),\beta(0))$$

$$=\int_{0}^{s}\left[2\left(\frac{t}{1-ts}\right)^3 + \left(\frac{t}{1-ts}\right)^2e^{-s\frac{t}{1-ts}}(t+1) + \left(\frac{t}{1-ts}\right)^2 + \frac{\left(\frac{t}{1-ts}\right)^3e^{-s\frac{t}{1-ts}}(t+1)}{\frac{t}{1-ts}+1} \right]ds + t^2+t$$

The problem is that $$\int_{0}^{s}\frac{\left(\frac{t}{1-ts}\right)^3e^{-s\frac{t}{1-ts}}(t+1)}{\frac{t}{1-ts}+1}ds$$ doesn't seen to have a solution

答:
From the equation

\begin{align}
x^{2} u_{x} - xy u_{y} = 2x^{3} + x^{2} y + x^{2} + \frac{x^{3} y}{x + 1}
\end{align}

you have the relations

\begin{equation}
\frac{dx}{x^{2}} = \frac{dy}{-x y} = \frac{du}{2x^{3} + x^{2} y + x^{2} + \frac{x^{3} y}{x + 1}}
\end{equation}

From the first equality, we find the characteristic curve

\begin{equation}
xy = C_{1}
\end{equation}

From the first and third ratios, we find

\begin{align}
\frac{dx}{x^{2}} &= \frac{du}{2x^{3} + x^{2} y + x^{2} + \frac{x^{3} y}{x + 1}} \\
&= \frac{du}{2x^{3} + C_{1} x + x^{2} + \frac{C x^{2}}{x + 1}} \\
\implies du &= \left( 2x + \frac{C_{1}}{x} + 1 + \frac{C}{x + 1} \right) dx \\
\implies u &= x^{2} + x + C_{1} \ln(x) + C_{1} \ln(x+1) + C_{2}
\end{align}

Using the functional relationship $C_{2} = f(C_{1})$

\begin{equation}
u = x^{2} + x + x y \ln(x) + x y \ln(x + 1) + f(x y)
\end{equation}

which you can check satisfies the PDE. Using $u(t, t+1) = t^{2} + t$, we get

\begin{align}
u(t, t + 1) &= t^{2} + t + t (t + 1) \ln(t) + t (t + 1) \ln(t + 1) + f(t (t + 1)) \\
&= t^{2} + t \\
\implies f(t (t + 1)) &= - t (t + 1) \ln(t (t + 1))
\end{align}

and so we have $f(x y) = - x y \ln(x y)$. Putting it all together

\begin{equation}
u = x^{2} + x + x y \ln(x) + x y \ln(x + 1) - x y \ln(x y)
\end{equation}

Title: Help solving this PDE


答:
$$x^2u_x-xyu_y=2x^3+x^2y+x^2+\frac{x^3y}{x+1}$$
Charpit-Lagrange systhem of characteristic ODEs :
$$\frac{dx}{x^2}=\frac{dy}{-xy}=\frac{du}{2x^3+x^2y+x^2+\frac{x^3y}{x+1}}$$
A first characteristic equation comes from solving $\frac{dx}{x^2}=\frac{dy}{-xy}$ :
$$xy=c_1$$
A second characteristic equation comes from solving
$\frac{dx}{x^2}=\frac{du}{2x^3+x^2y+x^2+\frac{x^3y}{x+1}}=
\frac{du}{2x^3+x^2\frac{c_1}{x}+x^2+\frac{x^3\frac{c_1}{x}}{x+1}}$

$du=\left(2x+\frac{c_1}{x}+1+\frac{c_1}{x+1}\right)dx$

$u=x^2+x+c_1\ln(x(x+1))+c_2$
$$u-x^2-x-xy\ln(x(x+1))=c_2 $$
The general solution of the PDE on the form $c_2=F(c_1)$ is :
$$\boxed{u(x,y)=x^2+x+xy\ln(x(x+1))+F(xy)} $$
$F$ is an arbitrary function to be determined according to the specified condition $\quad u(t,t+1)=t^2+t, t>0$ .
$$t^2+t+t(t+1)\ln(t(t+1))+F(t(t+1))=t^2+t$$
$$F(t(t+1))=-t(t+1)\ln(t(t+1))$$
Let $X=t(t+1)$ with $t>0\quad\implies\quad X>0$ .
$$F(X)=-X\ln(X)$$
The function $F(X)$ is known now. We put it into the above general solution where $X=xy$ :
$$u=x^2+x+xy\ln(x(x+1))-xy\ln(xy)$$
$$\boxed{u(x,y)=x^2+x+xy\ln(\frac{x+1}{y})} $$

Title: Help solving this PDE

Body:
I'm starting to study PDEs and i'm having problem solving this one:

$$x^2u_x-xyu_y=2x^3+x^2y+x^2+\frac{x^3y}{x+1}$$
$$u(t,t+1)=t^2+t, t>0$$

What i've done so far:

Using the method of characteristics, we can say (defining $x(s)=\alpha(s)$ and $y(s)=\beta(s)$: $$\alpha'(s)=\alpha^2(s)$$ $$\beta'(s)=-\alpha(s)\beta(s)$$

from this, we have that: $$\alpha(s)=\frac{1}{c_1-s}$$ $$\beta(s)=e^{-s\frac{1}{c_1+s}+c_2}$$
for us to have $\alpha(0)=t$ and $\beta(0)=t+1$ we need to have $c_1=\frac{1}{t}$ and $c_2=ln(t+1)$, then, we have: $$\alpha(s)=\frac{t}{1-ts}$$ $$\beta(s)=e^{-s\frac{t}{1-ts}}(t+1)$$

then (and here is where i'm stucked):

$$u(\alpha(s),\beta(s))=\int_{0}^{s}\frac{d}{ds}u(\alpha(s),\beta(s))ds + u(\alpha(0),\beta(0))$$

$$=\int_{0}^{s}\left[2\left(\frac{t}{1-ts}\right)^3 + \left(\frac{t}{1-ts}\right)^2e^{-s\frac{t}{1-ts}}(t+1) + \left(\frac{t}{1-ts}\right)^2 + \frac{\left(\frac{t}{1-ts}\right)^3e^{-s\frac{t}{1-ts}}(t+1)}{\frac{t}{1-ts}+1} \right]ds + t^2+t$$

The problem is that $$\int_{0}^{s}\frac{\left(\frac{t}{1-ts}\right)^3e^{-s\frac{t}{1-ts}}(t+1)}{\frac{t}{1-ts}+1}ds$$ doesn't seen to have a solution


问:
Understanding proof of $\operatorname{PSL}_2(\mathbb{F}_2)\cong S_3$ and $\operatorname{PSL}_2(\mathbb{F}_3)\cong A_4$
Given proof:

Consider the action of $\operatorname{PGL}_2(\mathbb{F}_q)$ on the projective line $\mathbb{P}^1(\mathbb{F}_q)$. This action is faithful and 3-transitive, so $$ \phi:\operatorname{PGL}_2(\mathbb{F}_q) \to S_{q+1} $$ is a monomorphism (note: $q+1 = |\mathbb{P}^1(\mathbb{F}_q)|$, $S_n$ is the symmetric group). Therefore, $\operatorname{im} \phi$ has to be a 3-transitive subgroup of $S_{q+1}$. So for $q=2$ or $q=3$, this is only possible if $\operatorname{im}\phi = S_{q+1}$
...

Question: I don't see why the last sentence is true. For $q=2$, $\operatorname{im}\phi$ should map every 3-tuple of elements in $S_3$ to every other 3-tuple of elements in $S_3$; so we have to consider all possible permutations of 3 elements, which would explain $\operatorname{im}\phi = S_3$. But why does it hold for $q=3$ as well?

Thanks.

答: The number of $3$-tuples of distinct elements of $\{1,2,3,4\}$ is ${}_4P_3=24$, so the image of $\mathrm{PGL}_2\mathbb{F}_3$ must have size divisible by $24$, forcing it to be all of $S_4$. Thus $\mathrm{PSL}_2\mathbb{F}_3$ must be an index $2$ subgroup, which means it must be $A_4$. You can either use a theorem about index $2$ subgroups of symmetric groups or you can verify a couple generators of $\mathrm{PSL}_2\mathbb{F}_3$ act as even permutations.


问:
Is $H^1(M) \subset L^2(M) \subset H^{-1}(M)$ a Hilbert triple for $M$ a manifold with boundary?
Is $H^1(M) \subset L^2(M)  \subset H^{-1}(M)$ a Hilbert triple for $M$ a manifold with boundary? What smoothness is required of the boundary? I would be grateful for some references to this.</p>

答:
I think a sufficient condition is that (it is highly probable this is not necessary): Whenever the smoothness of $\partial M$ supports a Poincaré type inequality:
$$
\|u\|_{L^2(M)}\leq c\|\nabla u\|_{L^2(M)},
$$
then the Hilbert triple
$$
H^1(M)/\mathbb{R}\subset L^2(M) \subset H^{-1}(M),\quad \text{ or }\quad H^1_0(M)\subset L^2_0(M) \subset H^{-1}(M)
$$
holds.

The first inclusion is dense due to the fact that $C^{\infty}$ is dense in $L^2$.

Second inclusion holds if Poincaré inequality on this manifold holds for $H^1(M)$: for $\phi\in L^2(M)$
$$
L_{\phi}(u) := \int_M \phi u \leq \|\phi\|_{L^2(M)}\|u\|_{L^2(M)} \leq c\|\phi\|_{L^2(M)}\|\nabla u\|_{L^2(M)},
$$
thus $\phi\in H^{-1}(M)$.

I checked Evans's PDE book, the proof of Poincaré inequality relies on Rellich–Kondrachov compact embedding theorem, which normally assumes the boundary is Lipschitz. In Sobolev met Poincaré, the assumption is weakened to twisted cone condition for a domain (see 9.2), I am guessing this condition should be able to be extended to a compact manifold.


问:
Difficult Calculus Question (Differentiation)
I stumbled across this question in my maths textbook and have no idea how to solve it. I have looked at the answers and don't understand how to get it from the question. It's in the chapter titled 'The Quotient Rule' but am not sure how to apply it to this question:

Find the normal to the curve $x = \frac{t}{t+1}$ and $y = \frac{t}{t-1}$ at the point T where t = 2.

I tried converting this into one equation ($y = \frac{x(t+1)}{t-1}$) but don't see how to get the answer which says:

$$\frac{dy}{dx} = \frac{-(t+1)^2}{(t-1)^2}$$

and that $ T = (\frac23,2)$, $3x - 27y + 52 = 0$

I'd appreciate any pointers.

答:
This seems to be a question of Parametric derivative.
$$(x,y)=\left(\frac{t}{t+1},\frac{t}{t-1}\right)$$
$$\frac{dx}{dt}=\frac{d(t(t+1)^{-1})}{dt}=t(-1)(t+1)^{-2}+(t+1)^{-1}=\frac{-t}{(t+1)^2}+\frac{1}{(t+1)^2}$$
Similarly,
$$\frac{dy}{dt}=\frac{d(t(t-1)^{-1})}{dt}=\frac{-t}{(t-1)^2}+\frac{1}{t-1}$$
Therefore,$$\frac{dy}{dx}=\frac{\frac{-t}{(t-1)^2}+\frac{1}{t-1}}{\frac{-t}{(t+1)^2}+\frac{1}{t+1}}=\frac{-t+t-1}{(t-1)^2}\times \frac{(t+1)^2}{-t+t+1}=-\left(\frac{t+1}{t-1}\right)^2$$
This is the slope of the tangent.

The slope of the normal is $\left(\frac{t-1}{t+1}\right)^2$. Since $t=2$, therefore the slope of normal is $\frac{1}{9}$.

Coordinates of point $T=\left(\frac{2}{2+1},\frac{2}{2-1}\right)=\left(\frac{2}{3},2\right)$

So, the equation of normal is $y=\frac{1}{9}x+c$. Since normal passes through the point T, therefore,
$$2=\frac{1}{9}\times \frac{2}{3}+c \implies c=\frac{52}{27}$$
Therefore the equation of normal is $y=\frac{1}{9}x+\frac{52}{27}\implies 3x-27y+52=0$.


问:
Why is $\mu$-measurable set contained in the Borel set?
I am reading the final chapter of baby rudin. I have trouble understanding the difference between   $\mu$-measurable sets and the Borel sets. The remarks on page 309 imply that the latter is contained in the former. But from the way I see it, they are both the collection of sets, which are the final forms of open intervals after undergoing intersections, taking complements and unions. So what exactly are $\mu$-measurable sets? And why are the Borel sets $\mu$-measurable for every $\mu$?

答:
For any measure $\mu$, remark (a) says that the open sets are $\mu$-measurable. A set $E$ is $\mu$-measurable simply if $\mu(E)$ is defined. Not every set is measurable though.

So the set of measurable sets $\mathfrak{M}(\mu)$ is a $\sigma$-algebra containing the open sets. Since the Borel set is the smallest $\sigma$-algebra containing the open sets, $\mathfrak{M}(\mu)$ must contain all the Borel sets. This is indeed regardless of what measure we start with.


问:
Inconclusive second partial derivative test
Consider the function $$f(x,y)=(x+y)^4$$ and determine whether $f$ has a maximum, a minimum or neither at the point $(0,0)$.

I thought I needed to use the second partial derivative test but how would I go about showing the point is neither a minimum nor a maximum if the test is inconclusive?

答:
The Second Derivative Test in single-variable calculus and its analogue for multivariate functions, the second partial-derivative index or Hessian determinant, is of limited help for such functions built on sums of terms using power-functions with exponents larger than $ \ 2 \ . $

One problem we have with the function $ \ f(x,y) \  = \ (x+y)^4 \ $ is that its surface is a sort of "flat-bottomed trench", in which the origin lies on the line of symmetry $ \ y \  = \ -x \ $ , with $ \ f(x,-x) \ = \ 0 \ . $  There is not a critical point, but a line in the plane on which the second derivative of the function equals zero.

The other issue arises if we rotate the coordinate system in the plane by $ \ \frac{\pi}{4} \ $ , using $ \ u \ = \ x  +  y \ $ to write the function as $ \ f(u,0) \ = \ u^4 \ , \ f(0,v) \ = \ 0 \ \ . $  We already know from single-variable calculus that the second derivative will be $ \ 12u^2 \ , $ which is zero at $ \ u = 0 \ $ and positive otherwise (the curve is always "concave upwards").  It is the second-derivative function that tells us about the behavior of the function, rather than simply the value of the second derivative at the single point.

Here, the Hessian determinant is $ \ 12(x+y)^2  ·  12(x+y)^2 \  -  \ [12(x+y)^2]^2 \ = \ 0 \ , $ which is rather unhelpful.  We have at the origin what some authors call a degenerate critical point.  In such situations, we must further investigate the properties of the function and its first derivative in the hope of better understanding these points.


问:
Question regarding number of non decreasing functions
I have a question that requires me to find the number of non decreasing functions $f: A \longrightarrow B$ where $A=\{1,2,3,4,5\}$ and $B = \{-2,-1,0,1,2,3,4,5\}$

I tried doing this by finding the Total number of functions, which according to me is $8^5$. Then, I found the number of decreasing functions to be ${8 \choose 5}$ and there is only one way to order each of those combinations so number of decreasing functions is ${8 \choose 5}$. Correct me if I'm wrong here.

But here's why I don't get the next part, the answer for the number of non decreasing functions isn't $8^5- {8 \choose 5}$

I don't get why this is, shouldn't the total number of functions $-$
number of decreasing functions $=$ number of non-decreasing
functions? I need a few counter examples to convince me otherwise and I can't seem to be able to come up with one, any help on this/visualizing it would be appreciated.

The number of decreasing functions I found to be ${8 \choose 5}$, which isn't asked in the question, just something that I calculated. However, the question also does ask for the number of increasing functions $f:A\longrightarrow B$, and that answer is stated as ${8 \choose 5}$, which makes me question whether my number of decreasing functions is valid. A confirmation here would be highly appreciated too.

Thanks in advance! (I know that there is a similar question from 2015, but since I had some trouble understanding the answers, and also had further questions of my own, I decided to repost rather than posting on a ~6 year old thread)

Here's the link to the old question: Number of non-decreasing functions?

答:
You are interpreting the phrase "non-decreasing function" to mean "a function that is not a decreasing function"; if this were the case, then your answer would be correct. However, the standard definition of a non-decreasing function is a function that is non-strictly increasing, that is, a function $f(x)$ that satisfies $f(x) \le f(y)$ whenever $x<y$. There are more non-decreasing functions than increasing functions—for example, every increasing function is non-decreasing and every constant function is non-decreasing, and there others as well.

It is true, by the way, that the number of (strictly) increasing functions is the same as the number of (strictly) decreasing functions, namely $\binom85$.


答:
$1$ - "Non-decreasing function" does not mean "A function which is not decreasing". It means $\forall i,j \ \ i>j \implies f(i) \geq f(j) $. So, for example your method counts $(1,3,2,4,5)$ as non-decreasing, but it is not.

$2$ - To find the number of decreasing functions, you just pick $5$ elements and order them in the only possible way (since they have to be decreasing). To find the number of increasing functions, you pick $5$ elements and order them in the only possible way again! That is why the answers are the same.

Bonus: To find the number of non-decreasing functions, you have to choose $5$ elements, but this time, repetitions are allowed! So, at the end you are choosing $5$ elements from $8$ elements with repetitions and again you order them in the only possible way. It is clear  that you can do this in $\binom{12}{5}$ ways.

Title: Question regarding number of non decreasing functions

Body:
I have a question that requires me to find the number of non decreasing functions $f: A \longrightarrow B$ where $A=\{1,2,3,4,5\}$ and $B = \{-2,-1,0,1,2,3,4,5\}$

I tried doing this by finding the Total number of functions, which according to me is $8^5$. Then, I found the number of decreasing functions to be ${8 \choose 5}$ and there is only one way to order each of those combinations so number of decreasing functions is ${8 \choose 5}$. Correct me if I'm wrong here.

But here's why I don't get the next part, the answer for the number of non decreasing functions isn't $8^5- {8 \choose 5}$

I don't get why this is, shouldn't the total number of functions $-$
number of decreasing functions $=$ number of non-decreasing
functions? I need a few counter examples to convince me otherwise and I can't seem to be able to come up with one, any help on this/visualizing it would be appreciated.

The number of decreasing functions I found to be ${8 \choose 5}$, which isn't asked in the question, just something that I calculated. However, the question also does ask for the number of increasing functions $f:A\longrightarrow B$, and that answer is stated as ${8 \choose 5}$, which makes me question whether my number of decreasing functions is valid. A confirmation here would be highly appreciated too.

Thanks in advance! (I know that there is a similar question from 2015, but since I had some trouble understanding the answers, and also had further questions of my own, I decided to repost rather than posting on a ~6 year old thread)

Here's the link to the old question: Number of non-decreasing functions?


问:
Composition of linear orders on the same set
I need to show an example of finite linear orders $K_1$ and $K_2$ are defined on the same set, such that their composition $K_1\circ K_2$ is not a transitive relation.

Whatever orders I choose, their composition is a transitive relation.

Any ideas?
Thank you for your time.

答: There is a general result: if $K$, $L$ are transitive relations and $KL=LK$ then  $KL$ is transitive.</p>


问:
Let $A\subseteq\mathbb{R}^k$, and let $A'=\{x\in\mathbb{R}^k \mid x\in\partial(A\setminus\{x\})\}$. How do I prove that $A'$ is closed?
I was asked this simple question a short while ago.

Let $A\subseteq\mathbb{R}^k$, and let $A'=\{x\in\mathbb{R}^k \mid x\in\partial(A\setminus\{x\})\}$. Prove that $A'$ is closed.

I think I was able to explain it, but I am not sure what I claim is true. (The following is not a formal solution because I am not sure the reasoning is correct) My explanation goes as follows:

Let $x\in A'$. Say $x$ is in the interior of $A$, then $x\notin\partial(A\setminus\{x\})$ because there exists a ball $B(x;r)\subseteq A$ by definition. By a smilar logic if x is in the exterior of $A$, I claim that $x\notin\partial(A\setminus\{x\})$. Thus $x\in\partial A$ (that is $A'\subseteq\partial A$). From here I am quite sure it is not too hard to explain that $A'$ is closed.

I doubt my reasoning, but I am not sure what have I missed.
Care to shed some light over the matter?

答: Suppose that $(x_n)_{n\in\Bbb N}$ is a sequence of elements of $A'$ which converges to some $x\in\Bbb R^k$; I will prove that $x\in A'$. Take $r>0$ and consider the ball $B(x;r)$. It contains some $x_n$. Since $x_n\in A'$, $x_n\in\partial(A\setminus\{x\})$. So, every ball $B(x_n;r')$ contains elements of $A$. So, take $r'$ so small that $B(x_n;r')\subset B(x;r)\setminus\{x\}$. Then $B(x_n;r')$ contains elements of $A\setminus\{x\}$. This proves that $x\in A'$. Therefore, since $A'$ contains the limit of any convergent sequence of its elements, it is a closed set.


问:
Randomised Response Technique

Context : To get answers to sensitive questions, we sometimes use a method called the randomized response technique. Suppose, for instance, that we want to determine what percentage of the students at a large university take ketamine. We construct 20 flash cards, write ‘I take ketamine at least once a week’ on 12 of the cards (where 12 is an arbitrary choice) and ‘I do not take ketamine at least once a week’ on the others. Then we let each student (in the sample interviews) select one of the 20 cards at random, and response yes or no without divulging the question.

Establish a relationship between P(Y), the probability that a student will give a yes response, and P(K), the probability that a student randomly selected at the university takes ketamine at least once a week.

I received the following question as an undergraduate Statistics student, and I am confused about the whole idea of the "randomised response technique". The process describes a student choosing the cards and they say yes or no, but how can we determine the percentage of students that may actually take ketamine? If I say that the probability a student gives a yes response is 0.5, this would imply that half the population in the university takes ketamine, although this would be incorrect. Could someone please explain how do I derive the probability more accurately?

答:
Let $A$ be the event that the card drawn by the student interviewed (supposed to be picked uniformly at random among all students) says "I take ketamine at least once a week" and $B$ be the event that it says "I do not take ketamine at least once a week"; so that, here,
$\Pr[A] = \frac{12}{20} = \frac{3}{5}
$
and $\Pr[B] = \frac{2}{5}$.

By an application of Bayes' rule, the probability $\Pr[Y]$ satisfies
$$\begin{align}
\Pr[Y] &= \Pr[Y\mid A]\cdot \Pr[A]
+ \Pr[Y\mid B]\cdot \Pr[B]
= \Pr[K]\cdot \Pr[A] + (1-\Pr[K])\cdot \Pr[B]\\
&= \frac{3}{5}\Pr[K]+ \frac{2}{5}(1-\Pr[K])\\
&= \frac{1}{5}\Pr[K]+\frac{2}{5} \tag{1}
\end{align}$$
assuming, of course, that the students answer truthfully (that is, say "Yes" iff the statement they read on their card is true).

Why would we do that? Well, our goal is to estimate $\Pr[K]$ without knowing for sure the response of any given student (as this would violate their privacy; it's sensitive information).

Here, we have added some randomness, so if a given student answer "Yes", we don't know whether it's because they're taking ketamine and got the first type of card, or because they don't take ketamine but got the second type of card. So we can't know for sure whether a given student takes ketamine: good!

But we can still estimate $\Pr[K]$! How? Because we can estimate $\Pr[Y]$: take sufficiently many students uniformly at random, get their answer, use that to estimate $\Pr[Y]$ (call the value of the estimate $p$). Now, compute $q = 5(p-\frac{2}{5})$: by (1), this $q$ is a suitable estimate for $\Pr[K]$. (Namely, if $|p-\Pr[Y]| \leq \varepsilon$, then $|q-\Pr[K]| \leq 5\varepsilon$.)

(In your example, if the probability that a student gives the answer "Yes" is $1/2$ (note that it must be between $2/5$ and $3/5$ because of (1)); then the probability that a student takes ketamine is $5(1/2-\frac{2}{5})=1/2$ as well. But if the probability of Yes was say 45/100, then the fraction of students taking ketamine would be $5(45/100-\frac{2}{5})=1/4$.)

So we can estimate $\Pr[K]$ to very good accuracy without ever determining for sure if any given student takes ketamine: their privacy is preserved, and we got the statistical info we wanted.

Title: Randomised Response Technique


问:
Maximize $\binom{N-m}{n-k}/\binom{N}{n}$ with respect to $N$
The question is actually in the title. I wish to maximize this function with respect to $N$, where $m, n$ and $k$ are fixed naturals less than $N$ and also $k\le \min(m,n)$. How do I go about proving it?

My attempt
I am having problem since this is not an analytic function (since I don't want a large scale approximation, I don't want to use Stirling's formula), otherwise I would have differentiated with respect to $N$ to solve this. Also, if I expand the binomial coefficients on the numerator and denominator, the number of terms above are $n-k$, whereas there are $n$ terms below. How should I go about proving it?

答:
Let $N$ be the integer for which the maximum is achieved, then we must have

$\frac{N -1-m \choose n-k}{N-1 \choose n} \leq \frac{N-m \choose n-k}{N \choose n} \geq \frac{N +1-m \choose n-k}{N+1 \choose n}$

$\frac{N -1-m \choose n-k}{N-1 \choose n} \leq \frac{N-m \choose n-k}{N \choose n}$

$\Rightarrow N \leq \frac{mn}{k}$

$\frac{N +1-m \choose n-k}{N+1 \choose n} \leq \frac{N-m \choose n-k}{N \choose n}$

$\Rightarrow N \geq \frac{mn}{k}-1$

Combining the above two inequalities, we have,

$\frac{mn}{k}\geq N \geq \frac{mn}{k}-1$

$\Rightarrow N=\lfloor \frac{mn}{k} \rfloor$


问:
When is the sine of radians and degrees the same?
I need to find the two smallest positive values of $x$ such that $$sin(x °) = sin(x)$$. This is what I did. First, I converted degrees to radians by doing $x ° = {x\pi \over 180}$. So our starting equation becomes  $$sin({\pi x \over 180}) = sin(x)$$, which becomes $$sin({\pi x \over 180}) = sin(x +2\pi n)$$ for an integer $n$. Now, I invert both sides to get $${x\pi \over 180} = x+2\pi n \implies x\pi=180x+360\pi n \implies x={360\pi n \over \pi -180} $$. Thus, we get our two smallest positive values of $x$ when we input $n=-1,-2$, to get $$x= {360\pi (-1) \over \pi -180},{360\pi (-2) \over \pi -180}$$, which are approximately $$x= 6.39, 12.79$$. However, a simple graph on desmos shows that my solutions aren't the smallest. Wolfram tells us that these solutions are given by $${-180 π\over180 + π} + {360 n π \over 180 + π}$$. I do not understand how they got this, though, any help is appreciated.

答:
If $\sin A=\sin B$, the identity
$$\sin A-\sin B=2\sin\left(\frac{A-B}2\right)\cos\left(\frac{A+B}2\right)\tag1$$
implies that
$$
 \sin\left(\frac{A-B}2\right)= 0\qquad\text{or}\qquad\cos\left(\frac{A+B}2\right)=0,\tag2
$$
which is the same as saying
$$
\qquad\frac {A-B}2=n\pi\qquad\text{or}\qquad\frac{A+B}2 = \frac\pi2+k\pi,\tag3$$
for some integer choices of $n$ and $k$.

In your case $A:=\frac{x\pi}{180}$ and $B:=x$. Your solution has got only the first half of (3). Actually the Wolfram solution is the other half (you need to replace $k$ with $n-1$ to get their solution), and is also incomplete. You'll need to plug various values of $n$ and $k$ into (3) to find the smallest two positive solutions.


问:
how can $\sqrt{2}/4=1/2^{3/2}$
$\sqrt{2}/4=1/2^{3/2}$ this is a subpart of a work example question I have but I don't understand how I can convert the first part into the second.

Title: how can $\sqrt{2}/4=1/2^{3/2}$

Body:
Rationalize the numerator $$\frac{\sqrt{2}}{4}=\frac{\sqrt{2}×\sqrt{2}}{4×\sqrt{2}}$$
$$=\frac{2}{4\sqrt{2}}=\frac{1}{2\sqrt{2}}=\frac{1}{\sqrt{2^2×2}}=\frac{1}{\sqrt{2^3}}=\frac{1}{(2^3)^\frac{1}{2}}=\frac{1}{2^\frac{3}{2}}$$


问:
Can we say that $T^n$ is a positive operator?

Let $\mathcal H$ be a complex Hilbert space and $T \in \mathcal L (\mathcal H)$ be a positive operator. Can we say that $T^n$ is also a positive operator?

I am able to prove that if $n$ is even then $T^{n}$ is a positive operator since positive operators on a complex Hilbert space are self-adjoint. What can I say about $T^n$ if $n$ is odd?

答:
Since $T$ is self-adjoint, we have
$$
\langle T^{2n+1}\xi,\xi\rangle=\langle T(T^n\xi),T^n\xi\rangle,
$$
and since $T$ is positive, this expression is positive.


答:
$(1)$

Total number of functions = decreasing functions $+$ non-decreasing functions $+$ functions that are neither

As an example of the third type, $\{1,4,3,5,2\}$

$(2)$

$8 \choose 5$ is not the number of decreasing functions. It's the number of strictly-decreasing functions.
I am inclined to believe that the textbook answer is wrong because increasing functions can be constant over an interval. The number of strictly-increasing functions is indeed $8 \choose 5$ because there is just one way to arrange them after choosing them.

Go through this link : https://en.m.wikipedia.org/wiki/Monotonic_function


问:
Does a matrix A and its reduced row echelon form R have the same nullspace?
I was thinking in terms of this. Consider the general solution to $Ax = 0$, represented in terms of free variables. Setting all but one of the free variables to 0, we get a vector in the nullspace of A. The vectors obtained this way are linearly independent and span the entire nullspace.

But how do I go about proving that both A and R have the same nullspace? I know that they do, but I was confused about how I should go about showing the same.

答: The matrices $A$ and $R$ are linked by $R=EA$, where $E$ is invertible. Now checking they have the same null space is easy.


答:
Let $p \notin A'$. This means that $\exists r>0: B(p,r) \cap A \subseteq \{p\}$.
If in fact $B(p,r) \cap A = \emptyset$, for any $x \in B(p,r)$ we have a ball $B(x,r') \subseteq B(p,r)$ (open balls are open sets) and this witnesses also that $B(x,r') \cap A= \emptyset$ and so $x \notin A'$. So $B(p,r) \subseteq \Bbb R^k \setminus A'$. If on the other hand, $B(p,r) \cap A= \{p\}$, we also have $B(p,r) \subseteq \Bbb R^k \setminus A'$: if $x \in B(p,r)$ and $x \neq p$ (WLOG) then there is a ball $B(x,r') \subseteq B(p,r)$ so that $p \notin B(x,r')$. Then $B(x,r') \cap A = \emptyset$ and so $x \notin A'$, as required. So any $p \in \Bbb R^k \setminus A'$ is an interior point of it, so the complement of $A'$ is open, and $A'$ is closed.

Title: Let $A\subseteq\mathbb{R}^k$, and let $A'=\{x\in\mathbb{R}^k \mid x\in\partial(A\setminus\{x\})\}$. How do I prove that $A'$ is closed?

Body:
I was asked this simple question a short while ago.

Let $A\subseteq\mathbb{R}^k$, and let $A'=\{x\in\mathbb{R}^k \mid x\in\partial(A\setminus\{x\})\}$. Prove that $A'$ is closed.

I think I was able to explain it, but I am not sure what I claim is true. (The following is not a formal solution because I am not sure the reasoning is correct) My explanation goes as follows:

Let $x\in A'$. Say $x$ is in the interior of $A$, then $x\notin\partial(A\setminus\{x\})$ because there exists a ball $B(x;r)\subseteq A$ by definition. By a smilar logic if x is in the exterior of $A$, I claim that $x\notin\partial(A\setminus\{x\})$. Thus $x\in\partial A$ (that is $A'\subseteq\partial A$). From here I am quite sure it is not too hard to explain that $A'$ is closed.

I doubt my reasoning, but I am not sure what have I missed.
Care to shed some light over the matter?


问:
Prove that a sequence is Cauchy using the definition
$\begin{cases}
U_{1} = 1
      \\
U_{n+1} = \sqrt{U_{n}²+\frac{1}{2^{^n}}}
   \end{cases}$

Given the following inequality prove that it's a Cauchy sequence:

$U_{n+1}\:\le \:U_n\:+\:\frac{1}{2^{^n}}$

I have proved before that it's an increasing sequence so we can do

$0\le \:\left|U_{n+1}\:-U_n\:\right|\le \:\frac{1}{2^{^n}}$

And now using Squeeze theorem I find that $\lim _{x\to \infty \:}\left|U_{n+1}\:-U_n\:\right|\:=\:0$

Which means that it's a cauchy sequence.

But how do I go about proving this using the (ε, N)-definition of cauchy sequence?

答:
$U_{n+1}-U_n \to 0$ is not the same  as as $(U_n)$ being a Cauchy sequence. For example $U_n=1+\frac  1 2+\cdots+\frac 1n$ satisfies this condition but it is not Cauchy.

$$|U_{n+m}-U_n|$$ $$\leq |U_{n+m}-U_{n+m-1}|+|U_{n+m-1}-U_{n+m-2}|+\cdots+|U_{n+1}-U_n|$$ $$\leq \frac 1 {2^{n+m-1}}+\frac 1 {2^{n+m-2}}+\frac 1 {2^{n}}< \frac 1 {2^{n+m-2}}.$$ Hence,  there exists $n_0$ such that $|U_{n+m}-U_n|<\epsilon$ for all $m$, for all $n >n_0$. This proves that $(U_n)$ is Cauchy.


问:
Joint probability mass function for series of coin flips
A fair coin is rolled until at least two heads and at least two tails are obtained. Let
the random variable X denote the roll in which the second head is obtained and let the
random variable Y denote the roll in which the second tail is obtained.

How do I find the joint probability mass function of X and Y?

I don't know how to start this question.

Any help is much appreciated.

答:
If the first two outcome is $HH$, then we have $X=2$, the smallest value that $Y$ can take is $4$.

Similarly, if the first two outcome is $TT$, then we have $Y=2$, the smallest value that $X$ can take is $4$.

If the first two outcome is $HT$ or $TH$,  then $\min(X,Y)=3$.

That is we have $\min(X,Y) \in \{2, 3\}$ and $\max(X,Y) \ge 4$.

Let $n \ge 4$.

To compute $P(X=2, Y=n)$, the first two tosses $H$, the last outcome is $T$, from the $3$rd to $(n-1)$-th outcome, there is exactly one $T$.

That is consider $$HHW\ldots WT$$

where we have $n-3$ $W$'s, we have to choose $1$ of them to be $T$. We have $\binom{n-3}{1} = n-3$. The probability to get a particular sequence is $\frac1{2^n}$, hence the corresponding probability is

$$P(X=2, Y=n) = \frac{n-3}{2^n}$$

We also have $P(X=n, Y=2) = P(X=2, Y=n)$ by symetry.

Now, let's compute $P(X=3, Y=n)$, in this case, the first two tosses is either $HT$ or $TH$, the third position must be $H$, the last position is $T$. The remaining position must be $H$.

Hence we have

$$P(X=3, Y=n)=\frac{2}{2^n}$$

We have $P(X=n, Y=3) = P(X=3, Y=n)$ by symmetry.


问:
Extendability of a continuous function on Comb Space
I have been asked to prove the following -

$C$ be the comb space in the Euclidean Space $R^2$.

$A =$ { $(a,0) : a=0 \ or \  a= \frac 1n, for \ some \ n \in N$}. Define $f:A \to C$ by $f(a,0)=(a,1)$. Then $f$ has no continuous extension over $C$.

I have no clue how to approach the problem. The facts that I know about comb space is that - comb space is contractible. But it is not contractible respect to the point $(1,0)$. But I don't think these facts will be helpful for the problem. Please provide some hints on how to approach the problem.

答:
HINT: Suppose that $g:C\to C$ is a continuous functionn such that $g\upharpoonright A=f$. Show that for each $n\in\Bbb N$ and each $y\in(0,1)$ there must be an $x_{n,y}\in\left(\frac1{n+1},\frac1n\right)$ such that

$$g(x_{n,y},0)=\left\langle\frac1n,y\right\rangle\,.$$

Use the fact that $\left[\frac1{n+1},\frac1n\right]\times\{0\}$ is connected.

Conclude that $\lim\limits_{n\to\infty}g(x_{n,y},0)=\langle0,y\rangle$ for each $y\in(0,1)$. On the other hand, $\lim\limits_{n\to\infty}\langle x_{n,y},0\rangle=\langle 0,0\rangle$ for each $y\in(0,1)$, so ... ?


问:
Prove that $\sigma(n)/n = \sum_{d\mid n} 1/d$
Prove that $$\frac{\sigma(n)}{n}=\sum_{d\mid n}\frac{1}{d} ,\forall n\in \mathbb N^*$$

It’s from a number theory textbook , i’ve tried to use to formula of $\sigma(n)$, which is :

$$\prod_{i=1}^{n} \frac{p_i^{a_i+1}-1}{p_i-1} \text{ Where } n=\prod_{i=1}^{n}p_i^{a_i}$$

But it failed.

答:
An example

Choose $n=6$. On the left hand side we get $\frac{12}{6}=2.$ Now how to we evaluate the right hand side?
First write the term out
$$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{6}{6}+\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\frac{1+2+3+4+5+6}{6}=\frac{12}{6}=2.$$

Do the same thing but in general.

The general case

We start with the sum. Define $d_j$ as a divisor of $n$.

$$\frac{1}{d_1}+\frac{1}{d_2}+\cdots+\frac{1}{d_{{m-1}}}+\frac{1}{d_{m}}.$$

We know that $d_1=1$ and $d_m=n$, where $d_m$ is the biggest divisor of $n$. Now Rewrite every single term like we did in our example.
$$\frac{d_m}{d_m}+\frac{d_{m-1}}{d_m}+\cdots+\frac{d_2}{d_{{m}}}+\frac{d_1}{d_{m}}.$$
Now you may add the terms on in the numerator to get
$$\frac{d_m+d_{m-1}+\cdots+d_2+d_1}{n}=\frac{\sigma(d_m)}{d_m}=\frac{\sigma(n)}{n}.$$
And this is your desired result.


答: The map $d\mapsto n/d$ is a bijection of the set of divisors of $n$ onto itself, hence $$\sigma(n)=\sum_{d\mid n}d=\sum_{d\mid n}\frac{n}{d}=n\sum_{d\mid n}\frac1d.$$

Title: Prove that $\sigma(n)/n = \sum_{d\mid n} 1/d$

Body:
Prove that $$\frac{\sigma(n)}{n}=\sum_{d\mid n}\frac{1}{d} ,\forall n\in \mathbb N^*$$

It’s from a number theory textbook , i’ve tried to use to formula of $\sigma(n)$, which is :

$$\prod_{i=1}^{n} \frac{p_i^{a_i+1}-1}{p_i-1} \text{ Where } n=\prod_{i=1}^{n}p_i^{a_i}$$

But it failed.


答:
The definition of $\sigma(n)$ is, that it is defined as the sum over all divisors of $n$

$$\sigma(n)=\sum_{d\mid n} d$$
But thats the same as saying

$$\sigma(n)=\sum_{d\mid n}\frac{n}{d}$$

Dividing by $n$ we get

$$\frac{\sigma(n)}{n}=\sum_{d\mid n}\frac{1}{d} $$

Edit:
The equation
$$\sigma(n)=\sum_{d\mid n}\frac{n}{d}$$
does hold, because for every divisor $k$ of $n$, there is another divisor $d$ such that $k\cdot d = n$

Therefore, if we sum over the divisors $k$ we can instead sum over $\frac{n}{d}$


问:
How to prove that $\neg (p \land \neg q) \lor (\neg p \land q) \equiv \neg p \lor q$?
This is what i got so far:

(~pvq) v (q∧~p)  ≡

I like using association property, but it's not possible in this case, because of (q∧~p), if I use Morgan's law on this part, this is what will happen

(~pvq) v ~(~qvp) ≡

But I do not know if we can do Distributive in this case since theres a negation mark.

How do i do this? I have to finish this ASAP. Help me please.

答:
Via Boolean form, we have
$$\neg\left ( p\land\neg q \right )\lor\left ( \neg p\land q \right )\quad\equiv\quad p\Rightarrow q\quad\equiv\quad\neg p\lor q$$


问:
how to calculate this limit: $\displaystyle \lim\limits_{n\to \infty} \sum_{i=1}^{n} \sin{\frac{2\pi}{i}} $
how to calculate this limit:
$\displaystyle \lim\limits_{n\to \infty} \sum_{i=1}^{n} \sin{\frac{2\pi}{i}}$

someone said that: $\displaystyle \sin \frac{2\pi}{i}=\frac{2\pi}{i}$, why?

Please help, thanks a lot in advance!

答:
$$\lim_{n\to\infty}\sum_{k=1}^{n}\sin{\Big(\frac{2\pi}{k}\Big)}=\sum_{k=1}^{\infty}\sin{\Big(\frac{2\pi}{k}\Big)}?$$

Since $$\sin{\frac{2\pi}{k}} \sim \frac{2\pi}{k}\text{ , in fact} \lim_{k\to\infty} \frac{\sin{\frac{2\pi}{k}}}{\frac{2\pi}{k}}=1$$
so
$$\sum_{k=1}^{\infty}\frac{2\pi}{k}\text{diverges}\implies  \sum_{k=1}^{\infty}\sin{\Big(\frac{2\pi}{k}\Big)} \text{ diverges from the limit comparison test}$$


问:
Find the area of the shaded region within the smaller equilateral triangle

WLOG, let the side length of one of the smaller equilateral triangles by $1$. So the overall area of one of the smaller equilateral triangles is $\frac{\sqrt{3}}{4}.$ To find the area of the shaded region (within the smaller triangle), I toyed with the idea of coord-bashing (I know, a bad method, but I didn't know what else to do), but quickly dismissed it. It would take way too much time.

I am at a loss for what to do, can someone please provide me with a hint (not solution) that will help me on my way?

答: As a possible helpful path, consider the (subdivided) rectangle shown, which is exactly half shaded overall and the subdivisions are $\frac 18, \frac 38, \frac 58$ and $\frac 78$ shaded.


问:
How to prove that $(p \land (\neg(\neg p \lor q))) \lor (p \land q) \equiv p$?
My attempt at this mathematical demonstration was as follows:

(p ∧ (¬(¬p ∨ q))) ∨ (p ∧ q) ≡ p
≡ (p ∧ (p ∨ q) ∨ (p ∧ q)
≡ ((p ∨ p) ∧ q) ∨ (p ∧ q)
≡ ((p ∨ p) ∧ q) ∨ (p ∧ q)
≡ (p ∧ q) ∨ (p ∧ q)
≡ p

答:
Via implies form, we have
$$\neg\left ( \neg p\lor q \right )\quad\equiv\quad\neg\left ( p\Rightarrow q \right )$$


问:
Minkowski's Integral Inequality when $p=\infty$
My version of the Minkowski's Integral Inequality is that, suppose $(X_1,\mathcal{M}_1,\mu_1)$ and $(X_2,\mathcal{M}_2,\mu_2)$ are complete $\sigma$-finite measure spaces and $f$ is non-negative measurable function on $X_1\times X_2$ with respect to the product measure $\mu_1\times \mu_2$, then for any $p\in [1,\infty]$ there is
$$\left\lVert \int f(x_1,x_2)dx_2\right\rVert_{L^p(X_1)}\leq \int \lVert f(x_1,x_2)\rVert_{L^p(X_1)}dx_2. $$

When $p=\infty $, this yields to prove that $\int f(x_1,x_2)dx_2\leq \int \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}dx_2$ almost everywhere $x_1$, so (I think) we need to show that for almost everywhere $x_1$,
$$ f(x_1,x_2)\leq \lVert f(x_1,x_2)\rVert_{L^{\infty}(X_1)},\ \ a.e. x_2.$$
What we have by hand is that for any fixed $x_2\in X_2$, $f(x_1,x_2)\leq \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}$ almost everywhere $x_1$, hence for any $x_2\in X_2$ there exists a nullset $Z(x_2)\subset X_1$ such that $f(x_1,x_2)\leq \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}$ for any $x_1\in X_1\setminus Z(x_2)$. It follows that for $Z= \bigsqcup_{x_2\in X_2}Z(x_2)\times \{x_2\}\subset X_1\times X_2$ we have $f(x_1,x_2)\leq \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}$ for any $(x_1,x_2)\in X_1\times X_2\setminus Z$. If we can show that $Z$ is measurable in $\overline{X_1\times X_2}$ the completion of $X_1\times X_2$, then by the Tonelli's Theorem $\overline{\mu_1\times \mu_2}(Z)=\int \mu_1(Z(x_2))dx_2=0$, whence for almost everywhere $x_1$ we have the inequality for almost everywhere $x_2$ as desired.

So the question reduces to show that $\bigsqcup_{x_2\in X_2}Z(x_2)\times \{x_2\}$ is measurable in the completion of $X_1\times X_2$ if each $Z(x_2)$ is of measure $0$. Is this true? If so, how to prove that? If not, then how to prove the Minkowski's Integral Inequality when $p=\infty$?

Thanks in advance.

答: As far as I am concerned, it suffices to show that $$\sup_{x_1\in X_1}\int f(x_1,x_2)d\mu_2\le \int \sup_{x_1\in X_1}f(x_1,x_2)d\mu_2$$I think you can just use a kind of Fatou's Lemma(use the $\sup$, not $\inf$) to prove that.


答:
A key point to understand here is that this randomized response survey technique does not come without a cost, which is increased variance in the  estimate for the parameter of interest.

To see why, let $p$ be the true proportion of students who take ketamine once a week, and suppose there are $a$ cards of type $A$ and $b$ cards of type $B$, where $a \ne b$ are known parameters.  We make the assumption that the card type drawn is independent of whether the student takes ketamine once a week.  So in a sample of $n$ students, let $X_i = 1$ if the $i^{\rm th}$ student takes ketamine once a week, and $0$ if they do not.  Then $X_i \sim \operatorname{Bernoulli}(p)$.  Now let $Y_i = 1$ if the answer to the the card they drew is yes, and $0$ if the answer to the card they drew is no.  Clearly $Y_i$ must also be Bernoulli, but what is its parameter?  Well, there is a third Bernoulli variable involved, call this $C_i$, for the card that they drew, where $C_i = 1$ if they drew card type $A$ and $0$ if card type $B$.  So $$C_i \sim \operatorname{Bernoulli}(a/(a+b)),$$ and $$Y_i = C_i X_i + (1-C_i) (1-X_i).$$  Therefore, $$\Pr[Y_i = 1] = \Pr[C_i = X_i = 1] + \Pr[C_i = X_i = 0] = \frac{a}{a+b} p + \frac{b}{a+b} (1-p) = \frac{ap + b(1-p)}{a+b}$$ is the Bernoulli parameter for $Y_i$.

In this sampling scheme, the data we have is only the $Y_i$:  we do not have access to $X_i$ or $C_i$.  How do we estimate $p$?  Because $$\operatorname{E}[\bar Y] = \frac{ap + b(1-p)}{a+b},$$ where $\bar Y$ is the sample mean of the $Y_i$, an "intuitive" point estimate (which is the method of moments estimator) is $$\tilde p = \frac{a+b}{a-b} \bar y - \frac{b}{a-b}.$$  We can also show that this is the same as the maximum likelihood estimate $\hat p$, which is obtained by computing the critical point of the log-likelihood function $$\ell(p \mid a, b, \bar y) \propto \bar y \log \frac{ap + b(1-p)}{a+b} + (1 - \bar y) \log \frac{a(1-p) + bp}{a+b}.$$
Hence, $\hat p$ is unbiased.  However, the variance of this estimator is what we need to consider:
$$\begin{align}
\operatorname{Var}[\hat p \mid a,b] &= \left(\frac{a+b}{a-b}\right)^2 \operatorname{Var}[\bar y] \\ &= \left(\frac{a+b}{a-b}\right)^2 \frac{1}{n} \frac{ap + b(1-p)}{a+b} \frac{a(1-p) + bp}{a+b} \\ &= \frac{p(1-p)}{n} + \color{red}{\frac{ab}{n(a-b)^2}}.\end{align}$$
Since the variance of the sample proportion based on the direct information $X_1, X_2, \ldots, X_n$ is $p(1-p)/n$, we see that the variance of the randomized responses is strictly larger by a term that does not depend on the true parameter, but rather, the extent of difference between $a$ and $b$ and the sample size $n$.

This illustrates several important principles:

When $b = 0$, then $Y_i = X_i$ identically and the sampling is not random, and we recover the usual binomial proportion sampling scheme; similarly, if $a = 0$, then $Y_i = 1 - X_i$ and there is no "additional randomness."

When $a = b$, the randomized response scheme fails because the responses $Y_i$ are completely noninformative of $p$, as seen by the fact that $ap + b(1-p) = a$.

The closer $a$ and $b$ are to each other--more specifically, the closer the ratio $a/(a+b)$ is to $1/2$--the larger the "excess" variance term for a fixed sample size, and this is the cost that is extracted from this survey methodology.

Title: Randomised Response Technique

Body:
Context : To get answers to sensitive questions, we sometimes use a method called the randomized response technique. Suppose, for instance, that we want to determine what percentage of the students at a large university take ketamine. We construct 20 flash cards, write ‘I take ketamine at least once a week’ on 12 of the cards (where 12 is an arbitrary choice) and ‘I do not take ketamine at least once a week’ on the others. Then we let each student (in the sample interviews) select one of the 20 cards at random, and response yes or no without divulging the question.

Establish a relationship between P(Y), the probability that a student will give a yes response, and P(K), the probability that a student randomly selected at the university takes ketamine at least once a week.

I received the following question as an undergraduate Statistics student, and I am confused about the whole idea of the "randomised response technique". The process describes a student choosing the cards and they say yes or no, but how can we determine the percentage of students that may actually take ketamine? If I say that the probability a student gives a yes response is 0.5, this would imply that half the population in the university takes ketamine, although this would be incorrect. Could someone please explain how do I derive the probability more accurately?


问:
Show that if $A$ is closed in $X$ and $B$ is closed in $Y$, then $A\times B$ is closed in $X\times Y$.
Show that if $A$ is closed in $X$ and $B$ is closed in $Y$, then $A\times B$ is closed in $X\times Y$.</p>

答:
HINT:

$$(X\times Y)\setminus(A\times B)=\Big((X\setminus A)\times Y\Big)\cup\Big(X\times(Y\setminus B)\Big)$$


问:
Confusion with set notation
If you wanted to state a set of every single number ever, would
$$ C $$
just suffice, or is it correct to say:
$$ \{ x \in C | -\infty \lt x \lt \infty \} $$
where 'C' is the complex set.

答:
The complex set $\mathbb C$ doesn't have an order relation between its numbers.

You can't compare $3+4i$ and $1-i$, the same way you can't compare vectors.

By abuse of notation, $\mathbb C \cong \mathbb R ^2$

So, if you wanted to write a set of every single number in existence, it depends on what you consider as a ''number''.

Title: Confusion with set notation


问:
Is the y axis on a PDF actually meaningless?
This idea popped in my head when I was reading this post on the normal distribution and the y-axis.

My question is (and taking advantage of a nearby computer), a PDF inputs one value and returns another, and this returned value is a probability. So, if we were using R, we'd do something like dnorm(0) and get 0.3989423. Fair enough.

However, the above post mentioned (all credit due to @Arkamis):

"By the fundamental theorem of calculus, the PDF is then the derivative of the CDF; that is, the PDF is the derivative of a function that returns a probability. So what is that intuitively? Honestly... it's not really anything. The "units" of the vertical axis in the PDF plot don't lead to anything intuitive; they are meaningful, but only in a derived, mathematical sense."

So, is the y-axis of a PDF returning a probability, or instead is it a mostly unintuitive construct?

答: Y axis represents probability density whereas probability is represented as the area. Y axis can be thought as probability/dx instead of probability. If you assume y axis as probability as you refine your range and get smaller and smaller the curve goes away. To avoid this paradox y axis is probability density.</p>


问:
Why $4b^3p=2a^2+qb$ has no solutions for integers $a,b$ and two odd primes $p<q$?
Empirically I do not find any integer solutions for $4b^3p=2a^2+qb$, where $a$ and $b$ are non-zero integers and $2<p<q$ are two odd primes.

If it is provable, why this equation does not provide a solution?

Body: Non-trivial solution: $p=3,q=47,a=1,b=2$