问:
What is the Hausdorff dimension of the set of irrationals on $\mathbb R^1$
I am sorry if this is a dump question but I can't find any reference for this. What is the Hausdorff dimension of the set of irrationals on $\mathbb R^1$?

I speculate this is $1$ but don't know how to prove it rigorously.

答:
$d$-dimensional Hausdorff measure for any $d > 1$ is nonatomic, so any countable set has $d$-dimensional Hausdorff measure $0$.  In particular the rationals have $1$-dimensional Hausdorff measure $0$, and the irrationals
have infinite $1$-dimensional Hausdorff measure.  On the other hand the $d$-dimensional Hausdorff measure of $\mathbb R$ is $0$ for $d > 1$.  So the irrationals have Hausdorff dimension $1$.


问:
Expected Value of the Square of Two Random Variables
If we let X be a random variable with expected value m and variance k, and let $X_1$ and $X_2$ be random variables that result from applying $X$ to two independent trials of this particular Bernoulli experiment, what is $E((X_1 − X_2)^2)$?

I learned linearity of expectation, but as far as I have gotten was $E((X_1)^2-2*X_1*X_2+(X_2)^2)$ after polynomial expansion.

I've no clue how to proceed from here.

答:
Since $X_1$ and $X_2$ are independent, we have $$\mathbb E[X_1X_2] =\mathbb E[X_1]\mathbb E[X_2] = \mathbf m^2.$$
(As an exercise, you might want to prove from the definition of independence that it indeed implies the expectation of the product is the product of the expectations - assuming both random variables have finite second moment.)
It follows then that
\begin{align}
\mathbb E[(X_1-X_2)^2] &= \mathbb E[X_1^2 -2X_1X_2 +X_2^2]\\
&= \mathbb E[X_1^2] - 2\mathbb E[X_1X_2] + \mathbb E[X_2^2]\\
&= 2(\mathbf m^2 +\mathbf k - \mathbf m^2)\\
&= 2\mathbf k.
\end{align}

ContentLicense: CC BY-SA 4.0


问:
Expected number of triples of friends
There are $n$ people are at a party. The probability that each pair of people is friends is $\frac{1}{2}$ (independent). Let $X$ be the number of triples of people that are friends. Find $E[X]$

For example, in the triple $X,Y,Z$,

$X$ is friends with $Y, Y$ is friends with $Z,$ and $X$ is friends with $Z$.

I think this is similar to the Birthday Paradox/Problem.

This is my approach, can someone tell me if I went wrong somewhere?

The total number of triples is $\binom{n}{3}$

Let $X_{i,j,k}$ be the random variable that is equal to $1$ when person $i$ is friends with person $j$, person $j$ is friends with person $k$ and person $i$ is friends with person $k$, and $0$ otherwise; where $1 \leq i < j < k \leq n$

I'm not sure if this is correct, but $E[X_{i,j,k}] = P(X_{i,j,k}) = (\frac{1}{2})^3$

$E[X]= \sum_{1 \leq i < j < k \leq n} E[X_{i,j,k}] = \frac{\binom{n}{3}}{2^3}$

Note that I'm not sure if $E[X_{i,j,k}] = P(X_{i,j,k}) = n(\frac{1}{2})^3$ because we need to sum over $n$ people

In that case, $E[X] = \frac{n \binom{n}{3}}{2^3}$

答:
The probability that any three selected people are mutual friends is $1/2^3$, due to the independence and identicallity of selection.   Let $X_{i,j,k}$ be the indicator function for this event, so the probability of the event is the expectation of the indicator, that is $\mathsf E(X_{i,j,k})$.

Thus the expected count of triples of mutual friends is the expectation of the sum of all indicators for selections of triples.

$$\mathsf E(X)=\sum_{1\leqslant i\lt j\lt k\leqslant n}\mathsf E(X_{i,j,k})=\binom n 3 \dfrac 1{2^3} = \dfrac{n!}{3!\,(n-3)!\, 2^3}$$

EG when $n=4$ then $\mathsf E(X)=\mathsf E(X_{1,2,3}+X_{1,2,4}+X_{1,3,4}+X_{2,3,4})=\tfrac 1{2}$

Title: Expected number of triples of friends


问:
Compute the infinite sum of product of sin and exponential function
I wonder can we find the special function related with limit of the following sum, let
$$f(t,a,b)=\sum_{n=1}^{\infty}\sin(na)\sin(nb)e^{-\frac{1}{2}n^2t}$$
it is said that: there exist functions $g, h$ and the number $\lambda>0$ such that for all $a, b,$ as $t \rightarrow \infty$
$$\lim_{t \rightarrow\infty}\frac{f(t,a,b)}{g(a)h(b)e^{-\lambda t}}=1$$

I want to find the function $g,h$ and $\lambda$ so that the above limit hold. I want to proceed in following way: find the upper and lower bound for $f(t,a,b)$, and then find $g,h$ and $\lambda$. But I don't know how to start. I also try WolframAlpha on specific value of $a,b,t$, it seems that the value related with $ab,a+b$.

Can anyone help on this limit?

答:
I have find the solution from Gregory F. Lawler book:"Random Walk and the Heat Equation" page 77.

$$
p(t, y, x, U)=\frac{2}{\pi} \sum_{k=1}^{\infty} e^{-k^{2} t / 2} \sin (k x) \sin (k y)
$$
Note that
$$
\left|\sum_{k=2}^{\infty} e^{-k^{2} t / 2} \sin (k x) \sin (k y)\right| \leq \sum_{k=2}^{\infty} e^{-k t}=\frac{e^{-2 t}}{1-e^{-t}}
$$
Hence, as $t \rightarrow \infty,$ the sum on the right-hand side is dominated by the $k=1$ term,
$$
p(t, y, x, U) \sim \frac{2}{\pi} e^{-t / 2} \sin (x) \sin (y), \quad t \rightarrow \infty
$$


问:
Are slices of Borel sets also Borel sets?

Suppose E is Borel in $\mathbb{R}^{a+b}$. Show that the slice $E^{x_1}$={$x_2\in\mathbb{R}^b|(x_1,x_2)\in E$} is Borel for each $x_1\in \mathbb{R}^a$.

What I have so far:
First, any sigma algebra containing the open sets contains every Borel set. Also, I know that given an open subset of $\mathbb{R}^{a+b}$, then for every $x_1\in\mathbb{R}^a$, $E^{x_1}$={$x_2\in\mathbb{R}^b|(x_1,x_2)\in E$} is also open. This implies that $E\in F$ and is Borel. I'm not sure if this reasoning is enough.

Show that this statement is not true if we replace both instances of "Borel" with "Lebesgue".

I thought I could use the Banach-Tarski Paradox. The unit ball is divided into 5 non-lebesgue measurable sections. However, take one of these sections and name it $A\in\mathbb{R}^3$. $A\times${$0$} is Lebesgue measurable (0 measure).

答:
Here are some hints:

To see why $E^{x_1}$ is borel, consider the (obviously borel) function
$f(x_2) = (x_1,x_2)$. Do you see why $E^{x_1}$ is the preimage of a borel set by a borel function?

Remember that the lebesgue measurable sets are exactly
$\{ \text{borel sets} \} \cup \{ \text{extra null sets} \}$, so we will probably have to use these extra null sets somehow.

More concretely: Take your favorite (lebesgue) nonmeasurable set $X$. Can you show that $X \times \{0\}$ is (lebesgue) measurable in $\mathbb{R}^2$? Remember you'll probably have to use nullsets!
Once you've done this, $(X \times \{0\})^0$ is obviously $X$, which was nonmeasurable by assumption.

I hope this helps ^_^


问:
If $(X,Y,Z)$ is jointly normal, what is the distribution of $(X,Y)\mid Z$?
Let $(X,Y,Z)$ be a jointly normal random vector. I am interested in the distribution of $(X,Y)\mid Z$. It is normal obviously, but what is its mean and variance matrix in terms of the mean and variance matrix of $(X,Y,Z)$ ?

答:
Suppose
\begin{align}
& \left[ \begin{array}{c} U \\ V \end{array} \right] \sim \operatorname N_{p+q} \left( \left[ \begin{array}{c} \mu \\ \nu \end{array} \right], \left[ \begin{array}{lc} A & B \\ B^\top & C \end{array} \right] \right) \\[8pt]
\text{where } & U,\mu\in\mathbb R^{p\times1}, \quad V,\nu\in\mathbb R^{q\times1}, \\
& A\in\mathbb R^{p\times p}, \quad B\in\mathbb R^{p\times q}, \quad C\in\mathbb R^{q\times q}.
\end{align}
Then
$$
\operatorname E(U\mid V) = \mu + B C^{-1}(V-\nu)
$$
$$
\operatorname{var}(U\mid V) = A - BC^{-1} B^\top.
$$
The way to prove this is somewhat interesting but more work than I'm going to do at this hour.

As applied to your three jointly normal random variables, it says this:
$$
\operatorname E \left( \left[ \begin{array}{c} X \\ Y \end{array} \right] \,\Big|\, V \right) = \left[ \begin{array}{c} \operatorname EX \\ \operatorname EY \end{array} \right] + \left[ \begin{array}{c} \frac{\operatorname{cov}(X,Z)}{\operatorname{var} Z} \\[4pt] \frac{\operatorname{cov}(Y,Z)}{\operatorname{var} Z} \end{array} \right] (Z-\operatorname EZ)
$$

\begin{align}
& \operatorname{var} \left( \left[ \begin{array}{c} X \\ Y \end{array} \right] \,\Big|\, Z \right) \\[10pt]
= {} & \left[ \begin{array}{cc} \operatorname{var} X & \operatorname{cov}(X,Y) \\ \operatorname{cov}(X,Y) & \operatorname{var}(Y) \end{array} \right] - \frac 1 {\operatorname{var}Z} \left[ \begin{array}{cc} \big(\operatorname{cov}(X,Z)\big)^2 & \operatorname{cov}(X,Z)\operatorname{cov}(Y,Z) \\ \operatorname{cov}(X,Z)\operatorname{cov}(Y,Z) & \big( \operatorname{cov}(Y,Z) \big)^2 \end{array} \right].
\end{align}


问:
What does the golden spiral converge to?
I am looking at pictures of spirals associated with the Fibonacci sequence and the golden ratio and I am seeing several different spiral diagrams.

For example one image I saw shows the spiral with the following numbers $8, 5, 3, 2, 1, 1$.

Then another has $34, 21, 13, 8, 5, 3, 2, 1$.

So, since the Fibonacci sequence continues forever is it possible to figure out where the golden spiral converges to in an image with Fibonacci sequence of $F_n, F_{n-1}, F_{n-2}, \dots$ ?

答: The Fibonacci sequence usually begins $0, 1, 1, 2, 3, 5, \ldots$  with $F_0 = 0$. The sequences above are this in reverse. The second one excludes $F_1$. It seems that $F_n$ where $n < 0$ are being excluded and you are asking for the coordinates of the the start of the spiral at $F_1$. That depends on the diagram, and the coordinate system. In general, the coordinates can be found.


问:
No of solutions for $x_1x_2x_3x_4 = 770$
Question:
Let $N$ be the set of all integral solutions of the equation $x_1x_2x_3x_4 = 770$. Find $N$

So, for this question I used cases. Using prime factorization we know that the factors are $7 , 2 ,5 ,11$.

Case 1:When $x_1,x_2,x_3,x_4$ are $7 , 11 , 5 ,2$ in all permutations then number of ways of arranging :$4!$

Case 2:When two of the numbers are already multiplied. Ex. $7,11,10,1$ and here first I have to choose two numbers and then arrange so:${4 \choose 2} \cdot 4!$

Case 3: When a pair of two numbers are multiplied. Ex. $77 , 10 , 1 , 1$ and here I have to choose two and then arrange:${4 \choose 2} \cdot \frac{4!}{2!}$

Case 4: When three numbers are multiplied. Ex.$7 , 110 , 1,1$ and here I have to choose three numbers then arrange:${4 \choose 3} \cdot \frac{4!}{2!}$

Case 5: When one number is $770$ and the others are $1$ then ways of arranging is $4$

Thus total no of ways ($N$) $= 292$. However the answer is given as $256$. Which case have I missed and is there a better method to approach this question. Any help will be appreciated!

答: Your case 3 is not correct.  $4 \choose 2$ is the number of ways to combine two prime factors, but you count $77,10,1,1$ twice, once when you choose $7,11$ for the two and once when you choose $2,5$.  This divides the cases by $2$.  You must have added wrong, because that correction reduces the total and the answer comes out $256$ as desired.


问:
What is a branch in Taylor polynomial (for someone with knowledge only in first-year calculus)?
I know that there has already many definitions of a branch, in the context of Taylor polynomial. Can someone describe to me, in a simplier term for someone with knowledge only in first-year calculus, what a branch actually is? For instance, let say I have the following equation:

$$y = \frac{1}{\sqrt x}$$

Is the branch point $x=0$ and $x = \infty$? If the point of expansion for a Taylor series is at the branch, how likely will the approximation fail?

答:
The equation $y^2=x$ has two solutions, $y=\sqrt x$ and $y=-\sqrt x$. These two solutions may be thought of as the two branches of the functional inverse of squaring, and they come together (at any rate, their graphs come together) at zero, the branch point. The question of behavior of the Taylor series at $x=0$ does not arise, since the functions are not differentiable at $x=0$ (so a fortiori there is no Taylor expansion at $x=0$).

But most of the time, I'm afraid you do need complex numbers to see what a branch point is, e.g., I don't think there's any way to explain branches of the logarithm without bringing in the complex plane.


问:
How to show that quantity quantifiers are not definable in first-order logic without equality?
In first-order logic with equality, we can define "there exists at least $n$ objects $x$ such that $P(x)$", for each specific positive integer $n$. However, I think this probably can't be done in first-order logic without equality, except for the trivial case $n=1$. But how to formally prove that it can't be done? More generally, is it true that if we add finitely many quantifiers $Q_2, Q_3, .... , Q_n$ to the language, where $Q_k$ means "there are at least $k$ objects such that", then we still will not be able to define all the remaining quantity quantifiers, $Q_{n+1}, Q_{n+2}, ... $ etc?

答:
Let $L_n$ be first-order logic without equality augmented by the quantifiers "$\exists^{\ge k}$" for $k\le n$. I'll take the approach that in $L_n$ we do have the symbol "$=$," but it can be interpreted as any equivalence relation; of course it doesn't affect the details.

To show inexpressibility of $\exists^{\ge n+1}$ in $L_n$ we just have to exhibit a pair of structures which are $L_n$-equivalent but are not $L_{n+1}$-equivalent. And this is easy to do: for each $i$, let $A_i$ be the "pure indiscrete set," that is the structure in the empty language where "$=$" is interpreted trivially ($x=y$ for all $x,y$) with $i$ elements. You can show$^*$ that $Th_{L_n}(A_i)=Th_{L_n}(A_j)$ whenever $i,j<n$ or $i,j\ge n$, but clearly if $i<n\le j$ then $Th_{L_n}(A_i)\not=Th_{L_n}(A_j)$ (the latter contains "$\exists^{\ge n}x(x=x)$" but the former doesn't).

$^*$There's probably a cute semantic way of doing this, but honestly it's probably easiest to just explicitly compute $Th_{L_n}(A_i)$ as a function of $n,i$. The point is that:

All atomic formulas are true under every variable assignment, and Booleans behave the same across all structures.

"$\exists^{\ge n}$" behaves like negation if $i<n$ and like affirmation (I guess we can call it that?) if $i\ge n$.

And those two facts inductively determine the whole $L_n$-theory of $A_i$.


问:
How do I answer this question on relations?
I am unable to understand what this question means (except that I is the identity relation). Would really appreciate a simple explanation of what it is saying, and also some hints/tips on how to answer it. Thank you.

$R$ is a binary relation on set $S$. $R_0$, $R_1$, $R_2$…. are defined as below:

$R_0 := I = \{(x,x) : x ∈ S\}$

$R_{n+1} := R_n ∪ (R;R_n)$ for $n >= 0$

There exists $i ∈ N$ such that $R_i = R_{i+1}$

Prove the following:

$R_j = R_i$ for all $j \ge i$

$R_j ⊆ R_i$ for all $j \ge 0$

答:
$R_j=R_i$ for all $j\geq i$

Let $B = \{n : n \in \mathbb{N} \land R_i=R_{i+n}\}$

First $1 \in B$ since, the exercise itself asserts "There exists $i \in \mathbb{N}$ such that $R_i=R_{i+1}$".

Now assume $n \in B$, that is $R_i = R_{i+n}$, by definition of $R$ we have  $R_{i+n+1} = R_{i+n} \cup (R; R_n)$, thus we have also,

$R_{i+n+1} = R_i \cup (R; R_n)$, but the right side is $R_i$ itself we can see this from:
$$R_{i+1}=R_i \cup (R; R_n) = R_i$$
Thus we got $R_{i+n+1} = R_i$, thus $n+1 \in B$ and we have shown that $n \in B \Rightarrow n+1 \in B$, therefore by mathematical induction $B=\mathbb{N}$.

Now we have just two cases to check because $(j \geq i \Leftrightarrow j \gt i \lor j=i)$, its obvious that if $i=j$ then $R_i=R_j$, but if $j>i$ then $(\exists u)(u \in \mathbb{N} \land i+u = j)$ from which follows $R_j = R_{i+u} = R_i$.

Therefore $j \geq i \Rightarrow R_j = R_i$

$R_j \subseteq R_i$ for all $j \geq0$

Here we will proceed by contradiction,

First by the previous part we have that if $j \geq i$ then $R_j=R_i$, from this follows that $j \geq i$ then $R_j \subseteq R_i$ since both are the same.

Now for the sake of contradiction assume that there is some $j \lt i$ such that $R_j \not\subseteq R_i$, that is there is some element $a$ in $R_j$ that is not on $R_i$, in symbols $(\exists a)(a \in R_j \land \notin R_i)$.

Since $j \lt i$ there is some $k$ such that $j+k = i$ we can reach the definition of $R_i$ by using the definition of $R_{n+1}$. We need to apply the definition $k$ times starting from $j$, but by the definition of $R_{n+1}$ it possible to see, that $R_{n+1}$ contains all element of $R_n$.

It suffices to see that for any element $a$, $a \in R_j \Rightarrow a \in R_{j+1}$, and therefore after $k$ times we will have $a \in R_i$ and therefore $R_j \subseteq R_i$ which contradicts our assumption.

Thus we conclude that $j \geq 0 \Rightarrow R_j \subseteq R_i$.


问:
Coloring the line
You probably know the Hadwiger-Nelson problem, which states:

What is the minimal number $k$ such that there is a $k$-coloring of the plane such that no two points at distance $1$ have the same color?

This problem is open. There are known upper ($k=7$) and lower ($k=4$) bounds.

My question is: is the same problem for the line solved? I can see the obvious bound below given by $2$ and $3$ as a bound above, given by the coloring by intervals of length $\frac{2}{3}$. But is there a $2$-coloring of the line solving the problem?

答: A colouring of the line using just two colours would be to colour the intervals $[n,n+1)$, where $n\in \mathbb Z$ alternatively using the two colours. Clearly, no two points of distance $1$ have the same colour. </p>

ContentLicense: CC BY-SA 3.0

Title: Coloring the line


问:
Finding a unique definition of eigenspace for degenerate eigenvalues
I am diagonalizing an Hermitian matrix numerically, and this results in a set of eigenvalues $\varepsilon_1, ..., \varepsilon_n$ and a set of eigenvectors $v_1, ..., v_n$.

If the eigenvalues $\varepsilon_i = \varepsilon_{i+1} = \varepsilon_{i+2}$ are degenerate this results in an eigenspace, spanned by $v_i, v_{i+1}, v_{i+2}$.

The Problem is, that unlike the eigenvalues, $v_i, v_{i+1}, v_{i+2}$ are not uniquely defined and they differ between different Lapack and ScaLapack implementations, which makes debugging very hard.

Is there an established algorithm to transform the eigenspace, so that it always results in a
presentation that is:

still orthonormal

unique regardless of the original state

I have been toying with different ideas, but I would prefer to use something established, that then is also numerically stable etc.

答:
Imagine that if we were given any vector space $V$, we had a heuristic by which to "order" the non-zero vectors of $V$. Now suppose (like the problem asks for) that we wanted a procedure so that, no matter what spanning set $S_1$ of $V$ we input, we will get the same (ordered) orthonormal basis $B_2$ of $V$. Let us call this process "finding the canonical basis" of $V$. We could devise a procedure to find the canonical basis of $V$ just by doing the following:

Start with the "first" vector $v_1 \in V$ (where "first" is defined by the order we give to the vectors in $V$), and normalize it. This vector will $\hat{v}_1$ will be the first element in our canonical basis.

Find the canonical basis of $V_2 = \{v ~|~ v \in V \text{ and } \langle v, v_1\rangle = 0\}$ (in other words, the subspace of $V$ orthogonal to $v_1$). Given $B_1$, finding a spanning set for $V_2$ is easy: we just subtract off the projection onto $v_1$ of every element.

Since in every iteration $V_2$ has dimension one less than $V$, we only need to iterate $\dim V$ times.

It remains to determine a procedure by which to find the "first" vector in some arbitrary subspace $V$, given any set of vectors $\{v_1, v_2, \cdots, v_n\}$ which span $V$. Fortunately, the RREF gives us a way of doing so! Let us construct the matrix $M$ where $v_1, \cdots, v_n$ are the rows. Then we can define the "first" vector in a subspace to be the first row of $\text{rref}(M)$. Since $\text{rref}(M)$ is the unique rref of any matrix whose row space is $V$, we can be guaranteed that "first" vector by our definition will be the same for any $\{v_1, v_2, \cdots, v_n\}$ that span $V$.

Some (rough) pseudocode for the algorithm:

Given: $S_1$, a set of vectors spanning $V$

Output: Canonical Basis of $V$

$S \gets S_1$

$B \gets [~]$

while there exists non-zero $v_i \in S$:

$\quad$Remove all zero vectors from $S$

$\quad$$M \gets $ the matrix whose rows are the elements of $S$

$\quad$$r_1 \gets $ first row of $\text{rref}(M)$

$\quad$Add $r_1 / \|r_1\|$ to $B$

$\quad$$S \gets$ all non-zero rows of $\text{rref}(M)$

$\quad$for each $v_i \in S$:

$\qquad$$v_i \gets v_i - \frac{\langle v_i, ~r_1 \rangle}{\|r_1\|^2} r_1$

return $B$

I am sure optimizations can be made (especially if the input is always an orthonormal basis of the subspace), but this is a working start!


问:
Finding $\int_{0}^{2\pi}\int_{0}^{\pi}\sin^3y \ e^{3\cos x\sin y+4\sin x\sin y}\,dy\,dx$
I am working on this double integral

$$\int_{0}^{2\pi}\int_{0}^{\pi}\sin^3y \ e^{3\cos x\sin y+4\sin x\sin y}\,dy\,dx$$

so far, I don't know how to start.
Can someone give a hint?

Thanks


问:
prove that a composite function is continuous
For these functions:

$$h(x)= \begin{cases}
      e^{-x} & x\geq  0 \\
      \sqrt{|x|} & x<0 \\
   \end{cases}$$

$$g(x)=(x−4)(x+1)^{2}$$

So obviously h(x) is discontinuous at x =0, but how do I prove that h(g(x)) is continuous? By the continuity definition you cannot prove h(g(x)) is continuous (since h(x) is discontinuous itself). I think a good place to start is to prove that g(x) is continuous, it is trivial how to do that. But what do I do next? I'm stuck.

答:
As the product of continuous functions, $g(x)$ is continuous everywhere.

As you indicated, $h(x)$ is only discontinuous at $x=0$.

Thus $h[g(x)]$ is only discontinuous when $g(x) = 0$.

Thus $h[g(x)]$ is only discontinuous when $x = 4$ or $x = -1$.


问:
$\int _a^{x_1} |\frac{\partial u}{\partial x_1}(s, x_2,...,x_n)|^p ds \le (x_1-a) |\frac{\partial u}{\partial x_1}(x_1, x_2,...,x_n)|^p$?
How  to show $\int _a^{x_1} |\frac{\partial u}{\partial x_1}(s, x_2,...,x_n)|^p ds \le (x_1-a) |\frac{\partial u}{\partial x_1}(x_1, x_2,...,x_n)|^p$, where $\frac{\partial u}{\partial x_1}(a, x_2,...,x_n)=0$ and $u(a, x_2,...,x_n)=0$.

This question is a step of proof of Poincare inequality, I fail to calculate it, and I feel it is not right,  so ask here, thanks for any help.

I have add the book, but it is written by Chinese.

答:
AH, found it. There was no way the inequality you wanted was true : I had to see the context and get the answer. There's a hanging variable.

It is this : let us write $dx = dx_1dx_2...dx_n$ for the integral over $Q$. Then the line before the question mark is actually (by definition of $Q$):
$$
d^{p-1}\int_Q \int_{a_1}^{a_1+d} |D_1u(s,x_2,...,x_n)|^p dsdx_1dx_2...dx_n  \\ =d^{p-1} \underbrace{\int_{a_1}^{a_1+d}\int_{a_1}^{a_1+d}\cdots\int_{a_1}^{a_1+d}}_{n \text{ times}} \int_{a_1}^{a_1+d} |D_1u(s,x_2,...,x_n)|^p dsdx_1dx_2...dx_n \\ = d^{p-1}\int_{a_1}^{a_1+d} \left[ \underbrace{\int_{a_1}^{a_1+d}\int_{a_1}^{a_1+d}\cdots\int_{a_1}^{a_1+d}}_{n \text{ times}} |D_1u(s,x_2,...,x_n)|^p dsdx_2...dx_n\right]dx_1 \\ = d^{p-1} \times d \times \left[\underbrace{\int_{a_1}^{a_1+d}\int_{a_1}^{a_1+d}\cdots\int_{a_1}^{a_1+d}}_{n \text{ times}} |D_1u(s,x_2,...,x_n)|^p dsdx_2...dx_n\right] \\ = d^p \int_Q |D_1u(s,x_2,...,x_n)|^p dsdx_2...dx_n
$$

where the fourth line follows from the fact that the inside integral doesn't depend on $x_1$. So the $d$ term comes from this variable. Finally, if we just call $s = x_1$ now, we get the line you mark $?$ in the author's notes.


问:
Is the set $X:=\{(x,y,z)^T \in \Bbb R^{3}|f(x,y,z)=0,g(x,y,z)\le 0\}$ convex?
Where $f(x,y,z):=x+z-1\text{ and }g(x,y,z):=y^2-xz$

I tried to prove the convexity of the set $X$ by showing it is the intersection of two convex sets. $f(x,y,z) = 0$ is a hyperplane and $g(x,y,z)\le 0$ is a half-space. Since both of these types of sets are convex, the intersection is convex. But I think this pseudo-proof is not specific enough. Should I try to prove that $f(x,y,z)$ and $g(x,y,z)$ are convex functions?

答:
Let $(x,y,z)\in X$. Then since $z=1-x$, $y^2-x(1-x)=x^2-x+y^2\le 0$. Hence $$\left(x-\dfrac{1}{2}\right)^2+y^2\le 1, z=1-x$$
implies that $X$ is actually convex.


问:
Continuous action of topological group and embedding
Let $G$ be a topological group act continuously on a topological space $X$.
Why the continuity of the action of $G$ on $X$ implies that $G$ embedded as
topological group in $S_{X}$. Here $S_{X}$ is the symetric group on $X$ i.e
the group of all self bijection on $X$ equipped with the pointwise topology.

Why the map $G\longrightarrow S_{X}$ is injective if the action of G on X is
continuous? Also, why this map is an homeomorphism?

Thank for any help.

答:
That's the way an action $\alpha : G \times X \to X$ is defined.

For a topological group, the group operation must be continuous. We define $\phi : G \to S_X$, where $\phi : g \mapsto \alpha_g$, by $\alpha_g(x) := \alpha(g,x)$ for all $g \in G$ and $x \in X$. Since $\alpha$ is a group action, we have $\alpha(g,\alpha(h,x)) = \alpha(gh,x)$ for all $g,h \in G$ and $x \in X$. Hence $\alpha_g \circ \alpha_h \equiv \alpha_{gh}$.

This tells us that $\phi$ is a group homomorphism.


问:
Find the Minimum of $x^2+y^2+z^2$, if $ax+by+cz=d$ without using inequality.
Find the Minimum of $x^2+y^2+z^2$, if $ax+by+cz=d$  without using inequality.

I saw similar question and solution was using inequality and hyper plane and least-norm. I want an elementary method.

答:
$ax+by+cz=d$

$A=x^2+y^2+z^2$

Solution: we find z in terms of x and y:

$z=\frac{ax+by-d}{c}$. . . . . . . .(1)

$$A=\frac{(a^2+c^2)x^2+2abxy+(b^2+c^2)y^2-2d(ax+b)y)+d^2}{c^2}$$

Let the numerator of this fraction is N, then we can write:

$N=\frac1{a^2+c^2}[(a^2+c^2)^2x^2+2a(a^2+c^2)(by-d)x]+(b^2+c^2)y^2-2bdy+d^2-
\frac1{a^2+c^2}[(a^2+c^2)x+a(by-d)]^2-\frac{a^2(by-d)^2}{a^2+c^2}+(b^2+c^2)y^2-2bdy+d^2=
\frac1{a^2+c^2}[(a^2+c^2)x+a(by-d)^2+\frac{c^2}{a^2+c^2}[(a^2+b^2+c^2)y^2-2bdy+d^2]$

We also have:

$(a^2+b^2+c^2)y^2-2bdy+d^2=(a^2+b^2+c^2)\big(y-\frac{bd}{a^2+b^2+c^2)}\big)^2+\frac{d^2(a^2+c^2)}{a^2+b^2+c^2}$

Therefore:

$$A=x^2+y^2+z^2=\frac{[(a^2+c^2)x+a(by-d)]^2}{c^2(a^2+c^2)}+(a^2+b^2+c^2)\frac{\big(y-\frac{bd}{a^2+b^2+c^2}\big)^2}{a^2+c^2}+\frac{d^2}{a^2+b^2+c^2}$$

A has three positive terms of them first and second terms are not constant and A is minimum when these two terms are equal to 0.So we must have:

$\begin{cases} y=\frac{bd}{a^2+b^2+c^2}\\(a^2+c^2)x+a(by-d)=0 \end{cases}$

Which gives:

$x=\frac{ad}{a^2+b^2+c^2}$ and $y=\frac{bd}{a^2+b^2+c^2}$ and considering relation (1): $z=\frac{cd}{a^2+b^2+c^2}$

With these relations for x, y and z minimum of A is:$\frac{d^2}{a^2+b^2+c^2}$


问:
Proof of the Hopfs Lemma
In his book Milnor proof the following Lemma:

If $v:X\to \Bbb{R}^m$ is a smooth vector field with isolated zeroes, and if $v$ points out of $X$ along the boundary, then the index sum $\sum{\iota}$ is equal to the degree of the Gauss mapping from $\partial X$ to $S^{m-1}$.

The proof is given as: Removing an $\epsilon$-ball around each zero, we obtain a new manifold with boundary. The function $\overline{v}(x)=v(x)/\|v(x)\|$ maps this manifold into $S^{m-1}$. Hence the sum of the degrees of $\overline{v}$ restricted to the various boundary conditions is zero. But $\overline{v}|\partial X$ is homotopic to $g$, and the degrees on the other boundary conditions add up to $-\sum{\iota}$. Therefore $$\text{deg}(g)-\sum{\iota}=0$$

I don't understand 2 points in this proof:

Hence the sum of the degrees of $\overline{v}$ restricted to the various boundary conditions is zero. Why is this? It should be in relation with the following Theorem: If $f : M \to N$ extends to a smooth map $F : X \to N$, then $deg(f; y) = 0$ for every regular value $y$. But I don't see the connection between them.

Why is this true: $\overline{v}|\partial X$ is homotopic to $g$?

Many thanks for your help.

答:
$M$ here is the union of the boundaries of the balls and $\partial X$, properly oriented. The $X$ is not $X$; it's what's left when you remove the balls.

The fact that $\bar\nu$ is outward-pointing on the boundary makes it easy to write down the homotopy to $g$: Take the straight-line homotopy (which can never pass through the origin) and project to the unit sphere.

Title: Proof of the Hopfs Lemma


问:
One-Dimensional Invariant Subspace of the Right-Shift Operator on $\ell^\infty$
My text asks me to show that the right-shift operator on the space of bounded sequences $\ell^\infty$ has no one-dimensional invariant subspace. It's trivial to show that a constant sequence $x_n = k$ is invariant under a right-shift, so the span of the constant sequence $x_n = 1$ is in fact a one-dimensional invariant subspace of the right-shift operator. Is there some error in my reasoning or is the text incorrect?

答:
I think you are using the left-shift instead of the right-shift.

If $M$ is the one-dimensional subspace spanned by a non-zero sequence $(a_n)$ and if this is invariant under the right shift then $(0,a_1,a_2,...) =c (a_1,a_2,...)$ for some constant $c$. If $c=0$ we get $a_n=0$ for all $n$ immediately.  Otherwise we have  $0=ca_1,a_1=ca_2,...$ and these imply that $a_n=0$ for all $n$.


问:
GRE Probability question
How many 3-digit positive integers are odd and do not contain the digit 5 ?

My attempt: 100-999 3 -digit integers, 900/2=450 odd numbers. Now how to calculate odd numbers which do not contain digit 5 in it.

答:
The important aspect in these kinds of permutations and combinations questioins is the direction with which you are approaching the problem.

For some questions, starting from the left most digit and then moving towards right may be a good strategy. Ex-- How many numbers are greater than 500 or less than 800 kind of questions, where hundred's digit has a lot of importance.

But for these kind of questions, where units digit is important, it is a better idea to start from the right-most digit.

Note: When repetition is not allowed, the above two strategies play an important role.

As repetition is allowed in this problem, it is a straight forward one.

First units digit : odd-1,3,5,7,9     No five : 4 possibilities

Middle digit      : all 10 values     No five : 9 possibilities

Hundred's digit   : zero not possible- 9 values    8 possibilities

Total = 4 * 9 * 8 = 288

Title: GRE Probability question


问:
Are slices of Borel sets also Borel sets?

Suppose E is Borel in $\mathbb{R}^{a+b}$. Show that the slice $E^{x_1}$={$x_2\in\mathbb{R}^b|(x_1,x_2)\in E$} is Borel for each $x_1\in \mathbb{R}^a$.

What I have so far:
First, any sigma algebra containing the open sets contains every Borel set. Also, I know that given an open subset of $\mathbb{R}^{a+b}$, then for every $x_1\in\mathbb{R}^a$, $E^{x_1}$={$x_2\in\mathbb{R}^b|(x_1,x_2)\in E$} is also open. This implies that $E\in F$ and is Borel. I'm not sure if this reasoning is enough.

Show that this statement is not true if we replace both instances of "Borel" with "Lebesgue".

I thought I could use the Banach-Tarski Paradox. The unit ball is divided into 5 non-lebesgue measurable sections. However, take one of these sections and name it $A\in\mathbb{R}^3$. $A\times${$0$} is Lebesgue measurable (0 measure).

答:
Consider $\{A\subseteq \mathbb{R}^{a+b}:A^{x_1} \text{ is Borel in }\mathbb{R}^b\}$. This set contains open sets of $\mathbb{R}^{a+b}$ (because projection is open) and is stable under countable unions and complements, so it consists of the Borel $\sigma$-algebra on $\mathbb{R}^{a+b}$. The same technique is quite useful in other situations where you can exhibit that some class of sets behaves nicely, hence the $\sigma$-algebra generated by them also behaves nicely.

When it comes to Lebesgue measures, we have to include null sets. Take for example, the Vitali set $V\subseteq \mathbb{R}$. Then $V\times \{0\}$ is a null set in $\mathbb{R}^2$, hence Lebesgue measurable but its projection is not.


问:
Circle Least Squares Fit
So my question is this:

Find the equation of the circle that gives the best least squares circle fit to the points $(-1,-2), (0,2.4), (1.1,-4),$ and $(2.4,-1.6).$

So far I have this general equation:

$2xc_1+2yc_2+(r^2-c_1^2-c_2^2)=x^2+y^2$ where $r^2-c_1^2-c_2^2 = c_3$

So then I think I create matrix:

$\begin{pmatrix} 2x_1 & 2y_1 & 1 \\ . & . & . \\ 2x_n & 2y_n & 1 \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \\ c_3\end{pmatrix} - \begin{pmatrix} x_1^2+y_1^2 \\ ... \\ x_n^2+y_n^2\end{pmatrix}$

and after replacing $x_1 = -1$ and $y_1 = -2$ until $x_4$ and $y_4$, I have this matrix:

$\begin{pmatrix} -2 & -4 & 1 \\ 0 & 4.8 & 1 \\ 2.2 & -8 & 1 \\ 4.8 & -3.2 & 1 \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \\ c_3\end{pmatrix} - \begin{pmatrix} 5 \\ 5.76 \\ 17.21 \\ 8.32\end{pmatrix}$

Which i guess I try to solve by setting equal to 0, then moving the last matrix to the other side and make it look like this:

$\begin{pmatrix} -2 & -4 & 1 \\ 0 & 4.8 & 1 \\ 2.2 & -8 & 1 \\ 4.8 & -3.2 & 1 \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \\ c_3\end{pmatrix} = \begin{pmatrix} 5 \\ 5.76 \\ 17.21 \\ 8.32\end{pmatrix}$

And finally solve for $c_1$, $c_2$, $c_3$. But this system has no solution as shown here. Am I missing something huge here or what am I doing wrong?

答:
It is true that the given equation $Ac = B$ has no solution. That is why we are trying to get an approximate answer (least-squares fit).

An explanation on Khan Academy
will give you an understanding of the result after approximating it.
If we cannot find the solution to $Ac = B$, we find solution to  $A^TAc=A^TB$.
This equation is the connection missing.

Title: Circle Least Squares Fit


问:
Find the Center of a circle given the tangent line 9x+3y+6=0 in the point A=(1,-5), the radius of the circle is 4.
A circumference is tangent to the line 9x+3y+6=0 in the point A=(1,-5), the radius of the circle is 4.
I've been trying to find the solution to this problem, but I don't know what to do anymore.

every problem that I have seen that is similar to this one has a point in the circumference, not in the tangent line.

What can I do to solve this problem? (sorry if you have difficulties understanding this, English is not my first language)

答:
You're given the point on the circle($1,-5$), on which $9x + 3y + 6=0$ is a tangent.
Ultimately, it is given that $(1, -5)$ is the common point for Circle and tangent.

I guess you know how to find the center when you're given point on the circle, tangent and radius.


问:
Functional Equation (similar to Cauchy's): $ f ( x + y ) = f ( x ) f ( y ) + k x y ( x + y ) $

Solve the following functional equation: $ k \in \mathbb R $, $ f : \mathbb R \to \mathbb R $ and
$$ f ( x + y ) = f ( x ) f ( y ) + k x y ( x + y ) \text , \forall x , y \in \mathbb R \text . $$

I have only got $ f ( 0 ) = 1 $, by letting $ f ( 0 ) = a $ and solving for $ a $.

If I use this, everything cancels out and leaves me with $ f ( x ) = f ( x ) $ which is fruitless.

答:
Sketch:
Assume $k \ne 0$.

Note that
\begin{align}
f(x)f(-x)= f(0) = 1
\end{align}
then it follows $f$ is never zero and
\begin{align}
f(x) = \frac{1}{f(-x)}.
\end{align}
Also, we have that
\begin{align}
f(2x) = f(x)^2+2kx^3
\end{align}
which means
\begin{align}
1=&\ f(2x)f(-2x) \\
=&\ (f(x)^2+2kx^3)(f(-x)^2-2kx^3) \\
=&\ (f(x)^2+2kx^3)(\frac{1}{f(x)^2}-2kx^3)\\
=&\ 1+\frac{2kx^3}{f(x)^2}-2kx^3f(x)^2-4k^2x^6
\end{align}

Hence
\begin{align}
&0=2kx^3f(x)^4+4k^2x^6f(x)^2-2kx^3\\
&\implies f(x)^4+2kx^3f(x)^2-1=0
\end{align}

So, we have that
\begin{align}
f(x)^2=\frac{-2kx^3\pm \sqrt{4k^2x^6+4}}{2}= -kx^3\pm \sqrt{k^2x^6+1}
\end{align}
which means
\begin{align}
f(x) = \sqrt{\sqrt{k^2x^6+1}-kx^3}.
\end{align}

However, by plugging $f$ back into the functional equation, we immediately arrive at a contradiction. Hence $f(x)^4+2kx^3f(x)^2-1\ne 0$, which means $k = 0$. This reduces the functional equation to $f(x+y) = f(x)f(y)$, which means $f(x) = e^{cx}$.


问:
How to proof "(∧)→¬(¬∨¬)" is a tautology
That is part of my homework and I really have no idea about that. Really need some help.

There is the question:

In our proof system for Sentential, if there is a proof of ℬ from  then →ℬ is a tautology. Use a proof to show that the following sentence is a tautology.
(∧)→¬(¬∨¬)

答:
The easiest way to go about proving that something involving only a few number of independent statements is a tautology, is to make use of the truth table. The underlying principle which you are appealing to is the completeness theorem, which is well known.

So, start doing like this:

Form a truth table, where you have one column for the truth values of $P$, and one column for the truth value of $Q$.

Then find out what are the various truth values for the two things in question you get for various combinations of the truth values of $P$ and $Q$.

Observe whether these two truth values always match or not. If they do then you have a tautology, and otherwise you don't.


答: Oh, I took this yesterday in my discrete mathematics class. you could either use a truth table, or just use the double negation and distributive laws. Basically $\neg(\neg  \vee \neg )$ becomes $(\wedge)$ because $\neg(\neg P)= P$, $\neg (\neg Q)$ and negating the or operator makes it an and operator(because logic!). So the end result is $(\wedge )→(\wedge )$.

Title: How to proof "(∧)→¬(¬∨¬)" is a tautology

Body:
That is part of my homework and I really have no idea about that. Really need some help.

There is the question:

In our proof system for Sentential, if there is a proof of ℬ from  then →ℬ is a tautology. Use a proof to show that the following sentence is a tautology.
(∧)→¬(¬∨¬)


问:
Analogy between $\mathbb{S}^1, \mathbb{R}$ and $\mathbb{Z}/m\mathbb{Z}, \mathbb{Z}$?
When one first learns about covering spaces (e.g. in Ch.1 of Hatcher), one typically learns about $\mathbb{S}^1$ and its universal cover $\mathbb{R}$. When one does calculations $\mod 1$, one lifts to the universal cover and works there instead.

By analogy, I want to think about $\mathbb{Z}/m\mathbb{Z}$ and its "universal cover" $\mathbb{Z}$ as discrete analogues of what is discussed above. Indeed, when one works $\mod m$, one often lifts to the "universal cover" $\mathbb{Z}$, and does basic operation there before "projecting down" to the original space $\mathbb{Z}/m\mathbb{Z}$.

The above two examples are clearly in different categories. The former is in the category of  topological spaces.

My Question: Is there a way to make the analogy more rigorous by formalizing the second example? (In particular, what is the correct notion of "covering space" in this context, and the corresponding notion of "fundamental group"?)

答:
A simple framework that encapsulates both examples is the notion of a short exact sequence of (topological) groups; the first is

$$0 \to \mathbb{Z} \to \mathbb{R} \to \mathbb{R}/\mathbb{Z} \to 0$$

and the second is

$$0 \to \mathbb{Z} \xrightarrow{m} \mathbb{Z} \to \mathbb{Z}/m \to 0.$$

Writing the second example as

$$0 \to \mathbb{Z} \to \frac{1}{m} \mathbb{Z} \to \left( \frac{1}{m} \mathbb{Z} \right) / \mathbb{Z} \to 0$$

shows that the second sequence even embeds into the first, and taking the colimit over all $m$ produces the sequence

$$0 \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to 0$$

where $\mathbb{Q}/\mathbb{Z}$ can be thought of as the group of roots of unity, which is the torsion subgroup of $S^1$. So you can informally think of the second sequence as a "discrete approximation" to the first, which becomes increasingly good but not quite perfect as $m \to \infty$. This can be used to relate, at least informally again, the theory of Fourier series on $\mathbb{R}/\mathbb{Z}$ and the discrete Fourier transform on $\mathbb{Z}/m$.

Whether this exhibits $\mathbb{Z}$ as the "universal cover" of $\mathbb{Z}/m$ in any sense is unclear to me. It seems to me that there is no way of producing the surjection $\mathbb{Z} \to \mathbb{Z}/m$ from $\mathbb{Z}/m$ functorially; in fact a choice of such a surjection is equivalent to a choice of generator, of which there are $\varphi(m)$. Similarly to produce a surjection $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}/2 \times \mathbb{Z}/2$ would involve picking a basis. But the universal cover is functorial given a basepoint (that is, it's functorial on pointed (nice) path-connected spaces), which for a topological group is conventionally taken to be the identity.


问:
Why does $\tan(50) = \frac{1}{\tan(40)}$?
I notice that $50 + 40 = 90$, so it might have something to do with the complementary functions (thats what comes to mind) it $\sin(90 - x) = \cos(x) $. However, $\tan(90 - x) = \cot(x)$. When I put the title question in the calculator it is indeed true but I don't understand why. Could someone explain?


问:
$Q\le \prod \frac{5+2x}{1+x}\le P$ find $P,Q$

if $x,y,z,$ are positives and $x+y+z=1$ and $$Q\le \prod_{cyc} \frac{5+2x}{1+x}\le P$$ find  maximum value of $Q$ and minimum value of $P$

This is actually a question made up myself ,so i don,t know whether a nice solution exists!?.

Finding P seems easier:AM-GM results $$\prod \frac{5+2x}{1+x}\le \frac{{\left(\sum_{cyc} \frac{5+2x}{1+x} \right)}^3}{27}$$

we write $$\sum_{cyc} \frac{5+2x}{1+x}=6+\sum \frac{3}{1+x}$$

But neither Jensen nor the tangent line method help as sign of inequality is reversed.

In fact i am more interested in finding $Q$.

I am looking for  a solution that avoids using computational aid(SOS) or ,uvw.

答:
For $x=y=z=\frac{1}{3}$ we obtain a value $\frac{4913}{64}.$

We'll prove that it's a minimal value.

Indeed, after homogenization we need to prove that
$$\prod_{cyc}\frac{7x+5y+5z}{2x+y+z}\geq\frac{4913}{64}$$ or
$$\sum_{sym}(687x^3+489x^2y-1176xyz)\geq0,$$ which is true by AM-GM or by Muirhead.

Also, for $y=z\rightarrow0^+$ we obtain a value $\frac{175}{2}.$

We'll prove that it's a supremum of the expression.

Indeed, we need to prove that:
$$\prod_{cyc}\frac{7x+5y+5z}{2x+y+z}\leq\frac{175}{2}$$ or $$\sum_{sym}(135x^2y+94xyz)\geq0,$$ which is obvious.


问:
Why do we say dN/dt = kN?
So, I just started doing differential equations and I'm trying to understand the basic idea of dN/dt = kN. I understand the whole process of solving a differential equation (the separation of variables, isolating N, finding k etc...) and I'm getting proficient at it. However, the only thing I'm still having difficulty grappling with is why we say that dN/dt = kN to begin with.

The examples used in my textbook:

Radioactive decay. After 3 days, 50 percent of the radioactivity produced by a nuclear explosion has disappeared. How long does it take for 99 percent of this radioactivity to disappear? The rate of change of the mass of our substance is negative, and is proportional at each moment to the mass of the substance at that moment. This statement means that if x = x(t) is the mass of the radioactive substance at time t, then dx/dt = -kx  (k > 0).

Population growth. Consider a laboratory culture of bacteria with unlimited food and no enemies. If N = N(t) denotes the number of bacteria present at time t, it is natural to assume that the rate of change of N is proportional to N itself, or dN/dt = kN  (k > 0). If the number of bacteria present at the beginning is N_0, and this number doubles after 2 hours (the “doubling time”), how many are there after 6 hours? After t hours?

There's always this idea present in these problems that the rate of change of _____ is proportional at each moment in time to _______ at that moment, or dN/dt = kN. Both mathematically and intuitively, why is this relationship "natural to assume"? What (if any) mathematical property dictates this?

P.S. It might also be worth noting that I don't know anything about population growth outside the context of the ODEs I've been solving.

答:
If $N$ represents the population size, then how the population changes will depend on the population size, i.e. the rate of change is a function of $N$
\begin{align}
\frac{dN}{dt} = f(N).
\end{align}
In this model, we are also assuming that the rate of change does not explicitly depend on time because whether it is current time or far into the future the only factor that contributes to the change of population should only be determined by its size.

The simplest model is if the rate is proportional to the population, i.e. $f$ is linear.

Title: Why do we say dN/dt = kN?


答:
Consider a triangle with angles $40^\circ$, $50^\circ$, and $90^\circ$; let $a, b, c$ respectively be the length of the sides opposite these angles (for example, $c$ is the hypotenuse).

Note that $\tan(50^\circ)$ is defined to be $\frac{b}{a}$, i.e. the length of opposite side divided by the length of the adjacent (non-hypotenuse) side.

Note also that $\tan(40^\circ)$ is defined to be $\frac{a}{b}$, for much the same reason. Now the side $a$ is opposite the $40^\circ$ angle, and the side $b$ is adjacent.

Now note that $\tan(50^\circ) \times \tan(40^\circ) = \frac{b}{a} \times \frac{a}{b} = 1$, and so, dividing both sides by $\tan(40^\circ)$,
$$\tan(50^\circ) = \frac{1}{\tan(40^\circ)}.$$

This holds more generally too. If we have an angle of $\theta < 90^\circ$ in a right-angled triangle, then $90 - \theta^\circ$ is the other angle. Following the same logic, we see that
$$\tan(90 - \theta^\circ) = \frac{1}{\tan(\theta^\circ)}.$$

Title: Why does $\tan(50) = \frac{1}{\tan(40)}$?

Body: I notice that $50 + 40 = 90$, so it might have something to do with the complementary functions (thats what comes to mind) it $\sin(90 - x) = \cos(x) $. However, $\tan(90 - x) = \cot(x)$. When I put the title question in the calculator it is indeed true but I don't understand why. Could someone explain?


问:
Opening a cube does not change its homotopy type
Consider the (skeleton of a) cube. I want to compute the fundamental group.

Now I know that if you "open the cube" - like you would open a box - or in another pov you splat it onto a plane, what you get is something which is homotopy equivalent to a bouquet of five circles, hence has the free product of five copies of $\mathbb{Z}$ as fundamental group.

My question is

Why "opening" the cube does not change its homotopy type?

答:
As far as I can tell this is a coincidence, sort of. You can write down a homotopy that "opens" the cube by "unzipping" four edges - it would be a lot easier to draw diagrams describing this than to describe it verbally but I don't have an easy way to draw diagrams right now.

In any case you don't need this. The general result is that a connected graph with $V$ vertices and $E$ edges is homotopy equivalent to a bouquet of $E - V + 1$ circles (the standard proof is by contracting any spanning tree to a point; you can also just repeatedly contract edges until you can't anymore). For the $1$-skeleton of the cube, $E = 12, V = 8$, so $E - V + 1 = 5$.


问:
Length of life of a fire detector
The length of life of a flame detector is exponentially distributed with paramater $\lambda=0.1/year$. Die number of events which activate the flame detector in an interval with length $t$ (heat, smoke etc.) is poisson distributed with paramter $\mu t,\mu=0.1/year$

Question: What is the probability, that the flame detector never makes an alarm during his whole length of life.

My attempt: $L$ is the length of life until the first alarm and $X$ is the time until the first alarm. Now I would like to calculate the common density function of $L$ and $X$

The density function of the exponentially distributed event is $F(x)=1-e^{-\lambda x}$ for $x\ge 0$ for the possion distribution we get the relgularised gamma function $Q(n+1,\lambda)$ but I do not see how to reach the goal, may you help me with this probability exercise.

答:
You've got off to a good start, the random variable $X$ is what you should be looking at.

First, what's the probability that $X>t$?  Well the statement $X>t$ means that there were no incidents in the interval $[0,t]$. As the number of incedents in this interval is a Poisson mean $\mu t$ we may calculate
$$P[X>t] = e^{-\mu t} \frac{(\mu t)^0}{0!} = e^{-\mu t}$$

For the exponential random variable $L$ we have that $P[L>t] = e^{-\lambda t}$.

Now as $\mu = \lambda$ the distributions of the two random variables are equal. We may assume from the context of the question that they are independent.

So if $X$ and $L$ are independent and identically distributed the probability that $P[X>L] = P[L>X]$ by symmetry. As $P[X=L] = 0$ for continuous random variables we must have
 $$P[X>L] = P[L>X] = 0.5$$

Title: Length of life of a fire detector


问:
How do I find two intersection points from two parabolas?
Let's say I have two equations:
$4x+y^2=12$ and $x=y^2-1$

I need to find the two intersection points of these parabolas so I can then calculate the enclosed area. I took a peek at what it would look like on Desmos.

So if I isolate for the $y$ variable for both equations, I managed to calculate the first intersection which is $(\frac{11}{5}$, $\frac{4}{\sqrt5})$.

How do I calculate the second intersection point? Do I just assume that since the function is a sideways parabola, then the other intersection point would just have a negative $x$ coordinate value?

答: Subtract one equation from the other, to eliminate y, and solve the resulting equation for x. Both solutions have the same x. Then plug the value of x into either parabola, and solve for y using the quadratic formula.</p>


问:
Basis such that projection onto a plane is diagonal
Suppose I have been given a vector space $V=\mathbb{R}^n$, and a subspace $W$ (a hyperplane) of it. I wish to find a basis $B$ for $V$ such that we obtain a $B$-diagonal matrix when projecting onto the subspace $W$.

Should we proceed to the above problem by finding the basis of the space $W$ and then somehow extending it to the whole space? May be we should find independent vectors which are orthogonal to vectors in the basis of the hyperplane? Or should we find the projection matrix by using the formula $A(A^TA)^{-1}A^T$ where $A$ consists of columns being the basis of the space $W$ and then appropriately extend it? Any hints? Thanks beforehand.

答: Form $B$ by appending a (one-vector) basis for $W^\perp$ to a basis for $W$. Note that $W$ and $W^\perp$ are eigenspaces for the projection operator, corresponding to eigenvalues $1$ and $0$ respectively. Also note their dimensions sum to $\operatorname{dim} V$. Therefore, $B$ is a basis, and indeed a basis of eigenvectors, which diagonalises the projection operator.


答:
Let $x=y=z=\frac 13$ then $Q \leqslant \frac{4913}{64}.$ We will show that it's a maximum value, or
$$(5x+5y+7z)(5z+5x+7y)(5y+5z+7x) \geqslant \frac{4913}{64}(2x+y+z)(2y+z+x)(2z+x+y).$$
Let
$$\left\{\begin{aligned} & a = 2x+y+z\\& b = 2y+z+x \\& c = 2z+x+y\end{aligned}\right. \Rightarrow \left\{\begin{aligned} & x = \frac{3a-b-c}{4} \\& y = \frac{3b-c-a}{4} \\& z = \frac{3c-a-b}{4}\end{aligned}\right.$$
The inequality become
$$(11c+3b+3a)(3c+11b+3a)(3c+3b+11a) \geqslant 4913abc.$$
Using the AM-GM inequality, we have
$$(11a+3b+3c)(11b+3c+3a)(11c+3a+3b) \geqslant 17^3 \cdot \sqrt[17]{a^{11}b^3c^3} \cdot \sqrt[17]{b^{11}c^3a^3} \cdot \sqrt[17]{c^{11}a^3b^3} $$
$$= 4913abc.$$

Title: $Q\le \prod \frac{5+2x}{1+x}\le P$ find $P,Q$

Body:
if $x,y,z,$ are positives and $x+y+z=1$ and $$Q\le \prod_{cyc} \frac{5+2x}{1+x}\le P$$ find  maximum value of $Q$ and minimum value of $P$

This is actually a question made up myself ,so i don,t know whether a nice solution exists!?.

Finding P seems easier:AM-GM results $$\prod \frac{5+2x}{1+x}\le \frac{{\left(\sum_{cyc} \frac{5+2x}{1+x} \right)}^3}{27}$$

we write $$\sum_{cyc} \frac{5+2x}{1+x}=6+\sum \frac{3}{1+x}$$

But neither Jensen nor the tangent line method help as sign of inequality is reversed.

In fact i am more interested in finding $Q$.

I am looking for  a solution that avoids using computational aid(SOS) or ,uvw.


Title: How do I find two intersection points from two parabolas?

Body:
Let's say I have two equations:
$4x+y^2=12$ and $x=y^2-1$

I need to find the two intersection points of these parabolas so I can then calculate the enclosed area. I took a peek at what it would look like on Desmos.

So if I isolate for the $y$ variable for both equations, I managed to calculate the first intersection which is $(\frac{11}{5}$, $\frac{4}{\sqrt5})$.

How do I calculate the second intersection point? Do I just assume that since the function is a sideways parabola, then the other intersection point would just have a negative $x$ coordinate value?

Body:
We have $$4(y^2-1)+y^2=12$$ or
$$y^2=\frac{16}{5},$$ which gives $$y=\frac{4}{\sqrt5}$$ or $$y=-\frac{4}{\sqrt5}.$$
Also, $$x=\frac{16}{5}-1$$ or $$x=\frac{11}{5}$$ and we got two intersection points:
$$\left(\frac{11}{5},\frac{4}{\sqrt5}\right)$$ and $$\left(\frac{11}{5},-\frac{4}{\sqrt5}\right)$$


问:
Determining power series for $\frac{3x^{2}-4x+9}{(x-1)^2(x+3)}$
I'm looking for the power series for $f(x)=\frac{3x^{2}-4x+9}{(x-1)^2(x+3)}$

My approach: the given function is a combination of two problems. first i made some transformations, so the function looks easier.

$$\frac{3x^{2}-4x+9}{(x-1)^2(x+3)})=\frac{3x^{2}-4x+9}{x^3+x^2-5x+3}$$

Now i have two polynomials. i thought the Problem might be even easier, if thinking about the function as:

$$\frac{3x^{2}-4x+9}{x^3+x^2-5x+3)}= (3x^{2}-4x+9)\cdot \frac{1}{(x^3+x^2-5x+3)}$$

Assuming the power series of  $3x^{2}-4x+9$ is just $3x^{2}-4x+9$ itself. I hoped, i could find the power series by multiplying the series of the both easier functions.. yeah, i am stuck.

$\sum\limits_{n=0}^{\infty}a_{n}\cdot x^{n}=(3x^{2}-4x+9)\cdot \ ...?... =$ Solution

答:
You can use the partial fraction decomposition:
$$
\frac{3x^{2}-4x+9}{(x-1)^2(x+3)}=
\frac{A}{1-x}+\frac{B}{(1-x)^2}+\frac{C}{1+\frac{1}{3}x}
$$
and sum up the series you get, which are known.

If you do the computation, you find $A=0$, $B=2$ and $C=1$, so

$$
\frac{3x^{2}-4x+9}{(x-1)^2(x+3)}=
\frac{2}{(1-x)^2}+\frac{1}{1+\frac{1}{3}x}
$$

The development of $(1-x)^{-2}$ can be deduced from the fact that
$$
\frac{1}{1-x}=\sum_{n\ge0}x^n
$$
so, by deriving, we get
$$
\frac{1}{(1-x)^2}=\sum_{n\ge1}nx^{n-1}=\sum_{n\ge0}(n+1)x^n
$$

The power series for the other term is again easy:
$$
\frac{1}{1+\frac{1}{3}x}=\sum_{n\ge0}\frac{(-1)^n}{3^n}x^n
$$
so your power series development is

$$
\frac{3x^{2}-4x+9}{(x-1)^2(x+3)}=
\sum_{n\ge0}\left(2n+2+\frac{(-1)^n}{3^n}\right)x^n
$$


问:
Suggesting a form of the equation of the given curve
I am really struggling with this problem for a long time.

I think it is easy to understand, but difficult to solve;

We have a curve that has a lower limit as $x \rightarrow -\infty$, and it has another lower limit as $x \rightarrow \infty$. This curve increases only before the peak and decreases only after the peak. The rate at which this curve increases may not be the same rate at which this curve decreases (apart from the sign). What I mean is: the function may increases rapidly but decreases slowly, or vice versa.

Also, it is not a piecewise function. Moreover, it should be integerable at any domain.

If the form of that function contains something like this: $ae^{-x^2}$ (Gaussian Distribution), then it will not be integerable.

I was trying to combine two expression (piecewise) to make it (not piecewise), but no luck: like combining

$y=a-be^{cx}$ for $-\infty <x \le j$, and $y=A-Be^{Cx}$ for $j < x < \infty$; where $a,b,c,A,B,C,j$ are constants.

The following curve is plotted (using piecewise function) just to illustrate the problem.

So the goal is to find the form of the equation of the curve with the following criteria:

has a lower limit before the peak.

has another lower limit after the peak.

The rate of increasing may be different from the rate of decreasing (apart from the sign).

Must be integerable in any domain.

Must not be a piecewise function.

Your help would be appreciated. THANKS!

答:
There are many functions which satisfy your requirements. Perhaps there are more requirements that you have not stated? One solution to the problem as stated is

$$y=a+b\frac{\frac{x}{c}}{\sqrt{(\frac{x}{c})^2+1}}+d^2\frac{1}{(\frac{x}{c})^2+1}$$
It has a lower limit both before and after the peak, and it is smooth, both integrable, and not "piecewise".

You may play around with the parameterized function at https://www.desmos.com/calculator/o0ja1e0c0o


问:
Prove that if polynomial's $f(x)=x^6+ax^3+bx^2+cx+d$ , all roots are real, then $a=b=c=d=0$.
$$f(x)=x^6+ax^3+bx^2+cx+d$$

I have some trouble proving it and all I know is that I should be using Vieta's formulas. Can anyone give me a hint?

答:
Call $x_1 \ldots x_6$ the roots of $f$.

You have to show that if $x_1+ \ldots+ x_6 = x_1x_2+ x_1x_3 + \ldots + x_{n-2}x_n + x_{n-1}x_n = 0$,

then $x_1 = x_2 = \ldots = x_6 = 0$. Or, put another way, that $x_1^2 + x_2^2 + \ldots + x_6^2 = 0$ (because the roots are real)


问:
Calculation on inner product space.
$V$ : vector space on $\mathbb{R}$ or $\mathbb{C}$

$ 〈 \cdot , \cdot 〉 $ : inner product on V

$x,y,z \in V$

Proposition;

If $y\neq 0$ and $z=\dfrac{〈x,y〉}{〈y,y〉} y$ hold, then $〈z,z〉=〈x,z〉$ holds.

Why this proposition holds? I tried to calculate

\begin{equation}
〈z,z〉=〈z,\dfrac{〈x,y〉}{〈y,y〉} y〉=\dfrac{〈x,y〉}{〈y,y〉} 〈z,y〉=\cdots
\end{equation}

I cannot lead to the conclusion.

答:
Let $c=\langle x, y \rangle$ and $d=\langle y, y \rangle$ The $\langle z, z \rangle =\langle (cy/d), (cy/d) \rangle =\frac {|c|^{2}} {d^{2}} \langle y, y \rangle= \frac {|c|^{2}} {d} $.

Also $\langle x, z \rangle =\frac {\overline {c}} d \langle x, y \rangle= \frac {|c|^{2}} {d} $.

Title: Calculation on inner product space.


问:
What is the expectation of number of rounds to win 3 scisssors-paper-stone in a row?
This question stems from the problem of tossing 3 heads in a row for a fair coin, but with draws. Assuming the game is fair, intuitively guessing, is the answer $14\div\frac 23=21$? How to deal with the infinite number of draws?

More generally, if the game is not fair, and the probability of winning, losing and drawing are $x$, $y$ and $1-x-y$ respectively, how would I calculate the expected time to win $n$ games in a row?

答:
Let $X_n$ and $Y_n$ be i.i.d. sequences of random variables, independent of each other, such that $X_1$ and $Y_1$ are uniformly distributed over $\{0,1,2\}$; the $X_n$ represent the player's choices at each time $n$ and the $Y_n$ the opponent's choices. (In this case $0$ denotes "rock," $1$ denotes "paper", and $2$ denotes "scissors.") Let $Z_n$ take values $1$, $0$, and $-1$ as according to when the player wins, draws, or loses, respectively, then we can write
$$
Z_n := \mathsf 1_{\{1,-2\}}(X_n-Y_n) - \mathsf 1_{\{-1,2\}}(X_n-Y_n).
$$
By symmetry, $Z_n$ is uniformly distributed on $\{-1,0,1\}$. Define
$$
\tau = \inf\{n\geqslant 3: Z_n=1, Z_{n-1}=1, Z_{n-2}=1\}.
$$
$\tau$ is the number of games played until the player wins three consecutive games. It has a phase-type distribution, i.e. it is the distribution of the hitting time of the absorbing state in an absorbing Markov chain with finitely many states. It seems that the closed form for $\mathbb P(\tau = k)$, for integers $k\geqslant 3$ is not practical to write explicitly. One can compute the value for a given $k$ by exponentiating a submatrix of the transition matrix of the absorbing Markov chain and left/right-multiplying it by two vectors: see here for more details https://en.wikipedia.org/wiki/Discrete_phase-type_distribution

Fortunately, there is a trick by which we can readily compute the expectation of $\tau$. Consider the following system of linear equations:
\begin{align}
\tau_0 &= 1+\frac23\tau_0 + \frac13\tau_1\\
\tau_1 &= 1+\frac23\tau_0 + \frac13\tau_2\\
\tau_2 &= 1+\frac23\tau_0 + \frac13\tau_3\\
\tau_3 &= 0.
\end{align}
Here $\tau_i$ is the expected number of remaining games played until winning three consecutively (hence $\tau_3=0$). Solving this system of equations gives
$$
\begin{align}
\tau_0 &= 39\\
\tau_1 &= 36\\
\tau_2 &= 27\\
\tau_3 &= 0.
\end{align}
$$
The quantity we are interested in is of course $\tau_0$, which is $39$. This agrees with the result obtained using the fundamental matrix of the absorbing Markov chain. I've left out the details of modelling this problem as a Markov chain for brevity; it would double or triple the length of the post without lending much more insight into the problem at hand.

If it is unclear as to how I came up with the system of equations to solve the problem, I recommend you construct the appropriate Markov chain and look at the transition probabilities. You should see a clear connection. If it is still not clear, try solving the problem for "number of rounds to win one game" (which would simply have a geometric distribution) and generalizing it to "number of rounds to win $n$ consecutive games," using the Markov chain approach.

As for generalizing to a game with the probability of winning being $p\in(0,1)\setminus\{1/2\}$, one would need only to construct the Markov chain with transition probabilities modified accordingly.


问:
Slope interpretation of a log model?
Why does a 1 unit change in $x$ in the following model equal to a $100\beta_2$% change in $y$?

$$\ln(y)=\beta_1 + \beta_2x$$

I know how to find the derivative  but my statistics textbook gives this alternative interpretation and I don't know why it holds.

答:
The change in $y$ should be

\begin{align}\frac{e^{\beta_1+\beta_2 (x+1)}-e^{\beta_1+\beta_2 (x)}}{e^{\beta_1+\beta_2 (x)}} \times 100 \% &=(e^{\beta_2}-1)\times 100\%\\
&=(\beta_2 + \frac{\beta_2^2}2+\ldots )\times 100 \%\end{align}

The approximation requires $\beta$ to be small.

Title: Slope interpretation of a log model?


答:
It doesn't change the homotopy type because it doesn't even change the homeomorphism type. Let $X$ be the skeleton of a cube and $Y$ be the planar graph with $6$ vertices and $12$ edges you have in mind. Define a bijection $X\to Y$ sending vertices of $X$ to vertices of $Y$ and corresponding edges of $X$ to edges of $Y$ (like a central projection from a point above the cube would). This map is a homeomorphism and in particular a homotopy equivalence.

Now you can contract edges in $Y$ to obtain homotopy equivalence with a bouquet of circles.

Title: Opening a cube does not change its homotopy type

Body:
Consider the (skeleton of a) cube. I want to compute the fundamental group.

Now I know that if you "open the cube" - like you would open a box - or in another pov you splat it onto a plane, what you get is something which is homotopy equivalent to a bouquet of five circles, hence has the free product of five copies of $\mathbb{Z}$ as fundamental group.

My question is

Why "opening" the cube does not change its homotopy type?


答:
Write $m$ for the expected number of games; it is easy to see $m$ is finite.

The expected number of games until the first win is $3$. From this point, one of three things can happen. You can win the next two games (probability $1/9$), finishing in $5$ games (in expectation); you can win the next one game (probability $2/9$); or you can fail to win the next game ($2/3$). In the latter two cases you are back where you started, and expect to play another $m$ games before winning three in a row; you have played $5$ and $4$ games respectively. So
$$m=\frac{1}{9}\times 5+\frac{2}{9}(m+5)+\frac{2}{3}(m+4).$$

Title: What is the expectation of number of rounds to win 3 scisssors-paper-stone in a row?

Body:
This question stems from the problem of tossing 3 heads in a row for a fair coin, but with draws. Assuming the game is fair, intuitively guessing, is the answer $14\div\frac 23=21$? How to deal with the infinite number of draws?

More generally, if the game is not fair, and the probability of winning, losing and drawing are $x$, $y$ and $1-x-y$ respectively, how would I calculate the expected time to win $n$ games in a row?


问:
A seemingly-complicated question about derivatives and continuity.
Question :

My problem lies in the last question. I think I have managed to solve (a), (b), and (c). I got $$f(0)=f'(0)=A=g'(0)=0$$ and $$g'(x)=f'(x)\sin\left(\frac{1}x\right)-\frac{f(x)\cos\left(\frac{1}x\right)}{x^2}$$ as my answers, if I am correct.

While for the last one I am quite confused. I think I have to follow the definition of being continuous, that is, showing that $$\lim_{x\to0^+}g'(x)=\lim_{x\to0^-}g'(x)=g'(0)=0$$ and same for $f'(x)$.

However I can't really see why this is related to the value of $L$. And I bumped into a lot of obstacles while attempting to solve it. For example, how should I deal with $\sin\left(\frac{1}x\right)$ and $\cos\left(\frac{1}x\right)$? Should I apply Squeeze Theorem?

Any hints or suggestions will be appreciated. Thanks a lot.

答:
Show $f'(x) \sin(1/x) \to 0$ as $x \to 0$. (Hint: $|\sin(1/x)| \le 1$.)

Thus $\frac{f(x)}{x^2} \cos(1/x) \to 0$ as $x \to 0$. If $L$ is not zero, then $\frac{f(x)}{x^2} \cos(1/x)$ will oscillate between [near] $-L$ to [near] $L$. If $L=0$, then all is well.


问:
Show that $dim\ rm{span} \{u_1,...,u_n\} = n-1 $ if and only if $dim\{(a_1,...,a_n) \in \mathbb{R}^n \vert a_1u_1+...+a_nu_n=0\} = 1$
Struggling with the following question:

Let $n \in \mathbb{N} $ with $n\geq 2$. Let $V$ be a vector space. Let $u_1,...,u_n \in V$. $\\$Show that $\dim\rm{span}  \{u_1,...,u_n\} = n-1 $ if and only if $\dim\{(a_1,...,a_n) \in \mathbb{R}^n \vert a_1u_1+...+a_nu_n=0\} = 1$$\\$

My "attempt" for the forward direction:

Let $\left \{(a_1,...,a_n) \in \mathbb{R}^n \vert a_1u_1+...+a_nu_n=0 \right \} = W$

I believe we need to show that $\exists v \in W s.t. v \ne 0$ and $\rm{span}\{v \} = W $

I formed a basis of dimension $n-1$ from $\dim \rm{span} \left \{u_1,...,u_n \right \}$, call it $\beta$.

Now, am I supposed to show $\beta \in W$ and that $\beta$ spans $W$? This is where I'm lost. Any pointers or corrections are greatly appreciated.

Edit: Not allowed to use rank-nullity theorem.

答:
Consider a subset of $I\subset \{1, \ldots, n\}$ that $\{u_i \ | \ i \in I\}$ is linearly independent and $I$ maximal with this property.  May assume that $I= \{1,2,\ldots, k\}$. Now, for every $j>k$ the system $(u_1, \ldots, u_k, u_j)$ is linearly dependent, so we have
$$- \sum_{i=1}^k c_{j i} u_i + u_j = 0$$

Therefore, the $n-k$ vectors

$$\phi_j = (-c_{j1}, - c_{j2}, \ldots, - c_{jk}, \delta_{j, (k+1)}, \ldots, \delta_{j n}) $$
for $j = k+1, \ldots, n$,  are in the space $\Phi\colon= \{(a_1, \ldots, a_n) \ | \ \sum a_i u_i = 0\}$. Let us show that they form a base of this space. It should be clear that they are linearly independent, since the $(n-k) \times n$ matrix formed by them has as  an identity block $I_{n-k}$. Let us show linear independence:  Consider $\phi\colon = (a_1, \ldots, a_n)$, $\sum a_i u_i = 0$.  The vector
$$\phi-\sum_{j=k+1}^n a_j \phi_j $$
has the last $n-k$ components $0$ and gives a linear dependence of $u_i$, $i=1, n$. Hence it also gives a linear dependence of $u_1, \ldots, u_k$. But these $k$ vectors are linearly independent. It follows that the vector
$\phi-\sum_{j=k+1}^n a_j \phi_j $  has all its compoents $0$, that is
$$\phi = \sum_{j=k+1}^n a_j \phi_j$$.


问:
Prove $QBQx = Qc$ has one, and only one, solution.
The problem

Let
$$ B = \left( \begin{array}{c c c}
1 & 0.5 & 0.2 \\
0.5 & 1 & 0.5 \\
0.2 & 0.5 & 1 \\
\end{array} \right) $$
Let $I = \left\{ 1,3 \right\}.$ We define $Q = (q_{i,j})$ s.t $q_{i,j} = 1$ if $i = j \in I$ and $q_{i,j} = 0,$ otherwise. Let $c = (1,1,1)$ and $x$ a vector.

Show $QBQx = Qc$, with $Qx = x$, has one, and only one, solution.

My works

I found
$$ Q = \left( \begin{array}{c c c}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{array} \right) $$
Then
$$ QBQ = \left( \begin{array}{c c c}
1 & 0 & 0.2 \\
0 & 0 & 0 \\
0.2 & 0 & 1 \\
\end{array} \right) $$
But $QBQ$ is not inversible. I don't understand how we can show the equation $QBQx = Qc$ has one solution...

Thank you

答:
According to your question, we have to satisfy 2 conditions:$QBQx=Qc$ and $Qx=x$.

Let $x=(x_1,x_2,x_3)'$, then from $Qx=x$, we have $x_2 =0$.

Now consider, $QBQx=Qc$, then $\begin{pmatrix} 1&0&0.2\\0&0&0\\0.2&0&1 \end{pmatrix}\begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix}=\begin{pmatrix} 1\\0\\1 \end{pmatrix} = \begin{pmatrix} x_1+0.2x_3\\0\\0.2x_1+x_3 \end{pmatrix}$

This gives $x_1=\frac{80}{96}, x_3=\frac{160}{192}$.


问:
How to expand a local search from linear to higher order trajectories if higher order differentials are missing?
Consider $$f(x,y,k) = kx-y$$ and its level set $f(x,y,k) = 0$

Now for example assume that I have found a point on this level set $(9,81,9)$, how can I then estimate $k$ in $(8,76,k)$? This translates to if we know $81/9=9$, how can we calculate $76/8 = ?$

I can calculate gradient $$\nabla f = [k,-1,x]$$

To remain on surface (locally), I shall go perpendicular to gradient. So $$[\Delta_x,\Delta_y,\Delta_k][k,-1,x]^T = 0$$

Now this lets us derive a wealth of algorithms which we can iterateively take small steps:

Take step along a line in the tangent plane.

Make correction in gradient direction to come back to surface.

But the problem I hit is when I want to go one step further into higher order models.

For example if I want to use a second order polynomial model...

All second order partial differentials are zero!

How can I modify $f$ or my algorithm to somehow add information of higher order than $1$ ?

Own work:

I've been thinking I can do $f \to f^3$. This is sure to generate a higher order power series expansion everywhere, right?

答:
As @TedShifrin writes, of course we do get second order partial derivatives:

$$\frac{\partial^2}{\partial k \partial x} f = \frac{\partial^2}{\partial x \partial k} f = 1$$

So we get a non-zero, but constant Hessian:

$${\bf H}\{f\} = \begin{bmatrix}0&0&1\\0&0&0\\1&0&0\end{bmatrix}$$

Perhaps even one of the simplest and sparsest Hessians possible.

We can now go on to devise (for example) a second order polynomial approximation as per this question:

$$f(\mathbf{x}^* + h\mathbf{y}) = f(\mathbf{x}^*) + h \nabla f(\mathbf{x}^*)^T \mathbf{y} + \dfrac{1}{2} h^2 \mathbf{y}^T \nabla^2 f(\mathbf{x}^*) \mathbf{y} + O(h^3)$$

If we explicitly write $\bf g$ and $\bf H$ for gradient and Hessian:

$$f(\mathbf{x}^* + h\mathbf{y}) = f(\mathbf{x}^*) + h {\bf g}^T \mathbf{y} + \dfrac{1}{2} h^2 \mathbf{y}^T {\bf H} \mathbf{y} + O(h^3)$$

Now we can use this however we see fit to pick out a better direction for estimation.


问:
Trying to understand numerical solutions using the iteration method
I am given the equation of a curve
$$ y = x\ln(8-x)$$
The slope of the curve is said to be 1 at only one point, $x = a$.
There are three parts to this question.

- Show that $a$ satisfies the equation $$x = 8- \frac{8}{\ln(8-x)}$$

The way I did this part is by finding the derivative of y (which is $-\frac{x}{8-x} + \ln(8-x)$ and equating it to 1, and solving for x. I get $$x = 8 - e^\frac{8}{8-a}$$ How do I get it into the form $x = 8- \frac{8}{\ln(8-x)}$? I suspect there is some exponent manipulation I can't think of.

- Verify by calculation $a$ lies between 2.9 and 3.1
Now I substituted 2.9 and 3.1 into y, but I did not get a sign-change. It seems to me as if I'm not looking for a sign-change here... I'm looking for some other characteristic. Can someone detail what I'm looking for, and why -- intuitively, so I can do this on any other problem I get like this?

Also, we are not looking for the x-intercept(s) of $y$, we are looking for where it cuts the point $x = a$, right? How does that change things? I think it means that I will have to use $x = y$?

I think I'll be able to do the third part myself, but I believe there are some conceptual gaps I need to fill first.

答:
For the first part, you don't need to switch to exponential form.

$$\begin{align}
-\frac{x}{8-x} + \ln (8-x) &= 1\\
\\
\ln (8-x) &= 1 + \frac{x}{8-x} \\
\\
 \ln (8-x) &= \frac{8-x}{8-x} + \frac{x}{8-x} \\
\\
\ln (8-x) &= \frac{8}{8-x} \\
\\
8-x &= \frac{8}{\ln (8-x)} \\
\\
x &= 8 - \frac{8}{\ln (8-x)}
\end{align}$$

For the second part, remember that $a$ is a solution to the above equation, not the original function. Just re-write the above equation as

$$x - 8 + \frac{8}{\ln (8-x)} = 0$$

Now evaluate at $x=2.9$ and $x=3.1$:

$$\begin{align}
2.9 - 8 + \frac{8}{\ln (8-2.9)} \approx -0.19 \\
3.1 - 8 + \frac{8}{\ln (8-3.1)} \approx 0.13
\end{align}$$

The sign change indicates $a$ must lie between $2.9$ and $3.1$.


问:
Prove that if $X^TX=X$, then $X=X^T=X^2$
If $X^TX=X$, show that $X=X^T=X^2$.

Can someone help me with this? I'm having trouble proving this question. Thank you.

答: From $X=X^TX$ it follows that $X^T=(X^TX)^T=X^T X^{TT}=X^TX.$


问:
Fermi-Dirac integral
I'm trying to do some integral calculation for Fermi-Dirac distribution, specifically for:

$$
\int_{0}^{\infty}{E^{2} \over
1 + \exp\left(E - \mu \over k_{B}\,T\right)} \, dE
$$

I know that it can be only solved numerically, but I got information that when we set $T = 0$ it can be solved analytically.

And here is my question: When we will do that directly, all is left is $0$. So, how can I approach such sort of problem $?$.

答: Applying $E=\mu+k_BTx$ for $E\ge\mu$,$$\int_0^\infty\tfrac{E^2dE}{1+\exp\frac{E-\mu}{k_BT}}=\int_0^\mu\tfrac{E^2dE}{1+\exp\frac{E-\mu}{k_BT}}+\ln2\cdot\mu^2k_BT+\frac{\pi^2}{6}\mu k_B^2T^2+\tfrac32\zeta\left(3\right)k_B^3T^3.$$The $T\to0^+$ right-hand limit is $\int_0^\mu E^{2}dE=\tfrac13\mu^3$.

Title: Fermi-Dirac integral


Body:
1)$y^2=12-4x;$

2)$y^2=x+1;$

Note: Both parabolas are symmetric about $x-$axis.

$12-4x=x+1;$

$x=11/5.$

Points of intersection:

$y>0:$ $y=4/\sqrt{5}$, e. g. $P_1(11/5, 4/\sqrt{5})$;

$y<0:$ By symmetry $P_2(11/5,-4/\sqrt{5})$.


问:
Solving $(x-y)u_x+u_y=x,~~~u(x,0)=x^2$ by the method of characteristics.
Q. Solve $$(x-y)u_x+u_y=x,~~~u(x,0)=x^2$$ by the method of characteristics.

I am getting the general solution from Lagrange-Charpit equations as $u=x+\frac{y^2}{2}+C$, where $C$ is an arbitrary constant. But it seems to be not coorporate with initial curve. How can I deal with this?

答:
$$(x-y)u_x+u_y=x$$
$$\frac{dx}{x-y}=\frac{dy}{1}=\frac{du}{x}$$
First characteristic equation from solving $\frac{dx}{x-y}=\frac{dy}{1}$ :
$$(x-y-1)e^{-y}=c_1$$
$x=c_1e^y+y+1$

Second characteristic equation from solving $\frac{dy}{1}=\frac{du}{x}=\frac{du}{c_1e^y+y+1}\quad\implies\quad\frac{du}{dy}=c_1e^y+y+1$ :
$$u-\frac{y^2}{2}-y-c_1e^y=c_2$$
$$u-\frac{y^2}{2}-y-((x-y-1)e^{-y})e^y=c_2$$
$$u-\frac{y^2}{2}-x+1=c_2$$
The general solution of the PDE expressed on the form of implicit equation $c_2=F(c_1)$ with arbitrary function $F$ is :
$$u-\frac{y^2}{2}-x+1=F\big((x-y-1)e^{-y} \big)$$
Explicitly the general solution of the PDE is :
$$\boxed{u(x,y)=\frac{y^2}{2}+x-1+F\big((x-y-1)e^{-y} \big)}$$
Condition :
$$u(x,0)=x^2=\frac{0^2}{2}+x-1+F\big((x-0-1)e^{-0} \big)$$
$$x^2-x+1=F(x-1)$$
Let $X=x-1\quad;\quad x=X+1$
$$(X+1)^2-(X+1)+1=F(X)$$
$$F(X)=X^2+X+1$$
Now the function $F$ is known. We put it into the above general solution where $X=(x-y-1)e^{-y}$ .
$$u(x,y)=\frac{y^2}{2}+x-1+\big((x-y-1)e^{-y}\big)^2+\big((x-y-1)e^{-y}\big)+1$$
The particular solution of the PDE which satisfies the specified condition is :
$$\boxed{u(x,y)=\frac{y^2}{2}+x+(x-y-1)^2e^{-2y}+(x-y-1)e^{-y}}$$


答: If $f$ has six real roots then $f'$ has five, $f''$ has four, $f'''$ has three.  Now $f'''(x) = 120 x^3 + 6 a$ and this only has three real roots if $a = 0$.  Then look at $f''$, $f'$ and $f$.</p>

Title: Prove that if polynomial's $f(x)=x^6+ax^3+bx^2+cx+d$ , all roots are real, then $a=b=c=d=0$.

Body:
$$f(x)=x^6+ax^3+bx^2+cx+d$$

I have some trouble proving it and all I know is that I should be using Vieta's formulas. Can anyone give me a hint?


问:
Types and nature of singularities of f(z)=1/z ln(1-z)
Consider the complex function f(z) = (1/z)×ln(1-z) ,
It seems like having a removable singularity (because,while comparing with corresponding real function,the limit exists.)

The function has a branch point at z=1.
Apart from my above findings, what more singularities are there?.

答: If $\ln z$ is the principal branch of logarithm then you have to remove the entire $[1,\infty)$ to get an analytic function. The function is not even continuous at points of this interval. Once you remove this part of the real axis you get an analytic function; the singularity at $0$ is actually removable.


问:
A closed set is intersection of countable collection of open sets
I am currently reading Apostol Mathematical Analysis. There was this question

Question : Prove that a closed set in $\mathbb{R}^1$ is the intersection of countable collection of open sets.

[Here $N_a(\varepsilon) =$ the open set $(a-\varepsilon, a+\varepsilon)$ ]

My Attempt : Let $A$ be a closed set. Let $A^c$ denote the complement of $A$. Take $G$ be a collection of neighborhoods where for each $N_a(\varepsilon)\in G $ where $a\in \mathbb{Q}\cap A^c=A'$(Let) and $\varepsilon$ is the minimum real number such that $A\subseteq N_a(\varepsilon) $

As the set of rational numbers are countable hence the set $G$ is also countable. Hence $$A=\bigcap_{S\in G}S$$

Can we write the last statement ?

答:
Let $C \subseteq \Bbb{R}$ be a closed subset, then $U := C^c$ is open and can be written as a countable union of open intervals with rational endpoints.

By De Morgan’s Laws, that means $C$ is a countable intersection of closed rays with rational endpoints.

So it suffices to show that any closed ray is a countable intersection of open sets. Since countably many closed rays intersect to form $C$, we would then have that $C$ can be expressed as a countable intersection of countable intersections of open sets, which is also a countable intersection of open sets.

But this is really easy: If our closed ray is unbounded above, i.e. of the form $R := [a, \infty)$ for $a \in \Bbb{Q}$, then our desired countable intersection of open sets would just be $$[a, \infty) = \bigcap_{n \geq 1} (a - 1/n, \infty).$$ You can mimic that construction in the case where the ray is unbounded below to get the same result, that the closed ray is a countable intersection of open sets. QED.


问:
Seems that I just proved $2=4$.
Solving $x^{x^{x^{.^{.^.}}}}=2\Rightarrow x^2=2\Rightarrow x=\sqrt 2$.

Solving $x^{x^{x^{.^{.^.}}}}=4\Rightarrow x^4=4\Rightarrow x=\sqrt 2$.

Therefore, $\sqrt 2^{\sqrt 2^{\sqrt 2^{.^{.^.}}}}=2$ and $\sqrt 2^{\sqrt 2^{\sqrt 2^{.^{.^.}}}}=4\Rightarrow\bf{2=4}$.

What's happening!?

答:
You have merely shown that the equation $\sqrt 2^y = y$ has more than one solution.

Then you assumed that $x^{x^{x^\ldots}}$ somehow made sense and tried to talk about it as if it meant "the solution $y$ of $x^y = y$". Which of course is nonsense when the equation has several solutions.

Title: Seems that I just proved $2=4$.


问:
The fastest way to multiply 2D matrices
I read in the article about the machine learning interview, there is a question:

"To multiply 2 matrices with the shape 10000x10000. How to speed up the calculation?"

The candidate said: "Multi-thread", but the interviewer told that there was another way to do that, the mathematics way.

Can someone give me a solution? Is it related to "eigenvalue"?

答:
The question was not fair, because if you don't know the answer it is hard to guess. Most probably, the interviewer was referring to the fast(er) multiplication method of Strassen and similar. https://en.wikipedia.org/wiki/Strassen_algorithm#:~:text=In%20linear%20algebra%2C%20the%20Strassen,algorithms%20for%20extremely%20large%20matrices.

The basic idea is that if you decompose both matrices each in four blocks, you can perform the product in 7 block multiplies instead of the expected 8. And by applying this principle recursively, you break the $O(n^3)$ barrier.

But I fail to see how this belongs to machine learning.

Title: The fastest way to multiply 2D matrices


问:
sequence/series limits
I am struggling a bit with the following two questions I got from a sister who is doing BSC in Applied Mathematics. I always try my best to help her but was stuck on these questions and so any answer and explanation is much appreciated:

Here's the question:

Decide which of the following sequences have limits when $n\rightarrow \infty$ , and find the relevant limits if they exist. In the case of $C(n)$ it is enough to show that the limit is bounded from above and below.

Sequence $B(n)$ such that
$B(0) = 0$ and $B(n) = B(n-1) + 1/2^n$ , for any positive integer $n$.

Sequence $C(n)$ such that $C(1) = 1$ and $C(n) = C(n-1) + 1/n^2$ , for any integer $n>1$

(Hint: Use the fact $\frac{1}{n-1} -\frac{1}{n} =\frac{1}{(n(n-1))}$ with the sandwich reasoning that if $C'(n)<C(n)<C''(n)$ for all $n$, then the limit of $C(n)$ is between the limits of $C'(n)$ and $C''(n)$, if such limits exists.)

On the first question , I tried to plug some values of n and got the limit of B(n) as 1, although I am not 100% sure. But on the second one, I couldn't do much and the hint didn't help me a lot.

Title: sequence/series limits

Body:
$B(0)=0$, $B(n)=B(n-1)+\frac 1 {2^{n}}$,
Thus $$B(1)= \frac 1 2,B(2)= \frac 1 2+\frac 1 4,B(3)=\frac 1 2+ \frac 1 4+\frac 1 8,\cdots B(n)= \frac1 2+\cdots \frac 1 {2^{n}} $$
Now see $$\frac{1}{2} B(n)= \frac 1 4+\frac1 8+\cdots \frac 1 {2^{n+1}}\\\implies \left(1-\frac 1 2\right)B(n)=\frac 1 2-\frac 1 {2^{n+1}}\\\implies \lim_{n\to +\infty}B(n)=1$$
$C(1)=1,C(n)=C(n-1)+\frac 1 {n^2}$

Similarly we get, $C(n)=1+\frac 1 {2^2}+\frac 1 {3^2}+\cdots +\frac 1 {n^2}$
Observe $$\frac 1 {2^2}+\frac 1 {3^2}<\frac 2 {2^2}=\frac 1 {2}\\\frac 1 {4^2}+\frac 1 {5^2}+\frac 1 {6^2}+\frac 1 {7^2}<\frac 4 {4^2}=\frac 1 {4}\\\cdots\\$$
Similarly we can say that $C_{2^{n+1}-1}<1+\frac 1 {2}+\frac 1 {4}+\cdots +\frac 1 {2^n}=B(n)+1$
Therefore $\lim C(n) <\lim(B(n))+1$, Using the fact that $C(n)$ is increasing you can conclude that $\lim C(n)$ exists.


问:
Them four primes
This is not a problem from any text, but a curiosity into the pattern of prime numbers.
It is known that 11, 13, 17, 19 are primes.
It is also known that 101, 103, 107, 109 are primes.
Is there any other pair of twin primes, that occur for the unit digit 1, 3, 7, 9 and the same preceding digits of the number? If yes, are they finite or infinite? Is there any pattern or logic to find them?</p>

答:
These are sometimes called prime decades, and they represent the totality of prime quadruplets with the exception of $\{5,7,11,13\}$. See also OEIS A007530.

It is unknown whether or not there are infinitely many. Certainly we don't have a proof of there being infinitely many, because that would imply the twin prime conjecture.

Title: Them four primes


问:
Quadratic covariation of two Compound Poisson Process.
This is not a homework problem but a curiosity question and hence the question may not be well built but I want the community to help me in understanding this concept and hence answer it. If you think the question can be framed better please do tell me. I would really really appreciate any help in the form of comments/answers.

I want to find out the Quadratic covariation of two semimartingales $Z_1,Z_2$ where both are compound poison processes and they are correlated by say a factor of $\rho_1$.

If $Z_1(t)=\sum^{N_1(t)}Y_i$ and $Z_2(t)=\sum^{N_2(t)}U_j$ are two compound poisson process then what will be the quadratic covariation of $Z_1,Z_2$ , i.e $[Z_1,Z_2](t)$ ?

In this $N_1$ and $Y_i$ are independent and $N_2$ and $U_j$ are independent and with $Y_i$ is an i.i.d and also $U_j$ is an i.i.d. We can assume that $E(Y_i)=\mu_1$ and $E(U_j)=\mu_2$. We can also assume that the jumps $N_1,N_2$ are correlated by $\rho_2$ and $Y_i,U_j$ are corrleated by a factor of $\rho_3$ (if needed).

One of the ways in which I thought was to calculate
$[Z_1,Z_2](t)=Z_1(t)Z_2(t)−Z_1(0)Z_2(0)−\int^t_0Z_1(s)dZ_2(s)−\int^t_0 Z_2(s)dZ_1(s)$
where we will have $dZ_1(s)=Y_iN_1(s)$ and $dZ_2(s)=U_jN_2(s)$.

I am open to other methods to calculate the same as I was unable to proceed much in above. Also if you can find a reference please mention the link below.

答:
The following is a simple model for correlated compound processes. Let $ N=\{N_t,t\ge 0\} $ be a Poisson process with $ \mathsf{E}[N_t]=\lambda t $
and $ Y=\{ ( Y_k^{(1)} ,Y^{(2)}_k), k\ge 1 \} $ be a  sequence of i.i.d. 2-d random variables, with
$$  \mathsf{E}[Y^{(i)}_k Y^{(j)}_k]=m_{ij},\qquad i,j=1,2,\quad k\ge 1 $$
Also, $ Y $ and $ N $ are mutual independent. Now define $X=\{X_t=(X^{(1)}_t,X^{(2)}_t), t\in\mathbb{R}_+\}$ as following:
$$ X^{(i)}_t=\sum_{k=1}^{N_t}Y_k^{(i)}, \qquad i=1,2. $$
Then $X^{(1)}=\{ X^{(1)}_t,t\in\mathbb{R}_+\}$, $X^{(2)}=\{ X^{(2)}_t,t\in\mathbb{R}_+\} $
are correlated compound Poisson Processes. Using the general results of  compound Poisson Processes we can get the quadratic covariation $[X^{(i)},X^{(j)}] $ as following:
$$  [X^{(i)},X^{(j)}]_t=\sum_{s\le t} \Delta X^{(i)}_s \Delta X^{(j)}_s
= \sum_{k=1}^{N_t}Y_k^{(i)} Y_k^{(j)},\qquad i,j=1,2 $$
And the predictable quadratic covariations are
$$  \langle X^{(i)},X^{(j)}\rangle_t=m_{ij}\lambda t,\qquad i,j=1,2
$$
About the general results of stochastic calculus and compound Poisson processes please refer to the following books:

J. Jacod and  A. N. Shiryayev, Limit Theory for Stochastic Processes, 2ed. Springer, 2003.

A. E. Kyprianou, Fluctuations of Lévy Processes with Applications, 2nd Ed., Springer, 2014.

D. Applebaum, Lévy Processes and Stochastic Calculus, 2nd Ed, Cambridge University Press, 2009.


问:
Theorem 2.17 from RCA Rudin

$\bf 2.17\ $ Theorem $\ $ Suppose $X$ is a locally compact, $\sigma$-compact Hausdorff space. If $\frak M$ and $\mu$ are as described in the statement of Theorem $\it 2.14$, then $\frak M$ and $\mu$ have the following properties:

$(a)\ \ $ If $E\in\frak M$ and $\epsilon>0$, there is a closed set $F$ and an open set $V$ such that $F\subset E\subset V$ and $\mu(V-F)<\epsilon$.

  $(b)\ \ $ $\mu$ is a regular Borel measure on $X$.

  $(c)\ \ $ If $E\in\frak M$, there are sets $A$ and $B$ such that $A$ is an $F_\sigma$, $B$ is a $G_b$, $A\subset E\subset B$, and $\mu(B-A)=0.$

$\rm P\scriptstyle{\rm ROOF}$

$\quad$ Every closed set $F\subset X$ is a $\sigma$-compact, because $F=\bigcup(F\cap K_n)$. Hence $(a)$ implies that every set $E\in\frak M$ is inner regular. This proves $(b)$.

I understood the proof of points $(a)$ and $(c)$. But I can't understand the proof of $(b)$. It's obvious that every closed set is $\sigma$-compact. But how Rudin applies $(a)$ here?

We have to show that if $\alpha>0$ then exists compact set $K\subset E$ such that $\mu(K)>\alpha$.

Can anyone explain it to me please?

答:
$\forall{E\in\mathfrak{M}}$, we can find closed $F$ such
that $\mu\left(E-F\right)<\epsilon$ (using part (a) of the theorem). Also $F$ is $\sigma-$compact
because $F=\cup\left(F\cap K_{n}\right)$. Note that $\left(F\cap K_{n}\right)$
is compact using Theorem 2.5 Corollary (b). Hence, $\mu\left(F\cap K_{n}\right)<\infty$
from Theorem 2.14 (b). Hence $F$ has $\sigma-$finite measure. Hence
${E}$ has $\sigma-$finite measure, since $E=F\cup\left(E-F\right)$,
and hence ${E}$ is inner regular (If $E$ has $\sigma$-finite measure, then $E$ is inner regular).

Title: Theorem 2.17 from RCA Rudin


问:
The probabilistic method exercise 2.7.4
This is exercise 2.7.2 of the book "the probabilistic method".

Suppose $p>n>10m^2$, with $p$ prime, and let $0<a_1<a_2<...<a_m<p$ be integers. Prove that there is an integer $x$, $0<x<p$, for which the $m$ numbers $(xa_i(\mod p))\mod n$, $1 \le i \le m$ are pairwise distinct.

And I think this problem in this way. Let $\chi_x$ be the indicator for the event that the m numbers $(xa_i(\mod p))\mod n$ are pairwise distinct. Then $P(\chi_x) = \frac{n\choose m}{n^m}$, The the expected number of $(xa_i(\mod p))\mod n$ are pairwise distinct is $\sum E[\chi_x] = (p-1)\frac{n\choose m}{n^m} > \frac{(\frac{n}{m})^m}{n^{m-1}} = \frac{n}{m^m}$. Obviously, this way can not get the result that the expected number is greater than 1. Could anyone help about this problem?

答:
I got an answer about this problem but I don't know if it is right.

For any $a_i,a_j$,let $A_x$ denote the event that $(xa_i(\mod p))\mod n = (xa_j(\mod p))\mod n$, then $P_r(A_x) = \frac{1}{n}$. The probability that all m numbers are pairwise distinct is $(1-\frac{1}{n})^{m \choose 2}$. Then the expected number of x is $(p-1)(1-\frac{1}{n})^{m \choose 2} \ge (10m^2 - 1)(1-\frac{1}{10m^2})^{\frac{1}{2}m(m-1)} \ge (10m^2 - 1)(1-\frac{\frac{1}{2}m(m-1)}{10m^2}) \ge 10m^2-1-\frac{1}{2}m^2 + \frac{1}{2}m > 1$, which yields the result when $m \ge 1$.

Title: The probabilistic method exercise 2.7.4


问:
Is this relation true for orthonormal matrices elements?
Let $R$ be a $3 \times 3$ proper orthogonal and normal matrix such that $RR^T = I$ and $det(R) = 1$, so basically a rotation matrix.

$R = \begin{bmatrix} r_{11} & r_{12} & r_{13} \\ r_{21} & r_{22} & r_{23} \\ r_{31} & r_{32} & r_{33} \\  \end{bmatrix}$

Is the following relation true?

$r_{13} = r_{32}r_{21} - r_{22}r_{31}$

答:
Let $\{e_1, e_2, e_3\}$ be the canonical basis and $u$ the rotation whose matrix is $R$. The given relation is equivalent to

$$\langle u(e_3), e_1 \rangle = \langle u(e_1) \times u(e_2), e_1 \rangle$$

$u(e_1), u(e_2)$ are unit orthogonal vectors. $u$ being a rotation, $u(e_1) \times u(e_2)$ is a unit vector equal to $u(e_3)$. Therefore the given relation is true. The very true relation is

$$u(e_3)= u(e_1) \times u(e_2).$$


问:
proper differentiable maps of manifolds
Is the following statement true?If yes, why?

Let $f: M\to N$ be a proper morphism between smooth manifolds.  Let $x$ be a point of $N$, and $U$ a nbhd of $f^{-1}(x)$ in $M$. Then there exists a nbhd $V$ of $x$ in $N$ such that $f^{-1}(V)\subset U$.

Here, proper means that the preimage of any compact set is compact.

It seems to me this was used in a expository article that I am reading.  If it is true, I expect it to be true for proper morphism of locally compact topological spaces. But for some reason I wasn't able to find a proof.

Thank you.

答:
You are quite correct that this is a purely topological statement. Proper maps  are closed and here is a proof that works for closed maps between topological spaces, not necessarily manifolds nor even locally compact .

Proof:  Since $U$ is open in $M$, $M \setminus U$ is closed, hence (by the assumption "closed map" )  so is $F\subset N$ defined by  $F=f(M\setminus U) $.   Just take $V=N\setminus F$. Done.

Edit: Since the issue of "properness" is a very interesting one, I hope you will excuse the following development, prompted by Theo's comment.

The principal property of proper maps is that they are closed and I think this should be built into the definition (obviously Algebraic Geometry makes its influence felt here).  So a nice definition  is that a continuous map $f:X\to Y$ between topological spaces is proper if it is closed  and if the inverse images of points in $y \in Y$ are compact subsets $f^{-1}(y) \subset X$ [technically some say "quasi-compact" subsets to stress that no Hausdorff condition is assumed].   The link with the usual notion of "properness" is then the following theorem (in which proper has the meaning just given), whose assumptions are pleasantly weaker than one might expect. The point I advocate is that one should learn it once and for all, and that then you can often give short elegant proofs in the context of properness.

Theorem   If $X$ is Hausdorff and $ Y $ locally compact, then $f$ is proper if and only if the inverse image of a compact subset $K\subset Y$ is a compact subset $f^{-1}(K) \subset Y$.

(The proof is in Chapter 1 of Bourbaki's book on General Topology, which adopts this point of view)

Title: proper differentiable maps of manifolds


答:
Following the idea of  @Rivers McForge:

Write every open subset as a union of open intervals.  We can write any open interval as
the union of those closed intervals with rational points contained inside it. Therefore,
every union of open intervals is also a union of some closed intervals with rational points. Now note that there are only countably many closed intervals with rational points.  Hence every open set is a countable union of closed intervals.

Title: A closed set is intersection of countable collection of open sets

Body:
I am currently reading Apostol Mathematical Analysis. There was this question

Question : Prove that a closed set in $\mathbb{R}^1$ is the intersection of countable collection of open sets.

[Here $N_a(\varepsilon) =$ the open set $(a-\varepsilon, a+\varepsilon)$ ]

My Attempt : Let $A$ be a closed set. Let $A^c$ denote the complement of $A$. Take $G$ be a collection of neighborhoods where for each $N_a(\varepsilon)\in G $ where $a\in \mathbb{Q}\cap A^c=A'$(Let) and $\varepsilon$ is the minimum real number such that $A\subseteq N_a(\varepsilon) $

As the set of rational numbers are countable hence the set $G$ is also countable. Hence $$A=\bigcap_{S\in G}S$$

Can we write the last statement ?


问:
Proving $\frac{7 + 2b}{1 + a} + \frac{7 + 2c}{1 + b} + \frac{7 + 2a}{1 + c} \geqslant \frac{69}{4}$.
Here's the inequality

For positive variables, if $a+b+c=1$, prove that
$$
\frac{7 + 2b}{1 + a} +
\frac{7 + 2c}{1 + b} +
\frac{7 + 2a}{1 + c} \geqslant
\frac{69}{4}
$$

Here equality occurs for $a=b=c=\frac{1}{3}$ which is not-so-usual, so I decide to write the inequality as
$$
\frac{21 + 2q}{3 + p} +
\frac{21 + 2r}{3 + q} +
\frac{21 + 2p}{3 + r} \geqslant
\frac{69}{4}
$$Where the constraint now is $p + q + r = 3$ and the equality occurs for $p = q = r = 1$. Now we are left to proving that just
$$
2\sum_{cyc}{\frac{q}{3 + p}} +
21\sum_{cyc}{\frac{1}{3 + p}} \geqslant \frac{69}{4}
$$
Now it is sufficient to prove that
$$
\sum_{cyc}\frac{q}{3 + p}\geqslant\frac{3}{4} \quad \textrm{and} \quad \sum_{cyc}\frac{1}{3 + p} \geqslant \frac{3}{4}
$$ The second is true but I can't prove that the first is true.

答:
Another way:flipping the inequality it suffices to prove $$\sum_{cyc}\frac{7a-2b}{1+a}\le 15/4...(3)$$.

Indeed by jensen  on $f(x)=\frac{x}{1+x}$ $$\sum_{cyc} \frac{a}{1+a}\le 3\sum_{cyc}\frac{\frac{(a+b+c)}{3}}{1+\frac{(a+b+c)}{3}}=3/4...(1)$$

This can also be done like this :
as $$\frac{x}{1+x}\le \frac{9x+1}{16}$$ which is just ${(x-1/3)}^2\ge 0$
$$\sum \frac{a}{1+a}\le \sum \frac{9a+1}{16}=3/4$$

Also similar to my previous answer (by Titu's lemma) $$\sum_{cyc}\frac{b}{1+a}=\sum \frac{b^2}{b+ab}\ge 3/4...(2)$$

By $(1)$ and $(2)$  inequality $(3)$ is completed.


问:
Checking solution to probability questions
Q1: There are four electronic components in a system named A,B,C and D.
Let P (A)=P(component A works)=0.6. Similarly, P (B) = 0.4, P (C) = 0.5, P (D) =
0.3. The system will work only if components A and B both work or components C and D both work. Find
the probability that the system works. All components work independently.

Q2: Assume that births of boys and girls are equally likely and occur independently. Suppose that a family has seven children and it is known that there are
at least 5 boys. What is the probability that there are exactly 6 boys?

My work for Q1:

$P(A \cap B)= (0.6)(0.4)=0.24$,

$P(C \cap D)= (0.5)(0.3)=0.15$

Required probability=$P((A \cap B) \cup (C \cap D))= P(A \cap B) + P(C \cap D) - P(A \cap B \cap C \cap D)$

$= 0.24 + 0.15 - (0.6)(0.4)(0.5)(0.3)$

$=0.354$ (Final answer)

My work for Q2:

$P(B)= 0.5,P(G)=0.5$

Required probability: $P(6B)| At least 5B)$ $= \frac{ P(6B \space \cap \space At least 5 B)}{P(At least 5B)}$

$= \frac{ P(6B \space \cap \space (5B \cup 6B \cup 7B))}{P(At least 5B)}$

$= \frac {P(6B)}{P(at least 5B)}$

Then $P(5B)= \binom{7}5 (0.5)^5 (0.5)^2 = \frac{21}{128}$

$P(6B)= \binom{7}6 (0.5)^6 (0.5)^1 = \frac{7}{128}$

$P(7B)= \binom{7}7 (0.5)^7 (0.5)^0 = \frac{1}{128}$

$ \frac {P(6B)}{P(at least 5B)} = \frac{P(6B)}{P(5B) +P(6B) + P(7B)} = \frac{7}{29}$ (Final answer)

Is my logic and reasoning correct for both of these questions?

答:
Your work is nice but I would suggest an easy way to solve both the exercises (useful to save time in the solutions)

Q1:

It's easier find the complement probability that both branches do not work

$$1-0.76\times 0.85=0.354$$

Q2:

It's easier  just counting the possible and favourable combinations (the probability of one single realization is always the same, thus it can be wasted in numerator and denominator)

$$\frac{\binom{7}{1}}{\binom{7}{2}+\binom{7}{1}+\binom{7}{0}}=\frac{7}{21+7+1}=\frac{7}{29}$$


问:
Proving that $(\sqrt 3+\sqrt 2)/(\sqrt 2)$ is irrational
Can you please help me prove this ? I can prove that $\sqrt p$, where p is prime is irrational, also that the sum $\sqrt 3 + \sqrt 2$ is irational, bud dont know how to prove that the whole fraction is irrational. Thanks for any answer.

答:
$$\frac{\sqrt3+\sqrt2}{\sqrt2}=1+\sqrt{3/2}$$
The whole fraction is irrational iff $\sqrt{3/2}=\frac12\sqrt6$ is irrational iff $\sqrt6$ is irrational. The last statement you should be able to prove.


答:
If $\frac{\sqrt3+\sqrt2}{\sqrt2}\in\mathbb Q$ so $\sqrt{\frac{3}{2}}\in\mathbb Q.$

Now, let $\sqrt{\frac{3}{2}}=\frac{m}{n},$ where $m$ and $n$ are naturals such that  $\gcd(m,n)=1$, and get a contradiction.

Can you end it now?

Title: Proving that $(\sqrt 3+\sqrt 2)/(\sqrt 2)$ is irrational

Body: Can you please help me prove this ? I can prove that $\sqrt p$, where p is prime is irrational, also that the sum $\sqrt 3 + \sqrt 2$ is irational, bud dont know how to prove that the whole fraction is irrational. Thanks for any answer.


问:
What's the solution to $ 2|2x - a| < |x + 3a|$ (found critical values)
$$ 2|2x - a| < |x + 3a|$$

I found the critical values to be $x = \frac{5a}{3}$ and $x = \frac{-a}{5}$ by removing the abs val signs and equation the left side to the +/- right side.

Now this is an inequality, and I have an arbitrary constant $a$. I'm not sure how to find the range.

P.S.: I'd prefer to learn how to solve it algebraically in particular. Also, if I knew a, I'd take those critical values and test inputs between and beyond those values. Not sure what to do here.

答:
For $a=0$ our inequality has no solutions.

But for $a\neq0$ it's equivalent to $$4(2x-a)^2<(x+3a)^2$$ or
$$15x^2-22ax-5a^2<0$$ or
$$\frac{11a-14|a|}{15}<x<\frac{11a+14|a|}{15}.$$