问:
A box contains 10 marbles: 6 white ones and 4 red ones. A person picks 3 marbles at once.
Question: A box contains 10 marbles: 6 white ones and 4 red ones. A person picks 3 marbles at once. What is the probability that the person picks at least one white marble?

So what confuses me here is picking the marbles at once. Isn't this the same as picking one-by-one without replacement? I came up with ((6/10) + (6/9) + (6/8)) / 3 but am not sure this is the right answer.

答:
Probability that the person picks only red marbles (no white marbles),

$P(3R) =  \displaystyle \frac{4}{10} \times \frac{3}{9} \times \frac{2}{8} = \frac{1}{30}$

This is same as saying $P(3R) = \displaystyle  \frac{4 \choose 3}{10 \choose 3}$

The desired probability $ = 1 -  P(3R) = \frac{29}{30}$


答:
It's easiest to consider the complement: the probability that the person draws no white marble, so only red ones. There are $\binom{10}{3}$ ways to draw three marbles form $10$ in one go and $\binom{4}{3}$ ways to draw $3$ red ones (out of the four available ones).

So the answer is $$1- \frac{\binom{4}{3}}{\binom{10}{3}}$$

It indeed is the same as picking one-by-one without replacement. But in your answer you compute the chance of $3$ white ones, not at least one white one, so much smaller.

ContentLicense: CC BY-SA 4.0

Body:
Question: A box contains 10 marbles: 6 white ones and 4 red ones. A person picks 3 marbles at once. What is the probability that the person picks at least one white marble?

So what confuses me here is picking the marbles at once. Isn't this the same as picking one-by-one without replacement? I came up with ((6/10) + (6/9) + (6/8)) / 3 but am not sure this is the right answer.


问:
Orthogonal Complements of null space and row space
I am reading a section of linear algebra book. And I am not sure whether my understanding is correct or not for the sentence that I highlight.

DEFINITION: The orthogonal complement of a subspace $V$ contains every vector that is perpendicular to $V$. This orthogonal subspace is denoted by $V^\perp$. (pronounced "V perp").

By this definition, the nullspace is the orthogonal complement of the row space. Every $x$ that is perpendicular to the rows satisfies $Ax = 0$, and lies in the nullspace.

The reverse is also true. If $v$ is orthogonal to the nullspace, it must be in the row space. Otherwise we could add this $v$ as an extra row of the matrix, without changing its nullspace. The row space would grow, which breaks the law $r + ( n - r) = n$.

We conclude that the nullspace complement $N(A)$ orthogonal is exactly the row space $C(A^T)$.

I understand every sentences separated. I see adding $v$ that breaks the law. I don't understand why suddenly jump to "We conclude.....".

答:
From the second paragraph (the paragraph after the definition), we know that all elements of the column space are orthogonal to the nullspace. That is, we can deduce that $C(A^T) \subseteq N(A)^\perp$.

From the third paragraph, we know that every $v$ that is orthogonal to the nullspace is also an element of the row space. That is, $N(A)^\perp \subseteq C(A^T)$.

Because $N(A)^\perp \supseteq C(A^T)$ and $N(A)^\perp \subseteq C(A^T)$, it must be the case that $N(A)^\perp = C(A^T)$.


问:
The slope of the curve having the equation $x^2-2xy+y^2+2x-4=0$
Please can anyone help me solve the problem:

The equation of a curve is given by $x^2-2xy+y^2+2x-4=0$, find slope of the curve at the point $(2, 2)$.

This is new to me. How to find the derivative of this function?

答:
Approach 1:

Rewrite your equation as: $y^2 - 2xy = 4-2x-x^2$

Taking the derivative with respect to $x$, we have:

$$2 y y' -2x = -2 -2x$$

Can you take it from here?

Approach 2:

Your equation is a quadratic in $y$. Solve for $y$ and then take the derivative to find the slope.

ContentLicense: CC BY-SA 3.0


问:
If $A^p$ is ergodic, does that make $A$ ergodic? And possible circular proof?
I am reading Walters' Introduction to Ergodic Theory and am struggling to make sense of the proof of Theorem 1.10. I have an issue with understanding the overall structure of the proof, and with a component of the proof itself.

Issue regarding the proof:

The sentence

So if $a_n \neq 0$... $\quad$ ... and so $f$ is constant a.e.

in my mind just doesn't follow. Doesn't this show that $A^p$ is ergodic, not $A$?

Issue regarding the structure:

At the beginning of the proof, an assumption is made and is later invoked to make a conclusion. How does saying

Suppose that whenever $\gamma A^n = \gamma$ for some $n \geq 1$ we have $\gamma \equiv 1$.

allow us to use that later on to prove that having the trivial character is the only $\gamma \in \hat{G}$ that satisfies $\gamma \circ A^n = \gamma$. It seems rather circular?

Appreciate any help.

答:
If $A^p$ is ergodic, then $A$ must be ergodic: the contrapositive of this statement says that if $A$ is not ergodic, then $A^p$ is not ergodic, which is easy to verify - every $A$-invariant function is also $A^p$ invariant.

Regarding the second issue: the assumption that "whenever $\gamma A^n=\gamma$ for some $n\geq 1$ we have $\gamma = 1$" is the second part of the "if and only if" assertion in the theorem, so it must be assumed to prove one direction of the "if and only if".


问:
Conditional Expectation for Mean Excess Loss
Let $X$ be a random variable with support on $[0,\infty)$. The mean excess loss can be defined as:

$$E(X-d|X>d)$$

prove that the mean excess loss is equal to:

$$\frac{\int_d^\infty(x-d)f_X(x)dx}{1-F_X(d)}$$

I understand that the denominator since that's just $P(X>d)$. I don't understand the numerator. How is the numerator obtained? What is its interpretation? I don't understand how to prove this.

答:
Lets start from the definition of conditional probability.

Let A and B be two random events with $Pr(A)>0$, than the probability that B happens given that A happened is defined as $$Pr(B\mid A)=\frac{Pr(A\bigcap B)}{Pr(A)}$$
whre $A\bigcap B$ is the event that A and B happen concomitantly.
Now let $A=X>d$ and $B=X>d+y$. It is easy to see that $A\bigcap B=X>d+y$. From these we can write $Pr(A)=Pr(X>d)=1-F(d)=S(d)$, where $S(d)$ is the survival function, and $Pr(A\bigcap B)=Pr(B)=Pr(X>d+y)=S(d+y)$.

Putting all together we get

$$Pr(X>d+y\mid X>d)=\frac{S(d+y)}{S(d)}$$

The event $X>d+y\mid X>d$ can be rewritten as $X-d>y\mid X>d$, which gives us
$$S(y\mid X>d)=\frac{S(d+y)}{S(d)}$$

From the last equation we derive the conditional distribution and the conditional pdf

$$F(y\mid X>d)=1-\frac{S(d+y)}{S(d)}$$

$$\boxed{f(y\mid X>d)= \frac{dF(y\mid X>d)}{dy}=\frac{f(d+y)}{S(d)}=\frac{f(d+y)}{1-F(d)}}$$
Finally from the last expression we can calculate the Mean Excess Loss given by
$$E(Y\mid X>d)$$ where $Y= X-d$

$$E(Y\mid X>d)=\frac{1}{S(d)}\int_{0}^{\infty}yf(d+y)dy$$

making the change of variable $x=d+y$ in the last integral gives the final result

$$\boxed{E(X-d\mid X>d)=\frac{1}{S(d)}\int_{d}^{\infty}(x-d)f(x)dx}$$


答:
P(at least one white) = 1 - P(all red)

For the probability of all red you should count the number of ways you can choose three red marbles and divide that by the number of ways you can choose three marbles:

$P(AllRed) = (4*3*2)/(10*9*8)$

And the answer is:

$P(At Least One White) = 1 - (4*3*2)/(10*9*8) = 0.9666667$

Title: A box contains 10 marbles: 6 white ones and 4 red ones. A person picks 3 marbles at once.


问:
"Translating" one value $- \infty$ to $+ \infty$ to another ($+ \infty$ to $ \gt 0$)
Well even if I need to use the following in a computer game this is a math question. I have a world map which I can scroll with a scroll velocity $(f(x))$ with my mouse. And I have a zoom factor $(x)$ which is $0$ in its initial state and can be positive infinite and negative infinite.

So I'm searching for a function where:

$x$ can range from $- \infty$ to $+ \infty$

$f(x)$ should be $\gt 0$

$f(x)$ should tend to $0$ for $x$ tending to $+ \infty$

$f(x)$ should tend to $+ \infty$ for $x$ tending to $- \infty$

I bet the solution is somewhat so obviously easy I'm just to blind to see in the moment.

答: By request, try $f(x)=e^{-x}$.  As @Andy points out, there are variations and tweaks one can make to the curve, such as $f(x)=e^{-Ax+B}$, where $A>0$ and $B\in \mathbb{R}$, which will give it different shape and move it around.</p>


问:
Proving a point is interior

Let $D = \left \{ (a,b,c,d,e) : ax^4+bx^3+cx^2+dx+e=0 \text{ has a real root} \right \}$

Show that (1,2,-4,3,-2) is an interior point in $D$.

My only idea was trying to somehow use the implicit function theorem, but I can't find how.

Hints appreciated.

答:
Here is a hint: is there any way you can take advantage of the structure of the polynomial, specifically with the coefficients you have been given to show that polynomials with close coefficients also have real roots?

ANSWER BELOW:

First we notice that $c = (1,2,-4,3,-2)$ is an element of $D$. To show that it's an interior point, we need to find an open ball around $c$ where each element of the ball is in $D$.

So let $B=B_{1/2}(c)$ be the open ball around $c$ of radius $1/2$. Then a point $d \in B$ may be written:

$$d = c+b = (1+b_1, 2+b_2, -4+b_3, 3+b_4, -2+b_5)$$

where $||b|| < 1/2$. Now notice that each component of $b$ can only be of magnitude less than $1/2$, so the fifth component of $d$ will always be negative. Moreover, the first component of $d$ will always be positive. Then any polynomial $d(x)$ with coefficient vector $d$ will have a negative value at $0$ and a positive value for large enough $x$. As polynomials are continuous, this implies that $d(x)$ has a real root.

Then every $d \in B$ is in $D$, hence we have an open ball centered at $c$ entirely within $D$, therefore $c$ is interior to $D$.

Title: Proving a point is interior


问:
How many N-digit numbers have the sum of odd places digits equal the sum of even places digits?
We define a N-digit number as a sequence $d_{1}d_{2}d_3...d_N$, where $d_k\in\{0,1,2,3,4,5,6,7,8,9\}$. So the first digits can be equal to zero. N is a positive even integer.
The problem is to find, how many N-digit numbers are there, which have their sum of the digits on odd places equal to the sum of digits on the even places. Example number would be 2563 for 4-digit numbers as 2 + 6 = 5 + 3.

My thought was to choose the odd digits, we can do that in $10^{N/2}$ ways and then for each of that combinations make even digits a permutation of them ($\frac{N}{2}!$), but this, if correct at all, is only a small fraction of all the possibilities. Why? Because the digits do not have to be the same. Only the sums have to be equal.

Then my second though was that each of such numbers consists of digits from other (shorter) numbers that satisfy the requirement, but it is not true since only the sums have to be equal.

We can for each sum of odd/even digits calculate how many ways we can express it on N/2 digits and then sum the squares of such numbers of combinations. However that leads to very unfriendly calculations.

After you find a nice way, please provide some example calculations for N=4 and N=6.

答:
The problem is considered in the series of papers here (in Russian), where the required number $C_N$ is called the number of $N$-digital lucky tickets. It is known that

$$C_N=\frac 1{\pi}\int_0^\pi \left(\frac {\sin 10 x}{\sin x}\right)^N dx\approx \frac{10^N}{\sqrt{33\pi n}}.\label{1}\tag{1}$$
The relative error of the approximation is no more than $4\%$ and quickly decreases when $N$ grows.

For small $N$ in the second paper are provided calculations of $C_N=\sum _{k=0}^{9N}\mathcal{A}_k^2$, mentioned by Phicar. Namely, for given $n$, $\mathcal{A}_k$ is denoted by $N_n(k)$, where $n=N/2$ and is provided a recurrent formula

$$N_n(k)=\sum_{l=0}^9 N_{n-1}(k-l), \label{2}\tag{2}$$
$$\begin{matrix}
\mbox{where if $k<9$ then $N_{n-1}(k-l)=0$ for $l>k$;}\\
\mbox{$N_1(k)$ equals $1$, if $0\le k\le 9$ and equals $0$, otherwise.}
\end{matrix}\label{3}\tag{3}$$

The recurrence allows to build a table for $N_n(k)$ for $n\le 4$ and then calculate $C_2=10$, $C_4= 670$, $C_6=55 252$, and $C_8=4 816 030$.

Moreover, following the observations from the second and third papers, we easily see that a map
$$d_{1}d_{2}d_3\dots d_N\mapsto d_1(9-d_2)d_3(9-d_4)\dots d_{N-1}(9-d_N)$$ is a bijection between the set of $N$-digit numbers which have their sum of the digits on odd places equal to the sum of digits on the even places and a set of $N$-digit numbers with the sum of the digits $9N/2$. That is, the size $C_N$ of the first set equals the size $N_{N}(9n)$ of the second set.

In the tiny third paper by means of generating functions is shown that
$$N_n(k)=C_{10}(n,k)=\sum_{i=0}^{\min\{n,k/10\}} (-1)^i {n\choose  i}{n+k-10i-1\choose  n-1}.$$

As an illustration, there is shown that
$$C_6=N_6(27)=C_{10}(6,27)= \sum_{i=0}^{2} (-1)^i {6\choose  i}{32-10i\choose  5}=$$ $${32\choose 5}-{6\choose 1}{22\choose 5}+{6\choose 2}{12\choose 5}=55 252.$$

I am reading the papers and updating the answer.


问:
Unique maximal ideal and ring epimorphism kernel with prime numbers equivalence
Let $A$ be a ring. Let $f: \mathbf{Z} \to A$ be a surjective ring homomorphism. Prove that $A$ has a unique maximal ideal iff there exists $n\in \mathbf{N}$ and $p\in\mathbf{N}$ a prime number such that $\ker f =(p^{n})$.

$(p^{n})$ denotes the ideal generated by $p^{n}$, I'm not sure if that's the standard notation.

We had to prove that in a recent exam and nobody could prove it. I haven't found it online neither. Thank you!

答: I assume by 'ring epimorphism' you mean 'surjective ring homomorphism'. Since $f$ is surjective we have $A\cong \Bbb Z/\ker f$. We know that the maximal ideals of $\Bbb Z/\ker f$ correspond bijectively to the maximal ideals of $\Bbb Z$ containing $\ker f$, hence $A$ has a unique maximal ideal iff there is a unique maximal ideal of $\Bbb Z$ containing $\ker f$. Now use the fact that $\Bbb Z$ is a PID.


答:
Hint By the first isomorphism theorem $A$ is isomorphic to $\mathbb Z / \mbox{Ker}(f)$.

$\Rightarrow$ Note that $\mbox{ker}(f)=(m)$ for some integer $m$. Assume by contradiction that $n$ is not a power of a prime. Since $n \neq 1$, there exist two primes $p \neq q$ which divide $m$. Show that $(p)/(m)$ and $(q)/(m)$ are maximal ideals in $\mathbb Z / \mbox{Ker}(f)$.

$\Leftarrow$: Show that $(p)/(p^n)$ is the only maximal ideal in $\mathbb Z/(p^n)$.

Second solution If you do not want to use the FIT, you can proceed as follows:

Since $\mathbb Z$ is a PID then $\mbox{ker}(f)= (m)$ for some integer $m$.

Claim $I$ is ideal in $A$ exactly when $I= f((k))$ for $k|m$.

Proof

$\Leftarrow$ follows from $f$ being surjective.

$\Rightarrow$ Let $I \subseteq A$ be any ideal. Then $f^{-1}(I)$ is an ideal in $\mathbb Z$ and hence
$$
f^{-1}(I)=(k)
$$
for some $k |m$.

[This comes from $(m) =\mbox{ker}(f) \subseteq f^{-1}(I) =(k)].

Since $f$ is onto, you get
$$
I=f((k))
$$

Now, your problem follows immediately from the claim.

Title: Unique maximal ideal and ring epimorphism kernel with prime numbers equivalence

Body:
Let $A$ be a ring. Let $f: \mathbf{Z} \to A$ be a surjective ring homomorphism. Prove that $A$ has a unique maximal ideal iff there exists $n\in \mathbf{N}$ and $p\in\mathbf{N}$ a prime number such that $\ker f =(p^{n})$.

$(p^{n})$ denotes the ideal generated by $p^{n}$, I'm not sure if that's the standard notation.

We had to prove that in a recent exam and nobody could prove it. I haven't found it online neither. Thank you!


问:
Verifying the solution to a vectorized matrix/vector equation
After struggling with a problem for some time I arrived at what I thought to be a solution. However, I could not verify the derived expression as a solution to my equation in numerical experiments. So my question is if it is the math or the code that is wrong, and to answer this I'm asking for your help to check my math.

For $\mathbf{Y}_n \in \mathbb{R}^{1 \times r}$, $\mathbf{U}_n \in \mathbb{R}^{1 \times r}$, $\mathbf{V} \in \mathbb{R}^{T \times r}$ and $\mathbf{S}_n \in \left \{ 0, 1 \right \}^{T \times T}$, my goal is to solve for $\mathbf{V}$ in
$$
\sum_n \left \{ - \mathbf{S}_n^\top \left ( \mathbf{Y}_n^\top - \mathbf{S}_n^\top\mathbf{V}\mathbf{U}_n^\top \right )\mathbf{U}_n + \mathbf{V} + \mathbf{D}\mathbf{V}  \right \} = 0.
$$
Note that $\mathbf{S}_n$ is really a shift matrix with ones on the superdiagonal/subdiagonal. The superdiagonal/subdiagonal may vary with each $n$ so usually $\mathbf{S}_n \neq \mathbf{S}_{n + 1}$. Expanding the LHS and rearranging I get
$$
\sum_n \left \{  \mathbf{S}_n^\top\mathbf{S}_n^\top\mathbf{V}\mathbf{U}_n^\top \mathbf{U}_n + \mathbf{V} + \mathbf{D}\mathbf{V} \right \} = \sum_n \left \{\mathbf{S}_n^\top \mathbf{Y}_n^\top \mathbf{U}_n  \right \}
$$
For future reference, let $\mathbf{\widetilde{S}}_n = \mathbf{S}_n^\top\mathbf{S}_n^\top$ and $\mathbf{\widetilde{U}}_n = \mathbf{U}_n^\top \mathbf{U}_n$. To solve for $\mathbf{V}$ I apply vectorization to each term in the summation using the following identities:
$$
vec(\mathbf{A}\mathbf{B}) = (\mathbf{I} \otimes \mathbf{A})vec(\mathbf{B}); \quad vec(\mathbf{A}\mathbf{B}\mathbf{C}) = (\mathbf{C}^\top \otimes \mathbf{A})vec(\mathbf{B})
$$
Here, $\mathbf{I}$ is the identity matrix and $\otimes$ represents the Kronecker product. The vectorization brings the equation I try to solve into the form
$$
\sum_n \left \{ \left [ (\widetilde{\mathbf{U}}_n ^\top \otimes \widetilde{\mathbf{S}}_n) + (\mathbf{I} \otimes \mathbf{I}) + (\mathbf{I} \otimes \mathbf{D}) \right ]vec(\mathbf{V}) \right \} = \sum_n \left \{ (\mathbf{U}_n^\top \otimes \mathbf{S}_n^\top)vec(\mathbf{Y}_n^\top) \right \}
$$
Now, I solve for $vec(\mathbf{V})$ and obtain
$$
vec(\mathbf{V}) = \sum_n \left \{ (\mathbf{U}_n^\top \otimes \mathbf{S}_n^\top)vec(\mathbf{Y}_n^\top) \left [ (\widetilde{\mathbf{U}}_n ^\top \otimes \widetilde{\mathbf{S}}_n) + (\mathbf{I} \otimes \mathbf{I}) + (\mathbf{I} \otimes \mathbf{D}) \right ]  ^{-1} \right \}
$$
However, reshaping this expression for $vec(\mathbf{V})$ it into a $T \times r$ matrix and plugging it into the first equation of this post, the output of my simulation is different from zero.

答:
You vectorized the equation correctly, but everything you did after that was incorrect. We have
$$
\sum_n \left \{ \left [ (\widetilde{\mathbf{U}}_n ^\top \otimes \widetilde{\mathbf{S}}_n) + (\mathbf{I} \otimes \mathbf{I}) + (\mathbf{I} \otimes \mathbf{D}) \right ]\operatorname{vec}(\mathbf{V}) \right \} =
\operatorname{vec}\left[ \sum_n \mathbf S_n^\top \mathbf  Y_n^\top \mathbf  U_n\right]
\\
\left[\sum_n \left [ (\widetilde{\mathbf{U}}_n ^\top \otimes \widetilde{\mathbf{S}}_n) + (\mathbf{I} \otimes \mathbf{I}) + (\mathbf{I} \otimes \mathbf{D}) \right ] \right]
\operatorname{vec}(\mathbf{V})
=
\operatorname{vec}\left[ \sum_n \mathbf S_n^\top \mathbf  Y_n^\top \mathbf  U_n\right].
$$
This is an equation of the form $\mathbf A \mathbf x = \mathbf b$, which has the solution $\mathbf x = \mathbf A^{-1}\mathbf b$. So, we have
$$
\operatorname{vec}(\mathbf{V})
=
\left[\sum_n \left [ (\widetilde{\mathbf{U}}_n ^\top \otimes \widetilde{\mathbf{S}}_n) + (\mathbf{I} \otimes \mathbf{I}) + (\mathbf{I} \otimes \mathbf{D}) \right ] \right]^{-1} \operatorname{vec}\left[ \sum_n \mathbf S_n^\top \mathbf  Y_n^\top \mathbf  U_n\right].
$$
This is not equivalent to the expression that you have presented.

You never state how many terms there are in the sum $\sum_n$. However, with this additional information, the sum that defines the matrix $\mathbf A$ can be simplified to
$$
\sum_{n=1}^N \left [ (\widetilde{\mathbf{U}}_n ^\top \otimes \widetilde{\mathbf{S}}_n) + (\mathbf{I} \otimes \mathbf{I}) + (\mathbf{I} \otimes \mathbf{D}) \right ] =
N \cdot \mathbf I \otimes (\mathbf {I + D}) + \sum_{n=1}^N (\widetilde{\mathbf{U}}_n ^\top \otimes \widetilde{\mathbf{S}}_n).
$$


问:
Analytic variety can be defined by finitely many equations
Let $(f_i)_{i\in I}$ be a family of holomorphic functions on $\mathbb C^n$. $V:=\{z\in\mathbb C^n\mid f_i(z)=0,\forall i\}$. I know that locally we could write down the equation with finitely many equations because the local ring $\mathscr{O}_0$ of the analytic sheaf is Noetherian.

But it this also true globally? First, we know that the ring of all holomorphic functions on $\mathbb C^n$ is not Noetherian.

I attempted like this assume $I_0(V):=\{f\in \mathscr{O}_0\mid f|_V=0\}$ is an ideal of  $\mathscr{O}_0$, so it is finitely generated say by $(g_1,...,g_m)$. Then when we localize any $f_i$ to $\mathscr{O}_0$, we can write $f_{i,x}=r_1g_1+...+r_mg_m$. This means they coninside on  an open neighborhood. Then apply identity theorem. But later when I reviewed it, it was not that correct, because the functions $g_1,r_1,...,r_m,g_m$ might actually not be globally defined. So maybe this claim is wrong? is there any counter example?

答:
Yes, this is true, even if you replace ${\mathbb C}^n$ by an $n$-dimensional complex manifold $M$. The precise statement is that if $V\subset M$ is the zero-set of a family of holomorphic functions $f_j, j\in J$, on $M$ then there are  holomorphic functions $g_1,...g_n$ on $M$ such that
$$
V=\{z\in M: g_i(z)=0, i=1,...,n\}.
$$

See Proposition 5.7 in

E.M. Chirka, "Complex analytic sets." Translated from the Russian by R. A. M. Hoksbergen, Mathematics and Its Applications: Soviet Series, 46. Dordrecht etc.: Kluwer Academic Publishers. xix, 372 p. Dfl. 195.00; (1989).


答: The formula is correct... but the values are not: For example, $b_1$ should be equal to $\frac{195}{13}=15$ rather than 13. Getting them right will  also resolve your problems with the inverses $b_1'=\frac{1}{15}\pmod{13}$ makes sense; unlike $\frac{1}{13}\pmod{13}$</p>

Title: CRT and systems of modular equations

Body:
The formula I found:

$$\sum_{i=1}^{k} a_{i}b_{i}b^{`}_{i} (\mod m_{i})$$

where:

$b_{i} = \frac{M}{m_{i}}$

$b_{i}^{`} = b_{i}^{-1} (mod m_{i})$

And for example:

$$x \equiv -7 \mod 13$$
$$x \equiv 39 \mod 15$$

$M = 13*15 = 195$

$b_{1} = 13$

$b_{2} = 15$

$b_{1}^{`} = \frac{1}{13} mod 13? = \frac{1}{13}?$

$b_{2}^{`} = \frac{1}{15} mod 15? = \frac{1}{15}?$

It would make $x = (a_{1} mod 13) + (a_{2} mod 15) = -7 + 9 = 2$, which is wrong. What am I missing? Maybe the formula is wrong?


问:
Poisson structure identity
I'm attempting a question under Hamiltonian Dynamics. We are given that $\omega^{ab}$ is an antisymmetric matrix such that it's components depend on coordinates $x^a$ and such that the Poisson bracket
$$ \{f,g\} = \omega^{ab} \frac{\partial f}{\partial x^a} \frac{\partial g}{\partial x^b}$$
satisfies the usual Jacobi identity.

In the first part of the question we prove $\{fg, h\} = f\{g,h\} + \{f,h\}g$. This was fine though I include it in case it is relevant in what follows.

Next we are asked to prove
$$\partial_a W_{bc} + \partial_b W_{ca} + \partial_c W_{ab} = 0 \tag{$*$} $$
where we denote $W = \omega^{-1}$ and $\partial_a = \frac{\partial}{\partial x^a}$. A given hint is that we note $\omega^{ab} = \{x^a, x^b\}$.

To prove $(*)$ I have tried differentiating $W_{ac}\omega^{cb} = \delta_a^b$. I have tried using the Jacobi identity, of which $(*)$ is reminiscent. I have got nowhere.

Any answers or suggestions are appreciated. Thank you.

答:
Ok so after a long break I finally got it - frustratingly its a just doing what I'd tried more carefully:

We take $W = \omega^{-1}$, then by definition $\delta_a^b = \omega^{ac} W_{cb}$ and differentiating this yields
$$
    0 = \partial_i (\omega^{ac}W_{cb}) = \omega^{ac} (\partial_i W_{cb}) + (\partial_i \omega^{ac})W_{cb}
        \quad \implies \quad
    \omega^{ac} \omega^{bd} (\partial_i W_{cb}) =  - W_{cb}\omega^{bd} (\partial_i \omega^{ac}) = - \partial_i \omega^{ad}\,.
$$
Next in preparation for use in the Jacobi identity we consider that
$$
    \{x^a, \{x^b, x^c\}\} = \{x^a, \omega^{bc}\} = \omega^{pq} (\partial_p x^a) (\partial_q \omega^{bc})
        = \omega^{aq} \partial_q\omega^{bc}
        = - \omega^{aq} \omega^{bp}\omega^{rc}(\partial_q W_{pr})
        = \omega^{ap} \omega^{bq} \omega^{cr} (\partial_p W_{qr})
$$
where the first equality is the hint, the second is a definition, the third is that $\partial_px^a = \delta_p^a$, the forth is from above, and the final uses the antisymmetry of $\omega$ and relabels $p\leftrightarrow q$.

Then from the Jacobi identity we have
\begin{align*}
    0 &= \{x^a, \{x^b, x^c\}\} + \{x^b, \{x^c, x^a\}\} + \{x^c, \{x^a, x^b\}\}\\
        &= \omega^{ap} \omega^{bq} \omega^{cr} (\partial_p W_{qr}) + \omega^{bp} \omega^{cq} \omega^{ar} (\partial_p W_{qr}) +\omega^{cp} \omega^{aq} \omega^{br} (\partial_p W_{qr}) \\
        &=\omega^{ap} \omega^{bq} \omega^{cr} (\partial_p W_{qr}) + \omega^{ap} \omega^{bq} \omega^{cr} (\partial_q W_{rp}) + \omega^{ap} \omega^{bq} \omega^{cr} (\partial_r W_{pq})\\
        &=\omega^{ap} \omega^{bq} \omega^{cr}  [\partial_p W_{qr} + \partial_q W_{rp} + \partial_r W_{pq} ]
\end{align*}
and hence applying $W_{ia} W_{jb} W_{kc}$ to each side then yields
$$
    0 = \partial_i W_{jk} + \partial_j W_{ki} + \partial_k W_{ij}\,.
$$

Title: Poisson structure identity


问:
Find Laurent Series of $f(z) = \frac{1}{z-1}$, $1<|z-2|<2$
I need to find the Laurent Series of $f(z) = \frac{1}{z-1}$, valid in the region $1<|z-2|<2$.

If I understand this, since $f(z)$ is analytic in $ A=\{z \in \mathbb{C}\ | 1 <|z-2|<2 \}$, we have that:

$$f(z)= \sum_{n=0}^\infty a_n(z-2)^n + \sum_{n=1}^\infty b_n(z-2)^{-n}, z \in A.$$

If $|z-2|>1$, then $\frac{1}{|z-2|}<1$, and with some manipulation we find that:

$$\frac{1}{z-1}=\frac{1}{1 + (z-2)}=\frac{1}{(z-2)(1 + 1/(z-2))}= \frac{1}{z-2}\cdot\frac{1}{1-(-1/(z-2))}.$$

Since $|\frac{-1}{z-2}| = |\frac{1}{z-2}|= \frac{1}{|z-2|}<1$, we have

$$\frac{1}{z-1}=\frac{1}{z-2}\cdot \sum_{n=0}^\infty \frac{(-1)^n}{(z-2)^n}= \sum_{n=0}^\infty \frac{(-1)^n}{(z-2)^{n+1}}= \sum_{n=1}^\infty \frac{(-1)^{n-1}}{(z-2)^n}.$$

If $|z-2|<2$, then $\frac{|z-2|}{2}<1$, but I can't seem to find a way to express $\frac{1}{z-1}$ in a suitable form $\frac{c}{1-(|z-2|/2)}$ to use the convergence of the geometric series.

Is this always possible? Or you just have to get lucky?

Since I also know that the coefficients of $a_n$ are given by:
$$a_n = \frac{1}{2\pi i}\int_{\gamma}\frac{dz}{(z-1)(z-2)^{n+1}}$$

I guess I could find them that way, but it seems too hard to do

Is using the above formula the only way to do it? What if there is no pattern to $a_n$? I can't infinitely calculate them!

If anyone could help or find any mistake in my reasoning I would appreciate it!

Body: If $1<|z-2|$, then\begin{align}\frac1{1+(z-2)}&=\frac1{1-\bigl(-(z-2)\bigr)}\\&=-\sum_{n=-\infty}^{-1}\bigl(-(z-2)\bigr)^n\\&=\sum_{n=-\infty}^{-1}(-1)^{n+1}(z-2)^n.\end{align}


问:
Let $T =\{X\in P(\mathbb{Q}) | X\cup \mathbb{N} = \mathbb{Q} \}$ the cardinality of $T$ is $\aleph$?
Let $T =\{X\in P(\mathbb{Q}) | X\cup \mathbb{N} = \mathbb{Q} \}$ the cardinality  of $T$ is $\aleph$ ? prove

attempt

we know that $X\in P(\mathbb{Q})$ so, $X\cup \mathbb{N} = \mathbb{Q}$ create an infinity numbers

答:
It helps to get a "picture" of what is going on.

$T = \{X \in P(\mathbb Q)| ......\}$ means that $T$ is a set of subsets of $\mathbb Q$.  Our question is just which of the subsets and how many of subsets are in $T$.

If $X$ is one of the elements of $T$, the $X$ is subset of $T$ with the following property:  $X\cup \mathbb N = \mathbb Q$.

Now as $\mathbb N$ contains no non-integer rationals, doesn't contain $0$, and no negative rationals, $X$ must "provide" those in order for $X\cup \mathbb N = \mathbb Q$.  And if $L = \{q\in \mathbb Q| q\not \in \mathbb N\}=\mathbb Q\setminus L$ then $L\subset X$ for all $X\in T$.

So what else is in $X$?  That's the choice we have.  Every $X\in T$ is a subset of $\mathbb Q$ that contains $L$ as a subset.  And $X$ may or may not contain some natural numbers.  Lets define $X_a = \{n\in \mathbb N| n= X\}=$ the set of all natural numbers in $X$.

Then $X = L \cup X_a= $(the set of all rational numbers that are not in $\mathbb N$) $\cup$ (the set of all natural numbers in $X$).

Now we can describe each of the $X \in T$ by identifying which naturals it contains (everything else it has will be had by everty other $Y\in T$ so we can take those for granted).

So if we define $f(X) =X_a$ that is a mapping for $X \in T$ to $X_a$ which is in $P(\mathbb N)$ as it is subset of $\mathbb N$.  So $f: T\to P(\mathbb N)$.

And further more $f$ is a bijection.  It's injective because for any two different $X,Y\in T$ they both will contain all the non-naturals so the only thing that distinguishes them are the natural numbers they contain.  In other words, $X \ne Y$ if and only if $X_a \ne Y_a$.

And $f$ is surjective.  For any $K \subset N$ then the set $L \cup K$ will one of the subsets in $T$.

So $f:T\to P(\mathbb N)$ is a bijection.

So $|T| = |P(\mathbb N)|$.

.....

Now I think in you class you have probably had it proven that $|P(\mathbb N)| = 2^{\mathbb N} = |\mathbb R| \ne  |\mathbb N|$.

so $T =  |\mathbb R| \ne  |\mathbb N|$.


问:
Elegant geometric proof for this conditional expectation formula instead of brute force expansion
I want to prove the following equation: Let $X$ be a real-valued RV with $E(X^2)<\infty$ and let $G, H\subseteq F$ be $\sigma$-algebras such that $H\subseteq G$. Then
$$E\left((X-E(X\vert H))^2\right) = E\left((X-E(X\vert G))^2\right) + E\left((E(X\vert G)-E(X\vert H))^2\right)$$

The exercise says that instead of just using the definitions of (conditional) expectation and forcing the problem, there is an interesting geometric perspective contained in this equation which can also be used to prove it. I was wondering what it is.

This seems to resemble Pythagoras, but I can't close the connection. Does anyone see this geometric perspective? How is it connected to a proof?

答:
One can interpret $E[X \mid H]$ as the $H$-measurable random variable $Y$ that minimizes the $L^2$ distance $E[(X-Y)^2]$. This can be viewed as an orthogonal projection of $X$ onto the space of $H$-measurable random variables. See the Wikipedia page for more info.

The geometric interpretation of your equation would indeed be a Pythagorean theorem involving a single projection onto $H$ (left-hand side) and two successive projections: first onto the larger sigma-algebra $G$, and then onto the smaller sigma-algebra $H$. Try drawing a picture of a projection of a 3-dimensional vector $x$ onto a line, along with the projection of $x$ onto a plane containing that line to visualize this.

In your algebraic proof, you probably used something like
$$E[(X-E[X \mid G]) (E[X \mid G] - E[X \mid H])] = 0,$$
which geometrically is an orthogonality statement that gives you the right angle in the Pythagorean theorem.


问:
Help me with z transform
So the question is basically z transform the given system.

$(y[n+2] + 3y[n+1] - 4y[n])=(x[n+2] - 5x[n+1])$

I've to find h[z] first then it's really easy to solve it. So that's what I got so far;

$z^2y(z) + 3zy(z) - 4y(z) = z^2x(z) - 5zx(z)$

$(z^2+3z-4)y(z) = (z^2 - 5z)x(z)$

$H(z) = \frac{y(z)}{x(z)} = \frac{z^2-5z}{z^2+3z-4}$

then

$H(z) = 1 - \frac{4}{5(z-1)}-\frac{36}{5(z+4)}$

I've to find Z transform pair but I'm stuck here. Thanks in advance!

Here is the z transform table from schaums;

答:
Using $H(z)=\frac{z^2-5z}{(z+4)(z-1)}=\frac{Az}{z-1}+\frac{Bz}{z+4}+C$ we get $A=-4/5,B=9/5,C=0$. Can you complete?

 From your list we see that $Z[-a^nu[-1-n]]=\frac z{z-a}$ and thus$$Z[-u[-1-n]]=\frac z{z-1}\\Z[-(-4)^nu[-1-n]]=\frac z{z+4}$$Thus, by the linearity of the inverse $Z$ transform, we get$$Z^{-1}[H(z)]=AZ^{-1}[z/(z-1)]+BZ^{-1}[z(z+4)]\\=\frac45u[-1-n]-\frac95(-4)^nu[-1-n]$$

Title: Help me with z transform


问:
How to write a sequence recursively
How can I write the sequence
1, 1/2,   1/2,   1/3,   1/3,   1/3,   1/4,   1/4,   1/4,   1/4,   1/5,   1/5,   1/5,   1/5,   1/5, …

in closed form or recursively?

I am trying to find a sequence that converges to 0 but the series diverges.

答:
Your sequence can be written as:

$$a_n = \frac{1}{\lfloor \sqrt{2n} + 0.5 \rfloor}$$

This is the inverse of OEIS sequence A002024.

Title: How to write a sequence recursively


问:
prove that $3x - x^3 < 2 \cdot \sin(\pi/2 \cdot x)$ $\forall x \in (0, 1)$
I faced problems of such kind before and usually I would just chech that on the ends of segment values of compared functions are equal and then find their derivatives and make sure that derivative of one function is always greater on some interval or segment. However, in this case it just didn't work. I found derivative but on some interval $A \subset (0, 1)$ one is greater than the other and on another interval $B \subset (0, 1)$ everything is vice versa.

Do you have any other meaningful approaches to this problem?

答:
Let $$
f(x) = 2 \cdot \sin(\pi/2 \cdot x) - 3x - x^3 $$
So you want to prove $f(x) \ge 0$ on $[0 \quad 1]$. Now $f(0) = f(1) = 0$ so for all values within the interval you want $f(x) > 0$.

By expansions, check the behavior of $f(x)$ in the vicinity of the interval boundaries. Near $x=0$, you have $f(x) \simeq  (\pi -3)x \simeq 0.14 x$, and near $x=1$, you have $f(x) \simeq  (6 - \pi^2/2) (x-1)^2 \simeq 1.07 (x-1)^2$, so indeed $f(x)$ is rising on both interval boundaries.

Now what is left to show is that $f(x)$ will not fall again below zero within the interval. This is achieved if $f(x)$ indeed only has one single extremal point in $(0,1)$. This can be checked by computing   $f'(x) = \pi \cos((\pi x)/2) + 3 x^2 - 3$. A standard curve discussion (below) gives the result that there is indeed only one solution  $f'(x) =0$ in $(0,1)$. We are done.

Remark: "Simple" Taylor expansion won't work since $f(0) = f(1) = 0$.

Curve discussion. The curve discussion is standard but a bit lengthy since we start identifing a derivative which doesn't change sign in $(0,1)$. With that information, and with the values of all the derivatives at the interval boundaries, we can argue backwards. We need to find the relevant maxima and minima only approximately, since all we want to establish is that  there is only one solution  $f'(x) =0$ in $(0,1)$, and we are not interested in its position.

We have
$$
f'(x) = \pi \cos((\pi x)/2) + 3 x^2 - 3\\
f''(x) =  6x - (\pi^2 \sin((\pi x)/2))/2 \\
f^{(3)}(x) = 6 - (\pi^3 \cos((\pi x)/2))/4 \\
f^{(4)}(x) = (\pi^4 \sin((\pi x)/2))/8
$$
Now we can argue backwards: we have in $(0,1)$ that $f^{(4)}(x) > 0$.

Therefore $f^{(3)}(x)$ is rising strictly monotonously from -1.75 to 6. So in  $(0,1)$, we have just one $x^*$ where $f^{(3)}(x^*) = 0$ which is roughly at  $x^* = 0.4365$.

This means that $f''(x)$, starting from $f''(0) = 0$, is first falling until $x^* = 0.4365$, then again rising to  $f''(1) = 1.0652$. This again means that in $(x^*, 1)$, there is just one  $x^{**}$ where $f''(x^{**}) = 0$ which is roughly at  $x^{**} = 0.77$.

Now we can describe the behavior of $f'(x)$, starting from $f'(0) = 0.1416$, is falling until  $f'(x^{**}) = -0.11$, and then rising until $f'(1) = 0$. Hence there is only one value $x^{***}$ where $f'(x^{***}) = 0$,  where $0 < x^{***} < x^{**}$.


问:
Explaining $\lim_{x\to0^+}\frac{\sin^{-1}x^{1/2}}{\cos(x^{1/4})-1}\approx\frac{x^{1/2}}{-\frac12(x^{1/4})^2}\to -2$
Can you explain to me if and why this is correct?
$$\lim_{x\to0^+}\frac{\sin^{-1}x^{1/2}}{\cos(x^{1/4})-1}\approx\frac{x^{1/2}}{-\frac12(x^{1/4})^2}\to -2$$

I don’t understand how to approx these functions and I don’t understand if I have to memorized them or there is a simple way to approx trigonometrics functions in every case.

答: Use the infinitesimal equivalences. Note that $\sin (x) \approx 0$ when $x\approx 0$ (then so does $\sin^{-1}$ as $\sin (0)=0$) and that $\cos (x) -1 \approx -x^{2}/2$ when $x\approx 0$. These equivalences come from the Taylor series of $\sin$ and $\cos$.


问:
Almost nilpotent groups of hyperbolic isometries are cyclic
Let $X$ be a complete, simply connected, Riemannian manifold of negative curvature.
Let $\Gamma$ be an almost nilpotent group of non-elliptic isometries containing a hyperbolic isometry $\gamma$. Then $\Gamma$ fixes a unique geodesic $\ell$ and is infinite cyclic.

It should follow from the fact that hyperbolic isometries fix a unique geodesic on each point of which they attain the minimum translation distance $d_\gamma=\min_{x\in X}d(x,\gamma(x))$.

I can prove the first part of the statement for nilpotent groups, because they have non-trivial center and if two isometries $\gamma_1,\gamma_2$ commute and one fixes a geodesic $\ell=\gamma_1(\ell)$, then also the other fixes it.

On the other hand I have some problem showing that $\Gamma$ is infinite cyclic. I was looking for counterexamples and came up with this: let $\gamma_1$ and $\gamma_{\sqrt{2}}$ two isometries of $X$ which translate the same geodesic $\ell$ of $1$ and $\sqrt{2}$. Their action on $\ell$ is the same of $\mathbb{Z}^2$. I know that there are no counterexamples, so I suspect that these two isometries generate a non-nilpotent group, and I would be able to detect it if I'd look at their action outside $\ell$.

So here are the questions: How can I prove the statement? Why the counterexample is in fact not a counterexample?

答:
In order to close this question. Yes, your example is a counter-example to the claim you are trying to prove. The key condition you are missing is that the group $\Gamma$ is discrete. Under this assumption the claim indeed holds. One proves this in three steps:

Suppose that $g_1, g_2$ are hyperbolic isometries which share exactly one fixed point $p$ in the sphere at infinity of $X$. Then $[g_1, g_2]$ is a parabolic isometry $f$ of $X$ fixing $p$.

If $g$ is a hyperbolic isometry of $X$ and $f$ is a parabolic isometry of $X$ such that $g, f$  have a common fixed point $p$ in the sphere at infinity of $X$, then either the sequence
$$
f_n:=g^n f g^{-n}
$$
or
$$
f_{n}=g^{-n} f g^{n}
$$
satisfies the property
$$
d(x, f_n(x))\to 0
$$
for (any) $x$ on the axis of $g$. Thus, a discrete subgroup of isometries cannot contain two hyperbolic elements which share exactly one fixed points, cf. this question.

Step 3. If $\Gamma$ is almost nilpotent and contains non (nontrivial) elliptic elements then $\Gamma$ either has exactly one fixed point $p$ in the ideal boundary of $X$ or preserves a 2-point subset $\{p, q\}$ in this ideal boundary of $X$.

a. Steps 1 and 2 show that the first case is impossible.

b. In the second case, $\Gamma$ will preserve the unique geodesic $c\subset X$ connecting $p, q$. Then, restricting $\Gamma$ to this geodesic we obtain an isomorphism of $\Gamma$ to a discrete isometry group of the real line. From this, you conclude that $\Gamma$ is cyclic (since it contains no elliptic elements).


答: Consider the vectors $\mathbf{a}=X-E[X|G]$ and $\mathbf{b}=E[X|G]-E[X|H]$ in $L^2(X)$. Since $E[X|G]$ is the orthogonal projection of $X$ onto the subspace of $G$-measurable functions we have that $\mathbf{a}$ is orthogonal to any $G$-measurable function, in particular it is orthogonal to $\mathbf{b}$. The pythogeran theorem for inner product spaces now says that $\|\mathbf{a}+\mathbf{b}\|^2=\|\mathbf{a}\|+\|\mathbf{b}\|^2$, i.e. your equation.

Title: Elegant geometric proof for this conditional expectation formula instead of brute force expansion

Body:
I want to prove the following equation: Let $X$ be a real-valued RV with $E(X^2)<\infty$ and let $G, H\subseteq F$ be $\sigma$-algebras such that $H\subseteq G$. Then
$$E\left((X-E(X\vert H))^2\right) = E\left((X-E(X\vert G))^2\right) + E\left((E(X\vert G)-E(X\vert H))^2\right)$$

The exercise says that instead of just using the definitions of (conditional) expectation and forcing the problem, there is an interesting geometric perspective contained in this equation which can also be used to prove it. I was wondering what it is.

This seems to resemble Pythagoras, but I can't close the connection. Does anyone see this geometric perspective? How is it connected to a proof?


问:
Regular functions and morphisms of varieties for field not algebraically closed
Let us work over a field that is NOT algebraically closed. Specifically, $K=\mathbb R$.

We can think of the affine line $\mathbb A_K^1$ as a ringed space $(\mathbb A_K^1,\mathcal O_{\mathbb A^1_K})$, where for an open set $U\subseteq\mathbb A^1_K$, we have $\mathcal O_{\mathbb A^1_K}(U)=(\text{the ring of regular functions over $U$})=\{\frac fg\mid f,g\in K[x] \,\&\,(\forall P\in\mathbb A^1_K)(g(P)\ne 0)\}$.

I think that according to this, $\frac1{1+x^2}$ should be a regular function over the entire $\mathbb A^1_K$. Moreover, it seems to me that $f:\mathbb A_K^1\to\mathbb A_K^1$ defined as $f(x)=\frac1{1+x^2}$ is a morphism from the ringed space $(\mathbb A^1_K,\mathcal O_{\mathbb A^1_K})$ to itself (in the sense that it pulls back regular functions to regular functions). Am I right?

I am probably confused by either different sources using different definitions or me seeing statements that only hold for $K$ algebraically closed. E.g. the Wikipedia article Morphism of algebraic varieties states

The regular functions on $\boldsymbol A^n$ are exactly the polynomials in $n$ variables [...]

,
which I only agree with for $K$ algebraically closed and not in the case I've shown above.

答:
If you define $\mathcal{O}(\mathbb{A}^1_{\mathbb{R}})$ as the set of functions with the following property (P1):

P1. $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ with the property that for any point $p\in \mathbb{R}^2$ there is an open subset $U(p)\subseteq \mathbb{R}^2$ and two polynomials $p(x),q(x)\in \mathbb{R}[x]$ such that the restriction $f_{U(p)}$ of $f$ to $U(p)$ satisfies $f_{U(p)}=\frac{p(x)}{q(x)}$,

then it follows (as you noted) the function $\frac{1}{x^2+1}$ is a regular function on $\mathbb{R}^2$. Hence $\mathcal{O}(\mathbb{A}^1_{\mathbb{R}})\neq \mathbb{R}[x]$.  If $z=a+ib \in \mathbb{C}$ with $b\neq 0$, it follows
the polynomial $p_z(x):=(x-z)(x-\overline{z})=x^2-2ax+a^2+b^2\in \mathbb{R}[x]$ is an irreducible polynomial - it has no real roots.
Hence if you use definition P1 to define the ring of regular functions, it follows the function $f(x):=\frac{1}{p_z(x)}$ is a regular function. If $S\subseteq A:=\mathbb{R}[x]$ is the set of polynomials on the form $p(x):=\prod_{z_i\in \mathbb{C}-\mathbb{R}} (x-z_i)(x-\overline{z_i})$ it follows $S$ is a multiplicatively closed subset, and the ring of fractions $S^{-1}A$ is a sub-ring of $\mathcal{O}(\mathbb{A}^1_{\mathbb{R}})$. Hence with definition P1 it follows the ring $\mathcal{O}(\mathbb{A}^1_{\mathbb{R}})$ is much larger than $A$.

Question: "The regular functions on $\mathbb{A}^n$ are exactly the polynomials in n variables [...], which I only agree with for K
algebraically closed and not in the case I've shown above."

Answer: You want (for many reasons) to recover the polynomial ring $\mathbb{R}[x]$ as the ring of global sections of the structure sheaf. This problem is "solved" by the affine spectrum $Y:=Spec(\mathbb{R}[x])$. What you define above is a "sheaf of rings" on the real algebraic variety $\mathbb{A}^1_{\mathbb{R}}$ in the sense of "real algebraic geometry", and this sheaf differ from the sheaf defined using the affine spectrum in the sense of Hartshornes book "Algebraic geometry". The topological space of the variety $X:=\mathbb{A}^1_K$ you consider above is the set of real numbers $r\in \mathbb{R}$. The topological space of $Y$ consists of all real numbers plus the set of all maximal ideals in $A:=\mathbb{R}[x]$ with residue field the complex numbers. Hence the two topological spaces differ. Your space $X$ is by definition the set of rational points $Y(K)$ with the induced topology.

Note: There is nothing "wrong" with the ringed space $(X, \mathcal{O}_X)$ - it has properties different from $Y$. In real algebraic geometry it follows projective space $\mathbb{P}^n_K$ may be realised as a real affine algebraic variety - there is a closed embedding into real affine space. Moreover any real projective variety is affine. The reason for this is that the structure sheaf has more local sections, as you have observed in your comment above. Hence there are "more maps". This does not happen with the Hartshorne definition: You cannot construct a closed embedding of $Proj(k[x_0,..,x_n])$ into
affine space $Spec(k[y_1,..,y_d])$.

As an example: The global section $s(x):=\frac{1}{p_z(x)}$ for $z\in\mathbb{C}-\mathbb{R}$ is a unit - the polynomial $p_z(x)$ is non-zero on $\mathbb{R}$, and with your definition of "regular map" it follows $s(x)$ is a global section - it lives in $\mathcal{O}_{\mathbb{A}^1_K}(\mathbb{A}^1_K)$. The function $s(x)$ is not a global section of $\mathcal{O}_Y(Y)\cong \mathbb{R}[x]$.

Real algebraic varieties are relevant for people studying complex algebraic varieties and their topology. You may for any complex projective manifold $X \subseteq \mathbb{P}^n_{\mathbb{C}}$ construct the Weil restriction $W_{\mathbb{C}/\mathbb{R}}(X)$. If $dim(X)=d$ it follows  $dim(W_{\mathbb{C}/\mathbb{R}}(X))=2d$ and $W_{\mathbb{C}/\mathbb{R}}(X)$ is a smooth scheme over $\mathbb{R}$ - the "underlying" real algebraic variety of $X$. The variety
$W_{\mathbb{C}/\mathbb{R}}(X)$ has the same topology as $X$ and you may realize $W_{\mathbb{C}/\mathbb{R}}(X)$ as a real affine algebraic variety. Hence the "underlying" real algebraic variety of $X$ is affine. Moreover: Any real affine algebraic variety may (topologically) be realized as a hypersurface. Hence if you study the topology of $X$ it follows there is a real affine hypersurface $Z(f)$ with the same topology as $X$ (here I am extremely "vague").

You find this discussed here:

https://math.stackexchange.com/posts/3957121/edit

and here:

Different definitions of regular map (between affine varieties)?


问:
Why must a ring homomorphism from $\mathbb Q[\sqrt n] \to R$ be the identity mapping?
Let $\mathbb Q[\sqrt d]:= \{a+b\sqrt d \, |\, a,b \in \mathbb Q\}$, and $d\in \mathbb N$ with the condition that $d$ is not a square number. I have shown that this is a subring of $\mathbb R$ and that for each $d$ there can only be two automorphisms, but this question utterly perplexes me. If $R$ is another ring which contains $\mathbb Q$ as a subring, then why must a homomorphism $\varphi : \mathbb Q[\sqrt d] \to R $ be the identity mapping? Ie, why must $\varphi (x) = x \quad \forall x \in \mathbb Q$?

答:
If $n \in \mathbb{Z}$, from $\varphi(1)=1$ it follows that $\varphi(n) = n$. Hence, if $q= a/b \in \mathbb{Q}$ where $a$ and $b$ are integers, we have
$$\varphi(q) = \varphi(ab^{-1}) = \varphi(a)\varphi(b)^{-1}= ab^{-1}= q$$

Moreover, consider the mapping
$$\varphi: \mathbb{Q}(\sqrt{d}) \to \mathbb{Q}(\sqrt{d}): a+ b \sqrt{d}\mapsto a - b \sqrt{d}$$
Check that this is a ring isomorphism. Hence, $\varphi$ need not be the identity.


问:
compute $\int_\gamma \sin(y-x)\ dx +2xy + \sin(x-y) \ dy$
The question is
$$ \int_\gamma \sin(y-x)\ dx +\left(2xy + \sin(x-y)\right) \ dy, \quad \gamma:y=\sqrt{x}, \ \ 0\leq x\leq 1$$
I've tried to solve it like this:
$$\int_{0}^{1}\sin(\sqrt{t}-t) +t+ \sin(t-\sqrt{t}) \ \frac{1}{2\sqrt{t}}\ dt \quad\text{with} \  \ \gamma:r(t)=(t,\sqrt{t}) $$

How should i proceed from here ?
Any suggestion would be great, thanks

Body:
hint

$$\int \sin(t-\sqrt{t})(\frac{1}{2\sqrt{t}}-1)dt=$$
$$\cos(t-\sqrt{t})$$


问:
$5\times 5$ Matrix $A$ such that $A^5 = I_5$ (in $R^5$)
I'm stuck on an exercise I got in my linear algebra class where you have to find a $5\times5$ matrix $A$ such that $A$ multiplied by itself five times is the identity matrix $I_5.$ $A$ must not be the identity matrix itself.

My problem is that I don't know how to go about this in a clever way. I know that it makes sense to put only ones on the main diagonal of A and only zeroes below the ones but I don't know how to construct the rest of the matrix. It seems tedious to just try out random examples and I want to do it systematically.
Any ideas would be appreciated!

答:
the companion matrix for polynomial $x^5 - 1$  is
$$
\left(
\begin{array}{ccccc}
0&1&0&0&0 \\
0&0&1&0&0 \\
0&0&0&1&0 \\
0&0&0&0&1 \\
1&0&0&0&0 \\
\end{array}
\right)
$$
or, if preferred, the transpose of this.


问:
Computing the integral $ \int_{1}^{\infty} \frac{x \,\mathrm{d} x}{\sqrt{2x^6+2x^3+1}}$
Compute the following integral
$$ \int_{1}^{\infty} \frac{x \,\mathrm{d} x}{\sqrt{2x^6+2x^3+1}}.$$
This integral obviously converges. However, is it computable?

答:
I have to share @RobertIsrael and @ProfessorVector's doubts.

Call it $I$. Following @WillJagy, take $2x^3+1=\sinh t$ so$$I=\frac{1}{3\sqrt[6]{2}}\int_{\operatorname{arsinh}3}^\infty(\sinh t-1)^{-1/3}\mathrm dt.$$Using this,$$I=2^{-5/3}[(\sinh t-1)^{2/3}\cosh t\cdot f(t)]_{\operatorname{arsinh}3}^\infty$$where$$f(t):=F_1(\tfrac23;\,\tfrac12,\,\tfrac12;\,\tfrac53;\,\tfrac12(1+i)(1-\sinh t),\,\tfrac12(1-i)(1-\sinh t)),$$with $F_1$ the Appell hypergeometric function. Assuming the bracketed expression vanishes in the $t\to\infty$ limit (hopefully an expert on $F_1$ can weigh in on that),$$I=-\sqrt{\tfrac52}F_1(\tfrac23;\,\tfrac12,\,\tfrac12;\,\tfrac53;\,-1-i,\,-1+i).$$If that has a nicer closed form, I'd love to see it.


问:
Justifying $\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$ for large $x$

I need to find a way to justify the inequality $$\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$$ that holds for large values of x.

As an instance, I will justify an example inequality just to show you the intended strategy. Consider the inequality
$$\left|\frac{5x + 21}{2x^2 - 7}\right| < \frac{5}{x}.$$ For example, $2x^2 - 7$ can be written as $x^2 + (x^2 - 7)$. Because $x^2 - 7$ is a positive value for all $x < -\sqrt{7}$, removing it from the denominator makes the absolute value of the fraction greater. Also note that when $x < -\frac{21}{10}$, the numerator $|5x + 21| < 5|x|$, and this happens for all $x < -\sqrt{7}$. So $$\left|\frac{5x + 21}{2x^2 - 7}\right| < \frac{5|x|}{x^2} = \frac{5}{x}$$ as long as $x < -\sqrt{7}$.

答:
Hint:

Set
$$
f(x)=\frac{5x^3 + 3x^2 + x}{x^3 - x^2 - 1}.
$$
Then you have
$$
\lim_{x\to+\infty}f(x)=5\tag{1}
$$

Now use the limit definition to show that $f(x)<10$ for sufficiently large $x$. So you get
$$
\frac{f(x)}{x}<\frac{10}{x}
$$
for sufficiently large $x$.

Notes: The limit (1) implies that
$$
|f(x)-5|<1
$$
for sufficiently large $x$. So you have
$$
4<f(x)<6<10.
$$


答:
$\int_\gamma \sin(y-x)\ dx +\left(2xy + \sin(x-y)\right) \ dy$

$= \int_\gamma \sin(y-x)\ dx +\left(\sin(x-y)\right) \ dy \ + \ \int_\gamma 2xy \ dy$

But please note that $\big(\sin(y-x), \sin(x-y)\big)  = \nabla f(x,y) \ $ where $\ f(x,y) = \cos(y-x)$ and $f(1,1) - f(0,0) = 1 -1 = 0$.

So the line integral for part of the vector field is zero and all you are left with is to find -

$\displaystyle \int_{\gamma} 2xy \ dy = \int_0^1 t \ dt = \frac{1}{2}$

Title: compute $\int_\gamma \sin(y-x)\ dx +2xy + \sin(x-y) \ dy$

Body:
The question is
$$ \int_\gamma \sin(y-x)\ dx +\left(2xy + \sin(x-y)\right) \ dy, \quad \gamma:y=\sqrt{x}, \ \ 0\leq x\leq 1$$
I've tried to solve it like this:
$$\int_{0}^{1}\sin(\sqrt{t}-t) +t+ \sin(t-\sqrt{t}) \ \frac{1}{2\sqrt{t}}\ dt \quad\text{with} \  \ \gamma:r(t)=(t,\sqrt{t}) $$

How should i proceed from here ?
Any suggestion would be great, thanks


问:
Find n where $(2n+1)2^{4n+5} = 3 \pmod{7}$
For $n$ normal number, the book solved it like this:

If $n$ can be divided by $3$ (which is $n = 3k$) then $n = 21L + 9$.

If $n$ can't be divided by 3(Which is either $n = 3k +1$ or $n = 3k + 2$) then $n = 21L + 1$ or $n = 21L + 2$ .

But I didn't solve it like this.

My logic is that since 3 is a prime number then $(2n+1)2^{4n+5} = 3 \pmod{7}$ means either $2^{4n+5} = 3\pmod{7}$ and $ 2n + 1 = 1\pmod{7} $ or the other way around.

But since there is no $n$ value that can make $2^{4n+5} = 3\pmod{7}$ then it means:

$ 2n + 1 = 3 \pmod{7}$ and by calculating we find in the end $n = 7k' + 1$

And $ 2^{4n+5} = 1\pmod{7}$ and by calculating we find that $n = 3k''$

So it means that $n = 7k' + 1$ AND $n = 3k''$

So, I know that my solution is faulty but can anyone explains to me the right solution or point out where I went wrong?

答:
Here is one way to solve it:

$(2n+1)\pmod 7\equiv 1,3,5,0,2,4,6,\cdots\ $ and cycling.

$2^{4n+5}\pmod 7\equiv 4,1,2,\cdots\ $ and cycling.

Both can be proved by induction on $n$.

$1\times 3\equiv 2\times 5\equiv 4\times 6\equiv 3\pmod 7\ $ verify there are no others.

Finally can you find $n$ for which the conditions are reunited ? (hint, Chinese theorem).


Title: Computing the integral $ \int_{1}^{\infty} \frac{x \,\mathrm{d} x}{\sqrt{2x^6+2x^3+1}}$

Body:
Compute the following integral
$$ \int_{1}^{\infty} \frac{x \,\mathrm{d} x}{\sqrt{2x^6+2x^3+1}}.$$
This integral obviously converges. However, is it computable?

Body:
The integral may be expressed in terms of elliptic integrals. Substitute $x=2^{-1/6}u$:
$$I=2^{-1/3}\int_{2^{1/6}}^\infty\frac u{\sqrt{u^6+\sqrt2u^3+1}}\,du$$
Then substitute $u+\frac1u=x$ (Byrd and Friedman 581.01) to get
$$I=2^{-4/3}\left(\int_y^\infty\frac1{\sqrt{(x-2)(x-a)(x-b)(x-c)}}\,dx+\int_y^\infty\frac1{\sqrt{(x-a)(x-b)(x-c)(x+2)}}\,dx\right)$$
where $y=2^{1/6}+2^{-1/6}$, $a=\sqrt2$, $b=\frac{-\sqrt2+\sqrt6}2$ and $c=\frac{-\sqrt2-\sqrt6}2$. Finally use B&F 258.00… or rather this small modification of B&F 258.00: if $y>a>b>c>d$,
$$\int_y^\infty\frac1{\sqrt{(x-a)(x-b)(x-c)(x-d)}}\,dx=J(a,b,c,d,y)=gF(\varphi,m)$$
where $m=\frac{(b-c)(a-d)}{(a-c)(b-d)}$ is the parameter, $g=\frac2{\sqrt{(a-c)(b-d)}}$ and
$$\varphi=\sin^{-1}\frac{\sqrt{(a-c)(b-d)}}{\sqrt{(y-a)(y-b)}+\sqrt{(y-c)(y-d)}}$$
Then
$$I=2^{-4/3}(J(2,a,b,c,y)+J(a,b,c,-2,y))$$

#!/usr/bin/env python3
from mpmath import *
mp.dps = 100

print(quad(lambda x: x/sqrt(2*x**6 + 2*x**3 + 1), [1, inf]))

def bf258_00(a, b, c, d):
    y = root(2,6) + root(2,-6)
    g = 2 / sqrt((a-c)*(b-d))
    m = (b-c)*(a-d)/(a-c)/(b-d)
    s1 = sqrt((a-c)*(b-d))
    s2 = sqrt((y-a)*(y-b))
    s3 = sqrt((y-c)*(y-d))
    vphi = asin(s1 / (s2 + s3))
    return g * ellipf(vphi, m)

ra = sqrt(2)
rb = (-sqrt(2)+sqrt(6))/2
rc = (-sqrt(2)-sqrt(6))/2
I1 = bf258_00(2, ra, rb, rc)
I2 = bf258_00(ra, rb, rc, -2)
print(root(16, -3) * (I1 + I2))

0.632421879503014798952234566937018710345295154697941062278011824742339961706287315740337102826246671...

$I$ with bounds $[2^{-1/6},\infty]$ (note that $2^{-1/6}<1$) evaluates to $2^{-4/3}(J(2,a,b,c,2)+J(a,b,c,-2,2))$. $I$ with bounds $[0,2^{-1/6}]$ evaluates to $2^{-4/3}(J(2,a,b,c,2)-J(a,b,c,-2,2))$. Therefore $I$ with bounds $[0,\infty]$ requires only one elliptic integral:
$$\int_0^\infty\frac x{\sqrt{2x^6+2x^3+1}}\,dx=2^{-1/3}J(2,a,b,c,2)$$


问:
Branch cut of the logarithm
I have a function $F$ holomorphic on some open set, and I have $ F(0) = 1 $ and $ F $ is non-vanishing. I want to show that there is a holomorphic branch of $ \log(F(z)) $.

Now, I'm getting confused. The principal branch of logarithm removes $ (-\infty, 0] $. But if the point 0 is missing from the plane, what happens when we take $ \log{F(0)} = \log{1} + 0 = 0 $? (I'm sure we can take the principal branch, because $ \exp(z) $ satisfies the conditions in question).

Any help would be appreciated. Thanks

答: That only means $F(z)$ cannot be 0, while $z$ can be 0.</p>

Title: Branch cut of the logarithm


问:
About the property of $m$: if $n < m$ is co-prime to $m$, then $n$ is prime
The number $30$ has a curious property:

All numbers co-prime to it, which are between $1$ and $30$ (non-inclusive) are all prime numbers!

I tried searching(limited search, of course) for numbers $\gt 30$ that have this property, but could not find any.

Are there any such numbers $\gt 30$?

答:
The reason why is that $30=2\cdot 3\cdot 5$, which are consecutive primes, so any number less than $30$ must not be divisible by these $3$ numbers to be coprime. If they are to be coprime but not prime themselves, then they would have to be divisible by another two primes (including multiplicities), and the least such one is $7$. But $2\cdot 3\cdot 5<7^2$.

Such numbers (products of consecutive primes) are called primorials, and they grow too quickly for this to happen again for any number above $30$ (for example, the next number is $210$, but $210=2\cdot 3\cdot 5\cdot 7$ and $121=11^2$ are coprime). You can see the OEIS listing to see the first few primorials, which would be candidates for the property you're searching for, to see just how large they get.


问:
Expected number of rolls until a number appears 2 times discussion
I have the following problem. How many times you have to roll a die until you get two 6 or two 5 in a row.

Problems similar to this one have been discussed in the forum in links like:

[1] Number of rolls until the same number appears 2 consecutive times

[2] Expected number of rolls for fair die to get same number appear twice in a row?

I "solved" this problem as follows:

Let A the event of two consecutives 6 or two consecutive 5 in a row, and B the event of two consecutive equal numbers in a row. Then, it follows that:

\begin{equation}
P(A) = P(A \cap B) = P(A/B)P(B)
\end{equation}

$P(A/B)=2/6$ because if you already know that there are two identical numbers in a row then A can happen in 2 ways (5,5) and (6,6) over the sample space (i,i), $1\leq i \leq 6.$

But P(B) is a little bit more tricky. I know that the answer is 1/7 (I run a simple program to compute this number) but for me is counter intuitive.

In that case the Expected value would be = 21 because I expect 1 "win" combination out of 21 rolls.

My confusion is that I would have answered 1/6 instead of 1/7 in $P(B)$ because, no matter what number came out from the first dice, the second has 1/6 probability to match that number. Why this intuition fails? I know the answer is in [2] but I still can not handle the intuition on this.

A generalization of this result can be found here:
Expected number of rolls until a number appears $k$ times

Moreover, this problem is often to get confused with the 1/36 probability of throwing two dices simultaneously.

Links where to the discussion of this topic:

[-] If you roll a fair six sided die twice, what's the probability that you get the same number both times?

[-] Why are the probability of rolling the same number twice and the probability of rolling pairs different?

答:
It seems you are trying to solve the problem by finding the probability that two consecutive throws are either both $5$ or both $6$, and then taking one over this probability. Similar to how the expected number of flips it takes to get heads when $P(heads)=p$ is $1/p$. However, this does not work here, since the sequences of pairs of flips are not independent.

Let $X$ be the number of rolls it takes to get two fives or two sixes in a row. Starting with zero rolls, three things can happen:

Your first roll is between $1$ and $4$. In this case, the number of subsequent roll has the same distribution as $X$.

You first roll is $5$ or $6$, and your second roll is different from the first roll. Again, the number of subsequent rolls is $X$.

You first two rolls are both fives or both sixes. Here, $X=2$.

We get that
$$
X\stackrel{d}=
\begin{cases}
X+1 & \text{with probability }\frac{2}{3} \\
X+2 & \text{with probability }\frac{1}{3}\cdot \frac56 \\
2 & \text{with probability }\frac{1}{3}\cdot \frac16 \\
\end{cases}\tag{*}
$$
Here is the interpretation of the above equality; $X$ has the same distribution as a random variable which is equal to one of the three options above, selected with the given probabilities independently of $X$.

Now, taking expectations of both sides, you get
$$
E[X] = \frac23 \cdot E[X+1] + \frac13\cdot\frac56 \cdot E[X+2] + \frac13\cdot\frac16 \cdot 2
$$
This lets you solve for $E[X]$.

As a side note, $(*)$ also implies that
$$
f(X)\stackrel{d}=
\begin{cases}
f(X+1) & \text{with probability }\frac{2}{3} \\
f(X+2) & \text{with probability }\frac{1}{3}\cdot \frac56 \\
f(2) & \text{with probability }\frac{1}{3}\cdot \frac16 \\
\end{cases}
$$
for any function $f$. Using $f(x)=x^2$, then taking expectations, lets you find $E[X^2]$, and therefore $\text{Var}(X)$. Using $f(x)=e^{tx}$ lets you find the full moment generating function of $X$.


问:
How can I find kernel of linear map?
What is the kernel of linear map $f\colon\mathbb{R}^{2\times2}\to \mathbb{R}^{3\times2}$ if:
$$f
  \begin{pmatrix}
    1 & 1\\
    1 & 1\\
  \end{pmatrix}
=
\begin{pmatrix}
    1 & 2 & 3\\
    1 & 2 &3\\
  \end{pmatrix},
$$
$$f
  \begin{pmatrix}
    0 & 1\\
    1 & 1\\
  \end{pmatrix}
=
\begin{pmatrix}
    3 & 3 & 5\\
    3 & 3 &5\\
  \end{pmatrix},
$$
$$f
  \begin{pmatrix}
    0 & 0\\
    1 & 1\\
  \end{pmatrix}
=
\begin{pmatrix}
    1 & 1 & 1\\
    1 & 1 &1\\
  \end{pmatrix},
$$
$$f
  \begin{pmatrix}
    0 & 0\\
    0 & 1\\
  \end{pmatrix}
=
\begin{pmatrix}
    1 & 0 & 1\\
    1 & 0 &1\\
  \end{pmatrix}.
$$
and how can I find it?

答:
To find the kernel of a linear map $f:V\to W$ is the same as solving the equation
$
f(x)=0\;.
$

What you know in your problem are the values of $f(\alpha_k)$ where $\alpha =(\alpha_k:k=1,2,3,4)$ is a basis of $V$.

You can then write out the matrix representation of $f$ with respect to the basis  $\alpha$ of $V$ and the standard basis of $W$. Then you can work with the representation matrix by standard procedures.

Title: How can I find kernel of linear map?


答:
Update:

I found that this solution provides a neat way in general to solve this kind of problems. I will apply the method here:

$$(2n+1)2^{4n+5} \equiv 3 \pmod 7 \\ \iff g(n)=(2n+1)2^n\equiv (2n+1) 2^{4n+6} \equiv 3\cdot 2 \equiv -1 \pmod 7$$

Suppose $n=3k+i, i=0,1,2$, then
$$-1 \equiv g(3k+i) = (6k+2i+1)2^{3k+i} \equiv (-k+2i+1)2^i=g(i)-k\cdot 2^i \pmod 7\\
\iff k\equiv 2^{3-i}(g(i)+1)\equiv (2i+1) 2^3 + 2^{3-i} \equiv 2^{3-i}+2i+1 \pmod 7 \\
(\text{ now write } k = 2^{3-i}+2i+1 + 7j)\\
\iff n=3k+i = 3(2^{3-i}+2i+1+7j)+i \equiv 3\cdot 2^{3-i}+7i+3 \pmod {21}\\
\iff n \equiv \begin{cases}3 \cdot 8 + 0 + 3 \equiv 6  \\ 3 \cdot 4 + 7 +3\equiv 1  \\ 3 \cdot 2 + 14+3\equiv 2 \\ \end{cases} \pmod{21}$$

Just some comments:

$f(n)=(2n+1)2^{4n+5}$, then $f(n+21)\equiv f(n) \pmod 7$ (notice that $2^3\equiv 1 \pmod 7$). So the most direct way is to plug in $n=1, 2, \ldots, 21$ into $f(n)$ and see which ones yield $3 \pmod 7$. No fancy theorems needed.

To save time, do as zwim showed by reusing partial results (you can make a $3 \times 7$ table).

Title: Find n where $(2n+1)2^{4n+5} = 3 \pmod{7}$

Body:
For $n$ normal number, the book solved it like this:

If $n$ can be divided by $3$ (which is $n = 3k$) then $n = 21L + 9$.

If $n$ can't be divided by 3(Which is either $n = 3k +1$ or $n = 3k + 2$) then $n = 21L + 1$ or $n = 21L + 2$ .

But I didn't solve it like this.

My logic is that since 3 is a prime number then $(2n+1)2^{4n+5} = 3 \pmod{7}$ means either $2^{4n+5} = 3\pmod{7}$ and $ 2n + 1 = 1\pmod{7} $ or the other way around.

But since there is no $n$ value that can make $2^{4n+5} = 3\pmod{7}$ then it means:

$ 2n + 1 = 3 \pmod{7}$ and by calculating we find in the end $n = 7k' + 1$

And $ 2^{4n+5} = 1\pmod{7}$ and by calculating we find that $n = 3k''$

So it means that $n = 7k' + 1$ AND $n = 3k''$

So, I know that my solution is faulty but can anyone explains to me the right solution or point out where I went wrong?


问:
Matrices - order of operations
If I have matrices $A, B,$ and $C$ and then I have to solve this:
$X = 3A − 2BC$
what is the order of operations for the $2BC$ part? Do I multiply matrices $B$ and $C$ first and then I multiply the result by $2$? Or do I multiply matrix $B$ by $2$ first and then multiply the two matrices? Thank you kindly.

答:
For any matrices $X$ and $Y$ that have compatible sizes (the number of columns of $X$ equals to that of the rows of $Y$), and any scalar $\lambda$, we have
$$
\lambda(XY)=(\lambda X)Y.
$$

(You can even do $X(\lambda Y)$ and get the same result.)

This is why we usually omit the paratheses and write $\lambda XY$.

Title: Matrices - order of operations


问:
What can I do this cos term to remove the divide by 0?
I was asked to help someone with this problem, and I don't really know the answer why. But I thought I'd still try.

$$\lim_{t \to 10} \frac{t^2 - 100}{t+1} \cos\left( \frac{1}{10-t} \right)+ 100$$

The problem lies with the cos term. What can I do with the cos term to remove divide by 0 ?

I found the answer to be $100$ (Google), but I do not know what they did to the $\cos$ term. Is that even the answer ?

Thanks!

答:
The cos term is irrelevant. It can only wiggle between $-1$ and $1$, and is therefore killed by the $t^2-100$ term, since that approaches $0$.  </p>

<p>For a less cluttered version of the same phenomenon, consider the function $f(x)=x\sin\left(\frac{1}{x}\right)$ (for $x\ne 0$). The absolute value of this is always $\le |x|$, so (by Squeezing) $f(x)\to 0$ as $x\to 0$. </p>

Body:
The cos term is irrelevant. It can only wiggle between $-1$ and $1$, and is therefore killed by the $t^2-100$ term, since that approaches $0$.

For a less cluttered version of the same phenomenon, consider the function $f(x)=x\sin\left(\frac{1}{x}\right)$ (for $x\ne 0$). The absolute value of this is always $\le |x|$, so (by Squeezing) $f(x)\to 0$ as $x\to 0$.


答:
Hint:
$$f(x)=f\left(\alpha\begin{pmatrix}
    1 & 1\\
    1 & 1\\
  \end{pmatrix}+\beta\begin{pmatrix}
    0 & 1\\
    1 & 1\\
  \end{pmatrix}+\gamma\begin{pmatrix}
    0 & 0\\
    1 & 1\\
  \end{pmatrix}+\delta\begin{pmatrix}
    0 & 0\\
    0 & 1\\
  \end{pmatrix}\right)=$$
$$=\alpha f\begin{pmatrix}
    1 & 1\\
    1 & 1\\
  \end{pmatrix}+\beta f\begin{pmatrix}
    0 & 1\\
    1 & 1\\
  \end{pmatrix}+\gamma f\begin{pmatrix}
    0 & 0\\
    1 & 1\\
  \end{pmatrix}+\delta f\begin{pmatrix}
    0 & 0\\
    0 & 1\\
  \end{pmatrix}$$

Title: How can I find kernel of linear map?

Body:
What is the kernel of linear map $f\colon\mathbb{R}^{2\times2}\to \mathbb{R}^{3\times2}$ if:
$$f
  \begin{pmatrix}
    1 & 1\\
    1 & 1\\
  \end{pmatrix}
=
\begin{pmatrix}
    1 & 2 & 3\\
    1 & 2 &3\\
  \end{pmatrix},
$$
$$f
  \begin{pmatrix}
    0 & 1\\
    1 & 1\\
  \end{pmatrix}
=
\begin{pmatrix}
    3 & 3 & 5\\
    3 & 3 &5\\
  \end{pmatrix},
$$
$$f
  \begin{pmatrix}
    0 & 0\\
    1 & 1\\
  \end{pmatrix}
=
\begin{pmatrix}
    1 & 1 & 1\\
    1 & 1 &1\\
  \end{pmatrix},
$$
$$f
  \begin{pmatrix}
    0 & 0\\
    0 & 1\\
  \end{pmatrix}
=
\begin{pmatrix}
    1 & 0 & 1\\
    1 & 0 &1\\
  \end{pmatrix}.
$$
and how can I find it?


答:
This sort of thing can easily be handled with states, and the symmetry makes it especially easy in this case.

There are three states where care about:  $ S_5, S_6,S_{\emptyset}$ according to whether the prior throw was a $5$, a $6$, or neither (we let $S_{\emptyset}$ include the starting state as well).  Symmetry tells us that we expect it to take the same number of rolls to finish from $S_5$ as it does from $S_6$, let $B$ denote that expected number.  Let $A$ denote the expected number starting from $S_{\emptyset}$, so the answer to the problem is $A$.

Considering the possible outcomes if you are in $S_{\emptyset}$ we see that $$A=\frac 13\times (B+1)+\frac 23\times (A+1)$$

Similarly, thinking about the possible outcomes if you are in $S_5$ or $S_6$ we see that $$B=\frac 16\times 1+\frac 16\times (B+1)+\frac 23\times (A+1)$$

Solving that system of equations yields $$A=21\quad \&\quad B=18$$ so the answer is $\boxed {21}$.

Title: Expected number of rolls until a number appears 2 times discussion

Body:
I have the following problem. How many times you have to roll a die until you get two 6 or two 5 in a row.

Problems similar to this one have been discussed in the forum in links like:

[1] Number of rolls until the same number appears 2 consecutive times

[2] Expected number of rolls for fair die to get same number appear twice in a row?

I "solved" this problem as follows:

Let A the event of two consecutives 6 or two consecutive 5 in a row, and B the event of two consecutive equal numbers in a row. Then, it follows that:

\begin{equation}
P(A) = P(A \cap B) = P(A/B)P(B)
\end{equation}

$P(A/B)=2/6$ because if you already know that there are two identical numbers in a row then A can happen in 2 ways (5,5) and (6,6) over the sample space (i,i), $1\leq i \leq 6.$

But P(B) is a little bit more tricky. I know that the answer is 1/7 (I run a simple program to compute this number) but for me is counter intuitive.

In that case the Expected value would be = 21 because I expect 1 "win" combination out of 21 rolls.

My confusion is that I would have answered 1/6 instead of 1/7 in $P(B)$ because, no matter what number came out from the first dice, the second has 1/6 probability to match that number. Why this intuition fails? I know the answer is in [2] but I still can not handle the intuition on this.

A generalization of this result can be found here:
Expected number of rolls until a number appears $k$ times

Moreover, this problem is often to get confused with the 1/36 probability of throwing two dices simultaneously.

Links where to the discussion of this topic:

[-] If you roll a fair six sided die twice, what's the probability that you get the same number both times?

[-] Why are the probability of rolling the same number twice and the probability of rolling pairs different?


答:
The input matrices form a basis of $\Bbb R^{2\times2}$.

Write $v\in\Bbb R^{2\times2}$ as $a_1\begin{bmatrix}1&1\\1&1\end{bmatrix}+a_2\begin{bmatrix}0&1\\1&1\end{bmatrix}+a_3\begin{bmatrix}0&0\\1&1\end{bmatrix}+a_4\begin{bmatrix}0&0\\0&1\end{bmatrix}$. Then $v\in\ker f\iff f(v)=0_{\Bbb R^{2\times3}}$ i.e.$$a_1\begin{bmatrix}1&2&3\\1&2&3\end{bmatrix}+a_2\begin{bmatrix}3&3&5\\3&3&5\end{bmatrix}+a_3\begin{bmatrix}1&1&1\\1&1&1\end{bmatrix}+a_4\begin{bmatrix}1&0&1\\1&0&1\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\end{bmatrix}$$from where you get the three independent equations$$a_1+3a_2+a_3+a_4=0\\2a_1+3a_2+a_3=0\\3a_1+5a_2+a_3+a_4=0$$yielding $v=a_1\left(\begin{bmatrix}1&1\\1&1\end{bmatrix}-\begin{bmatrix}0&1\\1&1\end{bmatrix}+\begin{bmatrix}0&0\\1&1\end{bmatrix}+\begin{bmatrix}0&0\\0&1\end{bmatrix}\right)$ and $\ker f=\text{Span}\left\{\begin{bmatrix}1&0\\1&2\end{bmatrix}\right\}$.

Title: How can I find kernel of linear map?


答: The function is bounded below by $\left|\frac{t^2-100}{t+1}\right|(-1)+100$ and bounded above by  $\left|\frac{t^2-100}{t+1}\right|(+1)+100$.  Each approaches 100, so by the squeeze theorem the original limit is 100.</p>

Title: What can I do this cos term to remove the divide by 0?

Body:
I was asked to help someone with this problem, and I don't really know the answer why. But I thought I'd still try.

$$\lim_{t \to 10} \frac{t^2 - 100}{t+1} \cos\left( \frac{1}{10-t} \right)+ 100$$

The problem lies with the cos term. What can I do with the cos term to remove divide by 0 ?

I found the answer to be $100$ (Google), but I do not know what they did to the $\cos$ term. Is that even the answer ?

Thanks!


答:
This is somewhere between a comment and an answer.

The simplest way to solve this is that the value of $(2n+1)2^{4n+5}$ is precisely a function of $n \mod 42 = 6 \times 7$. More precisely, $2n+1 \mod 7$ is precisely a function of $n \mod 7$ while $2^{4n+5} \mod 7$ is precisely a function of $n \mod 6$. For each $i \in \{0,1,2,3,4,5\}$, you can check directly what $n \mod 7$ must be--in particular, what $(2n+1) \mod 7$ must be, given $n \mod 6$--in particular, what is $2^{4n+5} \mod 7$, which is a function of $n \mod 6$.

For example, for the case where $n = 0 \mod 6$ , we see that $2^{4n+5}= 1 \times 2^5 = 32 = 4 \mod 7$. So $2^{4n+5}$ is $4 \mod 7$ for this case where $n = 0 \mod 6$. So for the case where $n = 0 \mod 6$, the integer $n$ must satisfy the equation $(2n+1) \times 4 = 3 \mod 7$, which gives $n = 6 \mod 7$. So for the case where $n = 0 \mod 6$, the integer $n$ must also be $6 \mod 7$ which gives $n = 6 \mod 42$. What about the case where $n = 1 \mod 6$. What about each of the cases where $n \mod 6 = i=2,3,4,5$.

This in fact would work if $2n+1, 4n+5$ were each replaced by any two polynomials in $n$. Can you see why this is.


答:
Notice that $2^{4n+5}\equiv 2^{4n}2^5 \equiv 16^n\cdot 32 \equiv 2^n\cdot 4\equiv 2^{n+2}$.

And notice that if $k=3$ then $2^3 \equiv 8 \equiv 1 \pmod 7$.  That implies if $n = 3j + r$ where $r$ is the remainder of $n$ then $2^{n}\equiv 2^{3j+r} \equiv 2^{3j}2^r\equiv (2^3)^j2^r \equiv 1^j2^r \equiv  2^r$.  And as $r$ can be $0,1,2$ we have $2^n\equiv 2^0,2^1,2^2\equiv 1,2,4$ depending on what $n\pmod 3$ is equivalent to.

....

So our answer will depend on if $n = 3k$ or $n = 3k + 1$ or $n = 3k + 2$ for some $k$.

Suppose $n = 3k$ for some $k$..

Then $(2n+1)2^{4n + 5}\equiv 3\pmod 7$

$(6k+1)2^{12k}32\equiv 3\pmod 7$ now $6\equiv -1$ and $32\equiv 4$ and $2^{12k}\equiv(2^3)^{4k}\equiv 1$ so

$(-k+1)4 \equiv 3\pmod 7$

$-4k + 4\equiv 3\pmod 7$

$-4k \equiv -1 \pmod 7$

$4k \equiv 1 \pmod 7$

If we multiply both sides by $2$ we get

$8k\equiv 2\pmod 7$ so

$k \equiv 2 \pmod 7$ and $k = 7L + 2$ for some $L$ nad $n = 3k = 21L + 6$.

(Not the same answer as the book; could someone check if I made an arithmetic error.)

Suppose $n = 3k + 1$ for some $k$

$(2n+1)2^{4n+5}\equiv 3\pmod 7$

$(6k + 3)2^{12k + 9}\equiv 3\pmod 7$

$(-k+3)(2^3)^{4k+3}\equiv 3\pmod 7$

$-k + 3 \equiv 3 \pmod 7$

$k \equiv 0 \pmod 7$.

so $k = 7L$ for some $L$ and $n = 3(7L) + 1 = 21L + 1$

(Ditto errors)

Suppose $n=3k+2$ for som $k$

$(3n+1)2^{4n+5}\equiv 3\pmod 7$

$(6k + 5)2^{12k+13}\equiv 3\pmod 7$

$(-k-2)2^{12k+12}2\equiv 3\pmod 7$

$(-k-2)2 \equiv 3\pmod 7$

$2k + 4 \equiv -3\equiv 4 \pmod 7$

$2k \equiv 0\pmod 7$

$k \equiv 0\pmod 7$

So $k = 7L$ for so $L$ and $n=3(7L) + 2 = 21L + 2$.

(Ditto)

=== below is my old work =====

So if $n = 3j + r$ and $r=\begin{cases}0\\1\\2\end{cases}$ then

$(2n+1)2^{4n+5}\equiv (2(3j+r)+1)2^{r+2}\equiv$

$6j2^{r+2} + 2r2^{r+2} + 2^{r+2} \equiv -j2^{r+2} + r2^{r+3} + 2^{r+2}\equiv$

$-j2^{r+2} + r2^{r} + 2^{r+2}\equiv$

$-j2^{\begin{cases}2\\0\\1\end{cases}} + r2^{\begin{cases}0\\1\\2\end{cases}} + 2^{\begin{cases}2\\0\\1\end{cases}}\equiv $

$-j\begin{cases}4\\1\\2\end{cases}+ \begin{cases}0\\1\\2\end{cases}\begin{cases}1\\2\\4\end{cases}+\begin{cases}4\\1\\2\end{cases}\equiv$

$-j\begin{cases}4\\1\\2\end{cases}+ \begin{cases}0\\2\\1\end{cases}+\begin{cases}4\\1\\2\end{cases}\equiv$

$-j\begin{cases}4\\1\\2\end{cases}+ \begin{cases}4\\3\\3\end{cases}\equiv3 \pmod 7$

So

$-j\begin{cases}4\\1\\2\end{cases}+ \begin{cases}1\\0\\0\end{cases}\equiv0 \pmod 7$

There are three cases to solve

I) $r = 0$ and $-4j +1 \equiv 0 \pmod 7$

Let $3j+1 \equiv 0\pmod 7$

So if $j= 7k + s; s= 0,1,2,3,4,5,6$ we get

$3s + 1\equiv 0\pmod 7$ and the only one that fits is $s=2$.

so we can have $j = 7k + 2$ and $n = 3j= 3(7k+2)= 21k + 6$ for any $k$.

II) $r =1$ and $-j \equiv 3\pmod 7$

$j \equiv -3 \equiv 4 \pmod 7$.

so we can have $j = 7k + 4$ and $n = 3j + 1 = 3(7k + 4)+1 = 21k + 13$ for any $k$.

II) $r =2$ and $-2j \equiv 3\pmod 7$

$2j \equiv -3 \equiv 4 \pmod 7$.

If $j = 7k + s; s=0,1,2,3,4,5,6$ then $s=2$ is the only one that works

So $j = 7k + 2$ and $n = 3j + 2 = 21k + 8$ for any $k$.

Those are the solutions.


问:
Can someone explain to me why the mass formula, the expected value formula, and the variance formula for Poisson distribution?
Can someone explain to me why the mass formula for Poisson is $P(X=x)=\cfrac{e^{-\lambda}\lambda^x}{x!}$? Also why the expected value is $E(X)=\lambda$ and Variance is $Var(X) = \lambda$?

My book only gives the formula for mass and no explanation. It also gives the expected value formula and variance formula without any explanation. Can someone explain to me where these formulas come from so that I can better understand them?

Thank You!

答:
"My book only gives the formula for mass and no explanation." The Poison distribution is defined by the pmf you mentioned.

The expectation and variance are calculated using the pmf and definitions. (There are other methods if you know what the characteristic function is.)

The calculations are very standard materials. See for instance this set of notes:

https://llc.stat.purdue.edu/2014/41600/notes/prob1804.pdf


答:
mrsamy's method is quite elegant, though it says nothing about how large $x\gg 1$ should be.

If you need a thighter inequality, the best way is to develop

$$f(x)=(5x^2+3x+1)x-10(x^3-x^2-1)$$

Graph it and notice it is negative for some $x$ close to (and below) $3$

Then just calculate $f(u+3)=-5u^3-32u^2-56u-5<0\quad$ for $\ u\ge 0$

Therefore the inequality is valid as soon as $x\ge 3$ (the actual value is $\approx 2.9057$).

Title: Justifying $\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$ for large $x$

Body:
I need to find a way to justify the inequality $$\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$$ that holds for large values of x.

As an instance, I will justify an example inequality just to show you the intended strategy. Consider the inequality
$$\left|\frac{5x + 21}{2x^2 - 7}\right| < \frac{5}{x}.$$ For example, $2x^2 - 7$ can be written as $x^2 + (x^2 - 7)$. Because $x^2 - 7$ is a positive value for all $x < -\sqrt{7}$, removing it from the denominator makes the absolute value of the fraction greater. Also note that when $x < -\frac{21}{10}$, the numerator $|5x + 21| < 5|x|$, and this happens for all $x < -\sqrt{7}$. So $$\left|\frac{5x + 21}{2x^2 - 7}\right| < \frac{5|x|}{x^2} = \frac{5}{x}$$ as long as $x < -\sqrt{7}$.


Body:
Hint for a more direct approach not involving limits (at least not explicitly): if $x$ is large enough, then you can conclude
\begin{align*}
3x & < (0.01) x^2; \\
1 & < (0.01) x^2; \\
x^2 & < (0.01) x^3; \mathrm{~and} \\
1 & < (0.01) x^3.
\end{align*}
For such $x$, you will have:
$$5x^2 + 3x + 1 < 5x^2 + (0.01) x^2 + (0.01) x^2 = (5.02) x^2$$
and
$$x^3 - x^2 - 1 > x^3 - 0.01 x^3 - 0.01 x^3 = (0.98) x^3.$$
Therefore, you will be able to conclude:
$$\frac{5x^2 + 3x + 1} {x^3 - x^2 - 1} < \frac{(5.02) x^2}{(0.98) x^3} = \frac{5.02}{0.98} \cdot \frac{1}{x} < \frac{10}{x}.$$


问:
Why does velocity along $x$ axis not change as a function of time during projectile motion?
$U_x$ will be equal to $u\cos\theta $ and $u=\sin\theta$. But in case of final velocity, it is $v\cos\theta$ where $-gt =0$ and $v\sin\theta - gt$.

Why is that ? I am not getting how to solve this.

答:
The generic equation that describes your problem is $$\begin{cases}x(t)=v_0\cos\theta_0t\\y(t)=v_0\sin\theta_0t-\dfrac12gt^2 \end{cases}\to\begin{cases}v_x=v_0\cos\theta_0\\v_y=v_0\sin\theta_0-gt\end{cases}.$$
Let's take a look at the final situation. $y(t)$ is clearly zero, so
$$\tan\theta_0x-\dfrac{gx^2}{2v_0^2\cos^2\theta_0}=0\implies x_{\text{final}}=\dfrac{2v_0^2\cos^2\theta_0\tan\theta_0}{g}=\dfrac{v_0^2\sin(2\theta_0)}{g}$$
$$\implies x_{\text{final,max}}=\dfrac{v_0^2}{g},\theta_0=\pi/4\text{ and }t_{\text{final}}=\dfrac{v_0^2/g\sin(2\theta_0)}{v_0\cos(\theta_0)}$$
$$v_y(t_{\text{final}})=-v_0\sin\theta_0,v_x(t_{\text{final}})=v_0\cos\theta_0$$


问:
Asymptotic distribution of MLE

Let $X$ be a random variable such that it's density is given by
$$
f_X(x) =c(\theta)I_{\{|x|\leq \theta\}}\left(1-e^{-|x|}\right)
$$
Find the $c(\theta)$ that makes this function a density, find the maximum likelihood estimator of $\theta$ and the asymptotic distribution of $n(\hat{\theta}-\theta)$.

Im having trouble with the last item in the question.
My whole attempt goes like this

$$
c(\theta)\int I_{\{|x|\leq \theta\}}\left(1-e^{-|x|}\right)dx =1 \\
c(\theta)\int_{-\theta}^{\theta} \left(1-e^{-|x|}\right)dx = 1
$$
Using the fact that the function inside the integral is even, $f(x)=f(-x)$, we get
$$
c(\theta)2\int_{0}^{\theta} \left(1-e^{-x}\right)dx = 1 \\
c(\theta)2(x+e^{-x})|_{0}^{\theta}=1 \\
c(\theta) = 2^{-1}(\theta+e^{-\theta}-1)^{-1}
$$

So we get the following likelihood
$$
2^{-n}(\theta+e^{-\theta}-1)^{-n} I_{\{\max_{i \in \{1,\dots,n\}}|x|\leq \theta\}}\prod_{i=1}^{n}\left(1-e^{-|x|}\right)
$$
We can see that the likelihood is a decreasing function of $\theta$ as
$$
\frac{d}{d\theta}(\theta+e^{-\theta}-1)= 1-e^{-\theta}
$$
This is positive for values greater than $0$ so the function is increasing and thus the likelihood is decreasing in $\theta$ making $\max_{i \in \{1,\dots,n\}}|x|$ the MLE of $\theta$.

Finally for the last part, I used the fact that
$$
P(|X|<x)=P(-x<X<x)
$$
and that the distribution of the maximum is
$$
P(X_{(n)}<x)=P^{n}(X<x)
$$
Getting the following CDF
$$
F_{\hat{\theta}}(x)=\left(\frac{x-1+ e^{-x}}{\theta+e^{-\theta}-1}\right)^n
$$
I tried calculating this probability to see if I notice something about it, however it led me nowhere
$$
P(n(\hat{\theta}-\theta)<x) = P(\hat{\theta} < xn^{-1}+\theta) =\left(\frac{xn^{-1}+\theta-1 +\exp(xn^{-1}+\theta)}{\theta + e^{-\theta}-1}\right)^n
$$
I would appreciate help.

答: You can use your cdf $F_{\hat{\theta}}$ to get the pdf for $\hat{\theta}$ (which you should notice is supported on $[0,\theta]$) by taking a derivative: $$f_{\hat{\theta}}(x)=n\Bigg(\frac{x+e^{-x}-1}{\theta+e^{-\theta}-1}\Bigg)^{n-1}\Bigg(\frac{1-e^{-x}}{\theta+e^{-\theta}-1}\Bigg)$$ Since $\hat{\theta}$ is supported on $[0,\theta]$ we have that $n(\hat{\theta}-\theta)$ is supported on $[-n\theta,0]$. Using the fact that $f_{n(\hat{\theta}-\theta)}(x)=\frac{1}{n}f_{\hat{\theta}}\big(\frac{x}{n}+\theta\big)$ we get $$f_{n(\hat{\theta}-\theta)}(x)=\Bigg(\frac{\frac{x}{n}+\theta-e^{-\big(\frac{x}{n}+\theta\big)}-1}{\theta+e^{-\theta}-1}\Bigg)^n\Bigg(\frac{1-e^{-\big(\frac{x}{n}+\theta\big)}}{\frac{x}{n}+\theta+e^{-\big(\frac{x}{n}+\theta\big)}-1}\Bigg)$$ The support of $f_{n(\hat{\theta}-\theta)}$ encompasses the entire negative real axis as $n \longrightarrow \infty$. The right hand term in the above expression approaches $\frac{1-e^{-\theta}}{\theta+e^{-\theta}-1}$ as $n$ gets large. Using the substitution $m=x/n$ along with L'Hopital's rule, you can easily show that $$\Bigg(\frac{\frac{x}{n}+\theta-e^{-\big(\frac{x}{n}+\theta\big)}-1}{\theta+e^{-\theta}-1}\Bigg)^n \longrightarrow \exp \Bigg\{\frac{x\big(1-e^{-\theta}\big)}{\theta+e^{-\theta}-1}\Bigg\}$$ Finally the limiting distribution of $\hat{\theta}$ is the pdf $f$ defined by $$f(x)=\frac{1-e^{-\theta}}{\theta+e^{-\theta}-1}\exp \Bigg\{\frac{x\big(1-e^{-\theta}\big)}{\theta+e^{-\theta}-1}\Bigg\} $$ which is supported on $(-\infty,0]$. In other words, the limiting distribution of $\hat{\theta}$ is $-\text{Exponential}\Bigg(\frac{1-e^{-\theta}}{\theta+e^{-\theta}-1}\Bigg)$.

Title: Asymptotic distribution of MLE


答:
Start by showing that the four $2\times 2$ matrices $A_1, A_2, A_3, A_4$ you know the images under $f$ of form a basis of $\mathbb R^{2\times 2}$. Then any $A\in\mathbb R^{2\times 2}$ can uniquely be expressed as $A = \lambda_1 A_1 + \dots + \lambda_4 A_4$. Using linearity, $f(A)=0$ is equivalent to
$$
\lambda_1 \begin{pmatrix}
    1 & 2 & 3\\
    1 & 2 & 3
  \end{pmatrix} +
\lambda_2 \begin{pmatrix}
    3 & 3 & 5\\
    3 & 3 & 5
  \end{pmatrix} +
\lambda_3 \begin{pmatrix}
    1 & 1 & 1\\
    1 & 1 & 1
  \end{pmatrix} +
\lambda_4 \begin{pmatrix}
    1 & 0 & 1\\
    1 & 0 & 1
  \end{pmatrix} =
\begin{pmatrix}
    0 & 0 & 0\\
    0 & 0 & 0
  \end{pmatrix}.
$$
This is a linear system of $6$ equations in $4$ variables, but it is rather redundant. It is equivalent to
$$
\lambda_1 \begin{pmatrix}
    1 \\ 2 \\ 3
  \end{pmatrix} +
\lambda_2 \begin{pmatrix}
    3 \\ 3 \\ 5
  \end{pmatrix} +
\lambda_3 \begin{pmatrix}
    1 \\ 1 \\ 1
  \end{pmatrix} +
\lambda_4 \begin{pmatrix}
    1 \\ 0 \\ 1
  \end{pmatrix} =
\begin{pmatrix}
    0 \\ 0 \\ 0
  \end{pmatrix}.
$$
Row reduction yields
$$
\begin{pmatrix}
1 & 3 & 1 & 1 \\
2 & 3 & 1 & 0 \\
3 & 5 & 1 & 1
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 0 & 0 & -1 \\
0 & 1 & 0 &  1 \\
0 & 0 & 1 & -1
\end{pmatrix}
$$
And hence the solution coefficient vectors are
$$
(\lambda_1, \lambda_2, \lambda_3, \lambda_4) \in \operatorname{span}\big((1,-1,1,1)\big).
$$
Thus, the kernel of $f$ is spanned by
$$
\begin{pmatrix}
    1 & 1\\
    1 & 1\\
  \end{pmatrix} -
\begin{pmatrix}
    0 & 1\\
    1 & 1\\
  \end{pmatrix} +
\begin{pmatrix}
    0 & 0\\
    1 & 1\\
  \end{pmatrix} +
\begin{pmatrix}
    0 & 0\\
    0 & 1\\
  \end{pmatrix}
=
\begin{pmatrix}
    1 & 0\\
    1 & 2\\
  \end{pmatrix}.
$$

Title: How can I find kernel of linear map?


问:
Connected components of a closed subset of $\mathbb R^2$ one of whose bounadary is a subset of the other
Let $K$ be a compact, connected subset of $\mathbb R^2$ which is the closure of its interior. Suppose that there are two connected components $U,V$ of $\mathbb R^2\setminus K$ whose boundaries satisfy $\partial V \subset \partial U$. Must it be the case that $ K$ has a cut point (i.e., there is a point $x \in K$ such that $K \setminus \{x\}$ is not connected)?

Intuitively, it seems like this should be the case (I was not able to think of any examples where it would fail) but I am unable to find a proof. A proof or a reference would be much appreciated.

答: $S^1$ is compact and connected and is the common boundary of itself and the two components of $\Bbb R^2\setminus S^1$, but it has no cut point.


问:
Probability for Union of Events when one of the probabilities is 0.
If, for example, $P(A \cup B)= P(B)$, does this automatically mean that $P(A)=0$ for all instances?

My understanding is that $P(A \cup B)= P(A) + P(B) - P(A \cap B)$ wherein $P(A \cap B)= P(A) \cdot P(B)$. The only instance that $P(A \cup B)= P(B)$ is if $P(A)=0$ which leads to $P(A \cap B)=0$. Is there an exception to this or is there any particular situation wherein this might not be the case?

答:
Simply consider $A=B=\Omega$. Then $P(A\cup B)=P(B)=P(A)=1$.

What goes wrong in your second paragraph is that $P(A\cap B)=P(A)P(B)$ is not true in general. It is only true when $A$ and $B$ are independent events.

If $A$ and $B$ are assumed to be independent, then
$$
P(A\cup B)=P(A)+P(B)-P(A)P(B)=P(B)
$$
implies that $P(A)(1-P(B))=0$ which implies $P(A)=0$ or $P(B)=1$.


问:
calculate $\lim_{t\rightarrow1^+}\frac{\sin(\pi t)}{\sqrt{1+\cos(\pi t)}}$
How to calculate $$\lim_{t\rightarrow1^+}\frac{\sin(\pi t)}{\sqrt{1+\cos(\pi t)}}$$? I've tried to use L'Hospital, but then I'll get

$$\lim_{t\rightarrow1^+}\frac{\pi\cos(\pi t)}{\frac{-\pi\sin(\pi t)}{2\sqrt{1+\cos(\pi t)}}}=\lim_{t\rightarrow1^+}\frac{2\pi\cos(\pi t)\sqrt{1+\cos(\pi t)}}{-\pi\sin(\pi t)}$$
and this doesn't get me further. Any ideas?

答:
I'll show you a trick: since $t\to 1^+$ you know that $\pi t\to \pi^+$, so $(\pi t -\pi)\to 0^+$.

First of all do the change of variable $u=\pi t-\pi$, so you get $\pi t=u-\pi$. Then $\sin(\pi t)=\sin(u-\pi)=-\sin u$ and $\cos(\pi t)=\cos(u-\pi)=-\cos u$. Thus your limit becomes

$$
\lim_{u\to 0^+}\frac{-\sin u}{\sqrt{1-\cos u}}=
-\lim_{u\to 0^+}\sqrt{\frac{\sin^2 u}{1-\cos u}}=
-\lim_{u\to 0^+}\sqrt{1+\cos u}=-\sqrt{2}
$$

The first equality is justified because $\sin u>0$ for $0<u<\pi$ (and you're interested in a right neighborhood of $0$).

Limits at zero are "psychologically" better, aren't they? There's really no difference with doing the limit at $\pi$ or using $\pi t$, but the region around $0$ is better known.


问:
Simple question on full differentials:
This one should be a quickie:

if we are calculating $\oint xdy+ydx$ around the curve $x^4 + y^4 = 1$, I can spot quickly that $\oint xdy+ydx = \oint df$ where $f = xy$. How would I then evaluate $\oint df$? Is it simple, or am I better off with a parameterisation approach ($x^2 = \mathrm{cos} \ t$ and so on)?

答:
Your form $\omega=ydx+xdy$ is defined in $\Omega=\mathbb R^2$, which is contractible, hence it's simply connected (the path $\gamma\equiv x^4+y^4=1$ is closed and continuous).

We can now solve the exercise using the fact that a form in $\Omega$ is closed $\iff$ the form is exact, so we observe that
$$\dfrac{\partial}{\partial_y}y=1=\dfrac{\partial}{\partial_x}x\overset{DF}{\implies}\omega\text{ is closed}\implies\omega\text{ is exact.}$$

Title: Simple question on full differentials:


问:
Vector division
I've been trying to understand how vectors work and this is my question so far: Is it possible to divide two parallel vectors? Is it possible to divide a vector by a scalar?</p>

答:
It is certainly possible to divide a vector by a nonzero scalar. In this case, you simply are dividing each component individually, e.g.:

$$(2,4,6)/2 = (1,2,3)$$

You can think about scalar division by $c$ as a scalar multiplication by $1/c$.

However, vector division by another vector is not usually well-defined. This is because cross products can not be uniquely inverted, in general.

Title: Vector division


问:
Affine mapping task
I'm currently refreshing my Mathematics knowledge to pass an exam, so I found a list of tasks and got stuck on this one. Could anyone please help me to understand, what am I supposed to even start with?

Here is the task:

A subset $Y$ of a vector space $V$ is called an affine subspace if there exists a $v \in V$ and

there exists a subspace $W$ of $V$ such that:

$$Y = v + W := \{u \in V | \exists w \in W : u = v + w \}$$

Show that:

For all $v' \in Y$ indeed $Y = v´ + W$

For another vector space $Z, f : V \to Z$ is linear, $z \in Z$ and $u \in f^{-1} ({z})$ is valid $f^{-1} ({z}) = u + \ker f$.

答:
For 1. start by showing $Y\subseteq v' + W$. For that, take $u = v + w \in Y$, note that $u = v' + (u - v')$ and check that $u-v'\in W$. The inclusion $Y\supseteq v' + W$ is shown in the same fashion.

For 2. note that any vector of the form $u+v$ with $f(u)=z$ and $f(v)=0$ also satisfies $f(u+v) = z$. Conversely whenever $f(u')=z$ you have $f(u'-u)=z-z=0$ so that $u'-u\in\operatorname{ker}(f)$ and hence $u'=u+(u'-u)\in u+\operatorname{ker}(f)$.

Title: Affine mapping task


答:
Hint:

Consider the standard bases of$\mathcal M_2(\mathbf R)$
and $\mathcal M_{2\times 3}(\mathbf R)$::
$$E_1=\begin{bmatrix}1&0 \\0&0\end{bmatrix},\enspace E_2=\begin{bmatrix}0&1 \\ 0&0\end{bmatrix},\enspace E_3=\begin{bmatrix}0&0 \\1&0\end{bmatrix},\enspace E_4=\begin{bmatrix}0&0 \\0&1\end{bmatrix} $$
$$e_1=\begin{bmatrix}1&0&0 \\0&0&0\end{bmatrix},\enspace e_2=\begin{bmatrix}0&1 &0 \\ 0&0&0\end{bmatrix},\enspace e_3=\begin{bmatrix}0&0 &0 \\ 0&0&1\end{bmatrix}, \cdots$$
The matrix of the linear map $f$ is
$$T_f=\begin{bmatrix}-2 & 2 & 0 & 1\\-1 & 2 & 1 & 0 \\-2 & 4 & 0 &1 \\ -2 & 2 & 0 & 1 \\ -1 & 2 & 1 & 0 \\ -2 & 1 & 0 & 1\end{bmatrix},$$
and, as it has obviously rank $\le 3$, you can simply consider the matrix
$$A=\begin{bmatrix}-2 & 2 & 0 & 1\\-1 & 2 & 1 & 0 \\-2 & 4 & 0 &1 \end{bmatrix}$$
to determine the kernel, which can easily be done by row reduction.

Title: How can I find kernel of linear map?


问:
How many N-digit numbers have the sum of odd places digits equal the sum of even places digits?
We define a N-digit number as a sequence $d_{1}d_{2}d_3...d_N$, where $d_k\in\{0,1,2,3,4,5,6,7,8,9\}$. So the first digits can be equal to zero. N is a positive even integer.
The problem is to find, how many N-digit numbers are there, which have their sum of the digits on odd places equal to the sum of digits on the even places. Example number would be 2563 for 4-digit numbers as 2 + 6 = 5 + 3.

My thought was to choose the odd digits, we can do that in $10^{N/2}$ ways and then for each of that combinations make even digits a permutation of them ($\frac{N}{2}!$), but this, if correct at all, is only a small fraction of all the possibilities. Why? Because the digits do not have to be the same. Only the sums have to be equal.

Then my second though was that each of such numbers consists of digits from other (shorter) numbers that satisfy the requirement, but it is not true since only the sums have to be equal.

We can for each sum of odd/even digits calculate how many ways we can express it on N/2 digits and then sum the squares of such numbers of combinations. However that leads to very unfriendly calculations.

After you find a nice way, please provide some example calculations for N=4 and N=6.

答:
Note: This is just a supplement to the answer from @AlexRavsky which I think is a most convenient approach. For those who are interested in a nice derivation of the problem written in English may consult section 1.1 The lucky tickets problem in*Lectures on Generating Functions* by S.K.Lando.

The author introcudes this section by mentioning that in the early 70s A.A. Kirillov usually opened his seminar with this problem :-)

Here is a variation based upon PIE
the principle of inclusion-exclusion. We consider digits in $\mathcal{V}=\{0,1,\ldots,9\}$ and words of even length $N=2m$ built from $\mathcal{V}$. We are looking for the number $a_N$ of words where the sum of digits at even positions is equal the sum of digits at odd positions:
\begin{align*}
d_1d_2\ldots d_{N-1}d_N\qquad\mathrm{with}\qquad\sum_{j=1}^{m} d_{2j}=\sum_{j=1}^{m} d_{2j-1}\tag{1}
\end{align*}
We replace the digits $d_{2j}$ at even positions with $9-d_{2j}$ and consider
\begin{align*}
d_1\left(9-d_2\right)\ldots d_{N-1}\left(9-d_N\right)\qquad\mathrm{with}\qquad\sum_{j=1}^{2m} d_{j}=9m\tag{2}
\end{align*}
We have an obvious bijection between the words in (1) and (2). We can consequently calculate $a_N$ by counting the number of words with $N$ digits from $\mathcal{V}$ which have sum $9m$.

PIE: We apply PIE as follows:

We consider the set of all distributions of non-negative integers at $N=2m$ positions which have sum $9m$. This number is given by
\begin{align*}
\binom{9m+2m-1}{2m-1}=\binom{11m-1}{2m-1}
\end{align*}

We denote with the property $p_j, 1\leq j\leq N$ those words where the number at the $j$-th position is at least $10$. The number of words with length $N=2m$ which have property $p_j$ is
\begin{align*}
\binom{11m-11}{2m-1}
\end{align*}

The number of words which have for $1\leq j\ne k\leq N$ at  positions $j$ and $k$  at least $10$, which is the number of words which have properties $p_{j}$ and $p_k$ is
\begin{align*}
\binom{11m-21}{2m-1}
\end{align*}
Continuing this way we observe the wanted number $a_N$ is
\begin{align*}
\color{blue}{a_N}&\color{blue}{=\binom{11m-1}{2m-1}-\binom{N}{1}\binom{11m-11}{2m-1}
+\binom{N}{2}\binom{11m-21}{2m-2}-+\cdots}
\end{align*}

Example: $N=2m=6$

We obtain:
\begin{align*}
\color{blue}{a_6}&=\binom{32}{5}-\binom{6}{1}\binom{22}{5}+\binom{6}{2}\binom{12}{5}\\
&=201\,376-6\cdot 26\,334+15\cdot 792\\
&\,\,\color{blue}{=55\,252}
\end{align*}


问:
Expansion of the Characteristic polynomial via exterior product
Looking at wiki, I want to

$p_A (t) = \sum_{k=0}^n t^{n-k} (-1)^k \operatorname{tr}(\Lambda^k A) $

where ${\displaystyle \operatorname {tr} (\Lambda ^{k}A)={\frac {1}{k!}}{\begin{vmatrix}\operatorname {tr} A&k-1&0&\cdots &\\\operatorname {tr} A^{2}&\operatorname {tr} A&k-2&\cdots &\\\vdots &\vdots &&\ddots &\vdots \\\operatorname {tr} A^{k-1}&\operatorname {tr} A^{k-2}&&\cdots &1\\\operatorname {tr} A^{k}&\operatorname {tr} A^{k-1}&&\cdots &\operatorname {tr} A\end{vmatrix}}~.}$

Wikipedia states this can be proven via the language of exterior algebra, How?

答:
Let $v_1,\dots,v_n$ be a basis of $V$, consisting of eigenvectors with eigenvalues $\lambda_1,\dots,\lambda_n$. Then $\mathrm{tr}(A) = \sum \lambda_i$.

As a basis for $\Lambda^k V$, you can take the collection of all $v_{i_1} \wedge \cdots \wedge v_{i_k}$, where $i_1 < \cdots < i_k$. Since by definition,
$$ (\Lambda^k A)(v_{i_1} \wedge \cdots \wedge v_{i_k}) = (Av_{i_1}) \wedge \cdots \wedge (Av_{i_k}) = \lambda_{i_1} \cdots \lambda_{i_k} \, v_{i_1} \wedge \cdots \wedge v_{i_k} $$
you see that this basis also consists of eigenvectors, with corresponding eigenvalues $\lambda_{i_1} \cdots \lambda_{i_k}$. So the trace of $\Lambda^k A$ is the sum of its eigenvalues, giving
$$ \mathrm{tr}(\Lambda^k A) = \sum_{i_1 < \cdots < i_k} \lambda_{i_1} \cdots \lambda_{i_k} $$
If you are familiar with symmetric polynomials, this is the "elementary symmetric polynomial" $e_k(x_1,\dots,x_n)$, evaluated at $x_i = \lambda_i$ for all $i$.

On the other hand, $\mathrm{tr}(A^k) = p_k(\lambda_1,\dots,\lambda_n)$, where these are the "power sum symmetric polynomials" $p_k(x_1,\dots,x_n) = x_1^k + x_2^k + \cdots + x_n^k$.

The determinant formula you quoted from Wikipedia follows from the well-known "Newton Identity", which tells how to express the two types of symmetric polynomials in terms of the other ones. See the Wikipedia page for Newton identity for more info.


答:
We define the quotient of two numbers as the number that, when multiplied with the divisor, produces the dividend - provided such a number exists and is unique.

So nothing prevents you from transfering this definition to vectors.
So the quotient of two vectors $\vec u$ and $\vec v$ should be a something $\xi$ such that $\vec u$ is $\xi\cdot \vec v$ (or perhaps $\vec v\cdot \xi$). What kind of object could $\xi$ be? Not a vector (unless in the special cas that you work in three dimensions and use the cross product as product). It could be a linear map. Or it could be  a scalar. The linear map would not be unique (except in case of one dimension), so let's go with the scalar. That would be unique as long as $\vec v\ne\vec 0$ (sounds familiar), but exist only when $\vec u$ and $\vec v$ are linearly dependent (or parallel). So here you have your division of vectors.

For dividing a vector by a scalar, things are easier: We want $\frac{\vec u}{a}$ to be some object $\xi$ with $\vec u=a\cdot \xi$ (or perhaps $\xi\cdot a$). The obvious choice is to let $\xi$ be the vector $\frac 1a\cdot \vec u$ (provided $a\ne 0$, of course).

Title: Vector division

Body: I've been trying to understand how vectors work and this is my question so far: Is it possible to divide two parallel vectors? Is it possible to divide a vector by a scalar?</p>