问:
Need help in proving that 3 roots of $z^4-6z+3=0$ lies in $|z|<2$
This question is from Ponnusamy and Silvermann Complex Variables with Applications (subsection: Argument principle) page 346.

Show that there exists $1$ root in $|z|<1$ and $3$ roots in $|z|<2$ of the equation $z^4 -6z +3=0$.

I have proved that there exists a root in $|z|<1$ but I am unable to prove the assertion for $|z|<2$. The problem arises as I have to look at a third degree equation and I am unable to find a suitable candidate .

So,  I am looking for your help!

答: Let $\varepsilon(z)=-6z+3$ and let $f(z)=z^4$. Then $z^4-6z+3=f(z)+\varepsilon(z)$. Furthermore, if $|z|=2$, then\begin{align}\bigl|\varepsilon(z)\bigr|&=|-6z+3|\\&\leqslant15\\&<16\\&=|z^4|\\&=\bigl|f(z)\bigr|.\end{align}So, your function has as many zeros on $D(0,2)$ as $f$, which has $4$ zeros there (if we count them with their multiplicities).


问:
Recursive relation induction problem: $c_n = \frac{1}{2^n} \cdot \sum_{k \text{ even}}^{0~ \leq ~k~ \leq ~n} {n \choose k}c_k$
This is part of problem 18 of chapter 13 of Michael Spivak's Calculus:

Define $c_n = \int_{0}^{1}x^n dx$

(b) Prove: $$c_n = \frac{1}{2^n} \cdot \sum_{k \text{ even}}^{0~ \leq ~k~ \leq ~n} {n \choose k}c_k$$

(c) Prove $c_n = \frac{1}{n+1}$ from this (without calculating the integral).

Note the actual question in the book is more than this, I am just condensing it to the parts I'm stuck on. I was able to do (b), but not (c). It seems to me to be another induction problem, but I am going in circles. So I assume it's true for $\forall n \leq p$, i.e. $ \forall n \leq p : c_n = \frac{1}{n+1}$. Then for $c_{p+1}$, we have:

$$c_{p+1} = \frac{1}{2^{p+1}} \cdot \sum_{k \text{ even}}^{0~ \leq ~k~ \leq ~{p+1}} {p+1 \choose k}c_k$$

Now already there's some technical issues as to whether the last term in the sum will be $c_{p+1}$ or $(p+1)c_p$, depending on whether $p+1$ is even or odd. So I split up the formula into those 2 separate cases, like so:

$$c_{n} = \frac{1}{2^{n} - 1} \cdot \sum_{k~ =~ 0}^{k ~= ~n/2-1} {n \choose 2k}c_{2k} ~~~: ~~\text{for n = even}$$

$$c_{n} = \frac{1}{2^{n}} \cdot \sum_{k~ =~ 0}^{k ~= ~(n-1)/2} {n \choose 2k}c_{2k} ~~~~~~: ~~\text{for n = odd}$$

Ok so we'll probably have to do 2 separate proofs by induction, one for when $p$ is odd, one when $p$ is even. Also we'll need to prove both the $n = 0$ and $n = 1$ base cases, but that's trivial, no problems there.

Let's try when $p$ is even. So we assume $p$ is even and $ \forall n \leq p : c_n = \frac{1}{n+1}$. Then $p+1$ is odd, so:

$$c_{p+1} = \frac{1}{2^{p+1}} \cdot \sum_{k~ =~ 0}^{k ~= ~p/2} {p+1 \choose 2k}c_{2k}$$

$$~~~~~~~~~~~~~~= \frac{1}{2^{p+1}} \cdot \sum_{k~ =~ 0}^{k ~= ~p/2} {p+1 \choose 2k}\frac{1}{2k+1}$$

And now I am stuck.

I need to somehow show $\frac{1}{2^{p+1}} \cdot \sum_{k~ =~ 0}^{k ~= ~p/2} {p+1 \choose 2k}\frac{1}{2k+1} = \frac{1}{p+2}$. I tried using the identity ${n+1 \choose k} = {n \choose k} + {n \choose k-1}$, which gives us:

$$c_{p+1} = \frac{1}{2^{p+1}} \cdot \sum_{k~ =~ 0}^{k ~= ~p/2} {p+1 \choose 2k}\frac{1}{2k+1}$$

$$~~~~~~~~~~~~~~~~~~~~~~~~~~= \frac{1}{2^{p+1}} \cdot \sum_{k~ =~ 0}^{k ~= ~p/2} \Bigr[{p \choose 2k} + {p \choose 2k-1}\Bigr]\frac{1}{2k+1}$$

$$~~~~~~~~~~~~~~~~~~~~~~~~~= \frac{1}{2^{p+1}} \cdot \biggr[c_p ~+~ \Bigr[ \sum_{k~ =~ 0}^{k ~= ~p/2} {p \choose 2k-1}\frac{1}{2k+1}\Bigr]\biggr]$$

But now how to deal with this new summation that runs on odd numbers; $2k - 1$, instead of the even numbers; $2k$, that we have in our definition of $c_p$ (which we need to incorporate somehow)? I tried to transform the ${p \choose 2k - 1}$ to ${p \choose 2k - 2}$ or ${p \choose 2k}$, using the identity ${n \choose k} = \frac{n - k + 1}{k}{n \choose k - 1}$ but the coefficients didn't cancel and looked very ugly so now I have no ideas.

Any?

答:
You say you are stuck on proving $$\frac{1}{2^{p+1}}\sum_{k=0}^{p/2}\binom{p+1}{2k}\frac{1}{2k+1}=\frac{1}{p+2}$$ when $p$ is even. We have
$$\frac{1}{2^{p+1}}\sum_{k=0}^{p/2}\binom{p+1}{2k}\frac{1}{2k+1}=\frac{1}{2^{p+1}(p+2)}\sum_{k=0}^{p/2}\binom{p+2}{2k+1}.$$ The sum computes the sum of coefficients of odd power terms in polynomial $f(x)=(1+x)^{p+2}$. That is equal to $(f(1)-f(-1))/2$ for any polynomial $f$. In this case it evaluates to $2^{p+1}$, which is what you want.

ContentLicense: CC BY-SA 4.0


问:
Uniqueness of the solution of $Ax=b$
Given a matrix equation
$$ {A}_{n\times m} x_{m\times 1}=b_{n\times 1},$$

how to find out if the solution vector x is unique?

答:
For any $b$, solution exists IFF   $C(A)=\mathbb{R}^n$   IFF   $rankA=n$

I suppose you found such $b$, then

$b$ is unique IFF $N(A)=\{0\}$ IFF number of free variable is zero.

Title: Uniqueness of the solution of $Ax=b$


问:
Given a $2\times 2$ matrix $A$, does two unique eigenvalues guarantee that $A$ is diagonizable?
This question relates to a seminar I've been working on, so I do not wish to disclose the whole question but simply ask how this would be handeled in theory.

$A = \begin{bmatrix}a & b \\ c & d \end{bmatrix}$
$Q=(\lambda I - A) = \begin{bmatrix}\lambda - a & -b \\ -c & \lambda - d \end{bmatrix}$

We then want to find the eigenvalues of $A$, which responds to solving $\det(Q)=0$. My question is, given that we would get two unique eigenvalues, does that mean we are guaranteed that matrix $A$ is diagonalizable? I have only found a theorem that says if we have two distinct eigenvectors for a $2\times 2$ matrix $A$, then $A$ is diagonalizable...

Help would be much appreciated!

答:
If $A$ is an $n \times n$ matrix the eigenvalues of which are distinct, there exist non-zero vectors $V_i$, $1 \le i \le n$, with

$AV_i = \mu_i V_i \tag 1$

the $\mu_i$ being the distinct eigenvalues of $A$.  It is well-known that the eigenvectors associated with distinct eigenvalues are linearly independent; thus the matrix

$S = [V_1 \; V_2 \; \ldots \; V_n ] \tag 2$

is non-singular and hence invertible, so there exists an $n \times n$ matrix $S^{-1}$ with

$S^{-1}S = SS^{-1} = I; \tag 3$

also,

$AS = [AV_1 \; AV_2 \; \ldots \; AV_n ] = [\mu_1 V_1 \; \mu_2 V_2 \; \ldots \; \mu_n V_n]; \tag 4$

thus

$S^{-1}AS = S^{-1} [\mu_1 V_1 \; \mu_2 V_2 \; \ldots \; \mu_n V_n] =  [\mu_1 S^{-1} V_1 \; \mu_2  S^{-1} V_2 \; \ldots \; \mu_n S^{-1}  V_n]; \tag 5$

now in accord with (2) and (3),

$S^{-1}S = S^{-1} [V_1 \; V_2 \; \ldots \; V_n ] = [S^{-1} V_1 \; S^{-1} V_2 \; \ldots \; S^{-1} V_n ] = I, \tag 6$

which shows that each $S^{-1} V_i$ is the column vector whose $i$-th entry is equal to $1$ with all other elements $0$; incorporating this observation into (5) we obtain

$S^{-1}AS = \text{diag}(\mu_1, \; \mu_2, \;  \ldots, \; \mu_n), \tag 7$

and we thus find that $A$ is diagonalized by $S$. $OE \Delta$.


问:
What does this mean: there exist an integer N such that $n\ge N$?
I'm reading Rudin's, Principles of Mathematical Analysis, and I keep tripping over this phrase. Usually the phrase by that it implies a some equation with n being the index, subscript, of a point. My guess is that the phrase means that there is two points and that 'n' is the limit. Not sure, I'll greatly appreciate any help with this.

(edit): I admit the question is vague without its implication but I wanted to clear with the premise first. Thanks for the fast and helpful responses.

答:
The phrase "there exist an integer N such that [for all] $n\ge N$, ..."

is quantified prepositional clause, (is not an assertion), itself, after which something is asserted about $n$ in relation to $N$.

That is, "there exist an integer N such that [for all] $n\ge N$ ... " [something is claimed to be  true about $n$ when for all $n \ge N$].

So, we say something like

"Let $\{s_n\}$ be a sequence of real numbers with the following property: For every real M, there exists an integer $N$ such that $n\geq N$ implies $x_n\leq M,$ we then write $s_n \to +\infty$."  (p. 55 Rudin PMA, Definition 3.15)

Here, indeed, $n$ refers to the index (position) of some point in the sequence, $n\geq N$ meaning $s_n$ appears later in the sequence than does $s_N$, after which the sequence diverges.

But $n$ can be a real number, as well, when the clause you are asking about is used in a different context.

ContentLicense: CC BY-SA 3.0


问:
Why is this set is closed in the subspace $([0,\infty)\times \mathbb{R}) \cup (\mathbb{R}\times \{0\})$ of $\mathbb{R}^2$
I am trying to show the set $\{(x,y)\in \mathbb{R}^2\ : xy=1, x>0\}$ is closed in the subspace $X := ([0,\infty)\times \mathbb{R}) \cup (\mathbb{R}\times \{0\})$ of $\mathbb{R}^2$.

To do this I need to show its complement equals $U \cap X$ for some open subset $U$ of $\mathbb{R}$. But the complicated structure is making it difficult.

If I'm not mistaken, the complement of $X$ is $\{(x,y)\in \mathbb{R}^2: x\leq 0$ or $y\leq 0$ or $xy\neq 0\}$

答:
The map $f \colon \mathbb{R}^2 \to \mathbb{R}$ defined by $f(x,y) = xy$ is a continuous function and the set $\{ (x,y) : xy = 1 \} = f^{-1}(\{1\})$ is the pre-image of a closed set $\{1\}$ so it's closed by continuity in $\mathbb{R}^2$.

Call your set $A = \{ (x,y) : xy = 1, x > 0 \}$.

Let $X$ be a subspace of $Y$. A set $A$ is closed in $X$ if and only if there exists a closed $C$ in $Y$ s.t. $C \cap X = A$.

Pick $Y = \mathbb{R}^{2}$ and $C = f^{-1}(\{1\})$.


问:
Draw a graph of periodic extension of a piecewise given function
Let $f(t)$ be a function given as follows:
\begin{equation*}
f(t)=\begin{cases}
          1 \quad &\text{if} \, x \in [0,1) \\
          2 \quad &\text{if} \, x \in [1,2) \\
     \end{cases}
\end{equation*}
Draw the graph of the odd extension of the previous function with period $4$ over the interval $(-4,4)$?
How to go ahead with such a function and draw it?

答: Given the piecewise function, you can graph the function from [0,2) (I assume you've already done that.) Then it tells you that the graph is odd, meaning it is symmetric around the origin. This tells you that f(t)= -1 if x∈[−1,0) and f(t)= -2 if ∈[−2,−1). Now you want to extend the periodic function from (-4,4). It is given that the periodic function has a period of 4. We have already graphed from [-2,2) so you can determine how the graph looks from [2,4) and (-4,2). f(t)= -2 if x∈[2,3), f(t)= -1 if ∈[3,4), f(t)= 2 if x∈[−3,-2), and f(t)= 1 if ∈(−4,-3). Sorry if this is hard to read, I'm also a little unsure about the interval notation I used.</p>


问:
(Hard Question) Calculate Number Of Vectors
What is the number of vectors of 4 where each digit is from 0 to 9 (10 possibilities) and X is the maximum digit?

I tried the following:

X is the maximum then once for sure is equal for X
Then the rest have X+1 possibilities, So in total I have: X*(X+1)^3 but that won't work.

Why? because using this method I am calculating duplicates

Any idea on how to solve this please?

答: The number of $k$-vectors where each element is an integer from $1$ to $n$ – i.e. the maximum is at most $n$ – is $n^k$. Thus the number of the same but with maximum exactly $n$ is $n^k-(n-1)^k$.


问:
Prove that if H is a normal subgroup of G and K is a normal subgroup of H, then K may not be a normal subgroup of G.
I was doing a course on algebra and had this question written in my notes.

Prove that if $H$ is a normal subgroup of $G$ and $K$ is a normal subgroup of $H$, then $K$ may not be a normal subgroup of $G$.

Now as I understand to prove a subgroup $K$ normal to $G$ I have to do $g^{-1}kg$ belongs to $K$. Clearly, This equation will be satisfied for all $g$ belonging to $H$ (as $K$ is a normal subgroup to H) but not necessarily for $g$ belonging to $G-H$.

Formally, I am at a loss how to show this that there may exist an element which will not satisfy this. Moreover I feel I am missing something as I have still not used $H$ is a subgroup.

答:
This helpful post by John Cook also spells out an equivalent counterexample. https://www.johndcook.com/blog/2011/09/07/normal-subgroups/

Hiding in plain sight, the symmetries of a square has at least TWO nonisomorphic normal subgroups of order 4. But the cyclic subgroup of order 4 is NOT the one you want. A useful H is the group generated by 180 degree rotation and by spinning about the y axis. This is a Klein 4 group, (grab a rectangular note book, start rotating and flipping to see the 4 configurations ). Moreover H is normal in D_8, (try conjugation by a rotation).

Of course one can also check by brute force, we are in the world of the eight 2x2 matrices under multiplication, all entries are 0 1 or -1, and all determinants are 1 or -1. So, as John points out, if K is the 2 element subgroup which includes flip about the y axis. Conjugate the flip by a 90 degree rotation, first rotate, then flip, then unrotate. Now your note book is not only flipped, but also upside down, kicking us out of K


答:
an example that might have nothing to do with what is in rudin but wich could make you understand the concept a bit is this question : does there exist a number $N$ such that for any number of person , call it $n$ with $n\geq N$ there are at least two person of the same sex among those people?
This number $N$ infact exists and is equal to $3$. So in this case : $\exists N : \forall n\geq N$ there are at least two person of the same sex among $n$ person.</p>

Title: What does this mean: there exist an integer N such that $n\ge N$?

Body:
I'm reading Rudin's, Principles of Mathematical Analysis, and I keep tripping over this phrase. Usually the phrase by that it implies a some equation with n being the index, subscript, of a point. My guess is that the phrase means that there is two points and that 'n' is the limit. Not sure, I'll greatly appreciate any help with this.

(edit): I admit the question is vague without its implication but I wanted to clear with the premise first. Thanks for the fast and helpful responses.


问:
Dealing twelve cards from a deck of 40
We have 40 cards. A, B, C extract resp. 5,4,3 cards. How many extractions can we make in which A and B don't extract swords while C extract at least 2? Solution: $C_{30,5}\cdot C_{25,4}\cdot C_{10,2}\cdot C_{29,1}$.

I wrote: $C_{30,5}\cdot C_{25,4}\cdot C_{10,2}\cdot C_{21,1}+C_{30,5}\cdot C_{25,4}\cdot C_{10,3}$. Why is my solution wrong?

答:
The reference to a $40$-card deck with swords as a suit means this is a Spanish deck with $10$ each of swords,  cups, coins and clubs, not the "standard" (French) $52$-card deck.

Your solution is the correct one, distinguishing between two-sword and three-sword cases. The given solution overcounts those cases where C's hand is all swords: it counts picking ($\{2,3\}$ and then $4$ of swords) and ($\{4,3\}$ and then $2$ of swords) as distinct, when they are not.

Title: Dealing twelve cards from a deck of 40


问:
Doubt in a 'probability problem
A lot of 50 spacing washers contains 30 washers that are thicker than the target dimension. Suppose that three washers are selected at random, without replacement, from the lot.

(a) What is the probability that all three washers are thicker than the target?

(b) What is the probability that the third washer selected is thicker than the target if the first two washers selected are thinner than the target?

(c) What is the probability that the third washer selected is
thicker than the target?

I had a doubt in the part (b) of the problem.

My solution :

Let A - Event that the washer is thicker

B - Event that the washer is thinner

Sample space S = {AAA , AAB , ABA , BAA , BBB , BBA , BAB , ABB }

So the only outcome which satisfies the condition in (b) is BBA so the answer should be

$(\frac{20}{50} . \frac{19}{49} . \frac{30}{48})$

But the answer given is

$(\frac{30}{48})$

Explanation given : Given that first two washers are thinner , the number of washers remaining would be 48(without replacement) , so the probability that the third washer is thicker is $(\frac{30}{48})$

I am not quite sure if this is correct , can anyone just help me out with it.

答: I agree with the answer 30/48. I thought about it the same way as the "correct answer". If the first two were thinner, there are 48 washing machines left to choose from for the third space. Of these 48 machines, 30 are thicker. Therefore, the probability to choose a thicker one is 30/48. Your answer answers the question what is the probability of choosing two thinner washing machines and then a thicker washing machine (part c). However the question fixes the fact that the first two were thinner. So using your expression, it would be (1 * 1* (30/48))</p>

Title: Doubt in a 'probability problem


答:
Of course the fastest way to solve the problem is to consider the remaining lot of 48 spacing washers 30 of which are thicker but your solution is not totally wrong.

They are requesting to calculate $P(A|B)$ and you correctly calculated $P(A\cap B)$. If you go on in your calculation you will find

$$\frac{\frac{20}{50}\frac{19}{49}\frac{30}{48}}{\frac{20}{50}\frac{19}{49}}=\frac{30}{48}$$

Title: Doubt in a 'probability problem

Body:
A lot of 50 spacing washers contains 30 washers that are thicker than the target dimension. Suppose that three washers are selected at random, without replacement, from the lot.

(a) What is the probability that all three washers are thicker than the target?

(b) What is the probability that the third washer selected is thicker than the target if the first two washers selected are thinner than the target?

(c) What is the probability that the third washer selected is
thicker than the target?

I had a doubt in the part (b) of the problem.

My solution :

Let A - Event that the washer is thicker

B - Event that the washer is thinner

Sample space S = {AAA , AAB , ABA , BAA , BBB , BBA , BAB , ABB }

So the only outcome which satisfies the condition in (b) is BBA so the answer should be

$(\frac{20}{50} . \frac{19}{49} . \frac{30}{48})$

But the answer given is

$(\frac{30}{48})$

Explanation given : Given that first two washers are thinner , the number of washers remaining would be 48(without replacement) , so the probability that the third washer is thicker is $(\frac{30}{48})$

I am not quite sure if this is correct , can anyone just help me out with it.


问:
Automorphism of Riemann Sphere without Picard's Theorem
When I am trying to get familiar with the contributions that Gauss made to the quaternions, the following question comes to my mind:

Question: Is there a known proof that every 1-1 comformal mapping of Riemann Sphere to itself is a Mobius Transformation, without using Picard's Theorem about essential singularities?

答:
There is. It uses the fact that the conformal automorphisms of the complex plane are the nonconstant linear polynomials $z\mapsto az+b,~a\neq0$. These are Möbius transformations.

Now let $\varphi:\overline{\mathbb C}\longrightarrow \overline{\mathbb C}$ be a conformal automorphism of the Riemann sphere. Let $a:=\varphi^{-1}(\infty)$. We consider two cases:

1: $a=\infty$

Then $\varphi\vert_{\mathbb C}$ is a conformal automorphism of the complex plane, so it is a Möbius transformation defined on $\mathbb C$. Since the conformal extension to $\overline{\mathbb C}$ is unique, $\varphi$ itself is also a Möbius transformation.

2: $a\neq\infty$

Then consider the Möbius transformation
$$\mu(z):=\frac{az+1}{z},$$
which sends $\infty$ to $a$. Then $\varphi\circ\mu$ is a conformal automorphism of $\overline{\mathbb C}$ which sends $\infty$ to $\infty$, so this automorphism belongs to case 1, and is thus a Möbius transformation. But then since Möbius transformations are a group, $\varphi\circ\mu\circ\mu^{-1}=\varphi$ is also a Möbius transformation.

The fact from the beginning (automorphisms of the complex plane) can be proved using Casorati-Weierstraß to show that in injective entire function must be a polynomial, and the fundamental theorem of algebra to show that to be injective, the polynomial must also be of degree 1.


问:
Why does singular value decomposition simultaneously diagonalize a symmetric matrix and its square?
So I took an online course on machine learning and in this course the instructor said that the eigenvectors of a covariance matrix (for principal components analysis) can be computed by a singular value decomposition.

Say the covariance matrix is $A$. The SVD yields $A = U \Sigma V^t$ and with $A^t A = A^2$ both $V$ and $U$ diagonalize $A^2$, i.e. $V^t A^2 V = U^t A^2 U = \Sigma^2$. What I do not understand is why U and V also diagonalize A directly, i.e. $V^t A V = U^t A U = \Sigma$.

Diagonalizable matrices that commute can be simultaneously diagonalized and so I understand that $A^2$ and $A$ can be simultaneously diagonalized. However, the dimension of each eigenspace can generally be $>1$ which means that a set of eigenvectors that diagonalizes $A^2$ does not have to diagonalize $A$ as well. So why does the SVD algorithm automatically find the eigenvectors of $A^2$ that are simultaneously eigenvectors $A$?

I am not a mathematician, please have mercy if the answer is kind of obvious.

答:
Covariance matrices are positive-semidefinite, and PSD matrices have unique PSD square roots (given by taking the unique nonnegative square root of each eigenvalue). This means that $V^T A V$ is the unique PSD square root of $V^T A^2 V$. We have $V^T A^2 V = \Sigma^2$ and the unique PSD square root of $\Sigma^2$ is $\Sigma$, so $V^T A V = \Sigma$.

We can give an alternative analysis in terms of eigenspaces as follows. Consider some eigenspace $E_{\lambda}$ of $A^2$. By definition we have $A^2 v = \lambda v$ for all $v \in E_{\lambda}$. Since $A$ commutes with $A^2$, it restricts to a map $A : E_{\lambda} \to E_{\lambda}$ which squares to $\lambda$. You are correct that in general it does not follow that $A$ acts by a scalar (and good job spotting this possibility!), but if $\lambda \ge 0$ and $A$ is PSD then $A$ must act by $\sqrt{\lambda}$. This is because $\sqrt{\lambda}$ is the only possible eigenvalue of $A$ here (and $A$ is diagonalizable by the spectral theorem).


问:
In what ways can the presidium of a meeting be elected?
The meeting of the Student Scientific Society was attended by $52$ students: with $13$ students each from $4$ faculties. In how many ways can four people be elected to the presidium of a meeting so that it includes students from $3$ faculties?

Attempt:  $4 \cdot 3 \cdot 13 \cdot 13 \cdot \dfrac{13 \cdot 12}{2!}$

答:
To find the four candidates, three of the four spots are fixed because you need at least 3 faculties. Each of these three spots can be occupied by 13 different students. The last spot can be a repeat or a different faculty. These two cases will be added.
Case 1: The fourth spot is occupied by the last faculty, there are 13 options
Case 2: The fourth spot is occupied NOT by the last faculty, there are 36 options

So now you have (13 * 13 * 13 * 13)+(13 * 13 * 13 * 36)=28,561+79,092=107,653

Now I think the question wants one of these candidates to be the presidium?
So you would do 107,653 choose 1 which is still 107,653

(The question is a little unclear to me)


问:
Help in a proof of a result in Hungerford's book
I need help to proof the last part of this corollary:

I didn't understand the part (IV) because the author proves just the canonical projection case and the statement says "every nonzero homomorphism of rings $R\to S$ is a monomorphism".

Thanks in advance.

答: Consider the fact that the kernel of the homomorphism is an ideal of $R$. If $R$ has no proper ideals, what are the choices for the kernel? What do these choices say about the homomorphism? Conversely, when would the projection $R \rightarrow R/I$ be a monomorphism, $I$ an ideal of $R$?</p>


问:
Is it hopeless to try and solve this equation analytically?
Can this equation be solved with analytical methods, or is it only numeric methods since current mathematical tools don't go that far? Its complex roots are the same as the roots of the zeta function on the critical strip.

I already know the positive odd integers $k$ are zeros, and the trivial zeros of the eta function ($1+2\pi i\,n/\log{2}$), $n$ integer, are also zeros. But I am more concerned with the complex zeros.

$$k-1=\frac{\pi}{2}\int_{0}^{\pi/2}\left(\sec{v}\,\text{csch}{\frac{\pi\tan{v}}{2}}\right)^2\left(1-\frac{\cos{k\,v}}{(\cos{v})^k}\right)\,dv$$

Perhaps a better form is:
$$k-1=\frac{\pi}{2}\int_{0}^{\infty}\left(\text{csch}{\frac{\pi\,u}{2}}\right)^2\left(1-(1+u^2)^{k/2}\cos{(k\arctan{u})}\right)\,du$$

and the referred paper shows how it reduces for integer $k$.

This is not a conjecture, I derived this result, perhaps will post the link here later. The help I need is with ways to solve the equation analytically. When I try to use the Weierstrass factorization theorem using only these 3 classes of zeros, the values don't add up, so something is amiss (maybe more zeros exist).

Edit: Paper A new approach to the Riemann hypothesis

Plot of $f(k)=1-k+g(k)$ for real $k$, where $g(k)$ is the integral:

It goes crazy near some points (explodes), it seems to always grow for negative $k$, though it's the non-real roots that are of interest.

答:
Long Comment:

Using the tangent half-angle substitution I get (with little help from Mathematica)
$$f(k)=(1-k)+\frac{ \pi}{2}  \int_0^{\infty } \text{csch}^2\left(\frac{\pi  t}{2}\right) \left(1-\frac{T_k\left(\frac{1}{\sqrt{t^2+1}}\right)}{\left(\frac{1}{\sqrt{t^2+1}}\right)^k}\right) \, dt \tag{1}$$

where $T_k$ is the Chebyshev polynomial of the first kind.

The function $ \left(1-\frac{T_k\left(\frac{1}{\sqrt{t^2+1}}\right)}{\left(\frac{1}{\sqrt{t^2+1}}\right)^k}\right)$ has some interesting divisibility properties involving prime $k>2$ and $t=1,2,3,...$

for $t=1$ for example, the resulting $1-2^{k/2} \cos \left(\frac{\pi  k}{4}\right)$ generates a sequence of integers (2) in $k$ that appear to be divisible by prime $k>2$ only.

$0,1,3,5,5,1,-7,-15,-15,1,33,65,65,1,-127,-255,-255,1,513,1025,... \tag{2}$

I can't find this sequence in OEIS. (Update 2: Add 1 to each number in sequence (2) and you will find it + others with sign differences etc.)

for $t=2$ the resulting terms in $k$ are also divisible other values of $k$ including powers of 2.

Alternatively integral (1) can be written

$$ \frac{\pi}{2} \int_0^{\infty }  \text{csch}^2\left(\frac{\pi  t}{2}\right) \left(1-\left(t^2+1\right)^{k/2} \cos \left(k \sec ^{-1}\left(\sqrt{t^2+1}\right)\right)\right) \, dt$$

Very quick Search of the internet reveals some hits to active academic research e.g. https://arxiv.org/abs/1703.09251

Update:

The above integral seems related to the slightly simpler integral below:

$$\frac{\pi}{2} \int_0^{\infty }    \text{sech}^2\left(\frac{\pi  t}{2}\right) \left(\left(t^2+1\right)^{k/2} \cos \left(k \sec ^{-1}\left(\sqrt{t^2+1}\right)\right)\right) \, dt\tag{3}$$

The first few Riemann Zeta Function non-trivial zero's I tried are also zero's for this integral function.

Update 2:
A simpler and perhaps more beautiful formula is equivalent to (3) in terms of  the first few non-trivial zero's

$$\frac{\pi}{4} \int_0^{\infty }  \text{sech}^2\left(\frac{\pi  t}{2}\right)  \left((1+i t)^k+(1-i t)^k\right)  \, dt$$

see https://oeis.org/A146559  (Paolo P. Lava, Nov 18 2008)

The conjecture is that for complex $k$ (assuming analytic continuation is valid for the Chebyshev polynomials of the first kind)

$$ \left(\frac{T_k\left(\frac{1}{\sqrt{t^2+1}}\right)}{\left(\frac{1}{\sqrt{t^2+1}}\right)^k}\right)=\left(\left(t^2+1\right)^{k/2} \cos \left(k \sec ^{-1}\left(\sqrt{t^2+1}\right)\right)\right)= \frac{1}{2}\left((1+i t)^k+(1-i t)^k\right) $$

where $i^2=-1$

Update 3: Just for Reference

$$\frac{2^{z-1}}{\Gamma (z+1)} \int_0^{ \infty } t^z \,\text{csch}^2(t) \, dt=\text{Li}_z(1)=\zeta(z)\;\; ;[z>1]$$

$$\frac{2^{z-1}}{\Gamma (z+1)} \int_0^{ \infty } t^z \,\text{sech}^2(t) \, dt=-\text{Li}_z(-1)=\eta(z)\;\; ;[z>0]$$


问:
Derivation of Fourier Series
Despite a lengthy search, I've been unable to turn up a derivation of the Fourier Series where the author doesn't begin by giving:

$$\frac{a_0}{2}+\sum_{n=1}^\infty \left(a_n \cos(\frac{2\pi}{T}nx) + b_n \sin(\frac{2\pi}{T}nx)\right)$$

And from this deriving coefficients $a_n$, $b_n$. Where does this first formula come from? I've been able only to find scraps of information, and what I have stumbled upon lacked much explanation and/or was very esoteric.

答:
There is a nice formal answer here, and the comment by @Qiaochu Yuan making reference to the solution to heat equation $$\frac {\partial u}{\partial t}=\alpha \left( u_{xx} + u_{yy} + u_{zz} \right)=\alpha \Delta u\tag 1$$ where $u$ is the temperature at a point and time  and $\Delta u$ the Laplacian (sum of second partial derivatives), which would be sketched out like this in one dimension with $P$ spanning from $[0,1)$ and $0$ at the boundaries:

The solution will be given by a function of space $f(x)$ $\times$ a function of time $T(t).$ Since on the LHS of $(1)$ we have the partial with respect to time $fT'=\alpha \Delta f\cdot T.$ And dividing by $Tf$ on both sides, $\frac{T'}T=\alpha \frac{\Delta f}f,$ a constant at all points and times, which has to be decreasing, and can be equated to a constant term $-\alpha \omega^2.$ This yields to differential equations: $T'=-\alpha \omega^2 T,$ which has an exponential solution $T(t)= C e^{-\alpha \omega^2 t},$ and an eigenfunction equation, $\Delta f(x)=-\omega^2 f(x),$ where $-\omega^2$ is the eigenvalue. This leads to any solution of the form $u(x,t) = \sum C_\omega f(x) \cdot e^{-\alpha \omega^2 t}$ that must satisfy the eigenfunction equation. Now, this eigenfunction has a second derivative on the LHS, and the solution will be of the general form $f= a \cos(\omega x) + b \sin(\omega x).$ This can be further solved noting that the initial condition is $0$ at the point $0$ of the interval $[0,1),$ eliminating the contribution of cosines. The fact that the boundary at the other extreme of the interval is also $0$ makes eigenvalues integer multiples of $\pi,$ resulting in a solution of the form

$$u(x,t)=\sum_{n=1}^\infty C_n \sin(n\pi)\cdot e^{-\alpha n^2\pi^2t}$$

The motivation behind the definition of the Fourier series is to express any minimally well behaved periodic function $s(x)$ as an infinity sum of sinusoid waves. This can be achieved just with sines or with cosines (earlier format of jpegs being an example of the latter), and introducing a phase shift: after all, cosines are just sine waves shifted by ninety degrees.

Each one of the terms in the sum, for example, $\cos\left(2\pi\frac{n}{P}x\right)$ and $\sin\left(2\pi\frac{n}{P}x\right)$ are elements in a basis function of a vector space (thank you @mjw), corresponding to the harmonic frequency $\frac n P.$ In here $P$ stands for the unit interval. The factor $2\pi$ can be understood as dividing one cycle around the unit circle by the length of an interval. $n$ can be interpreted as stops on a dial.

The idea is that the signal function that we wish to express as the sum of sinusoidal waves is "dotted" with each one the basis functions across the domain $P:$ The integral that will be used to solve for each coefficient $a_n$ and $b_n$ is an inner product;  a projection of the signal onto the basis function, much like the dot product of two vectors is maximal when the two vectors are aligned, the weight (coefficient) of a particular basis function in the (theoretical - clearly not in practice) infinite series will reflect the ability of that basis function to approximate the function $s(x).$

Picking up on @mjw comment, the basis functions $\left\{1,\cos\left(2\pi\frac{n}{P}x\right), \sin\left(2\pi\frac{n}{P}x\right) \right\}$ are orthogonal, rendering the determination of the coefficients feasible by simply assuming that $s(x)$ can indeed be represented as a FS, i.e.

$$s(x)=\small\frac{a_0}{2}+\sum_{n=1}^\infty \left(a_n \cos\left(2\pi\frac{n}{P}x\right) + b_n \sin\left(2\pi\frac{n}{P}x\right)\right)$$

and multiplying both sides by each one of the basis functions that will be considered, and proceeding to integrate (inner product) over the period $P$. The orthogonality of these basis vectors will nullify all summands, except for the one being extracted, which will be multiplied by $\small \frac 1 2 P, $ explaining the adjustment in front of the integral in the formula for the coefficients. If $P=2\pi,$ the coefficient of the cosine part of the equation for $n=1$ can be extracted by multiplying by $\cos(1x)$ and integrating  $a_{n=1}=\frac 2 {2\pi} \int_{-\pi}^\pi s(x) \cdot \cos\left(x\right)dx,$ for example, because in the expansion of $s(x),$ only $\cos(x)\cos(x)$ will be different to zero:

Title: Derivation of Fourier Series


问:
How to multiply time in different denominations?
I am writing some software that must do generalized time multiplication of various denominations. I can't wrap my head around how to generalize it.

[edit] The use case is this. You have a clock measuring time and you want to provide a mapping (ratio) to accelerate or decelerate the clock. For example, every seconds passed is equivalent to 1 days passed. Thus, $clocktime = xseconds * ndays$, with n days probably represented in seconds.

For example, lets say I have :

$1min * 2min = 2min\\
60s * 120s = 7200s^2\\
7200s^2 / 60s = 120s = 2min$

Now, if I multiply time in days, for example :

$1days * 2days = 2days$

And in seconds :

$1 day = 86'400s\\
86'400s * 172'800s = 14'929'920'000s^2\\
14'929'920'000s^2 / 60s = 248'832'000s$

But, that is incorrect.

$248'832'000s / 60 = 4'147'200min\\
4'147'200min / 60 = 69'120h\\
69'120h / 24 = 2'880days$

So my question is, how do you generalize time multiplication from/to various denominations? Ultimately, what am I missing here? Ty

答: This is a programming question, but ultimately you must encode the unit as well as its value. That way multiplying $ms\cdot ns = (m \cdot n) s$, and then handle conversions between units using a conversion function (with whatever templating mechanism you have in your language). For good examples see the time libraries in C++ and Rust (C++ is chrono and Rust is Duration I think)


问:
Calculate Operator norm of a given operator
I have the following operator on a Hilbert space:

$$T=\sum_{n=0}^{\infty}c_{n}\langle a_{n}\mid \cdot\rangle b_{n},$$

where the $c_{n}$ is a bounded sequence of complex numbers and $a_{n},b_{n}$ are two orthonormal families.

I have to show that the operator norm of $T$ is the maximum of the $c_{n}$.

Using the definition of the operator norm I have after some steps:

$$\Vert T\Vert=...=\sup_{\Vert x\Vert =1}\sqrt{\sum_{n=0}^{\infty}\vert c_{n}\vert^{2}\vert\langle a_{n}\mid x\rangle\vert^{2}}$$

I don`t know how to proceed. I should find that $\Vert T\Vert = \max_{n\in\mathbb{N}}\vert c_{n}\vert$. So I had the idea to show $\leq$ and $\geq$. Using the Bessel's inequality I get $\Vert T\Vert\leq \max_{n\in\mathbb{N}}\vert c_{n}\vert$, but I don't how to show the other direction ....

答:
You have that $\|T\|^2=\|T^*T\|$. And
$$
T^*Tx=\sum_{n,m}c_n\overline{c_m} \overline{\langle a_n,x\rangle}\,\langle b_n,b_m \rangle\, a_n
=\sum_n|c_n|^2\,\langle x,a_n\rangle\,a_n.
$$
From this you see that, since the maps $\langle\cdot,a_n\rangle a_n$ are a family of pairwise orthogonal projections,
$$
\|T^*T\|=\max\{|c_n|^2:\ n\}.
$$


答:
Let's consider a simple example.  Using the definition of the limit of a sequence, show that the sequence defined by

$$a_n=\frac{1}{n}$$

converges to $0$.  The definition says that no matter how small we choose a positive number $\epsilon,$ we should be able to find a big enough number $N$ so that as long as $n\ge N$, the terms $a_n$ are really close to $0$, that is, they are within a distance $\epsilon$ of $0$.

Let's start concretely and then move to the abstract.  Let's first choose $\epsilon=.1$.  Then, notice that $a_{10}=.1$, and that any term beyond $a_{10}$ is even closer to $0$.  This means that if we let $N=10$, we know that for any $n\ge 10$, the value of $a_n$ is within a distance $.1$ of $0$.  Notice we also could have chosen $N=11,12,\ldots$ because we would still have $a_n$ within $.1$ of $0$ so long as $n\ge 11,12,\ldots$.  Notice also that if $n<10$, then $a_n$ is not close enough to $0$.

Similarly, if we chose $\epsilon=.01$, you will see we could choose $N=100,101,102,\ldots$, and that for all $n\ge N$, $a_n$ is small enough:

$$|a_n-0|\le.01$$

Now, we can't continue to just choose different values for $\epsilon$ and prove it for those cases, because there are infinitely many possible values, and we don't have enough time to check all of them.  So let's just assume $\epsilon$ is any positive number.  For large enough values of $n$, we want the following inequality to hold:

$$|a_n|=\left|\frac{1}{n}\right|=\frac{1}{n}\le\epsilon$$

Rearranging the inequality, we see that we need $n\ge\frac{1}{\epsilon}$, hence, if we choose $N$ to be anything greater than $\frac{1}{\epsilon}$, then for all $n\ge N$, our desired inequality will hold, because then
$$\frac{1}{\epsilon}\le N\le n$$

Notice that as $\epsilon$ gets smaller, we must choose our $N$ to be larger and larger to obtain the desired closeness of $a_n$ to $0$, but it is always possible.

In this example, as in most, $N$ is a fixed large number such that a certain property holds for all $n\ge N$.  Here, the property is that the terms $a_n$ are all very close to the limit $0$ for all $n\ge N$.


问:
Simplifying exponential terms
How would I simplify the following equation to solve for $x$?

$$
\frac{\exp(\frac{1-x}{20}) + \exp(\frac{10-x}{20})}
     {\exp(\frac{1-x}{5}) + \exp(\frac{10-x}{5})} = 4
$$

What about if there were additional terms in each sum, i.e.

$$
\frac{\exp(\frac{1-x}{20}) + \exp(\frac{10-x}{20}) + \exp(\frac{20-x}{20})}
     {\exp(\frac{1-x}{5}) + \exp(\frac{10-x}{5}) + \exp(\frac{20-x}{5})} = 4
$$

答:
HINT

Note that
$$
\exp\left(\frac{1-x}{20}\right) = \frac{e^{1/20}}{e^{x/20}}
$$
so the numerator will simplify to expressions involving $e^{x/20}$, and the numerator will be in terms of $e^{x/5}$. Note that if we let $u = e^{x/20}$ then $e^{x/5} = \left(e^{x/20}\right)^4 = u^4$ and now plug in and simplify.

Title: Simplifying exponential terms


问:
How can a ring of polynomials with coefficients in a field $k$, and in infinitely many variables be a subring of $k[x,y]$?
I am trying to answer this question:

Let $k$ be a field and $k[x,y].$ Define the subring $A \subset k[x,y]$ by $A = k[x, xy, xy^2, xy^3, ...].$ Show that $A$ is not Noetherian.

But I have the following question:

How can a ring of polynomials with coefficients in a field $k$, and in infinitely many variables, in our case $A$ be a subring of $k[x,y]$ which is a ring of polynomials with coefficients in a field $k$, and in finitely many variables ? should not $A$ be at least a finite set? could anyone give me an example to clarify this point please?

答:
Consider the polynomial ring $A = k[xy^i \,|\, i \in \mathbb Z_{\geq 0}].$ By definition, an element of $A$ is of the form $$c_1 (xy^{i_{1,1}})^{n_{1,1}} \cdots (xy^{i_{1, \ell_1}})^{n_{1, \ell_1}} + c_2 (xy^{i_{2,1}})^{n_{2,1}} \cdots (xy^{i_{2, \ell_2}})^{n_{2, \ell_2}} + \cdots + c_k (xy^{i_{k,1}})^{n_{k,1}} \cdots (xy^{i_{k, \ell_k}})^{n_{k, \ell_k}}$$ for some scalars $c_j$ in $k,$ some distinct integers $i_{j, m} \geq 0,$ and some integers $n_{j, m} \geq 0.$ By the fourth sentence of the following link, "it is possible to speak of a ring of polynomials in an infinite set of variables if it is assumed that each individual polynomial depends only on a finite number of variables." Considering that the variables of $A$ are $x, xy, xy^2, xy^3,$ etc., it follows that a polynomial in $A$ is a finite $k$-linear combination of products of powers of the variables $x, xy, xy^2, xy^3,$ etc., and so, we arrive at a concrete description of a typical element of $A.$

Bearing this all in mind, it follows that $-x$ and $3(xy^2)^3 - \frac 1 7 (xy^9)^2$ and $(xy^2)^3 (xy^3)^2 + (xy^2)^2$ are all elements of $A.$ But as these are all polynomials in the variables $x$ and $y$ with coefficients in $k,$ these are all elements of $k[x, y],$ as well.

Ultimately, the fact that $A \subseteq k[x, y]$ follows from the fact that all of the monomial generators of $A$ belong to $k[x, y]$: $xy^i$ is a monomial in $k[x, y]$ for all integers $i \geq 0.$ Consequently, any $k$-linear combination of powers of the generators of $A$ belongs to $k[x, y]$ so that $A \subseteq k[x, y].$

We may now use the subring test to prove that $A \subseteq k[x, y]$ as rings. By the previous paragraph, we have that $A \subseteq k[x, y].$ Considering that $1 = 1x^0$ is an element of $A,$ it follows that $A$ contains the multiplicative identity of $k[x, y].$ Clearly, $A$ is closed under subtraction: the difference of two polynomials in $A$ is again a polynomial in $A,$ as it is a $k$-linear combination of products of powers of the generators of $A.$ Last, by the distributive property of $A$ (indeed, this holds because $A$ is by definition a ring), any product of elements in $A$ is itself another element of $A$ because the product $(xy^i)^{n_i} (xy^j)^{n_j}$ of any powers of any two monomial generators of $A$ is itself a polynomial in $A.$ By the subring test, therefore, we conclude that $A$ is a subring of $k[x, y].$


问:
Unbounded open set whose characteristic function is integrable in extended sense
Put simply, I would like to know if there is an unbounded open set (of $\mathbb{R}^n$) whose characteristic function is integrable in extended sense. I get the suspicion that something like the area between the graph of $f(x)=\frac{1}{x^2}$ for $1<x<\infty$ in $\mathbb{R}^2$ would do the trick, but I'm stuck extending this to higher dimensions.</p>

答: Clearly, for any continuous function $f:[a,\infty)\to\mathbb{R}^+$ with unbounded support and $\int_a^\infty f(x)dx <\infty$ you can define $U=\{(x,y)|a<x, 0< y< f(x)\}$. $U$ will be open and unbounded and clearly $\chi_U(x)$ will be integrable.</p>


问:
What are the most resourceful topics to write about in set theory?
I am a mathematics student in my last year of undergraduate studies and I am currently doing a project on set theory. I have so far explored Zermelo-Fraenkel Set theory with the Axiom of Choice and Elementry Theory of Category of Sets and would like to take my project further to investigate more topics. I was wondering whether there are any suggestions for specific topics or any particular topics that people have looked into and really enjoyed. I am quite new to set theory and this is the first time I've learned about it so I would really appreciate it if the answers are not too complicated.</p>

答:
One topic I quite like, which can be approached with very little background, is the study of cardinal characteristics of the continuum.

Roughly speaking (see also here), a CCC is a cardinal which measures how big a set of reals must be to be "sufficient" with respect to some relation: taking a relation $R$ between reals, or objects "morally equivalent" to reals such as functions $\mathbb{N}\rightarrow\mathbb{N}$, we can define the cardinal $$\kappa_R=\min\{\vert A\vert: \forall x\exists y\in A(yRx)\}.$$ For example:

The actual continuum is $\kappa_{=}$.

If we look at functions $\mathbb{N}\rightarrow\mathbb{N}$ and consider the domination relation $$fDg\iff \exists n\forall m>n[f(m)>g(m)],$$  $\kappa_D$ measures how many functions are needed to "grow as fast as every other function."

The "dual" relation to $D$, the escaping relation $$fEg\iff \forall n\exists m>n(f(m)>g(m))\iff \neg gDf,$$ measures how many functions we need in order to be "hard to dominate." It's easy to check that $\kappa_D\ge\kappa_E$.

Note that this notation is nonstandard. In particular, $\kappa_D$ and $\kappa_E$ are properly called "$\mathfrak{d}$" and "$\mathfrak{b}$" respectively.

Trivially we always have $\kappa_R\le 2^{\aleph_0}$. Conversely, diagonal arguments can be used to show that (as long as $R$ isn't silly - and we don't call silly $\kappa_R$s CCCs) each $\kappa_R$ is uncountable. For example, given any sequence $G=(g_i)_{i\in\mathbb{N}}$ of functions $\mathbb{N}\rightarrow\mathbb{N}$, the function $$i\mapsto g(i)+1$$ is not dominated by any member of $G$; this shows that $\kappa_D$ is uncountable. A similar trick will show that $\kappa_E$ is uncountable:

 Take $$i\mapsto 1+\sum_{j\le i}g(j)$$ instead.

Not all CCCs are so combinatorial in flavor. We can also consider more "analytic" ones, such as the cardinality of the smallest non-measurable set. There is a generally-agreed-upon set of ten fundamental CCCs, whose relationships are described by Cichon's diagram; verifying each of the relevant $\mathsf{ZFC}$-provable inequalities can be a good short project (if memory serves, none are particularly hard). And there are more cardinal characteristics out there, so you could continue the study of provable inequalities for quite some time.

Unfortunately (or excitingly), showing that a given inequality is not $\mathsf{ZFC}$-provable - and more generally that Cichon's diagram is complete in the relevant sense - is quite involved: we need to introduce the technique of forcing. Honestly if you're up for a project in forcing you should do that instead, perhaps tacking on a CCC-based observation at the end as a cute coda (e.g. proving the consistency of $\kappa_D>\kappa_E$). And it's worth noting that it was only recently discovered how to separate three or more CCCs simultaneously (see here).


问:
Limit of a difficult sequence
I stucked in one part of my task where i have to find a limit of given sequence. I have absolutely no idea how to even start solving this problem. Any hints?

$$a_n = \sqrt[n]{4^{2n} - 5^{n+2} - \sin(\sqrt{n})}  $$

答:
$$\lim_{n \to \infty} a_n = \lim_{n \to \infty}\sqrt[n]{4^{2n} - 5^{n+2} - \sin(\sqrt{n})}$$
as pointed out by @Ivan Neretin in the comments:
$$\sqrt[n]{4^{2n} - 5^{n+2} - \sin(\sqrt{n})}\approx\sqrt[n]{4^{2n} - 5^{n+2}}$$

$$= \lim_{n \to \infty}\exp(\frac{\ln(4^{2n} - 5^{n+2})}{n})=\exp(\lim_{n \to \infty}\frac{\ln(4^{2n} - 5^{n+2})}{n})$$
$$= \exp(\lim_{n \to \infty}\frac{2\ln(4)4^{2n} - \ln(5)5^{n+2}}{4^{2n} - 5^{n+2}})$$
notice that the the $5^{n+2}$ terms grow considerably smaller in proportion as n increases, this is due to the smaller exponent, therefore we can approximate both th top and bottom by eliminating the $5^{n+2}$ term:
$$= \exp(\lim_{n \to \infty}\frac{2\ln(4)4^{2n}}{4^{2n}})$$
$$= \exp(\lim_{n \to \infty}2\ln(4))$$
$$= \exp(2\ln(4)= 16$$

Title: Limit of a difficult sequence


问:
Galois group is solvable but $f$ is not solvable
Let $M$ be an algebraic closure of $F_p$, and let $F=M(x)$. Show that $f(t)=t^p-t-x$ is not solvable by radicals over $F$ but that the Galois group of the splitting field of $f$ over $F$ is cyclic.

It's an exercise from Patrick Morandi's Field and Galois Theory. I know that the Galois group of the splitting field of $f$ over $F$ is generated by $\sigma:\sigma(\alpha)=\alpha+1$, where $\alpha$ is a root of $f$. Thus, the Galois group is cyclic. However, I don't know how to prove that $f$ is not solvable by radicals.

Can anyone help me?

答:
Ok, I finally got through it.

Let $K$ be the splitting field of $f$ over $F$. $K/F$ is Galois with degree $p$.

If $K$ lies in a radical extension $L$ of $F$. Then we have
$$
F=F_0\subset F_1\subset F_2\ldots\subset F_r=L
$$
where $F_i=F_{i-1}(\alpha_i)$ and $\alpha_i^{n_i}\in F_{i-i}$. We may assume that $\alpha_i\notin F_{i-1}$ and $n_i$ are all primes.

Let $K_i$ be $K(\alpha_1,\ldots\alpha_i)$, then $F_i \subset K_i$.

By induction, we can prove that $K_i/F_i$ is Galois with degree $p$ as follows.

First, $K_0/F_0$ is Galois with degree $p$.

We assume $K_{i-1}/F_{i-1}$ is Galois with degree $p$. $K_i=K_{i-1}(\alpha_i), F_i=F_{i-1}(\alpha_i)$. If $\alpha_i\in K_{i-1}$, then $F_i=F_{i-1}(\alpha_i)=K_i=K_{i-1}(\alpha_i)=K_{i-1}$ and $n_i=p$, since $[K_{i-1}:F_{i-1}]=p$. Because $\alpha_i\notin F_{i-1}$, $g=(t-\alpha_i)^p=t^p-\alpha_i^p$ is irreducible over $F_{i-1}$. Then the minimal polynomial of $\alpha_i$ over $F_{i-1}$ is $g$. However, $K_{i-1}/F_{i-1}$ is Galois, so $\alpha_i$ is separable, but $g=(t-\alpha_i)^p$, which shows that $\alpha_i$ is not separable.

This contradiction shows that $\alpha_i\notin K_{i-1}$. Note that $\alpha_i^{n_i}\in F_{i-1}\subset K_{i-1}$ and all $n_i$th roots of unity is in $F$. Then we have $g=t^{n_i}-\alpha_i^{n_i}$ is irreducible over $K_{i-1}$ and $[K_i:K_{i-1}]=n_i$. Then we can conclude that $K_i/F_i$ is Galois with degree $p$.

By induction, $K_i/F_i$ is Galois with degree $p$ for all $i$.

On the other hand, $K_r=F_r=L$, so $K_r/F_r$ is of degree 1, which leads to a contradiction.


问:
Exponential Generating Function for the recurrence relation $a_n=\sum_{k=1}^n {n\choose k}a_{n-k}$
Here is the problem (this is problem 66 part iii) from page 280 of Principles and Techniques in Combinatorics):

Given the recurrence relation $a_n=\sum_{k=1}^n {n\choose k}a_{n-k}$, with $a_0=1$, show that the exponential generating function for $a_n$, $A(x)=\sum_{n=0}^{\infty}a_n\frac{x^n}{n!}$ is equal to $\frac{1}{2-e^x}$.

I have pretty much worked out a solution, but my final answer is missing a $1$. Here is my solution.

Plug in $a_n$ into $A(x)$:
\begin{equation}
A(x)=\sum_{n=0}^{\infty}a_n\frac{x^n}{n!}=\sum_{n=0}^{\infty}(\sum_{k=1}^n {n\choose k}a_{n-k})\frac{x^n}{n!}
\\ Note: a_n=\sum_{k=1}^n {n\choose k}a_{n-k}=(\sum_{k=0}^n {n\choose k}a_{n-k})-a_n
\\\implies A(x)=\sum_{n=0}^{\infty}((\sum_{k=0}^n {n\choose k}a_{n-k})-a_n\frac{x^n}{n!})
\\=\sum_{n=0}^{\infty}(\sum_{k=0}^n \frac{n!}{k!(n-k)!}a_{n-k})\frac{x^n}{n!}-\sum_{n=0}^{\infty}a_n\frac{x^n}{n!}=\sum_{n=0}^{\infty}((\sum_{k=0}^n \frac{a_{n-k}}{k!(n-k)!})x^n)-A(x)
\\=(\sum_{n=0}^{\infty}\frac{a_n}{n!}x^n)(\sum_{n=0}^{\infty}\frac{x^n}{n!})-A(x)=A(x)e^x-A(x)
\\\implies A(x)=A(x)e^x-A(x)
\end{equation}
Which, obviously does not give the desired result of $A(x)=\frac{1}{2-e^x}$.
I'm assuming I'm messing up some index or something, but I've picked over the computation over and over and haven't found any mistakes. Can anyone point my mistake out?

Body:
\begin{align}
A(x) &= \sum_{n \ge 0} a_n \frac{x^n}{n!} \\
&= a_0 + \sum_{n \ge 1} a_n \frac{x^n}{n!} \\
&= 1 + \sum_{n \ge 1} \sum_{k=1}^n \binom{n}{k} a_{n-k} \frac{x^n}{n!} \\
&= 1 + \sum_{k \ge 1} \sum_{n\ge k} \frac{n!}{k!(n-k)!} a_{n-k} \frac{x^n}{n!} \\
&= 1 + \sum_{k \ge 1} \frac{x^k}{k!} \sum_{n\ge k} a_{n-k} \frac{x^{n-k}}{(n-k)!} \\
&= 1 + \sum_{k \ge 1} \frac{x^k}{k!} A(x) \\
&= 1 + (e^x-1) A(x),
\end{align}
so $$A(x) = \frac{1}{2-e^x}$$


答:
As pointed out in comments, at $n=0$ the "note" fails; you need to separate out this case beforehand.
$$A(x)=\sum_{n=0}^\infty a_n\frac{x^n}{n!}=1+\sum_{n=1}^\infty a_n\frac{x^n}{n!}=1+\sum_{n=1}^\infty\sum_{k=1}^n \binom nka_{n-k}\frac{x^n}{n!}$$
$$=1+\sum_{n=1}^\infty\sum_{k=0}^n \binom nka_{n-k}\frac{x^n}{n!} - \sum_{n=1}^\infty a_n\frac{x^n}{n!}=2+\sum_{n=1}^\infty\sum_{k=0}^n \binom nka_{n-k}\frac{x^n}{n!} - A(x)$$
$$=1+\sum_{n=0}^\infty\sum_{k=0}^n \binom nka_{n-k}\frac{x^n}{n!} - A(x)=\cdots=1+A(x)e^x-A(x)$$

Title: Exponential Generating Function for the recurrence relation $a_n=\sum_{k=1}^n {n\choose k}a_{n-k}$

Body:
Here is the problem (this is problem 66 part iii) from page 280 of Principles and Techniques in Combinatorics):

Given the recurrence relation $a_n=\sum_{k=1}^n {n\choose k}a_{n-k}$, with $a_0=1$, show that the exponential generating function for $a_n$, $A(x)=\sum_{n=0}^{\infty}a_n\frac{x^n}{n!}$ is equal to $\frac{1}{2-e^x}$.

I have pretty much worked out a solution, but my final answer is missing a $1$. Here is my solution.

Plug in $a_n$ into $A(x)$:
\begin{equation}
A(x)=\sum_{n=0}^{\infty}a_n\frac{x^n}{n!}=\sum_{n=0}^{\infty}(\sum_{k=1}^n {n\choose k}a_{n-k})\frac{x^n}{n!}
\\ Note: a_n=\sum_{k=1}^n {n\choose k}a_{n-k}=(\sum_{k=0}^n {n\choose k}a_{n-k})-a_n
\\\implies A(x)=\sum_{n=0}^{\infty}((\sum_{k=0}^n {n\choose k}a_{n-k})-a_n\frac{x^n}{n!})
\\=\sum_{n=0}^{\infty}(\sum_{k=0}^n \frac{n!}{k!(n-k)!}a_{n-k})\frac{x^n}{n!}-\sum_{n=0}^{\infty}a_n\frac{x^n}{n!}=\sum_{n=0}^{\infty}((\sum_{k=0}^n \frac{a_{n-k}}{k!(n-k)!})x^n)-A(x)
\\=(\sum_{n=0}^{\infty}\frac{a_n}{n!}x^n)(\sum_{n=0}^{\infty}\frac{x^n}{n!})-A(x)=A(x)e^x-A(x)
\\\implies A(x)=A(x)e^x-A(x)
\end{equation}
Which, obviously does not give the desired result of $A(x)=\frac{1}{2-e^x}$.
I'm assuming I'm messing up some index or something, but I've picked over the computation over and over and haven't found any mistakes. Can anyone point my mistake out?


问:
How to prove that using the Sample Standard Deviation formula is more accurate than the population standard population formula on a sample dataset?
So, for an assignment for a Python class at college I have to demonstrate that the Sample Standard Deviation formula is more accurate than the population standard population formula on a sample data Set.

So the full original data Set is an array of numbers 5,7,8,3,10,21,4,13,1,0,0,9,17

I get a random sub set of those so, for example, it could be [5, 21, 7]

Using the population formula on that set gets a standard deviation of: 7.118052168020874

Using the sample formula on that set gets a standard deviation of: 8.717797887081348

So does this demonstrate the sample formula being more accurate? I feel like you need to know what the answer is beforehand to know which is more accurate for calculating it.

答:
The sample formula of the st. dev on $[5;21;7]$ is not the value you showed but it is $\approx 8.72$

The sample formula is more accurate because $S^2$ is an unbiased estimator of $\sigma^2$ (unbiased estimator of the population variance)

Proof:

$$\mathbb{E}[S^2]=\mathbb{E}\Bigg[\frac{1}{n-1}\Sigma_i(X_i-\overline{X}_n)^2\Bigg]=\frac{1}{n-1}\mathbb{E}[\Sigma_i(X_i-\mu)^2-n(\overline{X}_n-\mu)^2]=$$

$$=\frac{1}{n-1} \left[\sum_i\mathbb{E}(X_i-\mu)^2-n\mathbb{E}(\overline{X}_n-\mu)^2\right] = \frac{1}{n-1}\Bigg(n\sigma^2-n\frac{\sigma^2}{n} \Bigg) = \sigma^2$$


问:
Show something using the Mean Value Theorem
I've got an exercise about differentiability, mean value theorem and suprema.

To be honest I don't understand the structure of this question. Maybe you guys are so kind to help me out :)

Let $f: [0,1] \rightarrow \mathbb{R}$ be differentiable with $f(0) = 0$, and satisfying

$$|f'(x)|\le M|f(x)|, x\in[0,1]  $$
for some $M>0$

a.) Use the Mean Value Theorem to show that for all $x  \le x_0 \in[0,1], y\in[0,x_0]:$
$$ |f(x)|\le x_0 \text{ sup} |f'(y)|\le M x_0 \text{ sup} |f(y)|$$
b.) Use the previous part to show that f is the zero-function on [0,1].

(Hint: What happens if we choose $x_0$ such that M$x_0<1$?)

Known definitions, theorems:

Mean Value Theorem (There is a c for which f'(c) equals f(b)-f(a)/(b-a) if [a,b] is the domain, and f is continuous on [a,b], differentiable on (a,b)

Differentiable function means that that for all points c $\in$ A, the limit of f(x)-f(c)/(x-c) exists.

Interior Extremum Theorem (Intermediate Value theorem). States that if f attains a maximum or minimum value on a open interval, then at some point c $\in$(a,b), f'(c)=0

Darboux Theorem: If f is differentiable on an interval [a,b] and if $\alpha$ satisfies $f'(a) < \alpha < f'(b)$, then there exists a point c$\in$(a,b) where $ f'(c) = \alpha $.

答: Investigate the inequality in two different parts: right-hand side and left-hand side. This is for the solution of the first part. Let us first write what we know. $$x\leq x_0\in [0,1], y\in[0,x_0], f(0)\space=\space0$$ We can write mean value theorem$$\Biggl|\frac{f(x)\space-\space f(0)}{x-0}\Biggl|\space\leq\space \operatorname{sup}|f'(y)|\space \Rightarrow \space \frac{|f(x)|}{x}\space \leq \space \operatorname{sup}|f'(y)|$$ Since we know $x\space\geq\space 0$ we can multiply the inequality by $x$ on both sides $$|f(x)|\space\leq\space x \operatorname{sup}|f'(y)|\space \Rightarrow\space |f(x)|\space\leq\space x_0 \operatorname{sup}|f'(y)|$$ As seen we successfully covered the left-hand side. At this point we can start to cover right-hand side. Indeed right-hand side is relatively easy to cover. The answer is hidden in the question. Let us again write what we know. $$\operatorname{sup}|f'(x)|\space\leq\space M\operatorname{sup}|f(x)|$$ for some $M\gt 0$. If we change the variables and multiply by $x_0$ on both sides, we get $$x_0\operatorname{sup}|f'(y)|\space\leq\space Mx_0\operatorname{sup}|f(y)|$$ Now we can combine two part to obtain the whole inequality $$|f(x)|\space\leq\space x_0 \operatorname{sup}|f'(y)|\space\leq\space Mx_0\operatorname{sup}|f(y)|$$ For the second part we can start this way$$x_0\frac{|f(x)|}{x}\space\leq\space x_0\operatorname{sup}|f'(y)|\space\leq\space Mx_0 \operatorname{sup}|f(y)|\space\Rightarrow\space \frac{x_0}{x}\space|f(x)|\space\leq\space Mx_0 \operatorname{sup}|f(y)|$$ $$\frac{x_0}{y}\space|f(y)|\space\leq\space Mx_0\space\operatorname{sup}|f(y)|\space\Rightarrow\space \frac{1}{y}\space|f(y)|\space\leq\space M\operatorname{sup}|f(y)|\space\Rightarrow\space\frac{1}{y}\space\operatorname{sup}|f(y)|\space\leq\space M\operatorname{sup}|f(y)|$$ We reached the core of the question $\frac{1}{y}\space\leq\space M$, we can deduce that $$Mx_0\space\geq\space 1$$ Let us take the case $M\space\lt\space 1$ $$0\space\leq\space Mx_0\space\leq\space M\space\lt\space 1\space\Rightarrow\space Mx_0\space\lt\space 1$$ Since this is a contradiction, $f(x)$ should be zero-function on $[0,1]$.


问:
Does $Z[x]/(3,p(x))$ equal to $Z_3[x]/(p(x))$ for all integer coefficient polynomials?
Let's define the polynomial $p(x)$ as one of the following polynomials:  $x^n-1$ or $x^{n-1}+x^{n-2}+...+1$

Does  $Z[x]/(3, p(x))$  mean that we take the modulo of the elements of $Z[x]$ according to $p(x)$ and then take the modulo again according to 3? If so, does that mean  $Z[x]/(3,p(x))=Z_3[x]/(p(x))$?

Also does it make any difference if the polynomial differs? I mean is it true for any polynomials?

答:
I wouldn't say precisely that they are the same, but they are isomorphic, which is as good as it gets for such objects. To see this for yourself, consider the ring homomorphism:
$$\varphi: \mathbb{Z}[x]/(3,p(x)) \to (\mathbb{Z}/3\mathbb{Z})/(p(x))$$
given by:
$$
f(x) = a_0+a_1x+...+a_nx^n \mapsto (a_0\%3) + (a_1\%3)x+...+(a_n\%3)x^n
$$
Where by $a_i\%3$ I mean reduction modulo $3$. This map is clearly surjective, since any element of $(\mathbb{Z}/3\mathbb{Z})/(p(x))$ can be written as a polynomial with coefficients in $\mathbb{Z}$ by definition. Injectivity also follows rather easily by considering the kernel. This gives an isomorphism for a general $p(x)$ thus answering your second question.

Hope this helps.


问:
How to notate correctly a vector storing four 3-dimensional matrices in mathematics?
How to notate correctly a vector storing four 3-dimensional matrices in mathematics? For example, 4 images are stored in a vector and each of these matrices has three dimensions like  and RGB image (RED, GREEN, BLUE).</p>

答:
One way is to represent a $3\times 3$ matrix as an vector in $K^9$, i.e., by
$A=(a_1,\ldots ,a_9)$. Then we can represent $4$ matrices of size $3$ by a vector
$$
v=(a_1,\ldots,a_9,b_1,\ldots ,b_9,c_1,\ldots c_9,d_1,\ldots ,d_9)\in K^{36}.
$$


问:
Maximize the function $f(x) = (1-x) (1+x)^3$
Maximize the function $f(x) = (1-x)(1+x)^3.$

I took the first derivative of $f(x)$ and got $(1+x)^2 (2-4x)$

From there I got $2$ critical points of: $-1, \frac{1}{2}$

Then $I$ took the second derivative of $f(x)$ ($-12x^2 -12x$) and plugged in the critical points to get $0$ and $-6$.

I know I am doing it wrong can anybody help. The right answer is $\frac{27}{16}$.

Body:
You're on the right track.

Note that, when $x=\dfrac12, \;f(x)=(1-x)(1+x)^3=\dfrac12\dfrac{27}{8}=\dfrac{27}{16}$,

and $f'(x)=(1+x)^2(2-4x)=0$ and $f''(x)=-12x^2-12x<0$, so that's a maximum.


答: Why do you think you did something wrong? Yes, $-1$ and $\frac12$ are critical points. Besides, $f''\left(\frac12\right)=-9<0$. So, $f$ has a local maximum at $\frac12$. And it has no other local maximum. So, since $\lim_{x\to\pm\infty}f(x)=-\infty$, the maximum is attained at $\frac12$.

Title: Maximize the function $f(x) = (1-x) (1+x)^3$

Body:
Maximize the function $f(x) = (1-x)(1+x)^3.$

I took the first derivative of $f(x)$ and got $(1+x)^2 (2-4x)$

From there I got $2$ critical points of: $-1, \frac{1}{2}$

Then $I$ took the second derivative of $f(x)$ ($-12x^2 -12x$) and plugged in the critical points to get $0$ and $-6$.

I know I am doing it wrong can anybody help. The right answer is $\frac{27}{16}$.


问:
A faster way to find the term $x^0$ in the expansion of $(1-\frac{x}{3})^{5}(1+\frac{2}{x})^{3}$
Now I am actually looking for a method in the general case when you have a product of two binomials. The method I use is equating coefficients; I have provided my example solution to this question as an answer. Now this is a pretty good method I reckon but I wonder if someone out there has a faster, more efficient solution to these types of questions that they would like to share (please?). The thing that annoys me is plugging huge sums of numbers into the calculator that takes more time than the answer itself.</p>

答:
It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.

We obtain
\begin{align*}
\color{blue}{[x^0]}&\color{blue}{\left(1+\frac{2}{x}\right)^3\left(1-\frac{x}{3}\right)^5}\\
&=[x^0]\sum_{j=0}^3\binom{3}{j}2^jx^{-j}\left(1-\frac{x}{3}\right)^5\tag{1}\\
&=\sum_{j=0}^3\binom{3}{j}2^j[x^j]\left(1-\frac{x}{3}\right)^5\tag{2}\\
&\,\,\color{blue}{=\sum_{j=0}^3\binom{3}{j}\binom{5}{j}2^j\left(-\frac{1}{3}\right)^j}\tag{3}
\end{align*}
and the result can be manually calculated by summing up $4$ terms.

Comment:

In (1) we apply the binomial theorem to the left-hand binomial.

In (2) we use the linearity of the coefficient of operator and apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.

In (3) we select the coefficient of $x^j$.


答: One of the stationary points has first and second derivative both $0$, so it's a turning point if its third derivative vanishes and the fourth doesn't (which it won't, as $f$ is a quartic). It helps to write $y=1+x$ so $f=y^3(2-y)=2y^3-y^4$, and if $y=3/2$ we get a maximum with $f=27/16$, whereas $y=0$ is a point of inflection with $f=0,\,f^{(3)}=\left.(12-24y)\right|_{y=0}=12\ne0$.


问:
why measure theory for probability?
I studied elementary probability theory. For that, density functions were enough. What is a practical necessity to develop measure theory? What is a problem that cannot be solved using elementary density functions?</p>

答:
Simple answer: Tossing a coin.

Longer answer: You know that you treat discrete events like the above with probability mass functions or similar, but continuous things with probability density functions. Imagine you had $X$ which is randomly uniform on $[0,1]$ half the time and $5$ the rest of the time. Perfectly reasonable thing, could easily come up. Doesn't fit into either framework.

Measure theory provides a consistent language and mathematical framework unifying these ideas, and indeed much more general objects in stochastic theory. It removes any necessity to distinguish between fundamentally similar objects, and crystallizes the relevant points out, allowing much deeper understanding of the theory.

Title: why measure theory for probability?


问:
Calculate number of nodes in a tree from degrees
If I have a tree with the first $N$ nodes having degree $\{2, 3, 4,\dots, N+1\}$ and the remaining nodes having degrees of $1$, is it possible to calculate the total number of nodes?

I've no idea how to go about starting this.

I first thought I could try:

$ \sum_{k=1}^{N} k -1 = \frac{N(N+1)}{2} - 1$

for part of it before realising that degrees are the total number of edges that are incident to a vertex and not necessarily the number of edges that connect to its child nodes. So, there could be overlapping edges for the same nodes.

Edit: realised that I confused how you connect nodes in a tree with how you can connect them in a standard undirected graph. Ignore the final sentence.

答:
Let $n_d$ be the number of nodes of degree $d$.  Then the total number of nodes is
$$\sum_{d \ge 1} n_d = n_1 + \sum_{d=2}^{N+1} 1 = n_1 + N,$$
and the handshake lemma implies
$$2(n_1 + N-1) = \sum_{d \ge 1} d n_d = n_1 + \sum_{d=2}^{N+1} d = n_1 + \frac{N(N+3)}{2},$$
so
$$n_1 = \frac{N(N+3)}{2} - 2(N-1) = \frac{N^2-N+4}{2},$$
and the total number of nodes is
$$n_1 + N = \frac{N^2+N+4}{2}.$$


问:
If $x_n$ is strictly decreasing and $0\le x_n<\frac{1}{2}$, then $x_n \rightarrow 0$ as $n \rightarrow \infty $
Decide following statement is true or false if it true, prove if it false give a counterexample?

If $x_n$ is strictly decreasing and $0\le x_n<\frac{1}{2}$, then $x_n \rightarrow 0$ as $n \rightarrow \infty $

As I think this is true statement since $inf(\{x_n\})=0$ and decreasing sequence according to the Monotone convergence theorem it is convergence to $0$

Is it correct? if it is not give a counter example?

Thank you!!

答:
Nope! Consider $x_{n} = \frac{1}{ 2^{n + 2}} + \frac{1}{4}$. This sequence is strictly decreasing and has $0\leq x<\frac{1}{2}$ for all $n >0$, but $\displaystyle\lim_{n\to\infty}x_{n} = \frac{1}{4}$.

In this case, $x_{n}\geq 0$ is not enough to use the Monotone Convergence Theorem. Instead, we must have $\inf x_{n} = 0$.


答:
$x_n=\frac{1}{2}-\frac{1}{n+2}$, then $x_n$ is increasing and $0 \leq x_n < \frac{1}{2} $ , but converges to $1/2$

OR

$x_n=\frac{1}{100}$+$\frac{1}{n+5}$ is strictly decreasing and $0 \leq x_n < \frac{1}{2} $, but converges to $\frac{1}{100}$

In all cases we have a counterexample.

Body:
Decide following statement is true or false if it true, prove if it false give a counterexample?

If $x_n$ is strictly decreasing and $0\le x_n<\frac{1}{2}$, then $x_n \rightarrow 0$ as $n \rightarrow \infty $

As I think this is true statement since $inf(\{x_n\})=0$ and decreasing sequence according to the Monotone convergence theorem it is convergence to $0$

Is it correct? if it is not give a counter example?

Thank you!!


问:
What are the elements of this ideal $I = (x, xy, xy^2, xy^3, ...)$ look like in general?
I am trying to answer this question:

Let $k$ be a field and $k[x,y] \cong k^{[2]}.$ Define the subring $A \subset k[x,y]$ by $A = k[x, xy, xy^2, xy^3, ...].$ Show that $A$ is not Noetherian.

And I got the following hint:

Hint: Consider the ideal $I = (x, xy, xy^2, xy^3, ...).$ Assume $xy^{n+1}= f_{0}x + f_1xy + \dots + f_n xy^n$ for $f_i \in A.$ Divide by $x$ and evaluate at $x=0.$}

But I have the following questions:

1- What are the elements of this ideal $I = (x, xy, xy^2, xy^3, ...)$ look like in general? what confuses me while trying to write an element of this ideal, is that this ideal is generated by the following infinite set $x, xy, xy^2, xy^3, ...$, could anyone help me in figuring this out please?

答:
Let $K[x,y] = R$ for convenience. So you know that any element of $I$ is a linear combination of any of the generators, which as you rightly point out there are infinitely many of them. Nonetheless, they are very regular, so we can get some idea of what elements of this ideal look like. For example, it is clear that any element of $I$ must have degree in $x$ of at least $1$, since $x$ divides every generator. We can also see that there are no elements which have pure powers of $y$, since no such elements appear among the generators. These two facts tell you that the elements of $I$ are precisely polynomials with zero constant term and no terms composed just of powers of $y$ (including of course $0$, which is in every ideal).

To make this a bit more formal, we can write any element in $I$ as an $R$ linear combination of the generators, i.e.
$$\forall f(x,y) \in I, f(x,y) = f_1(x,y)x + f_2(x,y)xy+f_3(x,y)xy^2+...+f_n(x,y)xy^n = x(f_1(x,y) + f_2(x,y)y+f_3(x,y)y^2+...+f_n(x,y)y^n)$$
where $f_1(x,y),...,f_n(x,y)$ are polynomials from the original ring $R$. In this form, one can see that there can be no non-zero constant terms and no pure powers of $y$, since each term in the sum is divisible by $x$


问:
Do there exist $29$ consecutive integers so that every of them has exactly $2$ distinct prime factors?

Do there exist $29$ consecutive integers, denote $a,a+1,\cdots,a+28$, so that every of them has exactly $2$ distinct prime factors?

For example, $25$ has only one distinct prime factor, and $30$ has $3$ distinct prime factors.

These are my effors:

1.Since they are not divisible by $30$, so $a\equiv 1 \pmod {30}.$

2.I wrote a code (Mathematica 9.0) for this problem:

j = 0; i = 2; While[j < 29 && i < 10^8,
   If[Length[FactorInteger[i]] == 2, j = j + 1; i = i + 1, j = 0;
    i = i + 31 - Mod[i, 30]]]; Print[{j, i}]

After runing this program, i find that there are no such numbers when $a<10^8$, it takes about $4$ minutes.

Thanks in advance!

答:
(Hint) Since $a \equiv 1 \pmod {30}$, we have that $a + 5$, $a + 11$, $a + 17$, and $a + 23$ are each divisible by both 2 and 3.
Since they can only be divisible by 2 prime factors each, this should be enough restriction for you to prove there are no possible $a$.</p>


问:
subspace topology on lower limit vs standard
Let $(\mathbb{R},\tau)$ be the topological space of real numbers where $\tau$ is the standard topology on $\mathbb{R}$.

Let $(\mathbb{R},\tau_l)$ be the topological space of real numbers where $\tau_l$ is the lower limit topology. That is, elements of $\tau_l$ are subsets of $\mathbb{R}$ of the form $[a,b)$ where $a<b$ and $a,b \in \mathbb{R}$.

Given an example of a countably infinite subset of the reals say $A$, such that the subspace topology induced on $A$ when $\mathbb{R}$ has the standard is distinct than when the subspace topology induced on $A$ when $\mathbb{R}$ is endowed with $\tau_l$. So you get distinct open sets.

So my prime suspects when I hear countably infinite subset of $\mathbb{R}$ are typically $\mathbb{N}$, $\mathbb{Z}$, or $\mathbb{Q}$, but if I take $A=\mathbb{Q}$. Then $\mathbb{Q} \cap (a,b)$ is the same as $\mathbb{Q} \cap [a,b)$ whenever $a \in \mathbb{R} \backslash \mathbb{Q}$, right? Some help or possibly just a hint? Thanks in advance!

(of course, here I am taking the subspace topology to be the subspace intersect the open sets in the ambient or parent space)

答:
It is not true that all elements of $\tau_\ell$ are of the form $[a,b)$: those half-open intervals simply form a base for $\tau_\ell$.

$\Bbb Q$ will work: $0$ is the limit of the sequence $\left\langle-\frac1n:n\in\Bbb Z^+\right\rangle$ in the topology inherited from $\tau$, but not in the topology inherited from $\tau_\ell$, since $[0,1)\cap\Bbb Q$ is a $\tau_\ell$-open nbhd of $0$ that contains no term of the sequence.

In fact, you could simply use the subspace $\{0\}\cup\left\{-\frac1n:n\in\Bbb Z^+\right\}$.


问:
derivative for the following matrix
I have a question about the following derivative. Let us have $X\in\mathbb{R}^{m\times n}, z\in\mathbb{R}^{n}$ and I would like to find the derivative $$\frac{\partial (z^{T}X^{T}Xz)}{\partial z}. $$ Any idea? It gives me a hard time. Thank you.

答:
$$\frac{\partial (z^{T}X^{T}Xz)}{\partial z} = 2X^T X z.$$

Indeed:

$$z^{T}X^{T}Xz = \sum_{k=1}^n\sum_{h=1}^m \sum_{j=1}^n z_kx_{hk}x_{hj}z_j.$$

Fix $i \in \{1, \ldots, N\}$. Observe that:

$$\sum_{k=1}^n\sum_{h=1}^m \sum_{j=1}^n z_kx_{hk}x_{hj}z_j = \sum_{k=1}^n\sum_{h=1}^m \left(\sum_{j=1, j \neq i}^n z_kx_{hk}x_{hj}z_j + z_kx_{hk}x_{hi}z_i\right) = \\
= \sum_{k=1, k \neq i}^n\left[\sum_{h=1}^m \left(\sum_{j=1, j \neq i}^n z_kx_{hk}x_{hj}z_j + z_kx_{hk}x_{hi}z_i\right)\right] + \\
+\sum_{h=1}^m \left(\sum_{j=1, j \neq i}^n z_ix_{hi}x_{hj}z_j + z_ix_{hi}x_{hi}z_i\right).$$

As you can see, there are terms which depend on $i$, and other which do not depend on, i.e.

$$\sum_{k=1}^n\sum_{h=1}^m \sum_{j=1}^n z_kx_{hk}x_{hj}z_j = \sum_{k=1, k \neq i}^n\sum_{h=1}^m z_kx_{hk}x_{hi}z_i + \sum_{h=1}^m \sum_{j=1, j \neq i}^n z_ix_{hi}x_{hj}z_j +\sum_{h=1}^m z_ix_{hi}x_{hi}z_i + \text{terms which do not depend on}~i.$$

Taking the derivative with respect to $z_i$, one gets:

$$\frac{\partial z^{T}X^{T}Xz}{\partial z_i} = \sum_{k=1, k \neq i}^n\sum_{h=1}^m z_kx_{hk}x_{hi} + \sum_{h=1}^m \sum_{j=1, j \neq i}^n x_{hi}x_{hj}z_j + 2\sum_{h=1}^m x_{hi}x_{hi}z_i = \\
 = \sum_{k=1, k \neq i}^n\sum_{h=1}^m z_kx_{hk}x_{hi} + \sum_{h=1}^m \sum_{k=1, k \neq i}^n x_{hi}x_{hk}z_k + 2\sum_{h=1}^m x_{hi}x_{hi}z_i = \\
 = 2\sum_{k=1, k \neq i}^n\sum_{h=1}^m z_kx_{hk}x_{hi} +2\sum_{h=1}^m x_{hi}x_{hi}z_i = \\
 = 2\left(\sum_{k=1, k \neq i}^n\sum_{h=1}^m z_kx_{hk}x_{hi} +\sum_{h=1}^m x_{hi}x_{hi}z_i\right) = \\
 = 2\sum_{h=1}^m\sum_{k=1}^n x_{hi} x_{hk} z_k,
$$

which corresponds to twice the $i$-th component of $X^T Xz.$

Title: derivative for the following matrix


问:
The basis for the null space of a linear transformation
I am given the following linear transformation

$$f: \begin{bmatrix}x\\
y\\
z
\end{bmatrix}\rightarrow\begin{bmatrix}x\\
x
\end{bmatrix}$$

and asked to select all answers which provide a basis for the null space and which correctly identify the nullity of the map.

Here are the options:

$$\begin{array}{cccc}
B=\left\langle \begin{bmatrix}1\\
1\\
1
\end{bmatrix},\begin{bmatrix}0\\
1\\
1
\end{bmatrix}\right\rangle  &  &  & \text{null}\left(f\right)=2\end{array}$$

$$\begin{array}{cccc}
B=\left\langle \begin{bmatrix}1\\
1
\end{bmatrix}\right\rangle  &  &  & \text{null}\left(f\right)=1\end{array}$$

$$\begin{array}{cccc}
B=\left\langle \begin{bmatrix}0\\
1\\
0
\end{bmatrix},\begin{bmatrix}0\\
0\\
1
\end{bmatrix}\right\rangle  &  &  & \text{null}\left(f\right)=1\end{array}$$

$$\begin{array}{cccc}
B=\left\langle \begin{bmatrix}0\\
1\\
1
\end{bmatrix},\begin{bmatrix}0\\
1\\
-1
\end{bmatrix}\right\rangle  &  &  & \text{null}\left(f\right)=2\end{array}$$

We can immediately rule out option (2), as the null space is a subset of $\mathbb{R}^3$.

To rule out further options, we can find the nullity. To do this, we set the codomain equal to $0$, write the linear system as an augmented matrix, perform row operations to put the matrix into row echelon form, and solve for the leading variables. We can then parameterize the kernel, which results in the following.

$$\cal{K}\left(f\right)=\left\{ \begin{bmatrix}0\\
1\\
0
\end{bmatrix}y+\begin{bmatrix}0\\
0\\
1
\end{bmatrix}z\mid y,z\in\mathbb{R}\right\}$$

This tells us that $\text{null}\left(f\right)=2$. Therefore, we can rule out option (3).

To determine if option (1) and option (4) are valid, we need to show that (a) the candidate basis is linearly independent and (b) it spans the null space.

Unless I am mistaken, the null space is the same as the kernel.

I tested both (a) and (b) for options (1) and (4) and found that they both are linearly independent and span the kernel.

Here is my work for determining whether the candidate basis spans the kernel (I will omit the linear independence test because that is far simpler).

Option 1:

$$r\begin{bmatrix}1\\
1\\
1
\end{bmatrix}+s\begin{bmatrix}0\\
1\\
1
\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}$$

I wrote the linear system as an augmented matrix, performed row operations to put the matrix into row echelon form, and solved for the leading variables.

$$\begin{array}{rclccrclccrcl}
0 & = & c-b &  &  & s & = & b-a &  &  & r & = & a\end{array}$$

Unless I am mistaken, this shows that the candidate basis spans the kernel.

Option 4:

$$r\begin{bmatrix}0\\
1\\
1
\end{bmatrix}+s\begin{bmatrix}0\\
1\\
-1
\end{bmatrix}=\begin{bmatrix}a\\
b\\
c
\end{bmatrix}$$

I wrote the linear system as an augmented matrix, performed row operations to put the matrix into row echelon form, and solved for the leading variables.

$$\begin{array}{rclccrclccrcl}
0 & = & 0 &  &  & s & = & -\frac{1}{2}\left(c-b\right) &  &  & r & = & \frac{1}{2}\left(b-c\right)\end{array}$$

Unless I am mistaken, this shows that the candidate basis spans the kernel.

The answer:

Therefore, I selected both options (1) and (4). Apparently, this is incorrect.

Does anyone know where I went wrong? Thank you.

答: You should eliminate option 1 at first since it cannot span ker$f$. For example, $(0,1,5) \in kerf$, but it is not included in the set in option 1.


问:
How to make a unit step function?
I am trying to make a unit step function.

I have this function (the equation of an ellipse, not centered at the origin):

$$
f(x,y) = \frac{(x-X_c)^2}{a^2}+\frac{(y-Y_c)^2}{b^2}
$$

What I would like to, is present it as a unit step function with the attributes saying if the result of f(x,y) is between 0.95 and 1.05 then return 1, otherwise return 0. Is this possible? If so, how do I do it?

I hope you understand my question, and I am looking forward to help.

-David

答: Let $u(x)$ be the unit step function.  You want the function $g(z)=u(z-0.95)-u(z-1.05)$.  Replace z with $f(x,y)$</p>

Title: How to make a unit step function?


答:
Since $f(x) $ is non-negative only when $ |x| \leqslant 1$, then we only need to care when $|x|\leqslant 1$.

With that, we know that $ 1-x$ and $ 1 + x$ are non-negative numbers. Thus, we can apply the AM-GM inequality:

$$ AM\left(1 - x , \frac{1+x}3 , \frac{1+x}3, \frac{1+x}3 \right) \geqslant GM \left (1 - x , \frac{1+x}3 , \frac{1+x}3, \frac{1+x}3 \right) $$

Simplifying the above gives $$\frac12 \geqslant \left ( \frac{(1-x)(1+x)^3}{3^3} \right )^{1/4}  $$

Rearranging gives $ (1-x)(1+x)^3 \leqslant \boxed{\frac{27}{16}}.$

The maximum value is attained when $ 1-x = \dfrac{1+x}3 \Leftrightarrow x = \dfrac12. $


问:
Number of homomorphisms between two arbitrary groups
How many homomorphisms are there from $A_5$ to $S_4$ ?

This is how I tried to solve it.

If there is a homomorphism from $A_5$ to $S_4$ , then order of element of $S_4$ should divide the order of its preimage. Now what are the possible order of elements in $S_4$.1,2,3 and 4. Since $A_5$ contains (12345), which is of order 5.. what could be image of (12345). Definitely Identity element which is of order 1. Similarly all 5 cycles must be mapped to identity.  There are 24 elements of 5 cycles. 24 elements out of 60 are mapped to identity .. now only two types of homomorphisms are possible. either 30:1mapping  or 60:1 mapping.  Consider (12)(34) which belongs to $A_5$. It's image can be element of order 2 or identity.there are 15 elements of order 2 . suppose these 15 elements are mapped to some element 'g' of order 2 of $S_4$, you need another 15 elements to get mapped to 'g' to have 30 :1 mapping. Other type of elements left in $A_5$ is of order 3. None of them can be mapped to g. hence 15 elements of order 2 should be mapped to identity .. so ,  (24+15=39) elements mapped to identity.As mentioned earlier it should be 30 :1 or 60:1 mapping. So it must be 60:1 mapping.Hence a trivial homomorphism. Answer is 1.

I wanted to know is there any other technique which can be used to find number of homomorphism in the above question ?
In general, how to find number of homomorphism  between any two arbitrary groups ?

答: You can rely on another property of $A_5$, other than its simplicity, to get that the only homomorphism from $A_5$ to $S_4$ is the trivial one, namely the fact that $A_5$ has no subgroups of order $30$, $20$ and $15$ (see e.g. here). In fact, a homomorphism $\varphi\colon A_5\to S_4$ is equivalent to a $A_5$-action on the set $X:=\{1,2,3,4\}$. By the Orbit-Stabilizer theorem and the fact that the set of orbits forms a partition of $X$, the stabilizers can only have orders $60/k$, for $1\le k\le 4$; but the stabilizers are subgroups of the group which acts, and hence, by the abovementioned property of $A_5$, the only option $k=1$ is actually allowed (for every $i=1,2,3,4$). So, all the stabilizers must coincide with the whole $A_5$ and the only sought homomorphism has kernel $\bigcap_{i=1}^4{\rm{Stab}}(i)=A_5$, which precisely means that all the elements of $A_5$ are mapped to ${()}_{S_4}$.


问:
Differential vector divided by differential vector length
Does anyone know a general simplification: $\frac{d\vec{r}}{d||\vec{r}||}$ in terms of the unit vector $\hat{r}$. For a linear vectorfunction $\vec{r}$ the following is true of course: $\frac{d\vec{r}}{d||\vec{r}||}=\hat{r}$.

Is the following true for every $\vec{r}$ vector: $\frac{d\vec{r}}{d||\vec{r}||}=d\hat{r}$...?

答:
Let $\,r^2=(\vec r\cdot\vec r),\,r=\|\vec r\|,\,$ and $\,\hat r={\vec r}/{r}.\;$ Then the following derivatves are true
$$\eqalign{
\frac{dr}{d\vec r} &= \hat r, \qquad
\frac{d\hat r}{d\vec r} &= \tfrac{1}{r}(I-\hat r\hat r^T) \\
}$$
but the remaining variable combinations $\,\Big(\frac{d\vec r}{dr},\,\frac{d\vec r}{d\hat r},\,\frac{d\hat r}{dr},\,\frac{dr}{d\hat r}\Big)\,$ are nonsense.

Sometimes you can invert a derivative to get what you want, but the above derivatives do not have inverses.
They do have pseudoinverses, so depending on the application that you have in mind,  you could loosely assert something like
$$\eqalign{
\frac{d\vec r}{dr} &\approx \left(\frac{dr}{d\vec r}\right)^+
  =\; \hat r^+ \;=\; \hat r^T \\
\frac{d\vec r}{d\hat r} &\approx \left(\frac{d\hat r}{d\vec r}\right)^+
  =\; r\,(I-\hat r\hat r^T) \\
}$$
for two of the nonsensical derivatives.


问:
How to show that a equation is a vector space?
Let $W$ be the set $W := \{(p,q,r,s) \in \mathbb{R}^{4} \mid  x+3y+4q=0 \}$. How can I show that $W$ is a vector space? Is $(6,8,6,4)$ in $W$? How and why?
Please do all the works I am really confused about it.</p>

答:
First of all the question: "How to show that a equation is a vector space?" is meaningless. An equation is an equation, a vector space is a set together with operations and together with a field of scalars, so you can see these are very different things. Now, you can discuss wether the set of solutions of an equation is a vector space or not (after all it's a set, and we can sum and multiply solutions by scalars in the obvious way).

Now, this set $W$ is obviously a vector space. The reason is because it's the kernel of a linear transformation. Indeed set $L : \mathbb{R}^4 \to \mathbb{R}$ by the following $L(x_1, x_2, x_3, x_4) = x_1 + 3x_2 + 4x_4$. Now your set $W$ is just $W = \{x \in \mathbb{R} \mid L(x)=0$}, and since the kernel of a linear transformation is indeed a vector subspace of the domain, it's obviously a vector space itself because $W=\ker L$.

To check wether $(6,8,6,4)$ is in $W$ you compute $L(6,8,6,4) = 6+2\cdot8+4\cdot4\neq0$ so it doesn't satisfy the condition $L(x) = 0$ and so it's not in $W$.


问:
What does the caret symbol (^) mean in Feynman's water wave thought experiment?
A page from the "Quantum Behavior" chapter in Feynman's book "Six Easy Pieces":

He uses the following notation:

$h_1e^{i\omega t}$

But the "h" has a caret (^) on top. What does it denote in this context?

Side note: I used a picture instead of a MathJax-formatted quotation because the caret (^) is only used in the book format of his lectures, not in the online version.

答:
Feynman isn't really using something standard here, although I would be unsurprised if it's not unique, or not even truly original. The issue here is that, while the waves he's considering depend in general on both time and space, they are all of the same-$\omega$ form $f(x)e^{i\omega t}$ (or if they're not, separable basis elements of the relevant space of functions are).

So it makes sense to write all waves as $e^{i\omega t}$ times something time-independent, although it may still be complex. In fact, even if there's no overall space-dependence, this still makes sense. For example,$$he^{i(\omega t-\phi)}=\hat{h}e^{i\omega t},\,\hat{h}:=he^{-i\phi}$$is how this notation encodes the relative phase of multiple waves, even if $\phi$ is a constant for each wave. Doing all this lets Feynman reduce a lot of differential equations to characteristic polynomial equations.


问:
How is a Lie algebra called whose derived series vanishes after infinitely many steps?
Let $\mathfrak{g}$ be a Lie algebra and let
$$
  \mathfrak{g}^{(0)}
  \supseteq
  \mathfrak{g}^{(1)}
  \supseteq
  \mathfrak{g}^{(2)}
  \supseteq
  \mathfrak{g}^{(3)}
  \supseteq
  \dotsb
$$
be the derived series of $\mathfrak{g}$, given by $\mathfrak{g}^{(0)} := \mathfrak{g}$ and $\mathfrak{g}^{(i+1)} = [\mathfrak{g}^{(i)}, \mathfrak{g}^{(i)}]$ for every $i \geq 0$.

Question.
Does the property $\bigcap_{i \geq 0} \mathfrak{g}^{(i)} = 0$ have a name?

If $\mathfrak{g}$ is finite-dimensional, then this property means precisely that $\mathfrak{g}$ is solvable. But for infinite-dimensional Lie algebras this is in general not the case.

答:
Yes, it's called "residually solvable" (or "residually soluble" in British English).

This is part of a general terminology. Let P be a property of algebras that is stable under finite direct products and under taking subalgebras. An algebra $A$ is said to be residually P if for any pair  $(x,y)$ of distinct elements in $A$ there is a quotient $p:A\to B$ such that $p(x)\neq p(y)$ and such that $B$ has Property P. This is the same as asking that the intersection of (2-sided) ideals $I$ such that $A/I$ has Property P, is trivial.

For instance, if $(\mathfrak{g}^i)_{i\ge 1}$ is the lower central series, $\bigcap_i\mathfrak{g}^i=\{0\}$ means that $\mathfrak{g}$ is residually nilpotent.

Examples of residually solvable Lie algebras are free Lie algebras (they're even residually nilpotent). In particular, the property "residually solvable", unlike solvable, does not pass to quotients.


问:
Double integral in polar coordinates (trouble with boundaries)
Calculate double integral of $$\iint_\Omega \sqrt{x^2+y^2} \;\mathrm{d}x\mathrm{d}y,\;\mathrm{where} \;\Omega:(x-1)^2+y^2\leq 1, \; 0\leq y$$
I try to figure out the boundaries so $0\leq x \leq 2$ and $0 \leq y \leq \sqrt{1-(x-1)^2}$

I try to change into polar coordinates where I even substituted $\sqrt{x^2+y^2}=r$ and I got $8\pi/3$ (for $y=0 \mathrm{\;and\;} y=\pi$) but the result should be $16/9$.

答: The region that we are talking about here is the upper half of the circle with center $(1,0)$ and radius $1$. So, the possible values of $\theta$ lie in the range $\left[0,\frac\pi2\right]$. For each such $\theta$, let $t=\tan(\theta)$. One must determine where $y=tx$ intersects the circle $(x-1)^2+y^2=1$ (see the picture below). So, we solve the equation$$(x-1)^2+t^2x^2=1\left(\iff(t^2+1)x^2-2x=0\right)$$and we get that $x=0$ or that $x=\frac2{t^2+1}=2\cos^2\theta$. And, when $x=2\cos^2\theta$, then$$y=2\cos^2(\theta)\tan(\theta)=2\cos(\theta)\sin(\theta),$$and therefore $r=\sqrt{x^2+y^2}=2\cos\theta$. So, compute$$\int_0^{\pi/2}\int_0^{2\cos\theta}r^2\,\mathrm dr\,\mathrm d\theta.$$


问:
Recursive relation induction problem: $c_n = \frac{1}{2^n} \cdot \sum_{k \text{ even}}^{0~ \leq ~k~ \leq ~n} {n \choose k}c_k$
This is part of problem 18 of chapter 13 of Michael Spivak's Calculus:

Define $c_n = \int_{0}^{1}x^n dx$

(b) Prove: $$c_n = \frac{1}{2^n} \cdot \sum_{k \text{ even}}^{0~ \leq ~k~ \leq ~n} {n \choose k}c_k$$

(c) Prove $c_n = \frac{1}{n+1}$ from this (without calculating the integral).

Note the actual question in the book is more than this, I am just condensing it to the parts I'm stuck on. I was able to do (b), but not (c). It seems to me to be another induction problem, but I am going in circles. So I assume it's true for $\forall n \leq p$, i.e. $ \forall n \leq p : c_n = \frac{1}{n+1}$. Then for $c_{p+1}$, we have:

$$c_{p+1} = \frac{1}{2^{p+1}} \cdot \sum_{k \text{ even}}^{0~ \leq ~k~ \leq ~{p+1}} {p+1 \choose k}c_k$$

Now already there's some technical issues as to whether the last term in the sum will be $c_{p+1}$ or $(p+1)c_p$, depending on whether $p+1$ is even or odd. So I split up the formula into those 2 separate cases, like so:

$$c_{n} = \frac{1}{2^{n} - 1} \cdot \sum_{k~ =~ 0}^{k ~= ~n/2-1} {n \choose 2k}c_{2k} ~~~: ~~\text{for n = even}$$

$$c_{n} = \frac{1}{2^{n}} \cdot \sum_{k~ =~ 0}^{k ~= ~(n-1)/2} {n \choose 2k}c_{2k} ~~~~~~: ~~\text{for n = odd}$$

Ok so we'll probably have to do 2 separate proofs by induction, one for when $p$ is odd, one when $p$ is even. Also we'll need to prove both the $n = 0$ and $n = 1$ base cases, but that's trivial, no problems there.

Let's try when $p$ is even. So we assume $p$ is even and $ \forall n \leq p : c_n = \frac{1}{n+1}$. Then $p+1$ is odd, so:

$$c_{p+1} = \frac{1}{2^{p+1}} \cdot \sum_{k~ =~ 0}^{k ~= ~p/2} {p+1 \choose 2k}c_{2k}$$

$$~~~~~~~~~~~~~~= \frac{1}{2^{p+1}} \cdot \sum_{k~ =~ 0}^{k ~= ~p/2} {p+1 \choose 2k}\frac{1}{2k+1}$$

And now I am stuck.

I need to somehow show $\frac{1}{2^{p+1}} \cdot \sum_{k~ =~ 0}^{k ~= ~p/2} {p+1 \choose 2k}\frac{1}{2k+1} = \frac{1}{p+2}$. I tried using the identity ${n+1 \choose k} = {n \choose k} + {n \choose k-1}$, which gives us:

$$c_{p+1} = \frac{1}{2^{p+1}} \cdot \sum_{k~ =~ 0}^{k ~= ~p/2} {p+1 \choose 2k}\frac{1}{2k+1}$$

$$~~~~~~~~~~~~~~~~~~~~~~~~~~= \frac{1}{2^{p+1}} \cdot \sum_{k~ =~ 0}^{k ~= ~p/2} \Bigr[{p \choose 2k} + {p \choose 2k-1}\Bigr]\frac{1}{2k+1}$$

$$~~~~~~~~~~~~~~~~~~~~~~~~~= \frac{1}{2^{p+1}} \cdot \biggr[c_p ~+~ \Bigr[ \sum_{k~ =~ 0}^{k ~= ~p/2} {p \choose 2k-1}\frac{1}{2k+1}\Bigr]\biggr]$$

But now how to deal with this new summation that runs on odd numbers; $2k - 1$, instead of the even numbers; $2k$, that we have in our definition of $c_p$ (which we need to incorporate somehow)? I tried to transform the ${p \choose 2k - 1}$ to ${p \choose 2k - 2}$ or ${p \choose 2k}$, using the identity ${n \choose k} = \frac{n - k + 1}{k}{n \choose k - 1}$ but the coefficients didn't cancel and looked very ugly so now I have no ideas.

Any?

Body:
Note that $$\sum_{k \ge 0} a_{2k} = \sum_{k \ge 0} \frac{1+(-1)^k}{2}a_k.$$
Taking $a_k=\binom{p+1}{k}\frac{1}{k+1}$ yields
\begin{align}
\frac{1}{2^{p+1}}\sum_{k\ge 0} \binom{p+1}{2k}\frac{1}{2k+1}
&=\frac{1}{2^{p+1}}\sum_{k\ge 0} \frac{1+(-1)^k}{2} \binom{p+1}{k}\frac{1}{k+1} \\
&=\frac{1}{2^{p+1}(p+2)}\sum_{k\ge 0} \frac{1+(-1)^k}{2} \binom{p+2}{k+1} \\
&=\frac{1}{2^{p+2}(p+2)}\left(\sum_{k\ge 0} \binom{p+2}{k+1} - \sum_{k\ge 0} (-1)^{k+1} \binom{p+2}{k+1}\right) \\
&=\frac{((1+1)^{p+2}-(p+2)) - ((1-1)^{p+2}-(p+2))}{2^{p+2}(p+2)} \\
&=\frac{2^{p+2}}{2^{p+2}(p+2)} \\
&=\frac{1}{p+2}.
\end{align}


答:
HINT

Make the change of variables $x = 1 + r\cos(\theta)$ and $y = r\sin(\theta)$. Then the proposed integral becomes
\begin{align*}
\iint_{\Omega}\sqrt{x^{2} + y^{2}}\mathrm{d}x\mathrm{d}y = \int_{0}^{1}\int_{0}^{\pi}r\sqrt{1 + 2r\cos(\theta) + r^{2}}\mathrm{d}\theta\mathrm{d}r
\end{align*}

Title: Double integral in polar coordinates (trouble with boundaries)

Body:
Calculate double integral of $$\iint_\Omega \sqrt{x^2+y^2} \;\mathrm{d}x\mathrm{d}y,\;\mathrm{where} \;\Omega:(x-1)^2+y^2\leq 1, \; 0\leq y$$
I try to figure out the boundaries so $0\leq x \leq 2$ and $0 \leq y \leq \sqrt{1-(x-1)^2}$

I try to change into polar coordinates where I even substituted $\sqrt{x^2+y^2}=r$ and I got $8\pi/3$ (for $y=0 \mathrm{\;and\;} y=\pi$) but the result should be $16/9$.


答:
Let $u=x-1$ and $v=y$ to rewrite the integral

$$I= \iint_\Omega \sqrt{x^2+y^2}{d}x{d}y
= \int_{u^2+v^2\le 1, v>0}\sqrt{(u+1)^2+v^2}du dv
$$

Then, with polar coordinates

$$I=\int_0^{\pi}\int_0^1\sqrt{r^2-2 r\cos\theta+1}\>rdr d\theta$$

Fisrt, evaluate with $t=r-\cos\theta$

\begin{align}
f(\theta)& =\int_0^1\sqrt{r^2-2 r\cos\theta+1}\>rdr
= \int_{-\cos\theta}^{1-\cos\theta}\sqrt{t^2 +\sin^2\theta}\> (t+\cos\theta)dt\\
&= \frac13(8\sin^3\frac\theta2-1)+\frac12\cos\theta[4\sin^3\frac\theta2+\cos\theta+\sin^2\theta\ln(1+\csc\frac\theta2)]
\end{align}

Then, integrate over $\theta$ to obtain
$$I=\int_0^{\pi}f(\theta)d\theta
= \frac{16}9
$$


问:
Quadratic Equations word solving problems
I have three questions if you can solve them please show all of your work.

The height of Isosceles triangle is 2/3 of base. Find the sides and the height of this triangle if the area is 48cm^2.

答:
A = (1/2) base * height = 48

height = (2/3) base.

You have two equations and two unknowns.  From here, you “plug and chug”.


问:
Two cowboys A and B decide to solve a dispute with a duel.
Two cowboys $A$ and $B$ decide to solve a dispute with a duel.
Cowboy $A$ hits his target $\frac13$ of the time.
Cowboy $B$ hits the target $\frac23$ of the time.

It is decided that $A$ will take the first shot, cowboy $B$ will take the second shot (if still alive).
This will continue until there is only one left alive. Also, a cowboy can not shoot two times in a row. What are cowboy $A$ chances of winning the duel?

My logic is to simulate the cases where $A$ wins:

I Case:  $A$ wins: $\frac13$

II Case: $A$ loses, $B$ loses, $A$ wins: $\frac23\cdot\frac13\cdot\frac13$

...

But I don't know what to do from here. Any help?

答:
What we have is a geometric series starting from $\frac13$ and the next term $\frac29$ of the previous. The sum is
$$\frac13\cdot\frac1{1-2/9}=\frac37$$


答:
You are definitely on the right track. A hint might be that A wins at turn k ($A_k$) iff all the previous shots fail (both A's and B's), and the last one does not (and they are all mutually exclusive events).

The chance of both of them failing is ($\frac{1}{3}\frac{2}{3} = \frac{2}{9}$), and the chance of A hitting the target is 1/3.
Therefore the expression you look for is:
$$P(\text{A wins}) = P(\cup_{k\in\mathbb{N}}A_k) =  \sum_{k=0}^{\infty} (\frac{2}{9})^k (1/3)$$

Title: Two cowboys A and B decide to solve a dispute with a duel.

Body:
Two cowboys $A$ and $B$ decide to solve a dispute with a duel.
Cowboy $A$ hits his target $\frac13$ of the time.
Cowboy $B$ hits the target $\frac23$ of the time.

It is decided that $A$ will take the first shot, cowboy $B$ will take the second shot (if still alive).
This will continue until there is only one left alive. Also, a cowboy can not shoot two times in a row. What are cowboy $A$ chances of winning the duel?

My logic is to simulate the cases where $A$ wins:

I Case:  $A$ wins: $\frac13$

II Case: $A$ loses, $B$ loses, $A$ wins: $\frac23\cdot\frac13\cdot\frac13$

...

But I don't know what to do from here. Any help?


答:
If the first number is $30n+1$ we note that:

$$30n+6=6(5n+1)$$$$30n+12=6(5n+2)$$$$30n+18=6(5n+3)$$$$30n+24=6(5n+4)$$

So $5n+r$ is divisible only by $2$ or $3$ for $1 \le r\le4$. $5n+1$ and $5n+2$ are coprime, so one must be a power of $3$ and the other a power of $2$ (neither can be $1$). It is then easy to see that we can't also fit $5n+3$ and $5n+4$ into the same pattern.

Title: Do there exist $29$ consecutive integers so that every of them has exactly $2$ distinct prime factors?

Body:
Do there exist $29$ consecutive integers, denote $a,a+1,\cdots,a+28$, so that every of them has exactly $2$ distinct prime factors?

For example, $25$ has only one distinct prime factor, and $30$ has $3$ distinct prime factors.

These are my effors:

1.Since they are not divisible by $30$, so $a\equiv 1 \pmod {30}.$

2.I wrote a code (Mathematica 9.0) for this problem:

j = 0; i = 2; While[j < 29 && i < 10^8,
   If[Length[FactorInteger[i]] == 2, j = j + 1; i = i + 1, j = 0;
    i = i + 31 - Mod[i, 30]]]; Print[{j, i}]

After runing this program, i find that there are no such numbers when $a<10^8$, it takes about $4$ minutes.

Thanks in advance!