问:
Sobolev space of disjoint union of open sets
Let $\Omega_{1}$, $\Omega_{2}$ be two disjoint bounded open sets in $\mathbb{R}^{N}$, i.e. $\Omega_{1}\cap \Omega_{2} = \emptyset$

Is it true that one can express $H_{0}^{1}(\Omega_{1} \cup \Omega_{2})$ in the form $\displaystyle H_{0}^{1}(\Omega_{1} \cup \Omega_{2})\equiv \lbrace u+v \mid u \in H_{0}^{1}(\Omega_{1}),v\in H_{0}^{1}(\Omega_{2})\rbrace$?

I have kind of convinced myself that the R.H.S. is a subspace of the L.H.S but I'm not sure about the other way, or even if it should be true.

答:
Let me leave at least some notes of a proof for anyone who looks here. Let $\Omega_{1}$, $\Omega_{2}$ be disjoint  bounded open sets and $\Omega = \Omega_{1}\cup \Omega_{2}$.

Lemma 1. Suppose that $K\Subset \Omega$, then $K_{1}=K\cap \Omega_{1} \Subset \Omega_{1}$ and $K_{2}=K\cap \Omega_{2} \Subset \Omega_{2}$. ($\Subset$ means compactly contained here)

Proof. It suffices to show that $K_{1} \subset \Omega_{1}$ is closed. Let $x_{n}$ be a sequence in $K_{1}$ then by compactness it has a subsequence we denote $x_{n}$ also that converges to some $x\in K$. $x$ cannot be in $K_{2}$ since as each of $K_{1}$ and $K_{2}$ is at least some positive distance from the boundaries $\partial \Omega_{1}$ and $\partial \Omega_{2}$ respectively. Thus, $x\in K_{1}$ and hence $K_{1}$ is closed. $\square$

Lemma 2. We have that $C_{c}^{\infty}(\Omega) = C_{c}^{\infty}(\Omega_{1}) \oplus C_{c}^{\infty}(\Omega_{2})$

Proof. The inclusions $C_{c}^{\infty}(\Omega_{1}),C_{c}^{\infty}(\Omega_{2}) \subset C_{c}^{\infty}(\Omega)$ are clear and thus, noting that $C_{c}^{\infty}(\Omega_{1})\cap C_{c}^{\infty}(\Omega_{2}) = \lbrace 0\rbrace$, $C_{c}^{\infty}(\Omega_{1})\oplus C_{c}^{\infty}(\Omega_{2}) \subset C_{c}^{\infty}(\Omega)$.
Now we show that every $f\in C_{c}^{\infty}(\Omega)$ can be decomposed into $f=f_{1} +f_{2}$ where $f_{1}\in C_{c}^{\infty}(\Omega_{1})$ and $f_{2}\in C_{c}^{\infty}(\Omega_{2})$. Well let $S_{1} = \operatorname{supp}(f)\cap \Omega_{1}$ and $S_{2} = \operatorname{supp}(f)\cap \Omega_{2}$. Then notice that $S_{1}\cup S_{2} = \operatorname{supp}(f)$ and $S_{1},S_{2}$ are compactly contained in $\Omega_{1},\Omega_{2}$ respectively. Define $f_{1} = \left.f\right|_{\Omega_{1}}$ and $f_{2} = \left.f\right|_{\Omega_{2}}$. Then we have that $\operatorname{supp}(f_{1})=S_{1},\operatorname{supp}(f_{2})=S_{2}$. Notice that $f = f_{1}+f_{2}$ and thus we have the result. $\square$

Result 1. We have that $H_{0}^{1}(\Omega) = H_{0}^{1}(\Omega_{1})\oplus H_{0}^{1}(\Omega_{2})$.

Proof. $H_{0}^{1}(\Omega_{1})\oplus H_{0}^{1}(\Omega_{2}) \subset H_{0}^{1}(\Omega)$ similarly as to previous lemma. For $u\in H_{0}^{1}(\Omega)$  we can find some sequence $\phi_{n} = \phi_{n}^{(1)}+\phi_{n}^{(2)} \in C_{c}^{\infty}(\Omega)$ which converges to $u$ in the $H^{1}$ norm. Now we have that

$\displaystyle \begin{split} \lVert \phi_{n} - u \rVert_{H_{0}^{1}(\Omega)}^{2} & = \sum_{|\alpha| \leq 1}\int_{\Omega}(\phi_{n} - u)^{2} \\ & = \sum_{|\alpha| \leq 1}\int_{\Omega_{1}}(\left.\phi_{n}\right|_{\Omega_{1}} - u)^{2}+\sum_{|\alpha| \leq 1}\int_{\Omega_{2}}(\left.\phi_{n}\right|_{\Omega_{2}} - u)^{2} \\ &= \lVert \left.\phi_{n}\right|_{\Omega_{1}} - u \rVert_{H_{0}^{1}(\Omega_{1})}^{2}+\lVert \left.\phi_{n}\right|_{\Omega_{2}} - u \rVert_{H_{0}^{1}(\Omega_{2})}^{2} \end{split}$

Now observe that $\lVert \phi_{n} - u \rVert_{H_{0}^{1}(\Omega)}^{2} \to 0 \Leftrightarrow \lVert \left.\phi_{n}\right|_{\Omega_{1}} - u \rVert_{H_{0}^{1}(\Omega_{1})}^{2},\lVert \left.\phi_{n}\right|_{\Omega_{2}} - u \rVert_{H_{0}^{1}(\Omega_{2})}^{2} \to 0$ as $n\to\infty$. Thus, we have the desired result. $\square$

As an important remark we needn't make any assumptions on the whether $\overline{\Omega_{1}\cup \Omega_{2}}$ is connected.

As a point of curiosity notice that if $\lambda$ is a Dirichlet eigenvalue of $\Omega_{1}$ with normalised eigenfunction $u_{n}$ we have that for all $v\in H_{0}^{1}(\Omega)$

$\displaystyle  \lambda\int_{\Omega}u_{n}v =  \lambda\int_{\Omega_{1}}u_{n}\left.v\right|_{\Omega_{1}} = \int_{\Omega_{1}}\nabla u_{n} \cdot \nabla \left.v\right|_{\Omega_{1}} = \int_{\Omega}\nabla u_{n} \cdot \nabla v$.

So $\lambda$ is also a Dirichlet eigenvalue of $\Omega$ with eigenfunction $u_{n}$.

Best,
Cryptokyo

(SF)


问:
Two countable sets proof
If we know that $\mathbb{N} \times \mathbb{N}$ is equivalent to $\mathbb{N}$ (same cardinal), can we conclude that the cartesian product $A \times B$ of two countable sets $A$ and $B$ is countable? In particular I already proved that a bijection $\phi: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$ exists such that $\phi(x, y) = x +\frac{(x + y -1)(x + y - 2)}{2}$. So, presuming that I am writing a formal proof, how would I show that $A \times B$ is countable? Any assistance is much appreciated.

答: Certainly. A[n infinite] countable set is one which bijects with $\mathbb{N}$, so can you build a bijection $A \times B \to \mathbb{N} \times \mathbb{N}$ and then to $\mathbb{N}$?

ContentLicense: CC BY-SA 4.0

Title: Two countable sets proof


问:
Paul R. Halmos bases proof
Hi and thanks for reading. For those of you who know the book I am currently going through, Paul R. Halmos'Finite dimensional vector spaces: In the proof that all bases have the same number of elements, why has the author written that $n\geq{m}$, since, if $n=m$, then we would have replaced all of the X's with elements of Y, which is a contradiction to the original assumption, since at each stage of removing an $x_i$ and adding a $y_k$, the set of X's and Y's form a basis for V. But not if we remove all the X's.

For those not familiar with the book: I am having trouble with the following theorem, which states that any two bases of a vector space have the same number of elements.

The proof assumes that a set $X = \{x_1, x_2, ..., x_n\}$ has the property that, via a linear combination of the elements of $X$, we can create any element of the vector space under consideration. It also assumes there is another set $Y = \{y_1, y_2, ..., y_m\}$ of linearly independent vectors.

Since the set $Y$ is linearly independent, no linear combination of the vectors in a set $S\subset{Y}$ can be equal to any of the vectors in $Y\setminus{S}$. If we now take a new set $R$ formed by subtending the elements of $X$ to the last element of $Y$, we obtain the set $R=\{y_m, x_1, x_2, ..., x_n\}$. Since the X's form a basis of the vector space, $R$ is linearly dependent, which via the above, means that there is an $x_i$ that can be written as a linear combination of $y_m, x_1, ..., x_{i-1}$. This means that we can remove this $x_i$ from $R$ and still have a set that forms a basis for the vector space, since we can "add $x_i$" back into $R$ via a linear combination. This initial step has the effect of replacing $x_i$ with an element of $Y$. Since this new set is still a basis then, we can remove another $x_t$ from the set using the same logic, and we append the next element of $Y$ to the set, which gives the set

$\{y_{m-1}, y_m, x_1, ..., x_{i-1}, ..., x_t, x_{t+1}, ..., x_n\}$

Which is again a basis for the vector space. Continuing this way, we know that at each step, the new set is still a basis, since we can reconstruct the original base. It follows then, that we cannot remove all the X's since we would be left with the Y's; which by definition only fulfil one of the conditions of being a basis.

To conclude, the author states that this must imply that the number of X's must be greater than the number of Y's, ie $n\geq{m}$. My problem here is, however, that, by the previous statement, we cannot have $n=m$. We use a similar case where we swap the attributes of X and Y to show that $m\geq{n}$, and deduce that if each set has both conditions, $n=m$.

How is this possible, when $n=m$ contradicts the point of our proof?

Sorry for the long post.

答:
If $V$ is the vector space, the assumptions are that $X$ spans $V$, and $Y$ is linearly independent, where $|X|=n$ and $|Y|=m$. We keep replacing members of $X$ with members of $Y$ in such a way that after each replacement the resulting set still spans $V$. Suppose that $n<m$; then after $n$ replacements we will have a set $X'$ that spans $V$ and is a proper subset of $Y$. But that is impossible: $Y$ is linearly independent, so the vectors in $Y\setminus X'$ are in $V$ but not in the span of $X'$. Thus, it must be the case that $n\ge m$: no linearly independent set can be larger than any spanning set.

In particular, if $n=m$, the process ends when $X$ has been completely replaced by $Y$, and we discover that $Y$ not only is linearly independent, but also spans $V$ and is therefore a basis for $V$. You seem to have ruled out this possibility by misunderstanding the hypothesis: it is not true that $Y$ by definition fulfils only one of the conditions for being a basis. The hypothesis is simply that $Y$ is linearly independent, i.e., that it fulfils at least one of the requirements for being a basis; we are not told (and must not assume) that $Y$ does not span $V$.

Title: Paul R. Halmos bases proof


问:
Sieve for primes
In the sieve of Eratosthenes the erased composite numbers are identified by $ 5 $ equations:

$nc=2(a+1)$

$nc=3(2a+1)$

$nc=−1+6(6ab+a−b)$

$nc=1+6(6ab+a+b)$

$nc=1+6(6ab−a−b)$

with $ a\geq 1 $ and $b\geq 1$.

How a modified version of Eratosthenes sieve can be tested to find primes greater than $ 5 $, out of order, is as follows:

we write the numbers $-1\bmod6$ greater than $5$

```<code>11, 17, 23, 29, 35, 41, 47, 53, 59, 65, 71, 77, 83, 89, 95, 101, 107, 113, 119, 125, 131, 137, 143, 149, 155, 161, 167, 173, 179, 185, 191, 197, 203, 209, 215, 221, 227, 233, 239, 245, 251, 257, 263, 269, 275, 281, 287, 293, 299, 305, 311, 317, 323, 329, 335, 341, ...
</code>```

note that if we divide the multiples of $5$ by $5$ we get the numbers $1\bmod6$  (indicated in brackets)

```<code>11, 17, 23, 29, 35 (7), 41, 47, 53, 59, 65 (13), 71, 77, 83, 89, 95 (19), 101, 107, 113, 119, 125 (25), 131, 137, 143, 149, 155 (31), 161, 167, 173, 179, 185 (37), 191, 197, 203, 209, 215 (43), 221, 227, 233, 239, 245 (49), 251, 257, 263, 269, 275 (55), 281, 287, 293, 299, 305 (61), 311, 317, 323, 329, 335 (67), 341, ...
</code>```

if we delete multiples of $11, 17, 23, 29, 35, ...$

at the end delete the number $125$

and replace the muliples of $5$ with the same number divided by $5$,

only primes remain (considering also those in brackets).

Example of sieve procedure

The composite numbers eliminated with this method other than $125$ can be identified with the following Diophantine equation:

$$nc=-1+6 \cdot ( 6  \cdot a \cdot b +a-b)$$

with $ a\geq 2 $ and  $b\geq 1$

QUESTION

Is it possible to proof by contradiction that there are infinitely many prime numbers?

For example consider $p_n$ the greatest prime number any finite list of prime numbers

let's set $k_n=\frac{5p_n+1}{6}$  if  $p_n \equiv 1 \bmod 6$ or

$k_n=\frac{5p_n-1}{6}$  if  $p_n \equiv -1 \bmod 6$

if we assume that there are no prime numbers greater than $ p_n $ then it is the same as saying

$\forall  k> k_n$ $\quad \exists  a_k \geq 2 ,b_k \geq 1$ $ |$  $k =6 \cdot a_k \cdot b_k+a_k-b_k$

Is it possible to prove that this cannot happen so that it is a contradiction?

Is it possible to find in minimum value of $k$ for which $ k \neq 6 \cdot a_k \cdot b_k + a_k -b_k \quad \forall   a_k \geq 2 ,b_k \geq 1$ in order to determine an upper bound of the prime gap?

答:
The sieve implies something stronger than the existence of   infinitely many primes, namely $\sum_p 1/p=\infty$ as discovered  by Euler. Indeed the asymptotic density of the integers not divisible by the primes $p_1,p_2,\ldots,p_k$ is $\prod_{i=1}^k (1-p_i^{-1})$. These are exactly the integers not erased in the first $k$ stages of the sieve.
Since every integer greater than 1 is divisble by some prime, the product over all primes satisfies
$$\prod_p (1-p^{-1})=0 \,. \quad (*)$$
To see this analytically, expand each factor in the product as a geometric series
$1+1/p+1/p^2+\ldots$ and observe that the product $(*)$ yields the harmonic series $\sum_{n \ge 1} 1/n$ since each $n$ has a unique factorization as a product of prime powers. Finally $(*)$ implies $\sum_p 1/p =\infty$ because of the classical equivalence between infinite products and the sums of the differences of individual factors from 1.  https://en.wikipedia.org/wiki/Infinite_product

Title: Sieve for primes


问:
Can transitive graphs have non-integer growth dimension?
Say we have some transitive (edge and vertex) graph $G$ s.t $|B(v,n)| \leq c n^\alpha$ for some $\alpha\geq 0$.

Polynomial growth/dimension in this setting meaning that for some $d\geq 0$ we can pick two constants $a,b$ s.t $an^d\leq |B(v,n)| \leq bn^d$ for every integer $n$. Is it always the case that $d$ will be an integer when $G$ is transitive?

答:
I found the answer to your question in N.  Seifter, Properties of Graphs with Polynomial Growth (https://www.sciencedirect.com/science/article/pii/009589569190064Q)

[speaking about Gromov result on groups of polynomial growth] here always exist constants $c_l$, $c_2$ such that $c_1n^d \leq f_G(n) \leq  c_2n^d$ holds for some integer $d$. We call this well-defined integer the growth degree $d_G$ of $G$. A deep result of Trofimov (see [19, Theorem 2]) implies that the same also holds for the growth functions of graphs with polynomial growth.

sadly I could not get access to the english version of V.I. TROFIMOV, Graphs with polynomial growth, so I cannot elaborate, but this is apparently an highly non-trivial fact.


问:
$\sum_{n=1}^{\infty} \frac{n^2 +1}{2n^2 \sqrt{n} +5}$
How can I show this series diverges $\sum_{n=1}^{\infty} \frac{n^2 +1}{2n^2 \sqrt{n} +5}$. I tried doing a root test, and end up with the limit $$\lim_{n\to\infty} \frac{(n^2+1)^{1/n}}{(2n^2\sqrt{n}+5)^{1/n}}$$

which ends up as $\frac{1}{\sqrt{2}}$ after my calculations, which indicates convergence.

答:
Similar but alternative approach to Reveillark:

How can I show this series diverges $\sum_{n=1}^{\infty} \frac{n^2 +1}{2n^2 \sqrt{n} +5}$.

Since the goal is to show divergence:

Without Loss of Generality (WLOG) the numerator can be decreased to $(n^2)$.

Then, WLOG, the expression can be changed to
$\frac{1}{2\sqrt{n} + (5/[n^2])}.$

Then, WLOG, the denominator can be increased to $3\sqrt{n}$, with the understanding that in all but a finite number of terms, this change will be causing an increase in the denominator.

Then, WLOG, the factor of $(3)$ can be removed from the denominator, since if a series given by $\sum a_n$ diverges, so does the series $(1/3) \times \sum a_n.$

Finally, the rule can be invoked that the series $\sum_{k=1}^\infty k^s$ is divergent whenever $0 < s < 1.$


答: Since the purpose of the proof is to show that X and Y contain the same number of vectors, how does "m= n contradict the point of the proof"?</p>

Title: Paul R. Halmos bases proof

Body:
Hi and thanks for reading. For those of you who know the book I am currently going through, Paul R. Halmos'Finite dimensional vector spaces: In the proof that all bases have the same number of elements, why has the author written that $n\geq{m}$, since, if $n=m$, then we would have replaced all of the X's with elements of Y, which is a contradiction to the original assumption, since at each stage of removing an $x_i$ and adding a $y_k$, the set of X's and Y's form a basis for V. But not if we remove all the X's.

For those not familiar with the book: I am having trouble with the following theorem, which states that any two bases of a vector space have the same number of elements.

The proof assumes that a set $X = \{x_1, x_2, ..., x_n\}$ has the property that, via a linear combination of the elements of $X$, we can create any element of the vector space under consideration. It also assumes there is another set $Y = \{y_1, y_2, ..., y_m\}$ of linearly independent vectors.

Since the set $Y$ is linearly independent, no linear combination of the vectors in a set $S\subset{Y}$ can be equal to any of the vectors in $Y\setminus{S}$. If we now take a new set $R$ formed by subtending the elements of $X$ to the last element of $Y$, we obtain the set $R=\{y_m, x_1, x_2, ..., x_n\}$. Since the X's form a basis of the vector space, $R$ is linearly dependent, which via the above, means that there is an $x_i$ that can be written as a linear combination of $y_m, x_1, ..., x_{i-1}$. This means that we can remove this $x_i$ from $R$ and still have a set that forms a basis for the vector space, since we can "add $x_i$" back into $R$ via a linear combination. This initial step has the effect of replacing $x_i$ with an element of $Y$. Since this new set is still a basis then, we can remove another $x_t$ from the set using the same logic, and we append the next element of $Y$ to the set, which gives the set

$\{y_{m-1}, y_m, x_1, ..., x_{i-1}, ..., x_t, x_{t+1}, ..., x_n\}$

Which is again a basis for the vector space. Continuing this way, we know that at each step, the new set is still a basis, since we can reconstruct the original base. It follows then, that we cannot remove all the X's since we would be left with the Y's; which by definition only fulfil one of the conditions of being a basis.

To conclude, the author states that this must imply that the number of X's must be greater than the number of Y's, ie $n\geq{m}$. My problem here is, however, that, by the previous statement, we cannot have $n=m$. We use a similar case where we swap the attributes of X and Y to show that $m\geq{n}$, and deduce that if each set has both conditions, $n=m$.

How is this possible, when $n=m$ contradicts the point of our proof?

Sorry for the long post.


问:
What does it mean: “the closure of the axioms”?
«As we shall see, the logical axioms are so designed that the logical consequences (in the semantic sense, cf. p. 56) of the closure of the axioms of $K$ are precisely the theorems of $K$.» Page 60 “Introduction to Mathematical Logic“ SECOND EDITION by ELLIOTT MENDELSON The same is in fourth edition.

What does it mean: “the closure of the axioms”?

答:
Maybe: the closure of a formula means universally quantifying over all its free variables?  And perhaps this definition is somewhere before page 60?

Edit.  Look in the index under: closure, (universal) of a formula

ContentLicense: CC BY-SA 3.0


问:
The convergence of the series after integrating term-by-term
I am interested in the integral

$$ I = \int_{-\infty}^\infty e^{i A x} e^{- Bx^2 - \lambda x^4}, $$

for some real values $A$, $B>0$ and $\lambda >0$.

One way to tackle this integral is to expand in $\lambda$
and integrate term by term

$$ I = \int_{-\infty}^\infty e^{i A x} e^{- Bx^2} - \lambda \int_{-\infty}^\infty x^4 e^{i A x} e^{- Bx^2} + \ldots ,$$

which I can do analytically now and get simply

$$ I = G \left(1 - \frac{\lambda}{16} \frac{1}{B^4} (A^4 - 12 A^2 B + 12 B^2)+\ldots \right) ,$$

where $G$ is the Gaussian integrals $G = \sqrt{\pi/B} \exp(-A^2/4B)$.

However, I am not sure under which assumptions am I allowed to do all this.
Is this expansion formally convergent for any values of $\lambda >0$? And if so (or not) how can I see it or prove it?

答:
since $A\in\mathbb{R}$ you can say that:
$$\left|e^{iAx}\right|=1\,\,\,,\forall|x|<\infty$$
and so to see if your integral converges we just need the following integral to converge:
$$\int_{-\infty}^\infty\exp\left(-Bx^2-\lambda x^4\right)dx$$
which since $B,\lambda>0$ clearly converges.

In terms of whether you are allowed to expand to a series like that, you are since:
$$\exp(z)=\sum_{n=0}^\infty\frac{z^n}{n!}\,\,,\forall z\in\mathbb{C}$$
and so the series converges to the function for all values you are interested in.

In terms of the series and working out values, it may be nicer to make a substitution and get this as a series of Gamma functions?


问:
Prove that $C^{0}$ and $\mathbb{R}$ have equal cardinality
How to Prove that $C^{0}$ and $\mathbb{R}$ have equal cardinality ?

$C^{0}$ denote the set of all Continuous function $\mathbb{R} \rightarrow \mathbb{R}$

答:
Hint: To show that $\mathbb{R} \subset C^0$, consider $f \in C^0$, take $f(0)$.

For the converse,

Hint: $f\in C^0$ is uniquely determined by its values on a dense subset of points.

Hint: $|\mathbb{R^N}| = |\mathbb{R}|$.


问:
Coefficients of $p\circ p$ for polynomial $p(x)=\sum_{k=0}^d a_kx^k$
Let $p(x)=\sum_{k=0}^d a_kx^k$ be a polynomial of degree $d$. Then
\begin{align*}
(p\circ p)(x) &= \sum_{k=0}^da_k\left(\sum_{j=0}^da_j x^j\right)^k\\[3pt]
&=\sum_{k=0}^d \sum_{\substack{k_0+\cdots+k_d=k\\[3pt]k_0,\dots,k_d\geqslant0}}a_k\frac{k!}{k_0!k_1!\cdots k_d!} x^{k_1+2k_2+3k_3+\cdots+d k_d}\prod_{t=0}^da_t^{k_t}
\end{align*}
by the multinomial theorem.

Is there a nicer way I can express the coefficients of $p\circ p$? Where can I read more about this polynomial and study its properties?

答:
TL;DR: There are several ways to organize the computation of these coefficients.  None of them give a fully closed form expression (as expected), so which one is best depends on the application.  The most efficient for implementation on a computer is the Fourier method explained below.

Let me provide multiple approaches to this problem.

Approach 1: Direct assault.

Let's see what happens if we just try to evaluate the coefficient of $x^m$  in $\sum_k a_k \left(\sum_j a_j x^j\right)^k$ directly, starting with small values of $m$.

For $m=0$, only the $j=0$ term of the inner sum contributes, so the coefficient of $x^m$ is $\sum_k a_k (a_0)^k = p(a_0) = p(p(0))$.

For $m=1$, in each term of the outer sum we are looking to have one power of $x$.  When expanding out $\left(\sum_j a_j x^j\right)^k$, there are $k$ ways to pick a term $a_1 x^1$, and the other $k-1$ factors in the power must be $a_0 x^0$.  Hence the coefficient of $x^1$ is $\sum_k a_k \left(k a_1 a_0^{k-1}\right)$.  This can be simplified by noting that $\sum_k k a_k y^{k-1}$ is precisely $p'(y)$, so that the coefficient of $x^1$ becomes $a_1 p'(a_0)$.  Since $a_1=p'(0)$, this can in turn be written as $p'(p(0)) p'(0)$.

Do you see the pattern yet?  All we are doing is computing Taylor series coefficients.  In general, the coefficient of $x^m$ will be $$\left.\frac{1}{m!} \frac{d^m p(p(x))}{dx^m}\right|_{x=0}$$
So e.g. for $m=2$ we will have $\frac{1}{2}\left(p''(p(0)) p'(0)^2 + p'(p(0)) p''(0)\right)$.

Approach 2: Fourier methods.

(Note: This is a recapitulation of this post, also linked in comments above.)

When you multiply two polynomials, their coefficients get "convolved".  So if $c_p(j)$ denotes the $j^{th}$ coefficient of polynomial $p$, then the coefficient of a product $pq$ is given by $c_{pq}(j) = \sum_k c_p(k) c_q(j-k) = (c_p * c_q)(j)$, where $*$ denotes convolution.  If you take a power of a polynomial, like $p^k$, then its coefficients get convolved with themselves $k$ times, which is denoted $c_p^{*k}$.  So with this notation we can write $c_{p\circ p}(j) = \sum_k c_p(k) c_p^{*k}(j)$.

The most efficient way to handle convolutions is via the Fourier transform.  If you are unfamiliar with Fourier analysis, this may seem mysterious, but if you actually want to compute coefficients of an iterated polynomial e.g. on a computer, this is almost certainly the best way.  The main result is (as shown in the linked post above) that $$c_{p\circ p} (j) = \mathcal{F}^{-1} \left[ p\circ p\left(e^{-2\pi i k/N}\right)\right](j)$$ where $\mathcal{F}^{-1}$ is an inverse discrete Fourier transform of length $N$.  This gives an efficient, and easy to implement, way to compute coefficients of iterated polynomials.

A pseudocode recipe for implementing this would be something like the following:

```<code>m = ... # set this to the degree of polynomial p
N = m^2 + 1 # chosen to be total number of coefficients of p(p(x))
y = [p(p(exp(-2*pi*i*k/N))) for k=0:(N-1)]
coefficients = InverseDFT(y) # these are the coefficients of p(p(x))
</code>```

(Note that this assumes a conventional normalization of your DFT, so you may need to e.g. divide by $\sqrt{N}$ if your software uses a different normalization.)

Approach 3: Cauchy's formula.

Writing out the above Fourier method more explicitly yields a formula that might be easier to think about.  Let $\omega$ be a primitive $N^{th}$ root of unity, with $N \ge \deg(p\circ p)$.  Then the coefficient $c_{p\circ p}(j)$ can be written
$$c_{p\circ p}(j) = \frac{1}{N} \sum_{k=0}^{N-1} \omega^{-jk} p(p(\omega^k))$$
This is essentially Cauchy's integral formula specialized to the case at hand.  It follows directly from the Fourier formula above (cf. this related MO post).

Body:
TL;DR: There are several ways to organize the computation of these coefficients.  None of them give a fully closed form expression (as expected), so which one is best depends on the application.  The most efficient for implementation on a computer is the Fourier method explained below.

Let me provide multiple approaches to this problem.

Approach 1: Direct assault.

Let's see what happens if we just try to evaluate the coefficient of $x^m$  in $\sum_k a_k \left(\sum_j a_j x^j\right)^k$ directly, starting with small values of $m$.

For $m=0$, only the $j=0$ term of the inner sum contributes, so the coefficient of $x^m$ is $\sum_k a_k (a_0)^k = p(a_0) = p(p(0))$.

For $m=1$, in each term of the outer sum we are looking to have one power of $x$.  When expanding out $\left(\sum_j a_j x^j\right)^k$, there are $k$ ways to pick a term $a_1 x^1$, and the other $k-1$ factors in the power must be $a_0 x^0$.  Hence the coefficient of $x^1$ is $\sum_k a_k \left(k a_1 a_0^{k-1}\right)$.  This can be simplified by noting that $\sum_k k a_k y^{k-1}$ is precisely $p'(y)$, so that the coefficient of $x^1$ becomes $a_1 p'(a_0)$.  Since $a_1=p'(0)$, this can in turn be written as $p'(p(0)) p'(0)$.

Do you see the pattern yet?  All we are doing is computing Taylor series coefficients.  In general, the coefficient of $x^m$ will be $$\left.\frac{1}{m!} \frac{d^m p(p(x))}{dx^m}\right|_{x=0}$$
So e.g. for $m=2$ we will have $\frac{1}{2}\left(p''(p(0)) p'(0)^2 + p'(p(0)) p''(0)\right)$.

Approach 2: Fourier methods.

(Note: This is a recapitulation of this post, also linked in comments above.)

When you multiply two polynomials, their coefficients get "convolved".  So if $c_p(j)$ denotes the $j^{th}$ coefficient of polynomial $p$, then the coefficient of a product $pq$ is given by $c_{pq}(j) = \sum_k c_p(k) c_q(j-k) = (c_p * c_q)(j)$, where $*$ denotes convolution.  If you take a power of a polynomial, like $p^k$, then its coefficients get convolved with themselves $k$ times, which is denoted $c_p^{*k}$.  So with this notation we can write $c_{p\circ p}(j) = \sum_k c_p(k) c_p^{*k}(j)$.

The most efficient way to handle convolutions is via the Fourier transform.  If you are unfamiliar with Fourier analysis, this may seem mysterious, but if you actually want to compute coefficients of an iterated polynomial e.g. on a computer, this is almost certainly the best way.  The main result is (as shown in the linked post above) that $$c_{p\circ p} (j) = \mathcal{F}^{-1} \left[ p\circ p\left(e^{-2\pi i k/N}\right)\right](j)$$ where $\mathcal{F}^{-1}$ is an inverse discrete Fourier transform of length $N$.  This gives an efficient, and easy to implement, way to compute coefficients of iterated polynomials.

A pseudocode recipe for implementing this would be something like the following:

m = ... # set this to the degree of polynomial p
N = m^2 + 1 # chosen to be total number of coefficients of p(p(x))
y = [p(p(exp(-2*pi*i*k/N))) for k=0:(N-1)]
coefficients = InverseDFT(y) # these are the coefficients of p(p(x))

(Note that this assumes a conventional normalization of your DFT, so you may need to e.g. divide by $\sqrt{N}$ if your software uses a different normalization.)

Approach 3: Cauchy's formula.

Writing out the above Fourier method more explicitly yields a formula that might be easier to think about.  Let $\omega$ be a primitive $N^{th}$ root of unity, with $N \ge \deg(p\circ p)$.  Then the coefficient $c_{p\circ p}(j)$ can be written
$$c_{p\circ p}(j) = \frac{1}{N} \sum_{k=0}^{N-1} \omega^{-jk} p(p(\omega^k))$$
This is essentially Cauchy's integral formula specialized to the case at hand.  It follows directly from the Fourier formula above (cf. this related MO post).


答: Hint: A continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ is completely determind by its values on rational numbers $\mathbb{Q}$ (actually in any dense set of $\mathbb{R}$). the cardinality of the  range of a function from $\mathbb{Q}$ is at most countable.</p>

Body:
How to Prove that $C^{0}$ and $\mathbb{R}$ have equal cardinality ?

$C^{0}$ denote the set of all Continuous function $\mathbb{R} \rightarrow \mathbb{R}$


问:
proving supremum of sequence is infinity
For $j \in \mathbb{N}$ We have $$\|a_n\| \geqq
 \ln\frac{j^2+1}{j+1}$$
gives $\displaystyle
\sup_n\|a_n\|=\infty.$ Hence
$(a_n)_{n=1}^\infty$ does not converge.

Q1 How to prove $\displaystyle
\sup_n\|a_n\|=\infty?$

Q2 Why does $(a_n)_{n=1}^\infty$ not converge?

答:
Question 1 has already been answered in the comments. If the norm of $||a_n||$ is bigger than a number that converges to $\infty$ at $n \to \infty$ than the supremum has to be $\infty$ as well because any finite number $a$ being the supremum would imply that $||a_n|| \leq a$ for all $n$ which is not possible because there is an $N_0$ s.t. for all $n \geq N_0$: $\ln(\frac{n^2+1}{n+1}) > a$ because the logarithm is strictly increasing and not bounded and the input $\frac{n^2+1}{n+1}$ converges to $\infty$.

As for question 2 there is a theorem that states that a convergent sequence is bounded (which is actually not hard to proof). Now from Question 1 we know that $a_n$ is not bounded because $||a_n|| \leq c$ for all $n \in \mathbb N$ is not possible, so $a_n$ is not bounded.


问:
How can I find the domain of $ \int_{0}^{x} \frac{\log(1-t^{2})+t^{2}}{t}dt$?
Let $$ h(x) =  \int_{0}^{x} \frac{\log(1-t^{2})+t^{2}}{t}dt \ .$$
How can I find the domain of this function? I tried to integrate it, but it seems impossible; I also thought that it could be the domain of g(x): $$g(x) = \frac{\log(1-t^{2})+t^{2}}{t}dt$$ but on a second thought I realized that if some general function $f(x)$ has domain $D_{f}$ that doesn't implie that $f'(x)$ has this same domain (for example, if $f(x) = |x|$).

What do you suggest?
Any help is welcome.

答:
Observe that if $ h(x) $ exists, then $ h(-x) $ will also exist.

So, we can assume that $ x>0$.

Near $ 0^+$, we have

$$\ln(1-t^2)+t^2\sim -\frac{t^4}{2}$$
So,
$$\lim_{t\to 0}\frac{\ln(1-t^2)+t^2}{t}=0$$
and
$$\int_0 f(t)dt \; converges$$

where $ f $ is the integrand function, which is defined at $ (-1,0)\cup (0,1)$.

but
$$\int^1f(t)dt \text{ is convergent}$$
because $ \int^1\ln(1-t)dt $ is known to be convergent.

The domain is then $$D=[-1,1]$$


问:
Compunding Trading
I have $100$ dollars at start.

I invest $50 \%$ of what I have after every trade I make($50$ dollars in 1st trade and then compounding).

Every successful trade gives me $85 \%$ profit of the money invested.

Every loser trade deducts the entire money invested($100 \%$ loss of the money invested in each trade).

Total Win-rate(profitable trades) is $70 \%$.

After how many trades will I $100 \times$ my initial investment($ \$ 100 $) or is it mathematically not possible?

答:
We can think about this in terms of expected values, but there's no guarantee of where you'll end up when - that's what computer simulations are for.

Let's start with an initial wealth of $w_0$. Then, we invest half of it, and if the trade is profitable (with probability $0.7$), we end up with an $85 \%$ profit on the investment, i.e., the investment grows from $0.5w_0$ to $0.5w_0*1.85$. So, we end up with $0.5w_0 + 0.5w_0*1.85 = 1.425w_0$.

If the trade is not profitable (probability $0.3$), the investment goes from $0.5w_0$ to $0$, and we end up with $0.5w_0$. So, our wealth after that first trade will be:

$$ w_1 =
\begin{cases}
    1.425w_0 ,  & p = 0.7 \\
    0.5w_0 , & p = 0.3
\end{cases} . $$

The expected value of this trade is

$$ E[w_1] = 0.7*1.425w_0+0.3*0.5w_0 $$
$$ = 1.1475w_0 .$$

So, if you do this trade an infinite number of times, you can expect the long-run average growth rate in your wealth to be $14.75 \%$ per trade.

If you conduct this trade $n$ times, that gives you an expected wealth of $w_n = 1.1475^n w_0$, so you end up getting to $100$ times your initial wealth when:

$$ 1.1475^n w_0 = 100w_0 $$
$$ \Rightarrow 1.1475^n = 100 $$
$$ \Rightarrow n = \log_{1.1475} {(100)} $$
$$ \approx 33.47 . $$

So, on average, if you started with an infinite number of ensembles of wealth $w_0$ and ran this strategy, your 34th trade is the one that breaks 100 times initial wealth.

In reality, if you do this once, you could lose 34 times in a row, and end up at $0.5^{34}*w_0 \approx 5.8*10^{-11}$ of your initial wealth on trade 34. Or win 34 times in a row and end up at $ 1.425^{34}\approx 169709 $ times your initial wealth on trade 34. Or just about anywhere in between.

Title: Compunding Trading


问:
How to seat $2n$ people, where each person hates at most $n-1$ people, such that everyone sits besides his/her friends in a circle?
This question is a homework question. My professor says to make use of the extremal principle. I attempted to solve. I have seen a similar question on this website, but it's solution makes use of graph theory, and I know very little of graph theory.

Consider the person who hates the most amount of people, call him Bob. Bob hates at most $n-1$ people and thus has at least $n$ friends.

Suppose that that all of Bob's friends are friends of each other. But then the people outside of the friend group are hated by all the people in the 'in' group. If you are a member of the out group, you are hated by at least $n$ people; thus, we have arrived at a contradiction. So you can't have friends of a person liking each of Bob's friend.

I don't know how to continue this line of thinking. I feel it has to do with the selection of the friend who doesn't like the other friends and working from there. Other ideas I have considered is a friend group who have the most mutual hate, a group that has the most mutual friends, pairs of people, a person who has the most amount of friends (but I don't think that's useful).

Another idea is to consider the arrangement with minimal pairs of enemies who sit together. Suppose there is a pair of enemies who sit together. I'm working on this reasoning.

答:
Given a circular arrangement of the $2n$ people, there is some number neighbour pairs made of enemies.
Among all arrangements, consider one that minimizes this number.

Assume that in this arrangement, there is at least one neighbour pair of enemies, i.e., the circular arrangement looks
$$\tag1x_1,x_2,\stackrel\longrightarrow\ldots, x_{2n},(x_1) $$
where  $x_1,x_2$ are enemies.
Assume $x_1,x_k$ are friends where $3\le k\le2n-1$. Then we can flip part of  $(1)$ to obtain
$$\tag2 x_1,x_k,\stackrel\leftarrow\ldots,x_2,x_{k+1},\stackrel\longrightarrow\ldots, x_{2n},(x_1).$$
By minimality of $(1)$, we conclude that $x_2,x_{k+1}$ must be enemies.
Thus, the number of $x_2$'s enemies among $x_4,\ldots,x_{2n}$ is at least as big as the number of $x_1$'s friends among $x_3,\ldots,x_{2n-1}$. This latter number is at least $n-1$ so that $x_2$ has (including $x_1$) at least $n$ enemies, contradiction. Thus our minimal arrangement has no neighbour pair of enemies.


问:
The Variance Rule of $V(cX) = c^2V(X)$
As a statistics major, I learned a while back that the $V(cX) = c^2 V(X)$, where $V(X)$ is the variance of $X$. However, I never quite understood why that was the case. Consider this: We have an event $X$ such that each event is independent of any previous events. Take for instance, a roll of copper wire. We know that the mean length is $150 \text{m}$ and the standard deviation is $4\text{m}$. If we want to find the variance of $5$ rolls of copper wire, we could write this as $V(5X) = 25V(X) = 400\text{m}^2$. However, an alternative way to write the problem is doing $V(X + X + X + X + X) = V(X) + \ ... \ + V(X)$ - as they are all independent. Thus, we can simplify to $5V(X)$ and get $80\text{m}^2$. This second way seems like it should be the same as the first way, but they produce different answers. Can anyone explain this? I was helping a friend out with a similar problem and the second way actually produces the correct answer, so can anyone explain why this is the case?

答:
While the other answer do the computation very well, it is helpful to have the following intuition:

$5X$ only depends on one random thing, while $S_5:=X_1+X_2+X_3+X_4+X_5$ depends on $5$ (different) independent things. Imagine these $X_i$ are coin flips ($1$ for heads and $0$ for tails). If the first coin comes up heads ($X_1=1$), it is still possible for $S_5$ to be "small" (e.g., $=1$), if the others come up tails ($X_2=\cdots=X_5=0$). However, in the $5X$ case, there is no possibility of such "cancellation": $5X$ is guaranteed to be $5$ as soon as you know that $X=1$.

So it is intuitively obvious that $5X$ should have greater variance than $S_5$.

Title: The Variance Rule of $V(cX) = c^2V(X)$


问:
Is $\det(I+M\bar{M})\geq 0$ for any complex square matrix $M$?
Let $M$ be a complex $n\times n$ matrix and $I$ the $n\times n$ identity matrix. The overline $\bar{M}$ denotes the complex conjugate. The determinant of $I+M\bar{M}$ is obviously real, because $\det(I+M\bar{M})=\det(I+\bar{M}M)=\overline{\det(I+M\bar{M})}$.

I have no doubt that
$$\det(I+M\bar{M})\geq 0,$$
but I do not have a proof. Can someone help me?

Motivation: I need this to complete an answer at MO.

Body:
the 2 by 2 case with all letters real. We think the determinant is positive, attempt to write it as a sum of squares, I guess.

parisize = 4000000, primelimit = 500000
? m = [ a+b*I, c+d*I; e+f*I, g+h*I]
%1 =
[a + I*b c + I*d]

[e + I*f g + I*h]

? n = [ a-b*I, c-d*I; e-f*I, g-h*I]
%2 =
[a - I*b c - I*d]

[e - I*f g - I*h]

? matt = matid(2) + m*n
%5 =
[a^2 + (b^2 + ((e - I*f)*c + ((I*e + f)*d + 1))) (c - I*d)*a + ((I*c + d)*b + ((g - I*h)*c + (I*g + h)*d))]

[(e + I*f)*a + ((-I*e + f)*b + ((g + I*h)*e + (-I*g + h)*f)) (e + I*f)*c + ((-I*e + f)*d + (g^2 + (h^2 + 1)))]

? dett = matdet(matt)
%6 = (g^2 + (h^2 + 1))*a^2 +
((-2*g*e - 2*h*f)*c + (-2*h*e + 2*g*f)*d)*a +
 ((g^2 + (h^2 + 1))*b^2 +
 ((2*h*e - 2*g*f)*c + (-2*g*e - 2*h*f)*d)*b +
 ((e^2 + f^2)*c^2 + 2*e*c + ((e^2 + f^2)*d^2 + 2*f*d + (g^2 + (h^2 + 1)))))
?


问:
How do I find a number $\frac{a^4}{4}$ to be added to an odd integer $N$ to make it a Perfect Square?
Finding the least number to be added to an integer $N$ to make it a Perfect Square is simple:
https://www.geeksforgeeks.org/least-number-to-be-added-to-or-subtracted-from-n-to-make-it-a-perfect-square/

But how do we proceed if we require the number to be added must be a fourth power divided by four $\frac{a^4}{4}$?

Assume $N=p_1\cdot p_2$ is a product of two primes.
How do I find a number $a$ such that $\frac{a^4}{4}+N$ is a perfect square?

An example $N=p\cdot q=7\cdot3=21$ and we find $a=2$ which produces a perfect square $21+\frac{2^4}{4}=25$ too.

Another example $N=p\cdot q=7\cdot11=77$ and we find $a=2$ which produces a Perfect Square $77+\frac{2^4}{4}=81$.

An example including larger numbers is $N=p\cdot q=5\cdot13=65$ and we find $a=8$ which produces a perfect square $65+\frac{8^4}{4}=1089$.

For a given $N$, is there a way to find such an integer $a$ (without brute force)?

答:
Given $n=pq$ with $p$ and $q$ prime and $p\leq q$, if $a$ is an integer such that
$$n+\frac{a^4}{4}=c^2,$$
for some positive integer $c$, then $\tfrac{a^4}{4}$ is an integer and so $a$ is even, say $a=2b$. Then also
$$pq=n=c^2-\tfrac{a^4}{4}=c^2-4b^4=(c-2b^2)(c+2b^2).$$
Then either $c-2b^2=1$ and $c+2b^2=pq$, or $c-2b^2=p$ and $c+2b^2=q$, and correspondingly
$$(2b)^2=pq-1\qquad\text{ or }\qquad (2b)^2=q-p,$$
where of course $q-p$ is the smaller of the two. So

If $q-p$ is a perfect square, then $a=\tfrac12\sqrt{q-p}$.

Else, if $pq-1$ is a perfect quare, then $a=\tfrac12\sqrt{pq-1}$.

Else there is no such $a$.


问:
How to apply strong pigeonhole principle to this question?
For each $r\geq 1$, let $L(r,n)$ denote the integer for which given any sequence of length $L(r,n)$, and any $r$-coloring of the elements of the sequence, there exists a monochromatic increasing sequence of length $n+1$ or a monochromatic decreasing sequence of length $n+1$,  but there exists a sequence of length $L(r,n)-1$ and an $r$-coloring of that sequence which has no monochromatic increasing or decreasing sequence of length $n+1$. So, $L(1,n)=n^2+1$ for all $n$. Find $L(r,n)$ for all $r,n\geq 1$.

答: HINT: Given an $r$-coloring of $\langle a_1,\ldots,a_m\rangle$, you can consider each of the $r$ monochromatic subsequences independently. You need at least one of them to be long enough to ensure that it has a monotone subsequence of length $n+1$, and you know what $L(1,n)$ is.


问:
Simplifying to least non-negative residue when modulo is not prime
So I am attempting to simplify $3^{252} \bmod 610$ to its least non negative residue. However $610$ is not a prime modulus and I cannot split it into a product of two prime numbers so I'm quite lost on where to begin. A hint in the right direction would suffice.

答:
By Euler's theorem, $3^{240}\equiv 1\bmod 610$, so $3^{252}\equiv 3^{12}\bmod 610.$

$3\equiv1\bmod2$ and $3^4\equiv1\bmod5$, so $3^{12}\equiv1\bmod 10$.

$3^5=243\equiv-1\bmod61$, so $3^{10}\equiv1\bmod61$, so $3^{12}\equiv9\bmod61$.

So $3^{252}\equiv1\bmod10$ and $3^{252}\equiv9\bmod61$.

Can you use the Chinese remainder theorem to find $3^{252}\bmod 610$?


问:
Why we consider log likelihood instead of Likelihood in Gaussian Distribution
I am reading Gaussian Distribution from a machine learning book. It states that -

We shall determine values for the unknown parameters $\mu$ and
  $\sigma^2$ in the Gaussian by maximizing the likelihood function. In practice, it is more convenient to maximize the log of the likelihood
  function. Because the logarithm is monotonically increasing function
  of its argument, maximization of the log of a function is equivalent
  to maximization of the function itself. Taking the log not only
  simplifies the subsequent mathematical analysis, but it also helps
  numerically because the product of a large number of small
  probabilities can easily underflow the numerical precision of the
  computer, and this is resolved by computing instead the sum of the log
  probabilities.

can anyone give me some intuition behind it with some example? Where the log likelihood is more convenient over likelihood. Please give me a practical example.

Thanks in advance!

答:
Maybe a practical example for computer scientists: As @jojek mentioned

$$p(X|\Theta) = \prod_{i=1}^N p(x_i|\Theta)$$

and

$$\ln p(X| \Theta) = \sum_{i=1}^N \ln p(x_i| \Theta)$$

So if you execute a little python script you can directly see it:

```<code>>>> import numpy as np

# Create some small probability values.
>>> r = np.random.random_sample((100,)) * 0.000001

# Take the log of each (I ignore the logarithms basis, because it does not matter here).
>>> l = [np.log(x) for x in r]
# Compute product of probability values.
>>> np.product(r)
0.0

# Compute sum of log of probability values.
>>> np.sum(l)
-1472.245511811776

</code>```

Here you can see that it is more practical to work with logarithms. Especially in machine learning where you compute with many very small numbers (e.g. check the weights of a neural network) this can lead to much better results.

Body:
Maybe a practical example for computer scientists: As @jojek mentioned

$$p(X|\Theta) = \prod_{i=1}^N p(x_i|\Theta)$$

and

$$\ln p(X| \Theta) = \sum_{i=1}^N \ln p(x_i| \Theta)$$

So if you execute a little python script you can directly see it:

>>> import numpy as np

# Create some small probability values.
>>> r = np.random.random_sample((100,)) * 0.000001

# Take the log of each (I ignore the logarithms basis, because it does not matter here).
>>> l = [np.log(x) for x in r]
# Compute product of probability values.
>>> np.product(r)
0.0

# Compute sum of log of probability values.
>>> np.sum(l)
-1472.245511811776

Here you can see that it is more practical to work with logarithms. Especially in machine learning where you compute with many very small numbers (e.g. check the weights of a neural network) this can lead to much better results.


问:
Represent this ordered pair using a function

Suppose we have two boxes with Box 1 containing $p$ number of
chocolates and Box 2 containing $q$ number of chocolates, $p$ and $q$
being positive integers.

A game consists of taking one chocolate from Box 1 at every odd turn
and leaving Box 2 intact in that turn, and taking 1 chocolate from Box
2 at every even turn and leaving Box 1 intact in that turn.

Represent this setup using an ordered pair of suitable functions.

I tried taking the ordered pair as $(f(n), g(n))$, where it goes like,

$$ (f(n), g(n)) \xrightarrow{\text{odd turn}} (f(n)-1, g(n)) \xrightarrow{\text{even turn}} (f(n)-1, g(n)-1) \xrightarrow{\text{odd turn}} \ldots$$

And then tried setting $f(n)$ as some combination of $(-1)^n$, but it didn't work. I don't know what else to do now.

Any help will be appreciated.

答: $(-1)^n$ often a good idea when working with functions that act differently at odd and even inputs, but floor and ceil are the better choice in this case. $(p-\lceil n/2 \rceil, q-\lfloor n/2 \rfloor)$ should be what you want


问:
Puiseux series over a non algebraically closed field.
Suppose that $K$ is a field of characteristic $0$ and $L/K$ is an extension so that every degree $\leq n$ polynomial over $K$ splits in $L$. We do not assume that $L$ is algebraically closed.

It is a classical fact that the algebraic closure of $K[[t]]$ is $K[[t^{1/n} : n \geq 1]]$ when $K$ is algebraically closed. In our set up, is it true that every degree $n$ polynomial over $K[[t]]$ splits completely over $L[[t^{1/n!}]]$?

I think the classical proof of Newton-Puiseux expansions can be modified to prove this but I would be very happy if there were a reference.

答:
Thanks to this question, I found a proof that can be pretty easily modified to show what I want.

We will use a form of Hensel's lemma which says that if we have a polynomial $f$ over $K[[t]]$ that splits into $f = gh$ over $t = 0$ with $\gcd(g,h) = 1$, then we can find a lifting of the splitting to $K[[t]]$.

Now suppose we have a polynomial $f(x)$ of degree $d \leq n$ over $K[[t]]$. We show that it has a non trivial factorization over $L[[t]]$. First, we make a linear change of variables so that:
$$f(x) = x^d + a_2(t)x^{d-1} + \dots + a_0$$
and then define:
$$g(x) = t^{-d\delta}f(xt^{\delta}) = x^d + b_2(t)x^{d-1} + \dots + b_0$$
where we choose $\delta = \mathrm{val}_t(a_i(t)/i)$ so that the $b_i(t)$ have a positive $t$-valuation and are not all zero modulo $t^{1/n!}$. (This step is why we need to adjoin $t^{1/n!}$).

Then, we see that $g(x) \pmod{t^{1/n!}}$ is of the form $x^d + c_2x^{d-2} + \dots + c_0$
over $L$ with the $c_i \in K$ and not all of them $0$. Since the $x^{d-2}$ term is $0$ and one of the other terms is not, we see that $g(x) \pmod{t^{1/n!}}$ is not of the form $(x-\gamma)^d$ so it factors into two non trivial, co prime factors over $L$. (This is where we use the fact that every degree $\leq n$ polynomial over $K$ splits over $L$).

By Hensel's lemma, we can now lift our factorization to $L[[t]]$.


问:
Why is the map $T(x,y)=(x,xy)$ not ergodic on $S^1\times S^1$.
Here is my reasoning. I use multiplicative notation.

Let $f \in L^2$ then it has a Fourier series $$f(x,y)=\sum a_{n,m}x^ny^m$$ because $x^ny^m$ are characters and so they form an orthonormal basis. Now $$f(T(x,y))=\sum a_{n,m}x^{n+m}y^m.$$ If $f$ is $T$ invariant then it must be that $a_{n,m}=a_{n+m,m}=a_{n+2m,m}=...$ Thus unless $a_{n,m}$ are zero this the coefficents are not square summable. The series
$$\sum a_{n,m}x^{n+m}y^m$$ is a Fourier series as $x^{n+m}y^m$ are disinct charcters for all $n,m$.

Where is my mistake? My books says this map is not ergodic.

Edit: Is $m=0$ perhaps the problem here? since $a_{n,0}$ might still be non-zero. If so, I demand a refund of my time i spent staring at this

答:
Assuming you mean with respect to Lebesgue measure: $f(x,y) = x$ is a non-constant $T$-invariant function.

In the Fourier series for this $f$ you have all $a_{n,m}=0$ except for $a_{1,0}=1$. So you were correct when you guessed that the coefficients with m=0 create the problem with your original proof


问:
Determine whether propositions are logically equivalent
I have two propositions $(\lnot p\land q) \lor ( p \land q)$  and  $p \leftrightarrow q$.
By using truth table, I found that they are not equal. But I have to find that without truth table and this is what i have solved so far.

$$p \leftrightarrow q = (p\rightarrow q) \land (q \rightarrow p) = (\lnot p \lor q) \land (\lnot q∨p)$$

I want to confirm my next step. Should I just apply DE Morgan's law and then state that they are not equal?

答:
We can applied distributivity here,

$(\lnot p \lor q) \land (\lnot q \lor p) = [(\lnot p \lor q) \land \lnot q] \lor [(\lnot p \lor q) \land p] \\ = [(\lnot p \land \lnot q) \lor (q \land\lnot q)] \lor [(\lnot p \land p)\lor (q \land p)] = (\lnot p \land \lnot q) \lor (q \land p)$

From here, it should be clear why this is not equivalent to the other proposition you gave.


问:
Is there a fixed-point free involution of the Möbius strip?
Is there a fixed-point free continuous involution of the Möbius strip? (Meaning a function $f:M\to M$ such that $f\circ f={\rm id}$ and $f(x)\ne x$ for all $x$.)

The Lefschetz fixed-point theorem is useless here because the Möbius band is homotopic to a circle, which does have fixed-point free involutions. Beyond that, I don't know what tools to use.

(You can't just rotate it 180 degrees because doing that twice flips it along the other axis)

答:
This is essentially the argument Johannes Pauling mentioned in the comments, with a lot more detail.

First, since $f$ is a fixed-point-free involution, we really have a free action of $\mathbb{Z}/2\mathbb{Z}$ on $M$.  Thus, the quotient space $N:= M/(\mathbb{Z}/2\mathbb{Z})$ is naturally a manifold.

Proposition 1:  $N$ is non-orientable.

Sketch of Proof:   We show that more generally, given any covering map $X\rightarrow Y$ of manifolds, if $Y$ is orientable then so is $X$.  The idea is to pick a collection of charts witnessing the orientability of $Y$.  By restricting these charts to small enough open subsets (more specifically, by intersecting them with open sets in an evenly covering neighborhood of $Y$), we obtain a new cover of $Y$ by charts which a) still witness the orientability of $Y$ and b) all lie in an evenly covered neighborhood of the covering $X\rightarrow Y$.

Then pulling these charts back along the covering projection, we obtain disjoint copies of the charts on $X$.  It's easy to convince yourself that these charts on $X$ witness the orientability of $X$., because any overlap is a homeomorphic image of an overlap of charts on $Y$.  $\square$

Before we move on, let me note that it is not clear to me whether $M$ denotes the open Mobius band or the compact Mobius band with boundary, so I'll supply a proof for both cases.

Proposition 2A:  If $M$ is the Mobius band with boundary, then $M$ is homeomorphic to $N$.

Sketch:  Since $M$ has a boundary, then so does $N$.  In fact, since the boundary of $M$ is connected, the same must be true of $N$.  Also, Euler characteristic is multiplicative in coverings, so $\chi(N) = 0$.  Lastly, Proposition 1 tells us that $N$ is non-orientable.  According to wikipedia, this is enough to assert that $N$ is homeomorphic to $M$.  $\square$

Proposition 2B:  If $M$ is the open Mobius band, then $M$ is homeomorphic to $N$.

Sketch:  Using the double cover $M\rightarrow N$, the transfer homomorphism implies $H_1(N;\mathbb{Q})$ is at most one dimensional, since the same is true of $M$.  If $M$ is open, then so in $N$, and so $\pi_1(N)$ is free on some possibly infinite number of generators (see this MO question).  The Hurewicz theorem now gives that $H_1(N;\mathbb{Z})$ is a sum of $\mathbb{Z}$s, one for each generator, and the universal coefficients says the same thing for rational homology.  In particular, the number of generators is at most $1$, so we must have $\pi_1(N) \cong \mathbb{Z}$.  Thus, $H_1(N)\cong \mathbb{Z}$ as well.

According to this paper, we may obtain $N$ by starting with $S^2$, deleting points and disks, and then gluing the boundaries of the disks together.  Notice that deleting two disks and gluing the boundaries together in an orientation preserving way corresponds to connect summing with a torus.  In particular, this operation contributes $\mathbb{Z}^2$ to $H_1(M)$.  Since $H_1(N)\cong \mathbb{Z}$, we cannot use this operation.

Deleting two disks and gluing the boundary together by an orientation reversing map corresponds to connect summing with a Klein bottle.  Again, this creates too much homology.

If there is are any deleted disks, the boundary of each must be identified with itself.  The only way to do this can get a manifold is using a double covering of the boundary, so this process is essentially adding a cross cap (that is, connect summing with $\mathbb{R}P^2$.  If we do this two or more times, we have created too much homology (since $\mathbb{R}P^2 \sharp \mathbb{R}P^2$ is a Klein bottle), so we can only do this at most once.  On the other hand, if we don't do this, the only operation we have left is deleting points, which will give an orientable manifold.

Thus, $N$ is obtained by connecting summing $S^2$ with precisely one $\mathbb{R}P^2$ and then deleting points.  Lastly, if we delete more than one point, then there is once again too much homology, so $N$ is obtained by deleting a single point from $\mathbb{R}P^2$.  In other words, $N$ is an open Mobius band.  $\square$

Using Proposition $2$ to identify $N$ and $M$, we have found a double covering $M\rightarrow M$.  The next proposition says that the only double coverings of $M$ are orientable, giving our final contradiction.

Proposition 3:  There is, up to homeomorphism, a unique connected double covering of $M$.   This double covering is orientable.

Sketch of Proof:  If $M$ is open, then clearly $\mathbb{R}\times S^1$ double covers $M$.  If $M$ has boundary, then clearly $[0,1]\times S^1$ double covers $M$.  Both of these covers are orientable.  So, it's enough to show they are unique up to homeomorphism.  Covering space theory tells us that connected double covers of $M$ corresponds with an index $2$ subgroup of $\pi_1(M)\cong \mathbb{Z}$.  This group has a unique index $2$ subgroup ($2\mathbb{Z}$), so uniqueness follows. $\square$


问:
Prove that for any sets $A,B$, either there exists a surjection from $A$ to $B$, or vice versa
I would like a proof that for any pair of sets $A,B$, either there exists a surjection from $A$ to $B$, or vice versa. This would prove that the usual order on cardinals has the connexity property.

Also, I would like to know if choice is required.

答: Using axiom of choice, $A,B$ can be well ordered. Based on those well orders, one is an initial segment of the other one which enables to build a surjection.


问:
What does the region enclosed by $y = 1/x^5 , y=0, x=3, x=4$ look like?
I'm having trouble trying to see what the region I'm supposed to be computing looks like.

The volume of the solid obtained by rotating the region enclosed by

$$y = 1/x^5 , y=0, x=3, x=4$$

about the line $x=−4$ can be computed using the method of cylindrical
shells via an integral. What does the integrand look like?

Given all the parameters I decided to visualize this first with the graphs. Here are all the curves:

So everything is being rotated around the black line on the left, $x=-4$. But there are many things that just don't make sense to me:

The red function, $y=1/x^5$ isn't even well defined at $x=0$, as it approaches infinity. How is this region "enclosed" at all?

The purple line on the far right, $x=4$, doesn't seem to contribute anything to the enclosure, does it? It's just a step after $x=3$ which already limits the volume of the solid.

Whatever this solid is, it doesn't seem to be enclosed in any way whatsoever to me. I see that it is limited by $y=0$ at the bottom, but there's nothing bounding it at the top as all the curves just continue straight up.

I already know that the answer is

$$\int\frac{2\pi(4+x)}{x^5}dx$$

But I can't see the reasoning behind the question at all. What does the region look like, and how is it enclosed?

答:
You might find Calcplot3D useful for this type of problem.  You can add a surface of revolution, and with enough manipulation of the axes/scales you can produce plots like this:

It will also be useful to have the rotation and key shortcuts page for reference.

Body:
You might find <a href="https://c3d.libretexts.org/CalcPlot3D/index.html?type=volrev;volrev=x;visible=true;alpha=-1;view=0;top2d=1/x%5E5;bot2d=0;umin=3;umax=4;axisvar=x;axisval=-4;revsliderval=0;grid=15,40;polarform=t;polartop=2;polarbottom=0;polarmin=%CF%80/4;polarmax=3%CF%80/4&type=window;hsrmode=0;nomidpts=true;anaglyph=-1;center=5,3,10;focus=0,0,0,1;up=0,2,0;transparent=false;alpha=140;twoviews=false;unlinkviews=false;axisextension=0.7;xaxislabel=x;yaxislabel=y;zaxislabel=z;edgeson=true;faceson=true;showbox=true;showaxes=true;showticks=true;perspective=true;centerxpercent=0.5;centerypercent=0.5;rotationsteps=30;autospin=true;xygrid=false;yzgrid=false;xzgrid=false;gridsonbox=true;gridplanes=false;gridcolor=rgb(128,128,128);xmin=-15;xmax=5;ymin=0;ymax=0.0045;zmin=-10;zmax=10;xscale=2;yscale=1;zscale=2;zcmin=-4;zcmax=4;zoom=0.151467;xscalefactor=1;yscalefactor=2000;zscalefactor=1" rel="nofollow noreferrer">Calcplot3D useful for this type of problem.  You can add a surface of revolution, and with enough manipulation of the axes/scales you can produce plots like this:

It will also be useful to have the rotation and key shortcuts page for reference.


问:
Invariant measure for continuous kernel
Let $p\in (0,1)$ and define a kernel on $(0,1]$ as
$$p(x,\cdot) =p \lambda(\cdot) + (1-p) \frac{\lambda(\cdot \cap [0,x])}{x} $$
where $\lambda$ is lebesgue measure on $[0,1]$. What is the invariant measure?

I tried to solve $\mu = \mu p$: For all measurable $B$,

$\mu(B) = \int \mu(dx)p(x,B) = \int \mu(dx)(p \lambda(B) + (1-p) \frac{\lambda(B\cap [0,x])}{x} )$

but I have difficulty with $\int \frac{\lambda(B\cap [0,x])}{x}\mu(dx)$.

Another idea is to find the limit $\lim_n p^n(x,\cdot)$ for an initial state $x$, but again I am struggling with $\int \frac{\lambda(B\cap [0,x])}{x}\lambda(dx)$.

答:
To start write $\mu=\mu p$ as
$$
\mu(t,1]=(1-t)p+(1-p)\int_t^1\left(\int_s^1{\mu(dx)\over x}\right) ds,\qquad 0<t\le 1.
$$
This shows that $\mu$ has a density $m$ with respect to $\lambda$ satisfying
$$
m(t) = p+(1-p)\int_t^1{m(x)\over x} dx,\qquad 0<t\le 1.
$$
This in turn implies that $m$ is $C^1$ on $(0,1]$ and satisfies the ODE
$$
m'(t) = -(1-p){m(t)\over t},\qquad 0<t\le 1.
$$
Solving this and normalizing to be a probability density you get
$$
m(t) =pt^{-(1-p)},\qquad 0<t\le 1.
$$

Title: Invariant measure for continuous kernel


问:
How to evaluate $\sum_{i=0}^{k-1} 2^i\cdot 4(k-i-1).$
I'm currently having trouble evaluating the following sum to get a formula in terms of $k$:
$$\sum_{i=0}^{k-1} 2^i\cdot 4(k-i-1)$$

I know that $$\sum_{i=0}^n 2^i = 2^{n+1}-1$$
but since my $2^i$ is multiplied with another term inside my summation, I can't use that formula to evaluate it. Any suggestions?

答:
I would first let $\ell=k-1$ and rewrite it as

$$\begin{align*}
\sum_{i=0}^\ell 2^i\cdot 4(\ell-i)&=4\sum_{i=0}^\ell 2^i(\ell-i)\\
&=4\sum_{i=0}^\ell i2^{\ell-i}\\
&=4\cdot 2^\ell\sum_{i=1}^\ell\frac{i}{2^i}\\
&=2^{k+1}\sum_{i=1}^\ell\frac{i}{2^i}\,.
\end{align*}$$

That last summation is of a type that comes up quite often enough that it’s worth knowing how to derive a closed form for it. The accepted answer to this question gives a very easy, elementary derivation of the general formula for such summations. Just plug this summation into that general formula and simplify, and you’ll have your answer.


问:
trigonometric equations finding $x$, when $\cos(x)=0$ or some value
I'm practicing trigonometric equations where we need to find the value of $x$, however sometimes I keep getting a completely different solution than the one in the book and I don't really understand where I am making a mistake (and it's always in finding the $x$ in the end). For example:

Problem: $$\cos x  + \sqrt3\cos 2x + \cos3x=0$$ which gives: $$2\cos2x(\cos x + \frac{\sqrt3}{2})=0$$

and then:  $2\cos2x=0$ , where I get that $x=\frac{\pi}{4}+\frac{k\pi}{2}, k\in \Bbb Z$

and:  $\cos x= -\frac{\sqrt3}{2}$, where I get $x=\frac{5\pi}{6}+ 2k\pi, k\in \Bbb Z$ (which based on my book is the only solution I got right)

I believe I also have this: $\cos (2\pi - x)= -\frac{\sqrt3}{2} $, and the solution I got: $x=\frac{7\pi}{6}+ 2k\pi, k\in \Bbb Z$

I guess I'm getting something wrong since the solutions given are:
  $x=\frac{(2k+1)\pi}{2}$ ; $x=\frac{5\pi}{6}+2k\pi$; $x=-\frac{5\pi}{6}+2k\pi$, $k\in \Bbb Z$

Body:
hint

If $$\cos(2x)(\cos(x)\frac{\sqrt{3}}{2})=0$$

then

$$\cos(2x)=0=\cos(\frac{\pi}{2})$$
or
$$\cos(x)=-\frac{\sqrt{3}}{2}=\cos(\pi-\frac{\pi}{6})$$

then
$$2x=\pm \frac{\pi}{2}+2k\pi=\frac{\pi}{2}+k\pi$$
or
$$x=\pm \frac{5\pi}{6}+2k\pi$$

The solutions are
$$\frac{\pi}{4}+k\frac{\pi}{2}\;,\;$$
$$ \frac{5\pi}{6}+2k\pi=-\frac{7\pi}{6}+2(k+1)\pi\;$$
and
$$\; \frac{-5\pi}{6}+2k\pi=\frac{7\pi}{6}+2(k-1)\pi$$


问:
A equivalent condtion for existence of feasible circulation

A circulation in a directed graph $D$ is a function $g:E(D)\rightarrow\mathbb{R}$ satisfying the conservation condition at every vertex. Let $l,u:E(D)\rightarrow \mathbb{R}^{+}_{0}$ be a lower and upper capacity function, resp. And assume for each edge $e$, $l(e)\le u(e)$.

A circulation $g$ is feasible if $l(e)\le g(e) \le u(e)$ for every edge $e$.

Prove that there exists a feasible circulation iff $l^+(A)\le u^-(A)$ for every set $A\subset V(D)$.

Hint: Add a vertex $s$ and an edge from $s$ to every vertex of $D$, and a vertex $t$ and an edge from every vertex of $D$ to $t$. Define $c(sv)=l^-(v), c(vt)=l^+(v), c(e)=u(e)-l(e)$  for $v \in V(D)$ and $e \in E(D)$. Given a flow f of value $\Sigma_{e\in E(D)} l(e)$ in the network thus defined, consider $f+l$.

I proved the forward direction. In the backward direction, if there exists such a flow $f$, then $f+l$ is in the lower and upper capacity bound. But I can't prove such a flow $f$ exist. I want to use max-flow min-cut theorem, but I can't find a minimal cut. Because I don't know the relation between $u(e)-l(e)$ and $l^+(x)+l^+(y)$ where $e=xy$.

And even if such flow $f$ exists, I'm confused to find out $f+l$ is a circulation and where the condition "$l^+(A)\le u^-(A)$ for every set $A\subset V(D)$" is used.

答:
Here is a slightly shorter solution using a different notation.
We consider a directed graph $G = (V, E)$ with edge capacities given by $c\colon E \to \mathbb{R}^+$ and node demands given by $d\colon V \to \mathbb{R}$. A circulation is an assignment $f\colon E \to \mathbb{R}^+$ satisfying $c(e) \leq f(e)$ for all edges $e$ and $\sum_{e \text{ into } v} f(e) - \sum_{e \text { out of } v} f(e) = d(v)$ for all vertices $v$.

The forward direction — if there exists a circulation, then $\sum_{v \in V} d(v) = 0$ and for any partition $(A, B)$ of $V$, $\sum_{v_B} d(v) \leq \operatorname{cap}(A, B)$ — is quite direct. You just need to use the definition of a circulation and think about what edges cancel in the sums.

For the backward direction, we consider the contrapositive.
Suppose that there does not exist a circulation in $G$, and that $\sum_{v \in V} d(v) = 0$.
Let us convert the circulation network into a flow network $G'$ by adding a source $s$ and a sink $t$, and add edges $(s, v)$ of capacity $d(v)$ where $d(v) > 0$, and edges $(v, t)$ of capacity $-d(v)$ where $d(v) < 0$.
Running the Ford-Fulkerson algorithm on $G'$ should give us $\operatorname{max-flow}(G') = \operatorname{cap}(A, B) < \sum_{\substack{v \in V \\ d(v) > 0}} d(v)$ (for otherwise the max-flow gives a circulation).

But $A = \{s\} \cup A'$ and $B = \{t\} \cup B'$ with $A', B' \subseteq V$,
and
$$\begin{aligned}
\operatorname{cap}(A, B)
&= \operatorname{cap}(A', B') + \sum c(s \to B) + \sum c(A \to t) \\
&= \operatorname{cap}(A', B') - \sum_{\substack{v \in B \\ d(v) < 0}} d(v) + \sum_{\substack{v \in A \\ d(v) > 0}} d(v)
\end{aligned}$$
Now
$$
\operatorname{cap}(A, B)
< \sum_{\substack{v \in V \\ d(v) > 0}} d(v)
= \sum_{\substack{v \in B \\ d(v) > 0}} d(v) + \sum_{\substack{v \in A \\ d(v) > 0}} d(v)$$
so $$\operatorname{cap}(A', B') < \sum_{\substack{v \in B \\ d(v) < 0}} d(v) + \sum_{\substack{v \in B \\ d(v) > 0}}  d(v) = \sum_{v \in B} d(v)$$
where $(A', B')$ is a partition of $V$.


答:
If $-5\pi/6+2m\pi=7\pi/6+2n\pi$ where $m,n$ are arbitrary integers

$\iff m=n+1$

So, we have actually not missed anything

As $\cos(2r\pi\pm A)=\cos A,$

If $\cos B=\cos C,B=2p\pi\pm C$ where $r,p$ are Integers

Title: trigonometric equations finding $x$, when $\cos(x)=0$ or some value

Body:
I'm practicing trigonometric equations where we need to find the value of $x$, however sometimes I keep getting a completely different solution than the one in the book and I don't really understand where I am making a mistake (and it's always in finding the $x$ in the end). For example:

Problem: $$\cos x  + \sqrt3\cos 2x + \cos3x=0$$ which gives: $$2\cos2x(\cos x + \frac{\sqrt3}{2})=0$$

and then:  $2\cos2x=0$ , where I get that $x=\frac{\pi}{4}+\frac{k\pi}{2}, k\in \Bbb Z$

and:  $\cos x= -\frac{\sqrt3}{2}$, where I get $x=\frac{5\pi}{6}+ 2k\pi, k\in \Bbb Z$ (which based on my book is the only solution I got right)

I believe I also have this: $\cos (2\pi - x)= -\frac{\sqrt3}{2} $, and the solution I got: $x=\frac{7\pi}{6}+ 2k\pi, k\in \Bbb Z$

I guess I'm getting something wrong since the solutions given are:
  $x=\frac{(2k+1)\pi}{2}$ ; $x=\frac{5\pi}{6}+2k\pi$; $x=-\frac{5\pi}{6}+2k\pi$, $k\in \Bbb Z$


问:
The sum of consecutive strings of positive integers of length $2*n$
Let $a(n)$ is the sum of consecutive strings of positive integers of length $2*n$, starting with the integer $2$; and $a(0) = 1$.

So, we'll have strings like below:
$5=2+3$;
$22=4+5+6+7$;
$63=8+9+10+11+12+13$; etc.

I know that according to OEIS $a(n)=n(2n^2+3)$, but I am not sure how to prove it.
I thought about geometrical approach to get the formula.
Any hints are highly welcomed.

答:
I would start by trying to find a formula for the last digits of the $n$th series. The last digits are $1, 1+2*1, 1+2*1+2*2, ...$ which you can see is equal to
$$1+\sum_{k=1}^n 2k= 1+2\sum_{k=1}^n k=1+n(n+1)$$
(look up triangular numbers if you don’t understand the last step).

$a(n)$ is equal to the sum of the integers less than or equal to the $n$th last digit, minus the sum of the integers less than or equal to the $n-1$st last digit. In other words, the $1+n(n+1)$st triangular number minus the $1+n(n-1)$st triangular number, which is equal to
$$(1+n(n+1))(2+n(n+1))/2-(1+n(n-1))(2+n(n-1))/2$$
and simplifies to the formula you found.


问:
Is $K=\{f_\lambda(x)=e^{\lambda x}\mid \lambda \in [0,a], x\in [0,b]\}$ equcontinuous?
I am trying prove that the following set is equcontinuous.

$K=\{f_\lambda(x)=e^{\lambda x}\mid \lambda \in [0,a], x\in [0,b]\}$

I read two prove of equcontinuty in two theorem and twice of theme have the hypothesis of compactness. So I have no idea prove this without compactness. Please help me with Hints or similar.

答: Since $[0,a]\times[0,b]$ is compact the function $(\lambda,x) \longmapsto e^{\lambda x}$ is uniformly continuous. Hence we may choose $\delta$ so that $|x-y| + |\lambda_x-\lambda_y| < \delta$ implies $|e^{\lambda_x x} - e^{\lambda_y y}| < \epsilon$. Hence for any $\lambda$ take $\lambda_x=\lambda_y$ and if $|x-y|<\delta$ then $|f_\lambda(x)-f_\lambda(y)|<\epsilon$ which shows equicontinuity.</p>


问:
Comparing meaning of first order logical statements
I came across simple problem as follows:

For each pair of predicate calculus formulae below, either show that they are equivalent, or illustrate by choosing a
simple domain and meaning for the predicate symbols that the two statements have different meanings.

(a) $(∀xP(x)) ∨ (∀yQ(y)) ,\quad  ∀x(P(x) ∨ Q(x))$

(b) $(∃xQ(x)) → (∃yP(y)) , \quad ∃x(Q(x) → P(x))$

I am able to solve it as follows:

For (a):

$P(x):$ $x$ has height more than 6 ft.

$Q(x):$ $x$ has height less than 6 ft.

Then, $(∀xP(x)) ∨ (∀yQ(y))$ means all people "considered together" either has height more than 6 ft OR less than 6 ft. This does not allow John to have height more than 6 ft and Roger to have height less that 6 ft.

$∀x(P(x) ∨ Q(x))$ means everyone "considered individually" either has height more than 6 ft OR less than 6 ft. This allow John to have height more than 6 ft and Roger to have height less that 6 ft.
Thus statements in (a) are not equivalent.

For (b):

$Q(x):x$ is a husband

$P(x):x$ is a wife

Then, $(∃xQ(x)) → (∃yP(y))$ means for some husband $x$, there is wife $y$

$∃x(Q(x) → P(x))$ means for some husband $x$, he is also wife (of himself), which does not makes sense.

Q1. Are above reasoning correct?

More importantly, can I solve it algebraically, for example by reducing them to   conjunctive normal form (CNF) and then comparing?

I tried it as follows:

For (a):

$∀xP(x)) ∨  (∀yQ(y)\equiv P(x)\vee Q(y)$

$∀x(P(x) ∨ Q(x))\equiv P(x) \vee Q(x)$

Obviously above two are not the same.

Am I correct with above.

For (b):

$(∃xQ(x)) → (∃yP(y))$

$\equiv \neg (∃xQ(x)) \vee (∃yP(y)) $

$\equiv (\forall x \neg Q(x))\vee (\exists y P(y))$

$\equiv \neg Q(x)\vee P(B)$  (...skolemized)(equation 1)

$∃x(Q(x) → P(x))$

$\equiv ∃x(\neg Q(x) \vee P(x))$

$\equiv \neg Q(A) \vee P(B))$  (...skolemized)(equation 2)

Above I somewhat applied skolemization rules as I understood them. For existential quantifier, we have to introduce variable (written in caps) for its operand. We can eliminate universal quantifier without changing its operand.

Q2. With variables in caps and in small case, how can I makes clear sense that equations 1 and 2 are different?

Q3. Or am I completely wrong with above CNF approach?

答:
Just for Q1:
the reasoning in part (a) looks good, but not in part (b).

To express the two implications in English, the important word "if" is missing. The first would be "If there is a husband then there is a wife". But the example interpretation does not work in showing the formulas are not equivalent. You seem to imagine a domain where some individuals are husbands, some are wives, none are both, and possibly some are neither. In this case your second formula will be true because the $x$ in the second can be any individual who is not a husband.


问:
Most accurate way to calculate matrix powers and matrix exponential for a positive semidefinite matrix
I do need to numerically calculate the following forms for any $x\in\mathbb{R}^n$, possibly in python:

$x^T M^k x$, where $M\in\mathbb{R^n}$ is a PSD matrix, where $k$ can get quite large values $-$possibly up to order of hundreds. I prefer $k$ to be a real number, but it is ok if it can only be an integer, if that makes a considerable accuracy difference.

Similarly I am interested in $x^Te^{-tM}x$ where $t$ is a non-positive real value.
For case 1 I can either:

Use the scipy.linalg.fractional_matrix_power to calculate $M^k$ and then derive $x^TM^Kx$, or

Use scipy.linalg.svd to find SVD of $M$ as $U\Lambda U^T$ and then finally evaluate the desired value using $x^TU\Lambda^k U^Tx$.

Finally if $k$ is integer, and again based on SVD I can calculate $\|x^TM^{k/2}\|^2$.

For case 2

Again I can use off the shelf scipy.linalg.expm and then exponentiate the singular values of $M$

I can do SVD for $M$ and then go with $x^TU\exp(\Lambda) U^Tx$.

Finally since I am only interested in $x^T \exp(M) x$, and not exactly $\exp(M)$ it self, I can consider the Taylor expansion of $x^T{\rm expm}(M)x\approx \sum_{i=0}^{l} \frac{1}{i!}x^TM^ix$ for some $l$ that controls the precision, and $x^TMx$ can be calculated based on case 1.

Can anybody guide me about what is the most precise way to calculate either of these expressions, up to hopefully machine precision? Any of these methods, or they're better solutions out there? I would be happy also with references.

P.S. Not knowing if here or stackoverflow being a good place to share this question at, I will be cross-posting it on their with the same title and content.

答:
One standard approach to computing matrix functions times a vector $f(M)x$ or quadratic forms $x^Tf(M)x$ when $M$ is symmetric is via the Lanczos algorithm. Lanczos computes an orthonormal basis $Q_k = [q_1, \ldots, q_k]$ for Krylov subspace $\operatorname{span}(x,Ax,\ldots, A^{k-1}x)$ by a Gram-Schmidt like procedure. This results in a factorization $AQ_k =Q_kT_k + \beta_k q_{k+1}e_k^T$, where $e_k$ is the $k$-th canonical basis vector and $T_k$ is a $k\times k$ symmetric tridiagonal matrix.

The "Lanczos approximation" $f(M)x$ is defined as $Q_k f(T_k) Q_k^T x$, and the approximation to $x^Tf(M)x$ is defined as $x^TQ_k f(T_k) Q_k^T x$. There are software packages to compute $Q_k f(T_k) Q_k^T x$. This can easily be turned into the Lanczos approximation to $x^Tf(M)x$ by taking the inner product with $x$. The Lanczos approximation to the quadratic form can be viewed as a certain Gaussian quadrature approximation to $x^T f(M) x$ and converges extremely quickly in $k$ if $f$ is analytic (see Golub Meurant "Matrices Moments and Quadrature").

This approach has runtime $O(T_{mv}k + nk)$ (or $O(T_{mv}k + nk^2)$ with full reorthogonalization), where $T_{mv}$ is the cost of evaluating $v\mapsto Mv$. This will almost certainly be cheaper than an SVD or any $n^3$ algorithm since $k$ can typically be taken to be fairly small even for a machine precision accurate output.

Note that if $x$ is unit length $Q_k^Tx = e_1$. This means the Lanczos approximation to $x^Tf(M)x$ is given by $e_1^T f(T_k) e_1$ and that $Q_k$ doesn't need to be stored. There are also some subtleties about whether full reorthogonalization should be used or not. Whether or not it is used, the output will still be accurate, but without reorthogonalization $k$ may need to be larger. However, reorthogonalizaton is more computationally expensive, so there is some tradeoff.


问:
Arithmetically weighted sum of half of the combinatorial numbers
I would like to see a proof for the identity
$$\sum_{k=1}^n k\binom{2n}{n+k} = \frac{n}{2}\binom{2n}{n}$$ This was suggested by the OEIS sequence (https://oeis.org/A002457), but I haven't been able to prove it.

答:
$k=\dfrac{(n+k)-(n-k)}{2}$, so after cancelling either $n+k$ or $n-k$ we can write the left-hand side as
$$
\sum_{k=1}^{n}\frac{2n}{2}\left(\binom{2n-1}{n-k}-\binom{2n-1}{n-k-1}\right) = n\binom{2n-1}{n-1} = \frac{n}{2}\binom{2n}{n}.
$$

OwnerDisplayName: lara

OwnerDisplayName: Alexander Burstein


问:
List all elements of the cyclic group $\langle f\rangle$ generated by $f$
Let $f =
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 &7 & 8 & 9 \\
7 & 9 & 5 & 8 & 3 & 6 & 1 & 4 & 2
\end{pmatrix}
$ and $g =
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 &7 & 8 & 9 \\
9 & 8 & 5 & 1 & 2 & 4 & 7 & 3 & 6
\end{pmatrix}$.

How do I find elements of the cyclic group $\langle f\rangle$ generated by $f$? Similarly, how do I find elements of the cyclic group $\langle g\rangle$ generated by $g$?

First I wrote $f$ and $g$ as a product of disjoint cycles: $f=(17)(29)(35)(48)$ and $g=(1964)(2835)$. Then I can see that their orders are $2$ and $4$ respectively. Now I am stuck. Would appreciate any help.

答:
As $f$ is a product of disjoint transpositions the cyclic group generated by $f$ is equal to the set containing the identity and $f$ itself.

The cyclic group generated by $g$ is the group $\{Id, g, g^2, g^3\}$ as $g$ is the product of two disjoint $4$-cycles.


问:
Use property of Cauchy sequence to prove the convergence of the recurrence sequence
Given the fact that $0 < \alpha < 1$, the sequence $(a_n)$ satisfies
$$ \lvert{a_{n+2} - a_{a+1}}\rvert \leq \alpha\lvert{a_{n+1}} - a_{n}\rvert,\quad \forall n \in \mathbb{N}.$$
then $(a_n)$ is a Cauchy sequence and hence convergent.

How can I use this information to show that the sequences
$$a_1 = 0,\, a_2 = 1$$
$$a_n = \frac{1}{3} a_{n-1}+\frac{2}{3}a_{n-2}$$
and
$$c_1=1, c_{n+1} = \frac{3+2c_n}{3+c_n}$$
are convergent?
I have difficulties in proving the sequence satisfy the given inequality, I think use induction is possible but I have no idea how can I start..

Body:
hint

For $ (a_n)$.

we have

$$3a_{n+2}=a_{n+1}+2a_n$$

thus

$$3(a_{n+2}-a_{n+1})=-2(a_{n+1}-a_n)$$

So,
$$\alpha=\frac 23$$

For the sequence $ (c_n) $, use the canonical form

$$c_{n+1}=2-\frac{3}{3+c_n}$$


问:
Derivatives of integrals with Variable Bounds
I'm working through a Green's function problem for a second-order linear differential operator $L$ and I need to find the derivative of:
$$ u(x) = \frac{u_2(x)}{a_0(\xi)W(\xi)} \int_a^x u_1(\xi)f(\xi) \ d\xi \ + \frac{u_1(x)}{a_0(\xi)W(\xi)} \int_a^x u_2(\xi)f(\xi) \ d\xi $$
I am thinking this has to be done using Leibniz integral rule for variable bounds. Since $\frac{d}{dx}(x)=1$ and $\frac{d}{dx}(a)=0$ I think it should be
$$ u'(x) = \frac{u_2'(x)}{a_0(\xi)W(\xi)}\int_a^xu_1(\xi)f(\xi) \ d\xi + \frac{u_1'(x)}{a_0(\xi)W(\xi)}\int_a^xu_2(\xi)f(\xi) \ d\xi $$
where the $f(x,b(x))\cdot\frac{d}{dx}b(x) $ term from the first part cancels out with the $-f(x,a(x))\cdot\frac{d}{dx}a(x) $ of the second part. Is this correct? It comes out so simple in the end, I am also wondering if there was an easier way to come to this conclusion.

Additionally, when we take the second derivative will we see that term cancel out again? Or will it not because in the second derivative we will have $u_2'(x)u_1(x)f(x)-u_1'(x)u_2(x)f(x)$? Thanks for any help!

答:
By the Leibniz integral rule, you would have, for instance,

$$\begin{align}
\frac{\mathrm d}{\mathrm dx}\left[\int_a^xu_2(x)u_1(\xi)f(\xi)\,\mathrm d\xi\right]&=u_2(x)u_1(x)f(x)\cdot\frac{\mathrm d(x)}{\mathrm dx}-u_2(x)u_1(a)f(a)\cdot\frac{\mathrm d(a)}{\mathrm dx}\\
&\quad\quad+\int_a^x\frac{\partial}{\partial x}\left[u_2(x)u_1(\xi)f(\xi)\right]\,\mathrm d\xi\\[1ex]
&=\underbrace{u_2(x)u_1(x)f(x)}{}+u_2'(x)\int_a^xu_1(\xi)f(\xi)\,\mathrm d\xi
\end{align}$$


问:
How would I evaluate this series using telescopic summation?
The series is $$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} ... + \frac{n-1}{n!},$$ which I can write as the sum $$\sum_{i=2}^n \frac{i-1}{i!}$$ I will try to evaluate the partial sums to look for a pattern. The first partial sums I get are $\frac12, \frac56 , \frac{23}{24}$. Thus, I make the guess that the sum is equal to $\frac{n!-1}{n!}$. I have already proven the base cases through experimentation, so I work on proving the induction step. I assume that $\sum_{i=2}^n \frac{i-1}{i!}= \frac{n!-1}{n!}$, for some $n=k$ so I get $$\sum_{i=2}^k \frac{i-1}{i!}= \frac{k!-1}{k!}$$ Now I prove this holds for $n=k+1$ as well. I get $$\sum_{i=2}^{k+1} \frac{i-1}{i!}= \frac{k!-1}{k!} + \frac{k}{(k+1)!} = \frac{(k!-1)(k+1)}{(k+1)!} + \frac{k}{(k+1)!} = \frac{(k+1)!-k-1}{(k+1)!} + \frac{k}{(k+1)!} = \frac{(k+1)!-1}{(k+1)!} $$ Thus the induction step holds as well, proving my expression is indeed the sum. However, this problem was in the telescoping sums section of my mathbook, so I was wondering how you would solve this sum telescopically since these induction proofs take some time.

Body:
We have:

$$\sum_{i=2}^{n}\frac{i - 1}{i!} = \sum_{i=2}^{n}\frac{i}{i!} -\frac{1}{i!} = \sum_{i=2}^{n}\frac{1}{(i-1)!} - \frac{1}{i!}$$

$$=\bigg(\frac{1}{1!} - \frac{1}{2!}\bigg)+\bigg(\frac{1}{2!} - \frac{1}{3!}\bigg) +\bigg(\frac{1}{3!}-\frac{1}{4!}\bigg)+\dots+\bigg(\frac{1}{(n-1)!}-\frac{1}{n!}\bigg)$$

Telescoping:

$$=\frac{1}{1!}-\frac{1}{n!} = 1- \frac{1}{n!} = \boxed{\frac{n!-1}{n!}}$$


问:
Show $\frac{x_1}{x_n} + \frac{x_2}{x_{n-1}} + \frac{x_3}{x_{n-2}} + \dots + \frac{x_n}{x_1} \geq n$
I was recently asked this question which stumped me.

How can you show $\dfrac{x_1}{x_n} + \dfrac{x_2}{x_{n-1}} + \dfrac{x_3}{x_{n-2}} + \dots + \dfrac{x_n}{x_1} \geq n$ for any positive reals $x_1, x_2, \dots, x_n$?

答:
Dude.  You shouldn't need all those fancy things.  Just pair up each $\frac{x_i}{x_{n-i}} + \frac{x_{n-i}}{x_i}$ and examine each of those separately.  Take into account even $n$ and odd $n$, since there'll be an unpaired fraction when $n$ is odd.

Spoiler:

 When $n$ is odd, identify what the unpaired fraction is, then it's value will be obvious.


Title: How would I evaluate this series using telescopic summation?

Body: The series is $$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} ... + \frac{n-1}{n!},$$ which I can write as the sum $$\sum_{i=2}^n \frac{i-1}{i!}$$ I will try to evaluate the partial sums to look for a pattern. The first partial sums I get are $\frac12, \frac56 , \frac{23}{24}$. Thus, I make the guess that the sum is equal to $\frac{n!-1}{n!}$. I have already proven the base cases through experimentation, so I work on proving the induction step. I assume that $\sum_{i=2}^n \frac{i-1}{i!}= \frac{n!-1}{n!}$, for some $n=k$ so I get $$\sum_{i=2}^k \frac{i-1}{i!}= \frac{k!-1}{k!}$$ Now I prove this holds for $n=k+1$ as well. I get $$\sum_{i=2}^{k+1} \frac{i-1}{i!}= \frac{k!-1}{k!} + \frac{k}{(k+1)!} = \frac{(k!-1)(k+1)}{(k+1)!} + \frac{k}{(k+1)!} = \frac{(k+1)!-k-1}{(k+1)!} + \frac{k}{(k+1)!} = \frac{(k+1)!-1}{(k+1)!} $$ Thus the induction step holds as well, proving my expression is indeed the sum. However, this problem was in the telescoping sums section of my mathbook, so I was wondering how you would solve this sum telescopically since these induction proofs take some time.

Body:
Hint

$$\frac{n-1}{n!}= \frac{1}{(n-1)!}- \frac{1}{n!}$$ and therefore

$$\sum_{k=1}^n \frac{k-1}{k!} = \sum_{k=1}^n \left(\frac{1}{(k-1)!}- \frac{1}{k!}\right) = 1 - \frac{1}{n!}$$


问:
Given algorithm, find and solve the recurrence relation
Consider the following recurrence relation:

$T(2) = 5$

$T(n) = 2T\left( {\left\lceil {{{2n} \over 3}} \right\rceil } \right) + T\left( {\left\lfloor {{n \over 3}} \right\rfloor } \right) + 3$

I got this recurrence relation by analysing the following algorithm:

I got $T(2) = 5$ as the base case because of 2 comparisons in the if statement plus 3 statements in swap() function.

The question requests to solve it using the characteristic equation. However, I don't know how to proceed with the constant $1$ at the end.

My attempt:

$a_n - a_{2n/3} - a_{n/3} = 3
\\
x^n - x^{2n/3} - x^{n/3} = 3
\\
x^{n/3}(x^3 - 2x^2 - 1) = 3$

Also, if I analysed incorrectly the algorithm, please tell me!
All help is appreciated!

edit: I incorrectly deduced the recurrence equation. I didn't count the cost of first and second if statement.

答:
Too long for a comment.

A good way to simulate it in MATHEMATICA

```<code>t[n_] := 2 t[Ceiling[2 n/3]] + t[Floor[n/3]] + 1

t[0] = 0;
t[1] = 0;
t[2] = 5;

tk = Table[{k, t[k]}, {k, 3, 200}]

gr1 = ListPlot[tk]
nlm = NonlinearModelFit[tk, a x^b, {a, b}, x]
gr2 = Plot[Normal[nlm], {x, 0, 400}, PlotStyle -> {Red,Dashed}]
Show[gr1, gr2]
</code>```

Body:
Too long for a comment.

A good way to simulate it in MATHEMATICA

t[n_] := 2 t[Ceiling[2 n/3]] + t[Floor[n/3]] + 1

t[0] = 0;
t[1] = 0;
t[2] = 5;

tk = Table[{k, t[k]}, {k, 3, 200}]

gr1 = ListPlot[tk]
nlm = NonlinearModelFit[tk, a x^b, {a, b}, x]
gr2 = Plot[Normal[nlm], {x, 0, 400}, PlotStyle -> {Red,Dashed}]
Show[gr1, gr2]


问:
Show the eigenvalues of this submatrix are all less than $1$
Let $C$ be a positive definite correlation matrix partitioned as

$$C=\begin{bmatrix}
I_{k_1} & A \\
A' & I_{k_2} \end{bmatrix}$$

How can I show that the eigenvalues of $AA'$ are all less than $1$?

答: Using the Schur complement of the lower-right entry, we see that $C$ is positive definite if and only if $I_{k_2} - AA'$ is positive definite. This occurs if and only if the eigenvalues of $AA'$ are less than $1$.


问:
Prove $\{0^k10^k10^k1 \in \{0,1\}^* \mid k \geq 0 \}$ is not context free
I'm practising for my CS exam and got stuck on this problem

$$ \{0^k10^k10^k1 \in \{0,1\}^*\mid k ≥ 0\}. $$

I think I have good start however I don't know how to proceed. I assume the L is context free, and that pumping lemma applies with some pumping length $n$.

$$|w| = 3k,\quad w = uvxyz$$

Since $|vxy| \geq n$ it can't contain all characters. Considering $uv^2xy^2z$ would end up having more than $3$ zeros if anything is pumped in. However I'm not sure how t o move from here. Is that enough of a proof?

答:
Proof that $L$ is not regular

Assume that $L$ is regular, then there exists $p$ such that for all $|w|\geq p\:, \exists x,y,z, \: y\neq \varepsilon \text{ and } \forall k, \: xy^kz \in L$.

Take $n>p$ (we're not going to be subtle here. Then $w=0^n10^n10^n1 \in L$, so we can slice it into three part $x,y,z$. Let us consider $y$

if $1 \in y$ then $xy^kz \not\in L$, for it would contains to many $1$'s

otherwise, $y=0^d$ for some $d$, but then $xy^kz = 0^{n_1} 1 0^{n_2} 1 0^{n_3} 1 \not\in L$ because the $n_i$ won't be equal anymore.

Hence we cannot apply the pumping lemma for regular languages, $L$ is therefore not regular.

Proof that $L$ is not context-free

as suggested by @posilon, we can also prove that the language is not context free by applying this pumping lemma

There exists a $p$ such that, if $|w|\geq p$ then we can slice $w = uvwxy$
such that:

$|vx| \geq 1$

$|vwx| \leq p$

$uv^n wx^n y \in L$ for all $n \geq 0$.

here again, let us consider the number of $1$'s in $vx$

if $1 \in vx$ then $uv^n wx^n y  \not\in L$, for it would contains to many $1$'s

otherwise, $v=0^d, x=0^{d'}$ for some $d,d'$, but then $uv^n wx^n y  = 0^{n_1} 1 0^{n_2} 1 0^{n_3} 1 \not\in L$ because the $n_i$ won't be equal anymore.

Hence we cannot apply the pumping lemma for context-free languages, $L$ is therefore not context-free.


问:
Intuitive understanding of the derivatives of $\sin x$ and $\cos x$
One of the first things ever taught in a differential calculus class:

The derivative of $\sin x$ is $\cos x$.

The derivative of $\cos x$ is $-\sin x$.

This leads to a rather neat (and convenient?) chain of derivatives:

```
sin(x)
cos(x)
-sin(x)
-cos(x)
sin(x)
...
```

An analysis of the shape of their graphs confirms some points; for example, when $\sin x$ is at a maximum, $\cos x$ is zero and moving downwards; when $\cos x$  is at a maximum, $\sin x$ is zero and moving upwards.  But these "matching points" only work for multiples of $\pi/4$.

Let us move back towards the original definition(s) of sine and cosine:

At the most basic level, $\sin x$ is defined as -- for a right triangle with internal angle $x$ -- the length of the side opposite of the angle divided by the hypotenuse of the triangle.

To generalize this to the domain of all real numbers, $\sin x$ was then defined as the Y-coordinate of a point on the unit circle that is an angle $x$ from the positive X-axis.

The definition of $\cos x$ was then made the same way, but with adj/hyp and the X-coordinate, as we all know.

Is there anything about this basic definition that allows someone to look at these definitions, alone, and think, "Hey, the derivative of the sine function with respect to angle is the cosine function!"

That is, from the unit circle definition alone.  Or, even more amazingly, the right triangle definition alone.  Ignoring graphical analysis of their plot.

In essence, I am asking, essentially, "Intuitively why is the derivative of the sine the cosine?"

答:
One of the main ways that sine and cosine come up is as the fundamental solutions to the differential equation $y'' = -y$, known as the wave equation. Why is this an important differential equation?  Well, interpreting it using Newton's second law it says "the force is proportional and opposite to the position." For example, this is what happens with a spring!

Now that's a 2nd degree equation, so it has a 2-dimensional space of solutions. How to pick a nice basis for that space? Well, one way would be to pick $f$ and $g$ such that $f' = i f$ and $g' = -i g$. However, that involves too many imaginary numbers, so another option is $f' = -g$, and $g' = f$.

Thus if you're trying to find two functions which explain oscillatory motion you're naturally lead to picking functions that have $f' = g$, $g' = -f$, etc.

(On the other hand it's totally unclear from this point of view why Sine and Cosine should have anything to do with triangles...)


问:
If $(a^{n}+n ) \mid (b^{n}+n)$ for all $n$, then $ a=b$
I happened to receive this from my friend.

Let $a,b \in \mathbb{N}$, such that $a^{n}+n \: \bigl|\: b^{n}+n$ for all $n \in \mathbb{N}$. Prove that $a=b$. How do we proceed?

答:
Hint $ $ excerpted from my $ $ sci.math post $ $ 2006/4/4 $ $ here or here (see there for much motivation), we can choose $\rm\,\color{#c00} n\,$ and $\rm\, p > |\color{blue}{b-a}|\,$ satisfying the LHS below, so the RHS $\rm\Rightarrow \color{blue}{b = a}$

$$\rm\quad\begin{align}
  \rm p-1 &\mid\rm \color{#c00}{n-1} \\
 \rm p &\mid\rm \color{#0a0}{a+n} \\
\end{align}\! \Rightarrow\,\
p\mid \overbrace{a^{\large \color{#c00}n^{\phantom{I}}}\!\!\!-\!a+\color{#0a0}{a\!+\!n}}^{\Large a^n+n}\mid \overbrace{b^{\large \color{#c00}n^{\phantom{I}}}\!\!\!-\!b+\color{blue}{b\!-\!a}+\color{#0a0}{a\!+\!n}}^{\Large b^n +n} \ \,\Rightarrow\,\ p\mid \color{blue}{b-a}\qquad\qquad\quad  $$


问:
Importance of estimating $\sigma^2$ in linear Statistical model
Statistical model for Complete Randomized design

$y_{ij} = \mu + \tau_i + \epsilon_{ij}$

where, $i$ denotes treatment and $j$ denotes observation.

$i=1,2,...,k\quad and \quad j=1,2,..., n_i$

$y_{ij}$ be a random variable that represents the response obtained on the $jth$
observation of the  $ith$ treatment.

$\mu$ is the overall mean of the response $y_{ij}$

$\tau_i$ is the effect on the response of $ith$ treatment.

$\mu_i = \mu + \tau_i$

here $\mu_i$ denotes the true response  of the $ith$ treatment.

and $\epsilon_{ij}$ is the random error term represent the sources of nuisance variation that is, variation due to factors other than treatments.

the assumption is $\epsilon_{ij}\sim^{iid} N(0,\sigma^2)$

By Least Square Estimation Procedure we estimate $\mu\quad and\quad\tau_i$.

Why is it also important to estimate $\sigma^2$?
 If we do not estimate it , what will be the effect?

Any help including reference will be appreciated.

答: Without estimating the error variance you cannot estimate the variability in your parameters and hence cannot conduct hypothesis tests.</p>


问:
what is the final velocity of the combined mass
A 0.25 arrow with a velocity of 12 m/s west strikes and pierces the center of a movable 6.8 kg target.

What is the final velocity of the combined mass?

答:
Hint: Combine the mass of the arrow and target. $P = mv$.

$v=\frac{m_1v_1+m_2v_2}{m_1+m_2}$


问:
Finding the limit of the complex function
I am asked to find the limit $$\lim_{z \to \infty_{\mathbb{C}}} \frac{x^2y-x-2}{1-x}.$$ I have no clue how to solve this because we never covered limits like these?

答: Hint: what happens when $x$ is fixed (for example $x=2$) and $y$ goes to infinity? And what happens when $y=1/x$ and $x$ goes to infinity?


问:
trigonometric equations finding $x$, when $\cos(x)=0$ or some value
I'm practicing trigonometric equations where we need to find the value of $x$, however sometimes I keep getting a completely different solution than the one in the book and I don't really understand where I am making a mistake (and it's always in finding the $x$ in the end). For example:

Problem: $$\cos x  + \sqrt3\cos 2x + \cos3x=0$$ which gives: $$2\cos2x(\cos x + \frac{\sqrt3}{2})=0$$

and then:  $2\cos2x=0$ , where I get that $x=\frac{\pi}{4}+\frac{k\pi}{2}, k\in \Bbb Z$

and:  $\cos x= -\frac{\sqrt3}{2}$, where I get $x=\frac{5\pi}{6}+ 2k\pi, k\in \Bbb Z$ (which based on my book is the only solution I got right)

I believe I also have this: $\cos (2\pi - x)= -\frac{\sqrt3}{2} $, and the solution I got: $x=\frac{7\pi}{6}+ 2k\pi, k\in \Bbb Z$

I guess I'm getting something wrong since the solutions given are:
  $x=\frac{(2k+1)\pi}{2}$ ; $x=\frac{5\pi}{6}+2k\pi$; $x=-\frac{5\pi}{6}+2k\pi$, $k\in \Bbb Z$

答:
$$\cos(nx)=c$$ has the solutions

$$nx=2k\pi\pm\arccos(c)$$ or

$$x=\frac{2k\pi\pm\arccos(c)}n.$$

In the (frequent) event that $\arccos(c)$ is a rational fraction $\dfrac pq$ of $\pi$,

$$x=\frac{2kq\pm p}{qn}\pi.$$

In particular we have the special cases

$c=-1\to\dfrac11\to \dfrac{2k\pm1}n\pi=\dfrac{2k+1}n\pi$,

$c=0\to\dfrac12\to \dfrac{4k\pm1}{2n}\pi=\dfrac{2k+1}{2n}\pi$,

$c=1\to\dfrac01\to \dfrac{2k\pm0}n\pi= \dfrac{2k}n\pi$.


问:
Why $P(z)\text{sin}(z)+Q(z)\text{cos}(z)$ has only finitely many non-real zeroes?
Consider entire functions (defined and holomorphic on the whole complex plane) of the form:
$$f(z)=P(z)\text{sin}(z)+Q(z)\text{cos}(z),$$
where $P(z)$ and $Q(z)$ are polynomials with real coefficients. They seem to always have only finitely many zeros that don't lie on real axis. Namely, there is always some real $M>0$ so that if $|z|>M$ and $f(z)=0$, then $z\in\mathbb{R}$. How can I prove this?

I tried using Cauchy's integral theorem and integral formula, I tried inverting the function or using logarithms, but none worked. Intuitively, I think the point is that in order to split ("split" because the zeroes would obviously be symmetrical around the real axis) infinitely many zeroes that $\text{sin}$ and $\text{cos}$ have, you (in some way) have to introduce infinitely many saddles (zeros of the derivative), but polynomials can have only finitely many. Does this argument make sense and how could I turn it into a rigorous proof?

答:
I think that your conjecture is correct, and can be proved using Rouché's theorem.

Without loss of generality one can assume that $\deg Q < \deg P$, otherwise consider the function
$$
\begin{align}
 f(z +a) &= \left(  P(z) \cos(a) - Q(z) \sin(a) \right) \sin(z)
 + \left(  P(z) \sin(a) + Q(z) \cos(a) \right) \cos(z) \\
 &= \tilde P(z) \sin(z) + \tilde Q(z) \sin(z)
\end{align}
$$
with a suitably chosen $a \in \Bbb R$.

Then $f(z) = 0$ can be written as
$$ \tag{*}
 \tan(z) = r(z)
$$
where $r$ is a rational function with real coefficients and $\lim_{z \to \infty} r(z) = 0$. In particular we can find an $M > 0$ such that
$$ \tag{1}
|r(z)| < 1/2 \text{ for } |z| \ge M \, .
$$

We also need some estimates for the absolute value of the tangent. Using
$$
 \tan(x+iy) = \frac{\sin(2x) + i\sinh(2y)}{\cos(2x) + \cosh(2y)}
$$
from https://proofwiki.org/wiki/Tangent_of_Complex_Number we have
$$ \tag{2}
 |\tan(x+iy)| \ge \frac{|\sinh(2y)|}{\cosh(2y)+1} \text{ for } y \ne 0 \, ,
$$
and
$$ \tag{3}
 |\tan(x+iy)| = \frac{|\sinh(2y)|}{\cosh(2y)-1} > 1 \text{ for $x = \pi/2 + k \pi$ with $k \in \Bbb Z$.}
$$

Using $(2)$ and increasing $M$ if necessary, we have
$$ \tag{4}
 |\tan(x+iy)| > \frac 12 \text{for $|y| \ge M$.}
$$

It follows from $(1)$ and $(4)$ that $(*)$ has no solutions with $\operatorname{Im}(z) \ge M$.

Now consider the rectangles
$$
 R_k = \{ x+iy \mid \frac 12 \pi + k\pi \le x \le \frac 32 \pi + k\pi, |y| \le M \}
$$
with integers $k$ satisfying $\frac 12 \pi + k\pi > M$ or $\frac 32 \pi + k\pi < -M$.

Then $(1)$, $(3)$ and $(4)$ imply that $|\tan(z)| > 1/2 > r(z)$ for all $z$ on the boundary of $R_k$, and it follows from Rouchés theorem that $\tan(z) -r(z)$ and $\tan(z)$ have the same number of zeros in $R_k$, which is one.

On the other hand it follows from the intermediate value theorem that $\tan(x) - r(x)$ has one zero in the interval $\frac 12 \pi + k\pi \le x \le \frac 32 \pi + k\pi$. (This is where we use that the rational function is real on the real axis!)

So we have shown the following:

$\tan(z) -r(z)$ has no zeros with $\operatorname{Im}(z) \ge M$.

For sufficiently large $|k|$, $\tan(z) -r(z)$ has exactly one zero in each strip $\frac 12 \pi + k\pi \le x \le \frac 32 \pi + k\pi$, and that zero is real.

In particular, the equation $(*)$ has only finitely many non-real solutions.


问:
The integral over a subset is smaller?
In a previous question I had $A \subset \bigcup_{k=1}^\infty R_k$ where $R_k$ in $\Bbb{R}^n$ are rectangles I then proceeded to use the following inequality $\left|\int_A f\right| \le \left|\int_{\bigcup_{k=1}^\infty R_k} f \right|$ which I am not really certain of. Does anyone know how to prove it? If its wrong what similar inequality should I use to prove the result here.</p>

答: The inequality is not true in general (think of an $f$ that is positive on $A$ but such that it is negative outside $A$). But it does hold if $f\geq 0$. This is not an obstacle to you using it, because you would just have to split your function in its positive and negative part. </p>

Title: The integral over a subset is smaller?


问:
Prove that for any sets $A,B$, either there exists a surjection from $A$ to $B$, or vice versa
I would like a proof that for any pair of sets $A,B$, either there exists a surjection from $A$ to $B$, or vice versa. This would prove that the usual order on cardinals has the connexity property.

Also, I would like to know if choice is required.

答:
First of all, the statement is false, since if $A=\varnothing$ and $B\neq\varnothing$ then there are no surjection from either set onto the other. But let's put that problem aside. The statement that any two non-empty sets admit a surjection in one direction or the other is equivalent to the Axiom of Choice, so it cannot be avoided.

Having said that, let me point out that working in $\sf ZF$, the pre-order defined by surjections (it is not provably antisymmetric: Cantor–Bernstein works for injections, but for surjections it requires using $\sf AC$) is a bit awkward to work with, so in general when one defines and studies the order on cardinals, one should attempt to use injections whenever possible.

Now. Here are a few ways to prove this, working in $\sf ZFC$.

If you already know that cardinals are linearly ordered, simply show that there is a surjection from $A$ onto $B$ if and only if there is an injection from $B$ into $A$ (again, assuming it's not the case that exactly one set is empty).*

Well-order both sets, then by the well-ordering comparability theorem, one order is an initial segment of the other, which provides us with an injection, which in turn provides us with a surjection in the other direction.

Use Zorn's lemma/Teichmüller–Tukey lemma with the partial order of partial injections from $A$ to $B$. Then a maximal element in that family is an injection from $A$ to $B$ or vice versa, so it provides us with a surjection in the other direction.

Fix a choice function from non-empty sets of $A$ and from non-empty sets of $B$. Now, suppose there is no surjection from $B$ onto $A$, and by [transfinite] recursion define a function which chooses a point from $A$ and assigns it a point in $B$, such that as long as there are points in $B$ that were not used yet, we keep choosing new ones. We can show that this process must exhaust $B$ well before we exhausted $A$, otherwise a surjection from $B$ onto $A$ would exist. When you're done, either repeat some fixed value (e.g. the last chosen point), or just restart choosing points from $B$.

There are many many equivalents to the Axiom of Choice, so there are probably other ways to prove this as well.

(*) To get a surjection from an injection we do not need the Axiom of Choice, so in fact the "only" use the Axiom of Choice is in producing a linear order to begin with.