答:
Hint:

Use the inverse isomorphism of the Chinese remainder theorem and lil' Fermat:

$1160\equiv0\mod 5$ and $1160\equiv 5\mod 7$, so $1160^n\equiv 0\bmod 5$ for all $n>0$, and $$1160^{602}\equiv 5^{602\bmod 6}=5^2=4\bmod 7.$$

Now  the inverse isomorphism $\mathbf Z/5\mathbf Z\times\mathbf Z/7\mathbf Z\longrightarrow \mathbf Z/35\mathbf Z$ is deduced from a Bézout's relation between $5$ and $7$: as we have $3\cdot 5-2\cdot 7=1$, it is given by
$$(x\bmod 5,y\bmod 7)\longmapsto [y\cdot3\cdot 5-x\cdot 2\cdot 7]\bmod 35. $$


问:
problems about pigeonhole principle

The only thing I can figure out is that I need to use pigeonhole principle. But I don't know how to build the holes in such problems. Any suggestion or hint is appreciated. Thanks!

答:
The idea is to partition the nodes into groups of 2 or 3, so that any two nodes within a group are connected by an edge. For example, try grouping the nodes like the following (the nodes with same color are in the same group):

Can you argue from here that every group can have at most 1 node colored black?

ContentLicense: CC BY-SA 4.0

Title: problems about pigeonhole principle


问:
Using symmetry of the dot product, how can I prove this?
Context of this question is projection perspective of principal component analysis, suppose we have orthonormal basis vectors $b_1, ... , b_m$ of the principal subspace where we project a data point $x_n \in \mathbb{R}^D$ onto subspace $U \subseteq \mathbb{R}^D$ and $\text{dim}(U) = M$, and the projection of $x_n$ onto $U$ is denoted by $\widetilde{x}_n$ such that the euclideann distance between the two points is the minimum.

If we can represent $\tilde{x}_n$ as linear combination of $B = [b_1, ..., b_m]$ as: $$\tilde{x}_n = \sum_{m=1}^M z_{mn}b_m$$
Suppose that we know that $z_{in} = {x_{n}}^\intercal b_i = {b_i}^\intercal x_n$, so now we can write:
$$\tilde{x}_n = \sum_{m=1}^M z_{mn}b_m = \sum_{m=1}^M ({x_{n}}^\intercal b_m)b_m $$

My book says that exploiting the symmetry of the dot product we can write:
$$\tilde{x}_n = (\sum_{m=1}^M b_m {b_m}^\intercal)x_n$$

But I don't understand how? how can we take $x_n$ outside of the summation like this?

答:
We have
$$
\sum_{m=1}^M (x_n^\top b_m)b_m
= \sum_{m=1}^M b_m(b_m^\top x_n)
= \sum_{m=1}^M (b_mb_m^\top) x_n
= \left(\sum_{m=1}^M b_mb_m^\top\right) x_n.
$$


问:
Prove $\int_{1}^{\infty}\widetilde{B}_{2n+1}(x)x^{-(s+2n+1)}dx$ is analytic

Given $B_{n}(x)$ satisfies
  \begin{equation}
    \frac{ze^{xz}}{e^z-1}=\sum_{k=0}^{\infty}\frac{B_k(x)}{k!}z^k
\end{equation}
  which is call Bernoulli polynomial. And $\widetilde{B}_{n}$ is 1-periodic extension of $B_n(x)\vert[0,1)$

Prove that for each $n\in \mathbb{N}$, map $F_n:\{z\in \mathbb{C}: Rez>-2n\}, s \mapsto \int_{1}^{\infty}\widetilde{B}_{2n+1}(x)x^{-(s+2n+1)}dx$ is analytic.

This is an exericse from H.Amann's Analysis.

I have already known that

$B_n(x)=\sum_{k=0}^{k=n}\binom{n}{k}B_kx^{n-k}$

$\widetilde{B}_{2n+1}(x)\in C^{n-2}(\mathbb{R})$

$\int_{k}^{k+1}\widetilde{B}_n(x)dx=0$

By using the above formulas, I obtained which as follows, but I don't know how to continue.
\begin{eqnarray*}
% \nonumber % Remove numbering (before each equation)
  RHS &=& \sum_{n=0}^{\infty}\int_{n}^{n+1}\sum_{k=0}^{2n+1}\binom{2n+1}{k}B_{k}(x-n)^{n-k}x^{-(s+2n+1)}dx \\
  &=&\sum_{n=0}^{\infty}\int_{n}^{n+1}\sum_{k=0}^{2n+1}\binom{2n+1}{k}B_{k}\sum_{i=0}^{n-k}\binom{n-k}{i}x^i(-n)^{n-k-i}x^{-(s+2n+1)}dx \\
   &=&\sum_{n=0}^{\infty}\sum_{k=0}^{2n+1}\sum_{i=0}^{n-k}\binom{2n+1}{k}\binom{n-k}{i}(-1)^{n-k-i}B_{k}n^{n-k-i}\frac{(n+1)^{-s-2n+i}-n^{-s-2n+i}}{-s-2n+i}
\end{eqnarray*}

答:
Let $G(s)=\int_1^\infty \widetilde B_{2n+1}(x) x^{-(s+2n+1)}dx$ and
$G_m(s)=\int_1^m \widetilde B_{2n+1}(x) x^{-(s+2n+1)}dx$.

The outline is the following:

(a) $G_m(s)$ is convergent pointwise to $G(s)$,

(b) $G'_m(s)=-(s+2n+1)\int_1^m  \widetilde B_{2n+1}(x) x^{-(s+2n+2)}dx$,

(c) $G'_m(s)$ is convernget locally uniformly to $-(s+2n+1)\int_1^\infty \widetilde B_{2n+1}(x) x^{-(s+2n+2)}dx$, which is $G'(s)$,

(d) $G(s)$ is holomorphic, and hence analytic.


问:
Calculating the correct z score
The question is: Assume that GRE Verbal scores approximate a normal curve with a mean of 500 and a standard deviation of 100. A university admissions office has decided to admit only those applicants who score in the top 10 percent of the GRE Verbal distribution. In order to be admitted, a student must have received a score of ____ or higher?

I know that this is a problem of converting proportions to z scores and then using the z scores to calculate the GRE score. I did that but got a different z score from the textbook. I got a z score of 2.56 but the textbook inputted a z score of 1.28 i.e half of what I got.

My reasoning is that the top 10 percent of applicants would lie to the right of the area under the graph i.e 0.1 and so I used python to calculate the area under the graph for a proportion of 0.9 to the left (python calculates the left tail).

what am I doing wrong that my z score is twice that of the textbook's?  Help please.

答:
Top $10\%$ is clearly

$$z\geq\Phi^{-1}(0.9)\approx 1.28$$

Then passing to your rv you get

$$x=1.28\times 100+500=628$$

Z-table

Title: Calculating the correct z score


问:
The number of solutions for $|x^2-5|x|+6|=|\cos x|$
I wish to find the number of solutions for $|x^2-5|x|+6|=|\cos x|$ graphically

I can graph $|\cos x|$ to some accuracy but struggle in graphing $|x^2-5|x|+6|$

Is this question solvable graphically? If so could a method to sketch the graph be given.

Thanks in advance.

答:
Let $f(x) = x^2 - 5x + 6 \implies |x^2-5|x|+6| = |f(|x|)|$

So to graph $|x^2-5|x|+6|$:

Graph $f(x) = x^2 - 5x + 6$

Graph $f(|x|)$: Remove the graph on the left side of $y-axis$, then take symmetric with respect to $y-axis$ of the right side of $y-axis$ to the left

Graph $|f(|x|)|$: Take symmetric with respect to $x-axis$ of the below of $x-axis$ (where $y$ is negative) to the top

(Red is $f(x)$, green is $f(|x|)$, purple is $|f(|x|)|$

Now graph $|\cos x|$ and you will see that there are $4$ different solutions:


问:
Remainder of division when $p$ is not prime
Calculate the remainder of the division of $1160^{602}$ by $35$.

What I thought of doing right away is to factor out $602$ until I get to a point where I can apply Fermat's theorem. However, $35$ is not a prime number. How to proceed in this case?

答:
$\ \ \ \,5^{\large 2+6n}\!\bmod 35 = \color{#0a0}5(5^{\large 1+\color{#c00}6n}\!\bmod 7) = 5(5^{\large 1}),\,$ by $\,5^{\color{#c00}{\large 6}}\equiv 1\pmod{\!7}\,$ by Fermat, and

using $\, ab\bmod ac =  \color{#0a0}a(b\bmod c),\,$ the mod Distributive Law to factor out $\,a= \color{#0a0}{5}$ from mod.


问:
positive Integer value of $n$ for which $2005$ divides $n^2+n+1$

How Can I calculate positive Integer value of $n$ for which $2005$
  divides $n^2+n+1$

My try:: $2005 = 5 \times 401$

means $n^2+n+1$ must be a multiple of $5$ or multiple of $401$

because $2005 = 5 \times 401$

now $n^2+n+1 = n(n+1)+1$

now $n(n+1)+1$ contain last digit $1$ or $3$ or $7$

$\bullet $ if last digit of $n(n+1)+1$ not contain $5$. So it is not divisible by $5$

Now how can I calculate it? please explain it to me.

答:
If a number is divisible by $2005=5\cdot401,$ it must be divisible by $5$ and by $401$

Now, as you have identified $n^2+n+1$ is not divisible by $5,$ it can not be divisible by  any multiple of $5$ like  $2005=5\cdot401$

Alternatively,

$$n^2+n+1\equiv0\pmod {2005}\implies n^2+n+1\equiv0\pmod 5$$

$$\implies 4(n^2+n+1)\equiv0\pmod 5\implies (2n+1)^2\equiv-3\equiv2$$

Now, $(\pm1)^2\equiv1\pmod 5,(\pm2)^2\equiv4$

So, there is no solution to $ (2n+1)^2\equiv-3\pmod 5$

ContentLicense: CC BY-SA 3.0


问:
Let $|G|=p^n, p$ a prime, and let $|G:C_G(x)|\leq p$ for all $x \in G$. Then $|G'|\leq p$.

Hi: I could solve (a) and (b). As for (c): Let $\phi:G\to G', \phi(x)=[x,y]$ for $y$ fixed. Then $\phi(gh)=[gh,y]=[g,y]^h[h,y]$. By (b) $[g,y]^h=[g,y]$ and then $\phi(gh)=[g,y][h,y]=\phi(g)\phi(h)$ and $\phi$ is homomorphism. Also $ker(\phi)= C_G(y)$ (easy). Now if $\phi$ were onto, then $G/C_G(y)$ isomorphic to $G'$ and then $|G'|\leq p$. Is $\phi$ onto? If it is I can't prove it.

The two preceding problems in this book are these:

I think they can help in the solution.

答:
At the risk of appearing presumptuous, I would like to present a complete solution to the problem. The inspiration will of course be taken from David Craven's approach (which is - as I also suspect - intimately related to the original proof obtained by Knoche, however I unfortunately have no access to his article), however I would like to bring clearer explanations for statements which are not fully justified in the above.

Let $n \in \mathbb{N}$ be a natural number, $p$ be a prime and $G$ a group of order $p^n$ such that for each $x \in G$ we have $\left(G \colon \mathrm{C}_G(x)\right) \leqslant p$. As indices of $p$-groups are necessarily powers of $p$, this applies in particular to the centralisers, such that for each $x \in G$ we have $\left(G \colon \mathrm{C}_G(x)\right)=p^k$, for a certain $k \in \mathbb{N}$. The condition that all indices of one-element centralisers are at most $p$ finally entails $k \leqslant 1$ and thus $\left(G \colon \mathrm{C}_G(x)\right) \in \{1, p\}$ for any $x \in G$. Since in any nilpotent group - which includes $p$-groups - a subgroup of prime index is necessarily normal, we derive two important conclusions:

$\mathrm{C}_G(x) \trianglelefteq G$ for each $x \in G$ and

the quotient $G/\mathrm{C}_G(x)$ is abelian (either trivial or cyclic of order $p$) for each $x \in G$.

If we denote by $\rho_x \colon G \to G/\mathrm{C}_G(x)$ the canonical surjection for each centraliser of arbitrary element $x \in G$, we consider the diagonal direct product of these morphisms $f \colon G \to \displaystyle\prod_{x \in G}G/\mathrm{C}_G(x)$ and remark that $\mathrm{Ker}f=\displaystyle\bigcap_{x \in G}\mathrm{Ker}\rho_x=\displaystyle\bigcap_{x \in G}\mathrm{C}_G(x)=\mathrm{C}_G(G)=\mathrm{Z}(G)$. This means that $f$ induces an injective quotient $g \colon G/\mathrm{Z}(G) \to \displaystyle\prod_{x \in G}G/\mathrm{C}_G(x)$, on the grounds of which we infer the quotient $G/\mathrm{Z}(G)$ is abelian (being embedded in the right-hand side group, which is also abelian). Since $\mathrm{D}(G)$ is the smallest normal subgroup inducing an abelian quotient, we gather the relation $\mathrm{D}(G) \leqslant \mathrm{Z}(G)$.

For arbitrary $x, y \in G$ let us write $[x, y]\colon=x^{-1}y^{-1}xy$ for the (left) commutator between $x$ and $y$ and let us also employ the notations ${}^tx\colon=txt^{-1}$ respectively $x^t=t^{-1}xt$ for the left respectively right conjugates of $x \in G$ by $t \in G$. Let us also recall the general relation $[xy, z]=[x, z]^y[y, z]$ and introduce - as both the original poster and David Craven have indicated - for each $t \in G$ the map:
$$\begin{align}
h_t \colon G &\to \mathrm{D}(G)\\
h_t(x)&=[x, t].
\end{align}$$
Since the derived subgroup is central, we infer that $[xy, t]=[x, t][y, t]$ for any $x, y, t \in G$ which means that $h_t \in \mathrm{Hom}_{\mathbf{Gr}}(G, \mathrm{D}(G))$ is a group morphism. Let us set $C_t \colon=\mathrm{Im}h_t$ and notice that $\mathrm{Ker}h_t=\mathrm{C}_G(t)$, so that by virtue of the fundamental isomorphism theorem $|C_t| \in \{1, p\}$. In particular, if $t \in G \setminus\mathrm{Z}(G)$ it follows that $|C_t|=p$.

We move on to remark that $x, y \in G$ are such that $[x, y]\neq 1_G$ we automatically have $x, y \in G \setminus \mathrm{Z}(G)$ and $[x, y] \in C_y \setminus \{1_G\}$ respectively $[y, x] \in C_x \setminus \{1_G\}$. Since $[x, y]=[y, x]^{-1}$, we gather that $[x, y] \in \left(C_x \cap C_y\right) \setminus \{1_G\}$ and since cyclic groups of prime orders are generated by any of their nontrivial elements we infer that $C_x=C_y$. By introducing $A \colon=G \setminus \mathrm{Z}(G)$ together with the binary relations $P, R \subseteq A \times A$ given by $xPy \Leftrightarrow [x, y] \neq 1_G$ respectively $xRy \Leftrightarrow C_x=C_y$, we can formulate this conclusion by stating $P \subseteq R$. The case $A=\varnothing$ is trivial as far as our problem is concerned so we assume that $A \neq \varnothing$ in what follows. This means we can write $|\mathrm{Z}(G)|=p^m$ for a certain $m \leqslant n-1$.

We also remark that $R$ is an equivalence and that $P$ is symmetric and irreflexive. Furthermore, let us consider elements $x, y \in P$ such that $P\langle x \rangle \cap P\langle y \rangle \neq \varnothing$, in other words such that there exists $z \in A$ with $[x, z], [y, z] \neq 1_G$. As $P$ is symmetric, we then automatically have $(x, y) \in P \circ P \subseteq R$. We use this remark to show that $R=A \times A$, in other words that for any two elements $x, y \in A$ we have $C_x=C_y$ and we proceed by establishing that necessarily $P\langle x \rangle \cap P\langle y \rangle \neq \varnothing$.

Assuming the contrary we would derive the existence of particular $a, b \in A$ such that $P\langle a \rangle$ and $P\langle b \rangle$ are disjoint. However, we have by definition that $P\langle x \rangle=G \setminus \mathrm{C}_G(x)$, hence $|P\langle x \rangle|=p^n-p^{n-1}$ for each $x \in A$. It would follow that $|P\langle a \rangle \cup P \langle b \rangle|=|P\langle a \rangle|+|P\langle b \rangle|=2\left(p^n-p^{n-1}\right) \leqslant |A|=p^n-p^m$. This cardinal relation leads – after rearrangement and simplification – to $p^{n-m} \leqslant 2p^{n-m-1}-1$, variant which can not hold for the following general reason: for any $k \in \mathbb{N}$ one has $2p^{k}-1 <2p^k \leqslant p^{k+1}$, and this applies in particular to $k=n-m-1$.

The previous argument shows that – having fixed a certain $a \in A$ – we have $C_x=C_a$ for every $x \in A$. Thus, for arbitrary $x, y \in G$ we have either $[x, y]=1_G \in C_a$ for trivial reasons or $[x, y] \neq 1_G$ which automatically means that $xPy$ and hence $C_x=C_y=C_a$. Since by construction $[x, y] \in C_y$ we once again infer that $[x, y] \in C_a$. It is then immediate that $\mathrm{D}(G) \leqslant C_a$ and as $G$ is not abelian (the centre being a proper subgroup) $\mathrm{D}(G) \neq \{1_G\}$. Since $C_a$ is cyclic of order $p$, it follows with necessity that $\mathrm{D}(G)=C_a$ and hence – in this case – $|\mathrm{D}(G)|=p$.


问:
Conditions under which a $2\times2$ block matrix has complex eigenvalues
Consider a matrix $A$ in $\mathbb{R}^2$:

\begin{equation}
A=\begin{bmatrix}
0 & -c \\
1 & -b
\end{bmatrix}
\end{equation}

Then it can be shown that the matrix has complex eigenvalues if $b^2-4c < 0$. Can a similar relation be derived also for the elements of a similarly shaped 2x2 block matrix, ie:

\begin{equation}
M=\begin{bmatrix}
0_n & -C \\
I_n & -B
\end{bmatrix}
\end{equation}

where $B, C$ are matrices in $\mathbb{R}^n$, such that $\lambda_M$ are complex?

答:
First we note that the elements $(2,1)$ and $(2,2)$ of matrix $M$ commute. Then we consider:

\begin{equation}
(\lambda I_{2n}-M)v = 0
\end{equation}

which is the eigenvalue problem of $M$. The equation admits a solution $v \neq 0$ when $|\lambda I_{2n}-M| = 0$, so we resort to computing the determinant of the characteristic matrix:

\begin{equation}
\begin{bmatrix}
\lambda I_n & -C \\
I_n & \lambda I_n + B \\
\end{bmatrix}
\end{equation}

We assume that $(1,1)$ is invertible, i.e., that $0$ is not an eigenvalue of $M$. Then since $I_n$ commutes with any matrix, we can write:

\begin{equation}
|\lambda I_{2n} - M| = |\lambda ( \lambda I_n + B) + C | = |\lambda^2 I_n + \lambda B + C| = 0
\end{equation}

This is a Quadratic Eigenvalue Problem and has been intensively studied. Particularly, in (Tisseur, 2001), properties of the solutions $\lambda$ have been given in terms of the coefficient matrices $M, C, K$ of $\lambda^2 M + \lambda C + K = 0$. When the three matrices are Hermitian, then the eigenvalues are some solutions of:

\begin{equation}
\lambda = \frac{-c(x) \pm \sqrt{c(x)^2 - 4k(x)m(x)}}{2m(x)}
\end{equation}

with $m(x) = x^*Mx,c(x) = x^*Cx, k(x) = x^*Kx$. Then in this case it can be seen that a sufficient, but not necessary condition for $\lambda$ to be complex is that $c(x)^2 < 4k(x)m(x)$ $\forall x$, which in our case reduces to

\begin{equation}b(x)^2 < 4c(x) \text{  }  \forall x\end{equation}

with $b(x) = x^* B x, c(x) = x^* C x$ (pay attention to the different meaning of $c(x)$ in the problem of Tisseur and the answer's problem).

If $B, C$ are Hermitian (which, being them real, implies symmetry), then $\lambda_{C,n} \leq c(x) \leq \lambda_{C,1}$ and $\min(|\lambda_{B}|)^2 \leq b(x)^2 \leq \max(|\lambda_{B}|)^2$, so a (very) broad sufficient condition is that:

\begin{equation}
 \min(\lambda_C) > \bigg(\frac{\max(|\lambda_{B}|)}{2}\bigg)^2
\end{equation}

If they are not Hermitian, then the same condition can be stated in terms of their symmetric parts since $b(x) = x^*Bx = \frac12 x^*(B+B^T)x$ and similarly for $c(x)$.


问:
Null set of Bernoulli distribution
I have an assignment in which I am supposed to determine the null set of the Bernoulli distribution w.r.t. the Borel $\sigma$-algebra on $\mathbb{R}$. Specifically:

Let $(\mathbb{R}, \mathscr{B},P)$ be a measure space, where $\mathscr{B}$ denotes the Borel $\sigma$-algebra on $\mathbb{R}$ and $P$ the probability measure Bernoulli distribution i.e. P is characterized by $$P\left(\{0\}\right)=p,\quad P(\{1\})=1-p,\quad 0<p<1$$
Determine $N_P=\{N\subseteq\mathbb{R}| \text{There exists $A\in \mathscr{B}$ so $N\subseteq A$ and $P(A)=0$}\}$

I don't what to do. I don't understand how the Bernoulli distribution is defined on $\mathscr{B}$. How would you evaluate any open interval on the real number line for example.

答:
Hint:

For $\ A\in\mathscr{B}\ $, $\ P(A)\ $ must have one of four values:
$$
P(A)=\cases{0&if $\ 0\not\in A$ and $\ 1\not\in A$\\
p&if $\ 0\in A\ $ but $\ 1\not\in A$\\
1-p&if $\ 1\in A\ $ but $\ 0\not\in A$\\
1&if $\ 0\in A\ $ and $\ 1\in A\ $.}
$$

Title: Null set of Bernoulli distribution


问:
$(x^2 + 3)(x^3 - q) = 0$
I'm trying to show that $(x^2 + 3)(x^3 - q) = 0$ is a counter-example to the Hasse Principle when $q \equiv 1 $ (mod $27$) is prime.
That is the above equation has solutions in $\Bbb{R}$, $\Bbb{Q}_p$ for all primes $p$ (p-adic integers) but none in $\Bbb{Q}$.

It is clear that the equation has solutions in $\Bbb{R}$ and it's also clear it doesn't have any in $\Bbb{Q}$.

For the $\Bbb{Q}_p$ case, I have shown that it has solutions when $p \equiv 1$ modulo 3 as when $p \equiv 1$ (mod 3) then -3 is a quadratic residue, and so the equation has a solution in $\Bbb{F}_p$ which lifts to one in $\Bbb{Q}_p$ by Hensel's Lemma.

I have also show than the equation has a solution in $\Bbb{Q}_p$ when $p = 3$.

Thus I'm left to showing that it has solutions in $\Bbb{Q}_p$ when $p \equiv 2$ modulo 3.
I'm stuck on this part! I believe it will come down to showing there is some $a$ such that $a^3 \equiv q$ (mod p) for all such primes $p$, and then such a solution can be lifted by Hensel's lemma. However I've no idea how to do this.

Any help with showing this much appreciated.

答:
I have no idea how to show there is some $a$ such that $a^3 \equiv q$ (mod p) for all primes $p\equiv2\pmod3$

When $p\equiv2\pmod3$, we can say that $2p-1\equiv0\pmod3$, and then the solution

to  $a^3\equiv q\pmod p$   is $a\equiv q^{(2p-1)/3}\pmod p$, because, by Fermat's little theorem,

then $a^3\equiv(q^{(2p-1)/3})^3\equiv q^{2p-1}\equiv q^pq^{p-1}\equiv q\cdot q^{p-1}=q^p\equiv q\pmod p$.

Title: $(x^2 + 3)(x^3 - q) = 0$


问:
How can I rewrite $\frac{1}{(1-z)^4}$ as a power series?
I'm supposed to rewrite $\frac{1}{(1-z)^4}$ as a power series, using the Cauchy-Product. It is given that $|z| < 1$.

So far I have
\begin{aligned}
\frac{1}{(1-z)^4} &= \left( \frac{1}{(1-z)^2} \right)^2\\
&= \left( \sum_{n=0}^{\infty} (n+1)z^n \right)^2\\
&=  \sum_{n=0}^{\infty} \sum_{k=0}^{n} (n-k+1)z^{n-k}kz^k  \\
&=  \sum_{n=0}^{\infty} \sum_{k=0}^{n} k(k+1)z^{n}  \\
\end{aligned}
First of all, I'm not sure whether every step was correctly so far. Additionally, I don't know how I can reduce that to a power series.

答:
The third line should be
\begin{eqnarray*}
&=  \sum_{n=0}^{\infty} \sum_{k=0}^{n} (n-k+1)z^{n-k} \color{red}{(k+1)}z^k  \\
\end{eqnarray*}
So we need to perform the sum
\begin{eqnarray*}
 \sum_{k=0}^{n} (n-k+1)(k+1) = -\sum_{k=0}^{n} k^2+ n \sum_{k=0}^{n} k +(n+1) \sum_{k=0}^{n} 1.
\end{eqnarray*}
Recall that
\begin{eqnarray*}
 \sum_{k=0}^{n}  k^2= \frac{n(n+1)(2n+1)}{6}
\end{eqnarray*}
and we have
\begin{eqnarray*}
 \sum_{k=0}^{n} (n-k+1)(k+1) = \frac{(n+1)(n+2)(n+3)}{6}.
\end{eqnarray*}


问:
Deriving the gradient of a multivariable function taking matrices as input?
I've been tasked with deriving the gradient for the following non-convex function and I'd like to verify whether my solution / progress is correct so far (I'm not too confident with calculus):

$$F(U, V) = \dfrac{1}{2}\sum_{(i,j)\in{\Omega}}(M_{ij}-u_iv_{j}^T)^2 + \frac{\lambda}{2}(||U||^2_F + ||V||^2_F)$$

Where $M, U, V$ are matrices, $u_i$ and $v_j$ refer to the $i$th row and $j$th row of $U$ and $V$ respectively, and $\lambda$ is a variable denoting some learning rate.

I've obtained the following so far for the partial derivative with respect to $U$:

$$\dfrac{\partial F(U,V)}{\partial U} =-\sum_{(i,j)\in{\Omega}}(M_{ij}-u_iv_{j}^T)^2 + \lambda(||U||^2_F)$$

But I'm a bit unsure about taking the derivative for $\frac{\lambda}{2}(||U||^2_F + ||V||^2_F)$. I think what I have above is correct as $V$ would be treated like a constant (since we're doing the partial derivation with respect to $U$). I assume the approach for taking the partial derivative with respect to $V$ would be the same.

答:
The tricky part here is computing
$$
\frac{\partial}{\partial U}\sum_{(i,j)\in{\Omega}}(M_{ij}-u_iv_{j}^T)^2.
$$
One approach is to go entry by entry. Note that the $p,q$ entry of this partial derivative is equal to the partial derivative of the sum with respect to the $p,q$ entry of $U$. We note that
$$
\begin{align}
\frac{\partial}{\partial U_{pq}} \sum_{(i,j)\in{\Omega}}(M_{ij}-u_iv_{j}^T)^2 &=
\sum_{(i,j)\in{\Omega}} \frac{\partial}{\partial U_{pq}}(M_{ij}-u_iv_{j}^T)^2
\\ & =
\sum_{j \in S_p} \frac{\partial}{\partial U_{pq}}(M_{pj}-u_pv_{j}^T)^2
\\ & =
\sum_{j \in S_p} 2(M_{pj}-u_pv_{j}^T)\frac{\partial}{\partial U_{pq}}(M_{pj}-u_pv_{j}^T)
\\ & = \sum_{j \in S_p} 2(M_{pj}-u_pv_{j}^T)(-V_{jq})
= -2\sum_{j \in S_p} (M_{pj}-u_pv_{j}^T)V_{jq}.
\end{align}
$$
Here, $S_p = \{j: (p,j) \in \Omega\}$.

If the known entries form a submatrix, then let $J_1$ denote the matrix whose rows are the rows of the identity matrix corresponding to the selected rows of $M$, and let $J_2$ denote the matrix whose columns are the columns are the columns of the identity matrix corresponding to the selected columns. We can then write the sum as
$$
\sum_{(i,j)\in{\Omega}}(M_{ij}-u_iv_{j}^T)^2 = \|J_1(M - UV^T)J_2\|_F^2.
$$
We compute
$$
\begin{align}
\|J_1(M - &[U+H]V^T)J_2\|_F^2 =
\operatorname{tr}([J_1(M - [U+H]V^T)J_2][J_1(M - [U+H]V^T)J_2]^T)
\\ & =
\|J_1(M - [U+H]V^T)J_2\|_F^2
-2 \operatorname{tr}([J_1HV^TJ_2][J_1MJ_2]^T) + o(\|H\|)
\\ & =
\|J_1(M - [U+H]V^T)J_2\|_F^2
-2 \operatorname{tr}([V^TJ_2][J_1MJ_2]^TJ_1H) + o(\|H\|)
\\ & =
\|J_1(M - [U+H]V^T)J_2\|_F^2 +
 \operatorname{tr}([-2J_1^TJ_1MJ_2J_2^TV]^TH) + o(\|H\|).
\end{align}
$$
Correspondingly, we compute
$$
\frac{\partial}{\partial U}\|J_1(M - [U+H]V^T)J_2\|_F^2
= -2 J_1^TJ_1MJ_2J_2^TV.
$$
To make some sense of this: we can write this product in the form $-2 J_1^T[J_1MJ_2][J_2^TV]$. $[J_1MJ_2]$ is the known submatrix of $M$, and $V$ is the submatrix of $V$ obtained by taking only the rows of $V$ corresponding to known columns of $M$.


问:
Translation of a bounded uniformly continuous function is fixed

if $f$ is a bounded uniformly continuous real-valued function on $\Bbb R$, then for all $T>0$, there is a sequence $a_n$ converging to infinity that $f(x+a_n)$ converges on $[0,T]$

I am trying to prove this beautiful result which is saying the figure of the function is fixed on a interval under translation. But I have trouble constructing such $a_n$

答: I proved it, I guess. By showing $\{f(x+a): a\in \Bbb R\}$ is equicontinuous and bounded. Therefore, I can apply Ascoli on $C[0,T]$


答:
In below images, the six green triangles cover all vertices except the one marked red. Hence either one of the green triangles has two of the eight black nodes (and we are done), or the red node in each case is black.
But that gives us two adjacent black nodes!

Title: problems about pigeonhole principle

Body: The only thing I can figure out is that I need to use pigeonhole principle. But I don't know how to build the holes in such problems. Any suggestion or hint is appreciated. Thanks!


问:
Elliptic integrals, general form
According to https://mathworld.wolfram.com/EllipticIntegral.html, Elliptic integrals are integrals of the form of equation (1).

I am trying to convert an integral of the form

$$\int_{|x_+|}^\infty \frac{dx}{\sqrt{(x^2-x_-^2)(x^2-x_+^2)}}$$

To express as a function of elliptic integrals, where $|x_+|>|x_-|$, we can take $x_+>0$ for simplicity. We can easily see that this integral converges, and that it is real.

I have trying following the method outline in the webpage I linked, with no success, because of the arising of imaginary roots in the process. However, plugging this expression into mathematica with $x_-=1$, $x_+=2$, yields me the expression :

$$\int_{2}^\infty \frac{dx}{\sqrt{(x^2-1)(x^2-4)}}=\frac{i}{2}K(\frac{3}{4})+K(4)$$

Where $K(m) = \int_0^{\pi/2}\frac{1}{\sqrt{1-m\sin^2(\phi)}}d\phi$. Now clearly the integral I'm looking to re-express is well-defined, and real, but it would appear I get an complex number as an answer. However, looking closer, $K(4)$ will be complex because the parameter $m>1$. Thus, if we numerically compute the expression the imaginary parts cancel and we are left with a real answer, which is correct.

So my question is as follows : when we say that Elliptic integrals (as defined in my link) are expressible as a function of the three "kinds" of elliptic integrals (+ elementary functions), do we allow "complex" elliptic integrals ? In other words, is there any way to express my example as a function of "well defined" elliptic integrals, in the sense that we only integrate real numbers ?

I am unsure if I expressed myself clearly, I hope it makes sense.

答:
One way to express this integral as a function of the Elliptic Integral K, is as follows. Do the change of variables $x^2 = \frac{x_0+^2}{1-t^2}$. This converts the integral into :
$$
\frac{1}{\sqrt{x_+^2-x_-^2}}\int_0^1 \frac{dt}{\sqrt{(1-t^2)(1+\frac{x_-^2}{x_+^2-x_-^2}t^2)}}
$$

Which can be written formally as $\frac{1}{\sqrt{x_+^2-x_-^2}}K(-\frac{x_-^2}{x_+^2-x_-^2})$. This is fine, but there is an abuse of notation. Usually Elliptic functions are defined for $K(m), 0<m<1$, but here we have a negative $m$. Nonetheless, the integral is defined and convergent, but in principle it is not an Elliptic integral according to the definition.

Title: Elliptic integrals, general form


问:
Integral equation which is equal to $\pi$
$$ \int_{0}^{\frac{1}{\sqrt2}}{\frac{4 \sqrt{2}-8x^{3}-4\sqrt{2}x^4-8x^{5}}{1-x^{8}}} dx = \pi$$

I can use without a proof that $(y-1)(y^8-16)=(y^2-2)(y^2-2y+2)(y^5+y^4+2y^3-4)$

I guess that I could use the substitution: $x=\frac{y}{\sqrt2}$ and after it a partial fraction decomposition but as I have started to do it I have seen how can I continue it.

答:
Using, as you did, $x=\frac y{\sqrt 2}$,$$I=\int{\frac{4 \sqrt{2}-8x^{3}-4\sqrt{2}x^4-8x^{5}}{1-x^{8}}} dx=\int \frac{-y^5-y^4-2 y^3+4}{1-\frac{y^8}{16}}\,dy$$
The roots of the numerator are
$(1,1\pm i,\pm i \sqrt 2)$ and among those of denominator are $(\pm \sqrt 2,\pm i\sqrt 2)$. So, a lot of simplifications can be done to arrive at
$$I=\int \frac{ 16(y-1)}{y^4-2 y^3+4 y-4}\,dy$$  Now
$$y^4-2 y^3+4 y-4=\left(y^2-2\right) \left(y^2-2 y+2\right)$$ Now partial fraction decomposition
$$\frac{ 16(y-1)}{y^4-2 y^3+4 y-4}=\frac{4 y}{y^2-2}-\frac{4 (y-2)}{y^2-2 y+2}$$It does not look too hard now.


问:
Find all reall numbers $A$, $w$ and $\delta$ such that $\sin t+\cos t=A\sin(\omega t+\delta)$, for all $t$.
I need help with this exercise.

Find all reall numbers $A$, $w$ and $\delta$ such that $\sin t+\cos t=A\sin(\omega t+\delta)$, for all $t$.

I really don't have a clue where to start. I will appreciate any hint that helps me to start working on it.

答:
You can start by recognizing that

$$\sin t + \cos t = \sqrt2 \left(\sin t \cos \frac{\pi}4 + \cos t \sin \frac{\pi}4\right) = \sqrt2 \sin \left(t + \frac{\pi}4\right)$$

Now what can we say about $A$, $\omega$, and $\delta$?


问:
What's the hard part of zero?
A lot of textbooks said it was hard for human to accept zero when it was first introduced.

How could it be? It seems to me as natural as positive integer which represent there is no elements at all.

答:
There are actually five different kinds of zero.  Just because some parts of zero are not implemented, it does not mean that any have.  Egypt did have a zero, but only in the sense of 'empty bag'.

If you have stones in columns of your abacus, the number system allows you to convert these into some portable form, so someone else can repeat the pattern.  In token systems, one gives the name of a token for each stone.  Positions with no tokens need no name.  Here is the roman layout for such a thing.  A number like 56, is an L, a V and an I stone, ie LVI.

```<code>      D     L     V
   -------------------      The inca used a similar based on 1 and 5 in
     CC    XX    II         each column of 20.
     CC    XX    II
</code>```

The demotic system, repeated into alphabetic systems, use a system where there is a seperate symbol for 'seven stones in the tens-column'.  Like the tokenised system, there is no need to present an empty column.

You only need empty column-markers, when the symbols consist purely of 'six stones in this column'.  The usual implementations are to skip outside columns, and skip internal columns.

For example, the greekish system (borrowed by the arabs etc), is to use the '10' letter as a kind of zero (which is why it appears at the end of the row of numbers).

Leading zeros appear in division-bases, like our time system (division-base 60), where in 0001 it means 0 in the first column and 1 in the second column.  In base 60, there is an alternation between the top and bottom half of the abacus (which is presented as horizontal vs vertical arrows).  Zeros were represented by points (ie sentence stops).

```<code>     h    m    s
     0    0    0        00:01:00 and 00:01:  mean the same thing!
     0    1    0
</code>```

Trailing zeros appear in multiple-bases, where the units are the lowest column, so we have

```<code>     m    c    x    i
     4    0    3    0
</code>```

Both the 0's at c and i are needed, the first says an empty column lies between m and x, and the second says the last quoted digit is an x, rather than an i.

Semimedial zeros do not occur in decimal, but you can readily find these in base 60.  The actual implemetation of the sumerian digits is not eg '35' as a digit, but two different runs of digits, eg

```<code>                  Sumerian                                Mayan

                 C2                 C   2
   A-F         C = 30          C = 30      -             A-C  (5,10,15)
  ----------              vs                            ----
   1-9         2  = 2             -       2 = 2          0-4

                 32                30  :   02
</code>```

Both of these would be written C (three horizontal rays) followed by 2, two verticals, but we would write these as 32 vs 30.02  The usual is to space the C from the 2, to show that there are empty places, ie C 2 vs C2.

The Mayans never had this problem, for the reason that the lower position 0-4 is always written where the column is significant.  It's kind of like writing base 30 numbers as 0 to 9, with an optional ' and " for +10 and +20 respectively: eg 5 = 5, 5' = 15, 5" = 25.

Title: What's the hard part of zero?

Body:
There are actually five different kinds of zero.  Just because some parts of zero are not implemented, it does not mean that any have.  Egypt did have a zero, but only in the sense of 'empty bag'.

If you have stones in columns of your abacus, the number system allows you to convert these into some portable form, so someone else can repeat the pattern.  In token systems, one gives the name of a token for each stone.  Positions with no tokens need no name.  Here is the roman layout for such a thing.  A number like 56, is an L, a V and an I stone, ie LVI.

      D     L     V
   -------------------      The inca used a similar based on 1 and 5 in
     CC    XX    II         each column of 20.
     CC    XX    II

The demotic system, repeated into alphabetic systems, use a system where there is a seperate symbol for 'seven stones in the tens-column'.  Like the tokenised system, there is no need to present an empty column.

You only need empty column-markers, when the symbols consist purely of 'six stones in this column'.  The usual implementations are to skip outside columns, and skip internal columns.

For example, the greekish system (borrowed by the arabs etc), is to use the '10' letter as a kind of zero (which is why it appears at the end of the row of numbers).

Leading zeros appear in division-bases, like our time system (division-base 60), where in 0001 it means 0 in the first column and 1 in the second column.  In base 60, there is an alternation between the top and bottom half of the abacus (which is presented as horizontal vs vertical arrows).  Zeros were represented by points (ie sentence stops).

     h    m    s
     0    0    0        00:01:00 and 00:01:  mean the same thing!
     0    1    0

Trailing zeros appear in multiple-bases, where the units are the lowest column, so we have

     m    c    x    i
     4    0    3    0

Both the 0's at c and i are needed, the first says an empty column lies between m and x, and the second says the last quoted digit is an x, rather than an i.

Semimedial zeros do not occur in decimal, but you can readily find these in base 60.  The actual implemetation of the sumerian digits is not eg '35' as a digit, but two different runs of digits, eg

                  Sumerian                                Mayan

                 C2                 C   2
   A-F         C = 30          C = 30      -             A-C  (5,10,15)
  ----------              vs                            ----
   1-9         2  = 2             -       2 = 2          0-4

                 32                30  :   02

Both of these would be written C (three horizontal rays) followed by 2, two verticals, but we would write these as 32 vs 30.02  The usual is to space the C from the 2, to show that there are empty places, ie C 2 vs C2.

The Mayans never had this problem, for the reason that the lower position 0-4 is always written where the column is significant.  It's kind of like writing base 30 numbers as 0 to 9, with an optional ' and " for +10 and +20 respectively: eg 5 = 5, 5' = 15, 5" = 25.


问:
Partial Differentiation using chain rule
I am trying to express $\frac{\partial^2u}{\partial y^2}$ and $\frac{\partial^2u}{\partial x\partial y}$ of the function $u=f(x,y,g(x,y))$ with partial derivative notation of $f$ and $g$.

However I am having a hard time doing so because I am a bit confused with the notation.

I know that $\frac{\partial u}{\partial y}=\frac{\partial f}{\partial y}+ \frac{\partial f}{\partial g}\frac{\partial g}{\partial y}$. But the problem is that I am not sure how to express $\frac{\partial^2u}{\partial y^2}$. If I differentiate the $\frac{\partial u}{\partial y}$ by $y$, I am thinking I will get $\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 f}{\partial y^2}+ ...$ , but I am not quite sure how the second term of the right hand side ($\frac{\partial f}{\partial g}\frac{\partial g}{\partial y}$) will look like. Similarly, I have trouble in expressing $\frac{\partial^2u}{\partial x\partial y}$ using partial differentiation notation of $f$ and $g$.

How should it be done?

Thanks.

答:
Edit: As pointed out by Ted Shifrin, it's a bit misleading to write $\frac{\partial f}{\partial g}$ when in reality I mean $\frac{\partial f}{\partial z}$ because $f$ is a function of three variables: $x,y,z$.  I just so happens that everywhere it's relevant we evaluate $f$ at $(x,y,g(x,y))$.

It becomes clearer when you add back in the arguments.  You already have
$$\frac{\partial u}{\partial y}=\frac{\partial f}{\partial y}(x,y,g(x,y))+ \frac{\partial f}{\partial z}(x,y,g(x,y))\frac{\partial z}{\partial y}(x,y).$$  So, to find $\frac{\partial^{2} u}{\partial y^{2}}$, we compute the derivatives the same way again.
\begin{align}
\frac{\partial^{2} u}{\partial y^{2}} &=\frac{\partial}{\partial y}\frac{\partial f}{\partial y}(x,y,g(x,y))+ \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial z}(x,y,g(x,y))\frac{\partial g}{\partial y}(x,y)\right)\\
&=\frac{\partial^{2} f}{\partial y^{2}}(x,y,g(x,y)) + \frac{\partial^{2}f}{\partial z\partial y}(x,y,g(x,y))\frac{\partial g}{\partial y}(x,y)\\ &\quad + \left(\frac{\partial^{2}f}{\partial y\partial z}(x,y,g(x,y)) + \frac{\partial^{2}f}{\partial z^{2}}(x,y,g(x,y))\frac{\partial g}{\partial y}(x,y)\right)\frac{\partial g}{\partial y}(x,y)\\ &\quad+ \frac{\partial f}{\partial z}(x,y,g(x,y))\frac{\partial^{2} g}{\partial y^{2}}(x,y)
\end{align}

Suppressing the arguments again, we have
$$\frac{\partial^{2} u}{\partial y^{2}} = \frac{\partial^{2} f}{\partial y^{2}} + \frac{\partial^{2}f}{\partial z\partial y}\frac{\partial g}{\partial y} + \left(\frac{\partial^{2}f}{\partial y\partial z} + \frac{\partial^{2}f}{\partial z^{2}}\frac{\partial g}{\partial y}\right)\frac{\partial g}{\partial y}+ \frac{\partial f}{\partial z}\frac{\partial^{2} g}{\partial y^{2}} $$

which if you know that the second order partial derivatives are continuous becomes
$$\frac{\partial^{2} u}{\partial y^{2}} = \frac{\partial^{2} f}{\partial y^{2}} + 2\frac{\partial^{2}f}{\partial y\partial z}\frac{\partial g}{\partial y} + \frac{\partial^{2}f}{\partial z^{2}}\left(\frac{\partial g}{\partial y}\right)^{2}+ \frac{\partial f}{\partial z}\frac{\partial^{2} g}{\partial y^{2}}.$$  You can compute the other second order partial derivative in a similar manner.  Suppressing the arguments this time from the start:
\begin{align}
\frac{\partial^2u}{\partial x\partial y} &= \frac{\partial}{\partial x}\frac{\partial f}{\partial y}+ \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial z}\frac{\partial g}{\partial y}\right)\\
&=\frac{\partial^{2}f}{\partial x\partial y} + \frac{\partial^{2}f}{\partial z\partial y}\frac{\partial g}{\partial x} + \left(\frac{\partial^{2}f}{\partial x\partial z} + \frac{\partial^{2}f}{\partial z^{2}}\frac{\partial g}{\partial x}\right)\frac{\partial g}{\partial y} + \frac{\partial f}{\partial z}\frac{\partial^{2}g}{\partial x\partial y}.
\end{align}


问:
To show that $f$ is an orthogonal linear map, if $f(0) = 0$ and $f$ preserves the distance

Let $V$ be a finite-dimensional inner product space, and $f:V\to V$ be a map such that $f(0) = 0$ and $\forall x,y\in V$, $||f(x) - f(y)|| = ||x-y||$. Show that $f$ is an orthogonal linear map.

I've been able to show only a part of what's required, and my work is as follows:

Since $||f(x) - f(y)|| = ||x-y||$ holds for all $x,y$, put $y=0$. This gives $||f(x)|| = ||x||$ for all $x\in V$. $f:V\to V$ is a norm-preserving map, and so it is orthogonal if it is linear (proof of this is given at the end).

Now, how do I show that $f$ is linear? Is there a nice way of doing this? I'd appreciate any hints.

Proof of statement above:

I know that a map $T:V\to W$ is orthogonal if $\forall x,y\in V$, we have $\langle Tx,Ty\rangle = \langle x,y\rangle$.

$f$ is norm-preserving, so for arbitrary $x,y\in V$, we have $||f(x+y)|| = ||x+y||$ and hence $||f(x+y)||^2 = ||f(x)+f(y)||^2 =  ||x+y||^2$. Expanding, we get $||f(x)||^2 + ||f(y)||^2 + 2\langle f(x),f(y)\rangle = ||x||^2 + ||y||^2 + 2\langle x,y\rangle$ and using $||f(x)||=||x||$ and $||f(y)|| = ||y||$ we get $\langle f(x),f(y)\rangle = \langle x,y\rangle$ as desired.

答:
Hint

All isometry can be written as
$$f(v)=\ell(v)+b,$$
where $\ell:V\to V$ is a linear map.


问:
How do you prove that $e^{ia} + 1 = 2e^{i\frac{a}{2}}\cos(\frac{a}{2})$?
How do you prove $e^{ia} + 1 = 2e^{i\frac{a}{2}}\cos(\frac{a}{2})$ where a is a real number?

I attempted to solve it, by using the second side and I couldn't find a correct result, sadly. I'm not even sure if my attempt is correct.

My solution:

$2cos^2(\frac{a}{2}) + 2isin(\frac{a}{2})cos(\frac{a}{2})$

By using cos and sin identities:

$2cos^2(\frac{a}{2}) + isin(a)$

$2cos(a) + 2sin^2(\frac{a}{2}) + isin(a)$

$e^{ia} + cos(a) + 2sin^2(\frac{a}{2})$

And that's it. The other further attempts have gone nowhere. Like I tried to use (cos^2 + sin^2 = 1) but it resulted in going nowhere, too. I just have no idea how to get 1 out.

Any ideas?

EDIT: I kept trying more and I think I figured it out.

By using cos and sin identities:

$cos(a) = cos^2(\frac{a}{2}) - sin^2(\frac{a}{2})$

By placing it in the equation I found earlier:

Then we get this

$e^{ia} + cos^2(\frac{a}{2}) + sin^2(\frac{a}{2})$

Which is

$e^{ia} + 1$

Is this correct?

答:
hint

$$e^{ia}+1=1+\cos(a)\color{red}{+}i\sin(a)$$

$$=2\cos^2(\frac a2)\color{red}{+}2i\sin(\frac a2)\cos(\frac a2)$$

$$=2\cos(\frac a2)\Bigl(....\Bigr)$$

You need these identities
$$1+\cos(X)=2\cos^2(\frac X2)$$
$$\sin(X)=2\sin(\frac X2)\cos(\frac X2)$$
and
$$e^{iX}=\cos(X)+i\sin(X)$$


问:
infinite intersection of sets
I'm trying  to proof this identity: $ [a,b]\equiv \bigcap_{n=1}^\infty [a,b+\frac{1}{n}) \equiv \bigcap_{n=1}^\infty (a-\frac{1}{n} , b]  $

I already try to use De-Morgan's lows but with no success .

TIA

答:
You only need to expand the relevant definitions and use some observations about the real numbers. I leave it to you to combine everything because this is a homework exercise.

Definitions:

$A = B$ iff $x \in A \iff x \in B$; equivalently, $A = B$ iff $A \subseteq B$ and $B \subseteq A$.

$x \in \bigcap_{n=1}^\infty A_n$ iff $x\in A_n$ for all $n$.

$[a,b] = \{x \in \Bbb R: a \le x \le b\}$; $[a,b) = \{x \in \Bbb R: a \le x < b\}$; $(a,b] = \{x \in \Bbb R: a < x \le b\}$.

Observation:

What can we conclude if $x  b$.) Similarly for $x > a - \frac1n$.

Title: infinite intersection of sets

Body:
You only need to expand the relevant definitions and use some observations about the real numbers. I leave it to you to combine everything because this is a homework exercise.

Definitions:

$A = B$ iff $x \in A \iff x \in B$; equivalently, $A = B$ iff $A \subseteq B$ and $B \subseteq A$.

$x \in \bigcap_{n=1}^\infty A_n$ iff $x\in A_n$ for all $n$.

$[a,b] = \{x \in \Bbb R: a \le x \le b\}$; $[a,b) = \{x \in \Bbb R: a \le x < b\}$; $(a,b] = \{x \in \Bbb R: a < x \le b\}$.

Observation:

What can we conclude if $x < b + \frac1n$ for all $n$? (Hint: suppose $x > b$.) Similarly for $x > a - \frac1n$.


问:
Finding the expected value of a geometric distribution shifted by a constant
Suppose that for a random variable $X, \mathbb{P}(X = 0) = 0, k \geq c$ and $\mathbb{P}(X = k) = (1 - p)^{k - c}p$ otherwise. How can we find the expected value for $X$? Sure, $\mathbb{E}(X) = \sum_{k = 0}^{\infty}k\mathbb{P}(X = k) = \sum_{k = c}^{\infty}k\mathbb{P}(X = k) =  \sum_{k = c}^{\infty}k(1 - p)^{k - c}p$. But how do we continue from here? By ratio test the series does converge (makes sense), but I am not sure how to manipulate the series into a nice "clean" form.

答:
If the pmf is the following

$$P(X=k)=(1-p)^{k-c}p$$

With $k\geq c$ that is $k=c,c+1,c+2,...$

This evidently means that the  mean is simply

$$E[X]=\frac{1-p}{p}+c$$

This because if we consdider the Geometric rv

$P(Y=k)=(1-p)^{k}p$ defined in $k=0,1,2,3,...$ and with mean $E[Y]=\frac{1-p}{p}$ your rv is $X=Y+c$


问:
How do I find the Inverse of $f(x)$?
Given
$f(x+1)+f(x-1)=x^2$

I have subtituted $(a=x+1)$ and $(a=x-1)$ and got
$$f(x)+f(x-2)=(x-1)^2 \text{ and } f(x+2)+f(x)=(x+1)^2$$
Combining those equations, I got

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

I could not even find $f(x)$

Please help me

答:
Note that $f(x)=1$ leads to $f(x+1)+f(x-1)=2$,
$f(x)=x$ leads to $f(x+1)+f(x-1)=2x$,
$f(x)=x^2$ leads to $f(x+1)+f(x-1)=2x^2+2$. In order to obtain $x^2$ on the right hand side, we might therefore combine
$$ f(x)=\frac12x^2-\frac12$$
and verify that this indeed makes
$$ f(x+1)+f(x-1)=x^2.$$
However, $f$ is not unique. For example,
$$ f(x)=\frac12x^2-\frac12+\sin\frac{\pi x}2$$
is another solution

Title: How do I find the Inverse of $f(x)$?


答:
Hint:

$$e^{ia}+1=e^{\frac{ia}{2}}(e^{\frac{ia}{2}}+e^{-\frac{ia}{2}})$$

It might be useful to note that

$$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$

Title: How do you prove that $e^{ia} + 1 = 2e^{i\frac{a}{2}}\cos(\frac{a}{2})$?

Body:
How do you prove $e^{ia} + 1 = 2e^{i\frac{a}{2}}\cos(\frac{a}{2})$ where a is a real number?

I attempted to solve it, by using the second side and I couldn't find a correct result, sadly. I'm not even sure if my attempt is correct.

My solution:

$2cos^2(\frac{a}{2}) + 2isin(\frac{a}{2})cos(\frac{a}{2})$

By using cos and sin identities:

$2cos^2(\frac{a}{2}) + isin(a)$

$2cos(a) + 2sin^2(\frac{a}{2}) + isin(a)$

$e^{ia} + cos(a) + 2sin^2(\frac{a}{2})$

And that's it. The other further attempts have gone nowhere. Like I tried to use (cos^2 + sin^2 = 1) but it resulted in going nowhere, too. I just have no idea how to get 1 out.

Any ideas?

EDIT: I kept trying more and I think I figured it out.

By using cos and sin identities:

$cos(a) = cos^2(\frac{a}{2}) - sin^2(\frac{a}{2})$

By placing it in the equation I found earlier:

Then we get this

$e^{ia} + cos^2(\frac{a}{2}) + sin^2(\frac{a}{2})$

Which is

$e^{ia} + 1$

Is this correct?


问:
Can the following integral be computed efficiently without knowing that $\int {\frac{1}{x^2+1}dx}=\arctan(x)$?

Compute $\displaystyle \int{\frac{x^2}{1+x^2}}$.

My first thought was to use long division which resulted in $\int {1-\frac{1}{x^2+1}dx}$. However I then did not come any further. The solution did the same thing and just stated that the integral obtained after long division is an elementary integral and $\int {1-\frac{1}{x^2+1}dx} =x-\arctan(x)+C$ with $x\in\mathbb R$. So did I just struggle because I didn't know that $\int {\frac{1}{x^2+1}dx}=\arctan(x)$ - or should I have been able to come up with that solution (in an efficient way) without knowing this?

Body:
$$y=\frac{x^2}{1+x^2}=\frac{1+x^2-1}{1+x^2}=1-\frac{1}{1+x^2}=1-\frac{1}{(x+i)(x-i)}$$
$$y=1-\frac{i}{2 (x+i)}+\frac{i}{2 (x-i)}$$ Just finish


Body:
You could continue your way like this:
$$=e^{ia} + \cos(a) + 2\sin^2(\frac{a}{2})=e^{ia} + \cos(a) + 2 - 2\cos^2(\frac{a}{2})=$$
$$= e^{ia} + \cos^2(\frac{a}{2})-\sin^2(\frac{a}{2}) + 2 - 2\cos^2(\frac{a}{2})=$$
$$=e^{ia} - \cos^2(\frac{a}{2})-\sin^2(\frac{a}{2}) + 2= e^{ia} - 1 + 2= e^{ia} + 1$$


问:
Proving limit using definition
I need to prove $$ \lim_{x\rightarrow\ 0}\frac{x^2-8}{{x-8}} =1 $$ using epsilon-delta definition.
I know I need to show that for every $\epsilon >0$ there exist a $\delta >0$ such that if $|x|<\delta$, then $| {\frac{x^2-8}{x-8}}-1|<\epsilon$

WHat I did:

$|{\frac{x^2-8}{x-8}}-1|=|{\frac{x^2-x}{x-8}}|$, but Im having troubles now deciding what my delta should be.

I would really appreciate if someone could give me an explanation on how should I choose delta and why.
Thanks.

答:
We've solved the problem on scrap paper and now present a solution devoid of the process details.

It is easy to show that

$\quad \large |x| \lt \frac{1}{6} \implies |\frac{x-1}{x-8}|  \lt \frac{1}{7}$

For $\varepsilon \gt 0$ set $\delta = \text{min}(\frac{1}{6}, 7\varepsilon)$.

If $x \lt \delta$ then

$\quad \large |{\frac{x^2-8}{x-8}-1|=| \frac{x-1}{x-8}}|\,|x| \lt \frac{1}{7} \cdot 7\varepsilon = \varepsilon$

Title: Proving limit using definition


问:
Prove $\det(A+BC) = \det(A) (1+CA^{-1}B)$
Assume $A_{n \times n}$ is an invertible matrix, $B$ is a vector and $C$ is a row vector.

How can we prove that $\det(A+BC) = \det(A) (1+CA^{-1}B)$?

I tried proving it using the statement proved in this question $\det(M) = \det(A) \cdot \det(D-C A^{-1} B)$ (which was presented to me in the similar questions while posting this question), but had no success. I would be grateful if a proof can be made using the statement proved in the linked question.

答: $A + B C = A(I + A^{-1} B C)$.  If $S$ and $T$ are $n \times m$ and $m \times n$ matrices,  $ST$ and $TS$ have the same nonzero eigenvalues.


问:
Proving a Set is a Club Set
The question goes like this,

Let $\kappa$ be an uncountable regular cardinal and let $f:\kappa^{<\omega}\rightarrow\kappa$.

Prove that the set

$$ C = \left\{\alpha \ \big|\ \text{range}(f\restriction\alpha^{<\omega})\subseteq\alpha\wedge\alpha<\kappa  \right\} $$

is a club.

I managed to prove that it is closed. I'm having trouble proving the unbounded part.

The direction I took was to assume $\sup C=\gamma < \kappa $ and build a maximal function from $\omega\rightarrow\kappa$ using an infinite branch of $\kappa^{<\omega}$ to contradict the regularity of $\kappa$. But I don't think it is the right direction.

Any help would be appriciated.

答:
It's always easy to prove this sort of statement in the positive.

Take $\alpha$, now define $\beta_0=\alpha+1$, and let $\beta_{n+1}=\sup\operatorname{range}(f\restriction\beta_n^{<\omega})$. Finally, let $\beta=\sup\beta_n$. Then $\operatorname{range}(f\restriction\beta^{<\omega})\subseteq\beta$: if $\vec x\in\beta^{<\omega}$, then there is some $n<\omega$ such that $\vec x\in\beta_n^{<\omega}$, so $f(\vec x)<\beta_{n+1}$.

And since $\alpha<\beta_0\leq\beta_1\leq\dots\beta$, we are done.

Title: Proving a Set is a Club Set


问:
How to solve $x^{T}Ax = 0$?
Given matrix $A \in \mathbb{R}^{n \times n}$, how do I solve $x^{T}Ax = 0$ for $x \in \mathbb{R}^n$?

Obviously, a zero vector is always a solution and if $A$ is positive or negative definite there is no other solution. However, I'm interested in the cases, where $A$ is neither. Just from plotting a few examples, I believe the solution in the two dimensional case should usually describe one or two lines, but an analytical solution eludes me.

The question Solving quadratic equations of the form $x'(A-B)x = 0$ seems to be closely related, but it only asks, if there is a solution, not how it looks like and is asking for the complex case. And, truth be told, I don't quite understand the answer anyway.

答: Hint: Proof that $x^T A x = x^T A_+ x$ for all $x$, where $A_+ = \frac{1}{2}(A+A^T)$ is the symmetric part of $A$. Then you can apply the spectral theorem.

Title: How to solve $x^{T}Ax = 0$?


问:
Motivation behind this eccentric Ramanujan Identity
I just visited the MathJaX page due to the Math.SE website showing some problems while loading the page. I saw some demo math equations samples at this page, when this identity actually caught my attention:

$$ \dfrac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\dfrac{e^{-2\pi}} {1+\dfrac{e^{-4\pi}} {1+\dfrac{e^{-6\pi}} {1+\dfrac{e^{-8\pi}} {1+\ldots} } } } $$

My question is how does one come up with such an identity. Could there be any motivation behind this or is this just brute force calculations and then discovering it as an identity.

答:
There is a lot of motivation behind this identity (although I don't know how it relates to Ramanujan's motivation, if at all).  The function $R(q)$ described in
J.M.'s answer is an example of a modular function (see the wikipedia entry), and the general theory of complex multiplication says that if $\tau$ is a quadratic irrational algebraic number
(e.g. $i$) then the value of any modular function at $q = e^{2\pi i\tau}$ lies in an abelian extension of $\mathbb Q(\tau).$

In the particular case of your identity, $\tau = i$ (so $q = e^{-2\pi}$), and the abelian extension
is $\mathbb Q(i,\sqrt{5})$.   The reason that $5$ appears here is related to the fact that the level of the modular function $R(q)$ equals $5$.

For more details from this point of view, you can look at this nice article
of William Duke.


问:
How do we extend boolean algebra for predicates with only one argument?
Let's consider a logical system where all predicates can take only one argument: $A(x), B(x), \cdots$. Let us also assume that we can use all available logical (boolean) operators to combine those "atomic" predicates and, in this way, construct new predicates. For example:

$F(x) := [A(x) \wedge B(x)] \lor [\lnot C(x)]$

Finally, in each formula, we will have unbound variable $x$ and we can bind it by $\forall$ and $\exists$ quantifiers. For example:

$\forall x : [[A(x) \wedge B(x)] \lor [\lnot C(x)]]$

or

$\exists x : [\lnot A(x) \lor B(x)]$

So, one can say that this system is somewhat between proposition logic and predicate logic. Or, more precisely, it is a reduced version of predicate logic since any formula in this system is a valid formula of predicate logic.

So, my question is: What are the "transformation rules" for this system apart from boolean formulas for logical operations. For example, I can think of those rules:

$\lnot \forall x : [C(x)] = \exists x : [\lnot C(x)]$

$\lnot \exists x : [C(x)] = \forall x : [\lnot C(x)]$

$\forall x : [A(x) \wedge B(x)] = \forall x : [A(x)] \wedge \forall [B(x)]$

By the way, can it be that the described logical system has a special name?

答: This is called monadic predicate calculus or monadic first-order logic. See this article on deductive systems for first-order logic for the  transformation rules that apply to predicate calculus in general and to monadic predicate calculus in particular. However, monadic predicate calculus is particularly simple: an interpretation for a sentence involving $n$ monadic predicate symbols $A_1, \ldots A_n$ is essentially just a set $X$ with $n$ chosen subsets $X_1, \ldots, X_n$, where $x \in X_i$ iff $A_i(x)$ holds. Monadic predicate calculus can be shown to be decidable: a sentence involving $m$ monadic predicate symbols is valid iff it holds under every interpretation in a set $X$ with $2^n$ elements (and it is a finite process to enumerate and check all such interpretations).


问:
Find Spectral decomposition of $\mathrm{adj}\ A$ matrix
I'm stuck at this question and Don't have any clue how to proceed.

The Spectral decomposition is :

$$A=\sum_{i=1}^k c_iE_i\newcommand{\adj}[1]{\mathrm{adj}\ #1}$$
As we know, the eigenvalues of $\adj{A}$ is $\frac {|A|}{c_1}$, $\ldots$, $\frac {|A|}{c_i}$ (although I don't know how to use it in this question)

I tried using the formula $\adj{A}\cdot A=|A|\cdot I$, Multiplying by $A^{-1}$ from the right and get

$$\adj{A}=|A|\cdot A^{-1} = |A|\cdot(c_1E_1+ ... + c_kE_k)^{-1}$$

And here I'm currently stuck, any hints?

Thank you very much!

答:
From "spectral decomposition", I assume that each $E_i$ is a projection onto the eigenspace associated with eigenvalue $c_i$.

It suffices to note that $(c_1 E_1 + \cdots + c_n E_n)^{-1} = c_1^{-1}E_1 + \cdots + c_n^{-1} E_n$. You can verify that this is true by expanding the product
$$
(c_1 E_1 + \cdots + c_n E_n)(c_1^{-1}E_1 + \cdots + c_n^{-1} E_n)
$$
and simplify using the fact that $E_iE_j = 0$ for all $i \neq j$.


问:
Let $H\unlhd G$ and $a\in G$. If the coset $aH$ has order $3$ in $G/H$ and $|H| = 10$, what are the possible $|a|$ in $G$?

Context

I am an undergrad student taking Abstract Algebra I, and this is a problem on a homework assignment.

Problem

Let $H$ be a normal subgroup of $G$, and let $a ∈ G$. If the coset $aH$ has order 3 in
the factor group $G/H$, and $|H| = 10$, what are the possible orders of $a$ in $G$?

What I know

Since $H$ is a normal subgroup, its left coset should equal its right coset.

Since $H$ has 10 elements, the highest order of an element should be 10.

答: By definition, the order of the coset $aH$ in the quotient group $G/H$ is the least positive integer $n$ such that $(aH)^n=H$, being $H$ the identity of $G/H$. But, by normality of $H$ in $G$, it is $(aH)^n=a^nH$ (induction on $n$), so $a^nH=H$ or, equivalently, $a^n\in H$. So, in your case, $a^3\in H$ and $a,a^2\notin H$. Now, $e=(a^3)^{|a^3|}=a^{3|a^3|}$, and hence $|a|$ divides $3|a^3|$. Since $|a^3|$ divides $|H|=10$, $|a^3|$ can be any among $1,2,5,10$, whence $3|a^3|$ can be any among $3,6,15,30$. To sum up, $|a|$ can be any among $3,5,6,10,15,30$, being $1$ and $2$ ruled out by the conditions $a,a^2\notin H$.


答:
For any $x$ and $y$ in $V$, $(x+y)/2$ is the unique $z\in V$ such that $\|z-x\| = \|z-y\|= \|x-y\|/2$.  Since $f$ is an isometry, $\|f(z) - f(x)\| = \|f(z) - f(y)\| = \|f(x)-f(y)\|/2$, therefore $$f((x+y)/2) = (f(x) + f(y))/2$$  The case $y=0$ gives us $f(x/2) = f(x)/2$
so we have $$f(x+y) = f(x) + f(y)$$
Next prove that $f(n x) = n f(x)$ for any integer $n$, then for rational numbers,
then (by continuity) for reals.

Title: To show that $f$ is an orthogonal linear map, if $f(0) = 0$ and $f$ preserves the distance

Body:
Let $V$ be a finite-dimensional inner product space, and $f:V\to V$ be a map such that $f(0) = 0$ and $\forall x,y\in V$, $||f(x) - f(y)|| = ||x-y||$. Show that $f$ is an orthogonal linear map.

I've been able to show only a part of what's required, and my work is as follows:

Since $||f(x) - f(y)|| = ||x-y||$ holds for all $x,y$, put $y=0$. This gives $||f(x)|| = ||x||$ for all $x\in V$. $f:V\to V$ is a norm-preserving map, and so it is orthogonal if it is linear (proof of this is given at the end).

Now, how do I show that $f$ is linear? Is there a nice way of doing this? I'd appreciate any hints.

Proof of statement above:

I know that a map $T:V\to W$ is orthogonal if $\forall x,y\in V$, we have $\langle Tx,Ty\rangle = \langle x,y\rangle$.

$f$ is norm-preserving, so for arbitrary $x,y\in V$, we have $||f(x+y)|| = ||x+y||$ and hence $||f(x+y)||^2 = ||f(x)+f(y)||^2 =  ||x+y||^2$. Expanding, we get $||f(x)||^2 + ||f(y)||^2 + 2\langle f(x),f(y)\rangle = ||x||^2 + ||y||^2 + 2\langle x,y\rangle$ and using $||f(x)||=||x||$ and $||f(y)|| = ||y||$ we get $\langle f(x),f(y)\rangle = \langle x,y\rangle$ as desired.


Body:
First, let's show that
$$
f(x+y)=f(x)+f(y).
$$
We have
\begin{align*}
|| (f(x+y)-f(x))-f(y)||^2&=
||
f(x+y)-f(x)
||^2
+
||f(y)||^2
-2<f(x+y)-f(x),f(y)>\\
&=
2||y||^2 -2<f(x+y),f(y)>+2<x,y>\\
&=
||y||^2 +||x||^2-||x+y||^2+2<x,y>\\
&=0
\end{align*}
since
$$
||f(x+y)-f(y) ||^2=||x+y||^2 + || y||^2 -2<f(x+y),f(y)>=|| x||^2.
$$
Using, the same method we have
$$
f(\lambda x)= \lambda f(x).
$$
Indeed, we have
$$
|| f(\lambda x)- \lambda f(x) || ^2=
2\lambda^2||x ||^2-\lambda<f(\lambda x),f(x)>
$$
but
$$
||f(\lambda x)-f(x) ||^2=|| (\lambda -1)x||^2=(\lambda -1)^2||x||^2
=
(\lambda^2+1)||x||^2 - 2<f(\lambda x),f(x)>
$$
so
$$
2\lambda||x||^2
=
- 2<f(\lambda x),f(x)>
$$
thus
$$
2\lambda^2||x||^2
=
- 2\lambda<f(\lambda x),f(x)>.
$$
So
$$
|| f(\lambda x)- \lambda f(x) || ^2=
2\lambda^2||x ||^2-2\lambda^2||x||^2=0.
$$


答:
Note that we have
$$
\det(A + BC) = \det(A[I + A^{-1}BC]) = \det(A)\det(I + [A^{-1}B]C).
$$
with $U = A^{-1}B$ and $V = C$, it suffices to show that
$$
\det(I + UV) = 1 + VU.
$$
This can be seen as an instance of the W-A identity. One proof of this result is given in the linked page.

If you're still set on using the linked result, we can indeed prove this result as a consequence. Let $M$ denote the matrix
$$
M = \pmatrix{A & -B\\ C & 1}.
$$
If we apply the linked result directly to this matrix, then we have
$$
\det(M) = \det(A)\det(D - CA^{-1}(-B)) = \det(A)(1 + CA^{-1}B).
$$
On the other hand, we note that
$$
\det \pmatrix{A & -B\\ C & 1} =
(-1)^{n-1}\det\pmatrix{-B & A\\ 1 & C} =
[(-1)^{n-1}]^2\det\pmatrix{1 & C\\ -B & A}.
$$
That is, we have
$$
\det(M) = \det\pmatrix{1 & C\\ -B & A}.
$$
Applying the linked result to this matrix yields
$$
\det\pmatrix{1 & C\\ B & A} = \det(1)\det(A - (-B)1^{-1}C) = \det(A + BC).
$$
So, we have $\det(A)(1 + CA^{-1}B) = \det(A + BC)$ as desired.

Title: Prove $\det(A+BC) = \det(A) (1+CA^{-1}B)$

Body:
Assume $A_{n \times n}$ is an invertible matrix, $B$ is a vector and $C$ is a row vector.

How can we prove that $\det(A+BC) = \det(A) (1+CA^{-1}B)$?

I tried proving it using the statement proved in this question $\det(M) = \det(A) \cdot \det(D-C A^{-1} B)$ (which was presented to me in the similar questions while posting this question), but had no success. I would be grateful if a proof can be made using the statement proved in the linked question.


问:
How do you determine the surface area of a sphere given the circumference of a cross-sectional circle?
The question was: Sketch the intersection of a sphere and a plane that does not pass through the center of the sphere. If you know the circumference of the circle formed by the intersection, can you find the surface area of the sphere?

Of course, this question is searching for a theoretical method and solution.

My first thought was to set corresponding variables and allow the cross-section that contains the circle to be parallel to the plane that intersects the diameter of the sphere - or the great circle. Of course, this creates two circles that lie on parallel planes, and therefore we can construct several right triangles, and resort to Trigonometry to solve this problem.

I set an angle θ that is a central angle, and increases as the distance between the origin of the great circle and the cross-sectional circle decreases, and cannot fully reach a half rotation of 180 degrees. I then connected the collinear points that form the radius of the sphere, which consist of the center of the sphere, the center of the cross-sectional circle, and the endpoint of the radius. I then formed a kite by connecting the endpoint of the radius with the endpoints of the diameter of the cross-sectional circle.

Hopefully the above describes the graphics I created.

How do I do this? How do I prove or disprove that the surface area is determinable given the circumference of a cross-sectional circle that lies on a different plane?

答: Imagine a big sphere and several planes going through it at various distances from the sphere center. Some intersections will have smaller circumference than others (start with a plane nearly tangent to sphere). So not determined.</p>


问:
Find the number $m$ such that $m^2 + 1$ is divisible by $x$

Find the number $m$ such that $m^2 + 1$ is divisible by $x$ for $x = 474993$

So, I think it will be $m^2+1 \equiv 0$ (mod $474993$), I have no clue how to solve this, any hints would be appreciated. Thank you!

答:
Hint. There are no solutions to the given equation.

 Observe that $474993=3^2\cdot89\cdot593$. Thus, $m^2+1\equiv 0\bmod 474993\implies m^2\equiv-1\bmod 3$ . Yet it is well known that this is impossible.


问:
Null set of Bernoulli distribution
I have an assignment in which I am supposed to determine the null set of the Bernoulli distribution w.r.t. the Borel $\sigma$-algebra on $\mathbb{R}$. Specifically:

Let $(\mathbb{R}, \mathscr{B},P)$ be a measure space, where $\mathscr{B}$ denotes the Borel $\sigma$-algebra on $\mathbb{R}$ and $P$ the probability measure Bernoulli distribution i.e. P is characterized by $$P\left(\{0\}\right)=p,\quad P(\{1\})=1-p,\quad 0<p<1$$
Determine $N_P=\{N\subseteq\mathbb{R}| \text{There exists $A\in \mathscr{B}$ so $N\subseteq A$ and $P(A)=0$}\}$

I don't what to do. I don't understand how the Bernoulli distribution is defined on $\mathscr{B}$. How would you evaluate any open interval on the real number line for example.

答:
Hint:

If $N$ is a null set and $M\subseteq N$ then also $M$ is a null set.

Now be aware that  $\{0,1\}^{\complement}$ is evidently the largest null set in this context.

Title: Null set of Bernoulli distribution


答: The sphere can be represented as the points $(x,y,z) \in \mathbb{R}^3$ such that $x^2 + y^2 + z^2 = R^2$ . Then, assuming the plane is horizontal(it can be rotated by a change in coordinates because the sphere is symetrical in all dimensions) we can express the plane as ${z = k}$ for some $k \in \mathbb{R}$, which would be the distance of the plane to the origin. Then the intersection of the sphere with the plane is ${x^2 + y^2 = R^2 - k^2}$ For it to be a circle, $-R < k < R$. Because we are given the circumference of this circle(lets call it $c$) we have that $c = 2\pi\sqrt{R^2 - k^2}$. Then it is easy to get the value of $R$ with respect to $c$ and because the sphere is fully determined by its radius we can also get its surface area from $c$.

Title: How do you determine the surface area of a sphere given the circumference of a cross-sectional circle?

Body:
The question was: Sketch the intersection of a sphere and a plane that does not pass through the center of the sphere. If you know the circumference of the circle formed by the intersection, can you find the surface area of the sphere?

Of course, this question is searching for a theoretical method and solution.

My first thought was to set corresponding variables and allow the cross-section that contains the circle to be parallel to the plane that intersects the diameter of the sphere - or the great circle. Of course, this creates two circles that lie on parallel planes, and therefore we can construct several right triangles, and resort to Trigonometry to solve this problem.

I set an angle θ that is a central angle, and increases as the distance between the origin of the great circle and the cross-sectional circle decreases, and cannot fully reach a half rotation of 180 degrees. I then connected the collinear points that form the radius of the sphere, which consist of the center of the sphere, the center of the cross-sectional circle, and the endpoint of the radius. I then formed a kite by connecting the endpoint of the radius with the endpoints of the diameter of the cross-sectional circle.

Hopefully the above describes the graphics I created.

How do I do this? How do I prove or disprove that the surface area is determinable given the circumference of a cross-sectional circle that lies on a different plane?


问:
Insulated Heat equation
So it is well known that the heat equation for a rod with perfectly insulated ends at $x= 0$ and $x = l$ is given by:
$u_{t} = k u_{xx}$
It has the following BCs:

$\frac{\partial u}{\partial x}|_{(0,t)}=0 \ \hbox{ and  } \frac{\partial u}{\partial x}|_{(l,t)}=0$

and IC :

$\frac{\partial u}{\partial x}|_{(x,0)}= h(x)$

Just wondering how the BCs for a spherically symmetric sphere with radius $r=a$ that is perfectly insulated around the boundary would look like. Should it look like this?
If we only consider radial heat flow, the heat equation would be:

$u_t = k (u_{rr}+ \frac{2}{r} u_{r})$.

By substituting $v = ur$, we have:
$v_t = k v_{rr}$

with the BCs given by:
$v_{r}(0,t) = v_{r} (a,t) = 0$

and IC given by:
$v = r h(r)$.

答:
SO the heat equation is
$$\partial_t u=\lambda\nabla^2u$$
With $\lambda$ being some diffusivity constant. We can use the Laplacian in spherical coordinates and remove the $\theta,\phi$ dependency to get
$$\partial_t u=\frac{\lambda}{r^2}\partial_r\left(r^2\partial_ru\right)$$
Let's assume the equation is separable.
$$\frac{G'(t)}{G(t)}=\frac{\lambda}{r^2F(r)}\partial_r\left(r^2F'(r)\right)=-c$$
Let's focus on the spacial part. Make the substitution $F(r)=f(r)/r$ so
$$F'(r)=\frac{r\cdot f'(r)-f(r)}{r^2}$$
Which leads us to
$$f''(r)=\frac{-c}{\lambda}f(r)$$
$$f(r)=a_1\sin\left(\sqrt{\frac{c}{\lambda}}r\right)+a_2\cos\left(\sqrt{\frac{c}{\lambda}}r\right)$$
Imposing the boundary conditions, namely $f(0)=f(a)=0$ we require $a_2=0$ and we get constants
$$c_n=\lambda \frac{n^2\pi^2}{a^2}$$
Which gives our spacial eigenfunctions
$$F_n(r)=A_n\operatorname{sinc}\left(\frac{n\pi}{a}r\right)=A_nj_0\left(\frac{n\pi}{a}r\right)=A_n\sqrt{\frac{a}{2nr}}J_{1/2}\left(\frac{n\pi}{a}r\right)$$
Using Bessel functions (this will be important later)

The time component follows:
$$\frac{G'(t)}{G(t)}=-c=-\lambda \frac{n^2\pi^2}{a^2}$$
$$G_n(t)=B_n\exp\left(-\lambda \frac{n^2\pi^2}{a^2}t\right)$$

So we can write our solution as (using $A_n\sqrt{\frac{a}{2n}}\cdot B_n= C_n$)
$$u(r,t)=\sum_{n=1}^\infty C_n\frac{J_{1/2}\left(\frac{n\pi}{a}r\right)}{\sqrt{r}}\exp\left(-\lambda \frac{n^2\pi^2}{a^2}t\right)$$
Initial conditions:

We can see that
$$\sqrt{r}\cdot u(r,0)=\sum_{n=1}^\infty C_nJ_{1/2}\left(\frac{n\pi}{a}r\right)$$
This is a perfect time to use Fourier-Bessel series. Let $\alpha_{1/2,n}$ be the $n$th root of $J_{1/2}$, which in this case is the rather simple $\alpha_{1/2,n}=n\pi$.
So
$$\sqrt{r}u_0(r)=\sum_{n=1}^\infty C_nJ_{1/2}\left(\frac{\alpha_{1/2,n}}{a}r\right)$$
The literature then tells us that
$$C_n=\frac{\int_0^a r^{3/2}u_0(r)J_{1/2}\left(\frac{\alpha_{1/2,n}}{a}r\right)\mathrm{d}r}{(a/2)J_{3/2}(\alpha_{1/2,n})^2}$$

EXAMPLE: Let $u_0(r)=(r^2-1)\log(r)$. The first few coefficients are
$$\{C_n\}=\{0.624023,0.386661,0.10893,...\}$$
Here is a plot of the initial data $u_0(r)$ as well as its first five partial series-

And the Mathematica code used (so you can try yourself):

```<code>Subscript[u, 0][r_] := (r^2 - 1) Log[r]
c[n_] := 2*
  NIntegrate[
   r^(3/2) Subscript[u, 0][r] BesselJ[1/2, n*\[Pi]*r], {r, 0, 1}]/
  BesselJ[3/2, n*\[Pi]]^2
Subscript[u, 0][r_, N_] :=
 Sum[c[n]*(Sqrt[2] Sin[n*\[Pi]*r])/(\[Pi]*r*Sqrt[n]), {n, 1, N}]
Plot[{Subscript[u, 0][r], Subscript[u, 0][r, 1],
  Subscript[u, 0][r, 2], Subscript[u, 0][r, 3], Subscript[u, 0][r, 4],
   Subscript[u, 0][r, 5]}, {r, 0.01, 1},
 PlotLegends -> {"\!\(\*SubscriptBox[\(u\), \(0\)]\)(r)",
   "\!\(\*SubscriptBox[\(u\), \(0\)]\)(r;1)",
   "\!\(\*SubscriptBox[\(u\), \(0\)]\)(r;2)",
   "\!\(\*SubscriptBox[\(u\), \(0\)]\)(r;3)",
   "\!\(\*SubscriptBox[\(u\), \(0\)]\)(r;4)",
   "\!\(\*SubscriptBox[\(u\), \(0\)]\)(r;5)"}]
</code>```

Here $u_0(r;N)$ is the $N$th partial series. Another example when $u_0(r)=\sqrt{1-r}$ :

EDIT: What if we have Neumann, not Dirichlet, boundary conditions? That is, what if we want
$$\partial_r[u](0,t)=\partial_r[u](a,t)=0$$
Well, let's return to our spacial part of our solution:
$$F(r)=a_1\frac{\sin\left(\sqrt{\frac{c}{\lambda}}r\right)}{r}+a_2\frac{\cos\left(\sqrt{\frac{c}{\lambda}}r\right)}{r}$$
The derivative is
$$F'(r)=\frac{\cos\left(\sqrt{\frac{c}{\lambda}}r\right)\left(a_1\sqrt{\frac{c}{\lambda}}r-a_2\right)-\sin\left(\sqrt{\frac{c}{\lambda}}r\right)\left(a_1+a_2\sqrt{\frac{c}{\lambda}}r\right)}{r^2}$$
In order to get $F'(0)=0$ we require $a_2=0$. Then our expression for the derivative simplifies to
$$F'(r)=a_1\frac{\sqrt{\frac{c}{\lambda}} r\cos\left(\sqrt{\frac{c}{\lambda}}r\right)-\sin\left(\sqrt{\frac{c}{\lambda}}r\right)}{r^2}$$
What we're trying to do is find values of $c$ such that $F'(a)=0$ given $\lambda,a$. Unfortunately, the roots of $F(r)$ are not easily obtainable in closed form. Letting $\beta_{\lambda,n}$ be the $n$th positive root of $F'(r)$ given the constant $\lambda$ (we can find these via numerical algorithms) we require
$$\beta_n=\sqrt{\frac{c_n}{\lambda}}a\implies c_n=\lambda \left(\frac{\beta_n}{a}\right)^2$$
So our spacial eigenfunctions are
$$F_n(r)=A_n\operatorname{sinc}\left(\frac{\beta_n}{a}r\right)=A_n j_0\left(\frac{\beta_n}{a}r\right)=A_n\sqrt{\frac{\pi}{2\frac{\beta_n}{a}r}}J_{1/2}\left(\frac{\beta_n}{a}r\right)\to A_n\frac{1}{\sqrt{r}}J_{1/2}\left(\frac{\beta_n}{a}r\right)$$
Where of course we redefine $A_n$ as we go along for convenience. After this point, we follow the same process as before with the Fourier Bessel series and whatnot - just in this case we have to compute the roots $\beta_n$ beforehand.

Title: Insulated Heat equation

Body:
SO the heat equation is
$$\partial_t u=\lambda\nabla^2u$$
With $\lambda$ being some diffusivity constant. We can use the Laplacian in spherical coordinates and remove the $\theta,\phi$ dependency to get
$$\partial_t u=\frac{\lambda}{r^2}\partial_r\left(r^2\partial_ru\right)$$
Let's assume the equation is separable.
$$\frac{G'(t)}{G(t)}=\frac{\lambda}{r^2F(r)}\partial_r\left(r^2F'(r)\right)=-c$$
Let's focus on the spacial part. Make the substitution $F(r)=f(r)/r$ so
$$F'(r)=\frac{r\cdot f'(r)-f(r)}{r^2}$$
Which leads us to
$$f''(r)=\frac{-c}{\lambda}f(r)$$
$$f(r)=a_1\sin\left(\sqrt{\frac{c}{\lambda}}r\right)+a_2\cos\left(\sqrt{\frac{c}{\lambda}}r\right)$$
Imposing the boundary conditions, namely $f(0)=f(a)=0$ we require $a_2=0$ and we get constants
$$c_n=\lambda \frac{n^2\pi^2}{a^2}$$
Which gives our spacial eigenfunctions
$$F_n(r)=A_n\operatorname{sinc}\left(\frac{n\pi}{a}r\right)=A_nj_0\left(\frac{n\pi}{a}r\right)=A_n\sqrt{\frac{a}{2nr}}J_{1/2}\left(\frac{n\pi}{a}r\right)$$
Using Bessel functions (this will be important later)

The time component follows:
$$\frac{G'(t)}{G(t)}=-c=-\lambda \frac{n^2\pi^2}{a^2}$$
$$G_n(t)=B_n\exp\left(-\lambda \frac{n^2\pi^2}{a^2}t\right)$$

So we can write our solution as (using $A_n\sqrt{\frac{a}{2n}}\cdot B_n= C_n$)
$$u(r,t)=\sum_{n=1}^\infty C_n\frac{J_{1/2}\left(\frac{n\pi}{a}r\right)}{\sqrt{r}}\exp\left(-\lambda \frac{n^2\pi^2}{a^2}t\right)$$
Initial conditions:

We can see that
$$\sqrt{r}\cdot u(r,0)=\sum_{n=1}^\infty C_nJ_{1/2}\left(\frac{n\pi}{a}r\right)$$
This is a perfect time to use Fourier-Bessel series. Let $\alpha_{1/2,n}$ be the $n$th root of $J_{1/2}$, which in this case is the rather simple $\alpha_{1/2,n}=n\pi$.
So
$$\sqrt{r}u_0(r)=\sum_{n=1}^\infty C_nJ_{1/2}\left(\frac{\alpha_{1/2,n}}{a}r\right)$$
The literature then tells us that
$$C_n=\frac{\int_0^a r^{3/2}u_0(r)J_{1/2}\left(\frac{\alpha_{1/2,n}}{a}r\right)\mathrm{d}r}{(a/2)J_{3/2}(\alpha_{1/2,n})^2}$$

EXAMPLE: Let $u_0(r)=(r^2-1)\log(r)$. The first few coefficients are
$$\{C_n\}=\{0.624023,0.386661,0.10893,...\}$$
Here is a plot of the initial data $u_0(r)$ as well as its first five partial series-

And the Mathematica code used (so you can try yourself):

Subscript[u, 0][r_] := (r^2 - 1) Log[r]
c[n_] := 2*
  NIntegrate[
   r^(3/2) Subscript[u, 0][r] BesselJ[1/2, n*\[Pi]*r], {r, 0, 1}]/
  BesselJ[3/2, n*\[Pi]]^2
Subscript[u, 0][r_, N_] :=
 Sum[c[n]*(Sqrt[2] Sin[n*\[Pi]*r])/(\[Pi]*r*Sqrt[n]), {n, 1, N}]
Plot[{Subscript[u, 0][r], Subscript[u, 0][r, 1],
  Subscript[u, 0][r, 2], Subscript[u, 0][r, 3], Subscript[u, 0][r, 4],
   Subscript[u, 0][r, 5]}, {r, 0.01, 1},
 PlotLegends -> {"\!\(\*SubscriptBox[\(u\), \(0\)]\)(r)",
   "\!\(\*SubscriptBox[\(u\), \(0\)]\)(r;1)",
   "\!\(\*SubscriptBox[\(u\), \(0\)]\)(r;2)",
   "\!\(\*SubscriptBox[\(u\), \(0\)]\)(r;3)",
   "\!\(\*SubscriptBox[\(u\), \(0\)]\)(r;4)",
   "\!\(\*SubscriptBox[\(u\), \(0\)]\)(r;5)"}]

Here $u_0(r;N)$ is the $N$th partial series. Another example when $u_0(r)=\sqrt{1-r}$ :

EDIT: What if we have Neumann, not Dirichlet, boundary conditions? That is, what if we want
$$\partial_r[u](0,t)=\partial_r[u](a,t)=0$$
Well, let's return to our spacial part of our solution:
$$F(r)=a_1\frac{\sin\left(\sqrt{\frac{c}{\lambda}}r\right)}{r}+a_2\frac{\cos\left(\sqrt{\frac{c}{\lambda}}r\right)}{r}$$
The derivative is
$$F'(r)=\frac{\cos\left(\sqrt{\frac{c}{\lambda}}r\right)\left(a_1\sqrt{\frac{c}{\lambda}}r-a_2\right)-\sin\left(\sqrt{\frac{c}{\lambda}}r\right)\left(a_1+a_2\sqrt{\frac{c}{\lambda}}r\right)}{r^2}$$
In order to get $F'(0)=0$ we require $a_2=0$. Then our expression for the derivative simplifies to
$$F'(r)=a_1\frac{\sqrt{\frac{c}{\lambda}} r\cos\left(\sqrt{\frac{c}{\lambda}}r\right)-\sin\left(\sqrt{\frac{c}{\lambda}}r\right)}{r^2}$$
What we're trying to do is find values of $c$ such that $F'(a)=0$ given $\lambda,a$. Unfortunately, the roots of $F(r)$ are not easily obtainable in closed form. Letting $\beta_{\lambda,n}$ be the $n$th positive root of $F'(r)$ given the constant $\lambda$ (we can find these via numerical algorithms) we require
$$\beta_n=\sqrt{\frac{c_n}{\lambda}}a\implies c_n=\lambda \left(\frac{\beta_n}{a}\right)^2$$
So our spacial eigenfunctions are
$$F_n(r)=A_n\operatorname{sinc}\left(\frac{\beta_n}{a}r\right)=A_n j_0\left(\frac{\beta_n}{a}r\right)=A_n\sqrt{\frac{\pi}{2\frac{\beta_n}{a}r}}J_{1/2}\left(\frac{\beta_n}{a}r\right)\to A_n\frac{1}{\sqrt{r}}J_{1/2}\left(\frac{\beta_n}{a}r\right)$$
Where of course we redefine $A_n$ as we go along for convenience. After this point, we follow the same process as before with the Fourier Bessel series and whatnot - just in this case we have to compute the roots $\beta_n$ beforehand.


问:
If $f_{n}$ has a dense image, then $\bigcap (f_{1}\circ\cdots\circ f_{n})(X_{n})$ is dense
Let $\{(X_{n}, d_{n})\}_{n\in\mathbb{N}}$ be a sequence of complete metric spaces and $\{f_{n}: X_{n}\to X_{n − 1}\}_{n\in \mathbb{N}}$ a sequence of continuous functions. If $f_{n}$ has a dense image for each $n\in \mathbb{N}$, show that $\bigcap_{n\in \mathbb{N}}(f_{1}\circ f_{2}\circ \cdots \circ f_{n}) (X_{n})$ is dense at $X_{0}$.

My attemps:
if $f_{n}$ has a dense image then $\overline{f_{n}(X_{n})}=X_{n-1}$ I want to show that
$$\overline{\bigcap_{n\in \mathbb{N}}(f_{1} \circ f_{2} \circ \cdots \circ f_{n})(X_{n})} = X_{0}. $$
Let $F_{n}=(f_{1}\circ f_{2} \circ \dots ◦ f_{n})(X_{n})$. By induction, I see that $F_{n+1} ⊆ F_{n} ⊆ X_{0}$. Indeed, if $n=1$, then
$$F_{2}=(f_{1}\circ f_{2})(X_{2})=f_{1}(f_{2}(X_{2}))\subseteq f_{1}(\overline{f_{2}(X_{2}}))=f_{1}(X_{1})=F_{1}$$
if we assume that $F_{k+1}\subseteq F_{k}\subset X_{0}$. Then,
$$F_{k+2} = (f_{1} \circ f_{2} \circ \cdots \circ f_{k+1} \circ f_{k+2})(X_{k+2}) = (f_{1} \circ f_{2} \circ \cdots \circ f_{k+1})(f_{k+2}(X_{k+2}))\subseteq (f_{1} \circ f_{2} \circ \cdots \circ f_{k+1})(\overline{f_{k+2}(X_{k+2})}) = (f_{1} \circ f_{2} \circ \cdots \circ f_{k+1})(X_{k+1})= F_{k+1}\subseteq F_{k}\subseteq X_{0}$$
from the above we have
$$\cdots\subseteq F_{n+1}\subseteq F_{n} \subseteq \cdots \subset F_{2}\subseteq F_{1}$$
then

$$\cdots\subseteq \overline{F_{n+1}}\subseteq \overline{F_{n}} \subseteq \cdots \subset \overline{F_{2}}\subseteq \overline{F_{1}}=X_{0}$$
then
$$\overline{\bigcap_{n\in \mathbb{N}}F_{n} }\subseteq \bigcap_{n\in \mathbb{N}}\overline{F_{n}}\subseteq X_{0} .$$

is my reasoning correct?.I've been thinking that Biare's theorem could be used, but I don't know if it's possible since $F_{n}'s$ don't necessarily open.

If it is, how could the other containment prove?. I appreciate any help

答:
I think I found a solution but its rather technical.
The following is more of a sketch than a solution with full details.

Let $x_0\in X$ and $\varepsilon>0$ be given.

In a first step, one has to
construct points $y_{k,m}$ for $m,k\in\Bbb N_0$ with $k\leq m$
such that
$$
\begin{aligned}
y_{k,m} &\in X_k
\\
y_{0,0} &= x_0
\\
f_k(y_{k,m}) &= y_{k-1,m}
\quad\forall m\geq k\geq 1,
\\
d_k(y_{k,m},y_{k,m+1}) &< \varepsilon 2^{-m-1}
\quad\forall m\geq k\geq 0.
\end{aligned}
$$
(This requires continuity of these functions
and density of the images, and can be done recursively in $m$.
This step requires some technical details that I skipped here.)
Then, one can show that $\{y_{k,m}\}_{m\geq k}$ is a Cauchy sequence
for each $k$.
Thus, there exist points $z_k$ with
$$
\lim_{m\to\infty} y_{k,m} = z_k \quad\forall k\geq0.
$$
Due to continuity of $f_k$, one can show that
$f_k(z_k)=z_{k-1}$ holds.
Therefore, we have
$$
z_0\in F_k \quad\forall k\geq 1
\qquad\text{and therefore}\qquad
z_0 \in F_\infty:=\bigcap_{k\in\Bbb N} F_k.
$$
Finally, one can also show that
$$d_0(z_0,x_0)<\varepsilon$$
holds. Since $\varepsilon>0$ was arbitrary, this shows that
$F_\infty$ is dense in $X_0$.


问:
What are the elements of the quadratic ring $\mathbb{Z}[ \sqrt{15}]$?
Hi MathStackExchange,

I am having trouble understanding this quadratic ring: $$\mathbb{Z}[\sqrt{15}]$$ whose elements can be written as $$ \{a+b\sqrt{15} | a,b \in \mathbb{Z}\}$$
If anyone understands what these elements actually look like,and could provide a few examples, that would be very helpful.
Wikipedia describes the set as :
"The ring Z[ω] consists of all roots of all equations $x^2 + Bx + C = 0$whose discriminant $B^2 − 4C$ is the product of D by the square of an integer. In particular √D belongs to Z[ω], being a root of the equation $x^2 − D = 0$, which has 4D as its discriminant." link to page - see Quadratic Rings section

...which doesn't seem to match the other expression I have for the elements.

For example, is something like(just a random example) $999999 + (-2)\sqrt{15}$ in this ring? It seems like it could be from the way the elements can be represented?

My Answer (see approved answer for more detail):

My example $999999 + (-2)\sqrt{15}$ is indeed an element of the quadratic ring. In general, the motivation for mentioning rings of this type in Algebra textbooks/lectures is that they have so many interesting properties. For example, notice that $\mathbb[Z\sqrt{15}]$ is not a Unique Factorization Domain. You can find two different factorizations of irreducible elements for some elements in the ring.

Quadratic integer rings are explicitly defined like this, which makes it clear to see that elements of this ring in particular are indeed of the form $ \{a+b\sqrt{15} | a,b \in \mathbb{Z}\}$

答:
The definition of ${\mathbb Z}[\omega]$ is
$$\{a + b \omega \mid a, b \in \mathbb Z\}$$
or, more generally for $\omega$ that are not the root of a quadratic equation
$$\{\sum_{j=0}^{n} c_j \omega^j \mid n \in {\mathbb N}, c_0, \dots, c_n \in {\mathbb Z}\}.$$

What you quote from Wikipedia is a property that quadratic rings happen to have - and it is not hard to show, but is not immediately obvious -; the actual definition is in the line right before.


问:
In a rectangle $ABCD$ where $AB = 6$ , $BC = 3$ , point $P$ is chosen on $AB $ such that $\angle APD = 2 \angle CPB$. Find $AP$.

In a rectangle $ABCD$ where $AB = 6$ , $BC = 3$ , point $P$ is chosen on $AB $ such that $\angle APD = 2 \angle CPB$. Find $AP$.

What I Tried: Here is a picture :-

You can see I tried angle-chasing, but that really did not lead me to any useful info. Also I tried Pythagoras theorem, taking $AP = k$ and $PB = (6 - k)$ gives :-
$$k^2 + 9 = AP^2$$
$$(6 - k)^2 + 9 = PC^2$$
And that is $3$ variables in $2$ equations, so I need to find another value before trying this.

I think the fact that $AD = \frac{AB}{2}$ should be used somehow, maybe by some construction or something. Also in Geogebra I saw that $\angle APD = 65.7^\circ$ , $\angle CPB = 32.85^\circ$ , which are $2$ odd-looking angles, and $AP$ came to be $1.35$ .

Can anyone help me how is this the case? Thank You.

Body: $$\tan 2x=\frac{3}{k},\tan x=\frac{3}{6-k}$$ now use $$\tan 2x=\frac{2\tan x}{1-\tan^2 x}$$


问:
Deriving the gradient of a multivariable function taking matrices as input?
I've been tasked with deriving the gradient for the following non-convex function and I'd like to verify whether my solution / progress is correct so far (I'm not too confident with calculus):

$$F(U, V) = \dfrac{1}{2}\sum_{(i,j)\in{\Omega}}(M_{ij}-u_iv_{j}^T)^2 + \frac{\lambda}{2}(||U||^2_F + ||V||^2_F)$$

Where $M, U, V$ are matrices, $u_i$ and $v_j$ refer to the $i$th row and $j$th row of $U$ and $V$ respectively, and $\lambda$ is a variable denoting some learning rate.

I've obtained the following so far for the partial derivative with respect to $U$:

$$\dfrac{\partial F(U,V)}{\partial U} =-\sum_{(i,j)\in{\Omega}}(M_{ij}-u_iv_{j}^T)^2 + \lambda(||U||^2_F)$$

But I'm a bit unsure about taking the derivative for $\frac{\lambda}{2}(||U||^2_F + ||V||^2_F)$. I think what I have above is correct as $V$ would be treated like a constant (since we're doing the partial derivation with respect to $U$). I assume the approach for taking the partial derivative with respect to $V$ would be the same.

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}$
Capture the index constraint using the Hadamard product $(\odot)$ and a projection matrix
$$\eqalign{
P_{ik} &= \begin{cases}
1\quad{\rm if\,}(i,k)\in\Omega \\
0\quad{\rm otherwise}
\end{cases} \\
P &= P\odot P
 \qquad\; ({\rm projective\,property}) \\
\sum_{(i,k)\in\Omega}\!\!M_{ik} &= P\odot M
 \qquad ({\rm same\,size\,as\,}M) \\
}$$
Denote the standard vector basis as {$e_k$} and the standard matrix basis as $\big\{E_{ik} \doteq e_ie_k^T\big\}\;$
and the trace/Frobenius product using a colon, i.e.
$$A:B \;=\; \sum_{i=1}^m\sum_{k=1}^nA_{ik}B_{ik} \;=\; {\rm Tr}(A^TB)$$
For typing convenience, define the indexed matrix
$$A_{ik} \;=\; P\odot\Big(Ue_i(Ve_k)^T-M\Big) \;=\; P\odot\Big(UE_{ik}V^T-M\Big)$$
Assembling all of this with the Einstein summation convention allows us to write the objective function and calculate its differential.
$$\eqalign{
 F &= \tfrac 12A_{ik}:A_{ik} + \tfrac \lambda 2\big(U:U + V:V\big) \\
dF &= A_{ik}:dA_{ik} + \lambda\big(U:dU + V:dV\big) \\
 &= A_{ik}:\big(dU\,E_{ik}V^T+UE_{ik}\,dV^T\big) +\lambda\big(U:dU+V:dV\big) \\
 &= (VE_{ik}^TA_{ik}+\lambda U):dU + (A_{ik}^TUE_{ik}+\lambda V):dV \\
}$$
Holding $V$ constant (i.e. setting $\,dV=0$) yields the gradient with respect to $U$, while holding $U$ constant yields the gradient with respect to $V$
$$\eqalign{
\p{F}{U} &= VE_{ik}^TA_{ik}+\lambda U, \qquad
\p{F}{V} &= A_{ik}^TUE_{ik}+\lambda V \\
}$$

A quick note about the Frobenius and Hadamard products. They are mutually commutative, i.e.
$$\eqalign{
A:B &= B:A \\
A\odot B &= B\odot A \\
C:A\odot B &= C\odot A:B \\
}$$
In addition, the properties of the trace allow the terms in a Frobenius product to be rearranged in a number of equivalent ways, e.g.
$$\eqalign{
A:B &= A^T:B^T \\
CA:B &= A:C^TB = C:BA^T \\
}$$
The thing to keep in mind in these manipulations is that the matrix on each side of a Frobenius (or Hadamard) product must have exactly the same dimensions.


问:
What is a Fourier Symbol
I was wondering what exactly the term Fourier symbol, or sometimes said the symbol of the Fourier transform means. Also, are any books that cover this topic in detail?

Edit:
After the comments, I have come to realise I mean to ask either: what is the symbol of the Fourier transform operator, or also known as Fourier multiplier.

Are there are good recommended texts that cover the topic of symbols of operators, and the Fourier transform, well?

答: My guess is it's the $\mathcal{F}$ or $\tilde{}$ in definitions such as$$\mathcal{F}(f(x))(k):=\tilde{f}(k):=\int_{\Bbb R^n}f(x)\exp(-2\pi ik\cdot x)d^nx.$$

Title: What is a Fourier Symbol


问:
A combinatorial identity involving Stirling numbers of the second kind
Answering a recent question I came across the following interesting identity:
$$
\sum_{k=0}^m\binom mk{n+k+1 \brace k+1}k!=\sum_{k=0}^m\binom mk (-k)^{m-k}(k+1)^{n+k}.
$$

Is there a simple way to prove it?

答:
Consider the following problem. You want to count the number of functions $f$ form $[n+m]$ to $[m+1]$ such that for $y\in [m]\subseteq [m+1],$ if $f^{-1}(y)= \emptyset,$ then $f(y)=m+1$ (such elements $y$ will be called special).

In the LHS you have this counted in the following way. Either you will have $m+1$ in the image of the non-special elements or not. In both cases, select $k$ special elements $y\in [m]$ which can be done in $\binom{m}{k}$ ways, and then make a surjective function of the other elements in ${n+m-k\brace m-k}(m-k)!$ or ${n+m-k\brace m-k+1}(m-k+1)!$ ways (depending on $m+1$ being or being not in the image).
You will get
$$LHS = \underbrace{\sum _{k=0}^m\binom{m}{k}{n+m-k\brace m-k}(m-k)!}_{m+1\text{ not in the image}}+\underbrace{\sum _{k=0}^m\binom{m}{k}{n+m-k\brace m-k+1}(m-k+1)!}_{m+1\text{ in the image}}.$$
Which, using the recursion of Stirling numbers, can be seen to be equal to
$$LHS =\sum _{k=0}^m\binom{m}{k}\left ({n+k\brace k}k!+{n+k\brace k+1}(k+1)!\right )=\sum _{k=0}^m\binom{m}{k}k!{n+k+1\brace k+1}.$$

Meanwhile, in the alternative universe of the RHS, you can be think of it as
$$RHS = \sum _{k=0}^m\binom{m}{k}(-1)^{m-k}k^{m-k}(k+1)^{n+k}=\sum _{k=0}^m\binom{m}{k}(-1)^{k}(m-k)^{k}(m-k+1)^{n+m-k},$$
if you put apart the first term looks like
$$RHS = (m+1)^{n+m}-\sum _{k=1}^m\binom{m}{k}(-1)^{k-1}(m-k)^{k}(m-k+1)^{n+m-k},$$
you can think of this as functions from $[n+m]$ to $[m+1]$ with some particular property. The property being the same as for the LHS. Here, you will then construct sets $$A_i = \{f:[n+m]\rightarrow [m+1]:i\text{ is not in the image and }f(i)\neq m+1\}.$$ So, by the PIE principle, you can think of the RHS as
$$\left |[m+1]^{[n+m]}\setminus \bigcup _{i=1}^mA_i\right |.$$


问:
Finding $\frac{\cos(x-z)}{\cos(w-y)}$ if $11\sin x+7\sin y=3(\sin z+3\sin w)$ and $11\cos x+7\cos y=3(\cos z+3\cos w)$
So I was given this question:

$$11\sin x+7\sin y =3(\sin z+3\sin w)$$
$$11\cos x+7\cos y=3(\cos z +3\cos w)$$
Find $\dfrac{\cos(x-z)}{\cos(w-y)}$.

I tried squaring the equations and adding them, but there were too many variables. Could someone give me an insight on how to approach this?

答:
You have the right idea, moving some parts, squaring and adding equations will lead to:

$$121-66(\sin x \sin z+\cos x\cos z)+9=81-112(\sin w \sin y+\cos w\cos y)+49$$ or

$$130-66 \cos (x-z)=130-112 \cos (w-y)$$ Can you finish?


问:
$X_1,...,X_n \text{ iid }X_1\sim \operatorname{Ge}\big( \theta\big)\implies \mathbf X =(X_1,...,X_n)^T \in \text{ Exponential Family}$

$$\bullet \text{ Let} \quad X_1,X_2,...,X_n \text{ iid }X_1\sim \operatorname{Ge}\big( \theta\big)$$
$$\text{Goal: } \mathbf X =(X_1,...,X_n)^T \in \text{ Exponential Family of distributions.}$$

$$\text{ Its enough to show that :}$$

$$\exists \phi \in \Phi \subseteq \mathbb R^k,k\in\mathbb N, \text{parametric space s.t pmf:}$$

$$f(\mathbf X;\phi)=h(x)\cdot \exp{\bigg\{ \big< s(\mathbf X),\phi \big>_{\mathbb R^k} -K(\phi)\bigg\}},\mathbf X\in \mathcal X ,\phi\in\Phi . $$
$$\text{Note : $\big< \cdot \big>$ Inner product on  $\mathbb R^k$}.$$

$$\text{pmf of Geometric distr: }\Omega =\mathbb N,\operatorname{P}\big( X=x\big)=f_X(x)=(1-\theta )^{x-1}\theta .$$

$$\text{cdf: }F_X(x)=\operatorname{P}\big( X\le x \big)=1-(1-\theta )^x $$

$\text{So,}$
$$F_{\mathbf X}(\mathbf x)=\operatorname{P}\big(X_1\le x_1,X_2\le x_2,...,X_n\le x_n \big)=$$

$$ \color{black}{\underbrace{ \operatorname{P}\big(X_1\le x_1\big) \operatorname{P}\big(X_2\le x_2\big)\cdots \operatorname{P}\big( X_n\le x_n \big)=   }_{\text{independent}  }}$$
$$=\color{black}{\underbrace{ \operatorname{P}\bigg(X_1\le x_1 \bigg)^n   }_{ \text{identically }}}\implies F_{\mathbf X}'(\mathbf x)=n\operatorname{P}\big( X_1\le x_1\big)^{n-1}\cdot f_{X_1}(x_1) $$

$$=n\big( 1-(1-\theta )^{x_1}\big)^{n-1}\cdot \theta \big( 1-\theta \big)^{x_1-1}=f_{\mathbf X}(\mathbf x)$$

So, how can i continue the proof from this point?

This is as far as  i can get...

Any ideas?

Thank you.

答:
let $\theta = p$ and $1-\theta = q$, then
$$
P_{\mathrm{X}}(x_1,...,x_n) = \prod_{i=1}^n q^{x_i-1}p=q^{\sum x_i - n}p^n,
$$
applying $x = g(x) = \exp\{\ln x\}$ you have
$$
P_{\mathrm{X}}(x_1,...,x_n) = \exp\Big\{(\sum x_i - n)\ln q+n\ln p\Big\}
$$


问:
$\frac{b^{2n}+b^{n+1}+3b-5}{b-1}$ is square
Find all $b>5$ so that $x_n = \frac{b^{2n}+b^{n+1}+3b-5}{b-1}$ is square for all sufficiently so large integers n.
I think the only value of $b$ is 10.

If there is $p \in \mathbb{P}$ (prime), $p \neq 3, p \mid b-1$, then $v_p(b-1)=1$. If $v_p(b-1) > 1$, we can choose n.

If $b=3^m+1$, then m must be even.(If m is odd, $r_8(x_n)=3$)

What should I do next?

答:
COMMENT: May be using these ideas help:

1-Suppose $\frac{b^{2n}+b^{n+1}+3b-5}{b-1}= (b+a)^2$

If polynomial $f(b)=b^{2n}+b^{n+1}+3b-5$ must have a double root like $-a$ then we must have:

$f(-a)=f'(-a)=0$

This gives following system of equations:

$\begin{cases}\frac{(-a)^{2n}+(-a)^{n+1}-3a-5}{-a-1}=0\\2n(-a)^{2n-1}+(n+1)(-a)^n+3=0 \end {cases}$

Which gives a and n.Plugging these in fraction gives an equation in terms of b which gives b.

2- We must have:

$\frac{b^{2n}+b^{n+1}+3b-5}{b-1}=(b+a)^2$

Which gives:

$b^{2n}+b^{n+1}+3b-5=(b-1)(b+a)^2$

Or:

$b^{2n}+b^{n+1}-b^3-(2a-1)b^2-(a^2-2a)b+a^2-5$

Which is a Diophantine equation to be solved.


问:
If $0$ is the unique critical value of $f$, than, with $ab>0$, $f^{-1}(a)$ and $f^{-1}(b)$ are diffeomorphics
I've have a huge problem to solve (at least to me): Let $f(x,y,z) = x^3 + y^2 -zy^2$. In fact, $0$ is the only critical value of $f$. Here comes the real deal: How can I prove that, if $ab>0$, than $f^{-1}(a)$ and $f^{-1}(b)$ are diffeomorphics? To solve it, one can use the Inverse Function Theorem or the Implicit Function Theorem.

After that, one more question: How to write the equation of the tangent plane to $X = f^{-1}(a)$ in a point $p=\left(\alpha,\beta,\gamma \right)$?

Thanks in advance!

答:
This is basically the  First Fundamental Theorem of Morse theory. As in the comments, define

$X_x=\begin{cases} \frac{\nabla f(x)}{\|\nabla f(x)\|^2} \quad x\in f^{-1}((a-\epsilon,b+\epsilon))\\ 0\quad \quad \quad \text{otherwise}\end{cases}$

This is well-defined because there are no singularities of $f$ between $a$ and $b$ so there is some $\epsilon>0$ neighbrhood on which this is still true. (why?).

All this is is the normalized gradient vector restricted to a certain piece of the domain of $f$.

Notice that $\phi:\mathbb R\times \mathbb R^3\to \mathbb R^3$ defined by $(t,x)\mapsto \frac{1}{\|\nabla f(x)\|^2}(t\partial_1f(x),\cdots,t\partial_3f(x) )$ satisfies

$X_{\phi(t,x)}=\frac{d(\phi(t,x))}{dt}$ and is a diffeomorphism for each $t\in \mathbb R.$ Note also that $\phi_{-t}\circ \phi_t=\text{id}.$

The claim is now that $\phi_{b-a}$ is a diffeomorphism onto $f^{-1}(b)$ when restricted to $f^{-1}(a).$

We have $\frac{df\circ \phi(t,x)}{dt}\overset{\text{chain rule}}{=}\langle \frac{d\phi(t,x)}{dt},\nabla f(x)\rangle=\langle X_{\phi(t,x)},\nabla f(x)\rangle.$

It follows from this that $\frac{df\circ \phi(t,x)}{dt}=\begin{cases} 1 \quad x\in f^{-1}((a-\epsilon,b+\epsilon))\\ 0\quad \quad \quad \text{otherwise}\end{cases},\ $ and so

$f\circ \phi(t,x)=\begin{cases} t+f(x) \quad x\in f^{-1}((a-\epsilon,b+\epsilon))\\ f(x)\quad \quad \quad \text{otherwise}\end{cases}.$

Now, consider the diffeomorphism $\varphi:=\phi_{b-a}|_{f^{-1}(a)}.$ If $x\in f^{-1}(a)$ then $\phi_{b-a}(b-a,x)=b-a+f(x)=b-a+a=b$ so $\varphi$ maps  $f^{-1}(a)$ into $f^{-1}(b).$

Now suppose $y\in f^{-1}(b)$ and let $x$ be such that $\phi_{a-b}(y)=x.$ Then, $\varphi(x)=\phi_{b-a}\circ \phi_{a-b}(y)=y$ so in fact, $\varphi$ maps $\textit{onto}\ f^{-1}(b)$ and we are done.


问:
All isometries $f:V\to V$ can be written as $f = T_u\circ A$ - proof of uniqueness?

Let $f:V\to V$ be an isometry on a finite-dimensional vector space $V$. Then there exists a unique $u\in V$, and a unique orthogonal operator $A:V\to V$ such that $f = T_u\circ A$. This helps conclude that all isometries on $V$ can be written in this form.

I will denote the translation map which sends all $v\in V$ to $v+u\in V$ by $T_u$, i.e. $T_u(v) = u+v$.

Using this post as a Lemma, we see that the map $f - f(0)$  is orthogonal. Let's say $u = f(0) \in V$. So, $f-f(0) = T_{-u}\circ f = A$ where $A$ is an orthogonal map on $V$. So we have $f = T_u\circ A$.

How do I prove the uniqueness of $u\in V$ and $A\in\mathcal{L}(V,V)$?

Usually, we work with cases where we have to show the uniqueness of only one of the variables, however, it's a little confusing here since we have to show the uniqueness of $u$ and $A$ both. To begin, I would assume that the isometry $f$ on $V$ can be written as $f = T_{u_1}\circ A_1 = T_{u_2}\circ A_2$ for some $u_1,u_2\in V$ and orthogonal $A_1,A_2\in\mathcal{L}(V,V)$. What next?

Thanks!

Body:
If $f = T_{u_1} \circ A_1 = T_{u_2} \circ A_2$, then we have
$$
f(0) = T_{u_1}(A_1(0)) = T_{u_1}(0) = u_1,
$$
and similarly $f(0) = T_{u_2}(0) = u_2$. Thus, $u_1 = u_2$. From there, we have
$$
A_1 = T_{u_1}^{-1} \circ f = T_{u_2}^{-1} \circ f = A_2.
$$


问:
The number of solutions for $|x^2-5|x|+6|=|\cos x|$
I wish to find the number of solutions for $|x^2-5|x|+6|=|\cos x|$ graphically

I can graph $|\cos x|$ to some accuracy but struggle in graphing $|x^2-5|x|+6|$

Is this question solvable graphically? If so could a method to sketch the graph be given.

Thanks in advance.

答:
To Plot the Graphs

$x^2-5|x|+6$ and $\cos(x)$ are even functions, so plot $\left|\,x^2-5x+6\,\right|$ and $|\cos(x)|$ for $x\ge0$, and then reflect across the $y$-axis. Therefore, for $x\ge0$, plot $x^2-5x+6$ and $\cos(x)$ and reflect their graphs across the $x$-axis where they are negative (dashed graphs), and then reflect across the $y$-axis.

The graphs intersect at $4$ points.

To Find the Points of Intersection

Since the functions are even, we can find the points of intersection for $x\ge0$ and reflect them across the $y$-axis (as we did the graphs). Note that, for $x\ge0$, the graph of $x^2-5x+6$ (not its reflection) intersects the graph of $-\cos(x)$ (the reflection of $\cos(x)$) at two points. So using Newton's Method (or whatever root finding algorithm you wish), solve
$$
x^2-5x+6=-\cos(x)
$$
to get the roots
$$
(1.80540492,0.23246232)\quad\text{and}\quad(3.57574485,0.90722699)
$$
whose $y$-reflections are
$$
(-1.80540492,0.23246232)\quad\text{and}\quad(-3.57574485,0.90722699)
$$


问:
Proving limit using definition
I need to prove $$ \lim_{x\rightarrow\ 0}\frac{x^2-8}{{x-8}} =1 $$ using epsilon-delta definition.
I know I need to show that for every $\epsilon >0$ there exist a $\delta >0$ such that if $|x|<\delta$, then $| {\frac{x^2-8}{x-8}}-1|<\epsilon$

WHat I did:

$|{\frac{x^2-8}{x-8}}-1|=|{\frac{x^2-x}{x-8}}|$, but Im having troubles now deciding what my delta should be.

I would really appreciate if someone could give me an explanation on how should I choose delta and why.
Thanks.

答:
There are many ways to get an upper bound for $\left| \frac{x-1}{x-8} \right|$ when $x$ is sufficiently small. A couple of things you can do to make this easier: 1) Use the triangle inequality, as one of the answerers did; 2) separate a constant so that you only need to deal with one term involving $x$. So if $|x| < 1$ then

$$\left| \frac{x-1}{x-8} \right|=\left| 1+\frac{7}{x-8} \right| \le 1+\frac{7}{8-x} \le 1+\frac{7}{8-1}=2.$$

Then you have $\delta = \min \left(1, \frac{\varepsilon}{2}\right).$

Title: Proving limit using definition


问:
show that the strip is mappable to the half plane(conformal map)
This is part of another problem.
show that for any real numbers ${a<b}$, the strip ${\{ z \in {\bf C}: a < \mathrm{Re}(z) < b \}}$ is complex diffeomorphic to the disk ${D(0,1)}$

I tried to map the strip to the upper half plane using an exponential and a linear transformation as described in the hint. So I took $z \to \lambda e^{z}$ Then I have x going from $a+i \infty \to a - i\infty$ $f(z)$ goes from $\infty \to 0$. Similary as x goes from $a-i\infty$ to $b-i\infty$ we have f(z) going from $0 \to 0$. Similarly as x goes from $b-i\infty$ to $b+i\infty$ I have $f(z)$ go from $0 \to \infty$.as x goes from $b+i \infty$ to $b - i \infty$ f(z) goes from $\infty \to 0$. I have made a mistake. What could that be?

答:
HINT

$e^z$ maps the strip $0<\Im(z)<\pi$ to the upper half plane, so start by mapping the given strip to this strip.