问:
Iwahori versus Bruhat decompositions
I am faced with the following issue that I do not understand but seems contradictory, coming from the book of Roberts and Schmidt about $GSp(4)$. Consider a local non-archimedean field $F$, let $p$ be its maximal ideal, $\mathcal{O}$ its ring of integers and $G=GSp(4, F)$. We are interested in the following Klingen congruence subgroup
$$K =
\left(
\begin{array}{cccc}
\mathcal{O} & \mathcal{O} & \mathcal{O} & \mathcal{O} \\
p & \mathcal{O} & \mathcal{O} & \mathcal{O} \\
p & \mathcal{O} & \mathcal{O} & \mathcal{O} \\
p & p & p & \mathcal{O}
\end{array}
\right)
$$

(from now on all the subgroups written in this matrix-entries form is meant to be their intersection with $GSp(4, F)$. I am interested in computing the index of this subgroup in the maximal compact subgroup $K_0$ (where all the entries are integers).

Iwahori decomposition
We have
$$
K =
\left(
\begin{array}{cccc}
1 &  & &  \\
p & 1 & &  \\
p &  & 1 &  \\
p & p & p & 1
\end{array}
\right)
\left(
\begin{array}{cccc}
\mathcal{O}^\times & &  &  \\
 & \mathcal{O} & \mathcal{O} &  \\
 & \mathcal{O} & \mathcal{O} &  \\
 &  &  & \mathcal{O}^\times
\end{array}
\right)
\left(
\begin{array}{cccc}
1 & \mathcal{O} & \mathcal{O} & \mathcal{O} \\
 & 1 &  & \mathcal{O} \\
 &  & 1 & \mathcal{O} \\
 &  &  & 1
\end{array}
\right)
$$

so that in particular by decomposing the left subgroup we should obtain
$$
\left(
\begin{array}{cccc}
\mathcal{O} & \mathcal{O} & \mathcal{O} & \mathcal{O} \\
\mathcal{O} & \mathcal{O} & \mathcal{O} & \mathcal{O} \\
\mathcal{O} & \mathcal{O} & \mathcal{O} & \mathcal{O} \\
\mathcal{O} & \mathcal{O} & \mathcal{O} & \mathcal{O}
\end{array}
\right)
=
\bigsqcup_{a, b, c \in \mathcal{O}/p}
\left(
\begin{array}{cccc}
1 &  &  &  \\
a & 1  & &  \\
b & & 1 & \\
c & b & -a & 1
\end{array}
\right)
K
$$

(where the fact that the entries on the right are this way comes from the conditions of belonging to $GSp(4)$). So that in particular the index should be, writting $N(p)$ for the norm of $p$,

$$[K_0:K] =N(p)^3$$

Bruhat decomposition
On the other hand, introducing the subgroup
$$
Q =
\left(
\begin{array}{cccc}
\mathcal{O} & \mathcal{O} & \mathcal{O} & \mathcal{O} \\
 & \mathcal{O} & \mathcal{O} & \mathcal{O} \\
 & \mathcal{O} & \mathcal{O} & \mathcal{O} \\
 & & & \mathcal{O}
\end{array}
\right)
$$

the Bruhat decomposition yields that for any field $k$,
$$
GSp(4, k) = Q
\sqcup Qx
\left(
\begin{array}{cccc}
1 & k &  & \\
 & 1 &  & \\
 & & 1 & k \\
 & & & 1
\end{array}
\right)
\sqcup
Qxy
\left(
\begin{array}{cccc}
1 & & k  &  \\
 & 1 & k & k \\
 & & 1 &  \\
 & & & 1
\end{array}
\right)
\sqcup
Qxyx
\left(
\begin{array}{cccc}
1 & k & k & k\\
 & 1 &  & k\\
 & & 1 & k \\
 & & & 1
\end{array}
\right)
$$

where the transformations $x$ and $y$ are defined by
$$x=
\left(
\begin{array}{cccc}
 & 1&  & \\
1 &  &  & \\
 & &  & 1 \\
 & &1 &
\end{array}
\right)
$$
$$y =
\left(
\begin{array}{cccc}
1 &  &  &\\
 &  & 1 & \\
 & -1 & &  \\
 & & & 1
\end{array}
\right)
$$

In particular if $k$ is the finite field with $N(p)$ elements, the index we search for is exactly the cardinality of $GSp(4,k) / Q$, and this one is $(1+N(p))(1+N(p)^2)$, so that we should say

$$[K_0:K] =(1+N(p))(1+N(p)^2)$$

Here is the question following from this discussion:

Both results are different, what is
  happening?

答:
Your Iwahori-decomposition computation is a bit too free with quotient computations.  It would work fine for vector spaces, which is, in some sense, why you get the correct leading term in your count; but there are additional subtleties on the group level that it does not take into account—among other things, that some of the entries denoted by $\mathcal O$ can be $0$, but some cannot!

The same computation would suggest that the quotient of $\operatorname{GL}_2(k)$ by its Borel subgroup of upper-triangular matrices should have as coset representatives the lower uni-triangular matrices $u_-(c) = \begin{pmatrix} 1 & 0 \\ c & 1 \end{pmatrix}$, whereas actually there is another coset, the one containing $w_0 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$.  Actually a better way to think about it is that most coset representatives are of the form $u_+(c)w_0$, where $u_+(c)$ is the transpose of $u_-(c)$, and only the trivial coset is not of this form; it just happens that $u_+(c)w_0$ and $u_-(c^{-1})$ lie in the same coset when $c \ne 0$.

Your count via the Bruhat decomposition shows what’s going on here:  you get $(1 + N(p))(1 + N(p)^2) = 1 + N(p) + N(p)^2 + N(p)^3$, which suggests, correctly, that we have stratified $\mathbb P^3k$ by affine spaces of the obvious dimensions, corresponding to the Weyl-group elements $1$, $x$, $x y$, and $x y x$ (which are minimal-length representatives in the quotient by an appropriate parabolic subgroup, which is doubtless what led you to choose them!).

Title: Iwahori versus Bruhat decompositions


问:
Limit of the product of functions equaling the product of the limits
Suppose that $f,g: \mathbb{R} \to \mathbb{R}$ are continuously differentiable on $(-1,1)$. We are asked to show that the product $fg$ is continuous, but $f$ is a function of $x$, and $g$ is a function of $y$. We know that both f,g are continuous, but $f$ and $g$ are differentiable on different domains. What changes from the proof of this, and the proof of $\lim\limits_{x \to a}(fg)(x) = \lim\limits_{x \to a}(f(x))\lim\limits_{x \to a} (g(x)).$

答: If I understand correctly, not much changes. A sequence of points (x,y) will converge to a point, say (x0, y0) in the two-dimensional space if and only if each of x and y converge to x0 and y0 in their respective one dimensional spaces. So the two-dimensional result basically follows from the one-dimensional one in this case; in other words f(x)g(y) will converge to f(x0)g(y0).</p>

ContentLicense: CC BY-SA 4.0


问:
Intuition on the Cuntz-Krieger relations for Leavitt path algebras and graph $C^*$-algebras
Let $k$ be a commutative ring with unity and $E$ a quiver with source and target functions $E^1 \xrightarrow{s,t} E^0$. The Leavitt path algebra of $E$ is the quotient of the path algebra of the double graph $k(E \sqcup E^\ast)$ by the Cuntz-Krieger relations

$$
e^\ast f := \delta_{s,f} \ t(f) \tag{CK 1}
$$

for each pair of edges $e,f \in E^1$ and

$$
v = \sum_{e  \ : \ s(e) = v}\tag{CK 2}ee^\ast
$$

for each regular vertex $v \in E^0$. This means that $s^{-1}(v)$ is non-empty and finite, which means that $v$ is not a sink or an infinite emitter.

This is analogous to the construction of graph $C^*$-algebras, where the original Cuntz-Krieger relations arose, and include several algebras of interest. However, I would like to gain some intuition as to why these relations were considered in the first place.

The original paper of Cuntz seems to relate $(CK1)$ and $(CK2)$ to topological Markov chains and symbolic dynamics, but I've had a hard time following the motivation. I'd really appreciate an explanation for non-experts.

答:
If $A$ is an $n\times n$ matrix of zeros and ones,  the Markov space
$$
  \Sigma _A=\big\{x=(x_k)_{k=1}^\infty \in  \{1, 2, \ldots , n\}^\mathbb N: A_{x_k, x_{k+1}}=1, \text{ for all } k\in \mathbb N\big\}
  $$
may be partitioned by the so called cylinder sets
$$
  Z_i = \big\{x=(x_k)_k\in  \Sigma _A: x_1=i\big\},
  $$
defined for each $i$ in $\{1, 2, \ldots , n\}$.

The shift map
$$
  S: (x_1, x_2, x_3, \ldots ) \mapsto  (x_2, x_3, x_4, \ldots )
  $$
maps $\Sigma _A$ to itself, and it is evidently injective when restricted to each cylinder $Z_i$.  We may then consider the
inverse branches of the shift, namely the maps
$$
  \theta _i : (x_1, x_2, \ldots ) \in    S(Z_i) \mapsto    (i,  x_1, x_2, \ldots ) \in   \Sigma _A.
  $$

Thus, while $S$ deletes the first coordinate of $x$, we see that $\theta _i$ inserts $i$ as a first coordinate of $x$.
However insertion cannot always be done since  $(i, x_1, x_2, \ldots )$ may sometimes fall outside $\Sigma _A$.
In fact notice that the
domain of each $\theta _i$ turns out to be
$$
  S(Z_i) = \bigcup_{{\buildrel {1\leq j\leq n} \over {A_{i, j}=1}}} Z_j.
  \tag 1
  $$
On the other hand  the  range of $\theta _i$ is precisely $Z_i$.

The original Cuntz-Krieger paper may be considered as a quantization of the Markov shift, where  $\theta _1, \theta _2, \ldots , \theta _n$ are replaced
by partial isometric operators  $S_1, S_2, \ldots , S_n$  on a Hilbert space.   The fact that the range  projections
$$
  P_i:= S_iS_i^*
  $$
are
required to add up to one may be seen as an interpretation of the fact that  $\Sigma _A$ is partitioned by the ranges of the
$\theta _i$.
The Cuntz-Krieger relation
$$
  S_i^*S_i = \sum_{j=1}^n A_{i,j} P_j,
  $$
if written in the equivalent form
$$
  S_i^*S_i = \sum_{{\buildrel {1\leq j\leq n} \over {A_{i, j}=1}}}  P_j,
  $$
expresses that the source projections   $S_i^*S_i$ behaves somewhat like the domains of the $\theta _i$, as  in (1).

It is therefore perhaps unsurprising that one of the main early results about the Cuntz-Krieger algebras  is that two Markov shifts are
conjugate iff the corresponding Cuntz-Krieger algebras are isomorphic.


问:
Evaluate $\int \frac{1+\cos(x)}{\sin^2(x)}\,\operatorname d\!x$
I`m trying to solve this integral and I did the following steps to solve it but don't know how to continue.
$$\int \dfrac{1+\cos(x)}{\sin^2(x)}\,\operatorname d\!x$$

$$\begin{align}\int \dfrac{\operatorname d\!x}{\sin^2(x)}+\int \frac{\cos(x)}{\sin^2(x)}\,\operatorname d\!x &= \int \dfrac{\operatorname d\!x}{\sin^2(x)}+\int \frac{\cos(x)}{1-\cos^2(x)} \\
&=\int \sin^{-2}(x)\,\operatorname d\!x + \int \cos(x)\,\operatorname d\!x - \int \frac{\operatorname d\!x}{\cos(x)}\end{align}$$
Any suggestions how to continue?

Thanks!

答:
Hints:

$1+\cos x=2 \cos^2 \dfrac{x}{2}$

$\sin^2 x=(2\sin\dfrac{x}{2} \cos \dfrac{x}{2})^2$

You expression will be $\dfrac{1}{2} \int \dfrac{1}{\sin^2\dfrac{x}{2}}$

ContentLicense: CC BY-SA 3.0


问:
If a function $f : D \rightarrow \mathbb{R}$ is continuous for all $x \in D$, does that imply the domain is an interval?

Context

I am working through Advanced Calculus, Fitzpatrick. Fitzpatrick uses a different definition of continuous than most other books, such as Rudin.

A function $f : D \rightarrow \mathbb{R}$ is said to be continuous at the point $x_{0}$ in $D$ provided that whenever $\{x_n\}$ is a sequence in $D$ that converges to $x_0$, the image sequence $\{f(x_n)\}$ converges to $f(x_0)$. The function $f : D \rightarrow \mathbb{R}$ is said to be continuous provided that it is continuous at every point in $D$.

My question is, if a function $f : D \rightarrow \mathbb{R}$ is continuous for all $x \in D$, does that imply the domain $D$ is an interval? Why?

Attempt

Rough sketch of a proof idea I had, trying a proof by contradiction. This is not rigorous, just trying to sketch my thoughts

Assume $D$ is an interval $[a, b]$. Then "remove" some $x \in [a, b]$, so $f$ is not defined on $x$.

Let $\{x_n\}$ be some sequence in $[a, b]$. Then since it is bounded, by the sequential compactness theorem, there is some subsequence $\{x_{n_k}\}$ that converges to $x$. Then since $f$ is continuous, the image $\{f(x_{n_k})\}$ converges to $f(x)$. But $f$ is not defined on $x$, which is a contradiction. Therefore $D$ is an interval.

I doubt this is right, but it's where I'm at after reading a few similar proofs in the book (Extreme Value Theorem, Intermediate Value Theorem). Would be helpful if any explanations use the Fitzpatrick definition of continuity.

答: No it doesn't. Take $D=[0,1]\cup [2,3]$ and let $f(x)=1$ for all $x\in D$.


答: This is not true, in fact one can show e.g. that any function $f \colon \mathbb N \to \mathbb R$ is continuous. This would be because any sequence $(a_n)_{n \in \mathbb N}$ converging to some $x \in \mathbb N$ will already have $a_m = x$ for all $m$ larger than some finite threshold and hence, $(f(a_n))_{n \in \mathbb N}$ must converge to $f(x)$, making $f$ continuous.

Title: If a function $f : D \rightarrow \mathbb{R}$ is continuous for all $x \in D$, does that imply the domain is an interval?

Body:
Context

I am working through Advanced Calculus, Fitzpatrick. Fitzpatrick uses a different definition of continuous than most other books, such as Rudin.

A function $f : D \rightarrow \mathbb{R}$ is said to be continuous at the point $x_{0}$ in $D$ provided that whenever $\{x_n\}$ is a sequence in $D$ that converges to $x_0$, the image sequence $\{f(x_n)\}$ converges to $f(x_0)$. The function $f : D \rightarrow \mathbb{R}$ is said to be continuous provided that it is continuous at every point in $D$.

My question is, if a function $f : D \rightarrow \mathbb{R}$ is continuous for all $x \in D$, does that imply the domain $D$ is an interval? Why?

Attempt

Rough sketch of a proof idea I had, trying a proof by contradiction. This is not rigorous, just trying to sketch my thoughts

Assume $D$ is an interval $[a, b]$. Then "remove" some $x \in [a, b]$, so $f$ is not defined on $x$.

Let $\{x_n\}$ be some sequence in $[a, b]$. Then since it is bounded, by the sequential compactness theorem, there is some subsequence $\{x_{n_k}\}$ that converges to $x$. Then since $f$ is continuous, the image $\{f(x_{n_k})\}$ converges to $f(x)$. But $f$ is not defined on $x$, which is a contradiction. Therefore $D$ is an interval.

I doubt this is right, but it's where I'm at after reading a few similar proofs in the book (Extreme Value Theorem, Intermediate Value Theorem). Would be helpful if any explanations use the Fitzpatrick definition of continuity.


答:
In the definition we are discussing continuity at an $x_0\in D$, so we consider sequences in $D$ converging to $x_0$. If a point is not in $D$, i.e., if $f$ is not defined at a point, then it is meaningless to ask about the continuity of $f$ at that point.

Secondly, it is not true that any sequence in $[a, b]\setminus \{x\}$ has a subsequence converging to $x$, it could converge to other points as well.

Lastly, it doesn't mean that $D$ should be an interval, for example the identity function is continuous on $(0, 1)\cup (2, 3]\cup\{4\}$.


问:
diagonal matrices kernels
Reconsider A =(5,4,3;-1,0,-3;1,-2,1). We wish to find a basis which puts A in a nearly
diagonal form J = (4,1,0;0,4,0;0,0,2)

i. Select (your choice) a pair of non-zero vectors w ∈ E−2(A) and y ∈ Ker((A − 4I3))^2 \ Ker(A − 4I3). Compute the vector (A − 4I3)y based on your choice of y.

ii. Show that your choice of vectors {(A − 4I3)~y, ~y, ~w} in (i) forms a basis of R^3.
Hence,
create a 3 × 3 matrix P such that P^-1AP = J.

A and J are 3x3 matrix. E−2(A) is the eigenspace associated with eigenvalue -2. Please help with this question.

答:
I'll compute one of the kernels for you to get you started:  $E_{-2}(A)=\rm{ker}(A+2I_3)$.  Form the matrix and row reduce.  We get $\begin{pmatrix}7&4&3\\-1&2&-3\\1&-2&3\end{pmatrix}\to\begin{pmatrix}1&-2&3\\0&18&-18\\0&0&0\end{pmatrix}\to\begin{pmatrix}1&-2&3\\0&1&-1\\0&0&0\end{pmatrix}$.  Now that it's row-reduced, you can read off the kernel by "back-substitution".  You get $y=z$ and $x=2y-3z$.  So we can get a solution by setting say $y=z=1$, and $x=2(1)-3(1)=-1$. That is $\begin{pmatrix}-1\\1\\1\end{pmatrix}$.  So that can be $w$.

Etc...

Title: diagonal matrices kernels


问:
For what values of alpha does the series converge
For what values of $\alpha >0$ does the series $$\sum_{k=1}^{\infty} \frac{1}{(k+1)\ln(k+1)^\alpha}$$ converge?

In the case that $\alpha = 1$, we have the $$\sum_{k=1}^{\infty} \frac{1}{(k+1)\ln(k+1)}$$ Now we consider the function $$ f(x) = \frac{1}{(x+1)\ln(x+1)}$$ for $f:[1, \infty)\rightarrow \mathbb{R}$. This function is monotonically decreasing and continuous, however, for each index $n$, $$ \int_{1}^{n} f(x)dx = \int_{1}^{n} \frac{1}{(x+1)\ln(x+1)} dx = \ln[\ln(n+1)]-\ln(\ln2)$$ by the First Fundamental Theorem (Integrating Derivatives). This sequence $\ln[\ln(n+1)]-\ln(\ln2)$ is not bounded, and therefore by the integral test, the series $$\sum_{k=1}^{\infty} \frac{1}{(k+1)\ln(k+1)}$$ diverges. Now we will have two more cases where $0<\alpha<1$ and where $\alpha>1$. This is where I get stuck. The ratio and root test are inconclusive from my attempts of using them. I would like to use the comparison test but not sure what I would compare this series too. I was wondering if anyone could please help me out.

答:
You already resolved the case when $\alpha = 1$. So if $\alpha \neq 1$, let $u_k=\frac 1 {1-\alpha}\ln^{1-\alpha}k$. Then
$$\begin{split}
u_{k+1}-u_k &=\frac 1{1-\alpha}\left [\left( \ln k + \ln\left(1+\frac 1 k\right)\right)^{1-\alpha}-\ln^{1-\alpha}k\right]\\
&= \frac 1{1-\alpha}\left [\ln^{1-\alpha}(k)\left( 1 + \frac{\frac 1 k +\mathcal O\left(\frac 1 {k^2}\right)}{\ln k})\right)^{1-\alpha}-\ln^{1-\alpha}k\right]\\
&=\frac 1 {k\ln^\alpha k} + \mathcal O\left(\frac 1 {k^2\ln^\alpha k}\right)\\
\end{split}$$
Because $\sum_{k>2}\frac 1 {k^2\ln^\alpha k}$ always converges,  $\sum_{k>2}\frac 1 {k\ln^\alpha k}$ converges iff $\{u_k\}$ converges. That is, if and only if $\alpha > 1$.


问:
Derivation of Poisson from Binomial
I am not very well versed in statistics so any clarification would be appreciated. I understand the mathematical derivation of Poisson from Binomial.

I can see just from plotting various Binomial distributions where I keep p constant and increase n, that Binomial will eventually converge to a Poisson.

That fits with the derivation where n approaches infinity, but I don't see why p has to be very small? What is the intuition behind p approaching 0?

答: The idea is that the poisson distribution is a one-parameter distribution (often $\lambda$), and that $\lambda$ corresponds to $np$ in the binomial distribution. So the limit you want is for $n$ to get larger while $np$ stays fixed - and this is why $p$ has to get smaller. Both $\lambda$ and $np$ count the number of events, which is why the correspond in this way - and the limit is taken while keeping the mean number of events constant.</p>

Title: Derivation of Poisson from Binomial


问:
fixing dual basis to find basis of vector space
Given a finite-dimensional vector space $V$ with basis $\{v_1,...,v_n\} \subset V$, we also fix a basis for dual space $V^*$ by $\{v_1^*,...,v_n^*\} \subset V^*$ where:
$v_i^*(v_j) = 1$ if $j = i$ and $0$ otherwise. If on the other hand, we are given the vector space $V$ and dual space $V^*$ and a basis $\{f_1,...,f_n\}$ of $V^*$, is it possible to find $v_1,...,v_n \in V$ such that $v_i^* = f_i$?

答:
Yes, it is. The key point is that for finite-dimensional spaces, there is a natural isomorphism $\theta:V\to V^{**}$, given by $\theta(v)=\text{evaluation at $v$} = \left(f\mapsto f(v)\right)$. If you want, I can elaborate more on this, but this is a common isomorphism, so you should be able to read up more on this easily.

Now, taking this for granted, consider the dual basis $\{f_1^*,\dots, f_n^*\}$ of $V^{**}$. Then, because $\theta:V\to V^{**}$ is an isomorphism, we can consider the vectors $v_i:= \theta^{-1}(f_i^*)$. Now, it's a matter of unwinding definitions to see that $\{v_1,\dots, v_n\}$ is a basis for $V$, whose dual is $\{v_1^*,\dots, v_n^*\}=\{f_1,\dots, f_n\}$.


答: Yes. Consider the dual basis $\phi_1,\dots,\phi_n$ of $f_1,\dots,f_n$. Now, since $v \mapsto E_v$ is surjective, we have that $\phi_i = E_{v_i}$ for some $v_i \in V$ (for $i=1,\dots,n$). Finally, observe that $$\forall i, j \in \{1,\dots,n\} : \quad f_i(v_j) = E_{v_j}(f_i) = \phi_j(f_i) = \delta_{ij},$$ which means that $v_1,\dots,v_n$ is the desired basis for $V$.

Title: fixing dual basis to find basis of vector space

Body:
Given a finite-dimensional vector space $V$ with basis $\{v_1,...,v_n\} \subset V$, we also fix a basis for dual space $V^*$ by $\{v_1^*,...,v_n^*\} \subset V^*$ where:
$v_i^*(v_j) = 1$ if $j = i$ and $0$ otherwise. If on the other hand, we are given the vector space $V$ and dual space $V^*$ and a basis $\{f_1,...,f_n\}$ of $V^*$, is it possible to find $v_1,...,v_n \in V$ such that $v_i^* = f_i$?


问:
Coin toss problem, probability
A coin has the property that the probability that it will land heads up when tosses $1$ time is $0.6$.
If the coin is to be tossed $3$ times, what is the probability that it will land heads up at least $2$ times.

Not quite know how to approach this question, especially I'm always bad at interpreting the probability of "at least" events.

Any suggestions?

答:
In the present context, at least two times means that it will land heads up two or three times.

Since each trial is independent of the others, one has the probability:
\begin{align*}
p = 0.6\times 0.6\times 0.4 + 0.6\times 0.4\times 0.6 + 0.4\times 0.6\times 0.6 + 0.6\times 0.6\times 0.6 = 0.648
\end{align*}

If you want to formalize it further, you can think of $H_{j}$ as the event 'outcome equals head in th $j$-th trial' and $T_{j}$ as the event 'outcome equals tail in the $j$-th trial'. Moreover, if we assume independence, we are interested in the following probability:
\begin{align*}
p & = \mathbb{P}(H_{1}\cap H_{2}\cap T_{3}) + \mathbb{P}(H_{1}\cap T_{2}\cap H_{3}) + \mathbb{P}(T_{1}\cap H_{2}\cap H_{3}) + \mathbb{P}(H_{1}\cap H_{2}\cap H_{3})\\\\
& = \mathbb{P}(H_{1})\mathbb{P}(H_{2})\mathbb{P}(T_{3}) + \mathbb{P}(H_{1})\mathbb{P}(T_{2})\mathbb{P}(H_{3}) + \mathbb{P}(T_{1})\mathbb{P}(H_{2})\mathbb{P}(H_{3}) + \mathbb{P}(H_{1})\mathbb{P}(H_{2})\mathbb{P}(H_{3})
\end{align*}
which equals $0.648$ as previously mentioned.

Title: Coin toss problem, probability


问:
$\displaystyle{\lim \limits_{x \to \frac\pi4}\frac{(\csc^2x-2\tan^2x)}{(\cot x-1)}}$ without L'Hôpitals Rule.
The limit,

$\displaystyle{\lim \limits_{x \to \frac\pi4}\frac{(\csc^2x-2\tan^2x)}{(\cot x-1)}}=6$

is easily found using L'Hôpitals Rule. However, the exercise is to evaluate it without using this method. I've tried multiple substitutions, mainly pythagoric and reciprocal identities, but without success, the indeterminate form $\frac{0}{0}$ keeps appearing. I’ve also tried splitting the limit and doing the substitutions. I'm looking to simplify the expression in order to evaluate the limit directly.

Thanks in advance.

答:
Hint:$$\begin{align*}\frac{\csc^2 x-2\tan^2 x}{\cot x-1}&=\frac{\csc^2 x-2-2\tan^2 x+2}{\cot x-1} ]\\&=\frac{\csc^2 x-2}{\cot x-1}+2\tan x\frac{1-\tan^2 x}{1-\tan x}\\&=\frac{\cot^2 x-1}{\cot x-1}+2\tan x\frac{1-\tan^2 x}{1-\tan x}\\&=\boxed{\cot x+1+2\tan x(1+\tan x)}\end{align*}$$

It is easier to take the limit now.....


问:
How to show the one point compactification of a space is unique up to homeomorphism?

Suppose $X$ is a space with a one point compactification $f: X\to Y$, where $x_0$ is the point that gets added to $X$, and suppose we have another one point compactification $f': X\to Y'$ where $x_1$ is the point that gets added to $X$.

I think we can define a homeomorphism $g: Y\to Y'$ in the following way: $g(x_0)=x_1, g(x)=f'f^{-1}(x)$ for all other $x$.  Intuitively I'm thinking this should be a homeomorphism (if it's not correct me), but I'm not seeing how to prove it.  I think it's fairly clear that it should be bijective, but how to show continuity (of both it and its inverse) isn't coming to me.  If anyone could give a hint that would be swell.

答:
The homeomorphisms $f$ and $f'$ are continuous and their inverses are continuous; therefore $f'f^{-1}$ is continuous and its inverse $ff'^{-1}$ is continuous.

That is enough to show that $f'f^{-1}$ and $ff'^{-1}$ are continuous at every point except $x_0$ and $x_1.$

Continuity of a function $f$ at a point $x$ in its domain can be characterized by saying that for every open neighborhood $V$ of $f(x)$ there is some open neighborhood $U$ of $x$ such that the image of $U$ under $f$ is a subset of $V.$

To show that $f'f^{-1}$ is continuous at $x_0,$ suppose $V$ is some open neighborhood of $f(x_0) = x_1.$ An open neighborhood of the point $x_1$ that gets added is the complement (within $X\cup\{x_1\}$) of a closed set in $X.$ So $X\smallsetminus V$ is closed in $X;$ hence the complement of $X\smallsetminus V$ within $X\cup\{x_0\}$ is an open neighborhood of $x_0.$ It can readily be shown that the image of that open neighborhood of $x_0$ is that open neighborhood of $x_1.$

Continuity of the inverse $ff'^{-}$ is shown in the same way.


问:
What do we know about the series $\sum_{k=1}^\infty \frac{1}{x+k}$?
If possible, I would like to find a closed-form expression for
$$
f(x)=\sum_{k=1}^\infty \frac{1}{x+k}
$$
The series is so simple (possibly deceptively so) that I'm sure it has been studied before somewhere, but I don't know what to call it and where to find it.

What is this series called so I can learn more? I know I can expand it into a double series by expanding $\frac{1}{1-x/k}$ for $|x|<k$ and $\frac{1}{1-k/x}$ for $|x|>k$ and working on the partial series, which I did before in the special case of $|x|<1$, and the result yields a sum over Bernoulli numbers that I'm not familiar with.

Is a nice closed-form solution for this series known?

答: This series is also basically a part of the Hurwitz zeta function in the so-called ``s=1'' case. It was introduced in the later half of the 1800's and comes up in analytic number theory. There is no closed form and the theory is not elementary.


答:
First, the "at least" part can be interpreted as either the coin landing on head $2$ times or landing on head $3$ times so we should find both probabilities and add them together, we add them since it is Event 1 OR Event 2.

The probability of getting head $2$ times in $3$ tosses is the following:

$3C2*(0.6)^2*0.4=0.432$

Where $3C2$ means $3$ combination $2$, we use combination to account for the different places head can occur in, so for example one scenario would be getting the two heads in the first and third toss, another scenario would be getting it in the first and second toss, etc... The combination lets us take into account the different places. Also, we use $0.6^2$ as we need head twice and the probability of each one occurring is $0.6$ and we used $0.4$ once as getting two heads in three tosses means getting tail once and this happens with probability $0.4.$

Similarly, getting three heads in three tosses has probability:

$3C3*(0.6)=0.216$

Adding both probabilities we get $0.216+0.432=0.648$ which is the final answer.

I don't know which part of learning probability you are in and if you have learned about discrete probability distributions but what I used is called the binomial distribution (https://en.m.wikipedia.org/wiki/Binomial_distribution), this distribution can be used when you have multiple Bernoulli trials, also known as a Bernoulli process (a Bernoulli trial is an event that has two possible outcomes only, either "success" or "failure", here we can consider getting head as success and getting tail as failure) and the trials have to be independent (so the probability of getting head is always the same) and when we need an "n" number of successes in "N" trials regardless at which attempts the successes come.

Title: Coin toss problem, probability

Body:
A coin has the property that the probability that it will land heads up when tosses $1$ time is $0.6$.
If the coin is to be tossed $3$ times, what is the probability that it will land heads up at least $2$ times.

Not quite know how to approach this question, especially I'm always bad at interpreting the probability of "at least" events.

Any suggestions?


问:
Finding the amount of magic numbers $N$
Suppose $N$ is called a magic number if it is a positive integer and when you stick $N$ on the end of any positive integer, the resulting integer is divisible by $N.$ How many magic numbers are there less than $2100?$

答:
Hint

$n$ can have at most $4$ digits.

Further, it is presumed that the leftmost digit of $n$ can not be $0$.

If $n$ has $k$ digits, then $n$ must be a divisor of $10^k$ such that $n \geq 10^{(k-1)}.$

Applying the prime factorization theorem against $10^k$ allows you to immediately identify all satisfying values of $n$ that correspond to a specific value for $k$.

Title: Finding the amount of magic numbers $N$


问:
How to find inverse of a quadratic function in standard form?
I want to find the inverse of: $y = -10 x^2 + 290 x - 1540$.

I tried using 'completing the square' to find it, but it did not work. I realize that the inverse will not be a function, but I still need this inverse.

If you want the complete question, here it is:
The solar radiation varies throughout the day depending on the time you measure it. Say the radiation on a cloudy day can be approximated by the function $R(t) = - 10 t^2 + 290 t -1540$, where $R$ represents the solar radiation in $W/m^2$, and $t$ represents the time in hours from $0$ to $24$. What's the domain and range of the function? Find the inverse of this function. What does the inverse represent? Is the inverse a function?

答:
Given a quadratic equation in the form

$$ y=ax^2+bx+c$$

it can always be re-written in the form

$$ y=a(x-h)^2+k\tag{1} $$

where $h$ is the value

$$h=-\dfrac{b}{2a} \tag{2}$$ which is the value of $x$ at the vertex of the graph.

We can find the value of $k$ by observing that it is the constant term when equation (1) is expanded. So

$$ ah^2+k=c \tag{3} $$

For your particular exercise, $a=-10, b=290, c=1540$. So from equation (2) we find that $h=14.5$ and from equation (3) we find that $(-10)(14.5)^2+k=-1540$. From this we conclude that $k=562.5$.

Substituting into equation (1) we get

$$ y=-10(x-14.5)^2+562.5 $$

To get the inverse relation we solve this equation for $x$ in terms of $y$.

\begin{eqnarray}
-10(x-14.5)^2&=&y-562.5\\
(x-14.5)^2&=&\frac{562.5-y}{10}\\
x-14.5&=&\pm\sqrt{56.25-\frac{y}{10}}\\
x&=&14.5\pm\sqrt{56.25-\frac{y}{10}}
\end{eqnarray}

Some instructors prefer that you exchange the $x$ an $y$ variables at this stage to get

$$ y=14.5\pm\sqrt{56.25-\frac{x}{10}} $$


问:
Determine the linear function $f: R \rightarrow R$
There's a linear function $f$ whose correspondence rule is $f(x)=|ax^2-3ax+a-2|+ax^2-ax+3$. State the values of the parameter $a$ that fully define the function $f$.

According to the problem condition the function $ f (x) $ is linear by
so the coefficient that accompanies $ x ^ 2 $ must be equal to 0, this is
meets when:
\begin{align*}
  ax^2-3ax+a-2 & < 0\\
 a(x^2-3x) &<2-a \\
  \
    a\underbrace{\left(x^2-3x+\frac{9}{4}\right)}_{
  \left(x-\frac{3}{2}\right)^2≥ 0} &<2-a +\frac{9}{4}a\\
  a \left(x-\frac{3}{2}\right)^2 &<2+\frac{5}{4}a\\
  0 & <2+\frac{5}{4}a \\
  -\frac{8}{5} &
\end{align*}

What else is missing to analyze?

答:
If there's some $x_0$ such that $ax_0^2-3ax_0+a-2>0$, then there's some neighborhood around $x_0$ such that $\forall x \in (x_0-\delta, x_0+ \delta)$, $ax^2-3ax+a-2>0$, $f(x)$ is quadratic unless $a=0$. On the other hand if $a=0$ then $f(x)$ is indeed linear.

If $a\ne 0$, it follows that $ax^2-3ax+a-2 \le 0, \forall x \in \mathbb R$. Therefore $a<0$ and the discriminant is non-negative, i.e.

$$(-3a)^2 \le 4 a (a-2) \iff a(5a+8) \le 0 \iff -\frac{8}{5} \le a <0.$$

Therefore $-\frac{8}{5} \le a \le 0$.


答:
$$
\frac 1 {\lfloor x \rfloor + k+1} \le \frac 1 {x+k} \le \frac 1 {\lfloor x \rfloor + k}
$$
The two series whose terms are the first and the third expressions above are just tail ends of the harmonic series that diverges to $+\infty.$

Title: What do we know about the series $\sum_{k=1}^\infty \frac{1}{x+k}$?

Body:
If possible, I would like to find a closed-form expression for
$$
f(x)=\sum_{k=1}^\infty \frac{1}{x+k}
$$
The series is so simple (possibly deceptively so) that I'm sure it has been studied before somewhere, but I don't know what to call it and where to find it.

What is this series called so I can learn more? I know I can expand it into a double series by expanding $\frac{1}{1-x/k}$ for $|x|<k$ and $\frac{1}{1-k/x}$ for $|x|>k$ and working on the partial series, which I did before in the special case of $|x|<1$, and the result yields a sum over Bernoulli numbers that I'm not familiar with.

Is a nice closed-form solution for this series known?


问:
Problem on inequality with power sum
Let $m$ and $n$ are positive integer

Can it be shown that, For every $m\ge 5$

$$\sum_{i=1}^ni^m-m^i>0\iff n=2,3,...,m$$

Example: let $m=5$, choose any $n$ between $2$ to $5$, now let $n=2$ then $\sum_{i=1}^2i^5-5^i=(1^5-5^1)+(2^5-5^2)=-4+7=3>0$

More on observation, $\sum_{i=1}^ni^m-m^i=0$ holds only for $(m,n)=\{(1,1),(2,3),(2,4)\}$

Score Pari/GP

```<code>for(m=5,50,for(n=1,50,if(sum(i=1,n,i^m-m^i)>0,print([m,n]))))
</code>```

答:
You're asking about proving for every $m \ge 5$ that

$$\sum_{i=1}^{n}(i^m - m^i) \gt 0 \iff n = 2, 3, \, \ldots \, , m \tag{1}\label{eq1A}$$

First, consider the $\impliedby$ direction, starting at $n = 2$. Define

$$f_1(m) = 1 - m + 2^m - m^2 \tag{2}\label{eq2A}$$

As you've already shown, $f_1(5) = 3 \gt 0$. Differentiating $f_1(m)$ gives

$$f'_1(m) = -1 + \ln(2)2^{m} - 2m \tag{3}\label{eq3A}$$

You can easily verify $f'(5) \gt 0$ and $f'(m)$ is an increasing function of $m$ (e.g., by induction using $2m \gt m + 1$). Thus, $f(m) \gt 0$ for $m \ge 5$. Next, since the natural logarithm is a strictly increasing function, consider the values of $i$ where

$$i^m \gt m^i \iff m\ln(i) \gt i\ln(m) \tag{4}\label{eq4A}$$

Define

$$f_2(i) = m\ln(i) - i\ln(m) \tag{5}\label{eq5A}$$

It's fairly easy to confirm $f_2(3) \gt 0$ for all $m \ge 5$. Differentiating here gives

$$f'_2(i) = \frac{m}{i} - \ln(m) \tag{6}\label{eq6A}$$

With $m = 5$, this gives $f'_2(3) \approx 0.057$. Also, $g(m) = \frac{m}{3} - \ln(m) \implies g'(m) = \frac{1}{3} - \frac{1}{m} \gt 0$, so $f'_2(3) \gt 0$ for all $m \ge 5$. Note $f'_2(i)$ is a monotonically decreasing function, with $\lim_{i \to \infty}f'_2(i) = -\ln{m} \lt 0$. Thus, there's only one value of $i$ where $f'_2(i) = 0$.

We therefore have $f_2(3) \gt 0$, with $f_2(i)$ reaching a maximum at $i = \frac{m}{\ln(m)} \lt m$ and then decreasing. Since $f_2(m) = 0$, this means $f_2(i) \gt 0$ for all $3 \le i \le m - 1$. This shows \eqref{eq4A} is true, so the value being summed in \eqref{eq1A} is positive, for all these $i$. As such, the left side summation in \eqref{eq1A} is positive for all $3 \le n \le m$.

Combining this with $f_1(m) \gt 0$ in \eqref{eq2A} means the $\impliedby$ direction of \eqref{eq1A} is true for all $2 \le n \le m$.

With the $\implies$ direction, $n = 1$ gives a summation of $1 - m \lt 0$. As discussed above, with \eqref{eq5A} giving $f_2(i) \lt 0$ for $i \gt m$, this means each extra term being summed is negative for $n \ge m + 1$. Thus, if the summation is negative for $n = m + 1$, it'll be negative for all $n \ge m + 1$. To check on this, using the $m^i$ summation being of a geometric series gives

$$\begin{equation}\begin{aligned}
\sum_{i=1}^{m+1}(i^m - m^i) & = \sum_{i=1}^{m+1}i^m - \sum_{i=1}^{m+1}m^i \\
& = \sum_{i=1}^{m+1}i^m - \frac{m(m^{m+1} - 1)}{m - 1}
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$

Since $i^{m} \lt x^{m}$ for $i \lt x \le i + 1$ means $\int_{i}^{i+1}i^{m}dx \lt  \int_{i}^{i+1}x^m dx \implies i^{m} \lt \int_{i}^{i+1}x^m dx$, then using this and combining the integrals for all $1 \le i \le m - 1$ in the summation gives

$$\begin{equation}\begin{aligned}
& \sum_{i=1}^{m+1}i^m  - \frac{m(m^{m+1} - 1)}{m - 1} \\
& \lt \int_{1}^{m}x^m dx + m^m + (m + 1)^m - \frac{m(m^{m+1} - 1)}{m - 1} \\
& = \left. \frac{x^{m+1}}{m + 1} \right|_{1}^{m} + m^m + (m + 1)^m - \frac{m(m^{m+1} - 1)}{m - 1}\\
& = \frac{m^{m+1} - 1}{m + 1} + m^m + (m + 1)^m - \frac{m(m^{m+1} - 1)}{m - 1} \\
& = \frac{m(m^{m}) + (m + 1)(m^m) + (m + 1)^{m+1} - 1}{m + 1} - \frac{m(m^{m+1} - 1)}{m - 1} \\
& = \frac{(m - 1)((2m + 1)(m^m) + (m + 1)^{m+1} - 1) - (m + 1)m(m^{m+1} - 1)}{(m + 1)(m - 1)} \\
\end{aligned}\end{equation}\tag{8}\label{eq8A}$$

Define

$$\begin{equation}\begin{aligned}
f_3(m) & = (m - 1)((2m + 1)(m^m) + (m + 1)^{m+1} - 1) - (m + 1)m(m^{m+1} - 1) \\
& = (2m^2 - m - 1)(m^m) + (m - 1)(m + 1)^{m+1} - (m - 1) \\
& \; \; \; \; - (m^3 + m^2)(m^{m}) + (m^2 + m) \\
& = (-m^3 + m^2 - m - 1)(m^m) + (m - 1)(m + 1)^{m+1} + m^2 + 1
\end{aligned}\end{equation}\tag{9}\label{eq9A}$$

This gives $f_3(5) = -144600$. Next, simplify \eqref{eq9A} by using $0 \gt (-m - 1)(m^m) + m^2 + 1$ to get

$$\begin{equation}\begin{aligned}
f_3(m) & \lt (-m^3 + m^2)(m^m) + (m - 1)(m + 1)^{m+1} \\
& = (-m + 1)(m^{m+2}) + (m - 1)(m + 1)^{m+1} \\
& = (m - 1)((m + 1)^{m+1} - m^{m+2})
\end{aligned}\end{equation}\tag{10}\label{eq10A}$$

As done before using the natural logarithm,

$$(m + 1)^{m+1} \lt m^{m+2} \iff (m + 1)\ln(m + 1) \lt (m + 2)\ln(m) \tag{11}\label{eq11A}$$

Next, define

$$f_4(m) = (m + 1)\ln(m + 1) - (m + 2)\ln(m) \tag{12}\label{eq12A}$$

Note $f_4(5) \approx -0.5155$. Taking the derivative, plus using the logarithm property that $\ln(x) \le x - 1$ for $x \gt 0$, gives

$$\begin{equation}\begin{aligned}
f'_4(m) & = \ln(m + 1) + \frac{m + 1}{m + 1} - \ln(m) - \frac{m + 2}{m} \\
& = 1 - \left(1 + \frac{2}{m}\right) + \ln\left(\frac{m + 1}{m}\right) \\
& = - \frac{2}{m} + \ln\left(1 + \frac{1}{m}\right) \\
& \le - \frac{2}{m} + \frac{1}{m} \\
& = -\frac{1}{m} \\
& \lt 0
\end{aligned}\end{equation}\tag{13}\label{eq13A}$$

This shows $f_4(m) \lt 0$ for all $m \ge 5$, so $f_3(m) \lt 0$ in \eqref{eq10A}. This confirms $\sum_{i=1}^{m+1}i^m  - \frac{m(m^{m+1} - 1)}{m - 1} \lt 0$ in \eqref{eq7A}, which proves the $\implies$ direction in \eqref{eq1A} always holds.

Title: Problem on inequality with power sum


问:
Question on Reeb's theorem.
I am reading through the first couple of chapters of Milnor's Morse theory, and I've gotten to Reeb's sphere theorem (theorem 4.1),

If $M$ is a compact manifold and $f$ is a differentiable function on $M$ with only two critical points, both of which are nondegenerate, then $M$ is homeomorphic to a sphere.

Milnor states that for some small enough $\varepsilon > 0$, the sublevel sets $f^{-1}[0,\varepsilon]$ and $f^{-1}[1-\varepsilon,1]$ are closed $n$-cells, which follows by the Morse Lemma. But I don't quite follow this. I've read other related posts, but none of them quite flesh out why we need a "small enough" $\varepsilon$, or why exactly the Morse lemma implies that the aforementioned preimages are closed $n$-cells.

Now, I understand that the two critical points will have index $0$ and $1$, which correspond to the minimum and maximum, but why does applying the coordinate maps given to us my the Morse lemma tell us that we get $n$-cells, and not only containments? If anyone could clarify, that would be great.

答:
The neighborhood of a critical point where Morse Lemma applies is not necessarily a disk, but it contains a, (sufficiently small radius) closed disk.

Suppose that $p$ is a minimum of $f$ with $f(p)=0$. By Morse Lemma, there is a coordinate system $(U,y^i)$ of $p$ such that $y^i(p) = 0$ for all $ i=1,\dots,n$ and $f$ has representation
$$
f = (y^1)^2 + \dots + (y^n)^2, \quad \text{on }U.
$$
So, for sufficiently small $\varepsilon>0$ we have $D \equiv f^{-1}[0,\varepsilon] \subseteq U$. That is
$$
D = \{q \in U \, \mid \, f(y_q) = (y^1_q)^2+ \dots + (y^n_q)^2 \leq \varepsilon \},
$$
which is clearly diffeomorphic to a closed $n$-disk in $\mathbb{R}^n$, via coordinate map $y$.

Title: Question on Reeb's theorem.


答:
I use (at the end)

$\begin{array}\\
\cos x-\sin x
&=\sqrt{2}((1/\sqrt{2})\cos x-(1/\sqrt{2})\sin x)\\
&=\sqrt{2}(\sin(\pi/4)\cos x-\cos(\pi/4)\sin x)\\
&=\sqrt{2}\sin(\pi/4-x)\\
\end{array}
$

$\begin{array}\\
v
&=\lim \limits_{x \to \frac\pi4}\dfrac{\csc^2x-2\tan^2x}{\cot x-1}\\
&=\lim \limits_{x \to \frac\pi4}\dfrac{\dfrac1{\sin^2x}-2\dfrac{\sin^2x}{\cos^2x}}{\dfrac{\cos x}{\sin x}-1}\\
&=\lim \limits_{x \to \frac\pi4}\dfrac{\cos^2x-2\sin^4x}{\sin x\cos^3 x-\sin^2x\cos^2x}\\
&=\lim \limits_{x \to \frac\pi4}\dfrac{1-\sin^2x-2\sin^4x}{\sin x\cos x(1-\sin^2x)-\sin^2x(1-\sin^2x)}\\
&=\lim \limits_{x \to \frac\pi4}\dfrac{(1-2\sin^2x)(1+\sin^2x)}{\sin x(1-\sin^2x)(\cos x-\sin x)}\\
&=\lim \limits_{x \to \frac\pi4}\dfrac{(1-2\sin^2x)(3/2)}{(\sqrt{2}/2)(\cos x-\sin x)}\\
&=\dfrac{3}{\sqrt{2}}\lim \limits_{x \to \frac\pi4}\dfrac{1-2\sin^2x}{(\cos x-\sin x)}\\
&=\dfrac{3}{\sqrt{2}}\lim \limits_{x \to \frac\pi4}\dfrac{(1-\sqrt{2}\sin x)(1+\sqrt{2}\sin x)}{\cos x-\sin x}\\
&=\dfrac{3}{\sqrt{2}}\lim \limits_{x \to \frac\pi4}\dfrac{(1-\sqrt{2}\sin x)(1+\sqrt{2}(\sqrt{2}/2))}{\cos x-\sin x}\\
&=\dfrac{6}{\sqrt{2}}\lim \limits_{x \to \frac\pi4}\dfrac{(1-\sqrt{2}\sin x)}{\cos x-\sin x}\\
&=\dfrac{6}{\sqrt{2}}\lim \limits_{x \to \frac\pi4}\dfrac{1-\sqrt{2}\sin x}{\sqrt{2}\sin(\pi/4-x)}\\
&=3\lim \limits_{x \to \frac\pi4}\dfrac{1-\sqrt{2}\sin x}{\sin(\pi/4-x)}\\
&=3\lim \limits_{y \to 0}\dfrac{1-\sqrt{2}\sin (y+\pi/4)}{\sin(-y)}\\
&=3\lim \limits_{y \to 0}\dfrac{1-\sqrt{2}(\sin y\cos(\pi/4)+\cos(y)\sin(\pi/4))}{\sin(-y)}\\
&=3\lim \limits_{y \to 0}\dfrac{1-2(\sin y+\cos(y))}{\sin(-y)}\\
&=3\lim \limits_{y \to 0}\dfrac{-2\sin y}{-\sin(y)}\\
&=6\\
\end{array}
$

Title: $\displaystyle{\lim \limits_{x \to \frac\pi4}\frac{(\csc^2x-2\tan^2x)}{(\cot x-1)}}$ without L'Hôpitals Rule.

Body:
The limit,

$\displaystyle{\lim \limits_{x \to \frac\pi4}\frac{(\csc^2x-2\tan^2x)}{(\cot x-1)}}=6$

is easily found using L'Hôpitals Rule. However, the exercise is to evaluate it without using this method. I've tried multiple substitutions, mainly pythagoric and reciprocal identities, but without success, the indeterminate form $\frac{0}{0}$ keeps appearing. I’ve also tried splitting the limit and doing the substitutions. I'm looking to simplify the expression in order to evaluate the limit directly.

Thanks in advance.


问:
Proving that a function is differentiable and equal to a constant value for all x
Let $f(x)$ denote a strictly positive continuous function defined on all real numbers with the property that $f(2012)=2012$ and $f(x)=f(x+f(x))$ for all $x$. Prove that $f(x)=2012$ for all $x$.

I am trying to prove that f is differentiable before I can do $f'(x)=f'(x+f(x))$. How can I do that?
Is this the correct approach?

答:
Here's a counter-example to your assertion as it was originally. You've now added the requirement that $f$ is continuous, which invalidates this. But since I had already written this when I saw the change, I've decided to post it nevertheless $$
  f(x) = \begin{cases}
    2012 &\text{for $x \in \mathbb{N}$} \\
    2102 &\text{otherwise.}
  \end{cases}
$$

Obviously $f(x) > 0$ for all $x \in \mathbb{R}$.

If $x \in \mathbb{N}$ then $$
  f(x + f(x)) = f(\underbrace{x + 2012}_{\in \,\mathbb{N}}) = 2012 = f(x) \text{.}
$$

If $x \in \mathbb{R}\setminus\mathbb{N}$  then $$
  f(x + f(x)) = f(\underbrace{x + 2102}_{\in \,\mathbb{R}\setminus\mathbb{N}}) = 2102 = f(x) \text{.}
$$


答:
Hint:

If $\ N\,\big|\,10^rM+N\ $ for any positive integer $\ M\ $, where $\ 10^r\ $ is the smallest power of $\ 10\ $ exceeding $\ N\ $, what does that tell you about $\ N\ $?

Title: Finding the amount of magic numbers $N$

Body: Suppose $N$ is called a magic number if it is a positive integer and when you stick $N$ on the end of any positive integer, the resulting integer is divisible by $N.$ How many magic numbers are there less than $2100?$


问:
using eisenstein integers to solve diophantine equations
Earlier, I asked a question; this is a question regarding an answer I received to it.

Apparently when $u^2 - 12p^2 = -3$, with $p$ a three digit prime, there is some sort of recursion of the values of $p$, causing $P(x+2) = 14P(x+1) - P(x)$, causing the first few values to be $1,13, 181,..$

Where did this recursion come from?

Note : $P(i)$ is referring to the $i$-th answer for $p$

Body:
Let $u_1=3$ and $p_1=1$.  Then $u_1^2-12p_1^2=-3$.  I.e., $(u_1-2\sqrt3p_1)(u_1+2\sqrt3p_1)=-3$.

Now $7^2-12\cdot2^2=1$; i.e., $(7-2\sqrt3\cdot2)(7+2\sqrt3\cdot2)=1$.

Therefore, if $u_n^2-12p_n^2=-3$, then $(u_n^2-12p_n^2)(7^2-12\cdot2^2)=-3$.

I.e., $(u_n-2\sqrt3p_n)(7-2\sqrt3\cdot2)(u_n+2\sqrt3p_n)(7+2\sqrt3\cdot2)=-3$.

I.e., $([7u_n+24p_n]-2\sqrt3[7p_n+2u_n])([7u_n+24p_n]+2\sqrt3[7p_n+2u_n])=-3$.

I.e., $u_{n+1}^2-12p_{n+1}^2=-3$ with $u_{n+1}=7u_n+24p_n$ and $p_{n+1}=7p_n+2u_n$.

Furthermore, $p_{n+2}=7p_{n+1}+2u_{n+1}$ and $p_{n+1}=7p_n+2u_n$ and $u_{n+1}-7u_n=24p_n$ imply

$\dfrac{p_{n+2}-7p_{n+1}}2-7\dfrac{p_{n+1}-7p_n}2=24p_n$ or $p_{n+2}=14p_{n+1}-p_n$

(a recurrence involving $p$ and not $u$, which you seemed to want).


问:
Prove the determinant is $0$
This is Problem 16.17 from the book Exercises in Algebra by A. I. Kostrikin.

Prove that
$$
\left|\begin{array}{ccccc}
\dfrac{1}{2 !} & \dfrac{1}{3 !} & \dfrac{1}{4 !} & \cdots & \dfrac{1}{(2 k+2) !} \\
1 & \dfrac{1}{2 !} & \dfrac{1}{3 !} & \cdots & \dfrac{1}{(2 k+1) !} \\
0 & 1 & \dfrac{1}{2 !} & \cdots & \dfrac{1}{(2 k) !} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & \dfrac{1}{2 !}
\end{array}\right|=0, \quad k \in \mathbb{N}
$$

My Attempt: I tried to expand it by the first column, but it seemed to be more complicated when I did that. I also tried to add edges to the determinant(in the hope that it will be easier to calculate), but I still failed to work it out.

So, My Question is, how to calculate this determinant?

答:
We want to prove that the $(n-1)\times(n-1)$ matrix
$$
A_n=\pmatrix{
\dfrac{1}{2!} &\dfrac{1}{3!} & \cdots &\cdots & \cdots &\dfrac{1}{n!} \\
1 &\dfrac{1}{2!} &\dfrac{1}{3!} &\cdots &\cdots &\dfrac{1}{(n-1)!} \\
0 & 1 & \ddots & \ddots & \ddots & \vdots\\
\vdots & \ddots & \ddots & \ddots&\ddots & \vdots\\
\vdots & \ddots & \ddots &1& \dfrac{1}{2!} & \dfrac{1}{3!}\\
0 & \cdots & \cdots & 0 & 1 & \dfrac{1}{2!}
}
$$
is singular when $n\ge4$ is even. This can be proved by mathematical induction on $n$. The base case $n=4$ can be verified directly. In the inductive step, note that for any positive integer $m$,
$$
\sum_{k=0}^m\frac{(-1)^k}{k!}\frac{1}{(m-k)!}=\frac{1}{m!}\sum_{k=0}^m(-1)^k\binom{m}{k}=\frac{1}{m!}(1-1)^m=0.
$$
Move the first and the last summands out of the sum, we obtain
$$
\frac{1}{m!}+\sum_{k=1}^{m-1}\frac{(-1)^k}{k!}\frac{1}{(m-k)!}=\frac{(-1)^{m+1}}{m!}.
$$
Denote the $i$-th row of $A_n$ by $a_i$. The previous identity means that
\begin{aligned}
u&:=a_1+\sum_{k=1}^{n-1}\frac{(-1)^k}{k!}a_{k+1}\\
&=\left(-\frac{1}{2!},\,\frac{1}{3!},\,-\frac{1}{4!},\,\frac{1}{5!},\,\ldots,\,-\frac{1}{(n-2)!},\,\frac{1}{(n-1)!},\,\frac{1}{(n-1)!}-\frac{1}{n!}\right).
\end{aligned}
(Note that the last entry of $u$ is not $-\frac{1}{n!}$, because the last column of $A_n$ ends with $\frac{1}{2!}$, not $1$.) In other words, by some appropriate elementary row operations, we can modify the first row of $A$ to $u$. Therefore, by the multilinearity of the determinant function, $\det(A_n)$ remains unchanged if we replace the first row of $A_n$ by
$$
v:=\frac12(a_1+u)
=\left(0,\,\frac{1}{3!},\,0,\,\frac{1}{5!},\,\ldots,\,0,\,\frac{1}{(n-1)!},\,\frac{1}{2\times(n-1)!}\right).
$$
Now suppose $n\ge6$ is even. By Laplace expansion along $v$ and by induction assumption, we get
\begin{aligned}
\det(A_n)
&=-\frac{1}{3!}\det(A_{n-2})-\frac{1}{5!}\det(A_{n-4})-\cdots-\frac{1}{(n-1)!}\det(A_2)+\frac{1}{2\times(n-1)!}\\
&=-\frac{1}{(n-1)!}\det(A_2)+\frac{1}{2\times(n-1)!}\\
&=-\frac{1}{(n-1)!}\frac{1}{2!}+\frac{1}{2\times(n-1)!}\\
&=0.
\end{aligned}

Title: Prove the determinant is $0$


问:
Chance of getting $20$ three times in a row on a $20$-sided die.
My friend and I are having an argument and I would like some help proving I'm right or perhaps proving me wrong. My friend and I where playing dungeons and dragons the other night when discussing the chances of rolling a $20$-sided die and getting a $20$ three times in a row. I believe the answer should $\frac1{8000}$. He is saying the chances are $\frac1{8420}$ why is that? He says it has something to do with recurrence.</p>

答:
The probability of rolling a $20$-sided die three times and getting three $20$s is
$$
\left(\frac1{20}\right)^3=\frac1{8000}
$$
If we were to roll the die three times and compare the triple rolled with $(20,20,20)$, it would, on average, take $8000$ of these triples to get the triple $(20,20,20)$.

However, if we were to simply roll the die until three $20$s occurred in succession, it would take, on average, $8420$ rolls to get three $20$s in a row. Here are a couple of ways to compute this expected duration.

Generating Functions

The generating function for the probability of getting three $20$s in a row in exactly $n$ rolls is
$$
\begin{align}
f(x)
&=\sum_{k=0}^\infty\left(\vphantom{\frac{19}{20}}\right.\overbrace{\ \ \ \frac{19}{20}x\ \ \ }^\text{non-$20$}+\overbrace{\ \frac{19}{400}x^2\ }^{\substack{\text{non-$20$}\\\text{one $20$}}}+\overbrace{\frac{19}{8000}x^3}^{\substack{\text{non-$20$}\\\text{two $20$s}}}\left.\vphantom{\frac{19}{20}}\right)^k\ \ \overbrace{\ \ \frac{x^3}{8000}\ \ }^{\text{three $20$s}}\\[6pt]
&=\frac{x^3}{8000-7600x-380x^2-19x^3}
\end{align}
$$
The probability of getting three $20$s in a row eventually is
$$
f(1)=1
$$
The expected number of rolls to get three $20$s in a row is the derivative at $1$.
$$
f'(x)=\frac{24000x^2-15200x^3-380x^4}{\left(8000-7600x-380x^2-19x^3\right)^2}
$$
The expected number of rolls to get three $20$s in a row is
$$
f'(1)=8420
$$

Recursion

Let $E$ be the expected number of rolls to get three $20$s in a row.
$$
\begin{align}
E
&=\overbrace{\frac1{8000}\cdot3}^{\text{three $20$s}}+\overbrace{\frac{19}{8000}(3+E)}^{\substack{\text{two $20$s}\\\text{non-$20$}}}+\overbrace{\frac{19}{400}(2+E)}^{\substack{\text{one $20$}\\\text{non $20$}}}+\overbrace{\frac{19}{20}(1+E)}^\text{non $20$}\tag1\\
&=\frac3{8000}+\frac{57}{8000}+\frac{38}{400}+\frac{19}{20}+\left(\frac{19}{8000}+\frac{19}{400}+\frac{19}{20}\right)E\tag2\\
&=\frac{421}{400}+\frac{7999}{8000}E\tag3\\[6pt]
&=8420\tag4
\end{align}
$$
Explanation:

$(1)$: collect the contributions to the expected duration

$(2)$: separate constants and coefficients of $E$

$(3)$: simplify

$(4)$: $8000\cdot(3)-7999\cdot(1)_\text{left}$


问:
Inequality :$\left(\frac{x}{x+1}\right)^{\tan^2(x)}+\left(\frac{1}{1+3}\right)^{\tan^2(1)}+\left(\frac{3}{x+3}\right)^{\tan^2(3)}>\frac{40}{39}$
Claim :

Let $\frac{13}{10}\leq x<\frac{\pi}{2}$

$$f(x)=\left(\frac{x}{x+1}\right)^{\tan^2(x)}+\left(\frac{1}{1+3}\right)^{\tan^2(1)}+\left(\frac{3}{x+3}\right)^{\tan^2(3)}>\frac{40}{39}$$

The function in the RHS is close to a  line and the choice of $\frac{40}{39}$ is because it's a simple constant and rational .As you can see I'm inspired by the Nesbitt's inequality so it's not totally random .

To prove I have tried convexity with the second derivative of $f(x)$ wich is positive but I cannot prove it currently .

Any idea to prove it ?

Thanks !

答:
Hint:

Fact 1: If $0 < x < \frac{\pi}{2}$, then $g(x) = \tan^2 x \ln (\frac{x}{x+1})$ is strictly decreasing.

(Hint: Take derivative and use $\tan x \ge x$.)

Now, by Fact 1, we have $f(x) \ge f(\pi/2) > \frac{40}{39}$. We are done.


问:
Show that the sequence $(a_n)_{n\ge 0}$ defined recurrence relation $a_0=25, a_{n+1}=2\sqrt{a_n}$ converges.
Show that the sequence $(a_n)_{n\ge 0}$ defined recurrence relation $a_0=25, a_{n+1}=2\sqrt{a_n}$ converges. I showed that it is bounded from below by $1$, but I'm struggling to show that the sequence is non-increasing. I tried induction, but the induction hypothesis is not helpful at all.

答:
Letting
$b_n = \log(a_n)$,
this becomes,
taking logs,
$b_{n+1}
=\log(2)+b_n/2
=c+b_n/2
$
where $c = \log 2$.

Then

$\begin{array}\\
b_{n+1}
&=c+\frac12 b_n\\
&=c+\frac12(c+\frac12 b_{n-1})\\
&=\frac32 c+\frac14 b_{n-1}\\
&=\frac32 c+\frac14 (c+\frac12 b_{n-2})\\
&=\frac74 c+\frac18 b_{n-2})\\
&...\\
&=(2-\frac1{2^k}) c+\frac1{2^{k+1}} b_{n-k})
\qquad\text{(easily proved by induction)}\\
&\to 2c\\
\end{array}
$

Therefore
$a_n
\to e^{2c}
=e^{2\log(2)}
=2^2
=4
$.

Note that this does not
depend on the initial condition.


问:
What is a conormal vector to a domain intuitively?
I read that a conormal vector of a domain is a vector that is tangential to the domain and normal to its boundary.

If we consider an open disk in $\mathbb{R}^2$ what is a conormal vector at a point on the boundary? I can't picture it at all.

答:
The terminology "conormal" refers to covectors (covector fields), also known as 1-forms. When a field is conormal, the corresponding vector field should be normal. This correspondence is given by the duality between vector fields and 1-forms.

In the Euclidean spaces there is the standard Riemannian metric ("dot product") that allows to identify the tangent and the cotangent spaces (that is usually done in multivariable calculus).

The whole lot of intuitions can be found in this picture:

Here we have a domain $U \subset \mathbb{R}^n$ represented as $U := \{x \in \mathbb{R}^n | f(x) < 0 \}$, sot that its boundary is given as the zero set of a function $f \colon \mathbb{R}^n \to \mathbb{R}$.

Tangent vectors to the (hyper)surface $S$ arise as the velocities $\dot{\gamma}$ of smooth curves $\gamma \colon ( -\varepsilon, \varepsilon) \to S$.

Using the chain rule one easily shows that
$$
\mathrm{d}f(\dot{\gamma})=0  \tag{*}
$$
(This holds for any level surface!).

The differential $\mathrm{d}f \colon T\mathbb{R}^n \to T\mathbb{R}$ can be seen as a 1-form $\mathrm{d}f \colon \mathbb{R}^n \to \mathbb{R}$ due to the natural identification $T\mathbb{R}^n \cong \mathbb{R}^n$ that is the feature of the Euclidean spaces.

Using the standard metric in $\mathbb{R}^n$ we can convert 1-forms into vectors (that are tangent to $\mathbb{R}^n$, in fact).

Thus, the gradient $\nabla f := (\mathrm{d}f)^{\sharp}$ turns out to be a normal vector to $S$ according to (*).

Putting this the other way around, $\mathrm{d}f$ is the conormal to $S$ at all $p \in S$.

One visualizes 1-forms (at a point) as distributions of (hyper)planes so that its value on a vector corresponds to the "number" of (hyper)planes passing through this vector.


问:
Sequence both converging and diverging?
I have the sequence sum 1 to infinity of 10^n/n!

If I plug it into symbollab, it says it converges by the ratio test.
If I plug it into wolfram alpha, it says it diverges by geometric series test.

Which is it ?!

答:
That sum is
$e^{10}-1$,
so it definitely converges.

Title: Sequence both converging and diverging?


问:
Weak Convergence - similar to continuous mapping theorem
I recently revisited the following textbook exercise (Probability and Measures by Billingsley, Problem 25.8 in the third edition):

Let $X_n$ and $X$ be real-valued random variables and $h_n, h: \mathbb{R} \to \mathbb{R}$ be Borel functions. Let $E$ be the set of $x \in \mathbb{R}$ such that $h_n(x_n) \to h(x)$ fails for some sequence $x_n \to x$. Suppose that $X_n$ converges to $X$ weakly (a.k.a. in distribution), denoted by $X_n \Rightarrow X$, and that $E$ is measurable with $P(X\in E) = 0$. Show that $h_n(X_n) \Rightarrow h(X)$.

This exercise can be solved quite easily using Skorohad's theorem: on some probability space, there exist real-valued random variables $Y_n$ and $Y$ with the same distribution as $X_n$ and $X$ satisfying $Y(\omega) \to Y(\omega)$ pointwise.

But now I want to prove it without using Skorohad's theorem. I wonder if I can do so by using the common characterization in Portmanteau theorem: $\limsup_n \mu(C) \to \mu(C)$ for any closed set $C$ if and only if $\mu_n \Rightarrow \mu$.

My efforts:

Let $\mu_n:= P \circ X_{n}^{-1}$ and $\mu:= P \circ X^{-1}$ denote the probability measures induced by $X_n$ and $X$. Also, we let $\nu_n := P \circ X_n^{-1} \circ h_n^{-1}$ and $\nu:= P \circ X^{-1} \circ h^{-1}$ denote the probability measures induced by $h_n(X_n)$ and $h(X)$. The goal is to show $\nu_n \Rightarrow \nu$.

Let $C$ be an arbitrary closed subset of $\mathbb{R}$. Then we have
$$
  \limsup_n \nu_n(C) = \limsup_n \mu_n (h_n^{-1}(C) ) \leq \limsup_n \mu_n( \overline{ h_n^{-1}(C)} ),
 $$
where $\overline{ h_n^{-1}(C)}$ denote the closure of $h_n^{-1}(C)$. I would like to use Portmanteau theorem on $\mu_n \Rightarrow \mu$, but the set $\overline{ h_n^{-1}(C)}$ also depends on $n$. Is there a way to bypass this issue?

答:
First of all, recall that weak convergence is equivalent to $\int f\;d\mu_n\to \int f\;d\mu$ for every bounded Lipschitz function $f.$
With this in mind, our goal reduces to proving that for every bounded Lipschitz function $f,$ we have  $$\int f\circ h_n d\mu_n\to f\circ h\;d\mu.$$
With some simple algebra, it can be reduced to proving that $$\int (f\circ h_n-f\circ h)d\mu_n\to 0.$$

To this end, we use Erogoff's theorem. Fix $\epsilon>0,$ and let $A$ be a closed set such that $h_n\to h$ uniformly on $A$ and $P(A^c)<\epsilon/||f||_{\infty}.$
Now note that
\begin{align}
\int |f\circ h_n-f\circ h|d\mu_n &=\int_A |f\circ h_n(X_n)-f\circ h(X_n)|d\mathbb{P}+ \int_{A^c}|f\circ h_n(X_n)-f\circ h(X_n)|d\mathbb{P}\\
&\le ||f||_{\text{Lip}}||h_n-h||_{\infty, A}\mathbb{P}(A)+2||f||_{\infty}\mathbb{P}(A^{c})\\
&\le ||f||_{\text{Lip}}||h_n-h||_{\infty, A}+2\epsilon.
\end{align}
Since $h_n\to h$ unfiromly on $A,$ the norm $||h_n-h||_{\infty, A}\to 0$ as $n\to \infty.$


答:
Using the following definitions
$$
\psi(z) = -\gamma + \sum_{n=0}^\infty \left( \frac 1 {n+1} - \frac 1 {n+z} \right)
$$
$$
\gamma = \lim_{n\,\to\,\infty} \left( -\ln n + \sum_{k=1}^n \frac 1 k \right)
$$

one can conclude that
\begin{align}
& \sum_{k=1}^\infty \frac{1}{x+k}=\sum_{k=1}^\infty \frac{1}{k+1}-\gamma-\psi(x)+\frac{x-1}{x} \\[6pt]
= {} & \lim_{n \to \infty}  \ln(n)-\psi(x)+\frac{3x-2}{2x}.
\end{align}


问:
Risk neutral probability for stock with continuous dividend
Setting: binomial tree with one step over time $\Delta t$.  I'm trying to derive the risk neutral probability for a stock which pays a continuous dividend, say $\delta$. i.e. probability $p$ such that $$e^{r \Delta t} S_0 = S_u p + S_d(1-p)$$

where $S_u, S_d$ are the values of the stock in the up and down states respectively.  This immediately gives $$p = \frac{S_0 e^{r \Delta t} - S_d}{S_u - S_d}$$

Now if we assume $S$ has volatility $\sigma$, we should be getting $S_d = S_0 e^{-\sigma \sqrt{\Delta t} - \delta \Delta t}$ and $S_u = S_0 e^{\sigma \sqrt{\Delta t} - \delta \Delta t}$ so that $$p = \frac{e^{r \Delta t} - e^{- \sigma \sqrt{\Delta t} - \delta \Delta t} }{ e^{ \sigma \sqrt{\Delta t} - \delta \Delta t} -  e^{- \sigma \sqrt{\Delta t} - \delta \Delta t}} = \frac{e^{(r+ \delta)\Delta t} - e^{- \sigma \sqrt{\Delta t}} }{ e^{ \sigma \sqrt{\Delta t}} -  e^{- \sigma \sqrt{\Delta t}}}$$

but this is wrong because the formula that's given in my course's lecture notes on this is $$ p = \frac{e^{(r- \delta)\Delta t} - e^{- \sigma \sqrt{\Delta t}} }{ e^{ \sigma \sqrt{\Delta t}} -  e^{- \sigma \sqrt{\Delta t}}}$$

(the only difference is the $r-\delta$ in the numerator instead of the $r+ \delta$).  I don't understand why my assumptions on the values for $S_u$ and $S_d$ are wrong.  Any help would be massively appreciated.

MY POTENTIAL EXPLANATION: perhaps the value of $S_u$ should be $S_0 e^{\sigma \sqrt{\Delta t} + \delta \Delta t}$ (and similarly with $S_d$) because we work with the payoff of owning one unit of the stock, so if we increase with upward factor $e^{\sigma \sqrt{\Delta t}}$ we GAIN the value of the dividend, not lose it.

答:
With a continuous dividend yield $\delta$, the end-of-period expected value of the stock price under the risk-neutral probability measure should be the forward price

$$\tag{*}\mathbb{E}(S) = F= S_0e^{(r-\delta)\Delta t},$$

and then it easily follows that

$$p = \frac{ S_0e^{(r-\delta)\Delta t}-S_d}{S_u-S_d} =  \frac{ e^{(r-\delta)\Delta t}-d}{u-d}= \ldots $$

We can show that (*) is the correct forward price by an arbitrage argument.  If $F > S_0e^{(r-\delta)\Delta t}$ then we could borrow $S_0e^{-\delta \Delta t}$ in cash, buy $e^{-\delta \Delta t}$ shares of stock and enter into a forward contract to sell the stock at the end of the period for $F$.  At the end of the period with dividend reinvestment the number of shares grows to $1$.  Delivering the one share of stock we receive the delivery price $F$ and repay our borrowings with interest totaling $S_0e^{(r- \delta) \Delta t}$ for a risk free gain of

$$F - S_0e^{(r-\delta)\Delta t} > 0$$

This is impossible in the absence of abritrage opportunity.  By a similar argument we can show that it is impossible to have $F < S_0e^{(r-\delta)\Delta t}$.


问:
A non-entire function whose square and composition of itself are polynomials.
This post is closely related to Entire function whose square or composition with itself is a polynomial.

In the linked post, the OP asked to prove the following:

If $f:\mathbb{C}\longrightarrow\mathbb{C}$ is entire, and $f^{2}(z)$ is a polynomial, then $f$ is also a polynomial.

If $f:\mathbb{C}\longrightarrow\mathbb{C}$ is entire, and $f(f(z))$ is a polynomial, then $f$ is also a polynomial.

Several proofs were given: see Proving $f(z)$ and $g(z)$ are polynomials for a proof using Casorati-Weierstrass Theorem, and Two confusions on Great Picard Theorem for a proof using Great Picard Theorem.

My question is what if $f$ is not entire in the first place, does $f^{2}(z)$ being a polynomial still imply $f$ being a polynomial? and does $f(f(z))$ being a polynomial imply $f$ be a polynomial?

As the proof depends heavily on the entire property of $f$, I believe the answer for both of my questions is NO.

But I cannot find a counterexample.

For the first one, I tried $f(z):=|z|$, but then $f(z)\neq z^{2}$, because $|z|$ is a complex norm, so that $$f(z)=|z|^{2}=x^{2}+y^{2}\neq z^{2}=x^{2}+2ixy-y^{2}.$$

Without absolute value, I am not sure what else I can try since $f^{2}(z)$ must be a polynomial (so all the fraction without removable singularities seems not work well..)

For the second one, perhaps we can use $f(z):=\dfrac{1}{z}$ which has a pole at $z=0$, and $f(f(z))=z$ is a polynomial.

Is my second example correct? Any idea about the first example? Thank you!

答:
What do you mean with 'not entire' ?

$f(z)=1/z^n$ gives $f(f(z))=z^{n^2}$

with $\phi$ biholomorphic (or just bijective ?) on the right open set then
$g(z)= \phi^{-1}(1/\phi(z))$ gives $g(g(z)) = z$,

same for $h(z)= \phi^{-1}(1-\phi(z)),h(h(z))=z$.


问:
nonsingularity of diagonally dominant matrix
I have a nonnegative matrix which is sort of diagonally dominant but not quite.

I have $a_{ii}>a_{ij}$ and $a_{ii}>a_{ji}$ for any $i$ and $j$ not equal to $i$.

I wonder whether it must be nonsingular.

答:
Not when $A$ is at least $3\times3$. E.g. $\pmatrix{5&0&3\\ 0&5&4\\ 3&4&5}$ is singular.

More generally, pick any unit vector $u$ such that $0<u_i<1$ for each $i$. Pick also some $\epsilon\ge0$. The matrix $A$ below is then a singular nonnegative matrix:
$$
\underbrace{\left[\pmatrix{I&u\\ u^T&1}+\epsilon\pmatrix{uu^T&u\\ u^T&1}\right]}_A\pmatrix{u\\ -1}=0.
$$
($A$ is entrywise positive if $\epsilon>0$.) When $\epsilon\ge0$ is sufficiently small, we have $a_{ii}>\max(a_{ij},a_{ji})$ for all $i\ne j$.


问:
Generating function of a Fractional Part
I'd like to compute the following generating function
\begin{align}
g_{a,b}(x) = \sum_{k \geq 0} \lbrace \tfrac{a k}{b} \rbrace z^{k}
\end{align}
where $\frac{a}{b} \in \mathbb{Q}$ and $\lbrace \cdot \rbrace$ denotes the fractional part function. I can compute the aforementioned generating function for a few special cases. I believe the general case should be of the form
\begin{align}
g_{a,b}(t) = \frac{P(z)}{1-z^{b}}.
\end{align}
where $P$ is a $\mathbb{Z}$-polynomial with degree no greater than $b-1$. The $b$-periodicity of the fractional part above should play an essential role. Any help or hints in computing the general case is appreciated.

Thanks!

Update: After a little more work, I compute $P(z) = \sum_{k = 0}^{b-1} \lbrace \frac{a k}{b} \rbrace z^{k}$.

答: $\frac{P(z)}{1 - z^b}$ is nothing more than $P(z) + z^b P(z) + z^{2b} P(z) + ...$ so the coefficients of $P$ are directly determined by the first $b$ terms. </p>


问:
How to solve this integral for a hyperbolic bowl?
$$\iint_{s} z  dS $$ where S is the surface given by $$z^2=1+x^2+y^2$$ and $1 \leq(z)\leq\sqrt5$ (hyperbolic bowl)</p>

答:
Use polar cylindrical coordinates.  Here, the surface is given by $r(z) = \sqrt{z^2-1}$.  The surface integral is then

$$2 \pi \int_1^{\sqrt{ 5}}dz \, z\, r(z) \sqrt{1+\left(\frac{dr}{dz}\right)^2}$$

I leave it to the reader to derive the final form of the integral to be evaluated.  I get

$$\pi \int_1^5 du \, \sqrt{2 u-1} = \frac{26 \pi}{3}$$

ADDENDUM

Note that this answer agrees with the parametrization approach:

$$z=1+x^2+y^2 \implies \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2} = \sqrt{\frac{1+2 r^2}{1+r^2}}$$

so that the surface integral is

$$2 \pi \int_0^2 dr \, r z \sqrt{\frac{1+2 r^2}{1+r^2}} = 2 \pi \int_0^2 dr \, r\, \sqrt{1+r^2} \sqrt{\frac{1+2r^2}{1+r^2}}$$

Do a little algebra and see that the surface integral is

$$\pi \int_0^2 du \sqrt{2 u+1}$$

which agrees with the above result.


问:
lowest bound of the maximal eigenvalue of the adjacency matrix
if A is the adjacency matrix, I have to prove that $\lambda_1 \ge \sqrt{d_1}$ where, $\lambda_1$ is the maximal eigenvalue, and $d_1$ the maximal degree of the graph.

I cannot find a simple way to proove this. Any tips?

答:
This result only applies to connected undirected graphs (i.e., the adjacency matrix is symmetric).  The proof is as follows.

Let $v$ be the right eigenvector of $A$ corresponding to $\lambda_1$.  By the Perron-Frobenius theorem, it must be positive.  For any node $i$,
$$ \lambda_1^2 v_i = \left( A^2 v \right)_i = \sum_{j, k} A_{ij} A_{jk} v_k $$

Now consider the sum on the right hand side.  Split it into terms where $k = i$ and terms where $k \ne i$, then drop the latter.  All of the terms are positive, so we get
$$ \lambda_1^2 v_i \ge \sum_j A_{ij} A_{ji} v_i $$
Now $v_i$ can be pulled out of the sum, and $A_{ij} A_{ji} = A_{ij}$ since $A$ is symmetric and only contains zeroes and ones.  So we get
$$ \lambda_1^2 v_i \ge v_i \sum_j A_{ij} = d_i v_i $$
where $d_i$ is the degree of node $i$.  For a non-connected graph, it is not clear whether we can divide out $v_i$, because it might be zero, but for a connected graph then it must be positive.  So we get
$$ \lambda_1^2 \ge d_i $$
for any node $i$.  If we assume without loss of generality that node 1 has the largest degree, then we get
$$ \lambda_1 \ge \sqrt{d_1} $$
as desired.


问:
Probability of two independent events...
Suppose $A$ and $B$ are independent with $P(A \cap B') = 0.2$ and $P(A' \cap B) = 0.3$. Find $P(A \cup B)$.

This problem came with this hint: ''there are two correct answers.''

My attempt: $P(A \cup B) = 0.2+0.3+P(A \cap B)$.

Also,  $P(A \cap B) = P(A) \cdot P(B)$.

How can I proceed? I tried drawing the Venn diagram but that didn't get me anywhere.

答:
The following is pure hindsight and coincidence. The solution from player3236 is the right way to go.

I will use player3236's notation. We have
$p(1-q)=0.2, q(1-p)=0.3$. Since $0.2+0.3=0.5$, we can't help noticing that the other two "joint-event" probabilities also sum to one half, i.e.

$$pq+(1-p)(1-q)=0.5=p(1-q)+q(1-p) \\ \implies
0=pq-p(1-q)+(1-p)(1-q)-q(1-p)=(1-2p)(1-2q)$$

Therefore either $p=\frac 12$ or $q=\frac 12$, and
$$\Pr(A \cup B)= \begin{cases} \Pr(A)+\Pr(A'\cap B) = 0.8 & \text{ if } p=\frac 12; \\ \Pr(B) + \Pr(A \cap B') = 0.7 & \text{ if } q=\frac 12. \end{cases}$$

It's much easier to understand with the following picture:


问:
Locus of point at a fixed distance from midpoint of intercepts of a variable line segment with fixed distance

Let A and B be variable points on the x-axis and y-axis respectively such that the line segment AB is in the first quadrant and of a fixed length 2d. Let C be the mid-point of AB and P be a point such that

(a) P and the origin are on the opposite sides of AB and,

(b) PC is a line of length d which is perpendicular to AB.

Find the locus of P.

Source

In this problem, I have drawn the figure containing the line AB and point P and everything. I have also named point as(h, k),thereby applying all suitable equations.Inspite of all that, it is getting clumsy.
Can anyone provide a elegant solution to it?

答:
Observe that $\triangle APB$ is an isosceles right-angled triangle, hence $OAPB$ is cyclic.

Note that $\angle BAP = \angle BOP = 45^\circ$. Hence $P$ lies on the line $y=x$.

Now we need to find the range of the length of $OP$. Intuitively it cannot be too long or too short since $P$ is bounded in a circle of radius $d$ containing the origin.

By Ptolemy's Theorem: $AO \times PB + AP \times OB = AB \times OP$.

Hence $AO \times \sqrt2 d+ \sqrt2d\times OB = 2d \times OP$, giving $\sqrt 2OP = AO+ OB$.

We have $AO^2 + OB^2 = 4d^2$. By letting $x = AO$ and considering the function $x + \sqrt{4d^2 - x^2}$ or otherwise (there should be a better way to do this), we see that the maximum occurs when $AO = OB = \sqrt 2d$ and the minimum occurs when $AO$ or $OB = 0$, giving the range:

$$\sqrt 2 d \le OP \le 2d$$

hence the locus is the line segment connecting the points $(d,d)$ and $(\sqrt 2 d, \sqrt 2d)$.


问:
Zeros of a cubic polynomial with rational coefficients
While discussing a related problem, one of my friends came out with a question as follows:

Is it possible that a cubic polynomial $p(x) \in \Bbb{Q}[x]$ has all of its zeros to be both real and irrational? That is, can $p(x)$ be factored into the following form?

$$ p(x) = a (x - \alpha_1)(x - \alpha_2)(x - \alpha_3), \quad a \in \Bbb{Q}, \ \alpha_{i} \in \Bbb{R} \setminus \Bbb{Q} $$

We struggled with this problem for a moderate time but failed to find any clue for its validity or invalidity.  I guess that this is impossible, but currently I have no idea how to attack this. Can you enlighten me by showing a valid direction?

答:
It is possible. For example, given a prime number $p$ and $a,b\in\mathbb Z$ with $a+b\le-2$, let
$$f(x)=x^3+pax^2+pbx+p\in \mathbb Z[x].$$
Then by Eisenstein's criterion, $f$ is irreducible in $\mathbb Q[x]$, i.e. $f$ has no rational root. However, since $f(0)=p>0$ and $f(1)=1+(a+b+1)p<0$, $f$ has three distinct real root located in $(-\infty,0)$,  $(0,1)$ and $(1,+\infty)$ respectively.</p>


问:
Number of Zonotope Edges Parallel to Generator
Suppose we have a zonotope $Z$ that is the Minkowski sum of line segments $U_1+\dots +U_n$. All the edges of $Z$ are parallel to some $U_i$. Is it also true that the number of edges parallel to $U_i$ equals the number of edges parallel to $U_j$ for every $j\neq i$? Surely this question has been asked and probably answered before. I looked through a page of zonohedra and verified that they all answer the question true.

https://www.ics.uci.edu/~eppstein/junkyard/ukraine/ukraine.html

It is known that $Z$ is a zonotope iff $Z$ is the projection of a cube from $\mathbb{R}^n$. Of course, a cube in $\mathbb{R}^n$ is the sum of $n$ orthonormal vectors and has the same number of edges parallel to each of its generators. I suspect there may be a good induction proof down this way by adding one generator, $v$, at a time and  but haven't found one. The new generator divides the old zonotope into two pieces. The edges parallel to $v$ will stretch between the two halves.

Any help is appreciated!
Thanks! Aaron

答:
Following up Ethan's answer.

Let $S\subset \mathbb{R}^d$ be finite and generate the zonotope $Z$. By Ethan's argument, $Z$ will have the same number of edges per generator if all 2D faces are parallelograms, i.e. no three elements of $S$ are coplanar.

Seeing coplanarity in $S$ as degenerate, we can still decompose the boundary of $Z$ into parallelograms. This is done by "parallelogramizing" all the 2D faces. Now, count the number of edges per generator by distinguishing between legit edges (the 1D faces of $Z$) and illegit edges (the parallelogramizing segments inside a 2D face). We seek a condition * that determines when $Z$ has the same number of legit edges per generator.

We know $Z$ has the same number of edges per generator if we include the illegit ones. Hence, * must hold iff $Z$ has the same number of illegit edges per generator, so let's count those. Say a generator $s\in S$ is parallel to a 2D face $F\subset Z$. Suppose $F$ is a $2n$-gon. Each pair of generators parallel to $F$ determines a parallelogram in the parallelogramization of $F$, so there are $n-1$ parallelograms in the $s$-zone of $F$, meaning $n$ total edges parallel to $s$, $2$ of which are legit, leaving $n-2$ illegit edges.

The total number of illegit edges parallel to $s$ is then the total "surplus" of all the 2D faces in the $s$-zone, which counts how many generators after $2$, i.e. hexagons are surplus 1, octagons are surplus 2, etc. This turns into the following condition:

Let $\mathcal{H}$ be the set of all 2D subspaces in $\mathbb{R}^d$. The zonotope generated by $S$ will satisfy * iff the total nullity,

$$
T(s):=\sum_{s\in H\in\mathcal{H}} \mathrm{nullity}(H\cap S)
$$

does not depend on $s$. As expected, it's a sort of balancing condition.


问:
Solving equalities with absolute values
I was goofing around with programming some math functions the other day, and I came up with an equality that I can't prove/disprove.  It involves absolute values and that is where I am getting stumped.

$$ \lvert(x-y)\rvert(x+y) = \lvert x^2 - y^2 \rvert $$

Is it possible to work this out with algebra or otherwise?

答:
We have that $|x-y| \cdot (x+y) = |x^2-y^2|.$ But
$$x^2 - y^2 = (x-y) \cdot (x+y),$$
so
$$|x-y| \cdot (x+y) = |x-y| \cdot |x+y|.$$

Therefore either $|x-y| = 0,$ or we can divide by it to find that $|x+y| = x+y.$

Hence the two solutions to this inequality are $x=y,$ or $x+y \geq 0.$

Title: Solving equalities with absolute values


问:
What does the limit of the characteristic function $\varphi(x)\xrightarrow{\lvert x\rvert\rightarrow \infty} 0$ tell us about $\mu(\{a\})$?
If we know that $\varphi(x)\xrightarrow{\lvert x\rvert\rightarrow \infty} 0$ for the characteristic function $\varphi$ of the probability measure $\mu$ on $\mathbb{R}$, what does that tell us about $\mu(\{a\})$ for any $a$?

Because of the identity
$$\mu(\{a\}) = \lim_{T\rightarrow\infty}\frac 1{2T}\int_{-T}^Te^{-ita}\varphi(t) dt$$
I think it should be $\mu(\{a\}) = 0$ for all $a$. But I don't know how to formalize this idea, if it's true.

Edit: A TA confirmed it is true that under the assumption $\mu(\{a\})=0$ for all $a$, but I still don't even know how to start a possible proof/what the key insight is.

Edit2: I have successfully shown that
$$\lim_{T\rightarrow\infty}\frac 1{2T}\int_{-T}^T\varphi^2(t)dt = \sum_{x\in\mathbb{R}}(\mu(x))^2$$
But I need to show that the LHS is $0$. It is not trivial to show that the integral is even finite in the limit, how can I show that the LHS is finite?

答: If I understand this correctly, it tells us that mu([a}) is zero for every a. Because each a that has positive probability contributes a sinusiod to the characteristic function and the sum of these sinusiodal components cannot vanish in the limit. They will at worst be a so-called almost periodic function. Neither the absolutely continuous not the singular part of the distribution will counteract the effect of the discrete component. A theorem of Wiener is relevant, and a general reference is the book by Lukacs.</p>


问:
Composition of two permutations, expressed in cycle notation
I'm wondering if I did the problem correctly.

$[(1 2 3) (45)] \circ [(3 4)(1 2)]$

So, I started with (1

and $1 \to 2 \to 3$

then it becomes (1 3

and I did 3; $3 \to 4 \to 5$

then it becomes (1 3 5

now I did 5; $5 \to 4$

then it becomes (1 3 5 4

now to check 4; $4 \to 3 \to 1$ and 1 is being repeated.

the composition of two permutation is (1 3 5 4)

So, I am wondering if it is correctly done.

答:
Your answer is correct.  You applied the permutation on the right and then the one on the left.

You should be aware, however, that, when combining permutations,

some authors apply the one on the left first and then the one on the right.

It's a good idea to specify which convention is being followed.


问:
Normalise rest of list
I have an list of fractions as such:

```<code>sum(3/10, 1/2, 1/5) = 1
</code>```

and I multiply some element (the second in this case) by some known factor (1/2 here):

```<code>sum(3/10, 1/4, 1/5) != 1
</code>```

So the answer would be:

```<code>sum(????, 1/4, ???) == 1, where arr[0] and arr[2] are proportionate to their original values.
</code>```

How can I normalise the rest of the array, so that the array still sums to 1, but the second element remains the same (1/4 in this case) and the rest of the elements are proportionately changed? I've tried adding/multiplying the rest of the elements by the factor but it doesn't work.

答:
Suppose the changed number has initial value $k$ and has been multiplied by a factor $a$.

Then the sum is now $k(a)+(1-k)$. In particular this differs from $1$ by $k(1-a)$.

So to set the sum to $1$ again, only allowing to multiply all the other numbers by a constant factor, we need to multiply by
$$\frac{k(1-a)+(1-k)}{(1-k)}=\frac{1-ak}{1-k}$$

Sanity-checking, for your question we have $k=a=1/2$. The formula then gives $3/2$

The formula works for all combinations of inputs as long as $k\neq 1$

Title: Normalise rest of list


问:
Show that convex hull of set of all positive determinant matrices is $\mathcal{M}_n(\mathbb{R})$.
I know that set of positive determinant matrices is not a convex set as we can take $A = \mbox{Diag}(1, ..., 1)$ and $B = \mbox{Diag}(-1, -1, 1, 1, ..., 1)$. We can see $\frac{1}{2} A + \frac{1}{2}B$ is not a positive determinant matrix.

Then what is it's convex hull ? Is it $\mathcal{M}_n\mathbb{R}$?

答: Sure.  Take any matrix $A$.  Now let $M$ be a very big real number.  Claim $ A + MI $ and $ A - MI $ are invertible.   Indeed, if $ M $ is larger than the largest eigenvalue this is true by definition.


问:
Finding the remainder when $N$ is divided by $1000$
Suppose there are $N$ ordered lists of positive integers $(a_1,a_2,a_3,a_4,a_5,a_6)$ such that $a_i \geq i$ for $i = 1,2,\ldots,6$ and $a_1 + a_2 + a_3 + a_4 + a_5 + a_6 \leq 40.$ Find the remainder when $N$ is divided by $1000.$

答:
Hint

Let $b_i = a_i - i$

This problem is equivalent to

$$b_1 + b_2 + b_3 + b_4 + b_5 + b_6 \leq 19$$

$$ b_i \geq 0$$

This is a standard problem that you should know how to solve to get $N$


问:
Where is the negative sign coming from in the distributive step
I came across a proof in the Book Quantum Computer for Computer Scientists. It shows the derivation of the equation for the division of complex numbers. I am trying to follow. It starts with
$$
(x,y) = \frac{(a_{1}, b_{1})}{(a_{2}, b_{2})}
$$
where a1, b1 represent the real and imaginary part of a complex number c1. Likewise for a2 and b2. The next step in the proof says:

then by definition of division as the inverse of multiplication
$$
(a_{1}, b_{1}) = (x,y) \times (a_{2}, b_{2})
$$

which I understand fine, but then it says

or
$$
(a_{1}, b_{1}) = (a_{2}x - b_{2}y, a_{2}y + b_{2}x)
$$

My question is where does that (-) come from? everything is positive. I expect it to be:
$$
(a_{1}, b_{1}) = (a_{2}x + b_{2}y, a_{2}y + b_{2}x)
$$
I am guessing my understanding of how multiplication works here is wrong. Any help is much appreciated.

答:
That negative sign comes from $i^2=-1$.

$(a_2+ib_2)(x+iy)=a_2x+ia_2y+ib_2x+\color{blue}{i^2}b_2y=(a_2x\color{blue}-b_2y)+i(a_2y+b_2x)$.


问:
How to use Kullback-Leibler Divergence if probability distributions have different support?
I have two discrete random variables $X$ and $Y$ and their distributions have different support. Assume $X$ and $Y$ can both take on the same number of values. Lets say $X$ takes values in $\{10,13,15,17,19\}$ and $Y$ takes values in $\{12,14,16,18,20\}$.

I would like to use the Kullback-Leibler Divergence but it requires that Q dominates P. Is it possible to modify the support of each random variable so that they have the same support?

If not, are there any measures of statistical distance that do not require $X$ and $Y$ to have the same support?

One solution I have created is to make kernel density estimators with a gaussian kernel using the datasets collected on $X$ and $Y$. Now the densities $\hat{f}(x)$ and $\hat{g}(y)$ have support on $( -\infty, \infty)$ and with suitable bandwidth they are multimodal with modes centered around the support of the original random variables. It remains to be seen how wise or foolish of an idea this is.

Note: Since the KL divergence of a finite gaussian mixture does not have a closed form solution, I used monte carlo methods to estimate it.

答: Statistically, Kullback-Leibler divergence measures the difficulty of detecting that a certain distribution say H1 is true when it is thought initially that a certain other distribution say H0 is true.  It essentially gives the so-called Bahadur slope for this problem of discrimination (i.e. of testing). If, as in your example, each distribution has some support on a set for which the other distribution has no support then perfect inference becomes possible and the divergence is legitimately infinite. The more interesting case is where one support is fully contained in the other. In that case one of the hypotheses can be confirmed with certainty while the other only exponentially fast so the divergence will be infinite only in one direction.</p>


答:
This is almost the same answer as @Paresseux Nguyen's one. The main purpose of this answer is to convince the reader that Paresseux's answer is correct, by supplying some extra details. I will gladly delete my answer if Paresseux's answer is resurrected and needs to be credited.

Since OP already knows that

$$ \mu(\{a\}) = \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} e^{-ita}\varphi(t) \, \mathrm{d}t \tag{1} $$

holds for any $a \in \mathbb{R}$, it suffices to show that the integral in the right-hand side converges to $0$ under the assumption

$$ \lim_{\left|x\right|\to\infty} \varphi(x) = 0. \tag{A}$$

Using this assumption, for each fixed $\epsilon > 0$, we can pick $R > 0$ such that $\left| \varphi(x) \right| < \epsilon$ whenever $\left|x\right| \geq R$. Then

\begin{align*}
\left| \int_{-T}^{T} e^{-ita}\varphi(t) \, \mathrm{d}t \right|
&\leq \int_{-T}^{T} \left| e^{-ita}\varphi(t) \right| \, \mathrm{d}t \\
&\leq \int\limits_{\left|x\right|\leq R} \left| \varphi(t) \right| \, \mathrm{d}t + \int\limits_{R<\left|x\right|\leq T} \epsilon \, \mathrm{d}t \\
&\leq 2R + 2\epsilon(T-R).
\end{align*}

Dividing both sides by $2T$ and taking $\limsup$ as $T\to\infty$,

\begin{align*}
\limsup_{T\to\infty} \left| \frac{1}{2T} \int_{-T}^{T} e^{-ita}\varphi(t) \, \mathrm{d}t \right|
\leq \limsup_{T\to\infty} \frac{2R + 2\epsilon(T-R)}{2T}
= \epsilon.
\end{align*}

Since the limsup in the left-hand side is a fixed number which is independent of the choice of $\epsilon > 0$, letting $\epsilon \downarrow 0$ shows that this limsup is in fact $0$, which is equivalent to

$$ \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} e^{-ita}\varphi(t) \, \mathrm{d}t = 0. $$

So by $\text{(1)}$, we have $\mu(\{a\}) = 0$ for any $a \in \mathbb{R}$ as desired.

Remark. A slight modification of this proof immediately yields the following well-known result:

Theorem. (Cesàro mean) Let $f: [0, \infty) \to \mathbb{C}$ be locally integrable and $f(t) \to \ell \in \mathbb{C}$ as $t\to\infty$. Then
$$ \lim_{T\to\infty} \frac{1}{T} \int_{0}^{T} f(t) \, \mathrm{d}t = \ell. $$

Conversely, this theorem can be used to produce a much shorter answer.

Title: What does the limit of the characteristic function $\varphi(x)\xrightarrow{\lvert x\rvert\rightarrow \infty} 0$ tell us about $\mu(\{a\})$?

Body:
If we know that $\varphi(x)\xrightarrow{\lvert x\rvert\rightarrow \infty} 0$ for the characteristic function $\varphi$ of the probability measure $\mu$ on $\mathbb{R}$, what does that tell us about $\mu(\{a\})$ for any $a$?

Because of the identity
$$\mu(\{a\}) = \lim_{T\rightarrow\infty}\frac 1{2T}\int_{-T}^Te^{-ita}\varphi(t) dt$$
I think it should be $\mu(\{a\}) = 0$ for all $a$. But I don't know how to formalize this idea, if it's true.

Edit: A TA confirmed it is true that under the assumption $\mu(\{a\})=0$ for all $a$, but I still don't even know how to start a possible proof/what the key insight is.

Edit2: I have successfully shown that
$$\lim_{T\rightarrow\infty}\frac 1{2T}\int_{-T}^T\varphi^2(t)dt = \sum_{x\in\mathbb{R}}(\mu(x))^2$$
But I need to show that the LHS is $0$. It is not trivial to show that the integral is even finite in the limit, how can I show that the LHS is finite?


问:
A theorem in Takesaki's book (Vol 2, Chapter VIII)

The above screenshot is from Takesaki's book.

I have two questions.

In the proof of Theorem 3.2. The author mention that if $\psi$ is not faithful, then we can choose $\psi^{'}$ om $M$ such that $s(\psi^{'}= 1-s(\psi)$. How to prove the existence of the weight $\psi^{'}$?

The proof of the last statement of the theorem is not given. I wonder how to construct $\xi\in H_{\varphi} $ such that the $\omega(x)=(\pi_{\varphi}(x)\xi,\xi)$?

答:
You define $\psi'(x)=\varphi((1-s(\psi))x(1-s(\psi)))$.

Taking $\psi=\omega$, you have
$$
\omega(x)=\langle \pi_\omega(x)\,\Omega,\Omega\rangle,\ \text{ for some } \Omega\in\mathfrak H_\omega.
$$
Let $\xi=u\Omega$. Then
$$
\omega(x)=\langle u^*\pi_\omega(x)u\Omega,\Omega\rangle
=\langle \pi_\varphi(x)u\Omega,u\Omega\rangle
=\langle \pi_\varphi(x)\xi,\xi\rangle,
$$
where $\xi=u\Omega$.


答:
A general solution where you can have the ratio according to the array numbers.
Let the 3 numbers in your array be : $x,\ y,\ z$

Condition 1:
$$x+y+z = 1$$
where $y= \frac{1}{2}$

Condition 2:

You have scaled $y$ by ratio of $0.5$, let us scale the other 2 numbers by $\alpha$

$$\alpha x+ 0.25 + \alpha y = 1$$

So the 2 equations are:

$$x+z = \frac{1}{2}$$
$$\alpha x+\alpha z = \alpha(x+z) = \frac{3}{4} $$
$$\alpha = \frac{3\times2}{4}= \frac{3}{2}$$
Hope this helps...

Title: Normalise rest of list

Body:
I have an list of fractions as such:

```<code>sum(3/10, 1/2, 1/5) = 1
</code>```

and I multiply some element (the second in this case) by some known factor (1/2 here):

```<code>sum(3/10, 1/4, 1/5) != 1
</code>```

So the answer would be:

```<code>sum(????, 1/4, ???) == 1, where arr[0] and arr[2] are proportionate to their original values.
</code>```

How can I normalise the rest of the array, so that the array still sums to 1, but the second element remains the same (1/4 in this case) and the rest of the elements are proportionately changed? I've tried adding/multiplying the rest of the elements by the factor but it doesn't work.


问:
Probability that the connection broke down in this grid

See below chart. The yellow blocks are two islands. It's connected by
a grid of cables. There are 16 vertical cables and 9 horizontal cables
and 12 nodes (blue highlighted) in between.

Hurricane strike and each cable has 1/2 probability of breaking. Two
islands lost contact when there is no path to go from one island to
the other on the grid. What's the probability that these two islands
lost contact?

I am thinking of something like letting $f(1,1)$ be the probability that the top left node cannot be reached from top island. Then $f(1,1) = 0.5 \cdot (0.5 \cdot (1-f(1,2)) + f(1,2))$. Reason is you need the direct cable from top island to first node broke, and you either need cable from $(1,1)$ to $(1,2)$ broke, or $f(1,2)$ broke. But the equation starts to get a bit messy.

It looks like this might be able to expressed in matrix form or Markov Chain or some sort though..

答:
This is an old chestnut (though I couldn't readily locate an MSE  duplicate) and best tackled by a symmetry argument.

To illustrate, consider the following problem:  Say all the cables are low, so a boat wanting to go from left to right will get stopped by them.  Now, if enough cables are broken, there will be a path for such a boat to cross.  In fact the possible paths of the boat (marked in your diagram in red as below) is a rotated copy of the cable network itself.

Note the symmetry of the boat crossing problem with your island connectivity problem, and also consider that these events are disjoint and one of them is bound to hold true.  It follows that the answer to the probability of either occurring is $\frac12$.


答:
The condition for the cubic $x^3 + px + q$ to have 3 real roots is $4p^3+27q^2 \leq 0.$ This condition can be proved algebraically (search for Discriminants of a polynomial) or by calculus: A cubic has 3 real roots if and only if it has a local minimum and a local maximum and the product of those two values is $\leq 0.$

So as Potato mentioned in the comments, $x^3-3x+1$ has $3$ real roots, yet none of them are rational ( the rational root theorem says the only candidates are $\pm 1$ which are easily ruled out).

Title: Zeros of a cubic polynomial with rational coefficients

Body:
While discussing a related problem, one of my friends came out with a question as follows:

Is it possible that a cubic polynomial $p(x) \in \Bbb{Q}[x]$ has all of its zeros to be both real and irrational? That is, can $p(x)$ be factored into the following form?

$$ p(x) = a (x - \alpha_1)(x - \alpha_2)(x - \alpha_3), \quad a \in \Bbb{Q}, \ \alpha_{i} \in \Bbb{R} \setminus \Bbb{Q} $$

We struggled with this problem for a moderate time but failed to find any clue for its validity or invalidity.  I guess that this is impossible, but currently I have no idea how to attack this. Can you enlighten me by showing a valid direction?


问:
A question about integral transformation from the Unit sphere to Euclidean space.
I want to show the following identity about integral on spherical coordinate system:

Let $\mathbb{S}^{d-1}$ denote the unit sphere of the Euclidean space $(\mathbb{R}^d,||\cdot||)$ equipped with
the surface Lebesgue measure $\sigma$ normalized by $\sigma(\mathbb{\mathbb{S}}^{d-1}) = 1$.  Prove that for $1\leq q<\infty$:

$$\Big(\int_{\mathbb{s}^{d-1}}|x\cdot y|^qd\sigma(y)\Big)^{1/q} = ||x||_2\Big(\frac{2\Gamma(d/2)}{\Gamma(1/2)\Gamma((d-1)/2)}  \int_{0}^1 x_1^q(1-x_1^2)^{\frac{d-3}2} dx_1\Big)^{1/q},$$

where $||x||_2=(\sum_{i=1}^d|x_i|^2)^{1/2}$.

My understanding is that this identity transfers the integral about sphere to integral about one variable $x_1$. I feel that this  maybe using the rotation invariance of the measure $d\sigma$ and doing the spherical coordinate transformation. But I have no idea to prove this identity.

Any suggestions would be welcome and thank you!

答:
Here's a way to construct a proof. Your intuition is correct, but doing a rotation in spherical coordinates can be a bit problematic, albeit probably doable. A solution to this is to rewrite the expression as an integral over $\mathbb{R}^d$ and do the rotation there, where it's pretty easy to express explicitly.

Rewrite the original integral as
$$I=\int_{S^{d-1}}|x\cdot y|^qd\sigma_y=\mu^{-1}(S_{d-1})\int_{\mathbb{R}^d}d^d y~\delta(||y||_2-1)|x\cdot y|^q$$
Now perform an arbitrary rotation in $y$ expressed as a change of variables $y'=Ry~,~ R^TR=I~,~d^dy'=d^dy$. The integral now reads
$$I=\frac{1}{\mu(S_{d-1})}\int_{\mathbb{R}^d}d^d y'~\delta(||y'||_2-1)|(Rx)^T y'|^q$$
Keeping in mind that we want to introduce multidimensional spherical coordinates eventually, we pick $Rx=||x||_2\hat{e_1}$. This is always possible. With this choice $(Rx)^T y'=||x||_2||y'||_2\cos\varphi_1$ and finally reintroducing spherical coordinates we obtain
$$I=\frac{||x||^q}{\mu(S_{d-1})}\int_0^{\infty} dr  ~r^n\delta(r-1)\int d\Omega_{d-1}|\cos\varphi_1|^q$$
The radial integral is, as expected, trivial. The angular integral can be expressed in terms of the spherical surface element as
$$\int d\Omega_{d-1}|\cos\varphi_1|^q=\int d\Omega_{d-2}\int_{0}^{\pi}d\varphi_1|\cos\varphi_1|^q\sin^{d-2}\varphi_1=2\mu(S_{d-2})\int_0^{\pi/2}d\varphi_1\cos^{q}\varphi_1\sin^{d-2}\varphi_1=2\mu(S_{d-2})\int_{0}^1dx_1 ~x_1^q(1-x_1^2)^{\frac{d-3}{2}}$$
where the last step above has been achieved by setting $x_1=\cos\varphi_1$. Substituting in the known values for the surface area of the spheres we get the desired result
$$I=||x||^q\frac{2\Gamma\left(\frac{d}{2}\right)}{\sqrt{\pi}~\Gamma\left(\frac{d-1}{2}\right)}\int_{0}^1 dx_1~ x_1^q(1-x_1^2)^{\frac{d-3}{2}}$$
PS the last integral can be done in terms of Gamma functions as well!


问:
The second differential versus the differential of a differential form
I'm curious to know if the definition given below of the second
differential is actually used in mathematics. It is based on the following definition of the differential of a function:

Let $dx\in\mathbb{R}$ be an independent variable. Let $y:\mathbb{R}\to\mathbb{R}$
be a differentiable function of $x$. The differential $dy$ of $y=y\left[x\right]$
is defined as

$$
dy\equiv\frac{dy}{dx}dx,
$$

where $\frac{dy}{dx}$ is the derivative of $y$ with respect to $x$
by the standard definition.

The second differential $d^{2}y$ of $y$ is defined as

$$
d^{2}y\equiv\frac{d^{2}y}{dx^{2}}dx^{2}.
$$

I don't recall any case where this construct has been useful to me.
Nonetheless, I have it in my notes, and am confident that it reflects
a reputable, though dated, source.

Edit to provide source:
Link: Calculus and Analytic Geometry: With Supplementary Problems, Classic Edition, by George B. Thomas $\S$ 2-7, Problem 22.

Is the above definition of the second differential used today
in mathematics? This is the question for which I will accept an answer.

Can the above definition be brought into consonance with the definition of the differential of a differential form? Which, as I understand
it goes as follows:

Let $\mathfrak{r}\equiv\left\{ x,y\right\} $ be a position variable
in $\mathbb{R}^{2},$ the functions $P,Q:\mathbb{\mathbb{R}}^{2}\to\mathbb{R}$
be continuously differentiable over the neighborhood of interest,
and $dx,dy:\mathbb{\mathbb{R}}^{2}\to\mathbb{R}$ be the coordinate
projection mappings. That is, given a vector $\mathfrak{v}=\left\{ v^{x},v^{y}\right\} ,$
$dx\left[\mathfrak{v}\right]=v^{x}$ and $dy\left[\mathfrak{v}\right]=v^{y}.$
The the following product is defined
$$
dxdy\equiv-dydx,
$$

$$
dxdx\equiv dydy\equiv0.
$$
A mapping which associates with every point $\mathfrak{r}$ a linear
mapping of the form

$$
\omega_{\mathfrak{r}}=P\left[\mathfrak{r}\right]dx+Q\left[\mathfrak{r}\right]dy
$$

is a differential 1-form. The differential of this differential form
is defined to be
$$
d\omega=\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy.
$$

Given a continuously differentiable function $f:\mathbb{\mathbb{R}}^{2}\to\mathbb{R},$
its differential (form) is defined to be
$$
df_{\mathfrak{r}}\equiv\left(\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy\right)_{\mathfrak{r}}.
$$

The differential (exterior derivative) of $df$ is therefore

$$
d^{2}f\equiv ddf=\left(\frac{\partial^{2}f}{\partial x\partial y}-\frac{\partial^{2}f}{\partial y\partial x}\right)dxdy=0.
$$

答:
I think that it's worth distinguishing second derivatives in the context of calculus and the second differential in the context of differential forms.  In the latter case, the differential is defined with an antisymmetric wedge product that, as you noted, means that $d^2 y = 0$ for any $y$.  Although this identity seems useless and trivial, it actually has plenty of applications.

For example, physicists like myself learn all sorts of multivariable calculus identities for vector fields in $\mathbb{R}^3$ as undergraduates.  Two of these are $\nabla \times \nabla f = 0$ and $\nabla \cdot (\nabla \times \vec{f}) = 0$.  These are really useful in solving problems in electrostatics and magnetostatics.  They are also literally identical to the second differential equation above (applied to 0-forms and 1-forms, respectively).  There are more such identities in higher dimensions, and it's much easier to remember/understand them when you realize that they are all "the same thing."

Differential forms unify many different identities of multivariable calculus, and this is only one example.  For a physicist, the next most obvious example is that Green's Theorem and the various forms of Stokes' Theorem are all the "same thing."  Another example from pure mathematics is that the equation $d^2 = 0$ is completely analogous to the equation $\partial^2 = 0$ where $\partial$ is the boundary operator in simplicial homology.  This has a lot of deep implications.  One simple implication is that this can be used to prove Euler's theorem on vertices, edges, and faces of polygons.


答:
The convex hull is the whole matrix space when $n\ge2$. Since every square matrix $A$ is the average of two nonsingular matrices (consider $A=\frac12\left[(A+cI)+(A-cI)\right]$ for a sufficiently large $c$), it suffices to prove that every matrix with a negative determinant is the average of two matrices with positive determinants.

Let $D=\operatorname{diag}(-1,1,\ldots,1)$ and $P=\pmatrix{-1&-2\\ 2&1}\oplus I_{n-2}$. Then $\det(P)>0$ and $D=\frac12(P+P^T)$. So, if $\det(A)<0$, then $A=DDA=\frac12(PDA+P^TDA)$ is the average of two matrices with positive determinants.

Title: Show that convex hull of set of all positive determinant matrices is $\mathcal{M}_n(\mathbb{R})$.

Body:
I know that set of positive determinant matrices is not a convex set as we can take $A = \mbox{Diag}(1, ..., 1)$ and $B = \mbox{Diag}(-1, -1, 1, 1, ..., 1)$. We can see $\frac{1}{2} A + \frac{1}{2}B$ is not a positive determinant matrix.

Then what is it's convex hull ? Is it $\mathcal{M}_n\mathbb{R}$?


问:
Are these ODEs equivalent?
I have the following set of ordinary differential equations:
\begin{equation}
\left\{
\begin{array}{l}
\dot{a} = f_1(a, b, c, d) \\
\dot{b} = f_2(a, b, c, d) \\
\dot{c} = f_1(c, d, a, b) \\
\dot{d} = f_2(c, d, a, b)
\end{array}
\right.
\end{equation}

where $f_1$ and $f_2$ are two functions from $\mathbb{R}^4$ to $\mathbb{R}$.
Also, I have the following initial conditions:
\begin{equation}
\left\{
\begin{array}{l}
a(0) = \alpha \\
b(0) = \beta \\
c(0) = \alpha \\
d(0) = \beta
\end{array}
\right.
\end{equation}
and hence, under some suitable conditions, I have the unique solutions $a(t)$, $b(t)$, $c(t)$ and $d(t)$.

Then, I pose that:
$$x(t) = \frac{a(t) + c(t)}{2} ~ \text{and} ~ y(t) = \frac{b(t) + d(t)}{2} $$

Can I prove that there exists two function, $g_1$ and $g_2$, from $\mathbb{R}^2$ to $\mathbb{R}$, such that the solution of the Cauchy problem

\begin{equation}
\left\{
\begin{array}{l}
\dot{x} = g_1(x, y) \\
x(0) = \alpha \\
\dot{y} = g_2(x, y)\\
y(0) = \beta
\end{array}
\right.
\end{equation}

has the properties that $x(t) = a(t) = c(t)$ and $y(t) = b(t) = d(t)$?

答:
If you can express $f_1(a,b,c,d) + f_1(c, d, a, b)$ and $f_2(a, b, c, d) + f_2(c, d, a, b)$ as a function of $(a+c)$ and $(b+d)$, then yes. If you can't, then no.

Of course if $f_1(a, b, c, d)$ and $f_2(a, b, c, d)$ are both function of $(a+c)$ and $(b+d)$ then this condition will be satisfied.

Edit: corrected by @the_candyman in a comment.

Title: Are these ODEs equivalent?


问:
Proof the sequence $(x_n)=\frac{n}{n+1}$ is contractive
After proving $(x_n)=(\frac{n}{n+1}:n\in \Bbb N^+)$ is convergent, a bounded, monotone, and Cauchy sequence. Next I have to prove that this sequence is contractive. Though I have no idea if it is true or false right now. I read some examples on how to do it, most of it are about defining $x_{n+1}$ first, but I don't know how to do that for this sequence.

So I don't even know how to begin proving this.

答: Note that $$\vert x_n-x_m\vert=\dfrac{\vert n-m \vert}{(m+1)(n+1)}\le\dfrac{\vert n-m\vert}{4}.$$ Becouse the minimum value of $(m+1)(n+1)$ achieve at $n=m=1.$