问:
Binomial coefficients identity : $\sum_{k=1}^{n-m+1} k\binom{n-k+1}{m}=\binom{n+2}{m+2}$
For any positive integer m&n.$n\ge m$ , let $\left( {\begin{array}{*{20}{c}}
n\\
m
\end{array}} \right) = {}^n{C_m}$. Prove that $\left( {\begin{array}{*{20}{c}}
n\\
m
\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}
{n - 1}\\
m
\end{array}} \right) + 3\left( {\begin{array}{*{20}{c}}
{n - 2}\\
m
\end{array}} \right) + .. + \left( {n - m + 1} \right)\left( {\begin{array}{*{20}{c}}
m\\
m
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{n + 2}\\
{m + 2}
\end{array}} \right)$

My approach is as follow
$n=4, m=2$

$\left( {\begin{array}{*{20}{c}}
4\\
2
\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}
3\\
2
\end{array}} \right) + 3\left( {\begin{array}{*{20}{c}}
2\\
2
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
6\\
4
\end{array}} \right) = 15$

$6 + 6 + 3 = 15 \Rightarrow LHS = RHS$

But not able to solve it via property

答:
Here is a combinatorial approach. Consider the set
$$\{ 1,2,3,4,\ldots,n-1,n,n+1,n+2\}$$

We will pick $m+2$ distinct numbers. This can be done randomly in $\binom{n+2}{m+2}$ ways.

Another way to do the same would be prioritizing the second smallest element.

If second smallest number is $2$, we can choose $1$ as the smallest element and select $m$ out of remaining $n$ numbers. $\binom{n}{m}$ ways.

If second smallest number is $3$, we can choose the smallest in two ways out of $1,2$ and select $m$ out of remaining $n-1$ in $\binom{n-1}{m}$ ways.

In general, if second smallest choice is $i+1$, the smallest can be picked in $i$ ways and rest in $\binom{(n+2)-(i+1)}{m}=\binom{n-i+1}{m}$ ways.

Summing over all valid values of $i$, LHS=RHS is proved.


问:
Numerical solution of HJB (Hamilton-Jacobi-Bellman equations) in practice
I am struggling to understand what the use of HJB equations in order to solve optimal control problems is in practice. I found the following approach in the book "Stochastic Controls" by Jiongmin Yong and Xun Yu Zhou on p.163:

1.) Solve the HJB-equation to find the value function,

2.) Minimize/Maximize the Hamiltonian pointwise to find the optimal control,

3.) Solve the ODE, since the optimal control is known.

While I do understand the above, I don't understand why apart from certain specific cases where the Hamiltonian is explicitly known, solving the HJB should be more efficient than simply minimizing the cost functional with dynamic programming or some type of gradient descent.

As far as I understand it, in order to solve the HJB-equation numerically, I have to discretize all equations in time and space and then have to solve a minimization problem. Solving this minimization problem again as according to step 2 seems completely redundant.

Am I missing something? Why are HJB equations useful in practice?

答:
When you directly minimize the cost functional (for instance using a discretization in time so that the integral becomes a summation), you are solving the optimal control problem for a single initial point $x_0$. Moreover, you will find an open-loop control, i.e. a function $t \mapsto u^*(t)$, which is not robust to perturbations in the dynamics (see Open-loop vs closed-loop).

When you solve the HJB equation, instead, you are simultaneously solving the problem for all values of $x_0$, and you can even provide an optimal control in feedback form.

Regarding the question on the double minimization problem, you are right. There is a double minimization problem, but

The dimension of the minimization problem is the dimension of the control space, which is typically very small, so it not very expensive.

Simulations have shown that the first minimization, the one while solving the HJB in step 1, does not need to be solved accurately, so one can discretize the control space with few controls and minimize over a finite set of possible control values. The second minimization problem, however, must be solved more accurately.

I recommend this book for more details.

ContentLicense: CC BY-SA 4.0


问:
Differentiation in multivariable Calculus
We say that $f$ is differentiable at $x$ if the following are satisfied:

The partial derivatives $D_{1} f(x), D_{2} f(x), \cdots, D_{n} f(x)$ exist.

There exist $r>0,$ and a real valued function $g$ on $\{\|z\|<r\}$ such
$$
\begin{array}{l}
\text { that } \lim _{\|z\| \rightarrow 0} g(z)=0, \text { and } \\
\qquad f(x+z)=f(x)+\nabla f(x) \cdot z+\|z\| g(z)
\end{array}
$$

I understand that first condition which says partial derivative exist in $n$ directions.

Now for second condition it looks like $1$st order Taylor approximation. Can someone explain how it matches with most common definition of multivariable differentiability. How to look at it Geometrically ?

答: The most common definition of derivative of $f$ at $x$ is that it is a linear map $L\colon\Bbb R^n\longrightarrow\Bbb R$ such that$$\lim_{z\to x}\frac{f(z)-f(x)-L(z-x)}{\|z-x\|}=0.\tag1$$So, define$$g(z)=f(z)-f(x)-L(z-x)$$and then $\lim_{z\to x}g(z)=0$. On the other hand, the only linear map for which $(1)$ can possibly hold is the map$$\begin{array}{ccc}\Bbb R^n&\longrightarrow&\Bbb R\\z&\mapsto&\nabla f(x)\cdot z.\end{array}$$


问:
about law of total probability and conditional probability
I have a basic question about law of total probability and conditional probability. We know from conditional probability that
$ P[A \cap B] = P[A|B] P[B] = P[B|A] P[A] $ (1)

We also know from law of total probability that
$ P[B] = \sum\limits_{i=1}^N P[A_i \cap B] = \sum\limits_{i=1}^N P[B \cap A_i] $ (2)

If we combine the two, I think we can come up with

$ P[B] = \sum\limits_{i=1}^N P[A_i|B] P[B] = \sum\limits_{i=1}^N P[B|A_i] P[A_i] $ (1), (2)

$ P[B] = \sum\limits_{i=1}^N P[A_i|B] P[B] $

The part above disturbs me. It seems something is wrong with it but I can't understand what.

Someone to help me clarify this?

Thanks in advance.

答:
There is nothing wrong. Note that the formula
$$\sum_{i=1}^N P(A_i |B)P(B)= P(B)$$
is equivalent to
$$\sum_{i=1}^N P(A_i |B)= 1.$$
The last formula is just a consequence of the fact, that the mapping $A \mapsto P(A|B)$ is in fact a probability measure, and thus if $S$ is the entire sample space, then we do have that $P(S|B)=1$.


问:
When is a boundary mapped to a boundary?
In multiple integration exercises one often wants to change the domain of integration using a well behaved function. Say you want to evaluate $\int_D f$, a mapping $\varphi$ transforms $D$ into another domain $\varphi(D)$. From what I've seen, "most of the time", the boundary $\partial D$ is mapped into the boundary $\partial \varphi(D)$. What are the requirements for that statement to hold? That is, when do we have $\varphi(\overline{D} \setminus \mathring{D})=\overline{\varphi(D)} \setminus \mathring{\varphi(D)}$?</p>

答:
Your statement surely holds when $D \subseteq \mathbb{R}^n$ is simply connected and $\varphi$: $D \to \text{Im} \ \varphi(D)$ is an homeomorphism. The proof of this is by absurd:

Let $x_0 \in \partial D$ and $\varphi (x) \notin \partial \varphi{D}$.

Then $\left. \varphi(x) \right|_{x_0}$ is an homeomorphism and it induces an isomorphism between the two homology chains: $$ \varphi : D \to \text{Im}(D)$$ $$ H_q(\varphi): H_q(D) \to H_q(\text{Im}(D))$$

By hypothesis $H_q(D) = \lbrace 0 \rbrace \ \forall q \in \mathbb{Z}$. (*)

The second homology sequence contains, for some $i \in \mathbb{Z}$ (**), $H_{i} \cong \mathbb{Z}$, so we get a contradiction.

I think this statement it's valid for every topological variety with a finite number of holes. Using homology you are formalizing the idea that the number of holes is a invariant under homeomorphisms and in fact you are adding an hole with that missing inner point

(*)  (even without that point, I can retract the space to a point, or in other words, there aren't "holes" inside it)

(**)for example in a $2$-dimensional topological variety with an hole inside it the loop around the hole isn't trivial (= can't be retracted to the constant path). To generalize this reasoning you have to consider (using singular homology maybe), the $n-1$ cycle around the hole and note that it's equivalence class is not trivial.

ContentLicense: CC BY-SA 3.0


问:
Ideals of $\mathbb{C}[x,y]$ contains $x^2 + y^2 - 5$ and $xy - 2$?
This is a question from Artin's algebra Chapter 11, 9.8. I am trying to consider the quotient ring $\frac{\mathbb{C}[x,y]}{(x^2 + y^2 - 5, xy - 2)}$ which imposes the relation $x^2 + y^2 - 5 = 0$ and $xy - 2 = 0$ to the ring. I think there are 4 solutions to this system and the ideals containing $(x^2 + y^2 - 5, xy - 2)$ can be reverse constructed by picking one or more solutions to this system which gives 16 ideals in total (is this correct or there are repetitions?). Am I on the right track? Or am I completely in the wrong direction?

Note that the ideal does not have to be maximal.

答:
Comment: "I think there are 4 solutions to this system and the ideals containing (x2+y2−5,xy−2) can be reverse constructed by picking one or more solutions to this system which gives 16 ideals in total."

Answer: Let $A:=\mathbb{C}[x,y]/(x^2+y^2-5,xy-2)$ and let $k:=\mathbb{C}$.

Let $I$ be the system $x^2+y^2-5=0, xy-2=0$.

Given a solution $(a,b)\in \mathbb{C}^2:=k^2$ to the system $I$, you get a map of $k$-algebras

$$\phi_{a,b}:A \rightarrow k$$

defined by $\phi_{a,b}(x):=a, \phi_{a,b}(y):=b$. the map $\phi_{a,b}: A \rightarrow k$
is surjective hence its kernel $\mathfrak{m}_{a,b}:=(x-a,y-b) \subseteq A$ is a maximal ideal.
Conversely given a maximal ideal $\mathfrak{m} \subseteq A$ it follows the quotient
$A/\mathfrak{m}$ is a finite extension of $k$ - this is the "nullstellensatz". Hence since $k$ is algebraically closed it follows $p:A/\mathfrak{m}\cong k$ is an isomorphism. Let $a:=p(x), b:=p(y)$ it follows $(a,b)$ is a solution to the system $I$ and $\mathfrak{m}=(x-a,y-b)$. Hence there is a 1-1 correspondence between solutions to $I$ and maximal ideals in $A$. As you have observed: There is 4 solutions to the system $I$, hence there should be 4 maximal ideals in the ring $A$ - each corresponding to a solution $(a,b)$ of $I$. If $\mathfrak{m}_1,..,\mathfrak{m}_4$ are the maximal ideals in $A$ it follows any product $\mathfrak{m}_i\mathfrak{m}_j$ is a non-zero non-maximal ideal in $A$.  In order to determine all ideals in $A$ you must check explicitly what happens to higher powers $\mathfrak{m}_i^k\mathfrak{m}_j^l$ for $l+k > 2$ or triple products $\mathfrak{m}_i\mathfrak{m}_j\mathfrak{m}_k$. You may be aware of the Chinese remainder theorem (Atiyah-Macdonald Proposition 1.10): It says that if $I:=\cap \mathfrak{m}_i=(0)$ there is an isomorphism

$$ A \cong A/I \cong \oplus_i A/\mathfrak{m}_i \cong \mathbb{C}^4.$$

Hence if this is the case you can easily count the number of non-trivial ideals in $A$. You should check if $I=(0)$ holds in this case.

Example: Let us assume $I=(0)$ and let $A:=k \oplus k \oplus k\oplus k$
with $e_1:=(1,0,0,0),..,e_4:=(0,0,0,1)\in A$.

You get nontrivial ideals in $A$:  $(e_i), (e_i,e_j), (e_i,e_j,e_k)$ for all choices of $i,j,k$. Note that $e_i^2=e_i$. It seems to me yo get 14 non-trivial ideals but you should check this yourself.


问:
What is the partition for a dice, where the universe is $\Omega=\{1, 2, 3, 4, 5, 6\}$?
We launch a dice once. The dice has 6 possible results.

The universe is $\Omega=\{1, 2, 3, 4, 5, 6\}$.

What is the partition of the universe ?

Is it $\{1, 2, 3, 4, 5, 6\}$ ?

Is it $\{1\}$, $\{2\}$, ... $\{6\}$ ?

Is it $\{1\}$, $\{2\}$, ... $\{6\}$, $\{1, 2\}$, etc. $\{1, 2, 6\}$ ?

(Sorry for my basic question)

答:
"A partition is a grouping of its elements into non-empty subsets, in such a way that every element is included in exactly one subset." (Wikipedia)

There are many partitions of the same set. You have listed some of the possible ones:

$$\mathcal{P}_1 = \Big\{ \{1,2,3,4,5,6\}\Big\} $$
$$\mathcal{P}_2 = \Big\{ \{1\},\{2\},\{3\},\{4\},\{5\},\{6\} \Big\} $$
$$\mathcal{P}_3 = \Big\{ \{1,2\},\{3,4\},\{5,6\}\Big\} $$

But there are many others:
$$\mathcal{P}_4 = \Big\{ \{1,4\},\{2\},\{3,5,6\}\Big\} $$
$$\mathcal{P}_5 = \Big\{ \{1\},\{2\},\{3,4,5,6\}\Big\} $$
$$\mathcal{P}_6 = \Big\{ \{1,2\},\{3\},\{4\},\{5,6\}\Big\} $$
and so on.


问:
Find an asymptotic upper bound for $\prod_{p|n}(1+\frac{2}{p})$.
Let's define a function:
$$f:\mathbb{N}\rightarrow \mathbb{R}$$
$$f(n) = \prod_{p|n}(1+\frac{2}{p})$$
Where $p\in \mathbb{P}$

Find a differentiable function $g:\mathbb{R}\rightarrow\mathbb{R}$ such $f(n) = O(g(n))$ as $n$ goes to infinity. Where $O$ is a big O.

I have an intuition that $g(x) = \log(\log (x))$. I have calculated  $\frac{f(n)}{\log(\log (n))}$ on my computer up to $4000$ numbers, and it seems that this quotient is bounded from above by $18$ and from below by $0.4$. However i would like to be sure, whether $\log(\log(x))$ is a good function.

Regards.

-Edit-

As far i understand, i can do the following:
$$f(n)=\prod_{p|n,p\in \mathbb{P}}(1+\frac{2}{p})\le \prod_{1\le k\le \omega(n)}(1+\frac{2}{p_k})$$
Where $p_k$ is the $k$-th prime number and $\omega(n)$ is the number of prime divisors of $n$. Let's use a logarithm here. Notice that $\log(1+x)\le x$.
$$\log f(n) \le \sum_{k=1}^{\omega(n)}\log(1+\frac{2}{p_k})\le 2\sum_{k=1}^{\omega(n)}\frac{1}{p_k}$$
$$2\sum_{k=1}^{\omega(n)}\frac{1}{p_k}=2\log\log\omega(n)+O(1)$$
Hence:
$f(n) = \exp(\log(f(n)))\le \exp(2\log\log\omega(n)+O(1))=e^{O(1)}(\log\omega(n))^2$.
Because $\omega(n) = O(\log(n))$, now we see that $f(n)=O((\log\log(n))^2)$.

Am i right?
Please correct me if i am wrong.

答:
Anything you don't like with $3^n$ ? The better result is $f(n)\le f(\prod_{p\le k_n} p)$ with $ \prod_{p\le k_n} p$ the largest primorial $\le n$ so that (from the $\theta(2m)-\theta(m)\le\log {2m\choose m}\le \psi(2m)$ argument) $k_n =O(\log n)$ and $\log n=O(k_n)$ ie. $\log k_n\sim \log \log n$.

Mertens theorem gives
$\log f(\prod_{p\le k_n} p)=2 \log \log k_n+C+o(1)$. Whence $f(n)\sim e^C (\log\log n)^2$.


问:
Methods for identifying points at which matrices become degenerate
I have two Hermitian matrices, $A$ and $B$ and I am interested in analyzing the spectrum of the matrix,
$$A - \mu B$$
where $\mu$ is some real number. In particular, I want to know precisely the values of $\mu$ where the lowest two eigenvalues of the $A-\mu B$ matrix become equal, i.e. where the eigenspace of the lowest eigenvalue becomes degenerate. Doing this numerically, I'm taking what I'm guessing is a rather inefficient approach. I diagonalize $A-\mu B$ for different values of $\mu$ (searching over some range), and then compute the difference between the lowest two eigenvalues. If that difference falls within some tolerance, then I save the value of $\mu$ and the value of the difference. Then when the routine is complete, I run a function that finds local minima in the array of these difference values, and voila, I have identified the $\mu$ values at which this degeneracy occurs.

I'm wondering if there are methods I could use that are more sophisticated, or analytical that I could employ? An additional aspect of the problem is that these two matrices, $A$ and $B$ are actually both projections into a particular subspace. I construct them by taking,

\begin{align}
A &= U^\dagger Z U\\
B &= U^\dagger Z^2 U
\end{align}

for some third matrix $Z$ and some set of eigenvectors $U$ that span the subspace. I haven't come up with a way to use this fact at all as part of my approach.

答:
Disclaimer: this is not a complete answer, just some thoughts about the problem.

I assume that $Z$ is an $n \times n$ hermitian matrix and $A$ and $B$ are $m \times m$ matrices (also hermitian, by definition). Also $n \geq m$ (we're projecting to subspace).

Let $\tilde U$ be the projection matrix onto the orthogonal complement of the subspace spanned by $U$. So $Q = (U \,|\, \tilde U)$ is a square unitary matrix.
Let $W = Z - \mu Z^2$. Then
$$
\begin{pmatrix}
A - \mu B & \ast\\
\ast & \ast
\end{pmatrix}
=
\begin{pmatrix}
U^\dagger W U & U^\dagger W \tilde U\\
\tilde U^\dagger W U & \tilde U^\dagger W \tilde U\\
\end{pmatrix}
=
\begin{pmatrix}
U^\dagger\\
\tilde U^\dagger
\end{pmatrix}
W
\begin{pmatrix}
U &
\tilde U
\end{pmatrix}
= Q^\dagger W Q.
$$
$A - \mu B$ is a principal submatrix of $Q^\dagger W Q$. Note that the eigenvalues of $Q^\dagger W Q$ are the same as the eigenvalues of $W$ and the latter are $\lambda_i = \zeta_i - \mu \zeta_i^2$ where $\zeta_i$ are the eigenvalues of $Z$.

Doing some numerical experiments with small $n$ and $m$ suggests that the problem might not have solutions at all in general. Here's a typical plot for random hermitian $10 \times 10$ matrix $Z$ and $4 \times 4$ matrices $A$ and $B$. Below is a plot of eigenvalues of $A - \mu B$ as functions of $\mu$. The eigenvalues eventualy become closer, but never touch each other. The exception is $m = n$ case when eigenvalues are linear in $\mu$ and do intersect (and interchange after the intersection, see the second image).

Though I do not have a proof I suggest that equality of eigenvalues is possible only for very special choice of $U$ and in general does not occur. I tried applying Cauchy's eigenvalue interlacing theorem at least for $m = n - 1$ case, but it requires studing when inequalities are strict (so it allows to conclude that the lower pair of eigenvalues is separated and thus cannot be equal).


问:
An inequality in expected values of Random Variable with different PDFs

Problem. Prove that
$$E[\log f(X)] \ge E [\log q(X)], $$
if $X$ is a random variable with PDF $f(x)$ and $q(x)$ is any valid PDF.

Hint provided: Use Jensen's Inequality.

This means:

\begin{gather*}
E [\log f(X)]
= \int f(x) \log f(x) \, \mathrm{d}x
\ge E[\log q(X)]
= \int f(x) \log q(x) \, \mathrm{d}x \\[0.75em]
\implies
-\int f(x) \log \biggl(\frac{q(x)}{f(x)}\biggr) \, \mathrm{d}x \ge 0.
\end{gather*}

Now, I am unable to understand how to move forward to use Jensen's inequality and thus get according expression.

答:
$$ E\biggl[ \frac{g(X)}{f(X)} \biggr]
= \int \frac{g(x)}{f(x)} f(x) \, \mathrm{d}x
= \int g(x) \, \mathrm{d}x
= 1 .$$

By Jensen's inequality applied to the convex function $-\log x$ we get

$$ 0
= -\log E\biggl[ \frac{g(X)}{f(X)} \biggr]
\le - E \biggl[ \log \frac{g(X)}{f(X)} \biggr]
= - E [\log g(X)] + E [\log f(X)]. $$

Hence, $E[\log g(X)] \leq E[\log f(X)]$.


问:
Find a convex function $f$ which satisfies $\lim\limits_{x\to+\infty}f'(x)=a$ and $\lim\limits_{x\to+\infty}\Big[x\Big(f'(x)-a\Big)\Big]\neq0$
Question 1.

Find a function  which satisfies the following conditions:

$f:\mathbb{R}\to \mathbb{R} ,\quad f\  \cup$

$\lim\limits_{x\to+\infty}f'(x)=a$

$\lim\limits_{x\to+\infty}\Big[x\Big(f'(x)-a\Big)\Big]\neq0$

It relates to this  problem.
I'm unable to find a such function.

Thanks in advance.

Edit: One final question (Perhaps what I should have asked).

Question 2.

Find a function $g$ such that:

$g:\mathbb{R}\to \mathbb{R}, \  g \  \cup $

$\lim\limits_{x\to+\infty}\Big[g(x)-ax \Big]=\lambda\in\mathbb{R}\quad$

$\lim\limits_{x\to+\infty}\Big[x\Big(g'(x)-a\Big)\Big]\neq0$

Let's note that if $\lim\limits_{x\to+\infty}\Big[g(x)-ax \Big]=\lambda\in\mathbb{R}$ $\ $then $\lim\limits_{x\to+\infty}g'(x)=a$

Also it can be shown that $\exists\lim\limits_{x\to+\infty}x\Big(g'(x)-a \Big) $

Second Edit ((Question 2. is impossible)

set $\lambda = k+c, c\neq 0$
then

$\lim\limits_{x\to+\infty}\Big[g(x)-ax-k \Big]=c\neq 0\quad$

So,  $c=\lim\limits_{x\to+\infty}\Big[g(x)-ax-k\Big]=\lim\limits_{x\to+\infty}\Big[{x(g(x)-ax-k)\over x} \Big]\overset{{\infty}\over \infty}{=}\lim\limits_{x\to+\infty}\Big[{\Big(x(g(x)-ax-k)\Big) '\over \Big(x\Big) '} \Big]=\lim\limits_{x\to+\infty}\Big[g(x)-ax-k+x(g'(x)-a) \Big]=c + \lim\limits_{x\to+\infty}x\Big(g'(x)-a \Big)\\ \Longrightarrow \lim\limits_{x\to+\infty}x\Big(g'(x)-a \Big)=0$

答:
We can make the limit anything we want. Consider
$$
x(f'(x)-a)=b\tag1
$$
Solving $(1)$ gives
$$
f(x)=ax+b\log(x)+c\tag2
$$
To make $f''(x)\ge0$, we need $-\frac{b}{x^2}\ge0$. That is, set $b\lt0$ in $(2)$.

If you want this to be defined for all $x\in\mathbb{R}$, then use
$$
f(x)=\left\{\begin{array}{}
ax+b\log(x)+c&\text{if }x\ge1\\
c-b+(a+b)x&\text{if }x\lt1
\end{array}\right.\tag3
$$


问:
$\delta$-response for challenging the $\lim_{x \to 10} \frac{1}{[[x]]} = \frac{1}{10}$ with $\epsilon=\frac{1}{2}$
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$

I am self-studying Real Analysis from Stephen Abbott's Understanding Analysis. I'd like to ask if my conclusions pertaining to exercise (a), (b) of the below problem are correct.

Notation. $f(x)= [[x]]$ is the box function, the greatest integer less than or equal to $x$, for all $x \in \mathbf{R}$.

Exercise 4.2.4. Consider the reasonable but erroneous claim that

\begin{align*}
\lim_{x \to 10} \frac{1}{[[x]]} = \frac{1}{10}
\end{align*}

(a) Find the largest $\delta$ that represents a proper response to the challenge of $\epsilon = 1/2$.

(b) Find the largest $\delta$ that represents a proper response to $\epsilon = 1/50$.

(c) Find the largest $\epsilon$ challenge for which there is no suitable $\delta$ response possible.

Proof.

(a) We require

\begin{align*}
\frac{1}{10} - \frac{1}{2} &< \frac{1}{[[x]]} &< \frac{1}{10} + \frac{1}{2} \\
\frac{-4}{10} &< \frac{1}{[[x]]} &< \frac{6}{10}
\end{align*}

So, $[[x]] < \frac{-10}{4} < -2$ and $[[x]]>\frac{10}{6} > 1$. In other words, $x-10 < -12$ and $x-10>-8$. The absolute distance must satisfy the inequality $\absval{x - 10} < 8$. Thus, the largest $\delta-$response to the challenge $\epsilon=1/2$ appears to be $\delta = 8$.

(b) We require

\begin{align*}
\frac{1}{10} - \frac{1}{50} &< \frac{1}{[[x]]} &< \frac{1}{10} + \frac{1}{50} \\
\frac{4}{50} &< \frac{1}{[[x]]} &< \frac{6}{50}
\end{align*}

So, $[[x]] < \frac{50}{4} < 13$ and $[[x]]>\frac{50}{6} > 8$. In other words, $x < 13$ or $x >9$. The absolute distance must satisfy the inequality $\absval{x - 10} < 1$. Thus, the largest $\delta-$response to the challenge $\epsilon=1/50$ appears to be $\delta = 1$.

(c) We would like to have the distance

\begin{align*}
\absval{\frac{1}{[[x]]} - \frac{1}{10}} > \epsilon
\end{align*}

no matter what the open interval $(10-\delta,10+\delta)$ in which $x$ lies. Rearranging, I get:

\begin{align*}
\epsilon < \frac{\absval{[[x]] - 10}}{10 \absval{[[x]]}}
\end{align*}

I am not sure how to proceed from here. I know that, $\absval{[[x]] - 10}$. That yields,

\begin{align*}
\epsilon < \frac{\delta}{10 \absval{[[x]]}}
\end{align*}

I can further write, $[[x]] > \lceil{10 - \delta}\rceil$. But, this $\epsilon$ is dependent on the $\delta$-interval I choose.

答:
For part (c), we’re looking for a distance $\epsilon$ such that, when we look around x = 10, the function values aren’t all closer than $\epsilon$ to 1/10.

I feel like a more intuitive approach might be effective here. What value does the function take on to the left of x = 10? How far is that value from 1/10? That difference should be your $\epsilon$, no?


问:
Powers of logarithm question
I have got a question about formal simple mathematics are these expressions equivalent?

$$ \log^{n} (x/a) $$

and  $$ \log^{n} (x) -\log^{n} (a) $$

I know is a dumb question and they are equal only when $ x= a$ but need some help

答:
they are not equivalent, and you can see it directly for $n=2$ written as product.
you have $=(\log(x)-\log(a))^2$

Title: Powers of logarithm question


问:
Why does this inequality hold $|\int_T \frac {|x|} {\sqrt n} dx - \frac {1} {\sqrt 3}| \leq \sqrt 3 \int_T |\frac {|x|^2} {n} - \frac 1 3 | dx$?
why does this inequality hold?
$$\left|\int_T \frac {|x|} {\sqrt n} \mathrm{d}x - \frac {1} {\sqrt 3}\right| \leq \sqrt 3 \int_T \left|\frac {|x|^2} {n} - \frac 1 3 \right| \mathrm{d}x$$
with $T=[0,1]^n$

答: Apply triangle inequality for the integral then use the fact that $|\frac{|x|^2}{n}-\frac{1}{3}| = |\frac{|x|}{\sqrt{n}}-\frac{1}{\sqrt{3}}| \cdot |\frac{|x|}{\sqrt{n}}+\frac{1}{\sqrt{3}}|$.


问:
Solving exponential function with linear and quadratic term
I came along this equation in a book for higher education engineering:

$$ x\cdot2^x - x^2 = 135 $$
What I know is that there are ways to get a solution numerically. But I‘m pretty sure that there is also an analytical way (besides rather complicated stuff like lamberts function) since the result is (simply) $$ x=5$$
Did I miss a simple trick?

Best wishes!

答:
I don't know if this is an upgrade over guessing but you can try this,

$x(2^x -x) = 135$

The factors of $135$ are $1 \times 135 , 3 \times 45,5 \times 27,9 \times 15$
so now you have a guess at x, So check for $(2^x -x)$ and at $x = 5 ,$ we get $(2^x -x) = 27$


问:
Finding the equation of line given the two intersecting lines of planes
I have an idea that in order to get the equation given only the lines, it requires to do cross product and solving for the coordinates that satisfies the given two lines.

The problem is that only intersecting lines of planes are given. I have no problem for the cross product but I am confused for finding the coordinates since the answer would be infinite solution. What could be another way to find the points?

The original problem goes this way.

Determine an equation for each of the lines described below.

line of intersection of the planes 2x - y + z = -1 and x + 4y - z = 2.

答:
Crossing the normal vectors to the plane, you’ll get the direction vector of the line, i.e. $$(2,-1,1) \times (1,4,-1) =(-3,3,9) $$

Now, you need any one point that lies on the line. This will be any point that lies on both the planes. One variable is arbitrary. Let’s take $z=0$. Then you get $$2x-y=-1 \\ x+4y =2 \\ \implies x =-\frac 29, y= \frac 59$$ and thus, the equation of the line is $$\vec r = \left( -\frac 29, \frac 59, 0 \right) +\lambda(-3,3,9)$$


答:
Crossing the normal vectors to the plane, you’ll get the direction vector of the line, i.e. $$(2,-1,1) \times (1,4,-1) =(-3,3,9) $$

Now, you need any one point that lies on the line. This will be any point that lies on both the planes. One variable is arbitrary. Let’s take $z=0$. Then you get $$2x-y=-1 \\ x+4y =2 \\ \implies x =-\frac 29, y= \frac 59$$ and thus, the equation of the line is $$\vec r = \left( -\frac 29, \frac 59, 0 \right) +\lambda(-3,3,9)$$

Crossing the normal vectors to the plane, you’ll get the direction vector of the line, i.e. $$(2,-1,1) \times (1,4,-1) =(-3,3,9) $$

Now, you need any one point that lies on the line. This will be any point that lies on both the planes. One variable is arbitrary. Let’s take $z=0$. Then you get $$2x-y=-1 \\ x+4y =2 \\ \implies x =-\frac 29, y= \frac 59$$ and thus, the equation of the line is $$\vec r = \left( -\frac 29, \frac 59, 0 \right) +\lambda(-3,3,9)$$

Title: Finding the equation of line given the two intersecting lines of planes

Body:
I have an idea that in order to get the equation given only the lines, it requires to do cross product and solving for the coordinates that satisfies the given two lines.

The problem is that only intersecting lines of planes are given. I have no problem for the cross product but I am confused for finding the coordinates since the answer would be infinite solution. What could be another way to find the points?

The original problem goes this way.

Determine an equation for each of the lines described below.

line of intersection of the planes 2x - y + z = -1 and x + 4y - z = 2.


问:
Find $\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}$ exactly by hand
Show the exact value of $\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}$ is $\frac{-(1-\sqrt{17})}{4}$ (no calculator).

By and large these things require considerations of Euler's formula so I rewrote the expression as $$\frac{1}{2}(a+a^2+a^4+a^8)+\frac{1}{2}(a^{-1}+a^{-2}+a^{-4}+a^{-8})$$ where $a=e^\frac{2\pi i}{17}$. This looks really nice like you could sum it or something but it really just doesn't play nice when you try to. It is not geometric, so pretty much the only way forward is to find a function whose taylor series starts with $f=x+x^2+x^4+x^8+...$ or something workable to look like that, but this is not obvious at all either, and the next problem would be whether that function can be used to evaluate $f(e^\frac{2\pi i}{17})$ easily which at this point is hopelessly improbable. By now I have no where else to turn, it seems likely that the solution doesn't involve complex numbers.

答:
For anyone who stumbles across this problem, it is an utter waste of your time and you should go do something else. Nothing here can be motivated with honesty.

Call the expression $\xi$ and expand $\xi ^2$ as a binomial

Use Prosthaphaeresis Formulas and Cosine double angle and $\cos{(2\pi - x)}=\cos{x}$ and $\sum^{16}_{n=1} \cos{\frac{2n\pi}{17}}=0$ to get $\xi^2=2+\frac{\xi}{2}-\xi$

Solve quadratic.

Notice all but $\cos\frac{16\pi}{17}$ are positive in original expression but $\frac{2\pi}{17}<\frac{4\pi}{17}<\frac{\pi}{3}$ so certainly $\cos{\frac{2\pi}{17}+\cos\frac{4\pi}{17}}>1>-\cos{\frac{8\pi}{17}}$ and only one of the quadratic roots is positive.


问:
Finding the upper and lower bounds, positive and negative each if applicable, for a complex function
Given $|z^4 -8 + 3|, |z|=2 $

I need to find the positive upper and lower bounds of the above expression.

My attempt:

By triangle inequality, we have

$||z^4|+|-8z|+3|\geq|z^{4} -8z + 3| \leq  ||z^4 -8z| +3|???||z^4|+|3-8z|| \geq ||z^4|-|8z|+3|\geq||z^4+3|-|8z|| $

(I used ??? since I don't know which is larger than which, and I believe i got all the combinations possible)

At the left most, I believe that will give me the upper bound so $|2^4|+|-8(2)|+3| = 35$

At the right most, $||2^4 +3| -|8(2)| = 3|$

BUT, and here is my question, given that $z = r*e^{i\theta}, |e^{i\theta}| = 1$

assuming my triangle inequality is true, we know $|8z| = |8r*e^{i\theta}|=|8r|$, however $|z^4+3|$ is not necessarily equal to $|r^4 +3|$, I wonder if $|z^4+3| = |8z|$ such that $||z^4+3|-|8z||= 0$ ?

Alternatively if $|z^4 - 8z| = -3$ is possible??

答:
A point on the circle is characterized by $z=2e^{i\theta}$.

Let $g(z)=z^4 - 8z+3.$

We seek the minimum and maximum of

$\begin{aligned} |g(\theta)|^2 &= g(2e^{i\theta})\overline{ g(2e^{i\theta})}=g(2e^{i\theta})g(2e^{-i\theta})\\&=(16e^{4i\theta}-16e^{i\theta}+3)(16e^{-4i\theta}-16e^{-i\theta}+3)\end{aligned}$

Differentiatate with respect to $\theta$ to find extrema $\cdots.$


问:
Three digit number even and greater than $500$
I am reading the following problem:

The digits $1$,$2$,$3$,$4$,$5$ are randomly arranged to form a three
digit number (digits are not repeated). What is the probability that
the number is even and greater than $500$?

My solution:

There are $5\cdot 4 \cdot 3$ ways to form the number in total.

In order to be greater than $500$ the first digit must be $5$ and in order to be even the last digit must be either $2$ or $4$. So there is $1$ way to select the first digit, $2$ ways to select the last and $2$ ways to select the middle one (either $3$ or $4$) i.e. $2\cdot 2\cdot 1$ = $4$ ways

Therefore the probability is: $\frac{4}{60}=\frac{1}{15}$

But the solution mentions $\frac{1}{10}$.

What am I doing wrong here?

答:
The total space of events consists of $ 5*4*3 = 60  $  3-digit numbers, consisting of non-repeating digits $ 1, 2, 3, 4$ and $5 $.

Of them, only those that start with $5$ and end with $2$ or $4$ are even greater than $500$.

If the last digit is "$2$", then the second digit is any of $1, 3,$ and$ 4$  ($3$ possibilities).

If the last digit is "$4$", then the second digit is any of $1, 3,$ and $2$  ($3$ possibilities).

In all, there are $3+3 = 6$  of these "eligible" numbers.

So the probability will  be $ \frac{6}{60} $


问:
A nontrivial everywhere continuous function with uncountably many roots?
This is my first post on SE, forgive any blunders.

I am looking for an example of a function $f:\mathbb{R} \to \mathbb{R}$ which is continuous everywhere but has uncountably many roots ($x$ such that $f(x) = 0$). I am not looking for trivial examples such as $f = 0$ for all $x$. This is not a homework problem. I'd prefer a nudge in the right direction rather than an explicit example.

Thanks!

Edit: Thanks all! I've constructed my example with your help.

答: Why not $f(x)=x+|x|$? Looks quite nontrivial to me.</p>


问:
Numerical method to solve a trigonometric (cotangent) function - transient heat transfer problem
I was trying to develop a mathematical model for transient one-dimensional heat conduction of spheres using approximate analytical solution as mentioned in Cengel{refer page number 229 in that document}. To obtain the temperature gradient I will have to find the first few roots of the Eigen function (refer Table 4.1 in the above document). For the sake of the others who do not want to download the above PDF I am writing down the equation below.

$f(\lambda) = 1-\lambda_{n} cot \lambda_{n}-Bi=0$, where $Bi\geq 1$ (Biots number) and $\lambda>1$.

The issue with the above equation is that for various values of $Bi$ this equation will have roots and singularities over a range. A typical example for $Bi=5$ and $0\leq\lambda\leq10$ is shown in Recktenwald, 2006 {refer Figure 2 in page number 3 in this document}. Please keep in mind that $\lambda$ - the eigen value myself and Cengel have used is called $\zeta$ in the above figure taken from Recktenwald, 2006 linked above.

Can you please suggest me an efficient numerical way of finding roots which also avoids singularity. I am planning to implement this in C. As the program given in the appendix of Recktenwald, 2006 is in Matlab, I couldn't understand the underlying logic as I can't handle it well. I will be doing my code with gcc. Please help to write the algorithm.

答:
My advice (I am not going to write your algorithm) is to break the interval of interest up into subintervals between the poles of the function $\cot{x}$, which are at values of $x=k \pi$ for integer $k$.  Within each subinterval, there is a single root of the function.  To find that root in, say, the $k$th subinterval, you may simply use a bisection routine as follows.

Begin with bracket $[(k-1) \pi + \epsilon, (k-1/2) \pi, k \pi - \epsilon]$ for some $\epsilon = 0.01$ or something like that (you may want to scale it with, say, $Bi$.  Is $(k-1/2) \pi$ the same sign as the left endpoint or the right?  Replace that endpoint with $(k-1/2) \pi$, determine the midpoint of the new bracket, and iterate until you find the root to some desired tolerance.  That gets you the root for that subinterval.  Then move onto the next one, etc.


问:
Complex quartic factorisation
Answer: I'm really not sure how to factorise.  I understand that z.z* gives 2Re(z) but it's still not clear to me how it works.

答:
Hint:

$$\dfrac{w^5+1}{w+1}=?$$

For $c,$ use Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$

Title: Complex quartic factorisation


问:
Why is $\kappa(A) = \infty$ if $A$ is a singular matrix?
Al though not important for my course, I was curious as to why the condition number of a singular matrix is $\infty$. Which also begs the question of what the condition number of a non-invertible non-square matrix would be.

答:
The condition number is defined as the ratio between the highest and smallest singular value of the matrix.

If the matrix is singular, then the smallest singular value is $0$.

Thus, the condition numbers tends to $\infty$.


问:
Proof of $\operatorname{arccot}(x)= \arctan(\frac{1}{x})$
$\newcommand{\ac}{\operatorname{arccot}}$
Here is my proof:

$ \theta = \ac(x)$

$ \cot \theta = x $

$ \displaystyle \tan \theta = \frac{1}{\cot \theta} = \frac{1}{x}$

$ \displaystyle \theta = \arctan \frac{1}{x}$

But based on the graph, $\ac(x) = \arctan(\frac{1}{x})$ is only satisfied when $ x > 0$. What is wrong with my proof? What is the error occurred inside a proof?

答:
Long comment:

The first problem with $$\arctan\left(\frac1x\right)=\operatorname{arccot}(x)$$ is that $\arctan$ is an odd function, while $\operatorname{arccot}$ isn't.

Indeed, $\forall x\in\Bbb R$, the following holds:

$$\arctan(x)+\operatorname{arccot}(x)=\frac\pi2\implies\operatorname{arccot}(x)=\frac\pi2-\arctan(x)$$
Hence, if $x>0$, then
$$\arctan\left(\frac1{-x}\right)=-\arctan\left(\frac1x\right)=-\operatorname{arccot}(x)=-\operatorname{arccot}(-(-x))$$
and we cannot just move the sign under the argument. Therefore,
$$\arctan\left(\frac1x\right)=\begin{cases}-\operatorname{arccot}(-x), &x<0\\\operatorname{arccot}(x),&x>0\end{cases}$$


问:
About $\sum\limits_{i=1}^\infty\sum\limits_{j=1}^\infty a_{ij}= \sum\limits_{j=1}^\infty\sum\limits_{i=1}^\infty a_{ij}$
Without any reason, I suddenly realize that  I seemly  never see any theory about  the  series with two index. For example, what is the condition of following equation is  right?
$$
\sum\limits_{i=1}^\infty\sum\limits_{j=1}^\infty a_{ij}= \sum\limits_{j=1}^\infty\sum\limits_{i=1}^\infty a_{ij}
$$
I don't know how to deal it. So I check the book about series, but fail to find relative conference. So I ask here.

In fact, what I really want is a book about  series with interesting theory and the two index series, thanks for any help.

答:
From Rudin (Theorem $8.3$):

Given a double sequence $\{a_{ij}\}$ and suppsoe that $$\sum_{j=1}^\infty | a_{ij}|=b_i\;\;(i=1,2,3...) $$ and $\sum b_i$ converges. Then $$\sum_{i=1}^\infty \; \sum_{j=1}^\infty a_{ij}=\sum_{j=1}^\infty \; \sum_{i=1}^\infty a_{ij}$$


问:
For which $\alpha$ this sum converges? $\sum_{n=3}^{\infty} {\frac{1}{n \cdot \ln(n) \cdot \ln(\ln(n))^{\alpha}}}$
Given:

$$\sum_{n=3}^{\infty} {\frac{1}{n \cdot \ln(n) \cdot \ln(\ln(n))^{\alpha}}}$$
I am asked: For what values of $\alpha$ does this summation converge?

So I said, $f(n) = \frac{1}{n \cdot \ln(n) \cdot \ln(\ln(n))^{\alpha}}$. $f(n)$ is obviously monotonically decreasing function. Then, by using the integral test, this summation converges if and only if $I = \int_{3}^{\infty} {\frac{dn}{n \cdot \ln(n) \cdot \ln(\ln(n))^{\alpha}}}$ converges. (has a value)

But I am finding this very hard to go on with. Any direction will be appreciated!

答:
Peter's solution and mine are the same, but the idea is to make a u-substitution:
$$\int_3^\infty \frac{dn}{n\ln n (\ln \ln n)^\alpha}$$

The cleanest substitution is $u=\ln\ln n$, since then $du=\frac{1}{\ln n}\cdot \frac{1}{n} dn$ which nicely ties up your denominator into
$$\int_{\ln(\ln (3))}^\infty \frac{1}{u^\alpha} du$$ and the rest is obvious.

Even if you didn't see the substitution $u=\ln\ln n$, you could at least notice that if you subbed $u=\ln n$ you could get rid of the  $1/n$ and be left integrating $\frac{1}{u\ln u}$, which you could apply the same technique to.


问:
Why is the power series of $\tan(x)$ not convergent everywhere $\cos(x)$ is non-zero?
The Taylor series expansion of $\tan(x)$ centered at $0$ has a radius of convergence of $\pi/2$, which means the power series will not converge for $|x|>\pi/2$.

How can this be when you consider $\tan(x)=\sin(x)/\cos(x)$, so Taylor series of $\tan(x)$ is just the Taylor series of $\sin(x)$ divided by the power series of $\cos(x)$, both of which converge everywhere.

At some point $|x|>\pi/2$ where $\cos(x)$ is not equal to $0$, how can the power series of $\tan(x)$ diverge? It is simply the quotient of two convergent series at that point which seems to me like it shouldn't have any problems.

答:
While $\tan(z)=\frac{\sin(z)}{\cos(z)}$, it is not true that the Taylor series of $\tan$ is just the Taylor series of $\sin$ divided by the Taylor series of $\cos$! I can confidently say this even though I'm not even sure what you mean be dividing the two series. I can imagine two ways to do this, none of which result in the Taylor series of $\tan$.

1: Dividing term by term

This breaks essentially immediately, since $\sin$ is an odd function, while $\cos$ is an even function. Being even, the Taylor series of $\cos$ has vanishing coefficients at odd powers. Being odd, the Taylor series of $\sin$ has non-vanishing coefficients at odd powers. So we'd be dividing something non-zero by zero. So this clearly doesn't work at all.

2: Dividing the partial sums

Series are essentially just sequences. The series $\sum_{k=0}^\infty a_k$ is simply the sequence $b_n:=\sum_{k=0}^n a_k$, and the series converges to the limit of this sequence. So we could take two power series $\sum_{k=0}^\infty a_k z^k$ and $\sum_{k=0}^\infty b_k z^k$ and consider the sequence
$$c_n:=\frac{\sum_{k=0}^n a_k z^k}{\sum_{k=0}^n b_kz^k}.$$
In fact, wherever the denominator doesn't converge to $0$, this sequence converges, and it converges to
$$\frac{\sum_{k=0}^\infty a_k z^k}{\sum_{k=0}^\infty b_kz^k},$$
so if we divide the Taylor series of $\sin$ and $\cos$ this way, the sequence will, in fact, converge to $\frac{\sin(z)}{\cos(z)}$ at all points where this quotient exists. But! This sequence is not a power series. Power series are sequences of the form $b_n=\sum_{k=0}^n a_k (z-z_0)^k$. The above quotient is not of this form, and it cannot be transformed to take that form. The Taylor series of $\tan$ is not equal to this sequence of quotients.


问:
How many set partitions of [n] are there in which 1 is not in a block of size 1?
I'm not at all sure how to solve the following question:
How many set partitions of [n] are there in which 1 is not in a block of size 1?
Could someone please help? Thanks!</p>

答:
Here's what I understood ,

If $n = \{1,2,3\}$

Then the partitions of n are,

$P_1 = \{ \{1\}, \{2\}, \{3\} \}$

$P_2 = \{ \{1, 2\}, \{3\} \}$

$P_3 = \{ \{1, 3\}, \{2\} \}$

$P_4 = \{ \{1\}, \{2, 3\} \}$

$P_5 = \{ \{1, 2, 3\} \}$

But according to your question we cannot consider $P_1$ and $P_4$ as $1$ is inside a block of size $1$.

So what we can do is , remove $1$ from the set and then calculate the number of partitions of the new set .

$n' = \{2,3\}$

Then the partitions of $n'$ are,

$P'_1 = \{  \{2\}, \{3\} \}$

$P'_2 = \{ \{2, 3\}\}$

Now to these partitions we can plug in $\{1\}$ , and get all the partitions which have $\{1\}$ in a block of size $1$, i.e $P_1$ and $P_4$.

So the number of partitions which do not have $1$ in a block size of $1 = B_n - B_{n-1}$
where $B_n$ is the Bell number.


问:
A coupon collector related problem with distinct probabilities.
Consider the integers $\{1,\dots, N\}$ for some large $N$.  Let us suppose that for each $\{1, \dots, n\}$ there are associated probabilities $p_1, \dots, p_n$ with $n \ll N$.

We sample independently and repeatedly from  $\{1,\dots, N\}$. For each $i \leq n$ we sample the integer $i$ with probability $p_i$.  With probability $q= 1-\sum_{1}^np_i$ we sample some integer greater than $n$. We are not concerned with which integer in this case.

After $s$ samples, how can you compute the expected number of distinct integers less than or equal to $n$ that have been sampled?

If anything is unclear, please let me know.

答:
Let $X_i$ be the variable with two values: 1 if the number $i$ is sampled an $0$ if it is not sampled. Then the expected value of the variable after $s$ samples is just probability that the number $i$ is sampled:
$$
\mathbb E(X_i)=1-(1-p_i)^s.
$$

By linearity of expectation, the expected number of sampled numbers in the range $1\dots n$ is:
$$
\mathbb E\left(\sum_{i=1}^n X_i\right)=\sum_{i=1}^n\mathbb E(X_i)=n-\sum_{i=1}^n(1-p_i)^s.
$$


问:
Maclaurin series for sin(x) representation
The Maclaurin series for $\sin(x)$ is:

$$ \sin(x) =  x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} ... $$

Which according to wikipedia is:

$$ \displaystyle \sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n+1)!} x^{2n+1}$$

I however don't understand why this notation is correct at all. It only seems to work if you consider that $x$ is the zeroth term. Why is it the zeroth term?

答:
What do you mean by the zeroth term? Do you mean the term given by $n = 0$? In this case simple evaluation shows this term to be $$\frac{(-1)^{0}}{(2(0)+1)!}x^{2(0)+1} = \frac{1}{1!}x^{1} = x$$

If, however, you are wondering why the first non-zero term of the Maclaurin series is $0$, then it is as simple as observing the following:

Let $f(x) = \sin(x)$.

Then $f'(x) = \cos(x)$

... $f''(x) = -\sin(x)$

... $f'''(x) = -\cos(x)$

... $f^{(4)}(x) = \sin(x)$

and so on. It is easy to note that for $f^{(k)}(0) = 0$ for $k$ even, since for $k$ even, $f^{(k)}(x) = c\sin x$, where $c$ is $1$ or $-1$, and $\sin(0) = 0$.

On the other hand, this gives us an expression for odd $k$. Writing $k$ odd as $2n+1$ for some $n \in \mathbb{N}$, we have $f^{(2n+1)}(x) = (-1)^{n}\cos(x)$, and so $f^{(2n+1)}(0) = (-1)^{n}\cos(0) = (-1)^{n}$. The rest of the expression given by wikipedia, then, follows from the definition of a Maclaurin series.


问:
Simple applications of higher derived functors
I was surprised to not have found this question discussed before on this or any other forum, hence I am posting it myself.

From an abstract point of view of homological algebra derived functors are very appealing: given a right exact functor $F$, they provide a canonical way of "completing" the image of a short exact sequence $0\to A\to B\to C\to 0$ to a long exact sequence
$$\dots\to L^2F(C)\to L^1F(A)\to L^1F(B)\to L^1F(C)\to F(A)\to F(B)\to F(C)\to 0.$$
This makes potential applications of such functors relatively clear: $L^1F(C)$ can be used to help measure the failure of exactness of $F$, specifically injectivity of the map $F(A)\to F(B)$. Hence if we have a way of finding $L^1F$ (and computing the connecting homomorphisms), then we gain information about the way $F$ acts.

This only explains the usefulness of $L^1F$. The way the use of higher derived functors was explained to me is by "kicking the can down the road" - to understand $L^1F$ we may want to use $L^2F$ which from the long exact sequence somehow measures how non-left-exact $L^1F$ is, and so on. Having learned some more advanced mathematics (group cohomology in CFT) I can see this being the case, but discussing the topics recently with more novice friends I have realized I don't really know any examples which I could briefly explain to them!

In contrast, I can think of plenty of cases where first derived functor shows up its usefulness. My favorite is probably the following result:

Prop: Suppose $C$ is a flat $R$-module. Then for any exact sequence $0\to A\to B\to C\to 0$ and $R$-module $M$ the sequence $0\to A\otimes M\to B\otimes M\to C\otimes M\to 0$ is exact.

Proof: since $C$ is flat, we have $\newcommand{\Tor}{\operatorname{Tor}}\Tor^R_1(C,M)=0$. The result is now clear from the long exact sequence.

Of course, fully explaning this result would take some work (specifically balancing of $\Tor$) but at the very least we see $\Tor_1$ is a nice device which makes such results straightforward (and makes the result easy to remember!). Other simple ways in which $\Tor_1$ comes up are the Kunneth's Theorem and Universal Coefficient Theorem.

However, as already mentioned, I am not aware of any simple such applications of any higher derived functors. Hence I ask:

What are some nice uses of higher derived functors $L^nF,n\geq 2$ (or $R^nF,n\geq 2$) which could be explained to, say, an advanced undergraduate as a motivation or example of how they are useful?

Some kinds uses I can think of would be a use in computation of lower derived functors (including the starting functor) or as coming up in a proof or at least a statement of some result. I am interested in any of these, though for the latter I would appreciate some indication of how the higher functors come up in the result.

答:
Here are two natural, elementary problems:

Given a smooth, compact, closed $n$ dimensional manifold $M$, how many open sets to we need to cover $M$ with open sets $U_i$, such that each each $U_i$ is contractible, with all finite intersections $U_{i_1}\cap U_{i_2}...\cap U_{i_k}$ are finite unions of contractible spaces.

Similarly, given a smooth projective variety $X$, whats the minimal cardinality of an open affine cover of $X$?

For both these questions, we can easily obtain a lower bound of $\dim(M)$ and $\dim(X)$ respectively, using some higher derived functors. In the manifold case, we observe that the constant sheaf $\underline {\mathbb{Z}}$ has $H^n(M,\underline {\mathbb{Z}})\neq 0$, by Poincare duality, so we have a (high) nonvanishing cohomology group (derived functor of global sections). But then if we had a cover of open sets of that nice form with less than $n+1$ opens, this cohomology group would vanish by direct computation, via Cech cohomology.

In the projective variety case, we do the same thing, we can find a quasicoherent sheaf with nonvanishing cohomology in degree $\dim(X)$ (canonical sheaf, by Serre duality), so if we had such an affine cover, computing again with Cech cohomology, this group must vanish by the acyclicity of quasicoherent sheaves over affines, so this cover can't exist.

The ideas behind these proofs goes beyond these examples too, in lots of situations, one can detect complexity via the nonvanishing of certain derived functors.

For a final commutative algebra example, the nonvanishing of the local cohomology group $H_I^n(R)$ implies that the ideal $I$ cannot be generated by $n-1$ elements, and that we cannot even find another ideal $J$ generated by $n-1$ elements with the same radical as $I$. To my understanding, these derived functors are one of the few tools we actually have for proving assertions about generation of ideals up to radical in this manner, which is another entirely elementary problem.


问:
Locus of a point of intersecting lines
The perpendicular from the centre of the ellipse $$b^2x^2+a^2y^2=a^2b^2$$ to the tangent at any point $P$ meets the line joining  point $P$ to the focus $(ae,0)$ at $Q$. Prove that the locus of $Q$ is $$x^2+y^2-2aex=b^2$$

My attempt

Tangent to an ellipse at $P(\theta)$

$$bx\cos\theta +by\sin\theta=ab\tag{1}$$

slope $m_1$ of a normal line to $(1)$

$$m_1=\frac{a\sin\theta}{b\cos\theta}$$

Equation of the normal passing through origin

\begin{align*}
y=m_1x
&=\frac{a\sin\theta}{b\cos\theta}x\tag{2}
\end{align*}

Equation of a line through $P$ and $S$

$$\begin{align*}
y=m_2(x-ae)&=\frac{b\sin\theta}{\left(a\cos\theta-ae\right)}(x-ae)
\tag{3}
\end{align*}$$

Point $Q(h,k)$ satisfies $(2)$ and $(3)$.

How to eliminate $\sin\theta$ and $\cos\theta$ to obtain the required locus?

Title: Locus of a point of intersecting lines

Body:
$$x(a\sin t)-y(b\cos t)=0\ \ \ \  (1)\implies\tan t=\dfrac{by}{ax}\ \ \ \ (3)$$

$$x(b\sin t)+ya(e-\cos t)-abe\sin t=0\ \ \ \  (2)$$

$$\dfrac x{ab^2e\sin t\cos t}=\dfrac y{a^2be\sin^2t}=\dfrac1{a^2e\sin t(1-e\cos t)}$$

$$\implies\dfrac{1-e\cos t}{\cos t}=\dfrac{b^2}{ax}\implies\sec t=\dfrac{b^2}{ax}+e\ \ \ \ (4)$$

Use $\sec^2t-\tan^2t=1$ to eliminate $t$

$$\left(\dfrac{b^2}{ax}+e\right)^2-\left(\dfrac{by}{ax}\right)^2=1$$

$$\iff(ax)^2+(by)^2=(b^2+aex)^2$$

$$\iff b^4+2ab^2ex=b^2y^2+a^2x^2(1-e^2)$$

Use $b^2=a^2(1-e^2)$

$$b^2(b^2+2aex)=b^2(y^2+x^2)$$


问:
Considering all the permutations of the following word, what is the total number of words in which the the following condition is true? ${}$
Among all the permutations of the letters of the word " JANLOKPAL", the number of words in which "O" never lies between "J" and "N" is?

I first thought of subtracting the cases in which O does occur between J and N, and then subtracting it from total possible permutations, but I can't seem to arrive at the correct answer.

Thank you.

答:
Interpretation 1: Only JON and NOJ are disallowed

There are $9$ letters which include two doubles.

Unrestricted permutations $ = \dfrac{9!}{2!2!}= P\;say$

Now form $2$ blocks, $JON$ and $NOJ$ along with the remaining $6$ letters which have two doubles,

"bad" permutations = $2\times \frac{7!}{2!2!} = Q,\;say$

and "good" permutations = $P-Q = 88200$

Interpretation 2: O can't be anywhere between J..N or N...J

Unrestricted permutations $ = \dfrac{9!}{2!2!}= P\;say$

Put $JON/NOJ$ together and start inserting unlabelled letters one by one between interstices, then label them.(With each insertion, the number of interstices will rise by $1$), and  so "bad" words = $2\times\dfrac{4\cdot5\cdot6\cdot7\cdot8\cdot9}{2!2!}= Q,\;say$,

Finally good words $=P-Q = 60480$


问:
Prove that $f(x) = \sum_{n=1}^{\infty} {\frac{x^2 \sqrt{n} + 1 }{ 2n^2 + x } } $ is continuous in $(-1,2)$.
I was asked this question:

Prove that $f(x) = \sum_{n=1}^{\infty} {\frac{x^2 \sqrt{n} + 1  }{ 2n^2 + x   } } $ is continuous in $(-1,2)$, when $x \in \mathbb R$.

On first thought, I said that if we prove that the summation uniformly converges, and it is obvious that the series of functions $f_n(x)$ are continuous, then according to uniform convergence rules, $f(x)$ is continuous everywhere, especially in $(-1,2)$.
Now I wanted to prove that it is uniformly convergent. By trying to apply the Weistrass M-Test I only have been able to say that it is $\leq |\frac{x^2}{x}| = |x|$ which helped with nothing. Then I thought about finding the general $g(x)$ that it might converge to and check if $a_n = \lim_{n \to \infty} \sup |g_n(x) - g(x)| = 0$ but I couldn't find what $g_n(x)$ was since I got to a point where I get $\frac{0}{\infty}$.

Any help will be appreciated, thanks!

答:
You are correct in your approach. The series does converge uniformly on the interval $(-1,2)$. As you implied correctly, since each term of the given series defines a function that is continuous on the interval $(-1,2)$, it follow that the series defines a continuous function on $(-1,2)$.

And, you can use the $M$-test to prove the convergence is indeed uniform on $(-1,2)$.  However, your bound is not suitable.

But, if you notice that the terms of the series are positive, then an upper bound for $f_n(x)={x^2\sqrt n+1\over 2n^2+x}$, valid for all $x\in(-1,2)$, can be obtained simply by replacing the numerator by its largest value in the interval $(-1,2)$ and the denominator by its smallest value in $(-1,2)$. This  leads to the bound $M_n={4\sqrt n+1\over 2n^2-1}$ for $f_n$. To emphasise, for each positive integer $n$, we have $|f_n(x)|\le M_n$ for all $x\in(-1,2)$.

Fortunately, this rather sloppy approximation works for the $M$-test:

The only thing left to do is to show that $\sum\limits_{n=1}^\infty M_n$ converges. This can be done by applying the Limit Comparision Test to this series and the convergent $p$-series  $\sum\limits_{n=1}^\infty{1\over n^{3/2}}$.


问:
Does Euclid's demonstration that there are only five Platonic solids need to assume convexity?
At the end of the Elements Book XIII Euclid gives a demonstration that there are only five regular solids, i.e. 'no other figure, besides the said five figures, can be constructed by equilateral and equiangular figures equal to one another' e.g. here.  The demonstration given of this is essentially that any vertex comprised of a number of equal polygons can only be formed by 3, 4 or 5 triangles or 3 squares or 3 pentagons, the reason being that any more of any of these leads to a total angle at the vertex of 360° or more.

It's been pointed out that Euclid also needs the condition of convexity.  This was probably assumed by Euclid, not least because much of the previous discussion in Book XIII was about inscribing solids in spheres.  The essence of my question is that if we don't assume this then the argument appears potentially to fail, i.e. we can easily have more than 5 triangles meeting at a vertex if it can be nonconvex, e.g. as at one of the inner vertices of a solid similar to the small stellated dodecahedron

(from Wikipedia) but composed of 60 equilateral triangular faces rather than 60 isosceles triangle (not thinking of it as twelve intersecting pentagrams either).

The standard way to then show that there are only five Platonic solids is using Euler's formula $V-E+F=2$ applied to a polyhedron comprising $n$ $m$-gons, say, and showing that the only cases that work with the formula are those corresponding to the Platonic solids.

I note that without convexity then there are some other possible solids, e.g. the icosahedron with one vertex pushed in such as

(taken from Richeson's book 'Euler's Gem').  In a sense this is just a varied icosahedron, however, and loses the regularity that we probably want from any Platonic solid so I don't think anyone would call it a 'sixth' Platonic solid.

As above with the small stellated dodecahedron, I'm not thinking of cases where faces 'intersect' such as the Kepler-Poinsot polyhedra, thinking of these as better described by a set of smaller polygons, and also note that in such solids we don't have the same number of smaller polygons meeting at each vertex so they aren't candidates for a sixth Platonic solid.  I'm also not thinking of cases where there are holes in the solid, i.e. genus not equal to 2, or where a cube might be stuck into the middle of a face etc.

To phrase one definite question then: is there a quick, knock down argument as to why a set of nonconvex vertices formed from more than 6 equilateral triangles, or more than 4 squares or pentagons, as described above, could not form a regular solid, meaning a solid with the same number of the same regular polygons meeting at each vertex?

Of course, it doesn't 'feel' as though this would be possible, and we can justify this with Euler's formula, but Euler's formula was not known to Euclid / requires algebraic manipulation not known at that time - I'm wondering if there is a justification of convexity rather than just assuming it.

Any ideas very welcome (or requests for clarification of what I mean, of course).

答: As per Guy's post, Descartes's observation that the total angular deficit of solids which are homeomorphic to a sphere is 720 deg justifies it not being possible to form such a solid where more than five equilateral triangles meet at each vertex (or more than three squares or pentagons) even though such vertices are possible in polyhedra.  To justify the total angular deficit sum being 720 deg needs Euler's formula, or similar reasoning, and so would not have been know to Euclid.  An assumption of convexity therefore seems to be needed for Euclid's argument for only five Platonic solids at the end of Book XIII of the Elements to work.</p>


问:
PDF of a function of a bounded linear random variable
I'm sure I solved this correctly, but the bounds are a bit tricky.

There's an rv w/pdf
$$
f(y) = \frac{a+by}{c}
$$
bounded on $(1,3)$ and function $W=Y^2$, I need to compute $f(w)$.

I decided to take  a longer route, and start with CDF of $Y$, which comes out to
$$
F(y) = \frac{a}{c}(y-1) + \frac{b}{2c}(y^2 - 1)
$$
and, since $Y$ is bounded, and $W=Y^2, W$ is also bounded: $(1,9)$. Now, using the CDF approach,
$$
P(W<w)= P(Y^2<w) = P(-\sqrt{w}<Y<\sqrt{w}) = F_{Y}(\sqrt{w})-F_{Y}(-\sqrt{w})
$$
This is where my confusion started, becasue $-\sqrt{w}$ is certainly negative, and $F(y)$ is only defined for $y>1$. What should the  lower bound be then? I used $F(1)=0$ to get
$$
F_W(w) = \frac{a}{c}(\sqrt{w}-1) + \frac{b}{2c}(w-1), 1 <w<9
$$

Differentiating wrt $w$, I get

$$
f_W(w) = \frac{a}{2c \sqrt{w}} + \frac{b}{2c}
$$
which should be the correct answer, but I'm still hesitant about that lower bound $F_{Y}(1)$. Is it correct?

答:
Sorry but

$$f_Y(y)=\int_1^3\frac{a+by}{c}dy \ne 1$$

thus you have to normalize $f_Y$ first...

The problem is that I try to normalize it I get

$$f_Y(y)=\frac{a+by}{2(2b+a)}\cdot\mathbb{1}_{(1;3)}(y)$$

thus $c$ disappears...

So check you $f_Y$ first...I  can assume that $a,b$ are density parameters while $c$ is a constant to be derived. In this case, using the following formula

$$f_W(w)=f_Y[g^{-1}(w)]\cdot|\frac{d}{dw}g^{-1}(w)|$$

You immediately get your desired density

$$f_W(w)=\frac{a+b\sqrt{w}}{4(2b+a)\sqrt{w}}\cdot\mathbb{1}_{(1;9)}(w)$$


问:
Identity elements of semirings
When we define a group the identity element can be any suitable one. E.g.

$(\mathbb{N}, 0, +, -)$ or $(\mathbb{N}, 1, \cdot , N_i^{-1})$ are two groups with $2$ different identity elements.

Now it is not clear to me if in the definition of a semiring when it is stated that the identity element is $0$ and $1$ if that is meant literary. I.e. that the identity elements of the two semirings have to be $0$ and $1$.

I am asking because when reading about groups, I encountered also some definitions that used $1$ to denote the identity element but in reality they meant $I_m$ (whatever is the Identity of the set $M$).

Could someone please confirm?

答:
A (semi)ring has 2 basic operations, the addition and the multiplication.

Independently, both are required to be associative and to have an identity element (so that they are monoids).

The identity element for addition is denoted by $0$, and the identity element for multiplication is denoted by $1$.

Title: Identity elements of semirings


问:
Texas TI-alternatives
I've been using a Texas TI graphing calculator at school and am wondering if there are programs for the PC that function similarly (with similar features) as those calculators. I don't need an emulator (its illegal to use one without owning the calculator anyways) but want a program that actually looks like one of those machines and functions like one.

The program should be free (preferably open source), and I'm looking for suggestions as to what programs I should download.

答:
Ok, I'm now using Microsoft Mathematics (free program) which has a GUI, natural display and the CAS features I need.

https://en.wikipedia.org/wiki/Microsoft_Mathematics

Download:

https://www.microsoft.com/en-ca/download/details.aspx?id=15702

Title: Texas TI-alternatives


问:
$p(x) = x^3 - 2x^2 + 4x - 8$ and $x = 2$ is a $0$, use synthetic division to obtain factorization of $p(x)$
Here is what I have, and I would like to get it verified.

$$p(2) = 0 = x^3 - 2x^2 + 4x - 8\\
  \frac{(x - 2)} {(x^3 - 2x^2 + 4x - 8)}\\
    x^2 + 4 = x^3 - 2x^2 + 4x - 8$$

I think the synthetic division answer is $(x^2 + 4)$

Is this correct?

How do I take this to identify other solutions?

I need to factor the polynomial using synthetic division then I need to identify other solutions of $p(x) = 0$.

答:
Yes you factor is correct.

$x^3 - 2x^2 + 4x - 8 = (x-2)(x^2+4)$

Solutions for

$p(x) = 0$

$(x-2)(x^2+4) = 0$

$(x-2) = 0 \Rightarrow x=2$

$x^2+4 = 0 \Rightarrow x^2 = -4  \Rightarrow x = 2i ,x =-2i$

So three roots are $2,2i,-2i$


问:
On the multiplicity of the eigenvalue 0 of the adjacency matrix?
Preliminaries:

-Laplacian matrix of graph $G$ is defined as follows:

 $$L=D-A $$
where $D$ is the degree matrix and $A$ is the adjacency matrix of the graph.

-The algebraic connectivity of a graph $G$ is the second-smallest eigenvalue of the Laplacian matrix of $G$.

It well known that the multiplicity of the eigenvalue 0 of the Laplacian matrix (or algebraic connectivity) is equal to the number of connected components in the graph.

Question:

Is there any significance for the  multiplicity of the eigenvalue 0 of the adjacency matrix? In other words is there any relation with the parameters of the graph (like the independence number of the graph, number of connected components,...).

Any idea will be useful!

答:
As Chris Godsil points out, the multiplicity of zero as an eigenvalue of the adjacency matrix of a tree does have a graph theoretic significance. It can be understood as follows:

The determinant of an matrix is a sum over all permutations (of, essentially, graph vertices), of a product of matrix entries. Every such permutation can be decomposed into disjoint cycles.

Suppose such a cycle is of length three or more. Then if we consider the matrix entries corresponding to the permuted elements in this cycle, at least one of them must be zero? Why? Because the matrix entries are nonzero only for graph edges. If all the entries were nonzero, this would correspond to a (graph) cycle of length three or more, which we know we don't have, because it's a tree.

Therefore, the product of the matrix element for the whole cycle (in the permutation sense) is zero. And therefore so is the product for the whole permutation.

So the determinant is a sum over permutations, where the permutation consists only of swapped pairs and fixed points. Furthermore, the swapped pairs must correspond to two vertices joined by an edge. Therefore, any such permutation is a collection of graph edges and isolated vertices. In other words, a partial matching of the graph.

Now, given such a permutation, what does the corresponding term of the determinant look like? Each edge contributes two factors, which are off-diagonal matrix entries, so both of them 1. Each fixed point or isolated vertex contributes a single entry from the diagonal of the matrix, in other words $-\lambda$. So the number of isolated vertices is the multiplicity of the root 0 for that particular term.

When all the terms are added the multiplicity of the root 0 is the minimum multiplicity of 0 in any of the individual terms (since no amount of terms 1 can produce a sum $\lambda$ or vice versa). So this gives the smallest number of isolated vertices in any graph matching.

From this result we can perhaps speculate on what this measure would tell us about a graph with cycles, if anything. For each such cycle we would have to add in additional terms to the characteristic polynomial where the permutation includes that cycle.


问:
When is the Nilradical sequence of a ring split?
Suppose that $R$ is a commutative ring with unity and denote by $N_R$ the nilradical of $R$. When is the sequence of $R$-modules

$0 \to N_R \to R \to R_{\text{red}}\to 0$

split, i.e. $R\cong N_R\oplus R_{\text{red}}$? The ring I am interested in is not necessarily semisimple, however it is local and noetherian.

Is there a general criterion when an ideal yields such a split exact sequence?

答:
When the radical is zero, it does.

Otherwise, never.

It can’t ever split because the nilradical is a superfluous submodule, meaning it cannot be complemented by a proper submodule.

For the same reason, the Jacobson radical can’t split out as an $R$ submodule (except in the case where it is zero.)

Another way to see it: the thing splitting out  has to contain a nontrivial idempotent, and it clearly cannot for the nilradical (or Jacobson radical either.)


问:
Finding the minimizer of the variational problem $J[y]= \int_{0}^{2} \left( 1-y'^2 \right)^2dx $ with $y(0)=0$ and $y(2)=0$.
This is a question from a math contest.

Find the minimizer of $J[y]= \int_{0}^{2} \left( 1-y'^2  \right)^2dx$ with the conditions: $y(0)=0$ and $y(2)=0$.

Now as $F$ is $\geq 0$, so to minimize we let $y'^2=1$, $i.e. y'= \pm 1 $ $\implies y= \pm x +c$

but then the initial conditions are not satisfied. So is the only minimizer is $y \equiv  0$?

Also for finding the extremal, using the Euler's condition we have:

$$ \frac{\partial F}{\partial y}- \frac{d}{dx} \left[ \frac{\partial F}{\partial y'} \right] =0 $$
$$ i.e. y''(3y'^2 -1)=0 $$
But I am unable to solve the differential equation.

Any help is highly appreciated.

答:
$\delta{J}[y]= \delta\int_{0}^{2} \left( 1-y'^2  \right)^2dx=-2\int_{0}^{2} \left( 1-y'^2  \right)y'(\delta{y})'dx=2\int_{0}^{2} \frac{d}{dx}\Bigl(\left( 1-y'^2  \right)y'\Bigr)\delta{y}\,dx$

As $\delta{y}$ is arbitrary $\Rightarrow$
$\frac{d}{dx}\Bigl(\left( 1-y'^2  \right)y'\Bigr)=0$, or $\left( 1-y'^2  \right)y'=C=const$

We got a cubic equation for $y'$ which can be resolved.
All three roots are constants, so we get $y'=C_{1,2,3} = const$ and $y=C_{1,2,3}x+b\,$ ($b\,$ is another constant).

Due to the requirement $y(0)=y(2)=0$ $\,\Rightarrow$ $\,y\equiv0$ at $x\in[0,2]$


问:
Numerical method to solve a trigonometric (cotangent) function - transient heat transfer problem
I was trying to develop a mathematical model for transient one-dimensional heat conduction of spheres using approximate analytical solution as mentioned in Cengel{refer page number 229 in that document}. To obtain the temperature gradient I will have to find the first few roots of the Eigen function (refer Table 4.1 in the above document). For the sake of the others who do not want to download the above PDF I am writing down the equation below.

$f(\lambda) = 1-\lambda_{n} cot \lambda_{n}-Bi=0$, where $Bi\geq 1$ (Biots number) and $\lambda>1$.

The issue with the above equation is that for various values of $Bi$ this equation will have roots and singularities over a range. A typical example for $Bi=5$ and $0\leq\lambda\leq10$ is shown in Recktenwald, 2006 {refer Figure 2 in page number 3 in this document}. Please keep in mind that $\lambda$ - the eigen value myself and Cengel have used is called $\zeta$ in the above figure taken from Recktenwald, 2006 linked above.

Can you please suggest me an efficient numerical way of finding roots which also avoids singularity. I am planning to implement this in C. As the program given in the appendix of Recktenwald, 2006 is in Matlab, I couldn't understand the underlying logic as I can't handle it well. I will be doing my code with gcc. Please help to write the algorithm.

答: If you look at the plots, you need $\cot \lambda_n$ to be pretty small because the other terms are pretty small, but $\lambda_n$ can be large.  The singularities come at $\lambda_n=k\pi$, the zero crossings for $\cot$ are at $(k+\frac 12)\pi$.  This suggests starting your root finding at $(k+\frac 12)\pi$.  As the function is less than zero at those points, pick a value near $(k+1)\pi$ and you will have bracketed the root.  Now bisection, the secant method, or Brent's method should have no trouble at all.  Newton-Raphson may have trouble because the function is almost horizontal near $(k+\frac 12)\pi$, particularly for the lowest root, but once you get close you could switch to it if you want.</p>


问:
if P is sufficient for Q: (P implies Q) or (Q implies P) ??
for the given statement:

" A sufficient condition for network failure is that the central switch
goes down."

let Q be the network failure, and P be the central switch shutting down. Then symbolically, the statement would be: P is sufficient for Q

can I say that: "if P then Q" so that the antecedent is P and the consequent is Q? or is it the opposite??

答:
In essence, yes. Note that because something is a sufficient condition, does not mean that it is a necessary condition. Example: A sufficient condition for the road to be wet, is for it to be raining. But, it isn't a necessary one. It could be that I spilled a bottle of water on the road.

In your example, "if P then Q" is correct: If the central switch is goes down, then the network is down as well. But if the the central switch is up, it does not mean that the network is not down. "if not-P then not-Q" would then be incorrect, because it could be the case that there is no power. However, you could say "if not-Q then not-P" (i.e. the converse: If the network is up, then the central switch must be up as well)


问:
Converting a set of inequalities into a linear program

Consider a polyhedron defined on the following set of constraints:

$$x_1+2x_2-2 \leq 0$$ $$-x_1 \leq 0 $$ $$ -x_1+x_2-1 \leq 0$$ $$ x_1 +
 x_2 -1 \leq 0$$

I want to convert it the standard LP form (which is possibly):

$$ \min c^T x $$
$$ \text{s.t.} \ \  Ax \leq b$$
$$ x \geq 0$$

I know how to find the constraint $Ax\leq b$. But what is the objective function here?

答:
It's not given in the information provided.

If it is not of interest to optimize anything in particular, for example, if your interest is to check if the feasible set is non-empty, you can just pick $c$ to be the zero vector to make every feasible point optimal.


问:
What is exact consistency strength of this kind of ZFC-Power set axiom theory?
What is the exact consistency strength of $\sf ZFC - Power set $ plus adding a primitive total unary function $f$ and axiomatize that:

$\forall x \, \forall y: f(x)=f(y) \to x=y$

$\forall x \, \forall y: f(x)=y \to y \text{ is a von Neumann ordinal }$

In English: there exists a injective function from the universe to the ordinals.

And also axiomatize that: All sets are countable.

Is it equal to second order arithmetic?

答:
As you conjectured and Asaf confirmed, your theory is equiconsistent with Second-order arithmetic. Some textbooks even equate $\mathsf{ZFC^-}$ + 'every set is countable' with the second-order arithmetic! (e.g., A book by Cheng.) But I will assume that the second arithmetic is defined as we usually do.

Interpreting the second-order arithmetic inside $\mathsf{ZFC^-}$ is easy: just consider the model $(\omega,\mathcal{P}(\omega))$ as we do over $\mathsf{ZFC}$. Of course, $\mathcal{P}(\omega)$ is not a set in that case, but almost all part of the proof would carry over.

The main idea for the proof of interpreting $\mathsf{ZFC^-}$ with 'every set is countable' into the second-order arithmetic uses sets-as-tree construction.
You may find the concrete proof in Chapter VII of Simpson's Subsystems of Second Order Arithmetic. Let me briefly sketch its proof.

Work over the full second-order arithmetic. (In fact, $\mathsf{ATR_0}$ suffices for our definition: but it does not prove the full separation and replacement schemes.) A subset $X\subseteq \mathbb{N}^{<\omega}$ is a suitable tree if it is a nonempty tree (i.e., a subset that is closed under initial segments of finite sequences) which is well-founded. (That is, $X$ has no infinite branch.)

Now define the relation $=^*$ and $\in^*$ for suitable trees as follows:

$S=^*T$ if and only if there is a tree isomorphism between $S$ and $T$. (The isomorphism must preserve the root.)

$S\in^* T$ if and only if $S$ is isomorphic to a tree of the form
$$\{s\in\mathbb{N}^{<\omega} \mid n^\frown s \in T\}$$
for some $n\in\mathbb{N}$.

Now we can interpret set-theoretic formulas $\phi$ into that of second-order arithmetic $\phi^*$ by using $=^*$ and $\in^*$. Moreover, we have

Theorem. Let $\phi$ be a theorem of $\mathsf{ZFC}^-$ with 'every set is countable.' Then $\phi^*$ is a theorem of the second-order arithmetic.

(It follows from a combination of Lemma VII.3.20, Exercise VII.3.39, and Lemma VII.5.3. of Simpson's book.)

The above interpretation may not validate the global choice. However, this is easily resolved: the construction of $L$ carries over even without the axiom of power set, and it also validates the global choice.


问:
My son's Sum of Some is beautiful! But what is the proof or explanation?
My youngest son is in $6$th grade. He likes to play with numbers. Today, he showed me his latest finding. I call it his "Sum of Some" because he adds up some selected numbers from a series of numbers, and the sum equals a later number in that same series. I have translated his finding into the following equation:
$$(100\times2^n)+(10\times2^{n+1})+2^{n+3}=2^{n+7}.$$

Why is this so? What is the proof or explanation? Is it true for any $n$?

His own presentation of his finding:

Every one of these numbers is two times the number before it.
  $1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$.

  I pick any one of them, times $100$. Then I add the next one, times $10$. Then I skip the next one. Then I add the one after that.

  If I then skip three ones and read the fourth, that one equals my sum!

答:
You may observe that whenever in any geometric progression $a_n=\lambda^na_0$, one term satisfies a linear recurrence ($a_n=c_0a_0+c_1a_1+\cdots+c_{n-1}a_{n-1}$, where some coefficients $c_i$ may be zero and the corresponding terms dropped), then every further term of the sequence satisfies the same recurrence: $a_{i+n}=c_0a_i+c_1a_{i+1}+\cdots+c_{n-1}a_{i+n-1}$. This is because the relation has simply been multiplied by $\lambda^i$. Even every term in another geometric progression with ratio$~\lambda$ will satisfy the recurrence.

What make the example especially attractive is that the nonzero coefficients are all powers of$~10$, which makes the linear combination easy to compute; this is related to the fact that all (nonzero) digits of $2^7=128$ themselves occur as powers of $2$. The same phenomenon happens for $2^{10}=1024$: take $1000$ times some power of$~2$, add $10~$times the next power of two, and once the one after that; the result is present again $8~$places further in the same progression. (For the digits of $2^{11}=2048$ you basically get the same relation.)


问:
Demystifying the Magic Diagram
Vakil calls the following pullback diagram the magic diagram. I have also seen it being called the magic square. It often shows up in fiber product diagram chases such as those associated with separatedness assertions.
$\require{AMScd}$
\begin{CD}
 X_1\times_Y X_2 @>>> X_1\times_Z X_2\\
    @V V V @VV V\\
    Y @>>> Y \times_Z Y
\end{CD}

On the other hand, I'm a firm believer that mathematics is the art of demystifying magic. So I'm interested in some intuition of this diagram. Maybe there is some geometric intuition or maybe there is a nice toy example for this?

答:
The diagram looks much less mysterious in the special case that $Z$ is the terminal object. Then the fibered products over $Z$ are just ordinary products, i.e., we have the diagram: $\require{AMScd}$
\begin{CD}
 X_1\times_Y X_2 @>>> X_1\times X_2\\
    @V V V @VV V\\
    Y @>>> Y \times Y
\end{CD}
The fact that this is a pullback square can be explained intuitively as follows. Given maps $f_1\colon X_1\to Y$ and $f_2\colon X_2\to Y$, we have a natural map $(f_1,f_2)\colon X_1\times X_2\to Y\times Y$. Pulling back this map along the diagonal $\Delta\colon Y\to Y\times Y$ extracts those fibers which lie over the diagonal, which gives us the fiber product $X_1\times_Y X_2$.

Ok, now let's move the discussion to happen over an arbitrary object $Z$, i.e., in the slice category $\mathcal{C}/Z$. Then $Z$ (with the identity map) is the terminal object, and products in this category are fiber products over $Z$. So in the diagram above, we replace $X_1\times X_2$ and $Y\times Y$ with $X_1\times_Z  X_2$ and $Y\times_Z Y$, respectively. On the other hand, fiber products in $\mathcal{C}/Z$ agree with fiber products in $\mathcal{C}$, since a square is a pullback square in $\mathcal{C}/Z$ if and only if it is a pullback square in $\mathcal{C}$ after forgetting the maps to $Z$. So we don't need to modify $X_1\times_Y X_2$ in the upper left corner. Thus we get a pullback square:
\begin{CD}
 X_1\times_Y X_2 @>>> X_1\times_Z X_2\\
    @V V V @VV V\\
    Y @>>> Y \times_Z Y
\end{CD}
in $\mathcal{C}/Z$, which (as we just noted) is also a pullback square in $\mathcal{C}$ after forgetting the maps to $Z$.

Title: Demystifying the Magic Diagram


答:
As it turns out, my answer, $60480 = 2 \times 6 \times 7!$

I strongly suspect that there is a more elegant approach.

Anyway...

See the Addendum, for an embarassingly elegant Stars and Bars approach, that I just thought of.

First, see my comment, following the OP's query.  I am going to

use the direct approach

assume that $J$ precedes $N$, obtain the count, and then multiply it by $2$.

assume that the number of letters between $J$ and $N$ is $k$, where $k \in \{0,1,\cdots, 7\}$.

let $f(k)$ denote the number of ways that $O$ does not lie between $J$ and $N$, as a function of $k$.

Then the answer will be $2 \times \left[\sum_{k=0}^7 f(k)\right].$

It should be noted that if there was no constraint on where $O$ occurred, the number of 9 character sequences would be

$$\frac{9!}{2! \times 2!}.$$

This is because there are 2 $A$'s and 2 $L$'s in the sequence, and the enumeration of $9!$ double-counts each specific positioning of the 2 $A$'s and (separately) each specific positioning of the 2 $L$'s.

$\underline{f(0)}$

There are $8$ choices for where the $J$ occurs, with the $N$ immediately following it.  For each of these choices, since it is impossible for the $O$ to be between the $J$ and the $N$, there are $\frac{7!}{2! \times 2!}$ ways.

Therefore,

$\displaystyle f(0) = 8 \times \frac{7!}{2! \times 2!}.$

$\underline{f(1)}$

There are $7$ choices for where the $J$ occurs, with the $N$ separated from the $J$ by one slot.  Here, for convenience, I will internally use Inclusion-Exclusion, within this individual case.

For each of these $7$ choices:

There are $\frac{7!}{2! \times 2!}$ ways of forming a word, with the $J$ and $N$ so positioned.

Of these, the only way of violating the constraint is if $O$ is in the single slot in between the $J$ and the $N$.  This can happen in $\frac{6!}{2! \times 2!}$ ways.

Therefore,

$\displaystyle f(1) = 7 \times \frac{7! - 6!}{2! \times 2!}.$

The remainder of the analysis will be very similar to the analysis in computing $f(1)$.

$\underline{f(2)}$

There are $6$ choices for $J$**$N$

For each of these $6$ choices:

There are $\frac{7!}{2! \times 2!}$ ways of forming a word, with the $J$ and $N$ so positioned.

To violate the constraint, there are $2$ slots, that the $O$ can appear in.  For each of these slots, a dis-satisfying word can be formed in
$\frac{6!}{2! \times 2!}$ ways.

Therefore,

$\displaystyle f(2) = 6 \times \frac{7! - (2 \times 6!)}{2! \times 2!}.$

$\underline{f(3)}$

There are $5$ choices for $J$***$N$.

For each of these $5$ choices:

there are $\frac{7!}{2! \times 2!}$ ways of forming a word.

To violate the constraint, there are $3$ slots, that the $O$ can appear in.  For each of these slots, a dis-satisfying word can be formed in $\frac{6!}{2! \times 2!}$ ways.

Therefore,

$\displaystyle f(3) = 5 \times \frac{7! - (3 \times 6!)}{2! \times 2!}.$

The pattern is clear.

$\displaystyle f(4) = 4 \times \frac{7! - (4 \times 6!)}{2! \times 2!}.$

$\displaystyle f(5) = 3 \times \frac{7! - (5 \times 6!)}{2! \times 2!}.$

$\displaystyle f(6) = 2 \times \frac{7! - (6 \times 6!)}{2! \times 2!}.$

$\displaystyle f(7) = 1 \times \frac{7! - (7 \times 6!)}{2! \times 2!}.$

As expected, $f(7) = 0$.

The summation can be facilitated by simplifying each expression for $f(k)$, noting that $\frac{1}{2! \times 2!} = \frac{1}{4},$ and noting that $(7!) = (7 \times 6!)$.

$$f(k) = \frac{1}{4} \times (8 - k) \times (7 - k) \times 6!.$$

Note that

$$\sum_{k=0}^7 (8-k) \times (7-k) = 56 + 42 + 30 + 20 + 12 + 6 + 2 + 0 = 168.$$

Thus,

$$2 \times \left[\sum_{k=0}^7 f(k)\right]$$

$$ =~ 2 \times \frac{6!}{4} \left[\sum_{k=0}^7 (8-k)(7-k)
\right]$$

$$ =~2 \times 180 \times 168 = 60480.$$

If I had opted for the alternative (indirect) approach of employing Inclusion-Exclusion, from the start, my work would have been equally inelegant.  Others might have found a more elegant use of Inclusion-Exclusion here.

Addendum

Use Stars and Bars.

The scaling factor of $\frac{1}{2! \times 2!} = \frac{1}{4}$ will pertain, throughout the Addendum.  I will therefore omit it, until the final computation.

There are $9!$ ways of forming any 9 character string.

Consider the letters JON as bars, occurring in that specific order, placed as

$$ \text{*}J\text{*}O\text{*}N\text{*}$$

To enumerate the number of bad words that can be created, you first have to enumerate the number of solutions to

$$x_1 + x_2 + x_3 + x_4 = 6 ~~ \text{number of solutions equals}~~ \binom{6 + [4-1]}{4-1} = 84.$$

For each such solution, the remaining $6$ letters can be placed in $(6!)$ ways.

Further, by symmetry, you have to multiply this computation by $2$ to account for the
$ \text{*}N\text{*}O\text{*}J\text{*}$ words.

Therefore, the final computation is

$$\frac{1}{2! \times 2!} \times
\left[(9!) ~-~ 2 \times 84 \times 6!\right] = 60480.$$

Addendum-2

Hijacking the analysis at the end of cosmo5's answer, and transplanting it into the Stars and Bars flowerbed:

Consider:

$$ \text{*}O\text{*}J\text{*}N\text{*}$$

There are actually 4 such sequences, re
$OJN, ONJ, JNO, NJO$.

Therefore, for the direct approach, you would have

$$\frac{1}{2! \times 2!} \times 4 \times 84 \times 6! = 60480.$$

Title: Considering all the permutations of the following word, what is the total number of words in which the the following condition is true? ${}$

Body:
Among all the permutations of the letters of the word " JANLOKPAL", the number of words in which "O" never lies between "J" and "N" is?

I first thought of subtracting the cases in which O does occur between J and N, and then subtracting it from total possible permutations, but I can't seem to arrive at the correct answer.

Thank you.


问:
Conservative edgeweights, minimal path tree characterization
This is a task from an old exam I'm doing for learning:

Let $G$ be a Digraph with conservative edgeweights $c: E(G) \rightarrow \mathbb{R}$. Let $T$ be a spanning arborescence with root $s$. Proof that the following is equivalent:

$T$ is a shortest path tree with root $s$ in $G$

For every edge $(x,y) \in E(G)$ it is: $\text{dist}_T(s,y) \leq \text{dist}_T(s,x) + c((x,y))\quad\quad\quad(I)$

My proof is:

$1. \Rightarrow 2.$ is trivial. So let's proof the other implication.

We have to proof: $\forall y\in V(G): \text{dist}_T(s,y) = \text{dist}_G(s,y) \quad\quad\quad(II)$. Now choose such a $y$.

Assume that $\text{dist}_G(s,y)$ is given by a path $P$ in $G$, i.e.: $\sum_{e \in P(E)}c(e) = \text{dist}_G(s,y)$.

Now we use an inductive argument over the paths length $n = |V(P)|$.

For $n=0$ it's trivial that $(II)$ is true since the distances are $0$.

Now assume it's true for $|V(P)| = n$. Let $|V(P)| = n+1$. We can now simply define $P'$ as $(v_1,...,v_n)$ with $P = (v_1,...,v_{n+1})$ with $v_{n+1} = y$, so we just "take out" the last vertex and also the corresponding edge, let it be $e$. Now we can use the induction on $P'$ and $v_n$ instead of $y$.

Thus, we get $\text{dist}_T(s,v_n) = \text{dist}_G(s,v_n)$. From here, we can simply use $(I)$ and get:

$\text{dist}_T(s,y) \leq \text{dist}_T(s,v_n) + c(e) = \text{dist}_G(s,v_n) + c(e) = \text{dist}_G(s,y)$.

That's all we needed.

Is this correct?

It seems a bit too easy to me. And I'm not using that the weights are conservative. But without using this, we'd have (if there's a negative cycle) a path of length negative infinity. I guess I have to use this in the induction, but the induction seems fine to me though.

答: You're (implicitly) using the fact the the edge weight are conservative when you write $\text{dist}_G$. If the weights weren't conservative, this would be $-\infty$ for at least part of the graph. And with non-conservative edge weights, that distance isn't given by any path (nor walk), because (obviously) each path or each walk has a finite length.


问:
$(a,b)$ is polish space with induced topology
From Topology Without tears:

Prove that each discrete space and each of the spaces $[a,b],(a,b),(a,b],[a,b),(-\infty,a),(a,\infty)\ and\ \{a\}\ for\ a,b\in \mathbb{R}$ with its induced topology is polish space.

Polish spaces:A topological space is said to be Polish space if it is separable and completely metrizable.

I know $\mathbb{Q}\cap [a,b] \subseteq [a,b]$ and is dense in $[a,b]$ and countable so $[a,b]$ is separable and the topology induced by $\mathbb{R}$ is same the topology induced by the euclidean metric on $[a,b]$.So it is metrizable.And since closed subset of complete metric space is complete implies $[a,b]$ is also complete.So,our $[a,b]$ with euclidean metric is separable and completely metrizable hence polish space.

But I am having doubt proving for other spaces:
1)$(a,b)$.This space may be metrizable,separable.But complete?? Suppose Take example of $(0,1)$.I have a
sequence $\{\frac{1}{n}\}\subseteq (0,1)$This is Cauchy and converges to zero.But since the $0\notin (0,1)$ the sequence is not convergent in $(0,1)$.So doesn't it imply $(0,1)$ is not complete metric space and hence $(a,b)$.So how $(a,b)$ is polish space??

答:
$(a,b)$ can be given another metric that is complete and induces the same topology.

One such metric can come from a homeomorphism $f: (a,b) \to \Bbb R$, where we introduce the new metric $d(a,a')= |f(a) - f(a')|$ and makes $f$ an isometry from $((a,b), d)$ to $(\Bbb R, d_e)$.

In general, $X$ is Polish iff it can be embedded as a $G_\delta$ (countable intresection of open sets) of a complete separable metric space. So open sets of $\Bbb R$ and the irrationals $\Bbb P$ are also examples of such spaces.


问:
How to prove rigorously that, for any natural number $n$, the open interval $(n, n+1)$ contains no natural number?
A subset $E$ of $\mathbb{R}$, the set of real numbers, is said to be inductive if $1 \in E$, and for every $x \in E$ the number $x+1 \in E$ also.

The set $\mathbb{N}$ of natural numbers is defined to be the intersection of all the inductive sets of real numbers.

Since the sets $\{ x \in \mathbb{R} | x > 0 \}$ and $\{ x \in \mathbb{R} | x \geq 1 \}$ are inductive, therefore we can conclude that all the natural numbers are $\geq 1$.

Now using this machinery how can we (use induction to) prove the following statement?

For any natural number $n$, there exists no natural number in the open interval $(n, n+1)$.

答:
Since it is an open interval, any $x$ in $(n, n+1)$ satisfies $x > n$ and $x < n + 1$.

We can use these two inequalities to show that $x$ cannot be an integer. Assume for contradiction that $x$ is an integer:

$x > n$ is equivalent to $x \ge n + 1$ but this contradicts $x < n + 1$. QED.


问:
Microsoft Paint Spraypaint Tool
The following question is motivated by the spraypaint tool in Microsoft Paint.  In a sense, I'm asking how many clicks on average are necessary to fill each pixel within the range of the tool if we leave the cursor fixed.  The following is a more precise statement of the question:

Consider $n$ dice, each with $k$ sides.  A roll consists of rolling all $n$ dice.  Determine $E(n,k)$, the expected number of rolls needed so that all numbers $1,\ldots,k$ have appeared on some roll.

The translation to spraypaint can be seen by assuming the range of the spraypaint tool is $k$ pixels, and each click paints $n$ random pixels.

I remember looking at this problem years ago, and only partially solving it.  I believe I worked on the case $n=1$:

$$E(1,k)=kH_k=k\sum_{i=1}^k\frac{1}{i}$$

Here $H_k$ is the $k$th harmonic number.  I'm not entirely convinced of this result.  For $n=2$ I could only find a closed form for asymptotic behavior as $k$ got large.  My memory of the solution is a bit fuzzy though.

答: For $n \gt 1$ you can imagine the die rolls in a batch of $n$ are sequential.  Except for the last roll, the situation is the same, so $E(n,k) \approx \frac 1nE(1,k)$.  It will actually be a little higher than this, as you will (in effect) keep rolling up to the next multiple of $n$, when with $n=1$ you would quit.  The heuristic answer would be that on average you would waste half the rolls, so $E(n,k) \approx \frac 1nE(1,k)+\frac 12$.  To do better, I think you would have to assess the probability of completion at each multiple of $n$.</p>

Title: Microsoft Paint Spraypaint Tool


问:
Solving Laplace's equation using finite difference method
Suppose I use a 5-point approximation for the  Laplacian $u_{xx} + u_{yy}$ at a point $( x_i , y_j )$
\begin{align*}
\frac{-2 u_{ij} + u_{i+1,j} + u_{i-1,j}}{h_x^2} +
\frac{-2 u_{ij} + u_{i,j+1} + u_{i,j-1}}{h_y^2}
\end{align*}
to solve $u_{xx} + u_{yy}=f$ with Dirichlet boundary condition, can I use the Fast Poisson solver? If yes (I think the answer should be yes since the coefficient matrix is of tridiagonal symmetric toeplitz block type), what about imposing a nonzero Dirichlet boundary condition instead to solve $u_{xx} + u_{yy}=0$ using finite difference method? And why the solution vector $u$ cannot have a local maximum at any interior node $(x_{i},y_{j})$? Thanks for any hints.

答:
Consider the discrete Dirichlet problem for Laplace equation:
$$
\frac{u_{i-1,j} - 2u_{i,j} + u_{i-1,j}}{h_x^2} +
\frac{u_{i,j-1} - 2u_{i,j} + u_{i,j-1}}{h_y^2} = 0, \quad i = 1,\dots,M-1, \quad j=1,\dots,N-1\\ \tag{*}
u_{0,j} = g_{0,j}, \quad j=0,\dots,N\\
u_{M,j} = g_{M,j}, \quad j=0,\dots,N\\
u_{i,0} = g_{i,0}, \quad i=1,\dots,M-1\\
u_{i,N} = g_{i,N}, \quad i=1,\dots,M-1
$$
Let's reformulate it as equivalent discrete Possion problem by substituting $v_{ij} = u_{ij} - g_{ij}$.
$$
\frac{v_{i-1,j} - 2v_{i,j} + v_{i-1,j}}{h_x^2} +
\frac{v_{i,j-1} - 2v_{i,j} + v_{i,j-1}}{h_y^2} = f_{ij}, \quad i = 1,\dots,M-1, \quad j=1,\dots,N-1\\
f_{ij} \equiv -\frac{g_{i-1,j} - 2g_{i,j} + g_{i-1,j}}{h_x^2} -
\frac{g_{i,j-1} - 2g_{i,j} + g_{i,j-1}}{h_y^2}\\
v_{0,j} = 0, \quad j=0,\dots,N\\
v_{M,j} = 0, \quad j=0,\dots,N\\
v_{i,0} = 0, \quad i=1,\dots,M-1\\
v_{i,N} = 0, \quad i=1,\dots,M-1
$$
For this to work we need to define $g_{i,j}$ at the internal nodes $i=1,\dots,M-1,j=1,\dots,N-1$.

For direct solvers (direct sparse or FFT methods) the choice for internal $g_{i,j}$ may be arbitrary, for example, $g_{i,j} = 0$. This is equivalent to eliminating boundary unknowns $u_{0,j}$, $u_{M,j}$,$u_{i,0}$, $u_{i,N}$, from the system. After the elimintation the right hand side would contain extra terms containing boundary values of $g_{i,j}$.

For iterative solvers the choice of interval values for $g_{i,j}$ may affect the number of required iterations and it may be wise to take $g_{i,j}$ as a smooth function, for example:
$$
g_{i,j} = \tilde c_1 g_{0,j} + \tilde c_2 g_{M,j} + \tilde c_3 g_{i,0} + \tilde c_4 g_{i,N}\\
c_1 = \frac{N h_y}{h_x i}, \quad c_2 = \frac{N h_y}{h_x (M-i)}\\
c_3 = \frac{M h_x}{h_y j}, \quad c_4 = \frac{M h_x}{h_y (N-j)}\\
\tilde c_i = \frac{c_i}{c_1 + c_2 + c_3 + c_4}.
$$
This formula is a very rough approximation of Cauchy integral but it gives a pretty smooth $g_{ij}$ if boundary data is also smooth.

The property that $u_{i,j}$ attains its extrema at the boundary is known as discrete maximum principle and does not depend on the way you solve (*), as it is the property of the solution itself.
The proof is very simple. Rewrite the discrete Laplace equation as
$$
u_{i,j} = \frac{h_y^2 u_{i-1,j} + h_y^2 u_{i+1,j} + h_x^2 u_{i,j+1} + h_x^2 u_{i,j-1}}{2(h_x^2 + h_y^2)}.
$$
Assume that $u_{i,j}$ is strictly greater (strictly less) than each of $u_{i-1,j}$, $u_{i+1,j}$, $u_{i,j+1}$, $u_{i,j-1}$. Then the equality cannot hold. So the discrete solution cannot have local maxima or minima inside the domain (that is in points that are surrounded by four neighbours). The only possibility left it that extrema are located at some boundaries.


问:
$X$ is Hausdorff, $G$ is a finite group prove that $X/G$ is Hausdorff.
I want to start this off by saying I've seen the other questions that are identical to this one, but I didn't really understand the suggestions given in those.

So the question states: Let $G$ act continuously on $X$ i.e the map $G\times X\to X$ sending $(g,x)\mapsto gx$ is continuous. Then define an equivalence relation $x\sim y$ if and only if there exists a $g\in G$ such that $gx=y$. Then give $X/\sim :=X/G$ the quotient topology. Show that if $X$ is Hausdorff and $G$ is finite then $X/G$ is Hausdorff.

I have my attempt bellow, however I am stuck on the last part where I want to show $U$ and $V$ are disjoint. I'm not really sure how to come to a conclusion.

Choose two orbits, $Gx\neq Gy$, then we have
$x\neq y$. Define $U_g$ and $V_g$ to be open neighbourhoods of $gx$
and $gy$ respectively, such that $U_g\cap V_h=\emptyset$ for all
$h,g\in G$, which we can do as $X$ is Hausdorff and $gx\neq hy$ as
they are in different orbits. Now consider all the sets $p(U_g)$ and
$p(V_g)$, I claim that $U=~\bigcap_{g\in G}p(U_g)$ and
$V=\bigcap_{g\in G}p(V_g)$ are disjoint neighbourhoods of
$\overline{x}$ and $\overline{y}$ (The equivalence classes of $x$ and $y$ respectively).

Since each $p(U_g)$ and $p(V_g)$ are open by Lemma 2 (Lemma 2 just
says $p$ is an open map), furthermore $U$ and $V$ are finite
intersections of these open sets, as $G$ is finite, we have that they
are open in $X/G$. $U$ certainly contains $\overline{x}$ as each $U_g$
contains $gx$ for some $g\in G$ which is equivalent to $x$, and with
an identical argument $V$ contains $\overline{y}$. Suppose for a
contradiction that $U\cap V$ is non-empty, take an element
$z\in~U\cap~V$ then $z\in p(U_g)$ and $z\in p(V_g)$ for each $g\in G$.

Thanks in advance.

答:
Let $Gx\neq Gy$. The argument is very similar to what you have done so far!

Take $g\in G$. Since $g\cdot x \neq y$, there are open sets $U_{g},V_{g}$ such that $g\cdot x\in U_{g}$, $y\in V_{g}$ and $U_{g}\cap V_{g}=\emptyset$. Now choose the following open subsets:

$$
U=\bigcap_{g\in G}g^{-1}U_{g}, V=\bigcap_{g\in G}V_{g}.
$$

Since $G$ is finite and left translations are homeomorphisms, $U$ and $V$ are open neighbourhoods of $x$ and $y$. Notice that, for every $g$, $gU\cap V=\emptyset$.

Now, choose $p(U)$ and $p(V)$. We only need to prove that $p(U)\cap p(V)=\emptyset$.

Let $p(z)\in p(U)\cap p(V)$, so that $p(z)=p(a)=p(b)$ for some $a\in U$, $b\in V$. Therefore, for some $h\in G$, $b=ha\in V\cap hU$, which is a contradiction.

Hope this helps!


答:
You can use synthetic division, or simply use inspection, to find that
$$
(x^3-2x^2+4x-8)=(x-2)(x^2+4) \, .
$$
Notice how the parabola with the equation $y=x^2+4$ does not touch the $x$-axis:

Hence, the equation $x^2+4=0$ has no real solutions.

Title: $p(x) = x^3 - 2x^2 + 4x - 8$ and $x = 2$ is a $0$, use synthetic division to obtain factorization of $p(x)$

Body:
Here is what I have, and I would like to get it verified.

$$p(2) = 0 = x^3 - 2x^2 + 4x - 8\\
  \frac{(x - 2)} {(x^3 - 2x^2 + 4x - 8)}\\
    x^2 + 4 = x^3 - 2x^2 + 4x - 8$$

I think the synthetic division answer is $(x^2 + 4)$

Is this correct?

How do I take this to identify other solutions?

I need to factor the polynomial using synthetic division then I need to identify other solutions of $p(x) = 0$.


问:
Sum of a singular PSD matrix and a full rank indefinite matrix
Let's $A \in R^{n x n}$ singular PSD matrix, $B \in R^{n x n}$ full rank indefinite matrix and $C \in R^{n x n}$ full rank PSD matrix, I want to show that in order to fulfill the next equation:
$C=A+B$, $B$ must be PSD matrix.
I have an intuition behind that, but I can't prove it is correct (which I'm not sure 100%  it's correct).

My intuition is based on:
Let us assume that B hold a negative eigenvalue, and w.l.o.g let's
arrange the rows of the matrices by the size of the eigenvalues: $\lambda_1\le\lambda_2\le..\lambda_{n-1}\le\lambda_{n}$ hence we will have rows in C that will be negetive dominated or with a negative value on the diagonal, and C will not be a PSD matrix anymore.

Any Ideas? thoughts? if I'm in the right direction ? how to make it formal?

Thanks for your help!!!

答:
No. this is false. Let $A$ be the identity. Then take $B$ to be $\epsilon$ times an arbitrary diagonal matrix with diagonal entries $\pm 1.$ Then $A+B$ will be $PSD.$

EDIT You had asked for a singular $A,$ but the same example works - let $A$ be identity with one entry on the diagonal zeroed, and make sure that $B$ has a $1$ in that position.


问:
Graph a function $|\frac{\sin x}{x}|$
How can one graph a function $\left|\frac{\sin x}{x}\right|$,  $x\neq0$ without any program by hand? Without using derivatives.

答:
hint

The function is even.

You begin by the graph $ C $ of $ \; x\mapsto \frac{\sin(x)}{x} $ at $(0,+\infty) $ and then its absolute value.

You know that

$$(\forall x>0)\;\; -\frac 1x\le \frac{\sin(x)}{x}\le \frac 1x$$

So, the graph $ C $ oscillates between the two hyperbolas whose equations are $ y=\pm \frac 1x$.

they intersect at $ x $  satisfying

$$\sin(x)=\pm 1$$
or
$$x=\frac{\pi}{2}+k\pi$$

Title: Graph a function $|\frac{\sin x}{x}|$