问:
Finding the number of non-neg integer solutions?
How would I find the number of non negative integer solutions to this problem?
$$x_1 + x_2 + x_3 + x_4 = 12$$ if $0 \leq x_1 \leq 2$.</p>

答:
The number of solutions $\def\N{\mathbf N}(x_1,x_2,x_3,x_4)\in\N^4$ to $x_1+x_2+x_3+x_4=k$ is the coefficient of $X^k$ in the power series expansion of $\frac1{(1-X)^4}$, which coefficient is $(-1)^k\binom{-4}k=\binom{k+3}k$, which is also $\binom{k+3}3$ if $k\geq0$. If you want to limit one variable to being${}<m$ (with $m>0$), just observe that the number of solutions of the problem which have $x_1\geq m$ equals the number of (unrestricted) solutions to $x_1+x_2+x_3+x_4=k-m$. You need to subtract these, so you get as number of solutions for the restricted problem
$$
  \binom{k+3}k - \binom{k-m+3}{k-m}
 \qquad\text{which is also $\dbinom{k+3}3-\dbinom{k-m+3}3$ if $m\leq k$.}
$$
Now plug in $k=12$ and $m=3$ to obtain $\binom{15}3-\binom{12}3=235$.</p>


问:
If $A$ and $B$ are commuting matricies, does the basis of $B$ block diagonalise $A$?
Let $A,B\in \mathbb{C}^{n\times n}$ and $[A,B]=0$. Let the set of eigenvectors of $B$ with eigenvalue $\beta^{(i)}$ be denoted by $\{\underline{b}_j^{(i)}\}$. We can construct the $i$-th projector via the outer product
$$
P_i \doteq \sum_j^{g^{(i)}} \underline{b}_j^{(i)}\left(\underline{b}_j^{(i)}\right)^{\dagger},
$$
where $g^{(i)}$ denotes the degeneracy of the $i$-th eigenvalue.

Is it true that, with the projector defined as above
$$
\Lambda_i \doteq P_i^{-1} A P_i = \cdots \oplus A_i \oplus \cdots,
$$
i.e. $A_i$ is the $i$-th block of the block diagonal representation of $A$ such that
$$
\Lambda \doteq P^{-1} A P \doteq \sum_i P_i^{-1} A P_i = A_1 \oplus \cdots \oplus A_i \oplus \cdots \oplus A_n =
\begin{pmatrix}
A_1 & \cdots & 0 & \cdots & 0\\
\vdots & \ddots & \vdots & \ddots & \vdots \\
0 & \cdots & A_i & \cdots & 0\\
\vdots &\ddots & \vdots  & \ddots & \vdots\\
0 & \cdots & 0 & \cdots & A_n
\end{pmatrix},
$$
where $A_i \in \mathbb{C}^{g^{(i)}\times g^{(i)}}$?

答: Yes. If $B$ has the above eigenbasis, in that basis, $[A,B]=0$ gives $\lambda_j A_{ij} - \lambda_i A_{ij}=0,$ where $A_{ij}$'s are the block matrices, so the block is identically $0$ for $A_{ij}$ with different $i$ and $j$.

ContentLicense: CC BY-SA 4.0


问:
How can I estimate the number of ads a snack bar owner needs to use along a road to achieve 99/100 of driver's attention
I snack bar owner performed a research and noticed that with one ad on the road he achieved 1/2 of driver's attention (the ad was seen). He was unsatisfied he wanted a value above 99/100  how many more ads he must use at minimum?

```<code>a) 99
b) 51
c) 50
d) 6
e) 1
</code>```

I don't even now how to solve it. I promese I tried but probability has no middleground or you know how to approach it or you just stand still. How can I approach a problem like this?

I corrected the question it is not how many but how many more ads he must use.

答:
It might be assumed that each ad will get $\frac{1}{2}$ of drivers' attention as well, which means missing the other $0.5$. Also, assume that the events of any driver see one ad and another ad are irrelevant. Then the ratio of drivers haven't seen an ad after $n$ ads is $0.5^n$. The goal is to find the minimal $n$ such that $$0.5^n<0.01$$
$$n>\frac{lg(0.01)}{lg(0.5)}\approx6.64$$
Or check that $\left(\frac{1}{2}\right)^6 = \frac{1}{64}, \left(\frac{1}{2}\right)^7 = \frac{1}{128}, 7$ is enough.

The minimal number of ads should be 7.


问:
Proof involving subgraph of a tree
Let $T$ be a tree with even number of vertices. Prove that $T$ has "exactly" one spanning subgraph in which all vertices have odd degree.

I could only think of this result:
In a tree with even number of vertices, the number of vertices with an odd number of children is odd.

But I don't know how to proceed. I'm self-studying discrete math and I found this question in a discrete math book.

EDIT:

The post linked in the comments proves existence of such tree. This question also requires the proof of uniqueness.

答:
Here's a quick proof of uniqueness; the linked post already covers existence in multiple ways.

Let $G_1, G_2$ be two odd subgraphs of $T$. Then their symmetric difference $G_1 \mathbin{\Delta} G_2$ (which has every edge found in exactly one of $G_1$ or $G_2$) is an even subgraph of $T$: all vertices have even degree, possibly $0$.

Suppose that the even subgraph has any vertices of positive degree. Then start at one such vertex and take a walk, without repeating an edge, until you've returned to a vertex you've seen before. (You'll never get stuck, because no vertex of $G_1 \mathbin{\Delta} G_2$ has degree $1$. If you enter a vertex, you can leave it.)

But now we've found a cycle in $T$! This is impossible: trees don't have cycles. So $G_1 \mathbin{\Delta} G_2$ cannot have any vertices of positive degree - in other words, it's the empty graph. Therefore $G_1 = G_2$.

Title: Proof involving subgraph of a tree


问:
Given that matrix $A$ is diagonalizable and has eigenvalues of $0$ or $1$, show that $A^2 = A$
Suppose that $A$ is a diagonalizable $n \times n$ matrix such that the characteristic polynomial of $A$ is $p(λ)=λ^k(1−λ)^{n−k}$, where $k$ is a positive integer such that $0≤k≤n$.

I want to prove $A^2=A$. I know that for this, I'll need to prove that for any eigenvector $v$, $A^2 v=Av$. I tried setting up my proof like this:

Since $A$ is diagonalizable, there exists an invertible matrix $P$ such that $A=P^{-1}DP$, where $D$ has all the eigenvalues of $A$ on its diagonal. But since every eigenvalue is either $0$ or $1$, $D= \lambda I$. From there, we get:
$$A = P^{-1}DP = P^{-1}(\lambda I_n)P = (\lambda I_n)(P^{-1}P) = \lambda I_n$$
$$ \implies A \times A = (\lambda I_n) \times (\lambda I_n) = \lambda ^2$$

All I've succeeded in doing here is showing $A^2 = \lambda ^2$. How can I show also that $A^2 = A$? Any guidance is greatly appreciated!

答:
Since $ A $ is diagonalizable, you can rewrite it as $ A = PDP^{-1} $, so, by associativity of matrix multiplication
$$ A^2 = A \cdot A = (PDP^{-1}) \cdot (PDP^{-1}) = PD \cdot (P^{-1}P) \cdot DP^{-1} = PD^2 P^{-1} $$

Now it’s left to calculate what $D^2$ is.

I won’t go in the details, it depends on the level of formality required, but it’s not difficult to get convinced that the product of two diagonal matrices is still a diagonal matrix. Moreover, the following property holds: given two diagonal matrices $S$ and $D$, one can use the compact notation for diagonal matrices and write only the elements on the diagonal $$S = \mathrm{diag}(a_1, \dots, a_n), \quad D= \mathrm{diag}(b_1, \dots, b_n)$$ then, the product of the matrices is the product of the elements. $$ D\cdot S = \mathrm{diag}(a_1\cdot b_1, \dots, a_n \cdot b_n)$$

Specifically, in your case, A has eigenvalues of either 0 or 1, and your $ D $ is the spectral matrix, the diagonal matrix that has the eigenvalues on the main diagonal. So when you write D with the compact notation, the entries in that “$ \mathrm{diag} $ vector” will only be either 0 and 1, and when you calculate $ D^2 $, since the entries will be the same, in every entry you’ll either do $0\cdot0=0$ or $1\cdot 1 =1 $. So, in the end, $D^2 = D $.

We left ourselves saying that $ A^2 = PD^2 P^{-1} $, so  $ A^2 = PDP^{-1} $, but $ A = PDP^{-1} $by hypothesis. In the end $ A^2 = A $.


问:
Probability of seeing each face except one in $n$ throws of a die with 4 faces
Imagine that we throw a die with 4 faces $n$ times. The faces have probabilities of $(0.1, 0.4, 0.2, 0.3)$ of showing up.

What is the probability that we see all faces except a specific, pre-determined face $i$ (say $i = 1$) in $n$ throws?

PS: This isn't homework but a puzzle that came across that I couldn't solve analytically. Thanks.

答:
Any formula for this is going to be messy, but I can explain to you how to do the problem.  I'm assuming that the forbidden face is specified in advance.  If you mean to say all but one (unspecified) face, do the problem for each specific face and add up the answers.

The probability that face $i$ doesn't show up on a single toss is $(1-p_i)$ and the probability that it doesn't show up in $n$ tosses is $(1-p_i)^n.$  However, this includes cases where some other face doesn't show up either, so we must subtract $$\sum_{j\neq i}(1-p_i-p_j)^n$$  where it is understood that $1\leq j\leq k$.  (Sorry, I don't know how to get two lines in the subscript in MathJax.)  This isn't the end of the story.  Suppose in some sequence of tosses that doesn't include $i$ neither $j$ nor $m$ has shown up.  We've subtracted the probability of this sequence twice, and we only want to subtract once so we need to add it back.  That is, we must add $$\sum(1-p_i-p_j-p_m)^n$$ where the sum is over all $2$-element sets $\{j,m\}$, with $1\leq j,m\leq k$ with $i\neq j,\ i\neq m$.

Now of course, we have to consider sequences where $3$ faces other than $i$ don't show up.  We continue as above, alternately, adding and subtracting, by applying the principle of inclusion and exclusion.


问:
How to prove $\cos(x) \ge 1-\frac{2}{\pi}(x+\sin(x))$?
I need help in solving the following inequality---
$$\cos(x) \ge 1-\frac{2}{\pi}(x+\sin(x))$$

The inequality is used in proving Margolus Levitin theorem as can be seen here.

I tried using the Cauchy Mean Value theorem, in similar ways as mentioned in Proving $\frac2\pi x \le \sin x \le x$ for $x\in [0,\frac {\pi} 2]$, and rearranging the inequality to a more manageable form, but couldn't proceed further.

Essentially the same question has been asked on Physics SE here
but the currently accepted answer is a bit heuristical because it assumes a specific form of expression for $\cos(x)$ with adjustable parameters and then finds the parameters by several arguments.

I am looking for a more formal and direct proof probably using some well-known inequalities (like Cauchy-Schwarz inequality) or theorems (like Mean Value Theorem).

EDIT: The domain should be $x>0$ because clearly the inequality doesn't hold for $x<0$ as pointed out in a comment by Greg Martin.
This is physically justifiable while using in the Margolus Levitin theorem because there $x\propto t$ and time $t>0$.

答:
Alternative proof

$f(x)=\cos(x)-1+\frac{2}{\pi}(x+\sin(x))$. Note that $f(0)=0, f(x) \to \infty$ when $x \to \infty$. We only need to show $f(x) \ge 0$ at every stationary point.

$$f'(x)=-\sin(x)+\frac{2}{\pi}(1+\cos(x))=-2\sin\frac x2 \cos\frac x2+\frac{2}{\pi}\cdot 2\cos^2 \frac x2\\=2\cos\frac x2\left(-\sin \frac x2 + \frac{2}{\pi}\cos \frac x2\right)$$

$$f(x)=-2\sin^2 \frac x2 + \frac{2}{\pi} x + \frac{2}{\pi} 2 \sin\frac x2 \cos \frac x2=\frac{2}{\pi}x + 2\sin\frac x2 \left(-\sin \frac x2+\frac{2}{\pi} \cos \frac x2\right)\\
$$
Now if $x_0$ is a stationary point, i.e., $f'(x_0)=0$, we have two cases:

$\cos \frac{x_0}{2}=0$. Then $\sin \frac{x_0}{2} = 1, x_0 \ge \pi \implies f(x_0)=-2+\frac{2}{\pi}x_0 \ge 0.$

$-\sin \frac{x_0}{2}+\frac{2}{\pi} \cos \frac{x_0}{2}=0$. Then $f(x_0) = \frac{2}{\pi}x_0 \ge 0.\blacksquare$


问:
If $\sum a_n $ is a positive series that diverges, does $\sum \frac{a_n}{1+a_n}$ diverge?
Let $ \{a_n\}_{n=1}^\infty $ be a sequence such that $\displaystyle \sum a_n $ that is divergent to $+\infty$.

What can be said about the convergence of $\displaystyle \sum \frac{a_n}{1 + a_n} $?

Any hints, thoughts or leads would be greatly appreciated.

Thanks!

答:
Because $ \displaystyle \lim_{n \to \infty} a_n = \infty $, it is true that $ \displaystyle\lim_{n \to \infty} \frac{a_n}{1 + a_n} = 1 $. Hence, the terms of the series do not tend to zero so it certainly does not converge. </p>

<p>In response to the edit, here's a brief sketch:
Assume, for the sake of contradiction, that $ \sum \frac{a_n}{1 + a_n} $ converges. Hence, its general term $ 1 - \frac{1}{1 + a_n} $ goes to $ 0 $. This means that $ a_n $ goes to $ 0 $. Hence, $ \frac{1}{1 + a_n} = 1 - a_n + O\left(a_n^2\right) $. Hence, $ \sum a_n + O\left(a_n^2\right) $ converges so $ \sum a_n $ converges, a contradiction. </p>

ContentLicense: CC BY-SA 3.0

Body:
Because $ \displaystyle \lim_{n \to \infty} a_n = \infty $, it is true that $ \displaystyle\lim_{n \to \infty} \frac{a_n}{1 + a_n} = 1 $. Hence, the terms of the series do not tend to zero so it certainly does not converge.

In response to the edit, here's a brief sketch:
Assume, for the sake of contradiction, that $ \sum \frac{a_n}{1 + a_n} $ converges. Hence, its general term $ 1 - \frac{1}{1 + a_n} $ goes to $ 0 $. This means that $ a_n $ goes to $ 0 $. Hence, $ \frac{1}{1 + a_n} = 1 - a_n + O\left(a_n^2\right) $. Hence, $ \sum a_n + O\left(a_n^2\right) $ converges so $ \sum a_n $ converges, a contradiction.


问:
Borel sigma algebra and measurability of continuous functions
Let ($X, \tau_X$) and ($Y, \tau_Y$) be topological spaces. Let $\mathcal{B}(X)=\sigma(\tau_X)$ and $\mathcal{B}(Y)=\sigma(\tau_Y)$ be the corresponding Borel sigma algebras. Do we have that if $\mathcal{B} \subseteq \mathcal{P}(X)$ is a sigma algebra such that every continuous function $f:X\rightarrow Y$ is $\mathcal{B}/\mathcal{B}(Y)$ measurable, then $\mathcal{B}(X) \subseteq \mathcal{B}$?

If not, does it hold for the reals? i.e. when $X=\mathbb{R}^m,Y=\mathbb{R}^n$ for some $m,n$.

My attempt:

Let $f^{-1}(\tau_Y) \equiv \{f^{-1}(E)|E\in\tau_Y\}$. Then $f$ continuous iff $f^{-1}(\tau_Y) \subseteq \tau_X$, and $f$ Borel measurable iff $f^{-1}(\sigma(\tau_Y))=\sigma(f^{-1}(\tau_Y)) \subseteq \sigma(\tau_X)$.  Assume that $f$ is $\mathcal{B}/\mathcal{B}(Y)$ measurable, then $f^{-1}(\sigma(\tau_Y))=\sigma(f^{-1}(\tau_Y)) \subseteq \mathcal{B}$. We now want to show that $\sigma(\tau_X) \subseteq \mathcal{B}$, and I don't know how to proceed.

For the real case, I found the following answer, but it doesn't seem to make sense for me, could someone please elaborate further?

Prove that Borel sigma field on R(d) is the smallest sigma-field that makes all continuous functions f:R(d)->R measurable.

答:
Part 1: In the general case, you cannot prove that $\mathcal{B}(X) \subseteq \mathcal{B}$.

Here is a counter-example.

Let $X=\{a,b\}$, $\tau_X$ be the discrete topology (that is  $\tau_X=\mathcal{P}(X)$) and then $\mathcal{B}(X)=\mathcal{P}(X)$.

Let $Y=\{c\}$, $\tau_Y =\{\emptyset, Y\}$  and so $\mathcal{B}(Y)= \{\emptyset, Y\}$.

All functions $f:X\rightarrow Y$ are continuous and constant functions.

Let $\mathcal{B}=\{\emptyset, X\}$ be a $\sigma$-algebra in $X$.

All (continuous) functions $f:X\rightarrow Y$ are $\mathcal{B}/\mathcal{B}(Y)$ measurable, but $\mathcal{B}\subsetneq \mathcal{B}(X)$.

Part 2 For the case of $X=\mathbb{R}^m,Y=\mathbb{R}^n$, let us prove the following more general result.

Let ($X, d$) be a metric space,  $\tau_X$ the topology induced by the metric and  let $\mathcal{B}(X)=\sigma(\tau_X)$ the Borel sigma algebra of $X$. Let us  consider $\mathbb{R}^n$ with its usual topology and its usual Borel sigma algebra.  If $\mathcal{B} \subseteq \mathcal{P}(X)$ is a sigma algebra such that every continuous function $f:X\rightarrow \mathbb{R}^n$ is $\mathcal{B}/\mathcal{B}(\mathbb{R}^n)$ measurable, then $\mathcal{B}(X) \subseteq \mathcal{B}$.

Proof: Let us prove that for any $F$ closed set of $X$, $F \in \mathcal{B}$. It will follow immediately that $\mathcal{B}(X) \subseteq \mathcal{B}$.

Given any $F$ closed set of $X$, define the function $f:X\rightarrow \mathbb{R}^n$ such that for all $x \in X$, $f(x) = (d(x,F),0,\dots,0) \in \mathbb{R}^n$.

It is easy to see that $f$ is continuous, so $f$ is $\mathcal{B}/\mathcal{B}(\mathbb{R}^n)$ measurable. Since $\{(0,\dots,0)\}$ is a Borel set in $\mathbb{R}^n$, we have that
$$F=f^{-1}(\{(0,\dots,0)\}) \in \mathcal{B}$$
So we have proved that for any $F$ closed set of $X$, $F \in \mathcal{B}$, and so $\mathcal{B}(X) \subseteq \mathcal{B}$.


问:
Prove that, if $a, b, c \in \mathbb{Z}$ then $8a^{3}+8b^{3}+8c^{3}-12a^{2}+6a-24abc+12bc \neq 1$
Given that $a, b, c \in \mathbb{Z}$, then $$8a^{3}+8b^{3}+8c^{3}-12a^{2}+6a-24abc+12bc \neq 1$$
I tried to factor this somehow, but I don't think it's factorable. After writing it as $8a^{3}+8b^{3}+8c^{3}=1+(2a-1)(12bc+6a)$, I noticed that $a=\frac{1}{2}$, $b=-c$ verifies this equation. If we can prove that these are the only solutions, we are done.

After this observation, I tried dividing the polynomial in $a$ by $(2a-1)$, which yields $$(2a-1)(4a^2-4a-12bc+1)=8b^3+8c^3$$
I thought I was onto something when i subtracted $(2a-1)(4a^2-4a-12bc+1)=8b^3+8c^3$ from $8a^{3}+8b^{3}+8c^{3}=1+(2a-1)(12bc+6a)$, but that just gives me $(2a-1)(4a^2+2a+1)=8a^3-1$, which is just a true statement.

I would appreciate if asomeone could give me some insight in how to solve this.

答:
Hint:

All of the coefficients in $$8a^{3}+8b^{3}+8c^{3}-12a^{2}+6a-24abc+12bc$$ are even numbers.


问:
Why is this process a semimartingale? How is Itô's formula applied here?
I am reading chapter 24 of Kallenberg's modern probability. In the proof of theorem 24.1

Cauchy Problem: $CP(f,v):\dot{u}=Au-vu$ and $u(0,x)=f(x)$.

Theorem 24.1. (Cauchy problem, Feynman, Kac) Let $A$ with domain $D$ be the generator of a Feller diffusion $X$ in $\mathbb{R}^d$, and fix $u\in C_b(\mathbb{R}^d)$ and $v\in C_b^+(\mathbb{R}^d)$. Then any bounded solution $u:[0;\infty)\times\mathbb{R}^d\rightarrow\mathbb{R}$ to $CP(f,v)$, that is $C^1$ in the time-variable and $C^2$ in the space variable, is given by
$$
u(t,x)=\mathbb{E}_x[e^{-V_t}f(X_t)],
$$
where $V_t=\int_0^t v(X_s)ds$.
Conversely the function $u$ above solves $CP(f,v)$ if $f\in D$.

they define $M_s=e^{-V_s}u(t-s,X_s)$ for fixed $t$ and $s\in[0;t]$ and show that it is a local martingale. However they use Îto's formula, i.e. they try to express
$du(t-s,X_s)$ in terms of a second order expansion. But I don't understand why we are allowed to use Îto's formula, since we didn't show that $X_s$ is a semimartingale. So why is $X_s$ is a semimartingale?

They mention Dynkin's formula, Lemma 19.21,

Lemma 19.21 (Dynkin's formula). The processes $M^h_t=h(X_t)-h(X_0)-\int_0^t Ah(X_s)ds$ for $h\in D$ are martingales.

so do they say that by choosing a suitable $h$ we can see that $X_t$ is a semimartingale? We know that $C_c^\infty(\mathbb{R}^d)\subseteq D$ by definition of Feller diffusion. So we can try to plug in some $h$ that look close to $X_t$ like the identity and elsewhere are $0$. Since $A$ is local, $Ah$ will vanish there, so $M^h_t$ will look like $X_t$.

Edit: I also have difficulties in the proof of the above theorem to understand how Itôs formula is applied. They write
$$
dM_s = e^{-V_s}(du(t-s,X_s)-u(t-s,X_s)v(X_s)ds)
$$
by integration by parts. But then they mention Dynkin's formula and Itô's formula to write further
$$
dM_s = e^{-V_s}(Au(t-s,X_s)-\dot{u}(t-s,X_s)-u(t-s,X_s)v(X_s))ds + L_s,
$$
where $L_s$ is some local martingale. I don't understand at all, how $A$ comes into play when using Itô's formula on $u(t-s,X_s)$. When applying Itô's formula, we get in Einstein summation
$$
du(t-s,X_s)=\partial_ju(t-s,X_s)dX_s^j + \partial_i\partial_ju(t-s,X_s)d[X^i,X^j]_s.
$$
The local martingale part from the $dX_s^j$-integration can be absorbed into $L_s$, but how to relate all the other terms to $A$? What is $[X^i,X^j]$ in this case? I know that there is a formula relating a second order expansion with the generator of a Feller diffusion, Theorem 19.24 in Kallenberg, but nothing is said about uniqueness.

答: You're on the right track. For each $x\in\Bbb R^d$ and $r>0$ let $f$ be a smooth continuous function of compact support such that $f(y) = y$ for all $y\in B(x,r)$. Then $f$ is in $D$ and so $f(X)$ is a semimartingale under $\Bbb P_x$. Therefore $X$ is a semimartingale up until the exit time $\inf\{t: X_t\notin B(x,r)\}$.  As $r$ can be made as large as you please, this shows that $X$ is a local semimartingale, hence a semimartingale under $\Bbb P_x$.


问:
Closed form for $\int_0^\infty\left(\int_0^1\frac1{\sqrt{1-y^2}\sqrt{1+x^2\,y^2}}\mathrm dy\right)^3\mathrm dx.$
I need to find a closed form for these nested definite integrals:
$$I=\int_0^\infty\left(\int_0^1\frac1{\sqrt{1-y^2}\sqrt{1+x^2\,y^2}}\mathrm dy\right)^3\mathrm dx.$$
The inner integral can be represented using the hypergeometric function $_2F_1$ or the complete elliptic integral of the 1st kind $K$ with an imaginary argument:
$$\int_0^1\frac1{\sqrt{1-y^2}\sqrt{1+x^2\,y^2}}\mathrm dy=\frac\pi2   {_2F_1}\left(\frac12,\frac12;1;-x^2\right)=K(x\,\sqrt{-1}).$$
My conjecture is the integral $I$ has a closed-form representation:$$I\stackrel{?}{=}\frac{3\,\Gamma (\frac14)^8}{1280\,\pi^2}=7.09022700484626946098980237...,$$
but I was neither able to find a proof of it, nor disprove the equality using numerical integration. Could you please help me with resolving this question?</p>

答: The first thing I would try is to replace $\int_{0}^{\infty}$ by $\frac12\int_{-\infty}^{\infty}$, then to consider this as a complex integral and pull the integration contour to $i\infty$. Since the only singularity of the hypergeometric function $_2F_1(\ldots,z)$ on the main sheet is the branch point $z=1$, in the end we would have to integrate the jump of $K^3(ix)$ on the cut $B=[i,i\infty)$. The jump of a function on the cut is often simpler than the function itself (like $2\pi$ vs $\ln z$), so I would hope for a simplification of the integral (as the branch point $x=i$ here seems to be logarithmic).</p>


问:
Points $p_o, p_l, ... , p_k$ are affinely independent if and only if the rank of the matrix of their barycentric coordinates equals k + 1.
Points $p_o, p_l, ... , p_k$ are affinely independent if and only if
the rank of the matrix of their barycentric coordinates equals k + 1.

So I came across this theorem while reading about affine space and the proof for the same is given by the author as follow:

I am having confusion with a matrix where it says the rank does not change after adding all the column and then obtains a matrice where subtraction is done. Could someone please explain what the author has done?

Thanks in advance.

答:
The sum of all columns is the all ones vector. Replacing the first column with the sum of all columns yields
$$\begin{pmatrix}1 & x_{01} & \cdots & x_{0n} \\ 1 & x_{11} & \cdots & x_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & x_{k1} & \cdots & x_{kn}\end{pmatrix}.$$

Subtracting the first row from each of the other rows leads to the given matrix. (This also does not change the rank.)


问:
Rank of endomorphism
Let $f\colon V \to V$ be an Endomorphism of a finite-dimensional K-Vector space V.

Let $r_k$ be $rank(f^k)$, now I want to show that
$$1.) r_k ≥ r_{k+1}$$
$$2.) r_k - r_{k+1} ≥ r_{k+1}-r_{k+2} $$
3.) If 1.) and 2.) are true and if there is a $k≥ 0$ so that $r_k = r_{k+1}$, then $r_k =r_{k+l}$ for every $l≥0$

Now I already showed 1.) and with 1.) Ajd 2.) I can show 3.) by induction. The problem is 2.). I do not know how to proof this.

Thx for any help.

答:
Let $V_k$ be the image of $f^k$. This gives a decreasing chain of subspaces
$$ V \overset{f}{\to} V_1 \overset{f}{\to} V_2 \overset{f}{\to} \cdots \overset{f}{\to} V_k \overset{f}{\to} V_{k+1} \overset{f}{\to} V_{k+2} \overset{f}{\to} \cdots $$

By definition, $r_k = \dim V_k$, and the restriction $f \colon V_k \to V_{k+1}$ is surjective. So by the Rank-Nullity Theorem, the difference $r_k - r_{k+1}$ is the dimension of the kernel of $f$ (restricted to $V_k$). And so also $r_{k+1}-r_{k+2}$ is the dimension of the kernel of $f$, restricted to $V_{k+1}$.

But since $V_{k+1} \subseteq V_k$, the kernel of $f$ on $V_{k+1}$ is a subset of the kernel of $f$ on $V_k$. This gives the inequality.

Title: Rank of endomorphism


问:
Determinant of exterior power and symmetric power
I am self studying the tensor algebra, symmetric algebra and exterior algebra.

The exterior algebra brings a new definition of determinant to me.

I am curious what if  $f:V \to V$ is a linear endomorphism on $n$ dimensional vector space $V$also with two linear maps
$$\Lambda^i(f):\Lambda^i(V) \to \Lambda^i(v)\text{ and } S^i(f):S^i(V) \to S^i(V), i \geq 1$$

then what is the determinant of these maps?

The determinant of $f$ is just the map $\Lambda^n(f)$. So the determinant of $\Lambda^i(f)$ is just the map $\Lambda^{\binom{n}{i}}(\Lambda^i(f))$.

Is there any way to find $\Lambda^{\binom{n}{i}}(\Lambda^i(f))$ without involving matrices? I am looking for a general way to solve it. However, I am frustrated in the first step expanding

$$\Lambda^{\binom{n}{i}}(\Lambda^i(f))(w_1 \wedge \cdots\wedge w_{\binom{n}{i}}).$$

Is there any mistake?

Also I have no idea on $\Lambda^{\binom{\binom{n+k-1}{k}}{i}}(S^i(f))$. Is is possible to find them without involving matrices?

答:
Pass to the algebraic closure of the ground field if necessary. Then:

$\det(f)$ is the product $\prod_{i=1}^n \lambda_i$ of the eigenvalues of $f$ (with algebraic multiplicity).

The eigenvalues of $\wedge^k(f)$ are products of unordered $k$-tuples of distinct eigenvalues of $f$, and the eigenvalues of $S^k(f)$ are products of unordered $k$-tuples of eigenvalues (not necessarily distinct) of $f$ (exercise).

So when computing $\det(\wedge^k(f))$ every eigenvalue $\lambda_i$ of $f$ appears the same number of times. A total of $k {n \choose k}$ eigenvalues appear, so each eigenvalue appears $\frac{k}{n} {n \choose k} = {n-1 \choose k-1}$ times and hence

$$\boxed{ \det(\wedge^k(f)) = \det(f)^{ {n-1 \choose k-1} } }.$$

Similarly, when computing $\det(S^k(f))$ every eigenvalue also appears the same number of times. Here a total of $k {n+k-1 \choose k}$ eigenvalues appear, so each eigenvalue appears $\frac{k}{n} {n+k-1 \choose k} = {n+k-1 \choose k-1}$ times and hence

$$\boxed{ \det(S^k(f)) = \det(f)^{ {n+k-1 \choose k-1} } }.$$

Abstractly, by functoriality both $\det(\wedge^k(f))$ and $\det(S^k(f))$ are polynomial multiplicative homomorphisms $\text{End}(V) \to F$ ($F$ the underlying field); it's possible to show that any such homomorphism must be a non-negative integer power of the determinant, so it only remains to compute which power.


问:
Does the equation $k^k+1=a^2$ have any positive integer solutions?
As the title of the question says, I want to know if the equation
$k^k+1=a^2$
has any integer positive solutions, (which means k and a are both positive integers.)

I'd really appreciate it if someone could provide the following:

If the given solutions exists, please provide the minimal solution. Are the solutions infinite or finite?

If the given solutions doesn't exist, please provide your proof.

答:
$k^k+1=a^2\Rightarrow k^k=(a-1)(a+1)$

$\gcd(a-1,a+1)=1\text{ or }2$

Case 1: $\gcd(a-1,a+1)=2$. Then $k$ is even, and $k^k$ is a perfect square. The only two squares that differ by $1$ are $0,1$, but $0$ is not a positive integer.

Case 2: $\gcd(a-1,a+1)=1$. Then both $(a-1)$ and $(a+1)$ are perfect $k$th powers that differ by $2$. This can only occur when $k=1$, but $1^1+1=2$ which is not a perfect square.


答:
It wouldn't win any beauty contests, but you could imagine a transition matrix between the sixteen possible states of combinations of faces that would come up.  For instance, if you imagined the problem with flipping a coin with a $\frac13$ probability of heads and a $\frac23$ probability of tails, your matrix would look like this:

$$
    M=\begin{pmatrix}
    0 & 0  & 0 & 0 \\
    \frac13 & \frac13 & 0& 0 \\
    \frac23 & 0 & \frac23 & 0 \\
    0 & \frac23 & \frac13 & 1 \\
    \end{pmatrix}
$$

where the first column represents not seeing any results, the second representing heads only, the third tails only, and the fourth both heads and tails.  Then you could just calculate $$M^n\begin{pmatrix}1\\0\\0\\0\end{pmatrix}$$ for any $n$ that interestsed you and read off the probability for the individual final state that interested you.  Your problem would invove a $16\times16$ matrix, but it wouldn't be hard to create it and load it into Mathematica or wherever to churn out the probability you wanted.

Title: Probability of seeing each face except one in $n$ throws of a die with 4 faces

Body:
Imagine that we throw a die with 4 faces $n$ times. The faces have probabilities of $(0.1, 0.4, 0.2, 0.3)$ of showing up.

What is the probability that we see all faces except a specific, pre-determined face $i$ (say $i = 1$) in $n$ throws?

PS: This isn't homework but a puzzle that came across that I couldn't solve analytically. Thanks.


问:
Integration problems
Can anyone help me with these:-

(a)Prove by induction: $\displaystyle\sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$

(b) By explicitly calculating upper and lower Riemann sums on a uniform partition and
passing to the limits, calculate $\displaystyle\int_0^1 x^2 \mathrm{d}x$.

答: Hint: Use part (a) to do part (b).</p>

Title: Integration problems

Body: Hint: Use part (a) to do part (b).


问:
How many ways can 7 seats and 4 seats car fill with 9 people at the same time?
I applied my combination skill and came up with bunch of answers that does not correspond to the correct answer please explain first where did I miss or break my logic

If I put 3 people in 4 seats car then there will be 7 people to pick for 7 seats car

9C2 * 7C7 = 36

If I put 3 people in 4 seat car then there will be 6 people to pick for 7 seats car

9C3 * 7C6 = 588

If i put 4 people in 4 seats car the there will be 5 people to choose for 7 seats car

9C4 * 7C5 = 4410

total = 5034

the correct answer is 246.

P.S:New to combinations.

答:
Suppose that you put $3$ people in the car that seats $4$. There are, as you say, $\binom93$ ways to choose them, but once you’ve chosen them, the $6$ who are left will go in the other car: you don’t have any more choosing to do. Similarly, there are $\binom94$ ways to choose $4$ for the smaller car, but then the other $5$ automatically all go in the larger car. Thus, there are really only

$$\binom92+\binom93+\binom94=36+84+126=246$$

different possibilities.

Note that this accounts only for the distribution of the people between the two cars: if we are concerned with how they arrange themselves in each car, the number of possibilities will be much larger.


问:
Show uniqueness of constants such that this limit is zero
Let $f$ be a smooth function (infinitely differentiable) with domain $\mathbb{R}$. The prompt is to find all real numbers $a, b, c \in \mathbb{R}$ such that

$$\lim_{x \to 0}\frac{f(x)-(ax^2+bx+c)}{x^2}=0$$

I have already shown that the numbers $a=\dfrac{f''(0)}{2}, b=f'(0), c=f(0)$ work. However, how can I show uniqueness? In other words, how can I show that no other numbers make the above limit equal to zero?

I can start off by supposing there are numbers $a', b', c'$ such that

$$\lim_{x \to 0}\frac{f(x)-(a'x^2+b'x+c')}{x^2}=0$$

And then I can show uniqueness by showing that $a'=\dfrac{f''(0)}{2}, b'=f'(0), c'=f(0)$. However, I am stuck. Any hints?

答:
If both limits equal zero, then the limit of their difference will also be zero. But their difference is

$$\frac{(a-a')x^2 + (b-b')x + (c-c')}{x^2}.$$


问:
How to re-write Complex Equations using the Argument Function?
According to wolfram alpha, we can rewrite the expression $\left(1+it\right)^{x}$ as mentioned below. How would one go about deriving the following equations by hand? ($x,t\inℝ$)

$\Re\left(\left(1+it\right)^{x}\right)=\left(1+t^{2}\right)^{\frac{x}{2}}\cos\left(x\arg\left(1+it\right)\right)$

$\Im\left(\left(1+it\right)^{x}\right)=\left(1+t^{2}\right)^{\frac{x}{2}}\sin\left(x\arg\left(1+it\right)\right)$

Thank you very much.

答:
In polar/exponential notation, if $z=re^{i \theta}$, then
$$z^x = r^x e^{i \theta x} = r^x \Big( \cos(\theta x) + i \cdot \sin(\theta x) \Big)$$
So the magnitude is raised to the $x$ power, and the argument is multiplied by $x$. In this ntoation, $\theta = \mathrm{arg}(z)$, so that explains the $\cos(x \mathrm{arg}(1+it))$ part of your question. For the other part, just notice that
the norm of $1+it$ is $|1+it| = \sqrt{1+t^2}$, and raising this to the $x$ power gives the $(1+t^2)^{x/2}$ part.


问:
$ E\left( \left|\frac{1}{n}\sum_{j=1}^n X_j\right|^p \right) \le \left( \frac{1}{n}\sum_{j=1}^n E(|X_j|^p)^{1/p} \right)^p$
The following is problem 14 of section 3.2 from Chung's "A Course in Probability Theory".

If $p>1$, we have

$$\left| \frac{1}{n}\sum_{j=1}^n X_j \right|^{p} \le \frac{1}{n}\sum_{j=1}^n |X_j|^p$$

and so

$$ E\left( \left|\frac{1}{n}\sum_{j=1}^n X_j\right|^p \right) \le \frac{1}{n}\sum_{j=1}^n E(|X_j|^p)$$

we have also

$$ E\left( \left|\frac{1}{n}\sum_{j=1}^n X_j\right|^p \right) \le \left( \frac{1}{n}\sum_{j=1}^n E(|X_j|^p)^{1/p} \right)^p$$

I understand how the the first two inequalities follow from just using the well known power mean inequality, but how to show the last inequality?

答: The last equation is actually the triangle inequality in $L^p$: Let $Y_j := \frac{1}{n} X_j$, then $$ \| \sum_{j=1}^n Y_j \|_{p} \leq \sum_{j=1}^n \|Y_j\|_p \\ \Rightarrow \left(\| \sum_{j=1}^n Y_j \|_{p} \right)^p \leq \left(\sum_{j=1}^n \|Y_j\|_p \right)^p$$ by applying the triangle inequality. Using the definition of $Y_j$ and $\|\cdot\|_p$, one can easily see that this is equivalent to $$ \mathbb{E} \left( \left| \sum_{j=1}^n \frac{X_j}{n} \right|^p \right) \leq \left( \sum_{j=1}^n \frac{1}{n} \|X_j\|_p \right)^p = \left( \frac{1}{n} \sum_{j=1}^n \mathbb{E}(|X_j|^p)^{\frac{1}{p}} \right)^p $$</p>


问:
How to solve this four variables system of equations?
$$(x-1)^2+(y-1)^2+(z-1)^2=(t-1)^2,$$

$$(y+1)^2+(z+1)^2=(x+1)^2+(t+1)^2$$

Ηow to solve this for $(x,y,z,t)$ if $x<y<z<t$,  $2\leq x, y, z, t\leq 16$ and if $x, y, z$ and $t$ are all natural numbers?

I don't how to start to solve this. Please help!

Body:
A line of research Exhaustive search show that the only sums of two squares that can be written in two different ways for distinct numbers in $[3, 17]$ are

130 {(3, 11), (7, 9)}
185 {(4, 13), (8, 11)}
205 {(6, 13), (3, 14)}
221 {(5, 14), (10, 11)}
250 {(9, 13), (5, 15)}
265 {(3, 16), (11, 12)}
305 {(7, 16), (4, 17)}
325 {(10, 15), (6, 17)}


答:
You can write $f(x) = ax^2 + bx + c + r(x)$ for any real numbers $a,b,c$, and the property that you're looking for is

$$\lim\limits_{x\to0} \frac{r(x)}{x^2} = 0.$$

Hence, you can prove the claim by noticing that $r(x)$ is smooth and computing its first and derivatives at 0, wich is possible by the above limit.

Title: Show uniqueness of constants such that this limit is zero

Body:
Let $f$ be a smooth function (infinitely differentiable) with domain $\mathbb{R}$. The prompt is to find all real numbers $a, b, c \in \mathbb{R}$ such that

$$\lim_{x \to 0}\frac{f(x)-(ax^2+bx+c)}{x^2}=0$$

I have already shown that the numbers $a=\dfrac{f''(0)}{2}, b=f'(0), c=f(0)$ work. However, how can I show uniqueness? In other words, how can I show that no other numbers make the above limit equal to zero?

I can start off by supposing there are numbers $a', b', c'$ such that

$$\lim_{x \to 0}\frac{f(x)-(a'x^2+b'x+c')}{x^2}=0$$

And then I can show uniqueness by showing that $a'=\dfrac{f''(0)}{2}, b'=f'(0), c'=f(0)$. However, I am stuck. Any hints?


问:
How can I solve this absolute value inequality: $|-2x|> \frac{-x}{2} -3$?

$|-2x|> \frac{-x}{2} -3$

This inequality isn't from a textbook, I made it out because I felt like there must be some cases when the methods I learnt in class won't work.

These are the methods I used and that I only know:

First method

$-2x>\frac{-x}{2}-3$→ $x<2$

$-(-2x)>\frac{-x}{2}-3$→ $x>-1.2$

Answer: $-1.2<x<2$

And after checking by substituting any number smaller than $2$ or bigger than $-1.2$, I get correct answers.

Second method

$(|-2x|)^2 > (\frac{-x}{2}-3)^2$

Answer: $-1.2<x<2$

But this answer is wrong because looking at the graph I can see that the absolute value function is always above the straight line.

The answers I get are true if the straight line had an absolute value sign too or because of the parts under the x-axis of functions $-2x$ and $-(-2x)$ that no longer exist because of the absolute value sign.

So how can I solve this inequality without drawing? And how does I know that the answers I get are wrong or incomplete when solving anything similar?

答: $|-x|$ is the same as $|x|$. $$|2x| \gt -\frac x2 -3 $$ For $x\gt -6$, this will hold as the RHS is $\le 0$ (and also the LHS is not zero when the RHS is zero). For $x\le-6$, it is equivalent to $$-2x \gt -\frac x2-3 \implies \frac{3x}{2}\lt 3 \implies x\lt 2$$ which is always true in this case. Hence, all real numbers are solutions.


答:
If $x\ge0$ then $|-2x|=2x$, and if $x<0$ then $|-2x|=-2x$.

So we have (1) $x\ge0$ and $2x>-\frac x2-3\iff x>-\frac65$

or (2) $x<0$ and $-2x>-\frac x2-3\iff x<2$,

which is equivalent to (1) $x\ge 0$ or (2) $x<0$, which is equivalent to $x\in\mathbb R$.

Title: How can I solve this absolute value inequality: $|-2x|> \frac{-x}{2} -3$?

Body:
$|-2x|> \frac{-x}{2} -3$

This inequality isn't from a textbook, I made it out because I felt like there must be some cases when the methods I learnt in class won't work.

These are the methods I used and that I only know:

First method

$-2x>\frac{-x}{2}-3$→ $x<2$

$-(-2x)>\frac{-x}{2}-3$→ $x>-1.2$

Answer: $-1.2<x<2$

And after checking by substituting any number smaller than $2$ or bigger than $-1.2$, I get correct answers.

Second method

$(|-2x|)^2 > (\frac{-x}{2}-3)^2$

Answer: $-1.2<x<2$

But this answer is wrong because looking at the graph I can see that the absolute value function is always above the straight line.

The answers I get are true if the straight line had an absolute value sign too or because of the parts under the x-axis of functions $-2x$ and $-(-2x)$ that no longer exist because of the absolute value sign.

So how can I solve this inequality without drawing? And how does I know that the answers I get are wrong or incomplete when solving anything similar?


问:
Finding $E[X\mid Y]$ and $\operatorname{Var}(X\mid Y)$ given mean and variance of $X$ and $Y$
Suppose we have 2 normal distributions $X$ and $Y$ with mean $u_1$ and $u_2$ and variance $\sigma_1^2$ and $\sigma_2^2$; find $E[X\mid Y]$ and $\operatorname{Var}(X\mid Y)$.

I know
$$E[X\mid Y] = \mu_1 + \rho\sigma_1 \frac{Y - u_2}{\sigma_2} $$
and
$$\operatorname{Var}[X\mid Y] = \sigma_1 (1 - \rho^2)$$
but I can't prove it.

For $E[X\mid Y]$ I start with
$$E[X\mid Y] = \int_{-\infty}^{+\infty} x f_{X|Y}(x\mid y)\ dx$$
but that doesn't work because for calculating $f_{X\mid Y}(x\mid y)$ I need $f_{X,Y}(x,y)$ I don't have that.
Can any one help me?

答:
Let us take for granted that
$$\frac{X-\mu_1}{\sigma_1} = \rho \frac{Y-\mu_2}{\sigma_2} + \sqrt{1-\rho^2} Z \tag{$*$}$$
where $Z \sim N(0,1)$ is independent of $Y$. See the end of my answer for an explanation.

Then
\begin{align}
E[X \mid Y] &= \mu_1 + \sigma_1 E[(X-\mu_1)/\sigma_1 \mid Y]
\\
&= \mu_1 + \sigma_1 \left(
E[\rho (Y-\mu_2)/\sigma_2 \mid Y]
+ E[\sqrt{1-\rho^2} Z \mid Y]
\right)
\\
&= \mu_1 + \sigma_1 (\rho(Y - \mu_2) / \sigma_2) + \sqrt{1-\rho^2} E[Z]
\\
&= \mu_1 + \rho \frac{\sigma_1}{\sigma_2} (Y-\mu_2).
\end{align}
Note that Ottavio Bartenor fixed a typo in your original expression for $E[X\mid Y]$.

Similarly,
\begin{align}
\text{Var}(X \mid Y)
&= \sigma_1^2 \text{Var}((X-\mu_1)/\sigma_1 \mid Y)
\\
&= \sigma_1^2 \text{Var}(\rho(Y-\mu_2)/\sigma_2 + \sqrt{1-\rho^2} Z \mid Y)
\\
&= \sigma_1^2 \text{Var}(\sqrt{1-\rho^2} Z \mid Y)
\\
&= \sigma_1^2 (1-\rho^2) \text{Var}(Z)
\\
&= \sigma_1^2(1-\rho^2).
\end{align}
Note that the expression in your post has a typo.

Suppose $Y \sim N(\mu_2, \sigma_2^2)$ and $Z \sim N(0,1)$ are independent. Let $X$ satisfy the above equality ($*$). The claim is that $(X,Y)$ follows a bivariate normal distribution with parameters $\mu_1, \sigma_1, \mu_2 ,\sigma_2, \rho$.

You can check that $X$ has mean $\mu_1$ and variance $\sigma_1^2$. You can also check that the correlation between $X$ and $Y$ is $\rho$. You can also check that the marginal distribution of $X$ is normal, because it is a linear combination of independent normal random variables. Finally, to justify that $(X,Y)$ is jointly (bivariate) normal, you can appeal to an equivalent characterization of jointly normal distributions by noting that any linear combination of $X$ and $Y$ is normal because it can be written as a linear combination of the independent normal random variables $Y$ and $Z$


问:
Points in the plane - Pigeonhole Principle - Sharpen the result
Thirteen points are given in the plane so that among any three of them there is a pair whose distance is less than $1$.  Prove that it is possible to select seven of the points so that they are all interior to a circle of radius $1$.

This problem can be solved by using the Pigeonhole Principle. Let $A$ be any given point, and consider the circle $C_A$ of radius $1$ around $A.$ If there are at least seven points in this circle, we are done. If not, let $B$ be a point outside $C_A$, and consider $C_B$, a circle of radius $1$ around $B.$
By the condition in the problem, all points are in $C_A$ or $C_B$, so by the Pigeonhole Principle at least seven of the points lie in the same circle.

Now the question is whether the result could be sharpened to assert that it is possible to select seven of the points so that they are all interior to a circle of radius $\frac{1}{\sqrt{3}}$. If so, how can this be shown?

答:
The picture shows an incorrect distribution of seven points in the circle, but helps us find proof of the statement:

Under given condition, it is possible to select seven of the points so that they are all inside a circle of radius $\frac{1}{\sqrt3}.$

Assume that seven of the points lie in a circle $C$ of radius $1,$ and the remaining six lie outside a circle $C'$ concentric to $C,$ radius of $C'$ being at least $3.$

Ignore for a while the six points that lie outside.

The picture shows the most stretched relative position of the seven points: one is a center of the circle, four points $1,4,5,3$ lie near the vertices of a regular hexagon, two other points are close to the arcs $(1,4)$ and $(3,5),$ respectively.

HOWEVER, the distribution of the thirteen points as decribed does not fulfill the condition. The problem only occurs if one of the three points lies outside $C'$ and the two inside $C,$ but they are too far apart. Therefore, the distance between any two points inside $C$ must be less than $1.$ Consequently, the seven points lie inside a curvilinear triangle (such as in the picture, with vertices $1,2,4$). As the circumradius of this triangle is $\frac{1}{\sqrt3},$ we are done.


问:
In a random walk starting at A and then moving to any adjacent point with equal probability. Find P(reach B before E)
It seems to me that for symmetric reasons the answer should be $\frac{1}{2}$.
I tried to prove this by separating paths that did not include A and those that did.
The probability of the former is $\frac{3}{8}$ (using the sum of an infinite geometric sequence) and the probability of the latter seems to me to be $\frac{1}{2}$$\frac{1}{3}$P where P is the requested probability.(the $\frac{1}{2}$ comes from going from A to C or D and the $\frac{1}{3}$ from going from those back to A). This "recursive" equation results in $P=\frac{18}{40}$ which is not expected. Will be grateful for any thoughts on this

答:
The answer is "obviously" $\frac12$ by symmetry.

Let's try a longer way: let $P_i$ be the probability that you reach $B$ before $E$ when starting from point $i$.  Then

$P_A=\frac14 P_B + \frac14 P_C + \frac14 P_D + \frac14 P_E$

$P_B=1$

$P_C=\frac13 P_A + \frac13 P_B + \frac13 P_D$

$P_D=\frac13 P_A + \frac13 P_C + \frac13 P_E$

$P_E=0$

which is five equations in five unknowns.  You can solve them by substitution or otherwise and get the unique solution

$P_A=\frac12$

$P_B=1$

$P_C=\frac58$

$P_D=\frac38$

$P_E=0$


问:
What is the solution to $a^*_i = arg min_{a_i} \sum_{i=1}^n \left(b_i - \frac{\sum_{i=1}^n b_i a_i^2}{\sum_{i=1}^n a_i^2}\right)^2$
Let $b \in \mathbb{R}^n$ denote a vector, where $b = \begin{bmatrix} b_1 & b_2 & \ldots & b_n \end{bmatrix}^T$, and let $a \in \mathbb{R}^n$ denote a vector where $a = \begin{bmatrix} a_1 & a_2 & \ldots & a_n \end{bmatrix}^T$

I'm trying to find $a^* \in \mathbb{R}^{n}$, where
$$a^*_i = \operatorname{arg}\operatorname{min}_{a_i} \sum_{i=1}^n \left(b_i - \frac{\sum_{i=1}^n b_i a_i^2}{\sum_{i=1}^n a_i^2}\right)^2$$

My thought is that it might be easier to work with the above in matrix form. Let $B$ denote a diagonal matrix with $b$ on the diagonal, then we have

$$a^* = \operatorname{arg}\operatorname{min}_{a} ||b - (a^Ta)^{-1} a^T B a||^2_2$$

I know the above is minimized when
$$\bar{b} = (a^Ta)^{-1} a^T B a$$

My question is, is there a way to isolate $a$ from here?

答:
$
\bar{b} = (a^Ta)^{-1} a^T B a = \frac{1}{||a||^2}a^T B a = \frac{1}{||a||}a^T B \frac{1}{||a||}a = u^T . B . u
$ where $||u|| = 1$

From there it is easy to find that $u_i = 1$ if $b_i = \max_j{b_j}$ and $0$ else. You can take $a = u$ as a result


问:
Recurrence relation of sum with binomial coefficients
Is there any way of solving for S(n) to obtain a formula that isn't a recurrence relation for: $$S(n)=\frac{ 1}{4 }\sum_{m=0}^n\binom{ n}{m }S(m)$$
The initial condition is: $S(0)=\frac{\pi^2}{6 }$

答:
Hint: Take $S(x)=\sum _{m=0}^{\infty}S(m)\frac{x^m}{m!},$ then $$S(x)e^x=\sum _{n=0}^{\infty}\frac{x^m}{n!}\sum _{m+k=n}\binom{n}{m}S(m)=4(S(x)-S(0))+S(0)=4S(x)-3S(0),$$ so $$S(x)=\frac{3S(0)}{4-e^x},$$
can you conclude? If not, take a look at Eulerian Polynomials.


问:
How to demonstrate that there would exist a vertex with degree less than $ 5$?
Question:

```<code>                           Letting S be a planar graph.
                Demonstrate that there would exist a vertex of S whose degree
                       would be no greater than 5, ie, <span class="math-container">$\leq 5$</span>
</code>```

I have just gotten started with Discrete Math Graph Theory and it would be a great help if someone could assist me in working out the steps to understand the problem and work through the answer with a detailed explanation? Thanks.

答: Assume that every vertex of a planar graph $G$ has degree $6$ or more. Thus by the handshaking lemma, we have $2m\geq 6n$ where $n$ and $m$ denote the order and size of $G$. Since $G$ is planar we know that $m\leq3n-6$ and we have $3n\leq m$. Thus $3n\leq 3n-6$ which is a contradiction.


问:
Why truth table is not used in logic?
One day, I bought Principia Mathematica and saw a lot of proofs of logical equations, such as $\vdash p \implies p$ or $\vdash \lnot (p \wedge \lnot p)$. (Of course there's bunch of proofs about rel&set in later)

After reading these proofs, I suddenly thought that "why they don't use the truth table?". I know this question is quite silly, but I don't know why it's silly either (just my feeling says that).

My (discrete math) teacher says that "It's hard question, and you may not understand until you'll become university student," which I didn't expected (I thought the reason would be something easy).

Why people don't use truth table to prove logical equations? (Except for study logic (ex: question like "prove this logic equation using truth table"), of course.)

PS. My teacher is a kind of people who thinks something makes sense iff something makes sense mathematically.

答:
First, of course, mathematicians and logicians often do use truth tables, so it is incorrect to suggest that they are not used. (I won't comment on your reading of Principia Mathematica, beyond saying that that text is not meant to be taken as a piece of pedagogy.)

But second, the truth table method is not a feasible method for large propositions. This is because for a formula with $n$ free variables, the truth table is an object with $2^n$ rows, exponentially large in size. But many logical expressions nevertheless have comparatively short derivations showing them to be tautological.

For example, $$(p_0\vee \neg p_0)\wedge(p_1\vee \neg p_1)\wedge\cdots\wedge(p_{100}\vee\neg p_{100})$$ has a truth table with $2^{101}$ rows, but there is a simple, obvious derivation that allows us to see it as taugological without calculating this table.

Thus, although when there are just two propositional variables, I don't mind computing a truth table, or even for three, it is often easier with more to use more focused reasoning.

Title: Why truth table is not used in logic?


问:
All nested partial sums of a sequence tend to $0$. Is the sequence constant?
$S^0:\mathbb N\to \Bbb R$ is a function. For any $m\in \Bbb N$, we define

$$S^m:\Bbb N\to \Bbb R$$
$$S^m(n)=\sum_{k=1}^n S^{m-1}(k)$$

For each $m\in \Bbb N$, we have:
$$\lim_{n \to \infty}S^m(n)=0$$

Can we deduce $S^0\equiv 0?$

答:
No, there are functions such that all your nested partial sums converge to $0$, but that are nonzero :

Suppose you have constructed the first $k(n)$ terms for $S_0$, such that $S_1(k(n)) = S_2(k(n)) = \ldots = S_n(k(n)) = 0$, and let $\varepsilon > 0$.
You want to pick the next $k(n+1)-k(n)$ terms so that the next partial sums stay bounded by $\varepsilon$, and $S_m(k(n+1)) = 0$ for $0 \le m \le n+1$.

This is always possible : in fact you can uniquely choose the next $n+2$ terms such that $S_m(k(n)+n+2) = 0$ for $0 \le m \le n+1$, by solving an invertible linear system of $n+2$ equations. Then, in order to not go over $\varepsilon$ in the intermediate partial sums, find $N$ such that they stay under $N\varepsilon$. Divide those $n+2$ numbers by $N$, and apply them $N$ times to get what you wanted.

You can also have a more explicit construction :

Pick $S_0(0,1) = +1, 0$, so that $S_0(1) = 0$

Then pick $S_0(2+n) = - S_0(n \mod 2)/2$ for $0 \le n < 4$, so that $S_0(5) = S_1(5) = 0$

Then pick $S_0(6+n) = - S_0(n \mod 6)/3$ for $0 \le n < 18$, so that $S_0(23) = S_1(23) = S_2(23) = 0$

Then pick $S_0(24+n) = - S_0(n \mod 24)/4$ for $0 \le n < 96$, so that $S_0(119) = \ldots = S_3(119) = 0$.

And so on.


答:
Suppose that every vertex has degree $\geq 6$. By the handshake lemma,

$$|E| = \frac{1}{2} \sum_{v \in V} d(v) \geq 3|V|$$

However, it is a well-known result that in a planar graph $|E| \leq 3v - 6$ (https://en.wikipedia.org/wiki/Planar_graph#, Theorem 1 under other planarity criteria)

Title: How to demonstrate that there would exist a vertex with degree less than $ 5$?

Body:
Question:

```<code>                           Letting S be a planar graph.
                Demonstrate that there would exist a vertex of S whose degree
                       would be no greater than 5, ie, <span class="math-container">$\leq 5$</span>
</code>```

I have just gotten started with Discrete Math Graph Theory and it would be a great help if someone could assist me in working out the steps to understand the problem and work through the answer with a detailed explanation? Thanks.


问:
$V$ is a vector space, $W,W',U,U'$ are subspaces of $V$ prove or disprove the following:
A) if $U+W=U'+W'$ then $W=W' , U=U'$ or $W=U' , W'=U$

B) if $U+W=V$ then for all $Y\subset V$ :  $Y=(Y\cap U)+(Y\cap W)$

I need help of how to approach these types of questions, I have tried to bring a counter example for A but I didn't get it, but I still also don't know how to prove it or if I just failed bringing the example.

 any explanation, tips, hints are appreciated, thanks in advance.

答:
Your intuition was correct about part A: there is a counterexample! Learning to construct counterexamples is a skill that takes practice. I'll try to explain my thought process for questions like these; I hope it helps.

This statement in particular might seem daunting because the consequent of the implication is a disjunction of two conjunctions, so it might seem hard to understand when this consequent will be false (which is what we need to understand -- a counterexample to an implication is just a situation where the antecedent is true and the consequent is false).

So, the first step should be to try to understand what the implication is "really saying". By this I mean you need to digest and internalize the content of this implication, which often involves rephrasing the statement in a more intuitive way. For me, this implication is "really saying" that "it's impossible to write a subspace as a sum of two subspaces in two different ways". This sentence is not meant to be a precise mathematical statement, but it is meant to be more intuitive and easy to think about. Does it make sense that this is just an imprecise rephrasing of the implication? If not, can you come up with a rephrasing that's easier for you to think about?

Once you've internalized what the statement is trying to say, we can try to produce a counterexample. A counterexample to this claim, based on my intuitive understanding, would be produced by writing a subspace of a vector space as a sum of two subspaces is two distinct ways. Can you think of an example of this happening? Once you have such an example, you can go back and check that it really is a counterexample to the precise statement you were given.

As you get more experience, this process of internalizing the content of mathematical statements and producing counterexamples will become easier and easier.


问:
Convergence & absolute convergence of an infinite series
How can i find all possible values for $a > 0$ which make the series absolute convergent?

$$\sum_{n=1}^{\infty}\frac{5^n}{(4^n+a^n)}$$

答:
Well it is clear that for $a \leq 5$ the terms don't go to $0$ so the sum can't converge.

On the other hand, for $a > 5$, the ratio test gives you that the quotient of the terms tends to $|5/a| < 1$ so it's convergent.


答:
$\textbf{Hint}$: Use the root test and the fact that

$$\lim_{n\to\infty} \sqrt[n]{a^n+b^n} = \max(a,b)$$

Title: Convergence & absolute convergence of an infinite series

Body:
How can i find all possible values for $a > 0$ which make the series absolute convergent?

$$\sum_{n=1}^{\infty}\frac{5^n}{(4^n+a^n)}$$


问:
Definition of Noetherian Topological Space
I am trying to make sense of the definition of a Noetherian Topological Space. It is a topological space that satisfies the descending chain condition for closed subsets. Here are my questions:

Why 'closed' subsets?

Is there an example of a Noetherian topological space where one has a sequence of the form $Y_1 \supseteq Y_2 \supseteq \ldots$ which doesn't stabilize, where the $Y_i$ are not necessarily closed?

Is my understanding right that $\mathbb{R}$ with the standard topology is not Noetherian: take $Y_n = [0, \frac{1}{2} + \frac{1}{n}]$, all of these are closed, and $Y_n \supseteq Y_{n+1}$, and they don't stabilize.

Take affine-n space with the Zariski topology. Is there a sequence of non-closed sets here that do not satisfy the descending chain condition?

答:
I will answer these out of order. I'll assume $k$ is algebraically closed.

(3) Correct, $\mathbb{R}$ in the euclidean topology is not Noetherian since e.g., the sequence $X_n = [0, 1/n]$ is a descending chain which does not stablise.

(4) Affine $n$-space in the Zariski topology is Noetherian, this follows from the Nullstallensatz (there is an inclusion reversing correspondence between closed subsets correspond to radical ideals) and the Hilbert basis theorem (which shows that $k[x_1,...x_n]$ is Noetherian, hence ascending chains of ideals eventually stabilise).

(2) Yes, take $\mathbb{A}^1_{\mathbb{C}}$, by (4) this is Noetherian in the Zariski topology, but there is a the sequence I gave in (3) which does not stablise (it just has nothing to do with the topology). What you're asking is equivalent to asking for a finite set (just remove one point, then 2 points etc).

(1) You want your definition to have something to do with the universe you are working in. If the question is just subsets, you're just asking for finite sets which is neither very interesting, nor enlightening from a geometric perspective. The main reason for this definition is the observation in (4) above that closed subsets and (radical) ideals correspond, so you want your notions of Noetherian to line up. For affine space you can think that you can insert a hypersurface, then a codimension 1 subvariety into that, and so on... but this process will have to terminate at a point.

Edit: I misread your question (4). But see my answer to (2) - just take some balls that would be closed in the Euclidean topology and shrink them (they'll just have nothing to do with the Zariski topology).


问:
stable and unstable limit cycles
Given a system $$\dot x = -y + xF(r), \ \dot y = x + y F(r) $$ where $F(r):= r^4-3r^2+1$.

I need to find out the stable and unstable limit cycle for the given system.

My Approach

I used the polar coordinates to reduce the system as $$\dot r = r(r^4-3r^2+1), \ \dot \theta =1$$

The reduced system has a unique critical point at origin since $\dot \theta \neq 0$.

So to find the stable and unstable limit points all I need to check is the sign of $\dot r$ in the annulus. I am not able to follow up from here.

答:
Letting $$ r^4-3r^2+1=0$$
gives two positive roots
$$ r_{1,2}=\sqrt{\frac{3\pm\sqrt{5}}{2}}. $$
Then the equation becomes
$$ r'=r(r^2-r_1^2)(r^2-r_2^2). $$
and $$ x^2+y^2=r_{1,2}^2$$ are two solution curves. Consider the region
$$ R_k=\{r: r_k-\epsilon <r<r_k+\epsilon\}$$
for small $\epsilon>0$. Then on $r=r_1-\epsilon$, $r'<0$ and on $r=r_1+\epsilon$, $r'>0$ and hence $r=r_1$ is an unstable limit cycle. Similarly $r=r_2$ is another unstable limit cycle in $R_2$ too.

Title: stable and unstable limit cycles


答:
Here is how I would solve this problem:
$$
|-2x|=|2x|>-\frac{x}{2}-3 \, .
$$
If $x\geq0$, then the inequality becomes
$$
2x>-\frac{x}{2}-3 \\
\frac{5}{2}x>-3 \\
x > -\frac{6}{5} \, .
$$
However, it is only true that $2x>-\frac{x}{2}-3$ when $x\geq0$. Hence, not all of the solutions we have found are valid. We are left with $x\geq0$.

If $x<0$, the inequality becomes
$$
-2x>-\frac{x}{2}-3 \\
2x < \frac{x}{2}+3 \\
\frac{3}{2}x<3 \\
x<2 \, .
$$
The condition that $x<0$ and $x<2$ is equivalent to the condition that $x<0$—if a number is smaller than $0$, it is certainly smaller than $2$. Putting the two solutions together, we find that this is true for all values of $x$. Because of the difficulty of splitting the problem up into different cases, I think it is easier to solve this equation graphically than algebraically.


问:
Math word problem. Any help is appreciated.
A math student writes a proof of the derivative of a certain trigonometric function. The last line she writes before stating her conclusion is...

$\dfrac{d}{d\theta} \left(\sin\left(\theta +\dfrac{\pi}{2}\right) \right)=\cos\left(\theta +\dfrac{\pi}{2}\right)$

These are the graphs of $\sin\left(\theta +\dfrac{\pi}{2}\right)$ and $\cos\left(\theta +\dfrac{\pi}{2}\right)$

$\sin\left(\theta +\dfrac{\pi}{2}\right)$ = http://www.wolframalpha.com/input/?i=sin%28theta%2Bpi%2F2%29&lk=4

$\cos\left(\theta +\dfrac{\pi}{2}\right)$= http://www.wolframalpha.com/input/?i=cos%28theta%2Bpi%2F2%29&lk=4&num=1

Hence, write the statement of her conclusion beginning with the word, "Therefore...".

答:
You can "read" the statement $\frac d{d\theta} \sin\left(\theta+\frac \pi 2\right) = \cos\left(\theta + \frac\pi 2\right)$ as

"The derivative of sine theta plus pi over two by theta is cos theta plus pi over two"

You should never write it out like that, it's how it's pronounced, and the maths is how it is spelt.

So when you conclude your proof   you might want to add a word or two to make it sound more natural, especially if it's on its own at the end as a one sentance paragraph.

You might say

"Therefore, $\frac d{d\theta} \sin\left(\theta+\frac \pi 2\right) = \cos\left(\theta + \frac\pi 2\right)$."

or it might be better to say

"We have calculated: $\frac d{d\theta} \sin\left(\theta+\frac \pi 2\right) = \cos\left(\theta + \frac\pi 2\right)$.

It doesn't really matter there isn't a right answer. Notice that the punctuation is the same as it would be if you wrote out the sentence longhand. The only really important thing is to remember the full stop at the end.


答:
The fastest way for me is to recognize $f(x)=|-2x|+\frac x2 +3$ is piecewise linear, and the only turning point is $0$.

Now $f(-\infty)=+\infty, f(0)=3>0, f(+\infty)=+\infty$, therefore $f(x)>0,\forall x \in \mathbb R$.

For some other examples of this type, check here.


问:
Prove that the triangle is equilateral
Let D be the Orthogonal projection of the vertex A of a given triangle.If it stands that [AB]+[BD] ≅ [AC]+[CD] prove that the triangle is equlaterial.

I have tried to prove in the following way but I’m not sure if this is enough:
Let it be [AB]+[BD] ≅ [AC]+[CD]....(1)
I have supposed that the the triangle is not equlaterial for example let it be scalene traingle
Then [AB] ≢ [AC] and [BD] ≢ [CD] so if we add two of give relactions we get that [AB]+[BD] ≢ [AC]+[CD], which is not possible because it contradicts (1)
Therefore the triangle is equlaterial.
Thank you in advance.

答:
If $a \ne b$ and $c \ne d$, it is still possible that $a+c = b+d$. Take $(a,b,c,d) = (1,2,2,1)$, for example.

Now your proposition is true not only for equilateral triangles, but for any isosceles triangle with $AB=AC$. The proof is a slight modification of yours, by using inequalities instead of unequal signs:

Suppose $AB > AC$. Then by Pythagoreas' Theorem and the fact that $BC \perp AD$, we have:

$$BD = \sqrt {AB^2 - AD^2} > \sqrt{AC^2 - AD^2} = CD$$

Hence $AB + BD > AC + CD$.

If $AB < AC$, $AB + BD < AC + CD$ follows from a similar argument.

Therefore if $AB +BD = AC + CD$, it must be the case that $AB = AC$.

When $AB = AC$, $\triangle ABD \cong \triangle ACD$ by RHS, and thus $AB + BD = AC + CD$, so the implication goes both ways, and we have shown that this proposition is true for any isosceles triangle with $AB = AC$.

Title: Prove that the triangle is equilateral


问:
$A_1,...,A_n,...$ measurable sets such that: ... Then $ \lambda (\bigcup _{k=1}^{\infty} A_k) \geq 1 $.
$A_1,...,A_n,...$  measurable sets such that:

i) $\lambda (A_n) \geq 1/2 $ forall $k$

ii) $\lambda (A_s \cap A_k ) \leq 1/4 $ forall $k \neq s$

Then $ \lambda (\bigcup _{k=1}^{\infty} A_k) \geq 1 $.

Attemp:

We consider the metric Boolean algebra $^\mathcal{A } / _\sim$ defined from : $d(A,B) = \mu (A \triangle B)$ .

We know that $^\mathcal{A } / _\sim$ is complete and when $\mathcal{A}= \mathbb{R} $ then  $^\mathcal{A } / _\sim$ is also seperable.

Also we have that
$d(A_n,A_m) \geq \frac{1}{2}$ for all $n \neq m$.

答:
A quantitative approach to the problem:

Let $B_n = \bigcup_{k=1}^n A_k$, and suppose $\lambda(B_n) \leq 1$ for a given $n$. Let $f_k$ be the indicator function of $A_k$. Then by the Cauchy-Schwarz inequality, we have
$$\left(\int_{B_n} 1 \,d\lambda\right) \left(\int_{B_n} \left(\sum_{k=1}^n f_k\right)^2 \,d\lambda\right) \geq \left( \int_{B_n} \sum_{k=1}^n f_k \,d\lambda\right)^2.$$
The second integral on the left satisfies
$$\int_{B_n} \left(\sum_{k=1}^n f_k\right)^2 \,d\lambda = \sum_{k=1}^n \int_{B_n} f_k^2 \,d\lambda + \sum_{i \neq j} \int_{B_n} f_i f_j \,d\lambda \leq n + n(n-1)(1/4) = \frac{n(n+3)}{4}$$
while the integral on the right has $\int_{B_n} \sum_{k=1}^n f_k \,d\lambda \geq n/2$, and putting these together gives $\lambda(B_n) \cdot \frac{n(n+3)}{4} \geq \frac{n^2}{4}$, which implies $\lambda(B_n) \geq \frac{n}{n+3}$, so $\lambda(B_n) \geq \frac{n}{n+3}$ in any case.

Thus for each $n$, $\lambda(\bigcup_{k=1}^n A_k) \geq \frac{n}{n+3}$, from which it follows that $\lambda(\bigcup_{k=1}^\infty A_k) \geq 1$.


问:
How do I show that $\log_2(k)$ is an irrational number if $k$ is an odd integer greater than $1$?

Show that if $k$ is an odd integer grater than $1$, then $\log_2(k)$ is an irrational number.

So my approach for this problem was to use proof by contradiction so I wrote "Assume by contradiction if $k$ is an odd integer greater than $1$ than $\log_2(k)$ will be a rational number. Such that $\log_2(k)=a/b$, $\gcd(a,b) = 1$ and $a,b$ are integers in its lowest terms. But I don't know what to do after. Should I use the fundamental theorem of arithmetic? Euclid's lemma?

答:
$2^{(a/b)} = k \implies 2^a = k^b.$

But $2^a$ is even and $k^b$ is odd.


问:
What does a proof in an internal logic actually look like?
The nLab has a lot of nice things to say about how you can use the internal logic of various kinds of categories to prove interesting statements using more or less ordinary mathematical reasoning. However, I can't find a single example on the nLab of what such a proof actually looks like. (The nLab has a frustrating lack of examples in general.)

Can anyone supply me with some examples? I'd be particularly interested in the following kinds of examples:

I've heard in a topos one can internalize the real numbers and, in the topos $\text{Sh}(X)$ of sheaves on a topological space, this reproduces the sheaf of continuous real-valued functions $X \to \mathbb{R}$. Moreover, one can internalize "finitely generated projective $\mathbb{R}$-module" and in $\text{Sh}(X)$ this reproduces real vector bundles on $X$. What can you prove about vector bundles this way?

I'd also like to see examples of what you can do in the internal logic of Cartesian closed categories.

This MO question is related but it doesn't really satisfy my curiosity.

答:
You can find examples in the work by Christopher Mulvey, for example

Intuitionistic algebra and representations of rings. In Recent advances in the representation theory of rings and C*-algebras. Mem. Amer. Math. Soc., 148, 3-57 (1974).

develops (essentially) topos theory and its internal logic from scratch and discusses the example of vector bundles.


问:
Lagrange multipliers on the unit sphere
I have a function $f(x,y,z)=x^2+xy+y^2+yz+z^2$ which I'm supposed to maximize on the Unitsphere $B_1(0)=\{x \in \mathbb{R^3}:\Vert x\Vert_2 \le 1\}$ and I have a few questions concerning the process of the computation:

My first question is if I'm allowed to transform the condition into $\{x \in \mathbb{R^3}:\Vert x\Vert_2^2 = 1\}$ so that it's $g(x,y,z)=x^2+y^2+z^2=1 \quad$ I'd assume this would work since the function grows larger for growing $x,y,z$, so the Maximum under my condition will be on the edge of the sphere. Furthermore since it is a unit-vector the norm is equal to the square of the norm.

If this is possible/correct I have some trouble with the algebraic part of the computation. When setting up my functions I'm left with:
$$\begin{align}&I. 2x+y=\lambda2x \\&
II. 2y+x+z=\lambda2y \\&
III. 2z+y=\lambda2z \\&
IV. x^2+y^2+z^2=1
\end{align}$$

Whatever I do, I can't seem to rewrite the equations in such a way that this problem is solvable. The furthest I can show is that $x=z$. Any further rewriting I do yields me something which is dependent on $\lambda$ and one of the other three variables. What am I missing? Thanks in advance.

答: The first three equations are equivalent to$$\left\{\begin{array}{l}(2-2\lambda)x+y=0\\x+(2-2\lambda)y+z=0\\y+(2-2\lambda)z=0,\end{array}\right.\tag1$$which is a system of three linear equations in the unknowns $x$, $y$, and $z$, depending upon a parameter $\lambda$. The matrix of this system is$$\begin{bmatrix}2-2\lambda&1&0\\1&2-2\lambda&1\\0&1&2-2\lambda\end{bmatrix}.\tag2$$The system has a solution other than $(0,0,0)$ if and only if the determinant of $(2)$ is $0$. It is cleat that that occurs if $\lambda=1$. On the other hand, the determinant of $(2)$ is $-4(2\lambda^3-6\lambda^2+5\lambda-1)$ and we already know that $1$ is a root of this polynomial. Using this, it is easy to see that the other roots are $1\pm\frac1{\sqrt2}$. So, for each $\lambda\in\left\{1,1+\frac1{\sqrt2},1-\frac1{\sqrt2}\right\}$, there are vectors $(x,y,z)\in\Bbb R^3\setminus\{(0,0,0)\}$ such that $(x,y,z)$ is a solution of $(1)$. Computes them, divide them by their norms, and you will get the six solutions of your original system. For each of them, compute $f(x,y,z)$ and see where is that you get the maximal value.

Title: Lagrange multipliers on the unit sphere


答:
$2x+y = 2\lambda x$

$2y+x+z = 2\lambda y$

$2z+y = 2\lambda z$

$x^2+y^2+z^2=1$

From first equation $\lambda = \frac{2x+y}{2x}$

Now plugging it into $2$nd and $3$rd equations, we have $3$ equations in $3$ variables.

The solution comes to $x = z = \pm \frac{1}{2}, y = \pm \frac{1}{\sqrt 2}$. (I took help of Wolfram Alpha)

On your first point, given the function $xy + yz + x^2 + y^2 + z^2, $ you can absolutely argue that the maximum will be on the sphere and not inside. You see why?

Last but not least, while you need to maximize $f(x, y, z) = xy + yz + x^2 + y^2 + z^2$, as we know the maximum value of $x^2 + y^2 + z^2 = 1 \,$, I suggest you try and maximize the function $g(x, y, z) = xy + yz$. You get the same result as I mentioned above but it makes things much simpler to solve by hand.

Title: Lagrange multipliers on the unit sphere

Body:
I have a function $f(x,y,z)=x^2+xy+y^2+yz+z^2$ which I'm supposed to maximize on the Unitsphere $B_1(0)=\{x \in \mathbb{R^3}:\Vert x\Vert_2 \le 1\}$ and I have a few questions concerning the process of the computation:

My first question is if I'm allowed to transform the condition into $\{x \in \mathbb{R^3}:\Vert x\Vert_2^2 = 1\}$ so that it's $g(x,y,z)=x^2+y^2+z^2=1 \quad$ I'd assume this would work since the function grows larger for growing $x,y,z$, so the Maximum under my condition will be on the edge of the sphere. Furthermore since it is a unit-vector the norm is equal to the square of the norm.

If this is possible/correct I have some trouble with the algebraic part of the computation. When setting up my functions I'm left with:
$$\begin{align}&I. 2x+y=\lambda2x \\&
II. 2y+x+z=\lambda2y \\&
III. 2z+y=\lambda2z \\&
IV. x^2+y^2+z^2=1
\end{align}$$

Whatever I do, I can't seem to rewrite the equations in such a way that this problem is solvable. The furthest I can show is that $x=z$. Any further rewriting I do yields me something which is dependent on $\lambda$ and one of the other three variables. What am I missing? Thanks in advance.


问:
Solution of Wave Equation initial conditions
A vibrating string fixed at $ x=0 $ and $ x=L $ undergoes oscillations described by the wave equation $ \frac{\partial^2u}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2u}{\partial t^2} $ where $ u(x,t) $ represents the displacement from equilibrium of the string. Initially, the profile of the string is $ u(x,0) = \sin(\frac{4 \pi}{L}x) $ and its initial velocity is $ u_t(x,0)=8 \pi \sin( \frac{4 \pi}{L} x ) $.

My attempt at a solution:

Take $ u(x,t) = [ A \cos(kx) + B \sin(kx) ][ C \cos (kct) + D \sin(kct) ] $

$ u(x,0) = [A \cos(kx) + B \sin(kx)] \times C = \sin( \frac{4 \pi}{L} x) $.

So, $ A = 0; k = \frac{4 \pi}{L} $

Then, $ B \sin({ \frac{4 \pi}{L} x})[ C \cos (\frac{4 \pi}{L} ct) + D \sin( \frac{4 \pi}{L} ct )] $

$ u_t(x,t) = B \sin( \frac{4 \pi}{L} x)[ \frac{4 \pi}{L}c \times C \times -\sin( \frac{4 \pi}{L}ct) + \frac{4 \pi}{L} \times D \times  \cos( \frac{4 \pi}{L} ct) ] $

$ u_t(x,0) = \frac{4 \pi}{L} c BD \sin( \frac{4 \pi}{L} x ) = 8 \pi \sin ( \frac{8 \pi}{L}x ) $

How do I solve this? Can I take $ BD = \frac{L}{c} \times 4 \cos( \frac{4 \pi}{L} x ) $

答:
You won't get $\sin(\frac{8\pi}{L}x)$. You are actually dealing with $\cos(\frac{4\pi}{L}c 0)=1$.

Update: I see that you may have a typo in the initial velocity condition which should have contained $\sin(\frac{8\pi}{L}x)$.

To handle that begin with the solutions as $\sum_n a_n e^{i\frac{2\pi n}{L}(x-ct)}+b_ne^{i\frac{2\pi n}{L}(x+ct)}$. With the initial conditions you can easily recognize the only coefficients left would be $a_2,b_2,a_4,b_4$. With the constraints of getting real solutions, you can see it is
$\frac12 \sin(\frac{4\pi}{L}(x-ct))+\frac12\sin(\frac{4\pi}{L}(x-ct) + \frac{L}{2c}\cos(\frac{8\pi}{L}(x-ct))-\frac{L}{2c}\cos(\frac{8\pi}{L}(x+ct)).$


问:
Index of a Subgroup with normal subgroup
I have read in a book, about the index that
$|G:H|=|G/N:H/N|$, with  $G$ some group, $H\subset G$ a subgroup and $N$ is an normal
subgroup of $G$ and $H$ contains $N$.
So why this equation holds, if $G$ ist finite this is easy to proof by Lagrange but is there an argument for infinit groups?
Straight without proof directly $|G:H|\leq |G/N:H/N|$ and $|G:H|\geq |G/N:H/N|$.

答:
The index $|G : H|$ is the number of cosets of $H$ (assume this is finite, even if $G$ and $H$ are not).

$\overline{H} = H/N$ is a subgroup of $\overline{G} = G/N$, and $|\overline{G} : \overline{H}|$ is the number of cosets.

You want to show a bijection between the two sets of cosets. For a coset $gH$ (of $H$ in $G$), the natural obvious choice is to associate it with the coset $\bar{g} \overline{H}$ (of $\overline{H}$ in $\overline{G}$, where $\bar{g}$ is the image of $g$ in $G/N$). Again, you want to show this is a bijection (without appealing to the finiteness of $H$ or $G$).

It is obvious that this map is surjective, since the map $G \to G/N$ is surjective.

To show it's injective, suppose that $\bar{x} \overline{H} = \bar{y} \overline{H}$ (and we want to conclude that $xH = yH$). The fact that $\bar{x}$ and $\bar{y}$ represent the same coset means that $\bar{x} \bar{y}^{-1} \in \overline{H}$. In other words, if $\pi \colon G \to G/N$ is the quotient map, then $\pi(xy^{-1}) \in \overline{H}$. If $\pi(xy^{-1}) = \bar{h} \in \overline{H}$, then $xy^{-1} \in hN$ (where $h \in H$). Since $N \subset H$, this shows that $xy^{-1} \in H$, and so $xH = yH$.


问:
If $C\supseteq A$ is finite over $A$, then any subring $A\subseteq B\subseteq C$ is finite over $A$
Let $K$ be a field and let $A, B,C$ be $K$-algebras such that:

$A \subset B \subset C$

$A\subset C$ is a finite extension.

$B \subset C$ is a integral extension.

So $C$ is a finitely generated $A$-module (why?) and this implies (why?) $B$ is a finitely generated $A$-module.

I asked my teacher the second question and she replied that it is a result present in Atiyah-Macdonald.
However, I cannot find it in the Chapter 2 dedicated to modules (neither in paragraph 1 for modules in general, nor in paragraph 4 for finitely generated modules, nor in paragraph 9 for algebras). Can any of you help me understand this or can you tell me which result to refer to?

答:
This is not true in general. A remark first: as I noted in my comment, to say that $A\subseteq C$ is a finite extension means by definition that $C$ is a finitely-generated $A$-module, so this answers your first question. Also note that this makes condition $3$ in your question superfluous; if $C$ is a finitely generated $A$-module, then it is integral over $A$, and thus in particular is integral over $B\supseteq A$.

In any case, let $A$ be any non-Noetherian $K$-algebra, and let $I$ be any non-finitely generated ideal of $A$. (For instance, take $A=K[y_i]_{i\in\mathbb{N}}$ and $I=(y_i)_{i\in\mathbb{N}}$, although the specific choice of $A$ and $I$ is not important.) Then let $C={A[x]}\big/{(x^2)}$. Certainly $C$ is a finitely generated $A$-module, generated by the elements $\left\{\overline{1},\overline{x}\right\}$. Also, $I\overline{x}=\left\{\overline{ix}\right\}_{i\in I}$ is an ideal of $C$, and so, by the second isomorphism theorem, $$B:=A+I\overline{x}=\left\{\overline{a+ix}:a\in A,i\in I\right\}$$ is a subring of $C$. However, note that $A\cap I\overline{x}=0$, and so $B=A\oplus I\overline{x}$ as an $A$-module. Since $I\overline{x}\cong I$ is not a finitely-generated $A$-module, this means that $B$ is not a finitely generated $A$-module either, and thus we have a counterexample. (The projection of any generating set of a module onto a direct summand of the module is a generating set of the summand. In particular, if $B$ were finitely generated, then $I\overline{x}$ would be too, a contradiction.)

The desired result does hold if $A$ is Noetherian, and in that case is essentially an immediate corollary of the Artin-Tate lemma. In fact, I suspect your professor was probably thinking of this lemma in her comment to you; a proof of it may be found in proposition 7.8. of Atiyah-Macdonald, if you'd like a specific reference.

So, assume $A$ is Noetherian. Since $C$ is a finitely generated $A$-module, it is in particular a finitely generated $B$-module. Also, again since $C$ is a finitely generated $A$-module, it is in particular a finitely generated $A$-algebra. Thus, by Artin-Tate, $B$ is also a finitely generated $A$-algebra.

Now, as noted above, $C$ is integral over $A$, so likewise $B$ is integral over $A$. Then $B$ is an integral and finitely generated $A$-algebra, and thus is a finitely generated $A$-module, as desired. (Slogan: "integral and finitely generated as an algebra implies finitely generated as a module." This is a standard result and can be proved by induction, but if you need a proof of it let me know.)


答:
Result: If $k$ is an odd integer greater than $1$, then $\log_2(k)$ is irrational.

Proof: Suppose, to the contrary, that $\log_2(k)$ is rational. In other words, $$\log_2(k)=\frac{a}{b}$$ for some $a,b\in\mathbb{Z}$ with $b\neq 0$. Exponentiating both sides of the above equation base 2, we have
\begin{align*}
k &= 2^{a/b} \\
k^b &= 2^a \\
&= 2\cdot 2^{a-1}.
\end{align*}
Since $2^{a-1}\in\mathbb{Z}$, $k^b$ must be even. However, $k$ is odd, so this is impossible.$\ \ \ \ \ \blacksquare$

Title: How do I show that $\log_2(k)$ is an irrational number if $k$ is an odd integer greater than $1$?

Body:
Show that if $k$ is an odd integer grater than $1$, then $\log_2(k)$ is an irrational number.

So my approach for this problem was to use proof by contradiction so I wrote "Assume by contradiction if $k$ is an odd integer greater than $1$ than $\log_2(k)$ will be a rational number. Such that $\log_2(k)=a/b$, $\gcd(a,b) = 1$ and $a,b$ are integers in its lowest terms. But I don't know what to do after. Should I use the fundamental theorem of arithmetic? Euclid's lemma?


问:
let $A$ be a matrix $m\times n$ if The colums of $A$ Span $\mathbb{R}^m$ then the equation $Ax=b$ will have an answer for all $b\in \mathbb{R}^m$
I can't understand why this is true, Intuitively, I thought that it should be the rows of A span all $\mathbb{R}^m$ , so they don't become zeros while row deducting.
Can someone help me understand this?
Thanks

答:
Fix $\vec b\in\mathbb R^m$. Let $\vec a_1,\ldots,\vec a_n\in\mathbb R^m$ be the columns of $A$. If $\vec a_1,\ldots,\vec a_n$ span $\mathbb R^m$, then we can write $$\vec b=x_1\vec a_1+\cdots+x_n\vec a_n$$ for some $x_i\in\mathbb R$. But this just says that there exists $\vec x=(x_1,\ldots,x_n)\in\mathbb R^n$ such that $A\vec x=\vec b$.

Basically, $A\vec x$ is a linear combination of the columns of $A$ where the coefficients are the entries of $\vec x$.


问:
If we have multiple unique RSA signed messages produced from the same private key, can we always find new valid signatures?
Say Alice publishes $t$ unique signed messages with her private key $d$,

$$
m_1^d, m_2^d, m_3^d, ... , m_t^d \mod n
$$

For $t = 2$, we have $m_1^d, m_2^d \mod n$ and if we let $x \equiv m_1m_2 \mod n$ then we can produce a valid signature,

$$
x^d \equiv (m_1m_2)^d \equiv m_1^d\times m_2^d \mod n
$$

For $t$ signatures, I think we can have at most $\sum_{i=2}^{t}\binom{t}{i}$ new signatures. Is this correct?

This was inspired by a quiz question in which Alice published $t$ RSA signatures and we were asked to find a valid signature for $M \equiv m_1^{-1} \times m_2^{-1} \times m_3^{-1} \times ... \times m_t^{-1} \mod n$ and the solution was,

$$
M^d \equiv (m_1^{-1} \times m_2^{-1} \times m_3^{-1} \times ... \times m_t^{-1})^{d}
\equiv (m_1^{d} \times m_2^{d} \times m_3^{d} \times ... \times m_t^{d})^{-1} \mod n
$$

We assumed that all inverses existed. So, I was wondering why did this question involve inverses? Are they necessary to produce new signatures? Because as I showed above, I don't see why we could not do that otherwise.

P.S. I also asked this question the cryptography stack community and received an answer.

答:
For $t$ signatures, I think we can have at most $\sum_{i=2}^{t}\binom{t}{i}$ new signatures. Is this correct?

There is no restriction for repetitions like $m_1 \cdot m_1$ has a signature as $(m_1\cdot m_1)^d$. Also, once we forged a signature, the signature space in your hand is probably increased. We can forge maybe all of the messages on the input space of the RSA modulus and maybe not depending on the messages given and the modulus ( see important note). Remember RSA works in modulus so $m_1\cdot m_1 = x \bmod n$.

So, I was wondering why did this question involve inverses? Are they necessary to produce new signatures? Because as I showed above, I don't see why we could not do that otherwise.

No, the inverse is not necessary for forging signatures as you noticed. It seems that they asked it only for educative purposes for you to notice that the only messages that don't have the inverses are $0,p,$ and $q$. Remember, for an $x \in \mathbb{Z}_n^*$ to have an multiplicative inverse we need the $\gcd(m,x)=1$ ( defining $\gcd(0,m)=0$ is problematic).

Important note: The forgery is worked since you are working with the textbook RSA. It is not secure as you can see you can forge messages. First of all, we don't sign messages rather their hashes known as hash-and-sign paradigm. So as long as the used hash cryptographically secure the signature multiplication will not work, which may only for negligible probability. Actually, the hash-and-sign is necessary for the security of the signature and known from the first signature scheme the Rabin Signature Scheme. In practice, the RSA is used with a Probabilistic Signature Scheme (PSS) that is designed and the security proof is reduced to RSA problem by Mihir Bellare and Phillip Rogaway. It is standardized in RFC 8017 as RSASSA-PSS.


问:
Calculus Made Easy Exercise 10 Question 10
I am currently stuck on this question. The question and my steps so far are given below:

(10) Suppose it to be known that consumption of coal by a certain steamer may be represented by the formula:
$$y = 0.3 + 0.001v^3$$
where $y$ is the number of tons of coal burned per hour and v is the
speed expressed in nautical miles per hour. The cost of wages, interest on capital, and depreciation of that ship are together equal,
per hour, to the cost of 1 ton of coal. What speed will make the
total cost of a voyage of 1000 nautical miles a minimum? And,
if coal costs 10 dollars per ton, what will that minimum cost of
the voyage amount to?

I am stuck on the first part. Here are my steps so far:

Let price of one ton of coal $= c$

wages + interest + depreciaition = T

According to the question: $T \times t = c$
If distance = 1000, $v = \frac{1000}{time}$

basically $y = \frac{tonnes}{time}$ so :
$$\frac{tonnes}{time} = 0.3 + 0.001\frac{1000^3}{time^3}$$
I solved this equation but the wring answer for v. I also do not think I am doing it right.

Please do not give away the answer but guide me in the right direction.

Many thanks and stay safe!!

答:
Edit

Well, this is embarassing.  I went off the rails, and made a mistake in analysis.

I have corrected the mistake below.

Hint

Although your analysis was a good step in the right direction, you should capitalize on :

The cost of wages, interest on capital, and depreciation of that ship are together equal, per hour, to the cost of 1 ton of coal.

Edit

I misinterpreted the meaning of the above constraint.  Further, there is an ambiguity here.  Are you supposed to minimize the total of [wages + interest + depreciation], or are you supposed to minimize the total of $\{$ [wages + interest + depreciation] + coal used $\}?$

Under the first interpretation, you are supposed to minimize the time taken for the voyage.  It is doubtful that this would be the intended solution.

Under the second interpretation, the cost per hour is
$(1 + y)$, rather than $y$.  I suspect that this is the intent of the problem.  Contrast that with the mistake that I made below, which (wrongly, indirectly) presumes that the cost per hour is $y$, instead of $(1 + y).$

This means that you do not need to use any variables other than $y$ and $v$.

Distance = rate $\times$ time.

The time for the voyage will be $[(1000)/v],$ so the amount of coal consumed will be $[(1000/v] \times y.$

Take both the first and second derivatives of the function [in the sole variable $v$] that represents the amount of coal that will be consumed on the voyage.

You want the first derivative to be zero and the second derivative to be positive.  Then, you will have identified the value for $v$ that represents the minimum value of the function.  From this, you can routinely determine the time of the voyage.


问:
Examples of functions with a removable discontinuity
I recently came across the following example of a function which is discontinuous at a point:

The cost of a sushi buffet is calculated by weighing your plate, where you are then charged £10 per pound. As part of a promotion, if you manage to fill your plate so that the weight is precisely 1lbs, you get your meal for free.

This is a nice way to motivate a function like
$$f(x) = \begin{cases}10\,x\hfil&\text{$x\neq 1$}\\\hfil0\hfil&\text{$x=1$}\end{cases}$$
which are nice for illustrating the idea of a limit to students.

Are there any other real-world examples of functions like this, for the sake of explaining limits? Ideally one where you would actually want the limiting value, and not the assigned value, to motivate the concept further. (Apart from the obvious difference quotient.)

答:
For each math exercise with a single number as an answer, the function that gives you your grade depending in what you write is full marks if you get the number right, and $0$ if you don't. Students will probably be used to this one.

In speedrunning of games, where you have to beat a game the fastest way possible, there are sometimes tricks called pixel perfect, where you have to do an action in a precise pixel or it won't work. Say for example the game is Super Mario Bros, and jumping in a particular pixel will let you beat the stage in 10 seconds, while missing the jump will let you beat it in 20s. The function of how much time you spend in the stage depending in what pixel you jump has a removable discontinuity.

Consider two wi-fi routers that give you internet if you're a distance strictly less than 5 meters around them. In your house there is a long hallway of 10 meters, so you put one in each extreme of the hallway. The amount of internet you'll receive in each point will be the amount the routers give except exactly at the middle, where you'll get zero because you're 5 meters away from each one.

Some students (or anyone) could complain that these examples don't have a continuous variable (in the pixels example, or the original weight question if you take into account we can only measure a finite number of possible weights) or that the wi-fi routers don't usually behave like that, or that if you write the answer up to 20 decimals points you'll likely get a full mark in the exercise even if it's not the exact number. And I'd argue they're absolutely right, that's a consequence of living in a real world where space and almost all of our measurements of physical things can be discretized.


问:
Trying to find mass of volume: apparently my bounds are incorrect but I don't see why

This is what I got: $\int_{0}^{1} \int_{-1}^{1} \int_{0}^{1-x}  \,4dzdydx$

Looking at the domain D on the xy plane, the lower bound for x is 0, as given, and I set z = 0 in x + z = 1 to get x = 1 as the upper bound. Then I set z = 0 in z = 1-y^2 to get y = -1 and 1; these are y's bounds. Finally, we know z = 0 as the lower bound and the upper bound is x + z = 1 => z = 1-x. Apparently my answer is wrong: can someone explain?

答: For each $(x,y,z)$ in that region, $x$ can take any value from $0$ to $1$ and $z$ can take any value from $0$ to $1-x$. Finally, $y$ can take any value from $-\sqrt{1-z}$ to $\sqrt{1-z}$. So, compute$$\int_0^1\int_0^{1-x}\int_{-\sqrt{1-z}}^{\sqrt{1-z}}4\,\mathrm dy\,\mathrm dz\,\mathrm dx.$$


问:
How can I use the Weierstrass M-test to show that $f(x)=\sum_{n=1}^{\infty}\frac{x^n}{\sqrt{n}}$ is continuous at $x:0 \in (-1,1)$?
I know that $f(x)=\sum_{n=1}^{\infty}\frac{x^n}{\sqrt{n}}$ converges for every $x \in [-1,1)$ but not for $x=1$. How can I use Weierstrass M-test to show that $f$ is continuous at a fixed $x_0 \in (-1,1)$?

答:
Let $x_0\in(-1,1)$. If you find some interval around $x_0$ where the series above converges uniformly, then the limit must be continuous (as the uniform limit of continuous functions). To show uniform convergence, you can use the Weierstrass M-test - choose your small interval, and then find a uniform bound on the series in that interval.

Since $x_0\in (-1,1)$, there is some $\epsilon>0$ such that $(x_0-\epsilon,x_0+\epsilon)\subset(-1,1)$. Choose this to be your interval. Let $M=max\{|x_0+\epsilon|,|x_0-\epsilon|\}$. Then by comparing your series to the series $\sum_{n=1}^{\infty}\frac{M^n}{\sqrt{n}}$ (and recalling that $M<1$), you see by the Weierstrass M-test that your series of functions converges uniformly in the interval you chose, which means that the limit function of the series is continuous at $x_0$.

Note that you can use other arguments which are a bit simpler by using the notion of radius of convergence, but if you want to use Weierstrass, this is a way.


问:
Riemann-Stieltjes integral with respect to a discontinuous function
I have to find if the function $f(x)=x^2$ is Rieman-Stieltjes integrable respect to the function $g(x)=3x$ if $x\in[0,1)$ and $g(1)=4$. Now, because $f$ is continuous and $g$ is of bounded variation, the integral indeed exists. To find the value of the integral, I chose to find the integral:

$$\int_0^1f(x)d(g(x))-\int_0^1f(x)d(3x)$$

By definition of $g$, this is:

$$\int_0^1f(x)d(h(x))$$

Where $h(x)=g(x)-3x$, so $h(x)=0$ for $x\in[0,1)$ and $h(1)=1$

Finding it by definition, if $\epsilon>0$, and I chose the partition $P=\{0,1-\epsilon,1\}$, then:

$S(P,f,h)=f(0)(h(1-\epsilon)-h(0))+f(1-\epsilon)(h(1)-h(1-\epsilon))=f(1-\epsilon)=(1-\epsilon)^2$
From here, I could conclude that:

$$\int_0^1f(x)d(h(x))=1$$
Is this correct? Because intuitively I thought it was going to be $0$, or did I do something wrong?.

答:
Since $f$ is increasing, the lower and upper sums for the partition $\{0,(1- \epsilon),1\}$ are

$$L(P,f,h) = f(0)(h(1-\epsilon) - h(0)) + f(1-\epsilon)(h(1) - h(1-\epsilon))\\=0 \cdot (0 - 0) + (1-\epsilon)^2(1 - 0)= (1-\epsilon)^2, $$

and

$$\\U(P,f,h)=  f(1-\epsilon)(h(1-\epsilon) - h(0)) + f(1)(h(1) - h(1-\epsilon)) \\ (1-\epsilon)^2 (0 - 0) + (1)^2(1 - 0)= 1,$$

The integral must fall in between the lower and upper sums,

$$(1-\epsilon)^2 \leqslant \int_0^1 f \, dh \leqslant 1$$

Since the limits as $\epsilon \to 0$ of the LHS and RHS are both $1$, it follows that the integral has a value of $1$.

Thus,

$$\int_0^1f \, dg = \int_0^1 x^2\, d(3x) + \int_0^1 f \, dh  = 3 \int_0^1 x^2 \, dx + 1 = 2$$

In general, to compute the value of the integral with Riemann-Stieltjes sums, you should take the limit of a sequence of sums corresponding to partitions $P_n$ with $\|P_n\| \to 0 $ as $n \to \infty$.

Taking $P_n = \{0,\frac{1}{n},\frac{2}{n}, \ldots, \frac{n-1}{n},1\}$, we have

$$S(P_n,f,g)= \sum_{k=1}^{n-1} \left(\frac{k}{n} \right)^2\left(3\frac{k}{n} - 3\frac{k-1}{n} \right)+ (1)^2 \left(4 - 3\frac{n-1}{n} \right) \\ = \frac{3}{n^3}\sum_{k=1}^{n-1}k^2+ 1  + \frac{3}{n} ,$$

and we have

$$\int_0^1 f \, dg = \lim_{n \to \infty}\left(\frac{3}{n^3}\sum_{k=1}^{n-1}k^2+ 1  + \frac{3}{n}\right) = 3\cdot \frac{1}{3} + 1 = 2$$


问:
Integration problems
Can anyone help me with these:-

(a)Prove by induction: $\displaystyle\sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$

(b) By explicitly calculating upper and lower Riemann sums on a uniform partition and
passing to the limits, calculate $\displaystyle\int_0^1 x^2 \mathrm{d}x$.

答: Hint: Use $$\int_a^b f(x)dx=\lim_{n\to\infty}\sum_{i=0}^n f\left(a+i\frac{b-a}{n}\right)\frac{b-a}{n}$$ Here $f(x)=x^2, a=0,b=1$ and so $$f\left(a+i\frac{b-a}{n}\right)=x^2\Big|_{x=0+\frac{i}n}=i^2/n^2$$ and so...</p>

Title: Integration problems


问:
How can I plot my curve, so it can be seen that it is a spiral?
I want to plot the parametric curve $ [\exp(-1/(1-t) )\cos(1/(1-t)) , \, \exp(-1/(1-t)) \sin(1/(1-t)) ]$ for $ t \in [0,1] $ in Maple or Matlab. It should look like a spiral which winds infinitley often around $(0,0)$, but when I plot it, all I can get is an arc which goes to $ (0,0) $ and does not look like a spiral. Is there a way to plot this curve, to see the behaviour of a spiral?

答:
This works fine for me in MATLAB R2016b, as far as I know it doesn't use anything version-specific.  I don't know Maple well enough to give a Maple version.

```<code>t = linspace(0,1,1000);
x = exp(-1./(1-t)).*cos(1./(1-t));
y = exp(-1./(1-t)).*sin(1./(1-t));
plot(x,y)
</code>```

This gave me the following graph, which shows more of the spiral the more I zoom in.  The spiral shrinks so fast that you can't see more than about this much of it any any given zoom level.  For completeness sake, here is the link to the Desmos demo showing the same.

Body:
This works fine for me in MATLAB R2016b, as far as I know it doesn't use anything version-specific.  I don't know Maple well enough to give a Maple version.

t = linspace(0,1,1000);
x = exp(-1./(1-t)).*cos(1./(1-t));
y = exp(-1./(1-t)).*sin(1./(1-t));
plot(x,y)

This gave me the following graph, which shows more of the spiral the more I zoom in.  The spiral shrinks so fast that you can't see more than about this much of it any any given zoom level.  For completeness sake, here is the link to the Desmos demo showing the same.


问:
baire $1$ function
Here is a new definition of Baire Class one function. Suppose that $X$ is a complete separable metric space. A function $f:X \rightarrow \mathbb{R}$  is said to be Baire class one if for any $\epsilon>0$, there exists $\delta:X \rightarrow \mathbb{R}$ such that for any $x,y \in X$,
$$d(x,y)<\delta(x) \wedge \delta(y) \Rightarrow |f(x)-f(y)|<\epsilon$$

Suppose that $f:\mathbb{R} \rightarrow \mathbb{R}$ given by

\begin{cases}
      0 & 0 \leq x < 1 \\
      1 & x=1 \\
   \end{cases}

Clearly $f$ is a Baire one function, as it is the pointwise limit of the sequence of continuous functions $f_n(x)=x^n$. However, I fail to verify that it is a Baire one function using the definition given above, that is, unable to find $\delta:\mathbb{R} \rightarrow \mathbb{R}$. Can anyone help?

答:
Rather old and easy question to remain unanswered for this long.  I was searching for something else and this popped up.  May as well finish it off. I expect the poster  Mr Idonknow is, after all this time, probably Mr Iamuchbetterinformed and not so interested anymore.

Consider the function (similar to the one requested):
$$f(x) =  \begin{cases}
      0 &  x < 0\\
      1 & x\geq 0 \\
   \end{cases}$$

This is clearly Baire 1.   Easy to find a sequence of continuous functions converging pointwise to it.   Or it is easy to investigate the sets
$\{x:  f(x)<\alpha\}$.

Let us check how to use this equivalent definition.

Let $\epsilon>0$.    For each  $x\not= 0 $  choose  $ \delta(x)=|x|/2$  and then choose   $\delta(0)=1$.    Note that if

$$d(x,y)< \min\{\delta(x),  \delta(y) \}$$

then either both  $x$  and $y$ are negative or both $x$ and $y$ are positive.  It follows that   $|f(x)-f(y)| = 0 < \epsilon$.  From that criterion it follows that the function is Baire 1.

So why did I hesitate on this trivial little problem?    This particular expression using a guage $\delta$ in relation to Baire one functions (and other properties) occurs repeatedly in this paper:

A. M. Bruckner, R. J. O'Malley and B. S. Thomson. Path Derivatives: A Unified View of Certain Generalized Derivatives.
Transactions of the American Mathematical Society, Vol. 283, No. 1 (May, 1984),  97-125.

In fact Theorem 5.2 of that paper could also be used to show this function is Baire 1 (but that would be overkill).

Title: baire $1$ function


问:
Let $(x_n)_n$ be a sequence of positives st $\exists n_0$ st $0<\frac{x_{n+1}}{x_n}\leq c<1,\ \forall n\geq n_0$. Show that $\lim_{n\to\infty}x_n=0$
I can show that this sequence converges because it is monotonic and bounded, but I'm having trouble showing that the limit is 0.

My attempt:

Let $\lim_{n\to\infty}x_n = L$, thus $$\forall\epsilon>0\exists n_0\in\mathbb{N}\forall n\geq n_0 \ (|x_n - L|<\epsilon)$$
For each natural $k$, let $\epsilon=\frac{c}{k}$, hence there exist $n_{0,k}$ s.t. $\forall n>n_{0,k}$

$$L-\frac{c}{k} < x_n < L+\frac{c}{k}$$
and
$$L-\frac{c}{k} < x_{n+1} < L+\frac{c}{k}$$

but $$0<x_{n+1} < cx_n<x_n$$
so
$$L-\frac{c}{k} < cx_n < L+\frac{c}{k}$$
$$\frac{L}{c}-\frac{1}{k} < x_n < \frac{L}{c}+\frac{1}{k}\implies |x_n-\frac{L}{c}|<\frac{1}{k} ,\ \forall n\geq n_{0,k}$$

I thought of using this to somehow show that $L$ cannot be greater than 0 by making $k$ large enough, but don't know exactly how.

答:
This comes down to showing that
$c^n \to 0$.
To do this, let
$c=1/(1+b)$
where $b > 0$.

Then, by Bernoulli's inequality,
$(1+b)^n \ge 1+nb
\ge nb
=n(1/c-1)
= nc/(1-c)
$

so
$c^n < (1-c)/(nc)$.


答:
You can set it up as below -

$\displaystyle \int_0^1\int_0^{1-x}\int_{-\sqrt{1-z}}^{\sqrt{1-z}} 4 \, dy\, dz \, dx$

or

$\displaystyle \int_0^1\int_{-\sqrt{1-z}}^{\sqrt{1-z}}\int_{0}^{1-z} 4 \, dx \, dy \, dz$

The problem in setting it up wrt $dz$ first the way you did is that the upper bound of $z$ is the plane and that will not give the correct bound for $y$.

Title: Trying to find mass of volume: apparently my bounds are incorrect but I don't see why

Body:
This is what I got: $\int_{0}^{1} \int_{-1}^{1} \int_{0}^{1-x}  \,4dzdydx$

Looking at the domain D on the xy plane, the lower bound for x is 0, as given, and I set z = 0 in x + z = 1 to get x = 1 as the upper bound. Then I set z = 0 in z = 1-y^2 to get y = -1 and 1; these are y's bounds. Finally, we know z = 0 as the lower bound and the upper bound is x + z = 1 => z = 1-x. Apparently my answer is wrong: can someone explain?


问:
What is the Lipschitz constant of a quadratic form?
Let $f(x) = -x^{T} A x$ for a positive definite matrix $A$.  Let the domain of $x \in R^N$ be in a compact ball (with radius > $\sqrt{N}$ if necessary)

What is the Lipschitz constant for $f$ in terms of (presumably spectral) properties of $A$?  Is it the maximum eigenvalue of $A$?

答:
Any continuously differentiable function over a compact domain is Lipschitz continuous with Lipschitz constant equal to the maximum magnitude of the derivative.

The function $f(x) = x^TAx$ is differentiable with $\nabla f(x) = 2Ax$. It follows that over the ball of radius $R$, the Lipschitz constant of $f$ is equal to
$$
\max_{x \in B_R} \|2Ax\| = 2R\cdot \max_{\|x\| \leq 1} \|Ax\| = 2R\cdot \rho(A),
$$
where $\rho(A)$ denotes the spectral radius of $A$ (the maximal absolute value among the eigenvalues of $A$).


答:
The Riemann sum for that integral is

$$\frac{1}{n} \sum_{k=1}^n \frac{k^2}{n^2} = \frac{1}{n^3} \sum_{k=1}^n k^2$$

Apply the formula from part (a) and take the limit as $n \to \infty$.

For part (a), you need to show that

$$\frac{n(n+1) (2 n+1)}{6} + (n+1)^2 = \frac{(n+1)(n+2)(2 n+3)}{6}$$

Title: Integration problems

Body:
Can anyone help me with these:-

(a)Prove by induction: $\displaystyle\sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$

(b) By explicitly calculating upper and lower Riemann sums on a uniform partition and
passing to the limits, calculate $\displaystyle\int_0^1 x^2 \mathrm{d}x$.


问:
Why is $R/I$ cyclic?
The question is in the title, in Rotman's "An introduction to homological algebra", page 125 is the next proposition:

After the diagram, it says that $R/I$ is cyclic, why ?

答: Because it is generated by $1+I$.

Title: Why is $R/I$ cyclic?

Body: Because it is generated by $1+I$.