答:
The cleanest proof of this I know is to use a Mayer-Vietoris sequence. You can do directly on a square using relative homology, but it's a bit cleaner to note that the original statement is equivalent to asking that for any pair of open subsets $A$ and $B$ of the cylinder $S^1\times I$ such that $A$ contains $S^1\times \{0\}$ and $B$ contains $S^1\times \{1\}$, there exists some loop $A\cap B$ which is not contractible in $S^1\times I$. You can get this equivalence by gluing the top and the bottom of the square together and adding in a strip along this glue line which is in both $A$ and $B$. There's some little details to take care of, but it's possible to ensure that the subspace of the cylinder outside of the extra strip in both $A$ and $B$ is just your original square with its original subsets. Once you have your non-trivial loop, you can find that it must, at some point, traverse over the part of the cylinder that is your original square - which gives you your answer.

That taken care of, we can just write out one part of the Mayer-Vietoris sequence for $A$ and $B$:
$$H_1(A\cap B)\overset{f}{\rightarrow} H_1(A)\oplus H_1(B) \overset{g}{\rightarrow} H_1(S^1\times I)$$
where the first map $f$ is the sum of the two embeddings $A\cap B\rightarrow A$ and $A\cap B\rightarrow B$ and where the second map $g$ takes $h_1\oplus h_2$ to the difference of the embeddings of $h_1$ and $h_2$ into $H_1(S^1\times I)$.

The image of this first map must equal the kernel of the second. However, if we let $\alpha\in H_1(A)$ be the loop around $S^1\times \{0\}$ and $\beta\in H_1(B)$  be the similarly oriented loop around $S^1\times \{1\}$, it's clear that these two loops embed to the same thing in $H_1(S^1\times I)$, hence $\alpha\oplus \beta \in \ker g$. As a result, there is some class $h\in H_1(A\cap B)$ such that the embedding of $h$ into $H_1(A)$ gives $\alpha$. Clearly, any chain in the homology class of $h$ contains some loop which is non-trivial in $S^1\times I$, so we are done.

Title: Existence of a path between two sides of a square, which is contained in the intersection of two open subsets of it

Body:
Denote by $I$ the interval $[0,1]$. Let $A,B$ be two open subsets of the square $I\times I=I^2$ such that $A\cup B=I^2$, $\{0\}\times I\subseteq A$ and $\{1\}\times I\subseteq B$. In more geometric terms, the left edge of the square is contained in $A$ and the right edge is contained in $B$.

My question is: does there exist a continuous path $\gamma:I\to A\cap B$ such that $\gamma(0)\in I\times\{0\}$ and $\gamma(1)\in I\times\{1\}$? In other words, is there a curve connecting the two horizontal sides of the square, whose trace is contained in $A\cap B$ ?


问:
Show that $8^3>9^{8/3}$.
I am trying to show algebraically that $8^3>9^{8/3}$. This came from trying to complete the base case of an induction proof.

I have struggled because $8$ and $9$ cannot be manipulated to be the same base. Otherwise I could just argue that $3>\dfrac{8}{3}$.

I tried raising both sides to the third power and got $8^9>9^8$. I can rewrite this as $8^9>9^{9-1}$ but I am not sure if this is the right direction.

答:
An approach that is alternative to (but inferior to) David Lui's answer is if you happen to know that

$$\log_{10} ~2 \approx 0.301 ~~\text{and}~~ \log_{10} ~3 \approx 0.477.$$

Then, you simply compare

$$0.301 \times (3 \times 3) ~~\text{vs}~~ 0.477 \times [2 \times (8/3)].$$

ContentLicense: CC BY-SA 4.0

Title: Show that $8^3>9^{8/3}$.


问:
prove the unit interval has outer measure 1.
For a subset $E \subset \mathbb{R}^d$ define the outer measure

$$m_*(E)=\inf \{\Sigma \vert I_j \vert\}$$

Where the infimum is taken over all countable coverings of $E$ by closed intervals. So $E \subset \bigcup_{j=1}^\infty I_j$.

Show $m_*([0,1])=1.$

How do I go about this rigorously? What would my $I_j$ be?

How about for general $[a,b]$ how can I show its just $b-a$?

答:
Assuming that $d = 1$ to keep things simple, we have by definition $|[0,1]| = 1$. (More generally, $|[a,b]| = b - a$, and for an elementary set which is a finite union of disjoint intervals we sum up their lengths.) This is the pre-measure that leads to the Lebesgue measure. Note that $| \cdot |$ can be shown to be countably subadditive for elementary sets.

Since $[0, 1]$ is itself a closed interval, we can take $I_1 = [0,1]$, $I_j = \emptyset, \; j \geq 2$ as a particular covering. Then the outer measure satisfies $m^*([0,1]) \leq |[0,1]| = 1$ (as it is an infimum).

Now for the other direction. Consider any countable covering of $[0,1]$ by closed intervals $(I_j)_{j \geq 1}$ (i.e. elementary sets with $[0,1] \subseteq \bigcup_{j \geq 1} I_j$). By countable subadditivity, $\sum_{j \geq 1} |I_j| \geq |\bigcup_{j \geq 1} I_j|$. By monotonicity, $|\bigcup_{j \geq 1} I_j| \geq |[0,1]| = 1$. Putting this together shows that $m^*([0,1]) \geq 1$ (again, because $m^*$ is an infimum).

Hence $m^*([0,1]) = 1$. For general $[a,b]$, which is an interval/elementary set, the proof is essentially identical. (More generally, this says that the Lebesgue outer measure $m^*$ agrees with the Lebesgue measure $m$ on elementary sets, which we call $|\cdot|$ here).


问:
Evaluate $\int_{-\infty}^{\infty} \chi_{[0,1]}(x-y) \chi_{[0,1]}(y) \, \mathrm{d}y$
I'm trying to evaluate the integral $$\int_{-\infty}^{\infty} \chi_{[0,1]}(x-y) \chi_{[0,1]}(y) \, \mathrm{d}y$$
where $\chi_{[0,1]}(x)=1$ is the characteristic function, i.e. equals $1$ for $x \in [0,1]$ and $0$ otherwise. How do I evaluate this integral? So far I've done
$$\int_{-\infty}^{\infty} \chi_{[0,1]}(x-y) \chi_{[0,1]}(y) \, \mathrm{d}y = \int_{0}^1 \chi_{[0,1]} (x-y) \, \mathrm{d}y.$$
I've tried substitution but I don't find it yielding anything useful. The answer should be the function $$f(x)= \left\{
  \begin{array}{l l}
    x & \quad \text{if $x \in [0,1]$}\\
    2-x & \quad \text{if $x \in [1,2]$}\\
    0 & \quad \text{otherwise.}
  \end{array} \right.$$</p>

答:
Change variables $x-y=s$, thus the new domain is $(x,x-1)$. You have
$$
\int_0^1\Xi(x-y)dy=\int_{x-1}^x\Xi(s)ds.
$$
Now, if $x-1>1$ and if $x<0$ the integral vanishes. Moreover, you have
$$
\int_{x-1}^x\Xi(s)ds=\text{length}\left((x-1,x)\cap(0,1)\right).
$$
If $0<x<1$
$$
\text{length}\left((x-1,x)\cap(0,1)\right)=x.
$$
Otherwise, if $1<x<2$
$$
\text{length}\left((x-1,x)\cap(0,1)\right)=1-(x-1)=2-x
$$</p>

ContentLicense: CC BY-SA 3.0


问:
Every abelian normal subgroup of a connected and compact Lie group lies in the center

Show that every abelian normal subgroup $H$ of a connected and compact Lie group $G$ lies in the center of $G$.

It may be helpfull that if $f:G\to H$ is a surjective lie group homomorphism with $H$ a abelian group, then the restriction $f|_T$ of $f$ to any maximal torus $T$ is also surjective.

But here is what I am trying: take $h\in H$ and $g\in G$. As $H$ is normal, one has that there is a $h'\in H$ s.t.
$$gh=h'g$$
My attempt is to show that $h'=h$.

Let $T$ be a maximal torus in $G$. One has that there is $r\in G\setminus T$ and $t\in T$ s.t.
$$rtr^{-1}=h.$$
As $H$ is normal, $t\in H\cap T$.

Thus,
$$(gr)t(gr)^{-1}=h'.$$

But I'm stuck here. Also I'm not using that $H$ is abelian.

答:
To give a more elementary solution: Let $H^0$ be the identity component of $H$.
$H^0$ is compact, connected and abelian, hence it is a torus and is contained in a maximal torus $T$. Thus,
$$H^0 = g H^0 g^{-1} \subset gTg^{-1}$$
Since all maximal tori are conjugates of each other, $H^0$ is contained in all of them and hence in the center of $G$ (This is Theorem 2.3 part (iii) pg. 165 in Theodor Brocker's Representation of Compact Lie Groups)

In the case that $H$ is disconnected, let $h\in H\backslash H^0$ and let $T_h$ be a maximal torus that contains $h$. Observe that $H/H^0$ is a discrete normal subgroup of $G/H^0$ and hence it is abelian. From this we get that for all $g\in G$ there exists an $h_0 \in H^0$ such that:
$$g h g^{-1} = h\, h_0$$
Finally, this means that the set $c_G(h) = \{ g h g^{-1}: \ \forall g\in G\}$ is contained in $T_h$ because both $h$ and $h_0$ are contained in $T_h$. With the same argument as before,
$$c_G(h) = g\, c_G(h) \, g^{-1} \subset gT_h g^{-1}$$
and this means that $c_G(h)$ is contained in all maximal tori of $G$.


问:
A question about Sylow's theorems and index of groups.
I'm trying to prove this exercise:

Let be $G$ a finite group, p a prime number, and $H$ a normal subgroup. If $p \nmid [G:H]$, then $\{ x \in G : o(x)=p^{n_x}, \text{ with } n_x \in \mathbb{N} \} \subseteq H$.

I tried by contradiction. If $B=\{x \in G : o(x)=p^{n_x}, \text{ with } n_x \in \mathbb{N} \} \nsubseteq H$. Then, $\exists \; y \in B$ such that $y \notin H$ and $o(y)=p^{n_y}$. I take $\langle y \rangle \in Syl_p(n_y, G)$ and I tried to use Sylow's theorems but didn't get to much...

I think that $B$ is a Sylow $p$-subgroup, but I can't prove this.

I appreciate any help you can give me.

I use the next:

$o(x):=$ order of $x$

Definition: Let be $G$ a finite group, $p \in \mathbb{N}$ a prime and $n \geq 1$ such that $p^n \mid |G|$ and $p^{n+1} \nmid |G|$. For every $r \in [1,n]$ define:

$a)$ $Syl_p(r, G)= \{H \leq G : |H|=p^r \}$.

$b)$ the set of Sylow p-subgroups  $Syl_p(G):=Syl_p(n, G)$.

答: Why so complicated? If $x \in G$, then for $\overline{x} \in G/H$, we have $order(\overline{x}) \mid |G/H|$ and also $order(\overline{x}) \mid order(x)$. Hence, if $x$ is a $p$-element, $\overline{x}$ is a $p$-element, while $p \nmid |G:H|$, so $\overline{x}=\overline{1}$, that is $x \in H$.


问:
Solve for $x$, $3\sqrt{x+13} = x+9$
Solve equation: $3\sqrt{x+13} = x+9$

I squared both sides and got $9 + x + 13 = x^2 + 18x + 81$

I then combined like terms $x^2 + 17x + 59 = 0$

I then used the quadratic equation $x= -\frac{17}2 \pm \sqrt{\left(-\frac{17}2\right)^2-59}$

However, the answer is 3

答:
Given: $\boxed{3\sqrt{x+13} = x+9}$

Isolate the radical, x's cancel:

$\dfrac{3\sqrt{x+13}}{3} = \dfrac{x+9}{3}$

$\sqrt{x+13} = \dfrac{x+9}{3}$

Square both sides:

$\left(\sqrt{x+13}\right)^2 =\left(\dfrac{x+9}{3}\right)^2$

$x+13 = \left(\dfrac{x+9}{3}\right)\left(\dfrac{x+9}{3}\right)$

Move the 9 to the other side:

$9(x+13) = \dfrac{x^2+18x+81}{9}$

Set the equation equal to zero and group like terms:

$9x+117 = x^2+18x+81$

$0 = x^2+18x-9x+81-117$

Factor:

$0 = x^2+9x-36$

$0=(x+12)(x-3)$

$0=x+12\implies \boxed{x=-12}$
$0=x-3\implies\boxed{x=3}$

We can check to see if these are solutions:

$3\sqrt{-12+13} = -12+9$

$3(1)=-3$

$\boxed{3 \neq -3}$

Therefore, -12 is not a solution to the original equation.

$3\sqrt{3+13} = 3+9$

$3(4)=12$

$\boxed{12= 12}$

Therefore, 3 is the solution to the original equation.

Title: Solve for $x$, $3\sqrt{x+13} = x+9$


问:
What is the integration of this function?
$$\int_0^1 \frac{1}{({1-x^6})^{1/6}} dx $$
I have been trying to integrate this function on given limit but I am struggling to find a way to start. How do I integrate this and such type of functions.

Body:
Substituting $u=1-x^6$ gives
$$I=\frac16\int_0^1u^{-1/6}(1-u)^{-5/6}\,du=\frac16B(1/6,5/6)=\frac\pi{6\sin\pi/6}=\frac\pi3$$
More generally
$$\int_0^1\frac1{(1-x^n)^{1/n}}\,dx=\frac\pi n\csc\frac\pi n$$


问:
determining the formula of equations similar to ${2}^{x-1}$
the formula for

$1,2,3,4,5,6,7\space\text{is}\space{x}\\
1,2,4,7,11,16,22,29\space\text{is}\space\frac{x\left(x-1\right)+2}{2}\\
1,2,4,8,15,26,42,64,93\space\text{is}\space\frac{x\left(x\left(x-3\right)+8\right)}{6}\\
1,2,4,8,16,31,57,99,163,256\space\text{is}\space\frac{x\left(x\left(x\left(x-6\right)+23\right)-18\right)+24}{24}\\
1,2,4,8,16,32,63,120,219,382,638\space\text{is}\space\frac{x\left(x\left(x\left(x\left(x-10\right)+55\right)-110\right)+184\right)}{120}\\
1,2,4,8,16,32,64,127,247,466,848,1486\space\text{is}\space\frac{x\left(x\left(x\left(x\left(x\left(x-15\right)+115\right)-405\right)+964\right)-660\right)+720}{720}\\
\text{is there a way to calculate the formula of these sequences easily?}$

答:
You haven't said what these sequences are, but it looks to me like they are (up to some reindexing) the unique polynomials $P_n(x)$ of degree $n$ with the property that $P_n(k) = 2^k$ for the integers $0 \le k \le n$. These have closed form

$$P_n(x) = \sum_{i=0}^n {x \choose i}$$

which is not hard to prove using the calculus of finite differences, or just by computing that $P_n(k) = 2^k$ for $0 \le k \le n$ and arguing that this (together with the degree condition) uniquely determines $P_n(x)$.


答:
To any finite sequence of integers $(a_1,\cdots,a_n)$, you can associate a polynomial $P(x)$ such that $P(i) = a_i$. This has nothing to do with the fact that your first few values are of the form $f(x) = 2^{x-1}$. The formula is
$$
P(x) = \sum_{j=1}^n a_j \left( \underset{i \neq i}{\prod_{j=1}^n} \frac{x-i}{j-i} \right).
$$
This is the Lagrange polynomial associated to the sequence of pairs $((1,a_1), \cdots, (n,a_n))$. It's easy to see that $P(k) = a_k$; you just have to notice that in this sum from $1$ to $n$, if $1 \le k \le n$ satisfies $k \neq j$, then one of the linear terms on the top of the form $x-i$ will vanish when $k=i$, so the whole term will disappear. The only term left will be $\underset{i \neq i}{\prod_{j=1}^n} a_j\frac{x-i}{j-i}$ evaluated at $x=j=k$, and in this case the whole term reduces to $a_j$.

A simpler way to write this is to express it in terms of the Lagrange basis polynomials:
$$
\ell_j(x) = \underset{j \neq i}{\prod_{i=1}^n} \frac{x-i}{j-i}
$$
and write $P(x) = \sum_{j=1}^n a_j \ell_j(x)$. These polynomial functions satisfy
$$
\ell_j(i) = \delta_{ij} = \left\{ \begin{matrix} 1 & \text{ if } i = j \\ 0 &  \text{ if not} \end{matrix} \right.
$$
which is what makes them special. We can then see that $P(i) = \sum_{j=1}^n a_j \delta_{ij} = a_i$.

There is not even anything special about the choice of $1,\cdots,n$ for the initial sequence. You could have chosen $n$ arbitrary pairs $((x_1,a_1), \cdots, (x_n,a_n))$ and consider again the Lagrange basis polynomials
$$
\ell_j(x) = \underset{j \neq i}{\prod_{i=1}^n} \frac{x-x_i}{x_j-x_i}
$$
and then the polynomial $P(x) = \sum_{j=1}^n a_j \ell_j(x)$ satisfies $P(x_i) = a_i$ for $i=1,\cdots,n$; this follows from the fact that $\ell_j(x_i) = \delta_{ij}$, so $P(x_i) = \sum_{j=1}^n a_j \delta_{ij} = a_i$.

This entire process is called polynomial interpolation.

Hope that helps,

Title: determining the formula of equations similar to ${2}^{x-1}$

Body:
the formula for

$1,2,3,4,5,6,7\space\text{is}\space{x}\\
1,2,4,7,11,16,22,29\space\text{is}\space\frac{x\left(x-1\right)+2}{2}\\
1,2,4,8,15,26,42,64,93\space\text{is}\space\frac{x\left(x\left(x-3\right)+8\right)}{6}\\
1,2,4,8,16,31,57,99,163,256\space\text{is}\space\frac{x\left(x\left(x\left(x-6\right)+23\right)-18\right)+24}{24}\\
1,2,4,8,16,32,63,120,219,382,638\space\text{is}\space\frac{x\left(x\left(x\left(x\left(x-10\right)+55\right)-110\right)+184\right)}{120}\\
1,2,4,8,16,32,64,127,247,466,848,1486\space\text{is}\space\frac{x\left(x\left(x\left(x\left(x\left(x-15\right)+115\right)-405\right)+964\right)-660\right)+720}{720}\\
\text{is there a way to calculate the formula of these sequences easily?}$


问:
Show that $n! ≥ 2^{n-1}$ is true by induction for n being a positive integer
I start off by doing the base case:

Base Case: Let P(n) be $n! \ge 2^{n-1}$. Then $P(1)$ will be:

$$
1! \ge 2^0
$$
$$
1 \ge 1
$$
So the base case is true.

Induction: I assume $P(k)$ is true for some arbitrary positive integer $k$. Now I need to show that $P(k+1)$ is true. So assuming $k! \ge 2^{k-1} $ is true, I need to show that $(k+1)! \ge 2^{(k+1)-1} $ is true.

$$
2^{(k+1)-1} = 2 \cdot  2^{k-1}
$$

$$\le 2 \cdot k! \tag{By inductive hypothesis}$$

My question So I have solved this to this point but I don't really know where to take it from here. I would appreciate any help on completing this proof.

Body:
From $k! \ge 2^{k-1}$ and $k+1 \ge 1$ we get

$$(k+1)!=k!(k+1) \ge 2^{k-1}(k+1) \ge 2^{k-1} \cdot 2= 2^k.$$


问:
Form of a number not divisible by x or y
In number theory.
What is the general method for generating the numbers that are not divisible by x and y.

For example to find the numbers that are not divisible by 3 and 5.
For example (6n+- 1) is the generator of numbers not divisible by 2 or 3.

Also how can i get the form of a perticular number like not divisible by 5.[1,2,3,4,6,7..]

A explanation would be appreciated  thanks

答:
let $x,y \neq 1, -1$ and $k=lcd(x,y)$ then $kn+1$ is not divisble by both $x$ and $y$

proof: Assume that $kn+1$ is divisible with $x$ then terhe exists a $z \in Z$ such that

$kn+1=zx \Rightarrow zx-kn=1$

$k=lcd(x,y) \Rightarrow x|kn$

$x|k$ and $x| zx \Rightarrow x|zx-kn=1$ which means $x=1 $ or $-1$

Contradiction!!

proposition can be proved for $y$ by same method


答:
General method for $x$ and $y$:

Step 1: Calculate $C = lcm(x,y)$.

Step 2: List all numbers from $1$ to $C-1$ that are not divisible by $x$ or $y$. Let's call this list $L$.

Step 3: The numbers that are not divisible by $x$ or $y$ have the form
$Cn+k$ where $k$ is in the list $L$.

Example: $x=2,y=3$

Step 1: $C = 6$.

Step 2: $L = (1,5)$.

Step 3: $6n+1$, $6n+5$.

Example: $x=3,y=5$

Step 1: $C = 15$.

Step 2: $L = (1,2,4,7,8,11,13,14)$.

Step 3: $15n+1$, $15n+2$, $15n+4$, $15n+7$, $15n+8$, $15n+11$, $15n+13$, $15n+14$.

Title: Form of a number not divisible by x or y

Body:
In number theory.
What is the general method for generating the numbers that are not divisible by x and y.

For example to find the numbers that are not divisible by 3 and 5.
For example (6n+- 1) is the generator of numbers not divisible by 2 or 3.

Also how can i get the form of a perticular number like not divisible by 5.[1,2,3,4,6,7..]

A explanation would be appreciated  thanks


问:
Prove that if $\varphi$ is injective then $G \cong \varphi(G)$
Let $G$ and $H$ be groups and let $\varphi : G \to H$ be  a homomorphism  and  the image  of $\varphi$ , $\varphi(G)$ is a subgroup of $H$ .
Prove that if $\varphi$ is injective then  $G   \cong \varphi(G)$

My attempt:

If $\varphi$ is injective, then for every $a \in \varphi(G)$,there exist $b \in G$ such that $\varphi(b)=a$.

After that I'm not able to proceed any further.

答: Just in case you haven't got the first homomorphism theorem at hand. Any map $f\colon X\to Y$ gives rise to a surjective (by construction) map $\tilde f\colon X\to f(X)$ defined by $\tilde f(x):=f(x)$. Therefore, your $\varphi$ gives rise to a bijective homomorphism ($=$isomorphism) $\tilde \varphi\colon G\to \varphi(G)$, meaning $G\cong\varphi(G)$.


问:
Condition for a point to be a limit point of some set in general topology
I generated many topological spaces for certain finite sets.
{{}, {1}, {2}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 4}, {1, 2, 3}, {1, 2,
4}, {1, 3, 4}, {1, 2, 3, 4}};
{{}, {1}, {4}, {1, 2}, {1, 3}, {1, 4}, {1, 2, 3}, {1, 2, 4}, {1, 3,
4}, {1, 2, 3, 4}};
{{}, {1}, {2}, {1, 2}, {2, 3}, {2, 4}, {1, 2, 3}, {1, 2, 4}, {2, 3,
4}, {1, 2, 3, 4}}.

I noticed a pattern that when {a} isn't an open set, 'a' is a limit point of some subset of X.
I tried to prove it the following way using an example.
{{}, {1}, {2}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 4}, {1, 2, 3}, {1, 2,
4}, {1, 3, 4}, {1, 2, 3, 4}}
1,2,4 etc can't be limit points because {1}, {2}, {4} are open sets.

If {1,3} is open, {2,3} can't be open because their intersection is {3}(which isn't open).
Any higher open neighborhood of 3 will have to contain {1,3} in it, thus making 3 the limit point of any set containing 1.
This condition seems quite strong. I haven't seen it being mentioned anywhere.
So i want to be sure if my argument is correct.

答:
It's true indeed, and here is a simple argument (an example does not prove a general observation!) why:

In a finite topological space $(X, \mathcal T)$ every point $x$ has a minimal open neighbourhood $M_x = \{O \in \mathcal T\mid x \in O\}$ (open as a finite intersection of open sets).

If $\{x\}$ is not open this means that there is some $y \in M_x$ such that $y \neq x$. Then by definition $x$ is a limit point of any $A$ with $y \in A$ (because if $O$ is an open neighbourhood of $x$, $y \in M_x \subseteq O$ is a witness). In particular $x \in \{y\}'$.

In your example $M_3 = \{1,3\}$ so there any set containing $1$ will do. Minimal neighbourhoods are a nice concept in the study of finite topological spaces; they are a natural base and elicudate the limit point relations between sets very nicely.


问:
Prove difference between a set and the union of other two sets
This question asked which of the two following formulas was always right whilst the other one was sometimes wrong, and what was the necessary and sufficient condition for the formula which was sometimes incorrect to be always right:

(i) $\; A - (B - C) = (A - B) \cup C $

(ii) $\; A - (B \cup C) = (A - B) - C $

Using Venn diagrams, I guessed (ii) was always right, and (i) would be wrong if $A$ and $C$ are disjoint. Thus, the necessary and sufficient condition for (i) to be correct was $A$ and $C$ contain some elements in common.

My questions are:

Were my guesses correct?

How could I prove the two formulas above using words/expressions rather than using Venn diagrams?

Thanks a lot.

答:
To obtain

the necessary and sufficient condition for the formula

(i) $\; A - (B - C) = (A - B) \cup C $

which was sometimes incorrect to be always right

let's first use Venn diagrams:

As we may see, the results are different. For an easy comparison let's display and enumerate individual parts of every result:

From the picture follows that for obtaining the same results, the parts 4 and 5 must be empty.

But what represent parts 4 and 5 together? The set $C - A$, of course.

It means that this set must be empty, i.e. (from definition of set subtraction) there must not be an element in $C$ which is in the same time not in $A$.

In other words, every element of the set $C$ must belong to the set $A$, too, which means that the set $C$ must be a subset of the set $A$.

So we may state our hypothesis:

The necessary and sufficient condition for the formula

$$A - (B - C) = (A - B) \cup \tag 1C$$

to be always correct is

$$C \subset A.\tag 2$$

Now we prove it:

Sufficiency.

Let $C \subset A.\ $ Then:

The left-hand side of $(1)$ is a subset of its right-hand side, because:

If $x \in A−(B−C)$,

then $x \in A$, but $x \notin (B−C)$. From the definition of the set subtraction follows that either $x \notin B$, or $x \in C$.

If $x \notin B$, then $x \in (A-B)$, so indeed $x \in (A - B) \cup C$.

If $x \in C$, then $x \in (A - B) \cup C$, too.

So if $x$ belongs to the left-side hand of $(1)$, it belongs to the right-side hand of $(1)$, too, which means that $$A - (B - C) \subset (A - B) \cup C \tag 3$$

The left-hand side of $(1)$ is a superset of its right-hand side, because:

If $x \in (A - B) \cup C$,

then $x \in (A - B)$, or $x \in C$.

If $x \in (A - B)$, then $x \in A$, but $x \notin B$. But if $x \notin B$, $x \notin (B-C)$, too. So $x \in A - (B - C)$.

If $x \in C$, then (by presumption $(2)$) $x \in A$, too, but certainly it not belongs to $B-C$. So, again, $x \in A - (B - C)$.

So if $x$ belongs to the right-side hand of $(1)$, it belongs to the left-hand side hand of $(1)$, too, which means that $$A - (B - C) \supset (A - B) \cup C \tag 4$$

From $(3)$ and $(4)$ follows, that

$$A - (B - C) = (A - B) \cup C,$$

so the $C \subset A$ is the sufficient condition for this equation.

Necessity.

Let $A - (B - C) = (A - B) \cup C,$

and let $x \in C$. Then $x$ belongs to the right-hand side of this equation, which means that it belongs to the left-hand side of this equation, too, which in turn means that $x$ certainly belongs to $A$, too.

So $C \subset A$, so we prove it as a necessary condition for $(1)$.

OwnerDisplayName: Nemo


问:
Countability of $\mathbb{Z} \times \mathbb{Z}$ using isomorphism with $\Bbb Z[i]$ - how to make this rigorous?

Visualise the set $\mathbb{Z}\times \mathbb{Z}$  as points on the complex plane,  and take origin as the center of a circle of radius $r$. With $r$ going from $0$ to $\infty$ - as you encounter points on the circle, write them down in counterclockwise order. This way you have a bijective mapping, thus $\mathbb{Z}\times \mathbb{Z}$ is countable.

This proof about the countability of $\mathbb{Z}\times \mathbb{Z}$ doesn't look rigorous enough to me. It was proposed by one of my colleagues. The proposal is to write down elements of $Z[i]$ as we encounter them in some sort of a spiral order.

I need help making this more rigorous.

Is there an explicit bijection for this? It'd be great if someone could help me construct the same.

Perhaps we must show that every such circle has only a countable number of such points and that only countably many such circles, i.e. values of $r$ matter. Thanks!

答:
As the the radius of the circle goes to infinity, each $a+bi$ lies on some circle in the same place as the point $(a,b)$ on the Cartesian graph in $\mathbb{Z}$. I guess there might be a way to explicitly find the radii (which are the going to be same for each point $(a,b)$ in the cartesian plane to the $a+bi$ in the complex plane) and do something with that, but that just seems to make the argument longer. Maybe the point is that this same circle process can be repeated on the Cartesian graph to get that the exact same thing happens.

However, this argument seems to overcomplicate the whole thing by turning this discrete problem into one involving a continuous process of taking $r \to \infty$. It really seems to be saying the same thing as the explicit bijection, but with more words and unnecessary ideas. Perhaps I am missing something.


问:
On the binary expansion of the reciprocals of prime numbers
I have been thinking a little bit about the binary decimal expansion of reciprocal prime numbers; and I have a few questions.

I found this neat table (Wayback Machine) which lists the binary expansion of many fractions, and I was trying to find some patterns.

Here are my questions, for brevity I say a natural number has a period N if the binary decimal expansion of it's reciprocal has period N: (for example, 1/7 = .001001... has period 3)

Given an arbitrary natural number N, does there exist a prime number of minimum period N?

(By minimum period I mean to exclude the case that one prime has a period which is a multiple of the period of another prime.  For example, 1/3 = .0101... has period 2, and 1/5 = .00110011... has period 4; so while 1/3 has period 4, what I call it's "minimum period" is 2)

2.
Can two prime numbers have the same minimum period?

A useful result which I believe is well known, is that a natural number has a period N if and only if it is a factor of 2N - 1.

Does anyone know of a good reference that describes some theory behind the relationship between the period of the reciprocal of a natural number, and the prime factorization of that number?

答:
(In the below post there are several links with the apostrophes omitted; fill them in if the links don't work.)

The phenomenon you are studying is a phenomenon in modular arithmetic.  If a prime $p$ has period $n$, this means that there is some numerator $N$ such that $\frac{N}{2^n - 1} = \frac{1}{p}$.  This is equivalent to $Np = 2^n - 1$, or $p | 2^n - 1$, or $2^n \equiv 1 \bmod p$.  The smallest $n$ for which this is true is called the order of $2 \bmod p$, sometimes denoted $\text{ord}_p(2)$ (although this is confusingly also used to denote the greatest power of $p$ which divides $2$...).

Fermat's little theorem guarantees that $\text{ord}_p(2)$ always divides $p-1$; you have already observed this yourself.  However, predicting the exact order is very difficult to do in general.  For example, knowing that the order is actually equal to $p-1$ is equivalent to knowing that $2$ is a primitive root, and it is not currently even known whether this is true infinitely often.  In any case, you should be able to find basic information about order in any good textbook on elementary number theory.

With that background out of the way...

The answer to question 1 is no.  The only exception is $n = 6$ by Zsigmondy's theorem.

The answer to question 2 is yes.  If $p$ and $n$ are relatively prime, then $p$ has period $n$ if and only if $p$ divides $\Phi_n(2)$, where $\Phi_n(x)$ is the $n^{th}$ cyclotomic polynomial.  (This is more or less a restatement of the condition that $p | 2^n - 1$ but $p$ doesn't divide $2^k - 1$ for $k < n$.)  So it suffices to show that some number of this form has more than one prime factor relatively prime to $n$.  There are two cases here which are particularly classical:

$n$ is a prime $q$.  In this case $\Phi_q(2) = 2^q - 1$ is a Mersenne number, and $2^{11} - 1 = 23 \cdot 89$ is the smallest composite Mersenne number, hence $23$ and $89$ both have period $11$.

$n = 2^k$ for some $k$.  In this case $\Phi_{2^k}(2) = 2^{2^{k-1}} + 1$ is a Fermat number, and $\Phi_{64}(2) = 2^{32} + 1 = 641 \cdot 6700417$ is the smallest composite Fermat number, hence $641$ and $6700417$ both have period $64$.

The answer to question 3 is the following.

Lemma: If $n, m$ are relatively prime odd numbers, then $\text{ord}_{mn}(2) = \text{lcm}(\text{ord}_n(2), \text{ord}_m(2))$.

Proof.  $\text{ord}_{mn}(2)$ is the order of the element $2$ in the multiplicative group of $\mathbb{Z}/mn\mathbb{Z}$, which we will denote $U(mn)$.  By the Chinese remainder theorem, $U(mn)$ is isomorphic to the direct product $U(m) \times U(n)$, so the order of $2$ in $U(mn)$ must be the $\text{lcm}$ of the orders of $2$ in $U(m)$ and $U(n)$.

It follows that to compute $\text{ord}_m(2)$ for arbitrary $m$ it suffices to compute it for the odd prime power factors of $m$ and then to take the $\text{lcm}$ of the resulting numbers.  (Note that if $m$ is divisible by a power of $2$ this only contributes a leading string of zeroes to the binary expansion of $\frac{1}{m}$ and hence does not affect the computation of the period.)  Again, you can find a discussion of the Chinese remainder theorem in any good textbook on elementary number theory.  (I am particularly bad at recommending textbooks on elementary number theory because I learned mine through a summer program, not a textbook...)


问:
when does a functor map products into products?
Motivation: wikipedia claims, that in algebraic topology, there holds: $\pi_1(X\times Y)\cong\pi_1(X)\times\pi_1(Y)$ and $\pi_1(X\vee Y)\cong\pi_1(X)\ast\pi_1(Y)$. A similar statement holds for arbitrary products and one-point unions, making the (covariant) fundamental group functor $\pi_1:\mathrm{TOP}^0 / h-\mathrm{TOP}^0\rightarrow GRPS$ preserve products and coproducts.

I'm guessing the same holds for the functors $\pi_k$ (homotopy groups), $H_k$ (homology groups)?

Definitions:

Examples:
In the category of sets, groups, rings, $R$-modules, vector spaces, topological spaces,etc, the product is the cartesian product. In the category of sets and topological spaces, the coproduct is the disjoint union/topological sum. In the category of groups, the coproduct is the free product $\ast$. In the abelian group / $R$-modules / vector spaces category, it is the direct sum $\oplus$. In the topological pointed spaces category, it is the one-point union $\vee$.

Question: I would very much like to prove this in a general way, so I'd like to know the following:
Theorem???: Suppose $F:\underline{A}\rightarrow\underline{B}$ is a covariant / contravariant functor. What are some (reasonably general) sufficient conditions on $F,\underline{A},\underline{B}$, that make $F$ send (products to products and coproducts to coproducts) / (products to coproducts and coproducts to products ), i.e.
$$F \text{ covariant }\Rightarrow F(\prod_{i\in I}A_i)=\prod_{i\in I}F(A_i)\text{ and }F(\coprod_{i\in I}A_i)=\coprod_{i\in I}F(A_i);$$
$$F \text{ contravariant }\Rightarrow F(\prod_{i\in I}A_i)=\coprod_{i\in I}F(A_i)\text{ and }F(\coprod_{i\in I}A_i)=\prod_{i\in I}F(A_i)?$$

The sufficient conditions that I'm looking for are primarily intended for important standard functors, such as $\pi_k$, $H_k$, $H^k$, tangent and cotangent bundle functor, ... so that I can prove the results in one sweep.

Counterexample: In the category of division rings / fields, we have $\mathbb{Z}_2\times\mathbb{Z}_2 = \mathbb{Z}_2$, thus the forgetful functor to the category of sets doesn't preserve products.

答: This is a comment really.  I just want to point out that there is a fairly natural functor that exchanges products and coproducts even though it has no left or right adjoint and is not representable: the functor that assigns to each finitely generated augmented $k$-algebra $\Lambda$ its cohomology ring $\operatorname{Ext}^* _\Lambda (k,k)$.</p>


问:
Find distribution with probability generating function
The probability generating function of $X$ is $G_x(s)=\frac{1}{2}(s^9(1+s^2))$. Find $EX$ and probability distribution function.

$$EX=G_x^{'}(s)=\frac{1}{2}(9s^8+11s^{10})$$

How about pdf? Do I need to expand $G_x(s)$ function with Taylor series?

答:
HINT.

Note that if pdf is $f(k) = \mathbb{P}[X=k]$ for $k \in \mathbb{N}$, then

$$
G(s) = \sum_{k=1}^\infty f(k) s^k = \frac{s^9 \left(1+s^2\right)}{2} = \frac{s^9}{2} + \frac{s^{11}}{2}.
$$

That means that $f(k) = 0$ for lots of $k$ -- which ones? And what values does $f(k)$ take where it is non-zero?


问:
Find the equation of the normal line
Find the equation of the normal line to curve $y=2x^2+3$ that is perpendicular to the line $8x-y+3=0$

I have already solved the $m = 8$,  $m'= -\frac{1}{8}$.

But I don't know how to solve for the $x,y$ points.

答:
Let the normal be at $(h,k)$, then $$y=2x^2+3 \implies y'=4x=-\frac{1}{m} \implies h=\frac{-1}{4m}$$
The equation of normal is $$y-k=m(x-h) \implies y-2h^2-3=mx-mh \implies
y-\frac{1}{8m^2}-3=mx+\frac{1}{4}~~~(2)$$
For for any value of $m$ Eq. (2) represents normal to the given parabola.
Let $m=-1/8$, the
$y-8-3=-x/8+1/4 \implies 8y+x=90$ is the required normal to the given parabola.

Title: Find the equation of the normal line


问:
Question 6 Section Finite Fields Hungerford Algebra
Consider the following problem:

If |K| = q and (n ,q) = 1 and F is a splitting field of $x^n-1_K$ over K, then prove that k = [F : K]
is the least positive integer k such that $n | (q^k - 1)$.

Attempt : $|F| = q^k$ . Let p be a root of $x^n -1$=0 . But I am unable to think of any other details on how the question should be approached and I need help.

So, Can you please outline a solution.

Thank you.

答: Hint: The roots to $x^n=1$ form a multiplicative group. Under what hypothesis does this group fit within $F^*$?


问:
Calculate the length of curve $f(x)=\arcsin(e^x)$, check solution, please.
As in the topic, my task is to calculate the length of $f(x)=\arcsin(e^x)$ between $-1, 0$. My solution: I use the the fact, that the length of $f(x)$ is equal to $\int_{a}^b\sqrt{1+(f'(x))^2}dx$ $$f'(x)=(\arcsin(e^x))'=\frac{e^x}{\sqrt{1-e^x}}\rightarrow (f'(x))^2=\frac{e^{2x}}{1-e^x}$$Next I use binomial coefficient:$$ \left (1+ \frac{e^{2x}}{1-e^x}\right )^\frac{1}{2}=\sum_{n=0}^{\infty}\binom{\frac{1}{2}}{n}\frac{e^{2xn}}{(1-e^x)^n}$$Let's evaluate $\binom{\frac{1}{2}}{n}=\frac{1}{n!}\cdot(\frac{1}{2})(-\frac{1}{2})(-\frac{3}{2})\cdot\dots\cdot(\frac{-2n+3}{2})=\dots=\frac{(-1)^{n-1}}{(2n-1)(n!)^2}\cdot\binom{2n}{n}$Hence the length is $$\int_{-1}^0\sum_{n=0}^{\infty}\frac{(-1)^{n-1}}{(2n-1)(n!)^2}\cdot\binom{2n}{n}\cdot \frac{e^{2xn}}{(1-e^x)^n}dx$$Normally I would integrate it but it is nasty and I am not sure whether it is correct. Any hints, ideas? Thanks in advance. </p>

答:
I think your initial derivative is off, so:

$$f'(x)=\frac{e^x}{\sqrt{1-e^{2x}}}$$

Then:

$$(f'(x))^2 = \frac{e^{2x}}{1-e^{2x}}$$

So we have:

$$\int_{-1}^0 \sqrt{1+\frac{e^{2x}}{1-e^{2x}}} = \int_{-1}^0 \sqrt{\frac{1}{1-e^{2x}}}$$

Then the rest might make more sense.


问:
Maximizing the area of a cyclic pentagon made up of triangle centres.
Triangle $ABC$ has $\angle BAC=60^\circ$, $\angle CBA \le 90^\circ$, $BC=1$, and $AC \ge AB$. Let $H$, $I$, and $O$ be the orthocenter, incenter, and circumcenter of $\triangle ABC$, respectively. Assume that the area of the pentagon $BCOIH$ is the maximum possible. What is $\angle CBA$?

This is problem 25 from the 2011 AMC12A. My friend sent me the following solution:

Angle formulas give that $BCOIH$ is cyclic. Further $I$ and $H$ are both on the arc to the "left" of $O$, with $H$ further "left" than $I$. To maximize the area, we want to distribute $I$ and $H$ evenly over arc $BO$. This gives that $\angle{IBC}=\angle{IBO}+\angle{OBC}=10+30=40$. Multiplying by two gives that the answer is $80$.

I agree with this solution up to the point where they claim they want to distribute evenly abut BO. Why not distribute evenly over the arc BC, doesn't this make the area even greater? If anyone could explain why we would only distribute on arc BO that would be great.

答: Note that $O$ is fixed as midpoint of arc $BC$. Since the perpendicular bisector of chord $BC$ passes through $O$.


问:
Urn Question without Replacement
I'm trying to understand how this works.

I have an urn with 6 white balls and 5 black balls. I then draw all 11 balls from the urn.

Now, let X be a random variable that takes the value y if the first white ball appears on the y-th draw.

How do I calculate the expected value of X?

I had thought that I would need to sum the value and probabilities, so:

$$ 1*\frac{6}{11}+2*\frac{6}{10}+3*\frac{6}{9}+\ldots+6*\frac{6}{6}$$

But I know my probabilities aren't right because they add up to too much. I had assumed the probability of drawing the first white ball on the first draw is 6/11 (6 white balls, 11 total), and the first white ball on the second draw is 6/10 (still 6 white balls, but now only 10 total) etc, but I'm clearly missing something....

What am I missing here?

答:
Here's an example. If I pick my first white ball on draw $y = 3$ then I took a black ball on attempts $1$ and $2$. What's the probability of this happening? The chance of drawing a black ball at attempt $1$ is $5/11$. Then, the chance that the second ball is also blacks is $4/10$. Finally, we must draw a white ball on attempt $3$, which occurs with probability $6/9$. To compute $P(X = 3)$ you must multiply these values. Do you see the general pattern in computing $P(X = y)$?

Once you have that, it's a matter of computing $E(X)$ via the usual formula $\sum y \cdot P(X = y)$.

Title: Urn Question without Replacement


问:
A CS-complete set is bounded
I will recall two definitions. Let $X$ be a normed space. One says that a subset $C\subset X$ is CS-complete (or CS-compact) if every convex series of elements of $C$ converges and its sum is in $C$. That is, whenever $\{x_n\}_{n \in \mathbb{N}}\subset C$ and $\{\lambda_n\}_{n \in \mathbb{N}}\subset \mathbb{F}$ are scalars such that $$\sum_{n \in \mathbb{N}} \lambda_n =1,$$ then $$\sum_{n \in \mathbb{N}} \lambda_n x_n\in C.$$

One says that a subset $C\subset X$ is CS-closed if every convex series of elements of $C$ which converges has its sum in $C$.

I know and I can prove that a CS-closed, bounded set in a Banach space is CS-complete and that a CS-complete set is always CS-closed. How can I prove that:

a CS-complete set must be bounded;

a CS-closed and bounded set need not be CS-complete if $X$ is not Banach?

Idea for 1. If $C$ was unbounded I would have a sequence $x_n$ of elements of $C$ with $\lvert \lvert x_n \rvert \rvert\ge n$, then ...? Can I prove that the partial sums of $\sum \lambda_n x_n$ are not close one to each other?

答:
Answer for 1): If $C$ is not bounded then there exist points $x_n$ in $C$ with $\|x_n\|>2^{n}$. Now $\sum \frac 1 {2^{n}}=1$ but $\sum \frac 1 {2^{n}} x_n$ does not converge since $\frac {x_n} {2^{n}}$ does not tend to $0$.

Answer for 2): Consider the space $\ell_0$ of all sequences which are eventually $0$ with the $\ell^{1}$ norm. Let $C$ be the closed unit ball of this space. If $(x_n) \subset C$ and  $\sum \lambda_n x_n$ converges then $\|\sum \lambda_n x_n|| \leq 1$ so $\sum \lambda_n x_n \in C$. Thus $C$ is bounded and CS-closed. However, if $e_n$ has $1$ in the $n-$th palce and $0$ elsewhere then $\sum \frac 1 {2^{n}} e_n$ does not converge in $\ell_0$.

Title: A CS-complete set is bounded


问:
analytical geometry
We have an affine coordinate system and $3$ points given: $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$, $D=(1,1,1)$. I have to find a linear transformation, which depicts the points $A$, $B$, $C$ and $D$ accordingly into $B$, $C$, $A$ and $D$.</p>

答:
Besides observing that you need a permutation matrix, another option is to write your matrix $\mathbf{A} = \left[\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right]$ explicitly, and use the given data.

Using the fact that you should map your point $A$ to point $B$, you get

$\mathbf{A}\left[\begin{array}{c} 1 \\ 0 \\ 0\end{array}\right] = \left[\begin{array}{c} 0 \\ 1 \\ 0\end{array}\right]$

or equivalently,

$a_{11} = 0$, $a_{21} = 1$, $a_{31} = 0$.

Now use the rest of the data to find the other entries.

Title: analytical geometry


问:
Contrapositive of: If $a$ is a real number such that $|a| < r$ for every positive real number $r$, then $a=0.$
I wish to state the contrapositive of:

If $a$ is a real number such that $|a| < r$ for every positive real number $r$, then $a=0.$

First, I want to state the original statement symbolically.

$\exists a \in \mathbb{R} (\forall r \in \mathbb{R}^+ (|a| < r \implies a = 0))$

Afterwards, we then can take the contrapositive:

$\forall a \in \mathbb{R}(\exists r \in \mathbb{R}^+ (a \neq 0 \implies |a| \ge r))$

The thing I am most concerned about are the quantifiers of $a$. Please let me know what you think of my attempt.

答:
You have arrived at a common misconception.

The "If $a$ is a real number..." part (and variations thereof), are almost always equivalent to the phrasing (and variations thereof).

"Let $a$ be a real number..."

Then there is in fact no quantifier to change for that part and your statement becomes:

Let $a$ be a real number such that $|a|<r$ for every positive real number $r$, then $a=0$.

Then the contrapositive is:

If $a \neq 0$, then there exists a positive real number $r$ such that $|a| \geq r$.


答:
Imagine the you only have 2 white balls out of 11. The probability of selecting a white ball on the first draw is $\frac{2}{11}$.

What is the probability that you select the first white ball on the second draw? In words: it is the "intersection" of two events: event "1" is that you select a black ball on the first draw and event "2" is that you select a white ball on the second draw, given that you selected a black ball on the first draw. Denoting the event of a "first white ball on second draw" as $W_2$, we get:

$$\mathbb{P}(W_2)=\frac{9}{11}*\frac{2}{10}$$

So if you only had two white balls, the expected value of your variable would be:

$$\frac{2}{11}+\frac{9}{11}*\frac{2}{10}$$

Now you just need to adjust the numbers to 6 white balls.

Title: Urn Question without Replacement

Body:
I'm trying to understand how this works.

I have an urn with 6 white balls and 5 black balls. I then draw all 11 balls from the urn.

Now, let X be a random variable that takes the value y if the first white ball appears on the y-th draw.

How do I calculate the expected value of X?

I had thought that I would need to sum the value and probabilities, so:

$$ 1*\frac{6}{11}+2*\frac{6}{10}+3*\frac{6}{9}+\ldots+6*\frac{6}{6}$$

But I know my probabilities aren't right because they add up to too much. I had assumed the probability of drawing the first white ball on the first draw is 6/11 (6 white balls, 11 total), and the first white ball on the second draw is 6/10 (still 6 white balls, but now only 10 total) etc, but I'm clearly missing something....

What am I missing here?


问:
How to find the limit $\lim_{n\to \infty}\left(n-\frac{n}{e}\left(1+\frac{1}{n}\right)^n\right)$
To find the limit

$$\displaystyle\lim_{n\to \infty}\left(n-\frac{n}{e}\left(1+\frac{1}{n}\right)^n\right)$$

答:
Hint: Using the Taylor series of the logarithm and the exponential function
\begin{align*}
\left( {1 + \frac{1}{n}} \right)^n & = \exp \left( {n\log \left( {1 + \frac{1}{n}} \right)} \right)  = \exp \left( {1 - \frac{1}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)} \right) \\ & = e\exp \left( { - \frac{1}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)} \right) = e\left( {1 - \frac{1}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)} \right).
\end{align*}


问:
Induction Proof of Algorithm [Greedy Graph Coloring]
Having a $G = (V,E)$ with each vertex having a range $[a,b]$.

Every two vertices are connected only if $[a_1,b_1]$,$[a_2,b_2]$ have a common subrange.

Each range of vertex is:

$rV1 = [0,5]$

$rV2 = [1,3]$

$rV3 = [2,10]$

$rV4 = [4,9]$

$rV5 = [6,7]$

$rV6 = [8,12]$

$rV7 = [11,13]$

Graph based on above ranges.

The Algorithm:

```<code>ranges = rV1,rV2,rV3,rV4,rV5,rV6,rV7...

COLOR_INTERVAL_GRAPH(rV1,rV2,rV3,rV4,rV5,rV6,rV7...){
if(number of rVi > 0){
 C_m  = MAXIMAL_COLOR_CLASS(rV1,rV2,rV3,rV4,rV5,rV6,rV7...);
 //paint C_m vertices with color m.
 //new_ranges <- remove C_m from rV1,rV2,rV3,rV4,rV5,rV6,rV7...
 return {C_m} U COLOR_INTERVAL_GRAPH(new_ranges)
 else:
   return []

MAXIMAL_COLOR_CLASS(new_range){
C = []
i = 1
while(i <= new_range.size()){
  C = C U {Vi}
  j = i+1
  while(j <= new_range.size() AND rVi (not common subrange with rVj)){
  j = j+1
i=j
return C
</code>```

How to prove using induction that the algorithm uses the fewest possible colors.

After searching a bit i found that the MAXIMAL_COLOR_CLASS function in line 4 extends the C set.

I have to prove that the optimum coloring of any graph (of this type) can be transformed in order the first chromatic class is the same as the output of MAXIMAL_COLOR_CLASS.

Then using the above (by induction) i can show that every optimum coloring is the same as the output of the MAXIMAL_COLOR_CLASS proving the algorithm correct.

Not sure how to use induction to prove that.

答:
Induction proof proceeds as follows:

Is the graph simple? Yes, because of the way the problem was defined, a range will not have an edge to itself (this rules out one of the easiest ways to prove that a graph is not n-colorable). Does it hold for 0 or 1 vertex? Yes, trivially. How about two vertices (with one edge, because disconnected vertices do not affect chromatic number)? You should be able to work out the justification for this step, and it is arguably the first non-trivial case and should be used as your base case.

Suppose you know that it holds for n vertices. Those vertices correspond to particular ranges. You could choose to prove that it then holds for n+1 vertices by adding another range (and an unknown number of edges) but this seems hard and potentially complicated. Instead, suppose it holds true for n edges. There are two ways that we could add an edge: by extending the endpoint of a nearby range until they intersect or by adding a new range entirely. Consider the output of the algorithm in either scenario, and you will have your proof.

In general, the more of your work you post, the better help you will be able to receive. This site is not intended to solve homework problems without some demonstrated effort and understanding or lack thereof.


问:
Boy and Girl paradox
I am trying to understand the boy and girl paradox.  The paradox states that if a family has two children and one of them is a boy, then the probability of the other being a girl is 2/3.  When you write out the set of possible outcomes { bb, bg, gb, gg } it makes a little more sense.  My question is why does age/order matter?  The two possible outcomes boy/girl and girl/ boy are the same right?  </p>

答:
Try it with coins: First, flip a nickel and a dime. There quite clearly are four possibilities: (N=H, D=H), (N=H, D=T), (N=T, D=H), and (N=T, D=T). Each, also quite clearly, has a probability of 1/4; so the probability of two heads is 1/4, the probability of two tails is 1/4, and the probability of one of each is 1/2.

Then repeat the same experiment with two absolutely identical quarters. There now seem to be three combinations we can discern: (Both H), (Both T), and (one H and one T). But do you really think the odds change to 1/3 for each combination just because the coins are now identical? I certainly hope not. Even though you can't see the difference between the coins, they still are different coins, and the ways that the different combinations can occur still depend on that difference.

When we count possibilities for two-child families, age-order matters only because we have to acknowledge that the two children are different from each other. Even if we aren't given any information that distinguishes them, that difference still matters.

But as for the probability paradox you mentioned, the answer depends on how you "know" that one is a boy. If it is because you asked if the family had any boys, then every BB, BG, or GB family would answer "yes" and the answer is 1/3. But if you learned that fact incidentally, then you could have learned a BG of GB family has a girl. You can only count half of them, and the answer is 1/2.

Title: Boy and Girl paradox


问:
How to find the shaded angle of a right angled isosceles triangle?

Hey everyone,

Just another question. I’m trying to figure out how to find the shaded angle of an isoceles triangle. All I know is the ratios between the sides and of course the value of the angles (as you can see in the image).

Can anyone help with this question? Thank you very much!

答:
Here is another way using two helper lines:

We have

$$\gamma = \alpha + 90^{\circ} + \beta$$

Because of the given ratios we have

$$\tan \alpha = \frac 13 \text{ and } \tan \beta = \frac 12$$

Using the tangent-addition formula it follows

$$\tan (\alpha + \beta) = \frac{\frac 13 + \frac 12}{1-\frac 16} = 1$$

Hence,

$$\alpha+\beta = 45^{\circ} \Rightarrow \gamma = 135^{\circ}$$


问:
Convergence in Probability implies Boundedness in Probability
Let $X_n$ be a sequence of random variables such that $X_nc_n \rightarrow 0$ in probability for any sequence $c_n \to 0$. To show $\{X_n\}$ is a tight sequence.

DEFINITION : A family $\{X_n\}$ is tight if for any $\epsilon > 0$ we can find a bounded interval $K_\epsilon$ such that $\mathbf{P}(|X_n| < K_\epsilon) > 1 - \epsilon$.

The converse also holds:
Let $\{X_n\}$ be a family of tight random variables. Then for any $c_n \rightarrow 0$ we have $X_nc_n \rightarrow 0$ in Probability.

Proof : Since $\{X_n\}$ is tight, we have for any $\epsilon > 0$ a $K_\epsilon$ s.t. $\mathbf{P}(|X_n| > K_\epsilon) < \epsilon$.
Let $\{c_n\}$ be any sequence converging to $0$. We can find a $n_0$ s.t. $n>n_0 \implies |c_n| < \epsilon/K_\epsilon$.

Thus for all $n>n_0$ , $\mathbf{P}(|c_nX_n| > \epsilon) = \mathbf{P}(|c_nX_n| > \epsilon, |X_n| > K_\epsilon) + \mathbf{P}(|c_nX_n| > \epsilon, |X_n| < K_\epsilon)$
$$< \mathbf{P}(|X_n| > K_\epsilon) + \mathbf{P}(|X_n| > K_\epsilon , |X_n| < K_\epsilon) (\because |c_n| < \epsilon/K_\epsilon)$$

$$< \epsilon$$

So this is what convergence in probability asks us to find.

Any help is appreciated.

EDIT : So this is a direction that I am trying :

Since $\{X_nc_n\} \rightarrow 0$ in probability $\implies$
Every subsequence $\{X_{n_k}c_{n_k}\}$ has a further subsequence that converges a.s. to $0$ and hence in distribution. This implies $\{X_nc_n\}$ is tight. Now as $\{c_n\}$ converges it is bounded by say $C$. Thus for any given $\epsilon$ we had a $K_\epsilon$ s.t. $\mathbf{P}(|c_nX_n| > K_\epsilon) < \epsilon$ for all $n$ and using the bound of $C$, we have $\mathbf{P}(|X_n| > K_\epsilon/C) < \epsilon$ for all $n$.

Is this right? Any comments are appreciated.

答: Suppose $\{X_n\}$ is not tight.  Then there exists $\epsilon > 0$ and a subsequence $(X_{n_k})$ such that $\mathbb{P}(|X_{n_k}| > k) \geq \epsilon$.  Now define the sequence $(c_n)$ by $c_n = 1/k$ for all $n_{k-1} < n \leq n_k$.  We have $c_n \to 0$ but $\mathbb{P}(|X_{n_k} c_{n_k}| > 1) = \mathbb{P}(|X_{n_k}| > k) \geq \epsilon$ for all $k$, and therefore $X_n c_n$ does not converge to $0$ in probability, a contradiction.


问:
$5^m = 2 + 3^n$ help what to do
how to solve this for natural numbers $5^m = 2 + 3^n$ i did this

$5^m = 2 + 3^n \Rightarrow 5^m \equiv 2 \pmod 3 \Rightarrow m \equiv 1 \pmod 2$
now  if   i put it like  this
$   5^{2k+1} = 2 + 3^n $

what to do ??

now is this right another try :

$ m = n \Rightarrow 5^n - 3^n = 2 = 5 - 3 \Rightarrow (m_1,n_1)= (1,1) $

we can prove by induction :

$5^n - 3^n > 2 \quad \forall n > 1  \Rightarrow m = n = 1 $

another case :

$m > n \Rightarrow 5^m > 5^n \geq 3^n+2 \quad \forall n \geq 1 \Rightarrow 5^m - 3^n > 2 \Rightarrow \emptyset$

$m < n \Rightarrow no \ sol. $ by putting some values

答:
Assume there is a solution with $n>1$. Then $ 5^m \equiv 2 \pmod 9 \Rightarrow m \equiv 5 \pmod 6$.

Then $ 5^m \equiv 3 \pmod 7 \Rightarrow 3^n \equiv 1 \pmod 7 \Rightarrow n \equiv 0 \pmod 3.$

Also, $ 5^m \equiv 5 \text { or } 8 \pmod {13} \Rightarrow 3^n \equiv 3\text { or } 6 \pmod {13} \Rightarrow n \equiv 1 \pmod 3.$

This contradiction means that $n$ cannot be greater than $1$ and so $m=n=1$.

Title: $5^m = 2 + 3^n$ help what to do


问:
Nondegenerated algebraic set and global sections of twisted sheaves
I encountered the following argument in Eisenbud's geometry of syzygies (p75):

I don't understand the part about the middle vertical map. How does $X$ being nondegenerate relates to the global sections of $\mathcal{O}_X(1)$ and $\mathcal{O}_\mathbb{P}(1)$? Why is the map injective?

答:
By definition, a nondegenerate scheme is not contained in any hyperplane. If we suppose that $X$ is contained in some hyperplane $H$, the equation of $H$ generates the kernel of the restriction map
$$
H^0(\mathcal O_{\mathbb P^r}(1)) \to H^0(\mathcal O_X(1)),
$$

namely $H^0(\mathcal I_X(1))$. When $X$ is nondegenerate, each intersection $X \cap H$ is a subscheme of codimension $1$ in $X$, hence the zero locus of a non-zero section of $\mathcal O_X(1)$.


答:
Observe that, points $I$ and $H$ will always stay on one side of point $O$$[$ Drop perpendiculars from point $A$, $O$ and $I$ on side $BC$ and then calculate the distances from point $B$ to the feet of these perpendiculars.$]$ and thereafter stay on the arc $BO$.

Hence, to maximize the area, we need to evenly distribute points $I$ and $H$ over the arc $BO$.

Title: Maximizing the area of a cyclic pentagon made up of triangle centres.

Body:
Triangle $ABC$ has $\angle BAC=60^\circ$, $\angle CBA \le 90^\circ$, $BC=1$, and $AC \ge AB$. Let $H$, $I$, and $O$ be the orthocenter, incenter, and circumcenter of $\triangle ABC$, respectively. Assume that the area of the pentagon $BCOIH$ is the maximum possible. What is $\angle CBA$?

This is problem 25 from the 2011 AMC12A. My friend sent me the following solution:

Angle formulas give that $BCOIH$ is cyclic. Further $I$ and $H$ are both on the arc to the "left" of $O$, with $H$ further "left" than $I$. To maximize the area, we want to distribute $I$ and $H$ evenly over arc $BO$. This gives that $\angle{IBC}=\angle{IBO}+\angle{OBC}=10+30=40$. Multiplying by two gives that the answer is $80$.

I agree with this solution up to the point where they claim they want to distribute evenly abut BO. Why not distribute evenly over the arc BC, doesn't this make the area even greater? If anyone could explain why we would only distribute on arc BO that would be great.


问:
If $f$ is even and $y'=f(y)$ then $y$ is odd

Let $f\in C^1(\mathbb{R}, \mathbb{R})$ be an even function.

Consider the maximal solution $y\colon\left]\alpha ,\beta\right[\to \mathbb{R}$ of the IVP $$y'=f(y),\ y(0)=0$$

Prove that $y$ is an odd function and $\beta =-\alpha$.

To be able to prove that $y$ is odd, I first need its domain to be symmetric ($x\in \left]\alpha ,\beta\right[\implies -x\in \left]\alpha ,\beta\right[$), from here I can conclude that $\alpha=-\beta$. But how to prove the domain is symmetric?

And how to prove that $y(-x)=-y(x)$ for all $x\in \left]\alpha,\beta\right[$? I suspect it has something to do with the fact that the derivative of an even function is odd and $y'=f(y)$, but I can't see how to get the desired result.

答:
Clearly, $\alpha<0<\beta$.
Let $r=\min\{-\alpha,\beta\}$ so that we at least have a solution on $\left]-r,r\right[$.

Then show: If $y$ is a solution on $]\alpha,\beta[$, then $x\mapsto-y(-x)$ is a solution on $\left]-\beta,-\alpha\right[$.

By uniqueness, these coincide on $\left]-r,r\right[$.

Then you can combine $y$ with $x\mapsto- y(-x)$ to find a solution on $\left]-\max\{-\alpha,\beta\},\max\{-\alpha,\beta\}\right[$. By maximality, we conclude $\alpha=-\max\{\alpha,\beta\}=-\beta$.


问:
Mathematical rigor for engineers
An engineer with a master's degree, I would like to improve my level of mathematical rigor. For example, I would like to come to grasp

what measure theory has to do with the Fourier series and

what group theory has to do with the eigenvalue,

among others.  I would like to understand why Lebesgue integration is important (the idea seems sort of trivial to me, which naturally means that I don't yet really get it), why the formality of Weierstrassian tests is preferable to the mere evasion of the problem by, say, the extraction of Hadamard's finite part, and so on.

To this end, I have been reading Georgi E. Shilov's Elementary Real and Complex Analysis, which I like; but at the rate Shilov is going it would take like 10,000 pages to advance as far as the Fourier series.  Therefore, I have looked instead on Amazon and Google Books for the book Mathematical Rigor for Engineers.

Unfortunately, no one seems to have written that book.

What are my options?

(To clarify: I do not mean to ask the opinion-based question, "What should I do?" Rather, I mean to ask the experience-based question, "What are my options?" Maybe the only answer is that what I seek resembles Financial Analysis for Kindergarteners or Transoceanic Logistics for Retail Clerks, but since I happen to possess an applicationist's acquaintance with mathematical productions like the Fourier Series and the eigenvalue, not to mention Hadamard's finite part, I thought that I would ask.)

答:
For the Fourier series<->measure theory connection, you can perform some operations using infinite series as a black box but in order to describe a Fourier series corresponding to a discontinuous function you will (eventually) begin to move into measure theory. Please remind me to respond after this semester is over, as I am enrolling in a course on this exact question which starts next week (my undergrad didn't emphasize the "applied" side outside of the Physics department).
Group theory is connected to eigenvalues because the square matrices form a group and eigenvalues allow you to detect properties of an individual matrix (such as whether it can be decomposed in any of several ways, what the "long-term behavior" of a Markov chain will be, whether the matrix will be stable or unstable under exponentiation even if it lacks the Markov property, etc.) I would be happy to answer further questions on this subject next week.</p>

Title: Mathematical rigor for engineers


问:
How can we say topology on set is collection of open sets?
I have recently started studying topology on my own. Topology $\tau$ on set X is defined as some 3 properties

(1)$\phi, X \in \tau$

(2)Arbitrary union is closed

(3)Finite intersection is closed in $\tau$

But then how can we say that
(1) topology is collection of "open" sets?

(2)for every $x \in X, \exists U_x : x \in U_x, U_x \in \tau$

How to prove (1) and (2)?

答:
You define a topology $\tau$ to be a collection satisfying the three standard axioms. Then you define $O$ to be open iff

$$\forall x \in O: \exists U_x \in \tau: x \in U_x \subseteq O\tag{1}$$

Now the claim is $O$ is open iff $O \in \tau$.

To go from left to right: if $O$ is open according to $(1)$, we have $U_x$ as stated for every $x \in O$ and in fact

$$O = \bigcup\{U_x\mid x \in O\}\tag{2}$$

(to see $\subseteq$ in $(2)$ we just note that every $x \in O$ is in $U_x$, which is a subset of the right hand side by definition, and to see $\supseteq$ we just note that all $U_x$ are subsets of $O$ by definition hence so is their union).
As $(2)$ holds and one of the axioms of $\tau$ is that it is closed under unions and all $U_x$ are in $\tau$, so is $O$. This shows $x$ open according to $(1)$ implies $O \in \tau$.

OTOH, if $O \in \tau$ we can just take $U_x = O$ for every $x \in O$ so $O$ is open according to $(1)$ too.

So by the above, a set is open iff it is in $\tau$. BTW, it's more common to use $(1)$ when we have defined (or are given) a base $\mathcal{B}$ for the topology and then we can define open wrt to that base as $(1)$, but choosing $U_x \in \mathcal{B}$ instead. In that case there is a proof that's needed to show that the thus-defined "open" sets obey the three axioms of $\tau$ and we have indeed defined a proper topology on $X$. Here the axioms on $\tau$ are just given and the observation is boring, but in general you can often define openness from a smaller base collection (which is convenient in many proofs) and don't bother that much with $\tau$ as such.

So in your question you do not have to show $(1)$ and $(2)$, but you're asked to show "$O$ is open by $(1)$ iff $O \in \tau$", which is different.


问:
Show that $X$ is compact iff $X_n$ compact for all $n\in \mathbb N$.
Let $(X_n,d_n) $  a family of metric spaces and $(X=\prod_{n \in \mathbb N} X_n,\mathcal d ) $ a metric space with $\mathcal T_p$ the product topology is equal to the topology defined by the distance $d$, such that for all $x=(x_n)_n, y=(y_n)_n\in X$ we have  $d(x,y) = \sum_{n=0}^{\infty} \frac{1}{2^n}d_n(x_n,y_n) $ and $d_n(x,y) \leq M$ for all $x, y\in X_n$.

Show that $X$ is compact iff $X_n$  compact for all $n\in \mathbb N$.

For the first implication, I have used the the fact that $X$ is compact and the projection maps are continuous. For the other one, I get stuck with it. any help is appreciated.

答:
Since compactness is equivalent to sequential compactness in metric spaces, we can show instead that that $X$ is sequentially compact. Let $\sigma=\left\langle x^{(n)}:n\in\Bbb N\right\rangle$ be a sequence in $X$, where $x^{(n)}=\left\langle x_k^{(n)}:k\in\Bbb N\right\rangle$ for each $n\in\Bbb N$; we want to construct a convergent subsequence of $\sigma$. To begin, $X_0$ is sequentially compact, so there is an infinite $N_0\subseteq\Bbb N$ such that $\left\langle x_0^{(n)}:n\in N_0\right\rangle$ converges to some $y_0\in X_0$.

Suppose that $\ell\in\Bbb N$, and we have an infinite $N_\ell\subseteq\Bbb N$ and points $y_k\in X_k$ for $k\le\ell$ such that $\left\langle x_k^{(n)}:n\in N_k\right\rangle$ converges to $y_k$ for each $k\le\ell$. $X_{\ell+1}$ is sequentially compact, so there is an infinite $N_{\ell+1}\subseteq N_\ell$ such that $\left\langle x_{\ell+1}^{(n)}:n\in N_{\ell+1}\right\rangle$ converges to some $y_{\ell+1}\in X_{\ell+1}$. In this fashion we recursively construct infinite subsets $N_\ell$ of $\Bbb N$ and points $y_\ell\in X_\ell$ for each $\ell\in\Bbb N$ such that $N_{\ell+1}\subseteq N_\ell$ and $\left\langle x_\ell^{(n)}:n\in N_\ell\right\rangle$ converges to $y_\ell$.

Now let $n_0=\min N_0$, for each $k\in\Bbb N$ let

$$n_{k+1}=\min\{m\in N_{k+1}:m>n_k\}\,,$$

and let $N=\{n_k:k\in\Bbb N\}$. Then $\langle n_k:k\in\Bbb N\rangle$ is a strictly increasing sequence of non-negative integers, so $\sigma'=\left\langle x^{(n)}:n\in N\right\rangle=\left\langle x^{(n_k)}:k\in\Bbb N\right\rangle$ is a subsequence of $\sigma$.

Let $\ell\in\Bbb N$ and consider the sequence $\left\langle x_\ell^{(n_k)}:k\in\Bbb N\right\rangle$ in $X_\ell$. For all $k\ge\ell$ we have $n_k\in N_k\subseteq N_\ell$, and $\left\langle x_\ell^{(n)}:n\in N_\ell\right\rangle$ converges to $y_\ell$, so $\left\langle x_\ell^{(n_k)}:k\in\Bbb N\right\rangle$ also converges to $y_\ell$. I’ll leave the last step to you.

Let $y=\langle y_k:k\in\Bbb N\rangle\in X$. Use the result of the last paragraph to show that $\sigma'$ converges to $y$ in $X$ and hence that $X$ is sequentially compact (and therefore compact).


问:
Find the $3 \times 3$ matrix
Question is
$$
\text{adj} B =
\left[
\begin{array}{ccc}
         -45& 33& -28 \\
         32& -10& -16\\
        -4& -24& 2\\
\end{array}
\right]
$$

If $\det B = −202$, then find the $3\times 3$ matrix $B$, whose adjoint is given as above.

First, I wrote the $3x3$ matrix as $$\begin{bmatrix}a & b & c\\
                                 d & e & f\\
                                 g & h & i\end{bmatrix}$$. Then I found the cofactor matrix and typed its transpose. As a result,
$$
\begin{cases}
-fh&+ei&=-45 \\
 -id&+fg&=32 \\
 dh&-eg&=-4 \\
 -ib&+ch&=33 \\
 ia&-cg&=-10 \\
 -ah&+bg&=-24 \\
 bf&-ec&=-28 \\
 -af&+cd&=-16 \\
 ea&-bd&=2
\end{cases}
$$

came from the transactions I made. Then I applied the determinant formulas and replaced what I had just found. $$\begin{cases}-45a+32b-4c=-202 \\
                 33d-10e-24f=-202\\
                -28g-16h+2i=-202\end{cases}$$   But I got stuck here. How can I find $a, b, c, d, e, f, g, h, i$ here? I will be grateful if you could help me.

答:
$B^{-1} = \frac{adj(B)} {\det B} \implies B = \frac{(adj(B))^{-1}}{\det B} $

Can you do now?

Title: Find the $3 \times 3$ matrix


问:
Linearization: Selecting of state variables to make the Jacobian matrix square
I have 5 equations like:

My state variables are:

I cannot make Jacobi matrix square according the this state vector. What do you suggest for this operation? Should i change state vector, or can i produce 7 equations by using this vector?

Thanks,

NOTE: Arastas' approach.

答:
$\dot{w}_z$ and $\dot{w}_x$ are not state variables since you do not have equations with their derivatives, namely $\ddot{w}_z$ and $\ddot{w}_x$. Remove them, and you have five equations for five states.

UPD

Define $x = [u,\ v,\ w_z,\ w_x,\ \theta]^\top$. Then the system can be written as (check it, typos are possible)
$$
\begin{bmatrix}2 & 0 & 0 & 0 & 0\\ 0 & 5 & -1 & -1 & 0 \\ 0 & -10 & 4 & 3 & 0 \\ 0 & -1 & 14 & 10 & -1 \\ 0& 0& 0& 0& 1\end{bmatrix} \begin{bmatrix}\dot{x}_1 \\ \dot{x}_2\\ \dot{x}_3\\ \dot{x}_4\\ \dot{x}_5\end{bmatrix} = \begin{bmatrix}x_3^2 + x_3x_4 + 2x_3x_2 \\ -5x_1x_3 \\ 10x_1x_3 \\ x_1x_3 + x_5\\ x_4\end{bmatrix}
$$
that is
$$
T \dot{x} = \psi(x).
$$
The matrix $T$ is invertible, and thus your model is
$$
\dot{x} = T^{-1}\psi(x) = f(x).
$$
Then you can compute the Jacobian.


答:
I think it's difficult to beat "Real Analysis" by John. M. Howie from Springer. I come from an engineering background and only started reading this book after I had completed my undergrad & MSc in engineering: having realized that after 5 years of studying, I still completely lacked the rigorous basics of Mathematics.

The book above is "thin" enough to be read within within 1 year including exercises (I had gone through the entire book just using my weekends: took me less than 12 months). It most definitely provided the basics that then allowed me to progress further.

Some of the topics you mention (i.e. Rigorous Lebesgue Integration) are at what I would describe "Graduate-level Mathematics": the book suggested above is for udergrads. But the approach I took is to start with that and then selectively study graduate topics that were of interest to me (i.e. Lebesgue integrals, probability measure theory).

I think the above approach is better than just picking up a thick book with Graduate-level mathematics, because as you point out: it would take too much time to try to build up "general graduate-level mathematical knowledge" by self-learning.

After reading the book from Howie, I then for example went through the MIT open-courseware lecture notes here: I felt the notes start where the book from Howie "ends", i.e. the book allowed me to "pick up" Grad-level topics that would have been previously unapproachable to me.

PS: It is an interesting topic in itself why Engineering courses (including graduate-level) do not teach mathematics rigorously. I think it's a huge problem and should be addressed.

Title: Mathematical rigor for engineers

Body:
An engineer with a master's degree, I would like to improve my level of mathematical rigor. For example, I would like to come to grasp

what measure theory has to do with the Fourier series and

what group theory has to do with the eigenvalue,

among others.  I would like to understand why Lebesgue integration is important (the idea seems sort of trivial to me, which naturally means that I don't yet really get it), why the formality of Weierstrassian tests is preferable to the mere evasion of the problem by, say, the extraction of Hadamard's finite part, and so on.

To this end, I have been reading Georgi E. Shilov's Elementary Real and Complex Analysis, which I like; but at the rate Shilov is going it would take like 10,000 pages to advance as far as the Fourier series.  Therefore, I have looked instead on Amazon and Google Books for the book Mathematical Rigor for Engineers.

Unfortunately, no one seems to have written that book.

What are my options?

(To clarify: I do not mean to ask the opinion-based question, "What should I do?" Rather, I mean to ask the experience-based question, "What are my options?" Maybe the only answer is that what I seek resembles Financial Analysis for Kindergarteners or Transoceanic Logistics for Retail Clerks, but since I happen to possess an applicationist's acquaintance with mathematical productions like the Fourier Series and the eigenvalue, not to mention Hadamard's finite part, I thought that I would ask.)


问:
Inequality concerning a function :${(f(x))}^2=1+xf(x+1)$

If $f:[0,\infty]\rightarrow [0,\infty]$ is such that for all $x\in [0,\infty]$ :$${(f(x))}^2=1+xf(x+1)$$  Prove that for all $x\ge 1$ $$\frac{x+1}{2}\le f(x)\le  2(x+1)$$

First of all I have never had experience proving such inequalities and since nothing is said about the differentiability I don't know whether calculus would be a good approach. We can however draw the following conclusions for all $x\ge 1$

As $f(x)\ge 1$ we have $$f(x)^2=1+xf(x+1)\ge 1+x \implies f(x)\ge \sqrt{x+1}$$

By Bernoulli's inequality $$f(x)=\sqrt{1+xf(x+1)}\le 1+\frac{xf(x+1)}{2}$$

As $f(x+1)\le f^2(x+1)$ and $x\le x^2$ $$f(x)^2=1+xf(x+1)\le 1+x^2f^2(x+1)$$  $$(f(x)-xf(x+1))(f(x)+xf(x+1))\le 1$$

That's it, I couldn't get any more ideas ...

P.S. The problem is from here.

答:
A stronger lower bound

Define the functions $a_n: [0, \infty) \to \Bbb R$ recursively as
$$
\begin{align}
 a_0(x) &= 1 \, ,\\
 a_{n+1}(x) &= \sqrt{1+xa_n(x+1)} \, .
\end{align}
$$
In Nested Radical of Ramanujan it is shown that for all $x \ge 0$
$$
 \lim_{n\to \infty} a_n(x) = x+1 \, .
$$
On the other hand, $f(x) \ge a_n(x)$ follows from the given functional equation with induction. Therefore
$$
 f(x) \ge x + 1 \text{ for all } x \ge 0.
$$

There is no upper bound

Choose an arbitrary function $g:[0, 1] \to \Bbb R$ with $g(0) = 1$ and $g(x) \ge 1+x$. Define $f$ on $[0, 1]$ as $f(x) = g(x)$, and extend the definition to all intervals $(n, n+1]$, $n = 1, 2, 3, \ldots$, with
$$
 f(x) = \frac{f(x-1)^2 - 1}{x-1} \, .
$$
Note that $f(x-1) \ge x$ on $(n-1, n]$ implies that $f(x) \ge x+1$ on $(n, n+1]$, so that $f$ is well-defined for all $x \ge 0$ and satisfies the functional equation.

But $f(1) = g(1)$ can be arbitrarily large.

Remark: If $g$ is chosen as a differentiable function with $g(1) = 2g(0)g'(0)$ then $f$ is continuous at $x=1$ and therefore continuous everywhere. An example would be $g(x) = 1+kx + (k-1)x^2$ for any $k \ge 1$.


问:
Find the integer pair (a,b) that satisfies this
I need some help finding a pair of integers $(a,b)$ that satisfies this constraint:
$x^2 - ax - b = 0$ has integer solutions and $x^2 +ax+b =0$ has integer solutions. The pair $(5,6)$ satisfies this because $x^2 - 5x - 6 = 0$ has integer solutions i.e. x = {-1,6} and $x^2 + 5x + 6 = 0$ has integer solutions i.e. x = {-3,-2}.

If there is no general form can you help me to at least find another pair of integers that satisfies this constraint other than $(5,6)$? Thanks in advance.

EDIT: the roots of the quadratic equations $x^2 - ax - b = 0$ hand $x^2 +ax+b =0$ can be rational also. The need not be constrained to integers. Just make sure they aren't irrational. The pair (a,b) however must be integers.

答:
Let the discriminants of the two equations be $a^2+4b=k^2$ and $a^2-4b=l^2$. Note that

$a,k,l$ have the same parity, so the discriminants being perfect squares is enough to ensure integer solutions

the difference of two squares of the same parity is always a multiple of $4$, so we need not worry about $b$ being non-integral – it suffices to get $a,k,l$

Now $l^2,a^2,k^2$ are perfect squares in arithmetic progression, so $l^2+k^2=2a^2$ or $(l/a)^2+(k/a)^2=2$. All rational solutions for $l/a$ and $k/a$ may be parametrised as follows:
$$\frac la=\frac{m^2-2m-1}{m^2+1}\qquad\frac ka=\frac{m^2+2m-1}{m^2+1}$$
where $m\in\mathbb Q\cup\{\infty\}$.

Given a rational solution for $\frac la$ and $\frac ka$, write the fractions with a common denominator, from which $l,a,k$ are immediately determined up to sign (their signs must all agree). From there $b$ may be easily worked out.


答:
The product of the roots of one equation must be the negative of the product of the other pair of roots. Also, the sum of the roots of one equation must be the negative of the sum of the other pair of roots.

A formula for this is to let the pairs of roots be

$L^2-3LM+2M^2,LM$ and $M(L-2M),L(M-L)$

for integers $L$ and $M$.

Title: Find the integer pair (a,b) that satisfies this

Body:
I need some help finding a pair of integers $(a,b)$ that satisfies this constraint:
$x^2 - ax - b = 0$ has integer solutions and $x^2 +ax+b =0$ has integer solutions. The pair $(5,6)$ satisfies this because $x^2 - 5x - 6 = 0$ has integer solutions i.e. x = {-1,6} and $x^2 + 5x + 6 = 0$ has integer solutions i.e. x = {-3,-2}.

If there is no general form can you help me to at least find another pair of integers that satisfies this constraint other than $(5,6)$? Thanks in advance.

EDIT: the roots of the quadratic equations $x^2 - ax - b = 0$ hand $x^2 +ax+b =0$ can be rational also. The need not be constrained to integers. Just make sure they aren't irrational. The pair (a,b) however must be integers.


Title: How to find the limit $\lim_{n\to \infty}\left(n-\frac{n}{e}\left(1+\frac{1}{n}\right)^n\right)$

Body:
To find the limit

$$\displaystyle\lim_{n\to \infty}\left(n-\frac{n}{e}\left(1+\frac{1}{n}\right)^n\right)$$

Body:
Let $x=1/n$, then
$$L=\lim_{x\to 0}\left(\frac{1}{x}-\frac{1}{ex}(1+x)^{1/x}\right)$$
Use the McLaurin series: $$(1+x)^{1/x}=e-ex/2+11ex^2/24+...$$
$$L=\lim_{x \to 0} \left( \frac{1}{x}-\frac{1}{x}+\frac{1}{2}-\frac{11x}{24}+...\right)=\frac{1}{2}.$$


问:
Calculate the length of curve $f(x)=\arcsin(e^x)$, check solution, please.
As in the topic, my task is to calculate the length of $f(x)=\arcsin(e^x)$ between $-1, 0$. My solution: I use the the fact, that the length of $f(x)$ is equal to $\int_{a}^b\sqrt{1+(f'(x))^2}dx$ $$f'(x)=(\arcsin(e^x))'=\frac{e^x}{\sqrt{1-e^x}}\rightarrow (f'(x))^2=\frac{e^{2x}}{1-e^x}$$Next I use binomial coefficient:$$ \left (1+ \frac{e^{2x}}{1-e^x}\right )^\frac{1}{2}=\sum_{n=0}^{\infty}\binom{\frac{1}{2}}{n}\frac{e^{2xn}}{(1-e^x)^n}$$Let's evaluate $\binom{\frac{1}{2}}{n}=\frac{1}{n!}\cdot(\frac{1}{2})(-\frac{1}{2})(-\frac{3}{2})\cdot\dots\cdot(\frac{-2n+3}{2})=\dots=\frac{(-1)^{n-1}}{(2n-1)(n!)^2}\cdot\binom{2n}{n}$Hence the length is $$\int_{-1}^0\sum_{n=0}^{\infty}\frac{(-1)^{n-1}}{(2n-1)(n!)^2}\cdot\binom{2n}{n}\cdot \frac{e^{2xn}}{(1-e^x)^n}dx$$Normally I would integrate it but it is nasty and I am not sure whether it is correct. Any hints, ideas? Thanks in advance. </p>

答:
Your derivative is slightly off, because you've neglected the squaring in the denominator of $\arcsin$'s derivative. The integral winds up as $$\int_{-1}^0\sqrt{\frac{1}{1-e^{2x}}}\,dx$$ which actually has an elementary antiderivative, so there will be no need for an infinite series. Substituting $$e^{x}=\sin(u)$$ gives

$$\int_{\arcsin(e^{-1})}^{\pi/2}\sqrt{\frac{1}{1-\sin^2(u)}}\,\frac{\cos(u)\,du}{\sin(u)}$$

or just $$\int_{\arcsin(e^{-1})}^{\pi/2}\csc(u)\,du$$


问:
How can I know when to negate quantifiers when taking the contrapositive of a statement?
Continued from a question I asked here, since I believed this question deserves its own thread.

When taking the contrapositive, I was taught to negate the quantifiers as well. For example, if we have the statement $$\forall x \in S, \exists y \in S, P(x) \implies P(y),$$ where $P$ is a statement over the domain $S$. Taking the contrapositive of the above statement would give us:
$$\exists y \in S, \forall y \in S, \neg P(y) \implies \neg P(x),$$ correct? The logic behind this makes sense to me. Consider the simple example:

If every car has wheels, then there is a car with at least one wheel.

The contrapositive of this would be:

If there is a car which does not have at least one wheel, then not every car has wheels.

Symbolically, we could say that the original statement is:

$\forall c, \exists d, c \implies d$, and the contrapositive as $\exists c, \forall d, \neg d \implies \neg c.$

However, recently I've found out that this is not always the case. If we have something along the lines of:

If $a$ is a real number, such that $a < b$ for every positive real number $b$, then $a = 0$.

Expressing this symbolically, I would think we have that $$\exists a \in \mathbb{R}, \forall b \in \mathbb{R}^+, a < b \implies a = 0.$$

However, when we take the contrapositive of this statement, we get $$\exists a \in \mathbb{R}, \forall b \in \mathbb{R}^+, a \neq 0 \implies a \ge b,$$ and not $$\forall a \in \mathbb{R}, \exists b \in \mathbb{R}^+, a \neq 0 \implies a \ge b.$$
How can I know when to negate quantifiers, and when not to?

答:
It seems that what you are actually applying to your logical statements is inversion, rather than contraposition. If we have a logical statement $P \rightarrow Q$ (which reads "if $P$ then $Q$") then the contrapositive (which is logically equivalent to the original logical statement) is $\neg Q \to \neg P$ (which reads if not $Q$ then not $P$).

In the context of your last example, the $\exists$ quantifier means "there exists", whereas your intended meaning is that it holds for any (non-negative?) real number $a$ that you choose. So given a non-negative real number $a$, the following statement is true:
$$
\forall b \in \mathbb{R}^+ ,\, a < b \Rightarrow a = 0 .
$$
(In English: if $a$ is a non-negative real number such that for all positive real numbers $b$, $a$ is less than $b$, then $a = 0$.)

Then the contrapositive is
$$
a \ne 0 \Rightarrow \exists b \in \mathbb{R}^+ :\, a < b .
$$
(If $a$ is not zero, then there exists a positive real number $b$ such that $a < b$.) Note that we flip the implication (i.e. the arrow), and negate the propositions, which includes negating the quantifiers. Perhaps see if this interpretation aligns with what you were taught?


问:
How can I get a vector inside a square as a function of two others?
The problem is as follows:

The figure from below represents three vectors $\vec{u}$, $\vec{v}$ and
$\vec{x}$. Find $\vec{x}$ as a function of a linear combination of the

other two. Assume the figure is a square.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&3\vec{u}-2\vec{v}\\
2.&4\vec{u}-3\vec{v}\\
3.&\frac{3\vec{u}+2\vec{v}}{4}\\
4.&\frac{2\vec{u}+3\vec{v}}{4}\\
5.&\frac{3\vec{u}+2\vec{v}}{5}\\
\end{array}$

How exactly should this problem should be approached?. I attempted different ways to accomodate the vectors but I have ran out of ideas.

From what I can see, it seems that it is intended to assume that $M$ and $N$ are midpoints. Although it has not been indicated otherwise.

But if that were the case, I can see that the angle which vector $\vec{v}$ makes with the horizontal, assuming the horizontal is the bottom edge of the square, would be $\frac{53^\circ}{2}$ as it is in a proportion of $2:1$ ratio. But, other than that I don't know what else can be done to simplify or solve this thing.

Can someone help me here please?.  because I'm stuck.

答:
I skip writing the vector symbols to type it faster.

A possible way is using a secant-tangent relation.

For simplicity let's set the side length of the square to $1$. So you have

$$v = OR + \frac 12 u \text{ and } OM = u + \frac 12 OR$$
So, you get
$$OM = \frac 12 v+\frac 34 u$$

Now, $x = t\cdot OM \Rightarrow t = \frac{|x|}{|OM|}$. Using above mentioned secant-tangent relation, you get

$$|PM|^2=\frac 14 = (|OM|-|x|)|OM|$$

Pythagoras gives $|OM| = \frac{\sqrt 5}{2}$ and solving for $|x|$ gives $|x| = \frac 2{\sqrt 5}$. Hence,

$$x = \frac{|x|}{|OM|}\left(\frac 12 v+\frac 34 u\right) = \frac 45\left(\frac 12 v+\frac 34 u\right)=\boxed{\frac{2v+3u}5}$$


问:
Behavior of $|\Gamma(z)|$ as $\text{Im} (z) \to \pm \infty$
Let $\Gamma(z)$ be the gamma function.

In a paper I'm reading, the author states that $$ |\Gamma(z)| = |\Gamma(a+ib)|  \sim \sqrt{2 \pi} |b|^{a-\frac{1}{2}} e^{-|b|\frac{\pi}{2}}$$ as $|b| \to \infty$.

Can this asymptotic behavior be derived from Stirling's formula?

EDIT:

I think I have something.

Assume that $a,b >0$ and that $b$ is very large.

Then it would seem that
$$ \begin{align}|\Gamma(a+ib)|  &\sim \left|\sqrt{\frac{2 \pi}{a+ib}} \Big(\frac{a+ib}{e}\Big)^{a+ib} \right| \\ &= \sqrt{2 \pi}  \left|(a+ib)^{a-\frac{1}{2}} \right| \left| (a+ib)^{ib} \right| \left|e^{-a-ib} \right| \\ &= \sqrt{2 \pi} \left(\sqrt{a^{2}+b^{2}} \right)^{a- \frac{1}{2}} \ \left|\left(\sqrt{a^{2}+b^{2}} e^{i \arg \left(\frac{b}{a}\right)} \right)^{ib} \right| e^{-a} \\ &= \sqrt{2 \pi} \left(\sqrt{a^{2}+b^{2}} \right)^{a- \frac{1}{2}} e^{-b \arg \left(\frac{b}{a} \right)} e^{-a} \\   &\sim \sqrt{2 \pi} \, b^{a - \frac{1}{2}} e^{-b \frac{\pi}{2}} {\color{red}{e^{-a}}}. \end{align}$$

But apparently this is not quite correct.

答:
Let $z=a+ib$.  As $\vert  b \vert \to \infty$, the argument of $z$ does not approach $\pi$.  Thus Stirling's approximation applies, giving

$$\vert \Gamma(z) \vert \sim \left\vert \sqrt{\frac{2\pi}{a+ib}}\left(\frac{a+ib}{e}\right)^{a+ib}\right\vert\sim \sqrt{2\pi}\left\vert(ib)^{-1/2}\left(\frac{a+ib}{e}\right)^{a+ib}\right\vert$$
$$\sim\sqrt{2\pi}\vert b \vert^{-1/2} \left\vert \left(\frac{a+ib}{e}\right)^{a+ib}\right\vert \sim \sqrt{2\pi}\vert b \vert^{-1/2}e^{-a} \vert z^z\vert,$$

The $\vert z^z \vert$ term above satisfies

$$\vert z^z \vert = \vert e^{z \log z} \vert = \vert e^{(a+ib)(\log \vert z \vert + i \mathrm{arg}z)}\vert=e^{a \log \vert z \vert-b \mathrm{arg} z} \sim \vert b \vert^a e^{-b \mathrm{arg} z}$$
because $\vert z \vert \sim \vert b \vert$ as $b \to \infty$.  Thus, it remains to approximate $e^{-b \,\mathrm{arg z}}$ as $\vert b \vert \to \infty$.

Note: An easy mistake at this point is the false implication
$$\mathrm{arg}z \to \pm \pi/2 \quad \Longrightarrow \quad e^{b \mathrm{arg} z} \to e^{\pm b \pi/2}.$$
However, this need not hold, because the exponential is sensitive to non-dominant terms.

We compute
$$\lim_{\vert b \vert \to \infty} e^{-b\, \mathrm{arg} z}e^{-a+\vert b \vert \pi/2}=\mathrm{exp}\lim_{\vert b \vert \to \infty}\left(-b \arctan(b/a)-a+\vert b \vert \pi/2\right)$$
$$=\mathrm{exp}\lim_{\vert b \vert \to \infty}-b\left(\frac{\pm\pi}{2}-\frac{a}{b}+O\left(\vert b \vert^{-3}\right)\right)-a+\vert b \vert \pi/2,$$
in which we take the positive sign if $b \to \infty$ and the negative sign if $b \to -\infty$.  (This is simply the Taylor series to $\arctan(b/a)$ at $\pm \infty$.) Therefore we can replace said $\pm$ with $\vert b \vert/b$, at which point we see that our limit is $1$.

Thus
$$\vert \Gamma(z) \vert \sim \sqrt{2\pi}\vert b \vert^{-1/2}e^{-a}\vert b \vert^a e^{-b \,\mathrm{arg} z} \sim \sqrt{2\pi}\vert b \vert^{-1/2}e^{-a}\vert b \vert^a e^{a-\vert b \vert \pi/2}$$
$$\sim\sqrt{2\pi}\vert b \vert^{a-1/2} e^{-\vert b \vert \pi/2},$$
as claimed.


答:
The adjugate satisfies the relation
$$
B\operatorname{adj}B=(\det B)I\qquad(*)
$$
and therefore, if $B$ is invertible,
$$
B=(\det B)(\operatorname{adj}B)^{-1}
$$
Your final linear system is part of the relation $(*)$, if you notice, but you forgot some equations.

With some patience, you get
$$
(\operatorname{adj}B)^{-1}=
\begin{bmatrix}
-1/101 &  3/202 & -2/101 \\
     0 & -1/202 & -4/101 \\
-2/101 & -3/101 & -3/202
\end{bmatrix}
$$

Title: Find the $3 \times 3$ matrix

Body:
Question is
$$
\text{adj} B =
\left[
\begin{array}{ccc}
         -45& 33& -28 \\
         32& -10& -16\\
        -4& -24& 2\\
\end{array}
\right]
$$

If $\det B = −202$, then find the $3\times 3$ matrix $B$, whose adjoint is given as above.

First, I wrote the $3x3$ matrix as $$\begin{bmatrix}a & b & c\\
                                 d & e & f\\
                                 g & h & i\end{bmatrix}$$. Then I found the cofactor matrix and typed its transpose. As a result,
$$
\begin{cases}
-fh&+ei&=-45 \\
 -id&+fg&=32 \\
 dh&-eg&=-4 \\
 -ib&+ch&=33 \\
 ia&-cg&=-10 \\
 -ah&+bg&=-24 \\
 bf&-ec&=-28 \\
 -af&+cd&=-16 \\
 ea&-bd&=2
\end{cases}
$$

came from the transactions I made. Then I applied the determinant formulas and replaced what I had just found. $$\begin{cases}-45a+32b-4c=-202 \\
                 33d-10e-24f=-202\\
                -28g-16h+2i=-202\end{cases}$$   But I got stuck here. How can I find $a, b, c, d, e, f, g, h, i$ here? I will be grateful if you could help me.


问:
Find the probability that the number of defective boards is at least $5$?
When circuits boards used in manufacture of compact disc(CD) players are tested , the long-run percentage of defective is $5\text{%}$. Assume that there are $25$ boards in a sample. Find the probability that the number of defective boards is:

$a).$ At least 5:

I think since $20\leq n$ and $\theta \leq0.05$ it is good to use poisson approximation to binomial distribution

so $\lambda=25\times0.05=1.25  $,
$P(X=x)=\dfrac{e^{-1.25}(1.25)^x}{x!}$

we have to find

$$P(x\geq 5 )=1-P(x<5)$$

But I saw in many website as well as
Statistics - Probability of getting a number using combinations  used only binomial distribution so I got problem why it is not used poisson distribution?

Can anyone help me is there anything wrong here?

答: The Poisson approximation only gives an approximate answer. To get an exact answer you must use a $B(25,0.05)$ distribution. The probability can be easily found by $$1-\sum_{k=0}^4\binom{25}{k}(0.05)^k(0.95)^{25-k}$$


问:
Find equation that denotes straight line that intersects middle of chord that is expressed by another equation
$(x-2)^2 + (y + 1)^2 = 16$ is circle

$x - 2y - 3 = 0$ is equation that denotes chord of this circle

How do I find equation of straight line that intersects middle of this chord?

答:
Line $L_2$ perpendicular to $L_1 : x-2y-3=0$ at the midpoint of the chord passes through the center $(2, -1) $.

$$ L_2 : 2x+y=2(2) +(-1) =3$$

Midpoint is their intersection : $(9/5, -3/5) $

Required equation is $y+3/5=m(x-9/5) $


问:
Direct Summation of the same space , meaning?
In a course , there is the definition of the notation :(for V a G module over a finite group)
$$V^{\oplus n}=V\oplus V\oplus ...\oplus V\quad(\text{n times})$$
But i don't understand the sense of this object , for $x\in V^{\oplus n}$ there is no unicity for the writing x=$v_1+v_2+...+v_n$ so what's the meaning of the direct sum , i don't get it,

答:
This is called the exterior direct sum. As a vector space, it is the Cartesian product $V^n$ equipped with componentwise addition. Though there is a slight difference where infinite direct sums are concerned. In general,
$$\bigoplus_{i\in I} V_i=\{(v_i)_{i\in I}~\vert~v_i=0\textrm{ for all but finitely many }i\in I\}$$
equipped with componentwise addition and $\lambda(v_i)_{i\in I}=(\lambda v_i)_{i\in I}$ as multiplication.

But the idea is that the exterior direct sum is a vector space into which you can embed all the summands such that the space is the interior direct sum (the one you're familiar with) of the embedded spaces. In this way, we have, for instance $\mathbb R^2\cong \mathbb R\oplus\mathbb R$, since we can embed the first instance of $\mathbb R$ into $\mathbb R^2$ like this: $r\mapsto(r,0)$, and the second instance like this: $r\mapsto(0,r)$. The corresponding images are subspaces of $\mathbb R^2$, and $\mathbb R^2$ is the (interior) direct sum of those subspaces. And then we say that it is the exterior direct sum of two instances of $\mathbb R$ . Basically, we identify the exterior direct summands with fitting interior direct summands of their exterior direct sum.


答:
By intersecting, I suppose that you are looking for a line that it is a perpendicular bisector to the given chord (intersecting lines through the mid point are infinite, so I am assuming that you are referring to the line passing through the mid point and perpendicular to the given chord).

If that is the case, then you may proceed as follows ;

The perpendicular bisector will have a slope 'm', where m satisfies m*(1/2)=-1 (condition for lines with perpendicular slopes)

To find the mid point of the chord, find the coordinates of the foot of perpendicular of the center of the circle (2,-1) in the given chord ( using the foot of perpendicular formula )

Using the slope obtained in 1, and the point obtained in 2, one may easily write the equation of the required line.

I am also attaching a picture of the solution for the problem.

Title: Find equation that denotes straight line that intersects middle of chord that is expressed by another equation

Body:
$(x-2)^2 + (y + 1)^2 = 16$ is circle

$x - 2y - 3 = 0$ is equation that denotes chord of this circle

How do I find equation of straight line that intersects middle of this chord?


问:
let $a^2 >4bc $ such that $a,b ,c \neq 0$ then find $a , b , c$

let $a^2 >4bc , b^2>4ac ,c^2 > 4ab $ such that $a,b ,c \neq 0$ then find range of $a , b , c$

in fact I want that $ax^2+bx+c =0 , bx^2+ax+c=0,cx^2+ax+b=0$ have Two distinct roots
.

答:
If $a,b,c$ are all positive or all negative, then $a^2b^2c^2>64a^2b^2c^2$ which is impossible.

Without loss of generality suppose that two of the variables, $b$ and $c$ say, have the same sign. Then we simply require $|a|>2\sqrt {bc}$ with $a$ having the opposite sign to $b$ and $c$..


问:
Proper term for a sequence that is the sequence of sums of each positional element in two sequences?
Is there a proper English term for Z in this example?

a = [ 1, 2, 4 ]

b = [ 1, 1, 1 ]

Z = [ 2, 3, 5 ]

答:
You can say that $Z$ is the "pointwise sum of $a$ and $b$" or "coordinate-wise sum".

This can be seen as addition in a three dimensional vector space (or free module).


问:
How is $y'=x^2+1$ a differential equation?
From James Stewart Essential Calculus Early Transcendentals Textbook,

A differential equation is an equation that contains an unknown function and one or more of its derivatives.

But the equation
$y'=x^2+1$
doesn't contain the unknown function $y$, it only contains $y'$ and $x$. And yet my textbook says its a DE of order 1.

Even Wikipedia says a DE "relates one or more functions and their derivatives", but $y'=x^2+1$ doesn't contain the function $y$ so how is it a DE ?

答:
I think, the following definition may clear your doubts in all matter. This is taken from "An Introduction to Differential Equations and Their Applications" by Stanley J Farlow.

A differential equation is an equation that relates the derivatives of an unknown function, the function itself, the variables by which the function is defined, and constants.

If the unknown function depends on a single real variable, the differential equation is called an ordinary differential equation.

A partial differential equation is one that contains partial derivatives with respect to more than one independent variable.

An $n^\text{th}$-order ordinary differential equation is an equation that has the
general form $$F\left(x, y, y',\cdots,y^{(n)}\right) = 0\tag1$$where the unknown quantity $y = y(x)$ is called the dependent variable, the real variable, $x$, is called the independent variable, the primes $"~'~"$ denote differentiation with respect to $x$, that is, $y' = \frac{∂y}{∂x},~~ y'' = \frac{d^2y}{dx^2}$ and so on.

Now lets come to your equation, which can be written as $~y'=x^2+1\implies y'-x^2-1=0~.$
So here it contains derivatives of an unknown function $~y~$, which is $y'$, the variable $~x,~$ and the constant $1~.$ Hence by the definition given above it is a differential equation and in particular it is an ordinary differential equation.


问:
Finding the posterior PMF using Bayes Theorem
The problem is as follows

Nefeli. a student in a probability class, takes a multiple-choice test with 10 questions and 3 choices per question. For each question. there are two equally likely possibilities, independent of other questions: either she knows the answer, in which case she answers the question correctly. or else she guesses the answer with probability of success $1/3$.

Given that Nefeli answered correctly 6 out of the 10 questions, what is the posterior PMF of the number of questions of which she knew the answer?

The textbook solution is as follows:

Let $K$ be the r.v. she knows the answer and $C$ be the r.v. she answers the question correctly
$$P(K|C) = \frac{P(C|K)P(K)}{P(C|K)P(K) + P(C|K^C)P(K^C)} = \frac{\frac{1}{2}}{\frac{1}{2} + \frac{1}{3}\frac{1}{2}} = \frac{3}{4}$$
The probability that Nefeli knows the answer to a question that she answered correctly is 3/4, so the posterior PMF is binomial with $n = 6$ and $p = 3/4$.

I decided I would try to approach this problem using Bayes Theorem, that is let $X$ be the r.v. for the number of questions correctly answered and $Y$ be the r.v. for the number of questions she knows the answers to. Thus we have:

$$p_{Y|X}(y|6) = \frac{p_{X|Y}(6|y)p_Y(y)}{p_X(6)}$$

$p_X(6)$ is the probability of answering 6 questions correctly which is just a binomial with $n=10$ and $p=P(C)=\frac{1}{2} + \frac{1}{3}\frac{1}{2} = \frac{2}{3}$:
$$\binom{10}{6}\left(\frac{2}{3}\right)^6\left(1-\frac{2}{3}\right)^4 $$

$p_Y(y)$ is the probability that she knows the answer for $y$ questions which is a binomial with $n=10$ and $p=\frac{1}{2}$
$$p_Y(y) = \binom{10}{y}\left(\frac{1}{2}\right)^y\left(1-\frac{1}{2}\right)^{10-y} $$

$p_{X|Y}(6|y)$ is the probability of answering 6 questions correctly given that she knows the answer for $y$ questions. This means that there will be $10-y$ questions to choose from and she will only need to answer $6-y$ more correctly with probability $P(C)$:
$$p_{X|Y}(6|y) = \binom{10-y}{6-y}\left(\frac{2}{3}\right)^{6-y}\left(1-\frac{2}{3}\right)^{10-6} $$

However, when I plug in say $y=3$ just to get a number for this result, it is not the same as the result I get from the book solution. Does anyone know what I may be doing wrong here? I seem to be off by a factor of 2, as in my answer is half the probability of the answer from the binomial with $n = 6$, $p = 3/4$, $k = 3$.

答:
Correcting a small error in your last expression, you should have $$p_{X\mid Y}(6\mid y) = \binom{10-y}{6-y}\left(\frac{1}{3}\right)^{6-y}\left(1-\frac{1}{3}\right)^{10-6}$$

You then get $$p_{Y\mid X}(y\mid 6) = \frac{\binom{10-y}{6-y}\left(\frac{1}{3}\right)^{6-y}\left(1-\frac{1}{3}\right)^{10-6} \binom{10}{y}\left(\frac{1}{2}\right)^y\left(1-\frac{1}{2}\right)^{10-y}}{\binom{10}{6}\left(\frac{2}{3}\right)^6\left(1-\frac{2}{3}\right)^4}$$  and with cancellation this is $$=\frac{6!}{y!(6-y)!}\frac{3^y}{2^{12}} =\binom{6}{y}\left(\frac{3}{4}\right)^{y}\left(\frac{1}{4}\right)^{6-y} $$


问:
Prove that the language L is regular
Let $\Sigma$ be the Latin alphabet ({a, b, c, ..., x, y, z} - 26 letters).

Given the language L =
{$\alpha \in \Sigma$*| if $\alpha$ has the letter a, then $N_a(\alpha)$ = 4 and
if $\alpha$ has the letter b then $N_b(\alpha)$ = 8 and
...
if $\alpha$ has the letter z then $n_z(\alpha) = 2^{27}$
},  prove that L is regular.

How exactly should I go about writing a formal proof on this? I tried to play around with $a^{-1}L$ but failed to accomplish much.

Any help is deeply appreciated.

答: Note that the language is finite: "a" may only appear $0$ or $4$ times, "b" $0$ or $8$ and so on. For each of the $2^{26}$ possible choices of the counts of letters in $\alpha$ there are only a finite number of ways to permute them, leading to a finite number of words in $L$ for the choice made. All finite languages may be recognised by a finite-state machine; finite-state machines recognise precisely the regular languages.

Title: Prove that the language L is regular


问:
Let $A$, $B$ be positive operators in a Hilbert space and $\langle Ax,x \rangle=\langle Bx,x \rangle$ for all $x$, show that $A=B$
Let $A$ and $B$ be positive operators in a Hilbert space $H$, and suppose that $\langle Ax,x\rangle=\langle Bx,x\rangle$ for every $x$ in $H$. Show that $A=B$.</p>

答:
If they are positive then they are selfadjoint and there exists a unique square root. from the fact that $\langle Ax,x \rangle = \langle Bx,x \rangle$ we get that also $A - B$ is positive and
$$0 = \langle Ax - Bx,x\rangle = \|\sqrt{A - B}x\|^2$$
then $\sqrt{A - B} = 0$ and by uniquiness of the square root $A - B = 0$ proving the claim.</p>


问:
Determine the volume of a shape if its dimensions change uniformly
This is a problem I'm trying to solve while creating a simple app by importing a 3D model (of any shape) and applying some changes to it, namely, increasing the size of the model.

When a model is imported, its volume and dimensions are automatically calculated by the 3D model library.

I calculate the dimension increase or decrease in percentages manually e.g. model increase:

$perc = \frac{increase - initial}{initial} \times100$

If the model's dimension is at default 100% and the volume is say $1800cm^3$ what will the volume be if the dimension is increase by say 15% ?

I don't know how to determine a formula for this. Can anyone help me please?

答:
If each length is to increase by a 15% then it is to increase by a factor of 1.15. now if you have a unit cube for example the cube will be of volume $ 1.15^3 $ because the formula for a cube is $ l^3 $ where l is the length of a side and if they all increase by a factor 1.15 we have the new length of a side of the cube is 1.15l and thus the volume is $ (1.15l)^3 $
We can apply the same concept to other shapes based on which lengths you are increasing.

This is called the volume scale factor of enlargement for 3D shapes and would be the Area scale factor for 2D shapes. So if you are still confused searching up them terms will help.