问:
Infinite ways to represent $\pi$ as product of nested square roots of $2$ and $2^n$ and odd numbers
Polygon inscribed in a circle leads to famous infinite product by Viete's formula.here
One way to represent that as infinite nested radical of 2 as follows

$$\pi = \lim_{n\to\infty}2^n \times \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}$$ being total number of '2's inside the nested square roots is 'n'

After some manipulation in numbers some interesting findings I'm able to observe as follows
$$3\pi = \lim_{n\to\infty}2^n \times \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...\sqrt{2-\sqrt{2}}}}}}}$$
$$5\pi = \lim_{n\to\infty}2^n \times \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...\sqrt{2-\sqrt{2-\sqrt{2}}}}}}}}$$

In both these situations total number of '2's inside the nested square roots are 'n'.

As the n grows big each converge to respective multiples of $\pi$ more and more decimals

Python code to verify my findings

```<code>
from decimal import*

getcontext().prec = 100

# nested sqrts of 2 & <span class="math-container">$\pi$</span>, <span class="math-container">$3\pi$</span>,<span class="math-container">$5\pi$</span>

b = int(input("Enter number of iterations: "))

a = Decimal(2).sqrt()

for i in range(b):
    a = Decimal(2 + a).sqrt()

a = Decimal(2 - a).sqrt()

Pi = 2 ** (b + 2) * a

print("\n\n Pi derived from Square as initial regular Polygon is:\n\n ", Pi)

a = Decimal(2).sqrt()
a = 2 - a
a = Decimal(a).sqrt()

b = int(input("Enter required number of cycles to calculate Pi: "))

for i in range(b):
    a = 2 + a
    a = Decimal(a).sqrt()

a = 2 - a
a = Decimal(a).sqrt()

Pi_3 = 2**(b+3)*a
C = Pi_3/3

print("\n\n",C)

a = Decimal(2).sqrt()
a = 2 - a
a = Decimal(a).sqrt()

a = 2 - a
a = Decimal(a).sqrt()

b = int(input("Enter required number of cycles to calculate Pi: "))

for i in range(b):
    a = 2 + a
    a = Decimal(a).sqrt()

#print("\n\n",a)

a = 2 - a

a = Decimal(a).sqrt()

Pi_5 = 2**(b+4)*a
C = Pi_5/5

print("\n\n",C)
# for practical purposes I have divided by 3 & 5 respectively (next to regular derivation by Viete's method) to get the value of <span class="math-container">$\pi$</span>
</code>```

Viete's formula is provable by geometry i.e. square becoming octagon then 16gon and so on ...

Is there any way to prove above 2 formulas by geometric shapes or some other methods? Are there similar infinite ways like those above?(with the help of nested square roots of 2 apart from equilateral triangle, regular Pentagon). Please comment. Thanks in advance

答:
Similarly to formulas, related to Viète's formula, in  your formulas is considered a limit $\lim_{n\to\infty} 2^n\sqrt{2-a_n}$, where $a_n=\sqrt{2+a_{n-1}}$ with the initial conditions
$a_2=-\sqrt{2}$ for the first sequence and $a_3=-\sqrt{2-\sqrt{2}}$ for the second. Assuming $a_{n-1}=2\cos x_{n-1}$ for $x_{n-1}\in [0,\pi]$, we obtain $a_n=2\cos\tfrac {x_{n-1}}{2}$. Putting $x_n=\tfrac {x_{n-1}}2$ for all sufficiently big $n$, we obtain

$$\lim_{n\to\infty} 2^n\sqrt{2-a_n}=\lim_{n\to\infty} 2^n \sqrt{2-2\cos x_n}=\lim_{n\to\infty} 2^{n+1} \sin x_{n+1}.$$

For the first sequence we have $a_2=2\cos\tfrac{3\pi}4$, so $x_2=\tfrac{3\pi}4$ and

$$\lim_{n\to\infty} 2^n\sqrt{2-a_n}=\lim_{n\to\infty} 2^{n+1} \sin \left(\tfrac{3\pi}4\cdot \frac 1{2^{n-1}}\right)=3\pi.$$

For the second sequence we have $a_3=2\cos x_3=-\sqrt{2-\sqrt{2}}<0$. So $x_3\in \left(\tfrac {\pi}2, \pi\right) $ and
$4\cos^2 x_3=2-\sqrt{2}$. Thus $2\cos 2x_3=-\sqrt{2}$ and so $2x_3=\tfrac {5\pi}4$ and

$$\lim_{n\to\infty} 2^n\sqrt{2-a_n}=\lim_{n\to\infty} 2^{n+1} \sin \left(\tfrac{5\pi}8\cdot \frac 1{2^{n-2}}\right)=5\pi.$$


问:
How to model a simple flight schedule in Combinatorial Optimisation
I have trouble with the following modeling exercise in Combinatorial Optimisation:

You want to schedule $n$ flights with as few planes as possible. For each flight $i$ the travel time $t_i$ and the start time $d_i$ are given. Let further be $\delta(i,j)$ the time for all flights $i,j$ be the time required to prepare a plane for the flight $j$ after it is finished with flight $i$.

Find a flight schedule that ensures all flights are done with as few planes as possible.

We are currently learnig about network flows in the lecture, so I suppose we should use this here.

My current attempt would be as follows. Start with a directed graph $G(V,E)$ with $\lvert V(G) \rvert = n$, where each edge $(i,j)$ connects two nodes $i$ and $j$ if the corresponding flights are viable to be performed after another, i.e. $d_i + \delta_{ij} + t_i \le d_j$. To use a network flow algorithm it would now be natural to define a capacity fucntion on this graph, but I do not see how this could help to solve the problem.

Could you help me?

答: A flight schedule corresponds to a path cover in your directed (acyclic!) graph. Finding a schedule that uses the minimum number of planes then corresponds to the so-called minimum path cover problem. This can be solved by reduction to the maximum matching problem in bipartite graphs, as sketched here. If you want, you can unpack this and write the maximum matching problem as a network flow problem.</p>

ContentLicense: CC BY-SA 4.0


问:
Can I use this method in Lagrange interpolation problem?
I want to find third degree Lagrange polynomial for these four points:$(1,2),(2,9),(3,28),(4,65)$.

I know how to use formula to find that. but I want to know whether this alternative method is ok or not:

Suppose we have $y=f(x)$:

$$f(1)=1^3+1$$
$$f(2)=2^3+1$$
$$f(3)=3^3+1$$
$$f(4)=4^3+1$$

Therefore $f(x)=x^3+1$ passes through these four points. Can I consider this function as Lagrange polynomial if problem explicitly mention use Lagrange Interpolation?

答: Absolutely, as the Lagrangian polynomial is unique. You don't have to justify how you obtained the coefficients (provided the degree of the polynomial you exhibit does not exceed the number of points minus one).</p>


问:
Determine if the given set is a subspace of Pn for an appropriate value of n.
I have encountered the following problem that I am struggling with:

Prove that:

(i) The set of all polynomials $f(x)$ in the variable $x$ of the form $f(x) = ax^3$
for some constant $a$ is a subspace of the vector space of all polynomials.

(ii) The set of all polynomials $f(x)$ in the variable $x$ of the form $f(x) = a+x^3$
for some constant $a$ is not a subspace of the vector space of all polynomials.

The solutions in the textbook are:
(i) Since the set is $Span\{t^2\}$, the set is a subset.

(ii) The zero vector is not a subset.

I do not fully understand these explanations. In particular, I do not understand how to prove that the zero vector is not a subset. Can you provide more detailed explanations?

答:
(i) Let us denote by $P$ the set of polynomials in (i) and $\phi(x)=x^3$ . Then

$$P= span \{\phi\}.$$ The span of a set is always a subspace.

(ii) Let us denote by $Q$ the set of polynomials in (ii) and $\psi$ the null - polynomial. Then $ \psi \notin Q.$ Hence $Q$ is not a subspace.


问:
Does the transformation $Ax$ change the orientation of the vector $x$?
Let $x\in\mathbb{R}^N, x\neq 0$ be a vector and $A\in\mathbb{R}^{N\times N}$ be a square matrix. It is true that the orientation of $Ax$ does not change with respect to the oroientation of $x$?

Could anyone please help me to understand why (or not)?

Thank you in advance!

答:
The resulting vector of the multiplication $Ax$, will only have the same orientation of $x$ (angle relative to the base $\left(\begin{smallmatrix}1& 0 \\0 &1\end{smallmatrix}\right)$) if $Ax=\lambda x$, where $\lambda$ is a constant scalar, also know as eigenvalue.
So, both will have the same orientation, if $x$ is an eigenvector of $A$ associated with that $\lambda$.


问:
I need help with proofs using mathematical induction: $2+7+12+17+...+(5n-3)=(\frac{n}{2})(5n-1)$
I need help with this problem: $2+7+12+17+...+(5n-3)=(\frac{n}{2})(5n-1)$</p>

ContentLicense: CC BY-SA 3.0

Body:
$2+7+12+17+...+(5n-3)=(\frac{n}{2})(5n-1)$

This equality holds true for n=1,2,3....... $\ $so we suppose it will hold true for n=k

$2+7+12+17+...+(5k-3)=(\dfrac{k}{2})(5k-1)$

then we will check it for $n=k+1 $
$$2+7+12+17+...+(5k-3)+(5k+2)=(\frac{k+1}{2})(5k+4)$$
$$(\frac{k}{2})(5k-1)+(5k+2)=(\frac{k+1}{2})(5k+4)$$
now  we try to prove LHS=RHS

$$\frac{(5k^2-k)+(10k+4)}{2}$$
$$\frac{(5k^2+9k+4)}{2}$$
$$\frac{(5k^2+5k+4k+4)}{2}$$
$$\frac{5k(k+1)+4(k+1)}{2}$$
$$\frac{(k+1)(5k+4)}{2}$$

hence proved..


问:
Is the union of partially ordered sets also partially ordered?
I am in the process of doing a proof exercise for showing that a function $f : L_1 \to L_2$  between partially ordered sets $L_1 = (L_1, \sqsubseteq_1)$ and $L_1 = (L_1, \sqsubseteq_2)$ is monotone (or isotone or order-preserving) if $\forall l, l^\prime \in L_1 : l \sqsubseteq_1 l^\prime \Rightarrow f(l) \sqsubseteq_2 f(l^\prime)$. I want to use the idea that the union of partially ordered sets also partially ordered; however, I'm unsure if this is even true. Is it true that the union of partially ordered sets also partially ordered?

答:
It depends on what you mean by "a union of partially ordered sets". Naively, the answer is no: you can take $(\{a,b\}, \leq_1)$ and $(\{b,c\}, \leq_2)$, where $a \leq_1 b$ and $b \leq_2 c$ (and $a \neq b \neq c)$. Then in the union $(\{a,b,c\}, \leq)$ we have $a \leq b \leq c$, but $a \not \leq c$, contradicting transitivity.

In general, the union of the two order relations is reflexive and symmetric but not necessarily transitive (unless the posets are disjoint). You can rectify this by taking the transitive closure after taking union.


答:
I assume you mean disjoint union. That is, $(L, \sqsubseteq) = (L_1, \sqsubseteq_1) \sqcup (L_2, \sqsubseteq_2)$ has as underlying set $L = L_1 \sqcup L_2$ and we set $x \sqsubseteq y$ iff $x, y \in L_i$ and $x \sqsubseteq_i y$ for $i \in \{1,2\}$ (so for example $x \not \sqsubseteq y$ if $x \in L_1$ and $y \in L_2$).

Then yes, this is a partial order. You can easily verify this by checking the axioms of a partial order. For example, we can check reflexivity. Let $x \in L$ we have $x \in L_i$ for some $i \in \{1, 2\}$. Since $(L_i, \sqsubseteq_i)$ is a partial order we have $x \sqsubseteq_i x$. So $x \sqsubset x$, as required. The other properties follow similarly, and I leave that to you to check.

Edit. Just as I posted this answer, another answer By Dániel G appeared, which shows very nicely why we need to consider the disjoint union.

Title: Is the union of partially ordered sets also partially ordered?

Body: I am in the process of doing a proof exercise for showing that a function $f : L_1 \to L_2$  between partially ordered sets $L_1 = (L_1, \sqsubseteq_1)$ and $L_1 = (L_1, \sqsubseteq_2)$ is monotone (or isotone or order-preserving) if $\forall l, l^\prime \in L_1 : l \sqsubseteq_1 l^\prime \Rightarrow f(l) \sqsubseteq_2 f(l^\prime)$. I want to use the idea that the union of partially ordered sets also partially ordered; however, I'm unsure if this is even true. Is it true that the union of partially ordered sets also partially ordered?


问:
Find the smallest possible number of different numbers between $a + b, b + c, c + a, ab + 1, bc + 1, ca + 1, abc$

Find the smallest possible number of different real positive number
between $a + b, b + c, c + a, ab + 1, bc + 1, ca + 1, abc$

$a$, $b$, $c$ are real positive numbers and they're all
different. From that,  I can tell that $a+b \ne b+c \ne c+a$, and $ab + 1 \ne bc + 1 \ne ca + 1$, but I couldn't get any further with the exercise. How can I find the answer and prove it?

P. S. I am not very good at these kinds of exercises but couldn't find any good resources. If you know about any, I'd appreciate it if you told me about some.

答:
Consider the inequality chain:
$$ a+b =ab+1 < a+c = ac+1 = abc < b+c < bc+1,$$

which has general solutions $ a = 1,  1 < b < \frac{1+\sqrt{5}}{2} , c = \frac{1}{b-1}$.

As an explicit example, with $a = 1, b = 1.5, c = 2$, we get 4 distinct values: $2.5, 3, 3.5, 4.$

We will show that 4 distinct values is the best we can do.

Proof by contradiction. Suppose we can get 3 distinct values.

WLOG $ a < b < c$.

Notice that $ a+ b < a+c < c + b $, so this must be the 3 distinct values in order.

and $ ab + 1 < ac + 1 < bc + 1$, so this must be the 3 distinct values in order.

Thus, we have $ a + b = ab+ 1, a+c = ac+1, c+b = bc+1$.

This yields $(a-1)(b-1) = 0$, $(a-1)(c-1) = 0$, $(b-1)(c-1) = 0$.

This can be satisfied if and only if (at least) 2 of the variables are equal to 1, which contradicts the assumption that they are distinct.

To come up with the example of 4 distinct values, study the possible chains of intertwining the 2 inequalities, along with $abc$. Several of these lead to solutions. (This is not an exhaustive list)

The initial inequality chain of

$$ a+b =ab+1 < a+c = ac+1 = abc < b+c < bc+1,$$

with solutions $ a = 1,  1 < b < \frac{1+\sqrt{5}}{2} , c = \frac{1}{b-1}$.

We could have

$$ a+b =ab+1 < a+c = ac+1 < abc = b+c < bc+1,$$

with solutions $ a = 1, 1 < b <2, c = 1 + \frac{1}{ b - 1 } $.

We could have

$$ a+b = ab+ 1 = abc , a+c, ac+1, b+c = bc+1, $$

which has solutions $0 < a < 1, b=1, c = \frac{a+1}{a}$.

As an explicit example, $ a = \frac{1}{2}, b = 1, c = 3$.

We could have

$$ a+b < ab+1 = a+c < ac+1 = b+c < bc+1 = abc,$$

with solution $a\approx1.33826, b\approx1.61803, c\approx1.82709$ taken from Wolfram.

We could have

$$ a+b < ab+1 = a+c =abc < ac+1 = b+c < bc+1,$$

with solution $a\approx 1.27716, b \approx 1.42902, c \approx 1.54792$ taken from Wolfram

We could have

$$ a+b = abc < ab+1 = a+c < ac+1 = b+c < bc+1,$$

with solution $a\approx 1.26523, b\approx 1.39919, c\approx 1.50507$ taken from Wolfram.


问:
1/2/inf-Norms of A=kI
Let the matrix A of size N x N, given by A = k I, be the identity matrix of size N x N and k a strictly positive real.

How would I calculate;

-the norms (1, 2, inf) of the matrix;

-its condition number  in each of the standards 1, 2, inf

I'm truly lost. I do have the formulas of norms (1,2,inf) for a matrix, but I can't figure it out.

答:
It is true for any norm that $\|kI\| = |k| \|I\| = k \|I\|$ (because $k>0$). It is also true for any (induced) norm that
$$
\|I\| = \sup_{x \ne 0}\dfrac{\|I x\|}{\|x\|} = \sup_{x \ne 0} \frac{\|x\|}{\|x\|} = 1.
$$

So, for any of these norms, or any other induced norm, $\|k I\| = |k|$.

Title: 1/2/inf-Norms of A=kI


问:
Weak 2-coloring of a graph but one color is stong?
Given a planar graph how can I color its vertices with 2 colors, A and B, given that:

Color A cannot have a neighbor of color A and should have 1 or more neighbors of color B

Color B can have any number of neighbors of color B and should have 1 or more neighbors of color A

Anybody knows the name of an algorithm for such coloring or the name of a coloring like this or a way to solve it?

答:
Given such a coloring, the vertices of color A form an independent dominating set, and conversely, given an independent dominating set, you can color its vertices with color A and the rest of the vertices with color B to get a suitable coloring. This is assuming that the graph contains no isolated vertices, but this is clearly necessary for the existence of such a coloring.

If you just want to find any such coloring, you can greedily take a maximal independent set; this will give you an independent dominating set. Finding the smallest independent dominating set appears to be NP-hard, even for planar graphs. As for concrete bounds, Theorem 5.1. in this survey gives an algorithm that, given a $k$-coloring (in the usual sense!) of the graph, finds an indepent dominating set of size $(k−1)n/k−(k−2)$. Practically, this gives you an independent dominating set of size $4n/5 - 3$, since there are fast algorithms for finding a five-coloring of a planar graph (while finding a four-coloring is difficult at this point).


问:
All theorems that should be considered before proving Fermat's Little theorem?
Hello fellow mathematicians. I was allowed to choose any topic that we learned about in our proof-writing/number theory class. And I decided to go with Fermat's little theorem because I find it really interesting and its applications in cryptography are also cool.

The goal of this assignment is to basically write a 3 page paper proving the theorem but it must be extremely explicit. Meaning I need to provide and prove theorems, lemmas, etc, before actually proving Fermat's Little theorem.

For now all I have considered is proving unique multiplicative inverses (modulo). But that is really all that I can think of currently. Is there anything else I should consider showing to a reader who may not be the most mathematically adept?

My goal with this question is to build a flowchart of how my paper should be laid out. Your help and guidance would be super appreciated!

答:
Here is a roadmap for a nice proof that $a^{p-1} \equiv 1 \bmod p$ when $a \not\equiv 0 \bmod p$.

The map $x \mapsto ax$ is injective in the residue classes mod $p$

The map $x \mapsto ax$ is bijective in the residue classes mod $p$

$1\cdot 2 \cdots (p-1) \equiv (a\cdot1)(a\cdot2)\cdots(a\cdot(p-1))=a^{p-1}(1\cdot 2 \cdots (p-1)) \bmod p$

It's essentially the proof of Lagrange's theorem for finite abelian groups.


问:
Choose best option to lower probability of valuables theaft.
Note:
I constructed this problem for learning purpose. I am not a student and this is no homework.

The Problem:
Two persons 1, 2 have some valuable assets. They have access to 2 storage locations, location 1 and location 2.
Person B is worried about possible robbery of his valuables. Person A is not concerned. Valuables of person B amounts to $x.

Person B wants to allocated his valuables in such a way to minimize total lost value for both persons in case of a robbery takes place.

What amount should person B use in each location so that the total loss of the 2 persons is minimized in case of robbery?

My Attempt:

$P1  = 0.30$ for   Location $1$ is robbed

$P2  = 0.20$ for   Location $2$ is robbed

$P3  = 0.10$ for   Locations $1$ AND $2$ are robbed

$P4  = 0.40$ for   No robbery will occur

Person A uses Location 2 Always.

Person B has 3 main options:

1-Use Location $1$ to Store Value $X$, Expected Loss:

$(0.3+0.1=0.4) X$

2-Use Location $2$ to Store Value $X$, Expected Loss:

$(0.2) Y$

3-Divide valuables between Locations $1$ and $2$ (Place $X-Y$ in one location and $Y$ in another location)

3.A: Place $X-Y$ in Location $1$ AND $Y$ in Location $2$, Expected Loss:

$(X-Y) * 0.3 + Y*(0.2)$  or Should this be $X*(P3=0.1)$?

3.B: Place $X-Y$ in Location $2$ AND $Y$ in Location $1$, Expected Loss:

$(X-Y)*0.2+(X-Y)*0.3$ or Should this be $X*(P3=0.1)$?

For options $3.A$ and $3.B$, $X$ has to be greater than or equal to $Y$ of course.

EDIT
User YJT provided a good point which made me re-phrase the objective.

答:
You can significantly simplify your calculations by using the linearity of the expectation. Let $X_i=1$ if location $i$ is robbed ($0$ otherwise). Suppose he puts $X$ in $1$ and $x-X$ in $2$. Then his expected loss is $XE(X_1)+(x-X)E(X_2)=X(E(X_1)-E(X_2))+const$.

Since $E(X_1)=0.3+0.1$ and $E(X_2)=0.2+0.1$, the loss is $0.1 X+const$. To minimize the loss, he should put $X=0$ in location 1.

This is very intuitive - to minimize the loss, put everything in the safer place. The probability that both are robbed is not important as it loses everything regardless of the split.


问:
Solve: $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$

Solve:   $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$

My Try

$6\sin{x}\cos{x}+4(\cos^2{x}-\sin^2{x})-2\cos{x}+6\sin{x}-6=0$

I have expanded the equation, But I cannot proceed further, Any hint would be appreciated. Thank you!

答:
Abbreviating $\sin x$ and $\cos x$ to $s$ and $c$, and writing $\sin2x=2sc$ and $\cos2x=2c^2-1$, the equation becomes

$$6sc+4(2c^2-1)-2c+6s-6=0$$

Factoring out a $2$ and grouping terms, we rewrite this as

$$3s(c+1)+4c^2-c-5=0$$

But $4c^2-c-5=(4c-5)(c+1)$, so we have two solutions in $s$ and $c$:

$$c=-1\quad\text{and}\quad3s+4c=5$$

Reverting to $\sin x$ and $\cos x$ and recognizing the Pythagorean triple $3^2+4^2=5^2$, these become

$$\cos x=-1\quad\text{and}\quad\sin(x+\theta)=1$$

where $\sin\theta=4/5$ and $\cos\theta=3/5$, e.g. $\theta=\arcsin(4/5)$. So the solution set is

$$\{\pi+2n\pi\mid n\in\mathbb{Z}\}\cup\{\pi/2-\arcsin(4/5)+2n\pi\mid n\in\mathbb{Z} \}$$

Remark: The abbreviations $s$ and $c$ are merely a convenience. Once I get all the sines and cosines down to $\sin x$ and $\cos x$, I find it tedious to write the functions out in full over and over again.


问:
Let $\alpha$ be a root of $(x^2-a)$ and $\beta$ be a root of $(x^2-b)$. Provide conditions over $a$ and $b$ to have $F=K(\alpha+\beta)$.
QUESTION: Let $K$ be a field of characteristic different of $2$. Let $F$ be a splitting field for $(x^2-a)(x^2-b)\in K[x]$. Let $\alpha$ be a root of $x^2-a$ and $\beta$ be a root of $x^2-b$. Provide conditions over $a$ and $b$ to have $F=K(\alpha+\beta)$.

MY ATTEMPT:

Let $\alpha=\sqrt{a}$, $\beta=\sqrt{b}$ and $\gamma=\alpha+\beta$.
First of all, we have $F=K(\alpha, \beta)$ due to the definition of splitting field.
Defining $K(\alpha+\beta)=K(\gamma)$.

Let's show that $K(\alpha, \beta)\subset K(\gamma)$:

From $\gamma=\alpha+\beta$ follows that
\begin{align*}
\gamma^2&=(\alpha+\beta)^2\\
&=\alpha^2+2\alpha\beta+\beta^2\\
&=(\sqrt{a})^2+2\sqrt{a}\sqrt{b} +(\sqrt{b})^2\\
&=(a+b)+2\sqrt{a}\sqrt{b}\qquad (*)\\
\end{align*}

Now we are going to show that $\sqrt{b}\in K(\gamma)$

Indeed, multiplying both sides in $(*)$ by $\sqrt{b}$ we have:

$\gamma^2\sqrt{b}=(a+b)\sqrt{b}+2\sqrt{a}(\sqrt{b})^2$. Then $$\sqrt{b}=\frac{2b\sqrt{a}+(a+b)\sqrt{b}}{\gamma^2}\in K(\gamma)$$

Similarly, $\sqrt{a}\in K(\gamma)$, this is

$\gamma^2\sqrt{a}=(a+b)\sqrt{a}+2(\sqrt{a})^2\sqrt{b}$, then

$$\sqrt{a}=\frac{(a+b)\sqrt{a}+2a\sqrt{b}}{\gamma^2}\in K(\gamma)$$

MY DOUBT:
I guess there is no conditions over $a$ and $b$ such that $\alpha=\sqrt{a}$ and $\beta=\sqrt{b}$, however I'm not sure. And I don't know how to connect this with the hypothesis that $K$ has characteristic different of two. Would you help me please?

答:
Once you know that $[ K( \alpha,\beta ) : K ]=4$, with basis $\{1,\alpha,\beta,\alpha\beta\}$, you can proceed as follows:
$$
\begin{pmatrix} 1 \\ \gamma \\ \gamma^2 \\ \gamma^3 \end{pmatrix}
=
\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ a+b & 0 & 0 & 2 \\ 0 & a+3b & 3a+b & 0
 \end{pmatrix}
\begin{pmatrix} 1 \\ \alpha \\ \beta \\ \alpha\beta \end{pmatrix}
$$
The matrix has determinant $4(b-a)$ and so is invertible iff $a\ne b$ since the characteristic of $K$ is not $2$. Therefore, $\{1,\gamma,\gamma^2,\gamma^3\}$ is also a basis and so generates the same space, that is, $K( \alpha,\beta ) = K(\gamma)=K( \alpha + \beta )$.

Bottom line: The main condition is that $[ K( \alpha,\beta ) : K ]=4$, or equivalently, $\beta \not\in K( \alpha)$.

This approach does not work in characteristic $2$ because $[K(\gamma):K]\le 2$ since $\gamma^2 = a+b \in K$.


问:
what is $\lim_{z\rightarrow 0}{(\cos(z))^{1/z^2}}$
The suggestion was to use L'Hospital's rule but I couldn't see how to apply it.

I tried manipulating the term to $\lim_{z\rightarrow 0} \exp(1/z^2\ln|º\cos(z)|+i\arg(\cos(z)))=\lim_{z\rightarrow 0} \exp(i\arg(\cos(z))$ and I also don't know how to proceed.

Body:
$$y=\big[\cos(z)\big]^{\frac 1 {z^2}}\implies \log(y)=\frac 1 {z^2}\log\big[\cos(z)\big]$$

Now, using Taylor series
$$\cos(z)=1-\frac{z^2}{2}+\frac{z^4}{24}+O\left(z^6\right)$$
$$\log\big[\cos(z)\big]=-\frac{z^2}{2}-\frac{z^4}{12}+O\left(z^6\right)$$
$$\log(y)=-\frac{1}{2}-\frac{z^2}{12}+O\left(z^4\right)$$
$$y=e^{\log(y)}=\frac{1}{\sqrt{e}}-\frac{z^2}{12 \sqrt{e}}+O\left(z^4\right)$$


问:
Understanding this proof that $\lim\limits_{h\to 0}\frac{\cos(h)-1}{h}=0$
I need help understanding how this limit is proved? :

Show that $$\lim_{h\to 0} \frac{\cos (h)-1}{h}=0$$
Proof:

Using the half angle formula, $\cos h = 1-2 \sin^2(h/2)$

$$\lim_{h\to 0} \frac{\cos (h)-1}{h}\\=\lim_{h\to 0}( -\frac{2 \sin^2(h/2)}{h})\\=-\lim_{\theta \to 0}\frac{\sin \theta}{\theta} \sin \theta\  \  \  \  \  \  \  \  \  \  \  \  \  \  \ \text{(Let $\theta=h/2$)}  \\ = -(1)(0)\\=0$$
I have no idea how this proof is done, so I apologize for the lack of my own thoughts in this question. I understand limits and know sin, cos, tan, but I am just very lost as what they did in each step. Can someone please explain all the steps of the proof as well as the half-angle formula. Thanks!

答:
For convenience, let us double the argument:

$$\lim_{h\to0}\frac{\cos h-1}h=\lim_{2h\to0}\frac{\cos2h-1}{2h}=\lim_{h\to0}\frac{\cos2h-1}{2h}.$$

Then it is well-known that $\cos2h=1-2\sin^2h$ and we have

$$-\lim_{h\to0}\frac{2\sin^2h}{2h}=-\lim_{h\to0}\frac{\sin h\sin h}{h}=-\lim_{h\to0}\frac{\sin h}{h}\lim_{h\to0}\sin h=-1\cdot0.$$


问:
How to transform these coordinates by substitution in classical mechanics?
The goal is to transform the following coordinates:
$$x(t)= R(\Phi-\sin\Phi)$$ and
$$z(t)=R(2 +\cos\Phi)$$

with the substitution: $u=\cos\left(\Phi/2\right)$

in order to get:
$$x(t)=2R(\arccos(u)-u\sqrt{1-u^2})$$ and
$$z(t)=R(1+2u^2)$$

How do I go about solving this problem?
I already tried using the following addition theorems: $\cos(\Phi/2)=\sqrt{(1+\cos\Phi)/2)}$ and $\sin\Phi=\sqrt{1-\cos^2\Phi}$
and identities for $\arccos$, but somehow I failed to transform the coordinates.

答:
This is quite straightforward, I'm not sure where you're having a problem. I won't give the full answer since this is a classic "homework-style" problem, I'll just show you how to do one of them, and the other can be done in almost exactly the same way. Take the first equation, you want to go from:

$$x(t) = R(\Phi - \sin\Phi) \quad \quad \text{to} \quad \quad x(t) = R ( 2 \cos^{-1}u - 2u\sqrt{1- u^2}).$$

The most sensible substitution to start with is $$\Phi = 2 \cos^{-1}u \quad \quad \text{or}\quad \quad u = \cos\left({\frac{\Phi}{2}}\right).$$

Draw out a little triangle (or use trigonometric identities), and you should be able to show that this means that $$\sin{\left(\frac{\Phi}{2}\right)} = \sqrt{1-u^2}.$$

Then, using the simple result $$\sin\Phi = 2\sin{\left(\frac{\Phi}{2}\right)}\cos{\left(\frac{\Phi}{2}\right)},$$ you should be done. You should now be able to do the next one, after you find a relation between $\cos\Phi$ and $\cos(\Phi/2)$ and $\sin(\Phi/2)$.

OwnerDisplayName: NirvanicUniverse

OwnerDisplayName: Philip


答:
As I always have problems to memorize trigonometric identities, I prefer to start by a basic identity:

$$e^{i\phi}=\left(e^{i\phi/2}\right)^2$$

an use Euler's formula for real and imaginary parts:

$$\cos(\phi)=\left(\cos(\phi/2)\right)^2-\left(\sin(\phi/2)\right)^2$$
$$\sin(\phi)=2\cos(\phi/2)\sin(\phi/2)$$

From these two trigonometric identities and introducing the definition of $u$:
$$\cos(\phi)=u^2-(\sqrt{1-u^2})^2=...$$
$$\sin(\phi)=...$$

you will finish in the target equations.

Title: How to transform these coordinates by substitution in classical mechanics?

Body:
The goal is to transform the following coordinates:
$$x(t)= R(\Phi-\sin\Phi)$$ and
$$z(t)=R(2 +\cos\Phi)$$

with the substitution: $u=\cos\left(\Phi/2\right)$

in order to get:
$$x(t)=2R(\arccos(u)-u\sqrt{1-u^2})$$ and
$$z(t)=R(1+2u^2)$$

How do I go about solving this problem?
I already tried using the following addition theorems: $\cos(\Phi/2)=\sqrt{(1+\cos\Phi)/2)}$ and $\sin\Phi=\sqrt{1-\cos^2\Phi}$
and identities for $\arccos$, but somehow I failed to transform the coordinates.

OwnerDisplayName: pasaba por aqui


问:
Tensor product of separable but not normal extensions
It is a well know fact that if $L$ is a Galois Extension of Field $K$ of degree $n$ then we have :

$$ L \otimes_K L \simeq L^n $$

I am trying to get insight on what happens when L is separable but not normal

Thus I am considering $\mathbb{Q}[\alpha]$ with $\alpha = \sqrt[3]{2}$ and :

$$ \mathbb{Q}[\alpha] \otimes_\mathbb{Q} \mathbb{Q}[\alpha] $$
$$ \simeq  \mathbb{Q}[\alpha][X]/(X^3 - 2)$$
$$ \simeq \mathbb{Q}[\alpha][X]/(X-\alpha) \times \mathbb{Q}[\alpha][X]/(X^2+\alpha X + \alpha^2) $$
$$ \simeq \mathbb{Q}[\alpha] \times  \mathbb{Q}[\alpha,\omega] $$

is a product of $\mathbb{Q}[\alpha]$ and $M=\mathbb{Q}[\alpha,\omega]$ a splitting field of $X^3-2$.

Now, contrary to the Galois case, I guess we cannot say that every element of the Galois Group corresponds bijectively to a maximal ideal of the tensor product, since the Galois group only has one element here and the tensor product has more than one maximal ideal.

But can we still deduce something from the set of  maximal ideals ?

Does the fact that we only get one copy of $\mathbb{Q}[\alpha]$ correspond to the fact the Galois Group only has one element ?

Do the other Maximal Ideals tell us something about $\mathbb{Q}$-homomorphisms of $Q[\alpha]$ into an algebraic closure ?

答: Ok thanks, this was my understanding. In this case it looks to me as if there are 3 possibilities for the compositum each corresponding to a $Q$-homomorphism of $L$ into an algebraic closure with the last two being isomorphic : $\mathbb{Q}[\alpha]=\mathbb{Q}[\alpha].\mathbb{Q}[\alpha]$ corresponding to the identity, $\mathbb{Q}[\omega,\alpha]=\mathbb{Q}[\alpha].\mathbb{Q}[\omega\alpha]$ and $\mathbb{Q}[\omega,\alpha]=\mathbb{Q}[\alpha].\mathbb{Q}[\omega^2\alpha]$ corresponding to the two other non trivial homomorphisms. Is that correct ?


问:
Sign of dual variable in dual lagrange problems
When I consider a general minimization problem
\begin{align}
&\min f_0(\textbf{x})\\
&s.t \quad  g_i(\textbf{x}) \le 0 \qquad i=1,\dots ,p\\
& \qquad h_i(\textbf{x}) = 0 \qquad i=1,\dots ,m.
\end{align}

and consider the primal function $L(\textbf{x},\lambda,\nu)$, I would expect that the dual problem becomes something like this

\begin{align}
&\max g(\lambda,\nu)\\
&s.t \quad \lambda \le 0
\end{align}

(Since $\lambda$ is a vector, the inequality is intended componentwise), and $g$ is the dual function $g(\lambda,\nu) := \inf_{\textbf{x}} (L(\textbf{x},\lambda,\nu))$. I would expect this because I have a minimization problem with a $\le$ inequality constraints.

Why on Convex Optimization by Stephen Boyd, even if he clearly refers to a general problem of the form I put at the beginning, he says that in the dual problem we have $\lambda \ge 0$??

答:
A cornerstone in the construction of the dual problem is the weak duality theorem. Let $f, q$ be the primal and dual problems (respectively). Then by weak duality $q^*\leq f^*$. To prove week duality, we define the primal set $$S=\{\mathbf{x}\in X: g_i(\mathbf{x})\leq 0, h_j(\mathbf{x})=0, i=1,\dots,m,\; j=1,\dots, p\}.$$ Then for any $(\lambda,\mu)\in\mathbb{R}^m_+\times\mathbb{R}^p$ we have
\begin{aligned} q(\lambda,\mu)&=\min_{\mathbf{x}\in X} L(\mathbf{x},\lambda,\mu)\\
&\leq \min_{\mathbf{x}\in S} L(\mathbf{x},\lambda,\mu) \\
&=\min_{\mathbf{x}\in S} \Big[f(\mathbf{x}) +\sum_{i=1}^m\lambda_i g_i(\mathbf{x})+\sum_{j=1}^p\mu_j h_j(\mathbf{x})\Big]\\
&\leq \min_{\mathbf{x}\in S} f(\mathbf{x})
\end{aligned}
where the last inequality follows from the fact that $\lambda_i\geq 0$ and $g_i(\mathbf{x})\leq 0$. If you change this, then you don't have weak duality anymore.


问:
How to find all Orthogonal vectors to one vector
I'm doing home work and I'm not really understanding what I'm supposed to do with this problem.

I have vector $\mathbf{v}=(1,-1,-1,1)$ in $\mathbb{R}^4$. I'm supposed to find all orthogonal vectors. I dont really know how to create a matrix from this.

I started by making $\mathbf{u}=(u_1,u_2,u_3,u_4)$. And then multiplying $\mathbf{u}$ with $\mathbf{v}$ would just basically give me $(u_1,-u_2,-u_3,u_4)$ which seems too simple and wrong. Could someone help me with this problem?

答:
The inner product of $v$ and $u$ is given by

$$u_1-u_2-u_3+u_4.$$

Hence

$$ v \perp u \iff u_1-u_2-u_3+u_4=0.$$

Can you proceed ?


问:
Solve a cubic equation?
Need help with solving an equation:

Solve the equation $5x^3 - 24x^2 + 9x + 54 = 0$ given that two of it's roots are equal.

Any help would be greatly appreciated.
Thanks!

答:
Using Vieta's Formula,

$a\cdot a+a\cdot  b+a\cdot  b=\frac95\implies 2ab+a^2=\frac95$

$a+a+b=\frac{24}5\implies 2a+b=\frac{24}5\implies 4a^2+2ab=\frac{48}5a$

Comparing the values of $a\cdot b$ and on simplification we get, $5a^2-16a+3=0$

Solve for $a,$ find the corresponding $b$ which satisfies   $a\cdot a\cdot b=-\frac{54}5$

Title: Solve a cubic equation?


问:
Alternative notation for inverse function
We all known the problems that presents the notation of inverse/reverse/anti functions as $f^{-1}(x)$, being the most important one the confusion with ${f(x)}^{-1}$, as in the classical $\sin^{-1}(x)$, or the strange cases such as ${f^{-1}(x)}^{-2}$,etc.

There are any other notation convention for the inverse functions more or less widely used ? Something like $f^{\#}(x)$, $f^{r}(x)$, $rev\{f\}(x)$?

If not, why not ? It seems something interesting to improve.

答: This 1909 trig textbook mentions $\mathrm{inv}\sin{x}$ as an alternate form of $\arcsin{x}$ and $\sin^{-1}{x}$ and says it may be better than those because of its "general applicability". Applying this to $f(x)$ would presumably give either $\mathrm{inv\ }f(x)$ or $\mathrm{inv}f(x)$. Wikipedia also cites a 2009 textbook when it talks about putting $\mathrm{inv}$ in front of $f(x)$


问:
How are complex functions different from functions of the form $\mathbb{R}^2\rightarrow \mathbb{R}^2$?
I have been trying to intuitively understand complex analysis for over an year now and I still seem to get stuck at the basics. How are complex functions different from a function from $\mathbb{R}^2$ to $\mathbb{R}^2$? Does it have to do with the difference in product of vectors? Will I get the same results if I just stick to studying multivariable functions? And why are analytic functions so important? Why do we not talk about analytic functions from $\mathbb{R}^2$ to $\mathbb{R}^2$?
I apologise for asking so many questions but any help or tip would be highly appreciated. I know the theorems and all but I need to know what they mean (perhaps) geometrically. Please try to give an intuitive and basic explanation. Thank you.

答:
The main difference arrives when you try to define differentiability.

In the complex case, you can do in the same way as in the real case:
$$f'(z)=\lim_{w\to0}\frac{f(z)-f(w)}{z-w}$$
And you can divide by $z-w$ as $\Bbb C$ is a field.

But in $\Bbb R^2\to\Bbb R^2$ case, you cannot proceed in the same way, because you cannot divide by a vector. So, the solution is to divide by the modulus


问:
Problem book on functors and category theory?
I am trying to find a book I saw many years ago at a university library. Unfortunately, I remember so little of the book that I haven't been able to effectively search. What I do recall is that:

the book was like a problem book on functors and category theory

the first few pages, iirc, had a dialogue between 2 people

the hardcover I saw had a blue (green?) cover, and probably had around 200 pages.

答:
https://www.springer.com/gp/book/9780817641863

You're welcome! (this is added just to reach 30 characters)


答:
The difference between $\mathbb{R^2}$ and $\mathbb{C}$ is that in $\mathbb{C}$ we define multiplication of complex numbers. This operation turns it into a field, and allows us to define a derivative of a complex function like this:

$f'(z)=\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}$

We can't define it like this in $\mathbb{R^2}$, because there we don't have a multiplication, and thus the division by $h$ makes no sense. The complex derivative is really different from a derivative in $\mathbb{R^2}$. For a complex function $f=u+iv$ to be differentiable at a point, the functions $u,v$ must have partial derivatives at that point and also they have to satisfy the Cauchy-Riemann equations: $u_x=v_y, u_y=-v_x$. As you can imagine, the number of functions which satisfy this is very limited. So complex differentiability is a much more complicated term than a usual derivative in the sense of $\mathbb{R^2}$. And if you studied complex analysis you know that it leads to a very big and beautiful theory-analytic functions have very surprising properties.

Title: How are complex functions different from functions of the form $\mathbb{R}^2\rightarrow \mathbb{R}^2$?

Body:
I have been trying to intuitively understand complex analysis for over an year now and I still seem to get stuck at the basics. How are complex functions different from a function from $\mathbb{R}^2$ to $\mathbb{R}^2$? Does it have to do with the difference in product of vectors? Will I get the same results if I just stick to studying multivariable functions? And why are analytic functions so important? Why do we not talk about analytic functions from $\mathbb{R}^2$ to $\mathbb{R}^2$?
I apologise for asking so many questions but any help or tip would be highly appreciated. I know the theorems and all but I need to know what they mean (perhaps) geometrically. Please try to give an intuitive and basic explanation. Thank you.


问:
The limit of a limsup of a nonnegative sequence is 0
I am currently looking at a result that states, for some nonnegative function $g: \mathbb{R} \times \mathbb{R} \to [0,1]$, there exists a constant $c> 0$ such that
\begin{align}
\lim_{k \to \infty} \limsup_{n \to \infty} |g(n,k) - c | = 0. ~(*)
\end{align}
It seems to me that the $\limsup$ is unnecessary, and could have been replaced by a $\lim$. This is because I believe that the above equation states that $\forall \epsilon >0, \exists K_{\epsilon}$ such that $\forall k > K_{\epsilon}, \exists N_{k, \epsilon}$ such that
\begin{align}
\sup_{n > N_{k,\epsilon}} |g(n,k) - c| < \epsilon.
\end{align}
(I found a nice explanation of the meaning of a double limit in quantifiers here Doubt about double limit definition. ).

Doesn't the $\sup$ imply that $|g(n,k) - c| < \epsilon|$ for all $n > N_{k,\epsilon}$, and vice versa? Also, is there any difference between $(*)$ and the following:
\begin{align}
\lim_{k \to\infty} \liminf_{n \to \infty} \frac{g(n,k)}{c} = \lim_{k \to\infty} \limsup_{n \to \infty} \frac{g(n,k)}{c} = 1
\end{align}
? Thanks so much.

答: Consider $g(n,k)=\frac12+\frac1{k^2+5} \cos(\pi n)$ and $c=\frac12$. It satisfies the condition, but $\lim_{n\to\infty}\lvert g(n,k)-c\rvert$ does not exist for any $k$.


问:
Unable to form an equation for the question
You own a hamburger stand. Your specialty is delicious cheeseburger that you sell for USD $7.00$ each. The cost of ingredients for each cheeseburger, such as meat, cheese, buns, tomato etc., is USD $1.65$. You hire workers at USD $12$ an hour. The production function for cheeseburgers is $Q = 40 \sqrt{L}$ , where L is the number of labor hours employed and $Q$ the number of cheeseburgers produced. There is a fixed cost of USD $172$.

There is no way to produce delicious cheeseburgers without also producing lots of grease, which needs to be hauled away. In fact, every cheeseburger that you produce, you also produce 1 ounce of grease. Lately, cooking grease has become popular as an ingredient to produce alternative fuel. So, a company begins to buy grease from you every day at the rate of USD $0.80$ per pound, thereby eliminating your cost of hauling away grease and providing you another revenue source.

Calculate the daily profit maximizing number of hamburgers, the amount of grease produced, the number of labor hours used and the consequent level of profit.

I have actually come up with an equation and differentiated it to get an answer. Just want to be sure it is right. My equation : $(-3x^2 + 2140x - 68800) / 400x$.

Number of burgers for maximum profit are coming out to be $151/152$. Does this sound right or am I going wrong somewhere?

答:
Hint:

Suppose you want to produce and sell $Q$ hamburgers a day

On the cost side

How many working hours do you need?  How much does that cost?

How much do the ingredients cost?

What is the fixed cost?

What is the total cost?

On the income side

How much do you earn from the sales of hamburgers?

How much do you earn from the sales of grease? ($16$ ounces in a pound)

What is the total income?

Putting these together

What is your profit from $Q$ hamburgers?

Which $Q$ maximises your profit?

How much grease corresponds to this optimal $Q$?

How many working hours correspond to this optimal $Q$?

How much profit corresponds to this optimal $Q$?


答:
There are many things to address here, but I'll take a few, and maybe others will bring other important points to your attention.

The only structural difference between $\mathbb{C}$ and $\mathbb{R}^2$ is the operation of complex multiplication. $\mathbb{R}^2$ is not endowed with any operation that resembles this, and thus we can construct functions from $\mathbb{C}$ into $\mathbb{C}$ that cannot be reviewed as function from $\mathbb{R}^2$ to $\mathbb{R}^2$.

As Tito Eliatron has already commented, the notion of complex differentiability is very different from the notion of differentiability in $\mathbb{R}^2$. Complex differentiability is a much stronger condition than real differentiability. As a matter of fact, you can compare the notion of analyticity to the notion of a divergence and curle free vector field in $\mathbb{R}^2$. That is how restrictive the condition of complex differentiability is.


问:
cohomology with compact support
Where is the cohomology with compact support useful? It seems that, a part from proving Poincaré duality, we also use it to compute the top dimensional cohomology group of closed manifolds: isn't singular cohomology sufficient here?

Thanks

答: For a closed manifold, cohomology and cohomology with compact supports coincide, but for open manifolds they do not.  The top dimensional cohomology with compact support is always one-dimensional for a connected orientable manifold, regardless of whether or not the manifold is closed.  Furthermore, on any (connected orientable) manifold, closed or not, Poincare duality is true when expressed as a duality between cohomology and cohomology with compact support in the complementary dimension.  Thus cohomology with compact support is a natural tool when working with non-closed manifolds.</p>

Title: cohomology with compact support


问:
Basic independent probability question
This question is a homework question.

The question states:

An airline can seat 100 people. Historically, the airline has noticed that each customer shows up independently and with probability $0.80$. The airline sells 101 tickets. What is the probability that there will be at least one unhappy customer?

My guess is that because this is about independent events, you add probabilities together?

So, suppose the airline could seat one person, and the customer had a 80% chance of showing. Well, the chance of the customer showing would be 80% right?

Now supposing the airline could seat 100 people, and each customer has an independently $0.80$ chance of showing, wouldn't the probability of all 100 seats being filled be $(0.8 * 100) = 80 $?

Since the airplane can only seat 100 people, but there are 101 tickets sold, there would be one unhappy customer if 101 people bought the ticket because the last person wouldn't have a seat. So I interpreted the question to imply: what is the probability of 101 customers showing up?

So my answer is $(0.8 * 101) = 80.8$

Is my reasoning correct, and if not, why not? Thanks.

Shoot .. aren't probabilities supposed to be $< 1$? Maybe it's $0.8^{100}$ or $0.8^{101}$ but now I'm just guessing. Our teacher left a hint saying don't be surprised if the probabilities are "very, very small", so it must be the $^$ operator right? I don't really know why though...

答:
Hint: Think of a coin purse that can only hold 1 coin. I have 2 fair coins, and flip them, putting the coins that come up heads into the purse. What's the probability the purse won't hold them all? Note that the purse won't hold them all $\iff$ both coins come up heads, and this is equal to
$$P(1^{\text{st}}\text{ coin = H})\times P(2^{\text{nd}}\text{ coin = H})=\left(\frac{1}{2}\right)\times\left(\frac{1}{2}\right)=\frac{1}{4}.$$
(Remember that the definition of two events $A$ and $B$ being independent is that $P(A\text{ and }B)$ is equal to $P(A)\times P(B)$; flipping each coin is independent, since the outcome of one coin can't affect the other.)

Can you generalize this simple example to your case?

Title: Basic independent probability question


问:
Triangle inequality for an integral
I have the following here.

Fix real numbers $a<b$, let $M$ be a positive real number, and define $$B(M)=\left\{ f \in C[a,b] \bigg| \int_{a}^{b}f(x)^2 dx \leq M \right\}$$ Show that if $f,g\in B(M)$, then $tf+(1-t)g \in B(M)$ for all $t \in[0,1]$.

I'm given a hint on using the triangle inequality for the functions $tf$ and $(1-t)g$. I'm not really sure how to proceed though.

The triangle inequality says: $||\vec{u}+\vec{v}||\leq||\vec{u}||+||\vec{v}||$

I am not sure what to do here though. Can someone guide me through how I would proceed?

答:
Let $h:=tf+(1-t)g$. Then $h^2=t^2f^2+2t(1-t)fg+(1-t)^2g^2.$ Thus

$$\int_a^b h^2 dx \le t^2M+2t(1-t)\int_a^b fg dx +(1-t)^2 M.$$

By Cauchy-Schwarz

$$\int_a^b fg dx \le (\int_a^b f^2 dx)^{1/2} \cdot (\int_a^b g^2 dx)^{1/2} \le \sqrt{M} \sqrt{M}=M.$$

Therefore

$$\int_a^b h^2 dx \le t^2M+2t(1-t)M+(1-t)^2M=M.$$

Consequence: $h \in B(M).$

Remark: similar computations show that if $ts \ge 0$ and $|t+s| \le 1,$ then $tf+sg \in B(M).$

Title: Triangle inequality for an integral


问:
$A^3 + A = 0$ then $rank (A) = 2$
Let $A$ be a $3\times 3$ non-zero real matrix and satisfies $A^3 + A = 0$. Then prove that $rank (A) = 2$.

As $A$ is satisfying $A^3 + A = 0$, so $0$ is an eigen value of $A$.So $\operatorname{rank} (A) < 3$. So $\operatorname{rank} (A) = 0,1,\text{or}\, 2$. Clearly $\operatorname{rank}(A) = 0$ is not possible as $A$ is a non-zero matrix.How to show that $\operatorname{rank} (A) = 1$ is also not possible?

答:
Well, the other two possible eigenvalues are $i$ and $-i$, as they are roots of the polynomial
$$x^3+x=0\,.$$
Rank $1$ is possible, if you allow complex matrices. For example
$$ A:=\left(\begin{matrix}i&0&0\\0&0&0\\0&0&0\end{matrix}\right) $$
has rank $1$ and satisfies $A^3+A=0$. Otherwise, if you ask for real values matrices, then the set of eigenvalues will be conjugated, i. e. there will be both eigenvalues $i$ and $-i$, therefore the rank must be $2$ or $0$. </p>

Title: $A^3 + A = 0$ then $rank (A) = 2$


问:
Prove $\{(x,y,z)\in \mathbb{R^3}| x^2+y^2+z^2>1\}$ is not connected
Prove $\{(x,y,z)\in \mathbb{R^3}| x^2+y^2+z^2>1\}$ is not connected

I have tried 2 open set A and B as follows: $\{(x,y,z)\in \mathbb{R^3}| x^2+y^2+z^2>1\}=A\bigcup B$

$=\{(x,y,z)\in \mathbb{R^3}| x^2+y^2+z^2>1,1-x^2-y^2<0\}\bigcup \{(x,y,z)\in \mathbb{R^3}| x^2+y^2+z^2>1,1-x^2-y^2\geq 0\}$. Then prove A,B is open set. But I have a hard time doing this, hope you help me

答:
Let call $S$ to your set.

CLAIM: Any point in $S$ can be arc-connected with $(2,0,0)$.

PROOOF: Let $u\in S$, then the segment from $u$ to $\frac{2u}{|u|}$ always lies in $S$. Now take any continuous path form $\frac{2u}{|u|}$ to $(2,0,0)$ lying in the sphere centered at $(0,0,0)$ with radius $2$ (this path also lies in $S$). Join the two paths and you're done.

COROLLARY: $S$ is arc-connected, so connected.

PROOF Take any two points $u,v\in S$. connect each point with $(2,0,0)$. You're done.


答:
Total cost of ingredients $a$ =  $1.65Q = 66 \sqrt L$

Labor cost $b$ = $12L$

Fixed cost $c = 172$

Profit $ = 280 \sqrt L - (66 \sqrt L + 12L + 172) = 214 \sqrt L - 12 L - 172$

Grease produced = $Q$ = $40\sqrt L \,$ ounces = $2.5 \sqrt L$ pounds.

Profit from grease  $ = 0.80 \times 2.5 \sqrt L = 2 \sqrt L$

Total profit $P = 216 \sqrt L - 12 L - 172$

EDIT: expressing in terms of $Q$,

$P = \frac{216}{40} Q - \frac{12}{1600} Q^2 - 172$

i.e. $\, P = \frac{27}{5} Q - \frac{3}{400} Q^2 - 172$

Title: Unable to form an equation for the question

Body:
You own a hamburger stand. Your specialty is delicious cheeseburger that you sell for USD $7.00$ each. The cost of ingredients for each cheeseburger, such as meat, cheese, buns, tomato etc., is USD $1.65$. You hire workers at USD $12$ an hour. The production function for cheeseburgers is $Q = 40 \sqrt{L}$ , where L is the number of labor hours employed and $Q$ the number of cheeseburgers produced. There is a fixed cost of USD $172$.

There is no way to produce delicious cheeseburgers without also producing lots of grease, which needs to be hauled away. In fact, every cheeseburger that you produce, you also produce 1 ounce of grease. Lately, cooking grease has become popular as an ingredient to produce alternative fuel. So, a company begins to buy grease from you every day at the rate of USD $0.80$ per pound, thereby eliminating your cost of hauling away grease and providing you another revenue source.

Calculate the daily profit maximizing number of hamburgers, the amount of grease produced, the number of labor hours used and the consequent level of profit.

I have actually come up with an equation and differentiated it to get an answer. Just want to be sure it is right. My equation : $(-3x^2 + 2140x - 68800) / 400x$.

Number of burgers for maximum profit are coming out to be $151/152$. Does this sound right or am I going wrong somewhere?


问:
Lower bound for the modulus of general complex polynomial of degree $n$?
Today I tried to show that $\mid P(z)\mid>\mid a_n\mid\mid z\mid ^n(\frac{\mid z\mid -R}{\mid z\mid -1})$. I already thought that I got it, but while reading trough my 'proof' I noticed a mistake, marked below with $*$. I'm trying to prove the result to show that all the roots of the complex polynomial $P(z)$ of degree $n$ lie in the disc $B(0,R)$, where $R=1+M, M=max\{\mid\frac{a_j}{a_n}\mid\},\,0\leq j\leq a_{n-1}$. Additional assumptions are $\mid z\mid\geq R\geq 1$ and $a_n\neq 0$.

$P(z)=a_nz^n+a_{n-1}z^{n-1}+...+a_0$. So I wrote (triangle inequality and geometric sum, $R\geq M$)

$\mid P(z)\mid>\mid a_n\mid(\mid z\mid^n-M(\mid z\mid^{n-1}+..+1))=\mid a_n\mid(\mid z\mid^n-M\frac{\mid z\mid^n-1}{\mid z\mid-1})>\mid a_n\mid(\mid z\mid^n-R\frac{\mid z\mid^n-1}{\mid z\mid-1})\stackrel{*}{>}
\mid a_n\mid(\mid z\mid^n-R\frac{\mid z\mid-R}{\mid z\mid-1})>\mid a_n\mid(\mid z\mid^n-\mid z\mid^n\frac{\mid z\mid-R}{\mid z\mid-1})=\,\mid a_n\mid\mid z\mid^n(1-\frac{\mid z\mid-R}{\mid z\mid-1})>\mid a_n\mid\mid z\mid^n(\frac{\mid z\mid-R}{\mid z\mid-1}).$

I was thinking backwards at the $*$: $R\geq 1$ and $\mid z\mid^n\geq\mid z\mid$ so the estimate goes in wrong direction at $*$ because now the numerator is actually smaller and not greater than the nominator in the previous step. The rest of the proof is correct, $\mid z\mid^n\geq R$ and the last inequality can be computed:

$1-\frac{\mid z\mid-R}{\mid z\mid-1}>\frac{\mid z\mid-R}{\mid z\mid-1}\implies \frac{1}{2}>\frac{\mid z\mid-R}{\mid z\mid-1}\implies...\implies R>1$, which is correct by the definitions.

Oh, and there seems to be the special case when $P(z)=a_nz^n\implies M=0\implies R=1$. So maybe the estimate holds only with $\geq$ and not stricktly with $>$. I'm quite lost with this and I'd appreciate any advice a lot!

答:
Now I got it. $\mid P(z)\mid\,=\,\mid a_nz^n+a_{n-1}z^{n-1}+...+a_0\mid\,\geq\,\mid a_n\mid\mid z\mid^n-...-\mid a_1\mid\mid z\mid-\mid a_0\mid\,\geq\,\mid a_n\mid(\mid z\mid^n-M(\mid z\mid^{n-1}+...+1))=\,\mid a_n\mid(\mid z\mid^n-M\frac{\mid z\mid^n-1}{\mid z\mid-1})=\,\mid a_n\mid(\frac{\mid z\mid^n(\mid z\mid-1)-M(\mid z\mid^n-1)}{\mid z\mid-1})=\,\mid a_n\mid(\frac{\mid z\mid^{n+1}-\mid z\mid^n-M\mid z\mid^n+M}{\mid z\mid-1})=\,
\mid a_n\mid\mid z\mid^n(\frac{\mid z\mid-1-M+\frac{M}{\mid z\mid^n}}{\mid z\mid-1})>\,\mid a_n\mid\mid z\mid^n(\frac{\mid z\mid-R}{\mid z\mid-1}).$

$M=max\{\mid\frac{a_j}{a_n}\mid\},\,0\leq j\leq n-1,$
$\mid z\mid\geq R=M+1$ and $1+M-\frac{M}{\mid z\mid^n}<R\iff M-\frac{M}{\mid z\mid^n}<R-1\iff M(1-\frac{1}{\mid z\mid^n})<R-1\iff M<\frac{R-1}{1-\frac{1}{\mid z\mid^n}}<\frac{R-1}{1-\frac{1}{R}}=\frac{R-1}{\frac{R-1}{R}}=\frac{R(R-1)}{R-1}=R=M+1\iff 0<1.$


问:
Are any two maps from a contractible space homotopic?
Let $X$ be a contractible space and $Y$ be any topological space. For any two continuous maps $f, g : X \to Y$ from $X$ to $Y$, is it true that $f$ is homotopic to $g$?

I understand that it is true if $Y$ is path-connected.

答:
Actually, the statement is true if and only if $Y$ is path connected. For the second direction, suppose any two continuous maps $f,g: X\to Y$ are homotopic. Take any two points $y_0, y_1\in Y$, and we will show there is a path in $Y$ which connects them. By assumption, the constant maps $f,g:X\to Y$ defined by $f(x)=y_0, g(x)=y_1$ are homotopic. Thus, there is a continuous function $F:X\times I\to Y$ such that:

$F(x,0)=y_0$ and $F(x,1)=y_1$ for all $x\in X$

Now choose any point $x_0\in X$ and define $\gamma:I\to Y$ by $\gamma(t)=F(x_0,t)$. This is a indeed a path between $y_0$ and $y_1$.


答:
First of all, you are trying to prove a false result. Try to draw $S$: it is the exterior of the unit ball and it is clearly "visually-connected". As connectedness has been invented to give a rigorous definition of "visually-connected", this would be a strange fact if $S$ was not connected.

To prove $S$ is connected, you can show:

every path-connected topological space is connected

$S$ is path-connected

The first statement is proven this way: suppose $X$ is path connected, and let $x\in X$. Suppose $f : X \to \{-1,1\}$ is continuous. We want to show that $f$ is constant. If $y \in X$, there exists a continuous path from $x$ to $y$ by path-connectedness, say $\gamma : [0,1] \to X$. Then $f\circ \gamma$ is continuous on $[0,1]$ which is connected and $f\circ \gamma$ is constant. This show that $f(y)=f(\gamma(1))=f(\gamma(0))=f(x)$. The previous being true for all $y$, $f$ is constant equal to $f(x)$. Hence, every continuous function from $X$ to $\{-1,1\}$ is constant, and $X$ is connected.

The second statement is proven by a simple drawing and a little bit of effort to write it mathematically.

Body:
Prove $\{(x,y,z)\in \mathbb{R^3}| x^2+y^2+z^2>1\}$ is not connected

I have tried 2 open set A and B as follows: $\{(x,y,z)\in \mathbb{R^3}| x^2+y^2+z^2>1\}=A\bigcup B$

$=\{(x,y,z)\in \mathbb{R^3}| x^2+y^2+z^2>1,1-x^2-y^2<0\}\bigcup \{(x,y,z)\in \mathbb{R^3}| x^2+y^2+z^2>1,1-x^2-y^2\geq 0\}$. Then prove A,B is open set. But I have a hard time doing this, hope you help me


问:
Simplify $\log\left(1+\frac{x_i^2}{\nu}\right)$ with a $\log(1+x)$ rule?
How can the following logarithm be simplified? Is there a $\log(1+x)$ rule to use?

$$ \log\left(1+\frac{x_i^2}{\nu}\right) = ?$$

答:
If the $1$ was not there, we could have written the expression as
$$2\log x_i-\log\nu$$
provided that the variables are positive. But the $1$ prevents this, though it does mean we get a Taylor series expansion
$$\log\left(1+\frac{x_i^2}\nu\right)=\sum_{n=1}^\infty(-1)^{n+1}\frac{(x_i^2/\nu)^n}n$$


问:
Solving $5400 \cos(\theta) \sin (\theta) - 735.75 \sin (\theta) - 600 = 0$
$$5400 \cos(\theta) \sin (\theta) - 735.75 \sin (\theta) - 600 = 0$$

This is a static course. I understand the question. I already solved to this point, but in the book they go from here to final answer $\theta =14.5$ and $\theta = 82.5$.

答:
We want to solve for $\theta$ in the equation
$$
a \cos(\theta) \sin(\theta) = b \sin(\theta) + c
$$
for some constants $a,b,c$. Squaring both sides we get
\begin{align*}
&a^2 \underbrace{\cos^2(\theta)}_{\color{blue}{1-\sin^2(\theta)}} \sin^2(\theta) = b^2 \sin^2(\theta) + 2bc \sin(\theta) + c^2\\
\implies& a^2 \sin^4(\theta)+ \left(b^2-a^2 \right) \sin^2(\theta) + 2bc \sin(\theta) +c^2 = 0
\end{align*}
Notice here that if we rename $\sin(\theta)$ as $x$, we see that the problem reduces to finding the roots of the polynomial
$$
\left(a^2\right) x^4+ \left(b^2-a^2 \right) x^2 + \left(2bc \right)x+c^2  \tag{1}
$$
which, although not impossible to solve analytically (see here for information about the quartic formula), it's most definitely not advisable to do. If you really want to go this route, you can verify your steps by plugging the polynomial into WolframAlpha to get the explicit roots. Once you get these values, you simply need to take the $\arcsin$ function of the roots to get your angle, remembering that our substitution was $\sin(\theta) = x \implies \theta = \arcsin(x)$ (keeping in mind the interval on which you're working on).

For your case, if we plug $a = 5400$, $b= 735.75$ and $c=600$ into $(1)$ and solve for the roots, considering that since the question came from physics you're probably taking $\theta \in \left(0, \frac{\pi}{2}\right)$, we get that the positive roots of the polynomial are
\begin{align*}
x_1 &\approx 0.129913\\
x_2 &\approx 0.967978
\end{align*}
and taking the inverse sine of these values we get that the values of the angles are
\begin{align*}
\theta_1 &\approx 7.4645 \ ^\circ\\
\theta_2 &\approx 75.461\ ^\circ
\end{align*}
which you can easily verify by plugging these values into your original equation.

As a final note, it's clear that this method is highly inefficient since we still end up relying on an answer given by WolframAlpha to solve our original equation. Given that your PDF also uses WA to get the solutions by directly plugging the original equation into WA, I believe the intended method for the solution of this problem is to imitate this, which would give you the same results that we've previously calculated.


问:
Hausdorff distance between metric spaces
I started to read a bit Gromov's paper 'Metric Structures for Riemannian and Non-riemannian Spaces'. Tthe Hausdorf distance of two metric spaces $X$ and $Y$ is defined by using isometric embeddings $f:X\rightarrow Z$ and $g:Y\rightarrow Z$ into a third metric space $Z$ and then taking some infimum over all such maps $f,g$ and spaces $Z$. So I am wondering why/if such a space $Z$ and maps $f,g$ always exist?Maybe one can construct always a metric on the product space to obtain such a $Z$...</p>

答: For two metric spaces $(X_1,d_1)$ and $(X_2,d_2)$, define the distance $d$ on $X:=X_1 \times X_2$ by $d : ((x_1,x_2),(y_1,y_2)) \mapsto \max(d_1(x_1,y_1),d_2(x_2,y_2))$. Then $\phi_1 : x \mapsto (x,0)$ and $\phi_2 : x \mapsto (0,x)$ are embeddings of $X_1$ and $X_2$ respectively into $X$.</p>


答:
There is no realy usefull simplification... But if was about approximation, you could  use the fact that
$$
\log(1+x) = x- \frac{x^2}{2} + O(x^3)
$$

to say that, for small $x$,
$$
\log(1+\frac{x_i^2}{\nu}) \approx \frac{x_i^2}{\nu}-\frac 12 \left(\frac{x_i^2}{\nu}\right)^2 = \frac{x_i^2}{\nu}\left(1- \frac{x_i^2}{2 \nu}\right).
$$

Title: Simplify $\log\left(1+\frac{x_i^2}{\nu}\right)$ with a $\log(1+x)$ rule?

Body:
How can the following logarithm be simplified? Is there a $\log(1+x)$ rule to use?

$$ \log\left(1+\frac{x_i^2}{\nu}\right) = ?$$


问:
Let $a,b$ be real numbers. Suppose that inequality $|(ax+by)(ay+bx)|\leqslant x^2+y^2$ holds for any real $x,y$. Prove that $a^2+b^2\leqslant2$
I know that this question has already been asked, but date of sending solutions passed (30th Sep), so this part of contest is inactive.

$$|(ax+by)(ay+bx)|\le x^2+y^2$$
$$|ab(x^2+y^2)+xy(a^2+b^2)| \le x^2+y^2$$
$$|k(a^2+b^2)+ab| \le 1$$
where  $k=\frac{xy}{x^2+y^2}$

$-\frac12\le k \le \frac12$  since $x^2+y^2\ge \pm 2xy$

Then I tried to treat left hand side as quadratic, but it did not lead to anything useful.
On the other hand from https://www.desmos.com/calculator/ljmcqbpobn it follows that there are $a,b$ s.t $a^2+b^2>2$, so I guess there are some mistakes in my approach.

Body:
For $x=y=1$: $(a+b)^2\le 2$, so $a^2+b^2+2ab\le 2$.

For $x=1$ and $y=-1$: $(a-b)^2\le 2$, so $a^2+b^2-2ab\le 2$.

$$a^2+b^2=\frac{(a^2+b^2+2ab)+(a^2+b^2-2ab)}{2}\le\frac{2+2}{2}=2.$$


问:
Second Order diff equation
Find all solutions to the differential equation:

$4y''+\beta^2 y = cos \beta x$

Where $\beta$ is a positive constant.

Starting with the homogeneous part :

$4r^2+\beta^2=0$

$r = \frac{i\beta}{2}$ since we only want the positive root.

Giving the solution:

$C_1\cos(\frac{\beta x}{2})+C_2\sin(\frac{\beta x}{2})$

Is this the correct way to start off? How would you find $y_p$?

答: You are on the right track. You can search for a particular solution "similar" to the RHS. In this case, you should try something of the form $y_p(x)=k_1 \sin (\beta x) + k_2 \cos(\beta x)=$, and compute $k_1, k_2$ such that it is actually a solution. You'll see that $k_1=0$ and $k_2 = \frac{1}{3 \beta^2}$.

Title: Second Order diff equation


问:
When my subset of $G$ containing identity and elements of same order is a subgroup of $G$?
I check if subset $H$ is a subgroup on the following way:

$H$ is nonempty

$H$ is closed under binary operation of G

The identity $e$ of $G$ is in $H$

For all $a\in H$, $a^{-1}\in H$.

So I have that $e\in H$, so 3. and 1. hold.

Next, elements in $H$ have same order, so there exists $k$ such that $a^k = e$, for every $a\in H$.

If $a\in H$ then  $a^{-1}\in H$ because inverse of element has same order as that element, and 4. holds.

I have troubles with 2. I proceed at this way:
Pick two elements in $H$, $x$ and $y$. I know that $x^k = e$ and $y^k = e$. I want to show that $H$ is closed and goal of this task is to find $k$ when this happens.

My group $G$ is abelian, but I don't know how to use this here.

How to proceed further?

答:
Let $G$ be an abelian group and $k$ a positive integer. Let
$$H=\{e\}\cup \{\,g\in G\mid \operatorname{ord}(g)=k\,\}. $$

If $k$ is prime, then $H$ is the kernel of the homomorphism $G\to G$, $x\mapsto x^k$ and therefore a subgroup (and in fact a normal subgroup)

If $k=1$, then $H$ is trivial and hence a subgroup.

If  no element of $G$ has order $k$, then $H$ is trivial and hence a subgroup.

In all other cases, $H$ is not a subgroup: As we are not in the third bullit point,  we can pick $h\in H$ with $\operatorname{ord}(h)=k$. As we are not in the first or second bullit point, $k$ is neither prime nor $=1$, hence it is composite. So $k=rs$ with integers $r,s>1$. Then $\operatorname{ord}(h^r)=s$, which is neither $k$ nor $1$ and hence $h^r\notin H$.


答:
(Using your approach)  You're real close.

Show that if $ a^2 + b^2 > 2$, then we can find some $ k$ in that range such that $|k (a^2 + b^2) + ab| > 1$, which is a contradiction. Thus $ a^2 + b^2 \leq 2$.

Hint:

 It might be helpful to separately consider cases of $ ab \geq 0, ab < 0$.

Title: Let $a,b$ be real numbers. Suppose that inequality $|(ax+by)(ay+bx)|\leqslant x^2+y^2$ holds for any real $x,y$. Prove that $a^2+b^2\leqslant2$

Body:
I know that this question has already been asked, but date of sending solutions passed (30th Sep), so this part of contest is inactive.

$$|(ax+by)(ay+bx)|\le x^2+y^2$$
$$|ab(x^2+y^2)+xy(a^2+b^2)| \le x^2+y^2$$
$$|k(a^2+b^2)+ab| \le 1$$
where  $k=\frac{xy}{x^2+y^2}$

$-\frac12\le k \le \frac12$  since $x^2+y^2\ge \pm 2xy$

Then I tried to treat left hand side as quadratic, but it did not lead to anything useful.
On the other hand from https://www.desmos.com/calculator/ljmcqbpobn it follows that there are $a,b$ s.t $a^2+b^2>2$, so I guess there are some mistakes in my approach.


问:
Types of convergence of sequence of random variables

Consider the following sequence of random variables defined on the
probability space $(Ω, F, P)$, where $Ω={0, 1,2, ..., m-1}$, $F$ is the
collection of all subsets of $Ω$, and $P$ is the uniform distribution over
$F$.

$$X_n(ω)=\begin{cases} 1/n,& \text{ω=n mod m}\\ 0,& \text{otherwise} \end{cases}$$

Does it converges to 0, (a) with probability 1? (b) in mean square?

My understanding is that $X_n(ω)=\begin{cases} 1/n,& \text{w.p. 1/m}\\ 0,& \text{otherwise} \end{cases}$ since it's uniformly distributed.

Therefore, (a) as n goes to ∞, X converges to 0 with probability 1 since 1/n also goes to 0.

(b) $\lim_{x\to∞}E(X_n^2)=\lim_{x\to∞}\frac{1}{n^2m}=0$, X converges to 0 in mean square.

Am I understanding the sequence correctly?

答:
I dont understand exactly what you have done, but it is easy to see that $X_n\to 0$ uniformly. As point-wise convergence imply convergence in probability then $X_n\to 0$ in probability also.

For the second part you have that
$$
\operatorname{E}[X_n^2]=\int_{\Omega }X_n^2\mathop{}\!d P= \int_{\{\omega :X_n(\omega )\equiv n\bmod m\}}\frac1{n^2}\mathop{}\!d P=\frac1{n^2}\Pr [X_n\equiv n\bmod m]\leqslant \frac1{n^2}
$$
So $\operatorname{E}[X_n^2]\to 0$.


问:
Positivity of principal eigenvalue for $L\phi=-\triangle \phi + \nabla \cdot( u \phi )$
EDIT: This question is still unresolved as of April 18, 2014. The two answers provide useful work in the right direction, but neither resolves the question. A counterexample should have $u \in C^1(\overline{\Omega})$.

I am reading this paper and in section 2 the authors explicitly state "We do not assume in this section that the flow $u(x)$ is incompressible." They immediately proceed to consider $\mu_1[u]$ and $\eta(x)$ the principal eigenvalue and normalized positive eigenfunction of the problem
$$
-\triangle \eta - \nabla \cdot (u \eta) = \mu_1[u] \eta
$$
in $\Omega$ and $\eta=0$ on $\partial \Omega$. Here $\Omega$ is any smooth bounded domain.

The authors then make the cryptic remark: "Note that $\mu_1[u]>0$ as the operator $-\triangle + u \cdot \nabla$ has no zero order term."

I see that $-\triangle + u \cdot \nabla$ is the adjoint of the operator in question. But how exactly does this tell me that $\mu_1[u]>0$? I know that a sufficient condition is that $\nabla \cdot u \leq 0$, (e.g. here), but the authors of the paper I am reading make no mention of any sort of divergence assumption.

答:
Let $L$ denote the operator
$$L\eta:= -\Delta \eta - {\rm div}\big( u(x) \eta \big)$$
and consider the adjoint
$$L^t\xi = -\Delta \xi + u(x) \cdot \nabla \xi.$$

Consider now the eigenvalue problems
$$
(1)_{\lambda}
\left \{
\begin{array}{rccl}
L \eta &=& \lambda \eta & \text{ in } \Omega\\
\eta &=& 0 & \text{ on } \partial\Omega
\end{array}
\right.
\quad \text{and} \quad
(2)_{\lambda}\left \{
\begin{array}{rccl}
L^{t} \xi &=& \lambda \xi & \text{ in } \Omega\\
\xi &=& 0 & \text{ on } \partial\Omega.
\end{array}
\right.
$$
By the Fredholm alternative, we have that:

(A) $(1)_{\lambda}$ has a unique solution (which is $\eta \equiv 0$) if and only if $(2)_{\lambda}$ has only the trivial solution $\xi \equiv 0$.

(B) Moreover, $(1)_{\lambda}$ admits nontrivial solutions if and only if $(2)_{\lambda}$ admits nontrivial solutions, and the dimension of the solution space of each problem coincides.

Let $\Sigma_{L^{t}} \subseteq \mathbb{C}$ denote the set of eigenvalues of $L^{t}$ (that is, the set of $\lambda$'s such that $(2)_{\lambda}$ admits nontrivial solutions).

By the above, we have that $(1)_{\lambda}$ admits nontrivial solutions if and only if $\lambda \in \Sigma_{L^t}.$

From the elliptic theory we know that $L^{t}$ has a smallest real eigenvalue $\lambda_1$ such that the solution space of $(2)_{\lambda_1}$ is one dimensional and, moreover, is generated by a positive function
(for instance, see the book of Evans:Theorem 3, section 6.5, for details).
By the maximum principle, we even know that $\lambda_1 > 0.$

Then, by the Fredholm alternative,
for
$(1)_{\lambda}$
to admit solutions, we must necessarily have ${\rm Re} \lambda \geq \lambda_1,$
and we also know that $(1)_{\lambda_1}$ has a one-dimensional space of solutions.

It remains to show that this space of solutions may be generated by a positive function.
However, I do not know how to show this at the moment (in the self-adjoint case, one uses the variational characterization of the eigenvalue, and in the non self-adjoint case the proof in Evans uses the maximum principle, which is not available for $L$)

Edit: It seems that the positivity of the principal eigenfunction already follows from the Krein-Rutman Theory. I was aware that, for $L^{t}$ with a maximum principle, the theory implied the positivity of the first eigenfunction, but Theorem 1.1. here implies that it is also the case for its adjoint $L$.


问:
Cardinality of the set of prime numbers
It was proved by Euclid that there are infinitely many primes. But what is the cardinality of the set of prime numbers ?

Cantor showed that the sets $\mathbb{Q}$ and $\mathbb{Z}$ have the same cardinality as the natural numbers $\mathbb{N}$ by constructing a pairing of the two sets, or a bijective function $ \pi_{\mathbb{Z}} : \mathbb{N} \rightarrow \mathbb{Z}$ and $ \pi_{\mathbb{Q}} : \mathbb{N} \rightarrow \mathbb{Q}$.

Let $\mathbb{P}$ denote the set of prime numbers. Is it possible to construct such a pairing function, $ \pi_{\mathbb{P}} : \mathbb{N} \rightarrow \mathbb{P}$ ?

It's clear that $|\mathbb{P}| \leq |\mathbb{N}| = \aleph_0$ since $\mathbb{P} \subset \mathbb{N}$. Is it possible to show that $|\mathbb{P}| \geq |\mathbb{N}|$, or do we have $|\mathbb{P}| < |\mathbb{N}|$ ?

答:
It is not hard to show that every infinite subset of $\Bbb N$ is in fact of cardinality $\aleph_0$. Let $A$ be such set, and define the following function: $$f(a)=\big|\{n\in A\mid n<a\}\big|.$$

It's not very hard to see that this is a bijection between $A$ and $\Bbb N$.

So we have that the cardinality of $\Bbb P$ is $\aleph_0$ just as well.

Another point worth mentioning is that if $|A|<\aleph_0$ then by definition $A$ is finite (because $\aleph_0$ is a minimal cardinal above the finite cardinals), so if $\Bbb P$ has cardinality smaller than $\aleph_0$ it is finite, in contradiction to all those proofs given that it's not.

Title: Cardinality of the set of prime numbers


答: The answer to this question is very easy: $|X_n(\omega )| \leq \frac 1 n$ for every $\omega$. This implies that $X_n \to 0$ in probability, almost surely and in mean square.

Title: Types of convergence of sequence of random variables

Body:
Consider the following sequence of random variables defined on the
probability space $(Ω, F, P)$, where $Ω={0, 1,2, ..., m-1}$, $F$ is the
collection of all subsets of $Ω$, and $P$ is the uniform distribution over
$F$.

$$X_n(ω)=\begin{cases} 1/n,& \text{ω=n mod m}\\ 0,& \text{otherwise} \end{cases}$$

Does it converges to 0, (a) with probability 1? (b) in mean square?

My understanding is that $X_n(ω)=\begin{cases} 1/n,& \text{w.p. 1/m}\\ 0,& \text{otherwise} \end{cases}$ since it's uniformly distributed.

Therefore, (a) as n goes to ∞, X converges to 0 with probability 1 since 1/n also goes to 0.

(b) $\lim_{x\to∞}E(X_n^2)=\lim_{x\to∞}\frac{1}{n^2m}=0$, X converges to 0 in mean square.

Am I understanding the sequence correctly?


答:
$$4r^2+\beta^2=0$$
Note that:
$$4r^2-i^2\beta^2=0$$
$$(2r-i\beta)(2r+i\beta)=0$$
$$\implies r= \pm i\dfrac {\beta}{2}$$
This line in your attempt is correct:
$$y=C_1\cos(\frac{\beta x}{2})+C_2\sin(\frac{\beta x}{2})$$
For the particular solution:
$$4y''+\beta^2 y = \cos \beta x$$
Try:
$$y_p =A \cos (\beta x)$$

Title: Second Order diff equation

Body:
Find all solutions to the differential equation:

$4y''+\beta^2 y = cos \beta x$

Where $\beta$ is a positive constant.

Starting with the homogeneous part :

$4r^2+\beta^2=0$

$r = \frac{i\beta}{2}$ since we only want the positive root.

Giving the solution:

$C_1\cos(\frac{\beta x}{2})+C_2\sin(\frac{\beta x}{2})$

Is this the correct way to start off? How would you find $y_p$?


问:
consistency of MLE estimator example
Given the density function $f(x;\theta)=\theta x^{\theta -1}$ with $\theta \gt 0$ and $x \in (0,1)$, once found the MLE estimator $\hat{\theta}=\frac{-n}{\sum{\log{X_i}}}$ i want to show the consistency of such an estimator. I was suggested to apply the strong law of large numbers, although i am not sure how. Should I first obtain the expected value of the random variable $\log(X_i)$ and then apply the SLLN on them or is there another way?

答:
You can show that $Y = -\ln X \sim Exp(\theta)$, hence $\mathbb{E}[Y]=1/\theta$, thus by the SLLN and the continuous mapping theorem
$$
\hat{\theta}_n \xrightarrow{a.s.} (1/\theta)^{-1} = \theta,
$$
as $n \to \infty$.

Title: consistency of MLE estimator example


问:
How to find the distribution of cases in the AstraZeneca/Oxford Phase III study interim analysis?
I'm trying to figure out whether it's possible to find how the 131 cases from an interim analysis of the AstraZeneca/Oxford Phase III study are distributed among the two arms of two dosing regimens.

Vaccine Efficacy is abbreviated as VE.

Facts from the AstraZeneca November 23rd Press Release, as I understand them:

There are two dosing regimens. Let's call them Regimen A and Regimen B.

Regimen A (2741 participants) has a VE of 90%

Regimen B (8895 participants) has a VE of 62%

The combined VE across the two regimens is 70%.

There were a total of 131 cases in the two regimens.

This is how the AstraZeneca Phase III Study Protocol defines Vaccine Efficacy (VE):

"VE is calculated as RRR = 100*(1-relative risk), which is the incidence of infection in the vaccine group relative to the incidence of infection in the control group expressed as a percentage."

Side Note: I think they actually made a mistake with the word "relative risk" inside the formula. I think "relative risk" must be "risk ratio". I looked up the notion of "relative risk" and it''s far more complicated than their own description of the formula. The word "risk ratio" I got from Lesson 3: Measures of Risk - Principles of Epidemiology in Public Health Practice

I updated this question to reflect that the study uses 2:1 randomization. Thanks to Micah for pointing this out to me.

Furthermore, the protocol states that participants will be randomized in a 2:1 ratio. Two people receiving the vaccine for every one person getting the placebo. I think this done to increase the 'signal', as generally speaking, far less cases of symptomatic covid-19 will be expected in the vaccine group.

Approximately 30 000 participants will be randomized in a 2:1 ratio to receive 2 IM doses of either 5 × 1010 vp (nominal, ± 1.5 × 1010 vp) AZD1222 (n = approximately 20 000) or saline placebo (n = approximately 10 000) 4 weeks apart, on Days 1 and 29.

I found the following formula to compute vaccine efficacy in a book chapter called Evaluating Protective Effects of Vaccination. It makes sense to me, except for the characters in subscript, which probably are defined elsewhere.

We're not speaking of incidence per se, but incidence relative to the amount of people at risk.

Since the vaccine group can be expected to have twice as many people in it, this means twice as many people are at risk. This works out to VE = 1 - ((V/2)/P) = 1 - (V/(2*P)) , where V is the amount of cases in the vaccine group, and P is the amount of cases in the placebo group.

Given this definition of VE and that there were 131 cases total (P+V=131), I think it must follow that there were 49 cases in the vaccine group, and 82 cases in the placebo group, because this is the only proportion that results in a vaccine efficacy of 70%.

1-(49/(2*82)) = 0.701 = 0.70

Now the question is, how are these cases (49 in the vaccine group, 82 in the placebo group) spread along the two regimens?

I tried to write down the formulas, as best as I could. I used the following for the variable names:

P: The amount of cases in the placebo group P.

V: The amount of cases in the vaccine group V.

Three suffixes: t for Total, a for Regimen A, b for Regimen B.

Pt =  82 = Pa + Pb

Vt =  49 = Va + Vb

(1-Va/Pa) = 0.90 ---> Va/Pa = 0.1

(1-Vb/Pb) = 0.62 ---> Vb/Pb = 0.38

Is this making sense?
Can this be solved?

If so, is there one solution, or multiple solutions?

Thanks in advance!

I found the following solution by trial and error. My guess is it's the only solution, but I don't know for sure.

```<code>VE A:        1 - (5/(2*24))  = 0.895  = 0.90
VE B:        1 - (44/(2*58)) = 0.6206 = 0.62
VE Combined: 1 - (49/(2*82)) = 0.701  = 0.70
</code>```

答: All the equations are ultimately linear. Thus the solution you found must be unique.


答:
If you are not familiar with SLLN, there are several other ways to approach the  problem,

The simplest: a property of ML Estimators is that they are consistent.

Consistency you have to prove is $\hat{\theta}\xrightarrow{\mathcal{P}}\theta$

So first let's calculate the density of the estimator.

Observe that (it is very easy to prove this with the fundamental transformation theorem)

$$Y=-logX\sim Exp(\theta)$$

Thus $W=\Sigma_iY_i\sim Gamma(n;\theta)$ and $\frac{1}{W}\sim \text{Inverse Gamma}$ and consequently

$$\mathbb{E}[\hat{\theta}]=\frac{n}{n-1}\theta$$

$$\mathbb{V}[\hat{\theta}]=\frac{n^2}{(n-1)^2(n-2)}\theta$$

This grant the following are both true

$$\lim\limits_{n\to \infty}\mathbb{E}[\hat{\theta}]=\theta$$

$$\lim\limits_{n\to \infty}\mathbb{V}[\hat{\theta}]=0$$

This is an IFF condition for the convergence in $L^2$ which implies also the convergence in probability (the consistency of your estimator)

Calculate the convergence in probability by the definition

Using 2 is a better way because you proved also the convergence in quadratic mean, that it is stronger

Trying all the method we showed is an useful exercise (IMHO)

Title: consistency of MLE estimator example

Body: Given the density function $f(x;\theta)=\theta x^{\theta -1}$ with $\theta \gt 0$ and $x \in (0,1)$, once found the MLE estimator $\hat{\theta}=\frac{-n}{\sum{\log{X_i}}}$ i want to show the consistency of such an estimator. I was suggested to apply the strong law of large numbers, although i am not sure how. Should I first obtain the expected value of the random variable $\log(X_i)$ and then apply the SLLN on them or is there another way?


问:
An application of Donsker's theorem.
Let ${X_i}$ be iid with $E[X]=0$ and $Var(X)=\sigma^2$ Let $S_0=0$ and $S_n=X_1+...+X_n$ for all $n \ge 1$. Show that $\lim_{n\rightarrow \infty}P(S_k>0 \space for \space k=n,n+1,\dots,2n)=1/4 $.

I try to apply Donsker's theorem with the continuous function $\omega\mapsto\min_{1/2\leq t\leq1}\omega\left(
t\right)  $. After this, I got the following expression.

$$
\lim_{n\rightarrow\infty}P\left(  \min_{n\leq k\leq2n}\frac{S_{k}}{\sqrt{2n}%
}>x\right)  =P\left(  \min_{1/2\leq t\leq1}B_{t}>x\right)
$$

Anyone knows what should I do next?

答:
Anyone knows what should I do next?

An approach is to condition on $B_{1/2}$. Slightly more generally, fix some $s\gt0$ and let
$$
A=\left[\min\limits_{s\leqslant t\leqslant2s}B_t\gt0\right],
$$
then $P[A\mid B_s=x]=0$ for every $x\leqslant0$ and $P[A\mid B_s=x]=P[T_x\gt s]$ for every $x\gt0$, where, for every $x$,  $T_x=\min\{t\geqslant0\mid B_t=x\}$.

By the reflexion principle, $P[T_x\lt s]=2P[B_s\gt x]$ hence $P[T_x\gt s]=P[|B_s|\lt x]$. Considering a Brownian motion $W$ independent of $B$, one gets
$$
P[A]=P[|B_s|\lt W_s]=\tfrac12P[|B_s|\lt |W_s|]=\tfrac12\cdot\tfrac12,
$$
first because the distribution of $W_s$ is symmetric and then because $(B,W)$ and $(W,B)$ are equidistributed.

More generally, for every positive $s$ and $u$,
$$
P[\min\limits_{s\leqslant t\leqslant s+u}B_t\gt0]=\tfrac12P[|B_u|\lt |W_s|]=\tfrac12P[|B_1|\lt v\cdot|W_1|],\qquad v=\sqrt{s/u}.
$$
The ratio $W_1/B_1$ is standard Cauchy hence
$$
P[\min\limits_{s\leqslant t\leqslant s+u}B_t\gt0]=\int_v^{+\infty}\frac{\mathrm dx}{\pi (1+x^2)}=\frac1\pi \arctan v.
$$

Title: An application of Donsker's theorem.


问:
Without solving explicitly show that for the IVP $x'=x^{3}-x,x(0)=0.5$ the solution converges to $0$ for $t\rightarrow\infty$
I am trying to determine whether the solution to the following IVP converges to $0$ for $t\rightarrow\infty$. The IVP is $x'(t)=x^{3}-x=f(x)$,$x(0)=\frac{1}{2}$.

By looking at the "extended phase portrait" I would infer  that this is indeed the case. However I would like to have solid argument.

I would argue as following: Let $x(t)$ be the solution to the above IVP.

For $x\in(0,1)$ $x^{3}-x<0$. Therefore $x'(t)<0$ and $x(t)$ is decreasing. Since $\tilde{x}(t)=0$ is a (equilibrium) solution $x(t)>0$, which follows from the uniqueness theorem. Hence, $0$ is a lower bound. Because $x(t)$ is bounded below and monotonically decreasing $lim_{t\rightarrow\infty}x(t)$ exists and $lim_{t\rightarrow\infty}x(t)\geq 0$.

Here is the part were I am stuck. I have two approaches. One works and the other doesn't I think. What I want to show is that $\lim_{t\rightarrow\infty}x'(t)=0$. Because then

$0=\lim_{t\rightarrow\infty}x'(t)=lim_{t\rightarrow\infty}f(x(t))=f(lim_{t\rightarrow\infty}x(t))=f(c)=c^{3}-c\iff c=0,c=1,c=-1$.

$c=-1$ is not possible, because the $x(t)$ would intersect $\tilde{x}(t)=0$, hence by the uniqueness it follows that $x(t)=0$ and thus $0>x'{t}=0$.
If $c=1$, then we can find $t_{1}\in\mathbb{R}_{>0}$ such that

$|x(t_{1})-1|<\frac{1}{2}\iff x(0)=\frac{1}{2}<x(t_{1})$.

Hence, by the mean value theorem it follow that there is

$\eta\in(0,t_{1})$ such that $f'(\eta)=\frac{x(t_{1})-x(0)}{t_{0}-0}>0$, but $f'(\eta)<0$. Therefore $c=0$.

Approach 1. Here I am stuck:

From the inequality $x'(t)<0$ it follows that $0\geq lim_{t\rightarrow\infty}x'(t)$.

Since $x(t)$ is a solution and because of the continuity of the r.h.s. of the IVP it follows that

$0\geq lim_{t\rightarrow\infty}x'(t)=lim_{t\rightarrow\infty}f(x(t))=f(lim_{t\rightarrow\infty}x(t))=f(c)=c^{3}-c$. $(*)$

Therefore $c\in[0,1]$ or $c\in(-\infty,-1]$. If the latter is the case then $x(t)$ would intersect with at least one equilibrium solution. Hence, this is not possible. Therefore $c\in[0,1]$. I think I can rule out any $c\in[\frac{1}{2},1]$, since then by the mean value theorem. $f'(\eta)>0$, which is not possible.

Can I somehow show that $f(c)=c^{3}-c\geq 0$. Then it would follow that $\lim_{\rightarrow\infty}x'(t)=0$, wouldn't it?

But isn't it the case that $f(c)\leq 0$, since $c\in[0,1]$ in equation $(*)$? Therefore we would have $0\leq f(c)\leq 0\iff f(c)=0$. Hence, $\lim_{t\rightarrow\infty}x'(t)=f(c)=0$

Is there a shorter way?

Approach 2:

We know that $\lim_{t\rightarrow\infty}x(t)=c$. Then $\lim_{t\rightarrow\infty}x'(t)=\lim_{t\rightarrow\infty}\lim_{h\rightarrow 0}\frac{x(t+h)-x(t)}{h}=\lim_{h\rightarrow 0}\lim_{t\rightarrow\infty}\frac{x(t+h)-x(t)}{h}=\lim_{h\rightarrow 0}\frac{c-c}{h}=0$.
I think all requirements for changing the order of the limits to be legal are met...or at least I hope so.

I would really appreciate any help. Thank you very much in advance!

答:
This problem can be easily solved using Lyapunov functions. Consider the Lyapunov function
$$
V=x^2.
$$
Its derivative along the trajectories,
$$
\dot V= 2x\dot x= 2x^4-2x^2=-2x^2(1-x^2),
$$
is negative definite, so the origin is asymptotically stable. Furthermore,
$
\forall x\in\mathbb (-1,1)\setminus \{0\}$
$\dot V(x)<0.
$
This means that the set
$$
S=\{x:\; V(x)<1\}=(-1,1)
$$
is a basin of attraction of the origin, thus, any solution $x(t)$ satisfying the initial condition $x(0)=C$, $C\in (-1,1)$, tends to zero.


问:
Can this be proved using Combinatorics or generating functions?
What is the proof for this identity?

$\displaystyle\sum_{r=0}^{n}\binom{n}{r}\binom{m+r}{n}\ = \displaystyle\sum_{r=0}^{n}\binom{n}{r}\binom{m}{r}2^{r}$

I thought of changing $r$ to $n-r$ on left and right both. Then, I thought of building a set with atleast $n$ elements and atmost $2n$ elements with atmost $n$ elements from $m$ element set, and we can do this on LHS by choosing $n-r$ elements from n and out of remaining $r+m$ elements we can choose n more elements. On RHS, we can again choose $n-r$ elements from $n$ and $r$ elements from $m$, then remaining $r$ elements from n element set have 2 choices each. But I think it has a flaw. Can someone please help? If my proof is not correct, please suggest a combinatorial or a generating function proof.

答:
This is another version of Phicar’s solution.

There are $n$ students and $m$ professionals. All professionals are member of chess club while students can be in the club or not. Suppose that we want to choose $n$ people from the club’s members to join a competition.

In the LHS, $r$ students are in chess club, $\binom{n}{r}$. Therefore, there are $m+r$ chess club members to choose from, $\binom{m+r}{n}$. Hence $\sum_{r=0}^{n}{\binom{n}{r}\binom{m+r}{n}}$

In the RHS, $r$ professionals join the competition, $\binom{m}{r}$. Therefore, $n-r$ students must be members of the club and join the competition, $\binom{n}{n-r}=\binom{n}{r}$. The remaining $r$ students can be in the club or not, $2^{r}$. Hence $\sum_{r=0}^{n}{\binom{n}{r}\binom{m}{r}2^{r}}$


问:
Another limit points task.
Let $x_n, y_n$ -- two numerical sequences with pairwise distinct terms (i.e $x_i \neq x_j$ and $y_i \neq y_j$ for all $i \neq j$).

We decided that $\lim_{n \to \infty} (x_ny_n) = \lim_{n \to \infty} (x_n + y_n) = 5$.

The question is to find all limit points of the set $A$, where:
$A = \{e^{x_n} + e^{y_n} \ | \ n \in \mathbb{N}\}$.

Definition of limit point which I know: $a \in X, A \subset X$ is limit point if for every $\varepsilon > 0$ punctured ball $B_{\varepsilon}(a)$ contains an element from $A$.

In our case $X = \mathbb{R}$.

I found that:
$x = \lim_{n \to \infty} (x_n) = \frac{5 + \sqrt{5}}{2}$
$y =\lim_{n \to \infty} (y_n) = \frac{5 - \sqrt{5}}{2}$

With solving a system

$xy=5$

$x+y=5$.

(We've got case with swapped values of x and y, but it's easy to show that we can do the same for them if we find solution).

My idea is to find $z = \lim z_n$, where $z_n = e^{x_n} + e^{y_n}$. And then we can say that limit point is only $z$ (from the limit defenition).

But I've got some troubles with finding this limit.

答:
$x_n$ and $y_n$ do not necessarily converge. For example let $u_n$ be strictly decreasing to $x$ and let $v_n$ be strictly increasing to $y;\,$ let $(x_n,y_n)=(u_n,v_n)$ when $n$ is even;$\,$ let $(x_n,y_n)=(v_n,u_n)$ when $n$ is odd.

Let $S_n=x_n+y_n$ and $P_n=x_ny_n.$ Let $D_n= S_n^2-4P_n.$ Then  $\{x_n,y_n\}=\{(S_n+\sqrt {D_n})/2,\,(S_n-\sqrt {D_n})/2\}.$ Let $b_n=\max(x_n,y_n)$ and $c_n=\min(x_n,y_n).$ Then $b_n\to x$ and $c_n\to y.$ Therefore $$e^{x_n}+e^{y_n}=e^{b_n} +e^{c_n}\to e^x+e^y=^{def} z.$$

So $z$ is the unique limit-point of $A$ if $A$ is an infinite set, which it could be.

It is possible that $A$ is a finite set. For example if $x<x_n<\ln z$ for every $n,$ with $x_n\to x,$ and $y_n=\ln  (z-e^{x_n}).$ Then $A=\{z\}.$

Title: Another limit points task.


问:
Subgraphs of $K_6$ that satisfy $|V|=3$
How many subgraphs $H=(V,E)$ of $K_6$ satisfy $|V|=3$? (If two subgraphs are isomorphic but have diffrent vertex sets, consider them distinct.)

We can choose the vertices as $\binom63$. My question is can I for theese $3$ vertices count all the subgraphs of $K_3$ or should they be non-isomorphic? If the should be non-isomorphic I get $4$ subgraphs and if they can be isomorphic the answer is $8$. What is correct?

答: The statement that "isomorphic subgraphs with different vertex set are distinct" implies that the graph is labelled, i.e. you cannot permute vertices at all. It follows that the answer is $8\binom63$.

Title: Subgraphs of $K_6$ that satisfy $|V|=3$


问:
Prove that the subset W of M2, consisting of all matrices of a form is a subspace of M2.
I have this problem in my textbook:

Let M2 be the vector space of all 2 x 2 matrices.

(i) Prove that the subset $W$ of $M2$ (the set of all 2 x 2-matrices), consisting
of all matrices

$\begin{bmatrix}a & b\\c & d\end{bmatrix}$

with $a + b + c + d = 0$ and $a - b + c - d = 0$ is a subspace of $M2$.

(ii) Find a basis for $W$.

Prove your answers.

(Hint: the set $W$ is a null space in $R4$ ' $M2$ of an appropriate 2x4 matrix.)

I have put  a + b + c + d = 0 and and a - b + c - d = 0 in a system, to get $\begin{bmatrix}a & b\\a & b\end{bmatrix}$, but I am lost from there.

答:
Hint:
Check that the map $\begin{aligned}[t]f:\mathcal M_2 &\longrightarrow K^2\qquad (K\text{ is the base field})\\\begin{pmatrix}a&b\\c&d\end{pmatrix}&\longmapsto (a+b+c+d, a-b+c-d)\end{aligned}$ is linear, and observe $W=\ker f$.


问:
What is the derivative of $\ln(4^x)$?
What is the derivative of $\ln(4^x)$  (which I believe is also equal to $x\ln4$)?

Is it $\dfrac{1}{x\ln4}$?

答:
The thing you are supposed to understand here is that constants don't do anything to derivatives -- they just factor out of the expression.
More precisely, we have for any function $f$ and a constant $c$ not depending on $x$ that
$$
\frac{d}{d x} c f(x) = c \frac{d}{d x} f(x)
$$
For example, since $\frac{d}{d x} (x^4) = 4x^3$, I can multiply by any constant and the derivative is just as straightforward:
$$
\frac{d}{d x} \left[ 2.33\pi^2 \cos (4) \right] x^4
= \left[ 2.33\pi^2 \cos (4) \right] 4x^3
$$

This works since the $ 2.33\pi^2 \cos (4) $ is constant - it doesn't depend on $x$.

If you get this, you should be able to calculate derivatives like this easily.

Title: What is the derivative of $\ln(4^x)$?


问:
How to compute $\int_0^\infty \frac{\cos(ax)}{(1+x^2)\sqrt{x}}dx$.
$$
\mbox{How can I compute}\
\int_{0}^{\infty}\frac{\cos\left(ax\right)}{\left(1 + x^{2}\right)\,\sqrt{\,{x}\,}}\,\mathrm{d}x\ ?.
$$

I saw many questions about computations of that integral without the term $\,\sqrt{\,{x}\,}\,$ and that it can be solved by using various techniques ( contour integral, iterated integral, etc$\ldots$ ). But how can I compute my integral $?$.

答:
Let
$$
f(a) = \int_0^{+\infty} \frac{\cos(ax)}{(1+x^2)\sqrt{x}} \, \mathrm{d} x,
$$

First, we note that $f(a)=f(-a)$, hence $f$ is even and we will focus only on $a \ge 0$.

Second,
$$
\begin{split}
\left| f(a) \right| &\le \left|\int_0^{+\infty} \frac{\cos(ax)}{(1+x^2)\sqrt{x}} \, \mathrm{d} x \right|\\
&\le \int_0^{+\infty} \frac{\left|\cos(ax)\right|}{(1+x^2)\sqrt{x}} \, \mathrm{d} x \\
&\le \int_0^{+\infty} \frac{\mathrm{d} x}{(1+x^2)\sqrt{x}}  = f(0)
\end{split}$$

With some math one finds that
$$
f(0) = \int_0^{+\infty} \frac{\mathrm{d} x}{(1+x^2)\sqrt{x}} =2\int_0^{+\infty} \frac{\mathrm dt}{1+t^4} =  \frac{\pi}{\sqrt{2}} \approx 2.22
$$

(see here)

Now, if we differentiate twitce under integral symbol (we can do it because of Leibniz integral rule)
$$
\begin{split}
f''(a) &= -\int_0^{+\infty} \frac{\cos(ax) \cdot x^2}{(1+x^2)\sqrt{x}} \, \mathrm{d} x\\
&= -\int_0^{+\infty} \frac{\cos(ax) \left(x^2+1-1\right)}{(1+x^2)\sqrt{x}} \, \mathrm{d} x\\
&= - \left[\int_0^{+\infty} \frac{\cos(ax)}{\sqrt{x}} \, \mathrm{d} x -\int_0^{+\infty} \frac{\cos(ax) }{(1+x^2)\sqrt{x}} \, \mathrm{d} x \right] \\
&= f(a) - \int_0^{+\infty} \frac{\cos(ax)}{\sqrt{x}} \, \mathrm{d} x
\end{split}$$

Now,

$$
\begin{split}
\int_0^{+\infty} \frac{\cos(ax)}{\sqrt{x}} \, \mathrm{d} x &= \frac{1}{2}\int_\mathbb{R}\frac{\cos(ax)}{\sqrt{|x|}} \, \mathrm{d} x \\
&= \mathcal F \left(\frac{1}{\sqrt{|x|}}\right)(a) \\
&=\frac{1}{2} \sqrt{\frac{2\pi}{|a|}}\\
&= \sqrt{\frac{\pi}{2}}\sqrt{\frac{1}{|a|}}
\end{split}
$$
where we used $ \mathcal F \left(\frac{1}{\sqrt{|x|}}\right)(a) = \frac{1}{2} \sqrt{\frac{2\pi}{|a|}}$ (see Fourier Transform of $\frac{1}{\sqrt{|x|}}$)

So, the problem is now to solve

$$
\begin{cases}
f''(a) -f(a) = -\sqrt{\frac{\pi}{2}}\sqrt{\frac{1}{|a|}}\\
f(0) = \frac{\pi}{\sqrt 2}\\
\end{cases}
$$

From linear differential equations we know that a solution is

$$
f(a) = c_1 \mathrm{e}^a + c_2 \mathrm{e}^{-a} + f_\mathrm p(a)
$$

where $c_1$ and $c_2$ are real numbers and $f_\mathrm p$ is the complementary solution.

Wolfram Alpha helps us to know that

$$
\begin{split}
f(a) &= c_1 \mathrm{e}^a + c_2 \mathrm{e}^{-a} + \sqrt{\frac{\pi}{2}}\int_0^{a} \frac{\mathrm{e}^{-x+a}-\mathrm{e}^{x-a}}{2\sqrt{x}}\, \mathrm d x\\
&= c_1 \mathrm{e}^a + c_2 \mathrm{e}^{-a} + \sqrt{\frac{\pi}{2}}\int_0^{a} \frac{\sinh(a-x)}{\sqrt{x}}\, \mathrm d x
\end{split}
$$

Now, if we add the initial condition, we get
$$
c_1+c_2 = \frac{\pi}{\sqrt 2},
$$
hence, after renaming the constant $c_1 = C$, the solution is

$$
\begin{split}
f(a)&= C \mathrm{e}^a + \left(\frac{\pi}{\sqrt 2} - C\right) \mathrm{e}^{-a} + \sqrt{\frac{\pi}{2}}\int_0^{a} \frac{\sinh(a-x)}{\sqrt{x}}\, \mathrm d x\\
&= 2C \sinh(a) +  \frac{\pi}{\sqrt 2} \mathrm{e}^{-a} + \sqrt{\frac{\pi}{2}}\int_0^{a} \frac{\sinh(a-x)}{\sqrt{x}}\, \mathrm d x
\end{split}
$$

If you find some other initial value (for example $f(1)$), then you can eliminate the constant $C$ too.


问:
Question on nullity and set relations
If T and S are two transformations, if nullT>0 and nullS>0 does this imply null(T+S)>0?

And what about if rankT=0 and rankS=0, does rank(T+S)=0?

答:
I suppose that $T$ and $S$ are linear.

First question: let $T,S :\mathbb R^2 \to \mathbb R^2$ given by

$$T= \operatorname{diag}(1,0)$$

and

$$S = \operatorname{diag}(0,1).$$

Then $\operatorname{nul}(T)>0$ and $\operatorname{nul}(S) >0.$ But $T+S= \operatorname{diag}(1,1)=I_2$, hence $\operatorname{nul}(T+S)=0.$

Second question: if $T :V \to W$  is linear and $\operatorname{rank}(T)=0$, then $T$ is the zero-transformation. Does this answer the second question ?

Title: Question on nullity and set relations


问:
Finding the structure of a group without using sylows theorem.

If$ |G|=pq $ and $p $ doesnt divide $(q-1)$ and $p <q$ then $G$ is cyclic.

My proof stands as this .I have used the fact that

$(i)$ If $|G|=pq$ then I showed that there will only be one element of order $p$ and one element of order $q$.

My approach in proving this part has been to show that if there are two elements of order $p$ say $x_1$ and $x_2$ , then say $H_1$ is a group of order $p$ generated by $x_1$ and $H_2$ is a group of order $p$ generated by $x_2$ .We assume that the intersection is {e} if not then we can get $H_1$=$H_2$(by property of subgroup).

Proving that the elements are distinct .

We assume that the elements $(x_1)^{i}.(x_2)^j$ are not distinct then $(x_1)^{i}.(x_2)^j =(x_1)^{i'}(x_2)^{j'}$.From here we can arrive at a contradiction as $H_1 \cap H_2 =e$.So if there are $p^2$ elements then we can arrive at a contradiction as $p$ and $q$ are both primes.

Similar results will hold in the case of $q$.

$(ii)$ Now  there is only one subgroup of order $p$ and one subgroup of order $q$ so they are both normal

$(iii)$ let $H$ and $K$ be two subgroups of order $p$ and order $q$.Then we know that $H \cap K={e}$.$H$ and $K$ are both normal .Then I showed that $x^{-1}y^{-1}xy \in H \cap K$ and $xy=yx$. So the order of the element $xy$ is $pq$.Where am I going wrong in my proof and since I have not used the fact that $p $ doesnot $q-1$.

答: Consider the center $Z(G)$ of G. The center $Z(G)$ is a normal subgroup of $G$. In particular the order of $Z(G)$ divides the order of $G$ by Lagrange's theorem. So the possible orders of $Z(G)$ are $1,p,q $  or $pq$. Now look at the order of the quotient group $G/Z(G)$.


答:
Recall that for $a>0$, we have
$$\log(a^b) = b \log(a)$$
Also, note that
$$\dfrac{d}{dx}\left( cx\right) = c$$
I trust you can finish it from here.</p>

Title: What is the derivative of $\ln(4^x)$?

Body:
What is the derivative of $\ln(4^x)$  (which I believe is also equal to $x\ln4$)?

Is it $\dfrac{1}{x\ln4}$?


问:
Not understanding the solution to a multiple choice question on linear transformations

I know B and F do not contain the zero element required for them to be subspaces and I understand D, the others I am unsure about A, C and E. I have another post asking about nullity relations which I will try to link below too

Any input is greatly appreciated, I just want to understand this fully

答:
The sets from $A$ and $E$ are the same set. Actually, both sets are equal to the set $\mathcal L(\Bbb R^5,\Bbb R^4)$, since no linear map from $\Bbb R^5$ into $\Bbb R^4$ is injective. And $\mathcal L(\Bbb R^5,\Bbb R^4)$ is a vector space.

The set from $C$ is also a vector space, since it consists only of the null map.


问:
Numerator layout for derivatives and the chain rule
We have three matrices $\mathbf{W_2}$, $\mathbf{W_1}$ and $\mathbf{h}$ (technically a column vector):

$$
\mathbf{W_1} =
\begin{bmatrix}
    a & b \\
    c & d \\
\end{bmatrix}
\;\;\;\;\;\;\;\;\;
\mathbf{W_2} =
\begin{bmatrix}
    e & f \\
\end{bmatrix}
\;\;\;\;\;\;\;\;\;
\mathbf{h} =
\begin{bmatrix}
    h_1 \\
    h_2 \\
\end{bmatrix}
$$

And a scalar $y$, where:

$$
y = \mathbf{W_2} \mathbf{W_1} \mathbf{h}
$$

I'd like to compute the derivative of $y$ with respect to $\mathbf{W_1}$, assuming numerator layout.

Using the chain rule:

$$
y = \mathbf{W_2} \mathbf{u}
\;\;\;\;\;\;\;\;\;
\mathbf{u} = \mathbf{W_1} \mathbf{h} \\
$$

$$
\begin{align}
\frac{\partial y}{\partial \mathbf{W_1}} &=
\frac{\partial y}{\partial \mathbf{u}} \frac{\partial \mathbf{u}}{\partial \mathbf{W_1}} \\
&= \mathbf{W_2} \frac{\partial \mathbf{u}}{\partial \mathbf{W_1}} \\
&= \mathbf{W_2} \mathbf{h}^{\top} \\
\end{align}
$$

All well and good. Except - this isn't a $2x2$ matrix!! In fact, the dimensions don't match up for matrix multiplication, so something must be incorrect.

If we take the Wikipedia definition of the derivative of a scalar by a matrix, using numerator layout, we know that actually:

$$
\frac{\partial y}{\partial \mathbf{W_1}} =
\begin{bmatrix}
    \frac{\partial y}{\partial a} & \frac{\partial y}{\partial c} \\
    \frac{\partial y}{\partial b} & \frac{\partial y}{\partial d} \\
\end{bmatrix}
$$

Each element is just a scalar derivative, which we can calculate without any matric calculus. If we do that by hand and then factorise, we end up with:

$$
\frac{\partial y}{\partial \mathbf{W_1}} = \mathbf{h} \mathbf{W_2}
$$

Clearly, $\mathbf{h} \mathbf{W_2} \neq \mathbf{W_2} \mathbf{h}^\top $.

Can anybody suggest where I went wrong?

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}$
Define the trace/Frobenius product as
$$A:B \;=\; {\rm Tr}(A^TB) \;=\; \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij}$$
Using this product eliminates a whole category of transposition errors, which arise in other approaches.

The properties of the trace allow such products to be arranges in many equivalent ways, e.g.
$$\eqalign{
A:C &= C:A &= C^T:A^T \\
AB:C &= A:CB^T &= B:A^TC \\
}$$
Note that the matrices on the LHS and RHS of the colon must have the same dimensions.

The Frobenius product is similar to the Hadamard product in this respect.

Let's define some variables without the distracting subscripts
$$W = W_1, \qquad w=W_2^T$$
Write the scalar function in terms of these new variables.
Then calculate its differential and gradient.
$$\eqalign{
 y &= w:Wh \\&= wh^T:W \\
dy &= wh^T:dW \\
\p{y}{W} &= wh^T \\
}$$
The dimensions of this result equal the dimensions of the $W$ matrix, expressed as the outer product of two column vectors.

Writing this in terms of the original variables
$$\eqalign{
\p{y}{W_1} &= W_2^Th^T \\
}$$


问:
Linear homeomorphism of $E \bigoplus F$ to $E \bigoplus F'$ implies linear homeomorphism of $F$ to $F'$?
If all the following spaces are banach spaces, then the existence of a linearhomeomorphism of $E \bigoplus F$ to $E \bigoplus F'$ implies on the existence of a linear homeomorphism of $F$ to $F'$?

I'm trying to argue this is true to complete a prove, but i'm not able to do it.

One thing that i'm trying to do to advance on this is prove that if $\phi$ is the linear homemorphism, and $a \in E$, $b \in F$, then $\phi(a) \in E$ and $\phi(b) \in F'$, but i'm not even sure that this is true.
Any hints would be helpful!

答: This not true, let $E$ be a separable infinite dimensional  Hilbert space  and $F,F'$ finite dimensional, $E\oplus F$ is isomorphic to $E\oplus F'$.


答:
if your question is $ln(4^x)$ then its answer will be $\ ln4$ and if your question is $(\ln4)^x$ then use this formula
$$\dfrac {d}{dx}a^x=a^x\cdot\ln{a}$$
so
$$\dfrac {d}{dx}{(\ln4)}^x=(\ln4)^x\cdot\ln{(\ln4)}$$</p>

Title: What is the derivative of $\ln(4^x)$?