问:
Question about groups and opposite elements - A question I got stuck on during test preparation
Hello everyone this is my first time on the site, I am preparing for a test I will have in a few more days in the abstract algebra course.

The lecturer gave us a database of questions without solutions, I solved them all, but I could not solve this question. I do not have to submit this question, it's just a question he gave in order to repeat the material.

the question

I think the correct answer is e. Because I ruled out all the other options.

a is incorrect, if I choose for example a1 = 1 a2 = 1, both of which are the identity element. And the formula does hold.

b is incorrect, because for a1 = 1, and a2 = -i, the reduction gives 3 and 3 is not equal to the opposite of itself.

c is incorrect, for the same reason b, the formula does not always work.

d is incorrect, for a1 = 1 and a2 = 1, both are opposite of each other. The formula does exist

So e is correct in my opinion, but I'm not sure

答:
Here are my remarks:

Option (a) is indeed false, but your argument is not sufficient to show that.
Option (a) says that the formula holds only when $a_1 = 1$ or $a_2 = 1$. So, to show that (a) is incorrect, you need show that there are some $a_1$ and $a_2$, where none is the identity element, for which the formula holds.

For (b), I don't understand what "does not guarantee correctness of the result" means... Also, why are you looking at the opposite of 3 (or of any element whatsoever)?

Option (c) is what I would choose. Though, we need to show that the reduction is proper. Do you have a formal definition of what a proper reduction is? The reason I choose this option as correct is because the reduction respects identities and opposites.

Option (d) is indeed incorrect and your argument proves it.

Option (e) is incorrect, because option (c) is correct.


问:
If $x + y \in \mathbb Q$ and $x − 3 y \in \mathbb Q$, prove $x$ and $y$ are rational
Consider $,  ∈ ℝ$, such that $ +  ∈ ℚ$ and $ − 3 ∈ ℚ$. Then $,  ∈ ℚ$.

Hey all. I'm recently working through a course-book that's involved with the math course I'm taking next semester and have come across this problem, although it has no solution (this would be taught in the course but I'm just looking over the content to get ahead)

I'm really stumped on how to solve this problem. My initial guess was to subtract $ + $ from $ − 3$ to prove that $-4y$ would be rational provided both equations were rational, and therefore we can assume $y$ is rational, and could potentially do the same for $x$ by multiplying the first equation... However I don't think this would be valid and am not sure.

How would I go about solving this type of problem? Thanks.

答: Yes, it is valid. Since $-4y=q$, for some $q\in\Bbb Q$, $y=-\frac q4\in\Bbb Q$. And then $x=(x+y)-y\in\Bbb Q$.

ContentLicense: CC BY-SA 4.0


问:
2nd order homogenous linear recurrence
What will be the 2nd-order homogeneous linear recurrence of:

$$x_{n+1}+ x_n+ 6x_{n-1}=0 \text{?}$$

答: Let $ {\left( x_i\right)}_{i=0}^{\infty} $ be your sequence and consider the power series $$ h(z)=\sum_{j=1}^{\infty}z^jx_j$$ So your recurrence formulates to $$ \dfrac{h(z)-a_1z -a_0}{z^2} + \dfrac{h(z)-a_0}{z} + 6h(z) = 0$$ So $$ h(z)=\dfrac{x_1z+x_0z+x_0}{1+z+6z^2} $$ I hope you can extract the coefficient now or tell me I would add it.

Title: 2nd order homogenous linear recurrence


问:
Solve one system of equations
Solve the system of equations
$$ ax + by + cz = 0, $$
$$ bcx + cay + abz = 0, $$
$$ xyz + abc (a^3x + b^3y + c^3z) = 0 $$

I tried solving this using cross multiplication method  but got stuck at one point :
$$x/ab^2-ac^2 = y/bc^2-ba^2 = z/ca^2-cb^2 = k (say) $$
I substituted the values in the third equation, and after simplifying ended up being here :
$$ k(abc)[a^2c^2(b^2 + a^2c^2 - b^2c^2 - a^2)] $$

答:
Let us consider $b \neq c, a \neq b, a \neq c$. Then from the first two equation we get
\begin{align}
\left[\begin{array}{cc}b&c\\c&b\end{array}\right]\left[\begin{array}{c}y\\z\end{array}\right]=\left[\begin{array}{c}-a\\-\frac{bc}{a}\end{array}\right]x,
\end{align}
i.e.
\begin{align}
\left[\begin{array}{c}y\\z\end{array}\right]=\frac{1}{a(b^2-c^2)}\left[\begin{array}{c}b(c^2-a^2)\\c(a^2-b^2)\end{array}\right]x.
\end{align}
By substitution to the third Eq. we get either $x=0$ or
\begin{align}
x^2=\frac{a^2(b^2-c^2)\left[a^4(c^2-b^2)+b^4(a^2-c^2)+c^4(b^2-a^2)\right]}{(c^2-a^2)(a^2-b^2)}\equiv a^2(b^2-c^2)^2.
\end{align}

Let us check the solution. For $a=1,b=2,c=3$ we get solution $x=5,y=-16,z=9$ (and also $x=-5,y=16,z=-9$), which are the correct solutions of the system of equations given above. $\spadesuit$

Title: Solve one system of equations


答:
Let $x,y \in \Bbb{R}$ such that $x+y,x-3y \in \Bbb{Q}$. We want to prove that $x,y \in \Bbb{Q}$. Since $x+y,x-3y \in \Bbb{Q}$, then
\begin{align*}
x+y = \frac{a}{b} && \text{ and } && x-3y = \frac{c}{d},
\end{align*}
where $a,c \in \Bbb{Z}$ and $b,d \in \Bbb{N}$. Now, form the first equality, one obtains that $x = \frac{a}{b} - y$. By substituting in the second inequality, one has $\frac{a}{b} - y -3y = \frac{c}{d}$. From here we get that $y = \frac{ad-bc}{4db}$. Substituting $y$ in the first equality yields that $x = \frac{3ad+bc}{4db}$. Therefore, we have that $x,y \in \Bbb{Q}$. $\square$

Title: If $x + y \in \mathbb Q$ and $x − 3 y \in \mathbb Q$, prove $x$ and $y$ are rational

Body:
Consider $,  ∈ ℝ$, such that $ +  ∈ ℚ$ and $ − 3 ∈ ℚ$. Then $,  ∈ ℚ$.

Hey all. I'm recently working through a course-book that's involved with the math course I'm taking next semester and have come across this problem, although it has no solution (this would be taught in the course but I'm just looking over the content to get ahead)

I'm really stumped on how to solve this problem. My initial guess was to subtract $ + $ from $ − 3$ to prove that $-4y$ would be rational provided both equations were rational, and therefore we can assume $y$ is rational, and could potentially do the same for $x$ by multiplying the first equation... However I don't think this would be valid and am not sure.

How would I go about solving this type of problem? Thanks.


问:
Finding the two unknown constants of the logistic function $f(x) = \frac{158}{1+e^{s(x-c)}}+22$
I'm trying to find the two unknown constants of the following function:

$$f(x) = \frac{158}{1+e^{s(x-c)}}+22$$
Where s and c are constants, and 22 (or $y_{min}$) and 180 (or $y_{max}$) are the horizontal asymptotes. c is the integration constant.

This was integrated from the following ODE (Ricatti equation):

$$\frac{s}{158}\frac{dy}{dx}=a(y-22)(180-y) , f(0) = y_0 > 22 $$

I can assume that f(0) lies near the horizontal asymptote. Additionally, I have the following equations:

$$f(0) = y_0 = \frac{158}{1+e^{sc}}+22$$
Hence
$$c = \frac{1}{s} log \frac{180 - y_0 }{y_0 - 22}$$

How do I find both constants mathematically? I realize that I need two equations, but am unsure of how I can find them.
From plugging in this into Desmos, I have found that c is the inflection point of the graph, while s is responsible for the shape of the curve: https://www.desmos.com/calculator/hwg4vsewro

The broader context of this question is that I'm finding a logistic function to fit a set of data (x is time, y is angle).

For context of the equations, I followed this source: http://sciendo.com/article/10.2478/fman-2020-0007

This is my raw data. Left column is x value(seconds); right is y value(angle).

x   y

0   22

60  23

120 24

180 27

240 32

300 32

360 36

420 36

480 39

540 41

600 42

660 45

720 46

750 51

780 54

810 57

840 63

870 79

900 103

930 124

960 155

990 177

1020    180

1080    180

1140    180

1200    180

I did not find an explanation of exactly how they find s and c, although the source has some quite complex equations involving higher order derivatives.

(One that I think might be useful involves the x-value $x_0$, where the function's second derivative takes its maximum value.

$$f(x_0) = \frac{y_{max}-y_{min}}{1+e^{s(x_0-c)}}+y_{min}$$
$$f(x_0) = y_{min} + 0.211(y_{max}-y_{min}) = 22 + 0.211(180-22) =55.338$$
No explanation why $\frac{1}{1+e^{s(x_0-c)}}=22$. From which it is somehow calculated:
$$x_0 = c - \frac{1.319}{s}$$)

答:
$$f(x) = \frac{158}{1+e^{s(x-c)}}+22$$
The given data is :
$$(x_1,f_1),(x_2,f_2),..., (x_k,f_k),..., (x_n,f_n)$$
Let :
$$Y(x)=\ln\left(\frac{158}{f(x)-22}-1 \right)$$
So, for each $f_k$ compute :
$$Y_k=\ln\left(\frac{158}{f_k-22}-1 \right)$$
The original equation becomes :
$$Y=s\:x-s\:c$$
You can compute approximate values of $(s)$ and $(-s\:c)$ thanks to linear regression (Least Mean Squares).

NOTE: Some difficulties may arrise due to eventually negative argument of the logarithm. That is why a representative example of data would be useful to check if the difficulties appear in your case and eventually suggest what to do.

NOTE: If a particular criteria of fitting is specified, you need a non-linear regression method. Then use a specialized software involving itterative calculus and guessed initial values of the parameters. The approximate values found above can be used as initial values.

UPDATED ANSWER with the given data.

Drawing $Y(x)=\ln\left(\frac{158}{y(x)-22}-1 \right)$ shows this shape of curve :

The curve is far to be linear. This proves that the chosen model $(1)$ is not convenient to obtain a good fitting whatever the method of calculus of the parameters.

$$y=a+\frac{b}{1+e^{s(x-c)}} \tag {model 1}$$

As a consequence the above proposed method is of no interest in the present case.

We have to look for another model.

For example one can try with the sum of two logistic functions :

$$y=A+\frac{b_1}{1+e^{s_1(x-c_1)}}+\frac{b_2}{1+e^{s_2(x-c_2)}} \tag {model 2}$$

The result is much better. Below, the comparison between the two models.


问:
How to prove $\frac xy + \frac yx \ge 2$
I am practicing some homework and I'm stumped.

The question asks you to prove that

$x \in Z^+, y \in Z^+$

$\frac xy + \frac yx \ge 2$

So I started by proving that this is true when x and y have the same parity, but I'm not sure how to proceed when x and y have opposite partiy

This is my proof so far for opposite parity

$x,y \in Z^+ $ $|$ $x \gt 0,$ $y \gt 0$. Let x be even $(x=2a, $ $ a \in Z^+)$ and y be odd $(y=2b+1, $ $b \in Z^+)$. Then,

$\frac xy + \frac yx \ge 2$

$\frac {2a}{2b+1} + \frac {2b+1}{2a} \ge 2$

$\frac {2b^2 + 4a^2 + 4a + 1}{2b(2a+1)} \ge 2$

$4b^2 + 4a^2 +4a + 1 \ge 4b(2a+1)$

$4b^2 + 4a^2 + 4a +1 \ge 8ab + 4b$

$4b^2 - 4b + (2a + 1)^2 \ge 8ab$

$(2b-1)^2 + 1 + (2a+1)^2 \ge 8ab$

I feel like this is the not the correct way to go about proving it, but I can't think of a better way to do it. Does anyone have any suggestions? Just hints please, not a solution.

答: Without loss of generality, we assume $t:=\frac{x}{y}\geq 1,$ we see that the $t+\frac{1}{t}$ has its derivative non-negative, so $t+\frac{1}{t}\geq 2.$

ContentLicense: CC BY-SA 3.0


问:
Deriving and proving the upper bound for a function when $x$ tends to $\infty$

Let $f(x)$ be a real valued function defined on the interval $[1,\infty)$. The following equations hold:

$$f(1)=1$$
$$f'(x)=\dfrac{1}{x^2+(f(x))^2}$$
Prove that $\lim_{x\to \infty}f(x)<1+\dfrac{\pi}{4}$

I have no idea how to proceed with this question. Please help.

Thank You

Edit: Working on the hint given by Atmos, it goes as follows:

Observe that $f'(x)>0$. This means that $f(x)$ is a strictly increasing function.

So $f(x)\geq 1\space \forall\ x\in [1,\infty)\Rightarrow x^2+(f(x))^2\geq x^2+1\space \forall\ x\in [1,\infty)$

Thus $$f'(x)=\dfrac{1}{x^2+(f(x))^2}\leq \dfrac{1}{x^2+1}$$

Since inequality is preserved while integration, integrating both sides give:
$$f(x)\leq \arctan(x)+C$$

now we have $f(1)=1\Rightarrow C=1-\dfrac{\pi}{4}$

Also $f(x)\leq \arctan(x)+1-\dfrac{\pi}{4}\leq \dfrac{\pi}{2}+1-\dfrac{\pi}{4}=1+\dfrac{\pi}{4}$

Hence we are done.

答:
Hint :

What can you say about the variation of $f$. What lower bound for $f$ can you deduce from this ?

Answering this question, try to show that
$$
f'\left(x\right) \leq \frac{1}{1+x^2}
$$

I let you conclude.


问:
Linear transformations inner matrix always symmetric
Let's say we have some linear transformation $T$.

If we express it in the form: $B^{-1} A B$, where $B$ maps us into some aritrary basis, and $B^{-1}$ maps us back out.

Is $A$ always symmetric? Or can it at least be converted to a symmetric matrix?

I got to the conclusion that $A$ is symmetric by using this argument: Why do we assume that a matrix in quadratic form is Symmetric?

And working through some small matrix examples, but looking to see if there is a simpler way to show this.

答: Hint: See if that happens with $T \colon \Bbb{R}^2 \to \Bbb{R}^2$ such that $T(x,y) = (2x+4y,3x+5y)$ for all $(x,y)\in\Bbb{R}^2$ if you consider $\mathcal{B} =((1,0),(0,1))$ and $\mathcal{B}’ = ((1,1),(1,-1))$.


答:
Alternatively,

$$(x+y)-(x-3y)=4y \Longrightarrow y \in\mathbb Q$$

$$(x+y)+(x-3y)=2(x-y) \Longrightarrow (x-y) \in\mathbb Q$$

$$(x-y)+(x+y)=2x \Longrightarrow x \in\mathbb Q.$$


问:
Confusion regarding probability of microbe producing everlasting colony.
My question is about the given solution to problem 4 in Newman's book 'A Problem Seminar'. Note that the book is available online at Springer.

Problem 4

A microbe either splits into two perfect copies of itself or else
  disintegrates. If the probability of splitting is p, what is the
  probability that one microbe will produce an everlasting colony?

Solution:
  We can (almost) solve this problem by expressing the unknown
  probability, $x$, in terms of itself. For, after all, how can the one
  microbe multiply forever? Only in one way. It must at the first
  moment split, and then at least one of its two daughters must
  succeed in the task of the mother (i.e., lead to an eternal
  progeny). In short $x= p\cdot(1-(1-x)^ 2)$. We can solve for x and
  get the answer! ...

From my understanding, $(1-(1-p)^2)$ means the probability of at least one microbe splitting. However, I have no idea why $x= p\cdot(1-(1-x)^ 2)$ instead of $x= p\cdot(1-(1-p)^ 2)$. How does $x$ get into the $(1-x)^2$? What does  $(1-(1-x)^ 2)$ intuitively mean?

答:
Since $x$ is the probability that a microbe produces an everlasting colony, $1-x$ is the probability that a microbe’s line will die out. Now we start with one microbe; with probability $p$ it splits. The daughter lines survive or die out independently of each other, so the probability that both daughter lines die out is $(1-x)^2$. The probability that at least one daughter line is everlasting is therefore the complementary probability $1-(1-x)^2$.

Thus, the probability that the original microbe splits and has at least one everlasting daughter line is $p\big(1-(1-x)^2\big)$. But this is just the probability that the original microbe’s line is everlasting, so $x=p\big(1-(1-x)^2\big)$.


答:
Over the reals this need not be possible.

Let $V:=V_2(\mathbb{R})$, and let $T:(1,0)\mapsto (0,1)$, $T:(0,1)\mapsto (0,0)$.

The characteristic polynomial is $X^2$.

Now look at an arbitrary symmetric matrix $\begin{pmatrix} a & b\\ b & c\end{pmatrix}$. Its characteristic polynomial is $X^2-(a+c)X+(ac-b^2)$, so it can only represent $T$ if $a=-c$ and $ac-b^2=0$: so $a=b=c=0$ which won't do.

Title: Linear transformations inner matrix always symmetric

Body:
Let's say we have some linear transformation $T$.

If we express it in the form: $B^{-1} A B$, where $B$ maps us into some aritrary basis, and $B^{-1}$ maps us back out.

Is $A$ always symmetric? Or can it at least be converted to a symmetric matrix?

I got to the conclusion that $A$ is symmetric by using this argument: Why do we assume that a matrix in quadratic form is Symmetric?

And working through some small matrix examples, but looking to see if there is a simpler way to show this.


问:
Invertability of a Multivector
I am writing a general geometric algebra library (dealing exclusively with euclidean bases) for Clojure to gain a better understanding on the topic, but how to invert a general multivector continues to cause me trouble.

In the first answer to Calculating the inverse of a multivector the formula for the inverse of a general multivector $B$ is stated to be

$$ B^{-1}= \frac{B^\dagger}{BB^\dagger}. $$

If this is the case, then that indicates that $BB^\dagger$ must evaluate to a non-zero scalar for $B$ to be invertible, correct? Otherwise it appears as though the definition of an inverse must be recursive.

Expanding $BB^\dagger$ given $B = a + be_1 + ce_2 + de_1e_2$ in $G_2$ results in the equation

$$ BB^\dagger = a^2 + b^2 + c^2 + d^2 + 2(ab + cd)e_1 + 2(ac - bd)e_2. $$

Given the assumption that $BB^\dagger$ must be scalar for $B$ to have an inverse, it follows that $ ab = -cd, $ and $ ac = bd $.

Is it the case that the general multivector $B$ is only invertible when the above identities between the components hold? Or is there a more general way to derive the inverse of a multivector which works in all cases given a euclidean set of basis vectors?

答:
Sorry, I don't know so much about this myself, but I (google) discovered  a paper in the physics arxiv that deals with the topic explicitly and seems to contain a lot of other relevant references:

Inverse of multivector: Beyond p+q=5 threshold
A. Acus, A. Dargys

Another set of recipes are described on the following web page
but I'm not so confident of their accuracy:

https://www.euclideanspace.com/maths/algebra/clifford/algebra/functions/inverse/manfred.htm

(I would have added this as a comment, but don't have adequate reputation)

Title: Invertability of a Multivector


问:
If $a^3=20a^2+b^2+c^2-a-340$, $b^3=20b^2+c^2+a^2-b-340$, $a^3=20a^2+a^2+b^2-c-340$, what is value of $abc$?
Question : If $a^3=20a^2+b^2+c^2-a-340$,  $b^3=20b^2+c^2+a^2-b-340$, $a^3=20a^2+a^2+b^2-c-340$, what is value of $abc$?

I think I should add this and get close to $abc$, but I can't think about this. I know that answer is 19. Please help me

Body:
$$20a^2+b^2+c^2-a-340-a^3=0$$
$$20b^2+c^2+a^2-b-340-b^3=0$$
$$20c^2+a^2+b^2-c-340-c^3=0$$
Taking note of @Ishraaqparvez comments

Maybe before you find the product of $abc$, you should first know what their corresponding values are
$$(a-19)^4(a^2+1)^4(a^3-22a^2+a+340)(a^4+5a^2-38a-356)(a^4-41a^3+422a^2+283a-6459)^2=0$$
$$(b-19)^4(b^2+1)^4(b^3-22b^2+b+340)(b^4+5b^2-38b-356)(b^4-41b^3+422b^2+283b-6459)^2=0$$
$$(c-19)^4(c^2+1)^4(c^3-22c^2+c+340)(c^4+5c^2-38c-356)(c^4-41c^3+422c^2+283c-6459)^2=0$$
Note that the equations are symmetric, so $a,b,c$ are different solutions of the same equation
$$(x-19)(x^2+1)$$
$$x^3-22x^2+x+340$$
Then their real product is $-340$


问:
Is Newton’s Law of Cooling as special case of the CSTR problem?
Mathematically, it obviously is, but I would like to have the physical justification also, and I’m stuck on one certain step.

The solution to the CSTR problem begins like this:

$$f’(t) = \frac{qw – qf(t)}{V}$$

where $f(t)$ is the concentration at time $t$, $q$ is the flow rate, $w$ is the concentration of the inlet flow, and $V$ is the volume.

Applying this to the situation of a cooling body, it is easy to identify $qw$ as the amount of “coolness” arriving, but I am unable to justify the amount of “hotness leaving” as $qf(t)$, because the “mixing” present in the CSTR problem seems absent in the case of a cooling body. It seems like the cooling body is losing heat much like an onion might be peeled. So, a picture or analogy that would justify the $qf(t)$ term would be appreciated.

Background: CSTR stands for “Continuous Stirred Tank Reactor”, which is an abbreviation for “Continuous Flow Stirred Tank Reactor”. Finding an expression for the concentration in the tank at time $t$ is a textbook example in Differential Equations.

Regards,

Mike Jones

22.May.2011 (Beijing time)

答: Physically there is no such thing as "coolness", but there is heat.  Heat leaves the cooling body at a rate proportional to the temperature of the body (that corresponds to the term $-q f(t)/V$) and enters the body at a rate proportional to the temperature of the environment (that corresponds to the term $qw/V$).  </p>


问:
Why $\frac{y’}{y}$ is equivalent to $(\ln|y|)’$?
I just read a proof process from a textbook about differential equations, it has following content :
$$y’= ay\implies\frac{y’}{y}= a\implies (\ln|y|)’= a$$
I want to know why $\frac{y’}{y}$ is equivalent to $(\ln|y|)’$?

答:
$\frac{d}{dx}[\ln|y|]=\frac{d}{dx}[\ln|x|]\bigg|_y\cdot \frac{d}{dx}[y]=\frac{1}{y}\cdot \frac{d}{dx}[y] = \frac{y’}{y}$

By the Chain Rule, because $\frac{d}{dx}[\ln|x|]=\frac{1}{x}$.

As for the absolute value, this link will provide context, $\ln(y)$ is defined for $y>0$ only, the absolute value allows you to consider negative values of $y\neq 0$.


问:
Is there a Peter-Weyl-Theorem over the real numbers?
I'd like to know: Is there a Peter-Weyl-like Theorem over the real numbers? If so, can you provide a reference?

With "over the real numbers" I mean that the representations considered act on vector spaces over the real numbers.

Best,
Leon

答:
I think that the Peter-Weyl theorem can be formulated for complex and real representations simultaneously using the ideas from the book by Bröcker and tom Dieck "Representations of compact Lie groups", Chapter III, Section 6, Exercise 4.

Let $\mathbb{K}$ be either $\mathbb{C}$ or $\mathbb{R}$ and let $\lambda$ be an irreducible representation of a compact topological group $G$ in a finite-dimensional $\mathbb{K}$-linear space $E$ with $G$-invariant inner product. Let $D(E)$ be the set of linear operators in $E$ that intertwine $\lambda$ with itself. If $\mathbb{K}=\mathbb{C}$, then $D(E)=\mathbb{C}$, while if $\mathbb{K}=\mathbb{R}$, then $D(E)$ may take three values, $\mathbb{R}$, $\mathbb{C}$, and $\mathbb{H}$, the skew field of quaternions.

Let $E^*$ be the linear space, dual to $E$, that is, the set of $\mathbb{K}$-linear functionals over $E$. It is a right $D(U)$-linear space via composition. On the other hand, $E$ is a left $D(U)$-linear space via evaluation. The tensor product $E^*\otimes_{D(U)}E$ is defined. A simple bookkeeping shows that the dimension of this $\mathbb{K}$-linear space is equal to $\frac{(\dim E)^2}{\dim D(E)}$. If $\mathbb{K}=\mathbb{C}$, then the denominator is equal to $1$, if $\mathbb{K}=\mathbb{R}$, then it may take three different values: $1$, $2$, and $4$.

The map that maps the tensor product $x^*\otimes y$ to the continuous function $f\colon G\to\mathbb{K}$, $f(g)=x^*(\lambda(g^{-1})x)$, can be extended by linearity to all of $E^*\otimes_{D(U)}E$. The image of this map is a finite-dimensional subspace of $C(G)$, call it $E_{\lambda}$. It is exactly the subspace grnerated by the matrix entries of the representation $\lambda$.

The Peter--Weyl theorem becomes:

If $\lambda_1$ and $\lambda_2$ are not equivalent, then the intersection of the spaces $E_{\lambda_1}$ and $E_{\lambda_2}$ is equal to $\{0\}$.

The set of finite $\mathbb{K}$-linear combinations of the elements of the spaces $E_{\lambda}$ is dense in both $C(G)$ and $L^2(G)$.

The space $E_{\lambda}$ carries the direct sum of $\frac{\dim E}{\dim D(E)}$ copies of the representation $\lambda$, an the Hilbert direct sum of all these representations is equivalent to the left regular representation of $G$ in $L^2(G)$.


问:
Find $\{a_n\}$ such that ${a_n}^c > k \cdot e^{a_n}$

Given $c>0$ can we find $\{a_n\} > 0$, $N_0$,  such that

$$(a_n)^c > k e^{a_n} \quad \forall n>N_0, k>0$$.

what if $\{a_n\} > 0$ is relaxed?

I was required to show $f(x)^c \in O(e^{f(x)})$ for $f(x)$ monotonic, which I could do using Taylor expansion of $e^x$, but I didn't use the monotonicity condition. So I was wondering if there are sequences for which this breaks.

答:
The exercise depends is true even if $f(x)$ is not monotonic, it's enought to assume that $f(x)>0\, \forall x$ (within the definition area). This assumption should be given, since $c$ doens't have to be a natural number.

For the exercise, it is enough to show that $\dfrac{e^{f(x)}}{f(x)^c}\geq M$ for some $M>0$.

For a sketch proof, one can use the Taylor expansion to see that $\dfrac{e^{f(x)}}{f(x)^c}$ is equal to $k_1 f(x)^{d_1} + k_2 f(x)^{d_2}$ plus many more summands bigger than $0$, where $-1\leq d_1<0$ and $0<d_2\leq 1$, and $k_1, \, k_2 >0$. If $f(x)$ tends to $0$ or infinity, one of those two summands goes to infinity. For in between, it is easy to prove that the function is bounded from below. (Remark for everyone reading that: This is only a raw sketch of the proof)

Is that possibly similar to the way you approached?

This does also answer the question about the sequences: It doesn't exist.


问:
Covariant derivatives along curves proof explanation
I am curious about the following proof of uniqueness for covariant derivatives along curves.

Proposition: There is precisely one operation $\frac{D}{dt}$ from $C^{\infty}$ vector fields along $c: I=(a,b)\rightarrow M$ ( a curve in $M$) such that

$\frac{D(V+W)}{dt}=\frac{DV}{dt}+\frac{DW}{dt}$

$\frac{D(fV)}{dt}=\frac{df}{dt}V+f\frac{DV}{dt}$ for $f\in C^{\infty}(I)$

If $V_s=Y_{c(s)}$ for some smooth vector field $Y$ defined on a neighbourhood of $c(t)$, then $\frac{DV}{dt}=\nabla_{c'(t)}Y$, where $\nabla$ is a connection.

Proof goes like this;

Choose a coordinate system $(U,x^1,.......,x^n)$ about $c(t)$ then $V(t)=\sum_i^nV^i(t)\frac{\partial}{\partial{x}^i}|_{c(t)}$ where each $V^i\in C^{\infty}(I)$.

Then we must have $\frac{DV}{dt}=\sum_i\frac{D}{dt}(V^i(t)\frac{\partial}{\partial{x}^i}|_{c(t)})$. (Whatever is defined inside the bracket is a local expression, so why are we allowed to define it like that?)

My problems with this:

Why does the above expression make sense? It should be something along the lines of $\frac{DV}{dt}=\sum_i\frac{D}{dt}(V^jg_i)$ where $V^j$ is a smooth function $I\rightarrow \mathbb{R}$ and $g_i$ is a smooth function $I\rightarrow TM$ such that $g(t)=\frac{\partial}{\partial{x^j}}|_{c(t)}$. But in this case, why would $g$ be a smooth vector field along $c$?

答:
As Professor Theodore Shifrin pointed out to you, your question is not worded very clearly, because you say that $U$ is only a neighborhood around a point $c(t)$ for fixed $t\in I$ but then you say that the $V^i$'s are functions on $I$ which makes no sense. This is because, if $U$ does not cover the entire curve $c$, the $\frac{\partial}{\partial x^i}$'s are not defined on the whole curve and thus writting $V(t)=\sum_i^nV^i(t)\frac{\partial}{\partial{x}^i}|_{c(t)}$ only makes sense for $t\in I_U:=I\cap c^{-1}(U)$. Consequently you would have $V^i\in C^{\infty}(I_U)$.

To cover the whole curve, you would need other coordinate neighborhoods around other points in the curve. In the intersections, the components $V^i$ would transform as components of vector fields under coordinate transformations. Namely, if $\tilde{U}$ is another neighborhood around some point $c(\tilde{t})$ of the curve, with coordinates $(y^1,\ldots,y^n)$, you would have

$$V(t)=\sum_i^nW^i(t)\frac{\partial}{\partial{y}^i}|_{c(t)},\quad t\in I_{\tilde{U}},$$

where $W^i\in C^{\infty}(I_{\tilde{U}})$, with $I_{\tilde{U}}:=I\cap c^{-1}(\tilde{U})$. On the intersection $I_U\cap I_{\tilde{U}}$ you would have coordinate transformations $(y^1(x),\ldots,y^n(x))$ and thus

$$W^i(t)=\sum_{j=1}^n\frac{\partial y^i}{\partial x^j}V^j(t),\quad t\in I_U\cap I_{\tilde{U}}.$$

Hope this helps.


问:
Closed abelian subgroup of $\operatorname{SU}(n)$
I'm confused why every closed abelian subgroup of $\operatorname{SU}(n)$ is contained in a maximal torus. I think there must be an obvious way to see this since the conclusion is so strong, but I failed to see how at the moment. Any hints ?

答:
Let $G$ be such a group.

By the spectral theorem, any matrix of $SU(n)$ is diagonalisable in an orthonormal basis.

Since $G$ is a abelian, elements of $G$ commute with each other and are therefore simultaneously diagonalisable.

But that means that it exists $P\in SU(n)$ (this is where the "in an orthonormal basis" part comes in) such that $PGP^{-1} \subset T$ where $T$ is the standard torus of diagonal matrix.

This shows that any such $G$ is conjugate to a subgroup of $T$. The result follows immediately : just take $P^{-1}TP$ which is maximal because $T$ is.


问:
Percentage Problem
Trying to help a student with this, but I cannot see how I am doing it wrong. Here is the problem and the answer from the book.

Q. Tina invested $30,000$ in a stock. In the first year, the
stock increased in value by $10\%$. In the second year,
the stock decreased in value by $20\%$. What percent-
age gain is required in the third year for Tina’s stock
to return to its original value? (Round to the nearest
tenth of a percent.)

A. $13.6\%$

My work:
The first year as $30000*10\%=3000$,  $30000+3000=33000$

Second year as $33000*-20\%=-6600$, $33000-6600=26400$

Third year as $26400/30000=0.88,$ $1-0.88=.12=12\%$

By steps, please, what am I doing wrong?

答:
Those $30,000$ don't matter at all.  After the first period Tina has $1.1$-times as before; after the second period only $80$% of that amount, that is in total $1.1\cdot0.8=0.88$-times of the original amount: a loss of $12$ percent.

To come out even the change rate $r$ in the third period must satisfy $0.88\cdot r=1$, that is $r=1/0.88\approx 1.136$.

Title: Percentage Problem


答:
As alreay mentioned by Lemoine, start by solving the characteristic equation $x^2+x+6=0$.

This has the complex solutions $z_{1/2}=-\frac12\pm\frac{\sqrt{23}}2i$.
So the general complex solution is $az_1^n+bz_2^n$ with $a,b\in\mathbb C$.

To transform it in real form note that $z_{1/2}=re^{\mp\varphi i}$ where $r=\sqrt6$ and $\varphi=\arctan\sqrt{23}$. Hence, the general real solution is $cr^n\cos(n\varphi)+dr^n\sin(n\varphi)$ with $c,d\in\mathbb R$.

(Of course, $cr^n\cos(n\varphi)+dr^n\sin(n\varphi)$ with $c,d\in\mathbb C$ is another representation of the general complex solution.)

Title: 2nd order homogenous linear recurrence

Body:
What will be the 2nd-order homogeneous linear recurrence of:

$$x_{n+1}+ x_n+ 6x_{n-1}=0 \text{?}$$


问:
Finding inverse in a quotient ring
Say we define $E$ to be the field $Q[\alpha]$ where $\alpha$ is a root of $x^3-3x-1$. So, $E=\{a+b\alpha+c\alpha^2:a,b,c\in\Bbb{Q}\}$. I want to find the linear combination s.t $\sum_{i=0}^2a_i\alpha^i=\alpha^{-1}$.

I know that there exists an isomorphism $\phi:\Bbb{Q[x]}/I\rightarrow \Bbb{Q[\alpha]}$ (evaluation at $\alpha$). So the question is equivalent to finding inverse of $x+I$ in my quotient ring (which is a field).

Using Euclid's algorithm, one can easily note that $(x^2-3)x=(x^3-3x-1)+1$, so $(x+I)^{-1}=(x^2-3)+I$ and so $\alpha^{-1}=\phi(x^2-3+I)=\alpha^2-3$, but looking online for general inverses in such fields, my answer doesn't match.

Is my approach correct or am I missing something? any help would be appreciated.

答:
Your work is correct. You can also check it using SageMath:

```<code>sage: E.<a> = NumberField(x^3 - 3*x - 1)
sage: 1/a
a^2 - 3
</code>```

You can also write the linear map multiplication-by-$\alpha$ as a matrix $M_\alpha$ with respect to the basis $1,\alpha,\alpha^2$ and find a preimage of the vector $(1,0,0)$ by solving the linear system:

```<code>sage: a.matrix().transpose()
[0 0 1]
[1 0 3]
[0 1 0]
sage: E.one().vector()
(1, 0, 0)
sage: a.matrix().transpose() \ E.one().vector()
(-3, 0, 1)
sage: E(a.matrix().transpose() \ E.one().vector())
a^2 - 3
</code>```

Title: Finding inverse in a quotient ring

Body:
Your work is correct. You can also check it using SageMath:

sage: E.<a> = NumberField(x^3 - 3*x - 1)
sage: 1/a
a^2 - 3

You can also write the linear map multiplication-by-$\alpha$ as a matrix $M_\alpha$ with respect to the basis $1,\alpha,\alpha^2$ and find a preimage of the vector $(1,0,0)$ by solving the linear system:

sage: a.matrix().transpose()
[0 0 1]
[1 0 3]
[0 1 0]
sage: E.one().vector()
(1, 0, 0)
sage: a.matrix().transpose() \ E.one().vector()
(-3, 0, 1)
sage: E(a.matrix().transpose() \ E.one().vector())
a^2 - 3


问:
Element in compact connected Lie group
Well, I just saw a magical result that said given an element $g$ in a compact connected Lie group $G$ and a natural number $n$, then there always exist another element in $G$ such that $g=h^n$. Well, I really don’t see how can this be true, even don’t know which aspects of the theory of Lie group can bring this result. Any suggestions?

答:
For compact connected Lie groups, the exponential map $\exp : \mathfrak{g}\to G$ is surjective.

Let $g\in G$ and $X\in \mathfrak{g}$ such that $g=\exp(X)$.

For $n\in\mathbb{N}^*$, just take $h:=\exp(\frac{X}{n})$.

Clearly, $h^n=g$.

Title: Element in compact connected Lie group


问:
Set theory, intersection and union between $n$ sets $A_1 ⊆ A_2\ldots ⊆ A_n$
Let's say I have $n$ sets $A_1, A_2\ldots A_n$ and also I know $A_1 ⊆ A_2\ldots ⊆ A_n.$

I need to prove 2 things (in the picture)

I tried to do it by induction but my teacher said he doesn't want me to prove by induction (I need to stick to definitions of union and intersection).

I tried to put bracket between the sets and claim that the place of a bracket doesn't damage the unions and the intersections.

Do you think it will be enough ? If you have another idea, I will be glad to get some help.

答:
Double inclusion...

If $x \in \bigcup A_i$ then $x \in A_j$, for some $1 \le j \le n$. And thus $x \in A_n$.

Conversly, if $x \in A_n$ then $x \in \bigcup A_i$.

If $x \in \bigcap A_i$ then $x \in A_j$, for every $1 \le j \le n$, and thus also $x \in A_1$.

And if $x \in A_1$, then...


答:
It is clear that $A_n\subset\bigcup_{i=1}^nA_i$. On the other hand, if $x\in\bigcup_{i=1}^nA_i$, then $x\in A_i$, for some $i\in\{1,2,\ldots,n\}$. But $A_i\subset A_n$, and therefore $x\in A_n$. So $A_n\supset\bigcup_{i=1}^nA_i$.

Can you deal with the other equality now?

Title: Set theory, intersection and union between $n$ sets $A_1 ⊆ A_2\ldots ⊆ A_n$

Body:
Let's say I have $n$ sets $A_1, A_2\ldots A_n$ and also I know $A_1 ⊆ A_2\ldots ⊆ A_n.$

I need to prove 2 things (in the picture)

I tried to do it by induction but my teacher said he doesn't want me to prove by induction (I need to stick to definitions of union and intersection).

I tried to put bracket between the sets and claim that the place of a bracket doesn't damage the unions and the intersections.

Do you think it will be enough ? If you have another idea, I will be glad to get some help.


问:
Where does the Integral went wrong .... $ \displaystyle \int _{0}^{\pi} \frac {\ln ( 1 + \sin \alpha \cos x )}{\cos x} \ \mathrm {d}x $
This is how I solved :

Let $$ \displaystyle y = \int_{0}^{\pi} \frac {\ln ( 1 + \sin \alpha  \cos x )}{\cos x} \ \mathrm {d}x
 $$
Then I used expansion of $\displaystyle \ln (1+x) = \sum _{n=0}^{\infty} (-1)^n \frac {x^{n+1}}{n+1} $

Now, $\displaystyle y = \sum _{n=0}^{\infty} (-1)^n \frac {\sin ^{n+1}  \alpha }{n+1} \int_{0}^{\pi} \cos ^n x \ \mathrm {d}x \implies y = \sum _{n=0}^{\infty} [1+(-1)^n] \frac {\sin ^{n+1}  \alpha }{n+1} \int_{0}^{\frac {\pi}{2} } \cos ^n x \ \mathrm {d}x $

Replacing $n$ by $2k$ (as we will end up with $0$ for odd $n$) and using gamma function  :

$$ \displaystyle y =  2 \sum _{k=0}^{\infty} \frac {\sin ^{2k+1}  \alpha }{2k+1} \cdot \frac { \Gamma \left ( \frac {2k+1}{2} \right ) \sqrt {\pi} }{ \Gamma \left ( \frac {2k+2}{2} \right )} $$

Which can be simplified to $\displaystyle y =  2 \pi \sum _{k=0}^{\infty} \frac {\sin ^{2k+1} \alpha }{2k+1} \cdot \frac {(2k-1)!!}{k! \cdot 2^k}$

Modification :
$$ \displaystyle y = 2 \pi \sum _{k=0}^{\infty} \frac {\sin ^{2k+1} \alpha }{2k+1} \cdot \frac {(2k-1)!!}{k! \cdot 2^k} \cdot \frac {k! 2^k}{k! 2^k} \implies y = 2 \pi \sum _{k=0}^{\infty}  \frac {\sin ^{2k+1} \alpha }{2k+1} \binom {2k}{k} \frac {1}{4^k} $$

We get : $\displaystyle \frac {\mathrm {d}y }{\mathrm {d} ( \sin \alpha ) } = 2 \pi \sum _{k=0}^{\infty} \binom {2k}{k}  \left ( \frac {\sin ^{2} \alpha}{4} \right ) ^k$

As $\displaystyle (1-4x)^{-1/2} = \sum _{n=0}^{\infty}  \binom {2n}{n} x^n$

We will finally get : $\displaystyle y = 2 \pi \int \frac {\mathrm {d} (\sin \alpha )}{ \cos \alpha }   \implies  y = 2 \pi \alpha + C $

Now as $\alpha = 0 $ the summation is zero $\implies C = 0  $

Hence,

$$ \displaystyle \int_{0}^{\pi} \frac {\ln ( 1 + \sin \alpha  \cos x )}{\cos x} \ \mathrm {d}x = 2 \pi \alpha $$

But :

This suggests : $ \displaystyle \int_{0}^{\pi} \frac {\ln ( 1 + \sin \alpha  \cos x )}{\cos x} \ \mathrm {d}x = \pi \alpha $

答: Your mistake is thinking $\int_0^{\pi/2}\cos^nx\mathrm{d}x$ is a Beta function; it's only half of one, viz.$$\operatorname{B}(a,\,b)=2\int_0^{\pi/2}\sin^{2a-1}x\cos^{2b-1}x\mathrm{d}x.$$


问:
Show $ \int_0^{\frac\pi2} \frac{(1+\sec^2t)\sqrt{\sec t}}{ (1+\sec t)^2-2}dt= \frac\pi{\sqrt2}$
I try to figure out how to deal with the trigonometric integral

$$ \int_0^{\frac\pi2} \frac{(1+\sec^2t)\sqrt{\sec t}}{ (1+\sec t)^2-2}dt
$$
which is supposed to be equal to $\frac\pi{\sqrt2}$. Despite the simplicity of the result, it does not seem easy. What is troublesome is the square-root part in the integrand, which renders known techniques, such as half-angle substitution, ineffective.

I also threw it into WA, which churns out a complicated antiderivative expression in elliptical functions not all that helpful.

答: $$\int_0^{\pi/2}\frac{(1+\sec^2t)\sqrt{\sec t}}{(1+\sec t)^2-2}\,dt\stackrel{u=\frac{\sin t}{\sqrt{\cos t}}}=2\int_0^\infty\frac{du}{u^2+2}=\frac\pi{\sqrt2}$$


问:
Predetermine the middle term of an quadratic equation when there are two possibilities

Factorise  completely:

$2x^2+13x+15$

$3x^2-5x+2$

As i figured, there could be two ways to break the middle term. Either, $15x-2x$ or, $10x+3x$.

If I continue with  $15x-2x$:

$2x^2+13x+15$

$2x^2+15x-2x+15$

$2x(x-1)+15(x+1)$

But If I go down with $10x+3x$:

$2x^2+13x+15$

$2x^2+10x+3x+15$

$2x(x+5)+3(x+5)$

I know the former is incorrect and the later one is correct. But I would really like to know how to predetermine this problem and avoid continuing with the former one.

NB: No. 2 has the exact same approach.

答:
There is a way to factorize a general quadratic trinomial
$$Ax^2 + Bx + C = (Dx + E)(Fx + G),$$
that is, assuming that the trinomial can be factored in the first place.  (This method will also let you determine whether the trinomial cannot be factored at all.)

Set
$$Ax^2 + Bx + C = 0$$
and solve for $x$ using the Quadratic Formula:
$$x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}.$$
Note that, in order for this method to work, then the discriminant
$$d = B^2 - 4AC$$
needs to be a perfect square.

Now, let us apply this method to your first quadratic trinomial:
$$2x^2 + 13x + 15$$
Set $2x^2 + 13x + 15 = 0$, then solve for $x$ using the Quadratic Formula, where $A_1 = 2, B_1 = 13, C_1 = 15$:
$$x = \frac{-{B_1} \pm \sqrt{{B_1}^2 - 4{A_1}{C_1}}}{2{A_1}} = \frac{-13 \pm \sqrt{169 - 4\cdot{2}\cdot{15}}}{4} = \frac{-13 \pm \sqrt{49}}{4}.$$
Note that the discriminant
$$d_1 = {B_1}^2 - 4{A_1}{C_1} = 49$$
is a perfect square.  Hence, this means that the first quadratic trinomial is indeed factorable.

We obtain the $x$-values
$$x = \frac{-13 \pm 7}{4} = -5, -\frac{3}{2}$$
which means that we have the factors
$$(x + 5)(2x + 3) = 2x^2 + 13x + 15.$$

I leave it as an exercise for you to apply the Quadratic Formula Method for factoring your second quadratic trinomial:
$$3x^2 - 5x + 2$$


问:
Discarding apples - why is my reasoning wrong?
There are two types of apples - red and green. The ratio of red to green is $4:1$. They are mixed together before they are boxed. There is $\frac{1}{50}$ chance that a green apple is discarded, and $\frac{1}{100}$ chance that a red apple is discarded. What is the probability that an apple selected will be discarded?

Attempt: I have

Green apple discarded OR red apple discarded

$\frac{4}{5}\frac{1}{50}$ + $\frac{1}{5}\frac{1}{100}$

Question: Why is this incorrect? It seems I have to multiply by $\frac{2}{3}$ to get the correct answer.

答:
You seem to have confused red and green and your calculation should have been $$\frac{4}{5}\frac{1}{100}+\frac{1}{5}\frac{1}{50}=\frac{3}{250}$$

As a check, suppose you had $1000$ apples

You would expect to have $800$ red apples and $200$ green apples

You would expect to discard $8$ red apples and $4$ green apples

So you would expect to discard $12$ apples, $\frac{12}{1000}=\frac{3}{250}$ of the total


答:
Haven't you mixed it up a bit?

Let $P(G)$ be probability of a green apple being selected and $P(R)$ be probability of a red apple being selected.

Let $P(D|G)$ be probability of discard if green apple and $P(D|R)$ be probability of discard if red apple.

Probability of an apple being discarded $P(D) = P(G)\cdot P(D|G) + P(R)\cdot P(D|R) = \frac 15 \frac 1{50} + \frac 45 \cdot \frac 1{100} = \frac 3{250}$

Title: Discarding apples - why is my reasoning wrong?

Body:
There are two types of apples - red and green. The ratio of red to green is $4:1$. They are mixed together before they are boxed. There is $\frac{1}{50}$ chance that a green apple is discarded, and $\frac{1}{100}$ chance that a red apple is discarded. What is the probability that an apple selected will be discarded?

Attempt: I have

Green apple discarded OR red apple discarded

$\frac{4}{5}\frac{1}{50}$ + $\frac{1}{5}\frac{1}{100}$

Question: Why is this incorrect? It seems I have to multiply by $\frac{2}{3}$ to get the correct answer.


问:
Change of basis matrix for inner product space
If $B = (e_1,e_2,\ldots, e_n)$ is a basis for an inner product space $V$ and  $B' = (f_1,f_2,\ldots,f_n)$ is an orthonormal basis of $V$.

Is the change of basis matrix $P$ necessarily orthogonal?

答: Take any non-orthogonal invertible matrix $Q$. We wish to find a basis $e_1,\dots,e_n$ such that $Q$ is the change of basis from $(e_1,\dots,e_n)$ to $(f_1,\dots,f_n)$. But for this we may just set $e_i = Q_{1i} f_1 + \dots + Q_{ni} f_n$.</p>


问:
Simplifying the positive semidefinite constraint if closed-form expressions of the eigenvalues are found
I came across an article and I am not sure how the simplification is done. Please find my question and related information below.

[Information]

In the paper,

$$\begin{aligned} A_1 &= I_N-\lambda_1 h_1 h_1^H+\lambda_2 r_2 h_1 h_1^H+\lambda_3 r_2h_2 h_2^H \\ A_2 &= I_N-\lambda_2 h_1 h_1^H-\lambda_3h_2 h_2^H\end{aligned}$$

where $\lambda_1,\lambda_2,\lambda_3 \geq0$, $r_1,r_2 \geq 0$, $h_1,h_2 \in \mathbb{C}^N$, and $I_N$ is the $N \times N$ identity matrix.

Furthermore there is a lemma: For $h$, $g \in \mathbb{C}^N$, define $\cos^2(\theta)=\frac{h^Hgg^Hh}{\|{h}\|^2\|g\|^2}$. Then $h h^H+g g^H$ has at most 2 non-zero eigenvalues: $$\frac{1}{2}\left(\|{h}\|^2+\|g\|^2\right) \pm \sqrt{\left(\|{h}\|^2+\|g\|^2\right)^2-4 \|{h}\|^2\|g\|^2 \sin^2\left(\theta\right)}, \tag{a}$$
and similarly $h h^H-g g^H$ has at most two nonzero eigenvalues:
$$\frac{1}{2}\left(\|{h}\|^2-\|g\|^2\right) \pm \sqrt{\left(\|{h}\|^2-\|g\|^2\right)^2+4 \|{h}\|^2\|g\|^2 \sin^2\left(\theta\right)}. \tag{b}$$

(P.s. one may assume the lemma is correct. Ref: Eigenvalues of a sum of rank-one matrices? )

[Question]

Based on the given information, the authors stated that the constraints $A_1\geq0$ and $A_2\geq0$ can respectively be reduced to the following:
$$\|{k_1}\|^2-\|{k_2}\|^2 \leq \text{min} \{ 1-\|k_1\|^2 \|k_2\|^2 \sin^2(\theta),2 \}\tag{1}$$
$$\|{l_1}\|^2+\|{l_2}\|^2 \leq \text{min} \{ 1+\|l_1\|^2 \|l_2\|^2 \sin^2(\theta),2 \} \tag{2}$$

where
$$\begin{aligned} k_1 &= \sqrt{\lambda_1-\lambda_2 r_2} h_1\\ k_2 &= \sqrt{\lambda_3 r_3}h_2\\ l_1 &= \sqrt{\lambda_2}h_1\\ l_2 &= \sqrt{\lambda_3}h_2\\ \cos^2(\theta) &= \frac{h_1^Hh_2 h_2^Hh_1}{\|{h_1}\|^2\|h_2\|^2}\end{aligned}$$

My question is: given the eigenvalues as expressed in (a) and (b), how can we reduce the constraints $A_1\geq0$ and $A_2\geq0$ to (1) and (2), respectively.

[Extension of Rodrigo de Azevedo's answer]

Many thanks to Rodrigo de Azevedo's answer. As requested, I would extend his answer to fully address my question. To avoid confusion, the eigenvalues used in his answer are denoted as $\phi_1,\phi_2$. WLOG, I will start from the 2+1 inequalities. By further exploiting orthogonality, we thus obtain
$$\begin{aligned}
\phi_1 & \leq 1\\\\
\phi_2 & \leq 1\\\\
\phi_1 + \phi_2 & \leq 1 + \phi_1 \phi_2
\end{aligned}$$
Hence, we immediately have $A_1 \geq 0 \iff (1)$ and $A_2 \geq 0 \iff (2)$.

答:
Let

$$\begin{aligned} {\rm  A} &:= {\rm  I}_n - \begin{bmatrix} | & | \\ {\rm  v}_1 & {\rm  v}_2\\ | & | \end{bmatrix} \begin{bmatrix} \lambda_1 & 0\\ 0 & \lambda_2 \end{bmatrix} \begin{bmatrix} | & | \\ {\rm  v}_1 & {\rm  v}_2\\ | & | \end{bmatrix}^\top\\\\ &\,= {\rm  I}_n - \begin{bmatrix} | & | \\ \sqrt{\lambda_1} \, {\rm  v}_1 & \sqrt{\lambda_2} \, {\rm  v}_2\\ | & | \end{bmatrix} \begin{bmatrix} | & | \\ \sqrt{\lambda_1} \, {\rm  v}_1 & \sqrt{\lambda_2} \, {\rm  v}_2\\ | & | \end{bmatrix}^\top = {\rm  I}_n - {\rm  W} {\rm  W}^\top\end{aligned}$$

and ${\rm  w}_i := \sqrt{\lambda_i} \, {\rm  v}_i$. If $\color{blue}{{\rm  A} \succeq {\rm  O}_n}$, then

$$\begin{bmatrix} {\rm  I}_n & {\rm  W}\\ {\rm  W}^\top & {\rm  I}_2 \end{bmatrix} \succeq {\rm  O}_{n+2}$$

From the complementary Schur complement,

$${\rm  I}_2 - {\rm  W}^\top {\rm  W} = \begin{bmatrix} 1 - \| {\rm  w}_1 \|_2^2 & - \langle {\rm  w}_1, {\rm  w}_2 \rangle \\ - \langle {\rm  w}_1, {\rm  w}_2 \rangle & 1 - \| {\rm  w}_2 \|_2^2 \end{bmatrix} \succeq {\rm  O}_2$$

Using Sylvester's criterion for positive semidefiniteness, since both diagonal entries and the determinant must be non-negative, we obtain $2+1$ inequalities

$$\begin{aligned} \| {\rm  w}_1 \|_2^2 & \leq 1\\\\ \| {\rm  w}_2 \|_2^2 & \leq 1\\\\ \left( 1 - \| {\rm  w}_1 \|_2^2 \right) \left( 1 - \| {\rm  w}_2 \|_2^2 \right) - \left| \langle {\rm  w}_1, {\rm  w}_2 \rangle \right|^2 &\geq 0\end{aligned}$$

From which we conclude that

$$\begin{aligned} \| {\rm  w}_1 \|_2^2 + \| {\rm  w}_2 \|_2^2 &\leq 2\\\\ \| {\rm  w}_1 \|_2^2 + \| {\rm  w}_2 \|_2^2 &\leq 1 + \| {\rm  w}_1 \|_2^2  \| {\rm  w}_2 \, \|_2^2 \, \sin^2(\theta)\end{aligned}$$

and, finally,

$$\color{blue}{\boxed{\\ \,\\\qquad \| {\rm  w}_1 \|_2^2 + \| {\rm  w}_2 \|_2^2 \leq \min\left( 1 + \| {\rm  w}_1 \|_2^2  \| {\rm  w}_2 \, \|_2^2 \, \sin^2(\theta), 2\right) \qquad \\}}$$


问:
Probability question with similar/distinct dice.
The question:

We're throwing $3$ normal dice twice. (Suppose that every vector with length of $6$ has an equal probability). What is the probability to get two similar results from the two throws of the $3$ dice if:

a) The dice are different.

b) The dice are similar.

My Work:

for (a):
  $\Omega = \{(a_1,a_2,...,a_6) \mid a_i \in \{1,...,6\}  ,(i=1,2,...,6)\}$

if all the dice are different then: $|\Omega| = 6^6$

Let $A\subset\Omega$, ($A$- we get similar results from the two throws), then $|A|=6^3$ (first dice throw can be whatever, but the second one must be the same).

So the final answer I got is $P(A) = \frac {6^3}{6^6}$.

for (b): (and here comes the confusion):

Things I tried to do and didn't get the right answer:

I tried to choose $3$ of the dice (since they're all similar) throw them, and sort them, so what I got is $\binom{6}{3} \cdot 6^3 \cdot 3!$.

I tried to choose $1$ of the two throws, and find how many options there are and sort them (same as before, each dice has $6$ sides so $6^3$ for all), and got $\binom{2}{1} \cdot 6^3 \cdot 3!$.

Both of the two ways led me to wrong answers, I'm trying where are my mistakes and why my thinking led me to wrong answers, which can lead me to reach the true answer.

The Final answer is $\frac {996}{6^6}$

EDIT:

After deeply thinking about the problem I obtained a new approach (I couldn't reach the answer with it, but I'm not sure if it's my weak combinatorics or if it's a wrong approach).

By writing down a couple of examples, I noticed that there's something that I can also use to calculate the probability, we can divide this into $3$ possible cases:

First: Getting same number on all of the six dice. $(2,2,2, 2,2,2)$

Second: Getting $2$ different numbers. $(2,4,2, 2,4,2)$

Third: Getting $3$ different numbers. $(1,2,3, 1,2,3)$

So I could calculate the number of options each case has and divide by the total options $|\Omega|=6^6$.

I felt like I've gone too far but I can't see why it wouldn't be true. Any Feedback is really appreciated.

答:
With the help of N. F. Taussig in the comments, I think I have reached the right solution, So I decided to answer my own question, and would love to hear feedback.

Starting with why my first two approaches are wrong, The number of ways that the second roll could match the first is different whenever there's $3$ different numbers or $2$ different or all of them are the same, but in my first two approaches, I didn't pay attention to that.

Solution: (Approach 3):

The number of ways to get the same number on all dice, is just simply choosing a number which is $\binom{6}{1}=6$.

The number of ways to get two different numbers in the first roll is:
 $\binom{6}{2}$ choosing $2$ numbers, $\binom{2}{1}$ choosing the number that will appear twice, $\binom{3}{1}$ choosing which place is the single number in the first roll, $\binom{3}{1}$ choosing in which place is the single number in the second roll. So in total by multiplying all of them we get $\binom{6}{2}*2*3*3 = 270$ ways.

The number of ways to get three different numbers in the first roll is:

$\binom{6}{3}$ choosing $3$ numbers, $3!$ ordering the first roll, $3!$ ordering the second roll.  in total $720$.

Hence we reach the answer by adding them up and dividing by $|\Omega|$.


问:
An entire function which is onto
Prove that $\sin z - z$ is surjective, z$\in \mathbb C$

$\sin z - z$ is an entire function and hence by Little Picard its image can omit at most one complex number. I am unable to arrive at a contradiction from here on. Is there a way out! Please help.

答: $\sin(z) - z$ is odd, so if it omitted one non-zero point, it would omit two.

Title: An entire function which is onto


答:
Assume that $f(z) = \sin(z) - z$ omits the value $a \in \Bbb C$. Then
$$
 f(z) = \sin(z  +2\pi) -(z+2\pi) + 2\pi = f(z+2\pi) + 2\pi
$$
also omits the value $a+2\pi$, contradicting Picard's theorem.

(And “Picard's Great Theorem” implies that $\sin(z)-z=a$ has infinitely many solutions for any $a \in \Bbb C$).

An alternative solution is possible with the Hadamard factorization theorem. $\sin(z)-z-a$ has the order $1$. If it has no zeros then
$$
 \sin(z) = a+z+e^{cz+d}
$$
with complex constants $c, d \in \Bbb C$. But the right-hand side is unbounded for $z=x \in \Bbb R$, so this is not possible.

In the same way it can be shown that $\sin(z) - P(z)$ has infinitely many zeros for any polynomial $P$.

Title: An entire function which is onto

Body:
Prove that $\sin z - z$ is surjective, z$\in \mathbb C$

$\sin z - z$ is an entire function and hence by Little Picard its image can omit at most one complex number. I am unable to arrive at a contradiction from here on. Is there a way out! Please help.


问:
How to prove $A=\mathcal{O}_n$, use trace of matrix
Let $A=B+C$, and $B=B^T$, $C=-C^T$, $BC=\mathcal{O}_n$. Prove that if $A^2=\mathcal{O}_n$, then $A=\mathcal{O}_n$.
I want to prove $B=C=
\mathcal{O}_n$, and I got that
$$A^2=B^2+CB+BC+C^2\implies0=\mathrm{tr}(A^2)=\mathrm{tr}(B^2)+\mathrm{tr}(C^2)$$
So$$\mathrm{tr}(BB^T)=\mathrm{tr}(CC^T)\geqslant0.$$
But I don't know how to go on.

答:
I got it
$$BC=-B'C'=-(CB)'=O\implies CB=O\implies A^2=B^2+C^2=O.$$
$$B^2(B^2+C^2)=B^4+B^2C^2=O\implies B^4=O.$$
$$B^2(B^2)'=O\implies \mathrm{tr}(B^2(B^2)')=0\implies B^2=O\implies\mathrm{tr}(BB')=0\implies B=O.$$
$$C^2=O\implies\mathrm{tr}(CC')=O\implies C=O\implies A=B+C=O.$$


问:
Distance to a curve via Newton's Method
So I wanted a quick confirmation if my thinking is correct.
The particular problem I'm working on is a $2$D Bezier Curve, with at most $4$ control points and at a minimum of $2$ control points.
The more general question (also I'm going to use the distance squared because it is cleaner to look at);

If given an arbitrary coordinate, $(x,y)$, and $2$ arbitrary functions, $F_x(t)$ and $F_y(t)$, with the distance squared being

$D(t)=(x-F_x(t))^2+(y-F_y(t))^2$

The first order derivative of that equation with respect to $t$ being

$D'(t)=2*(x-F_x(t))(-F_x'(t))+2*(y-Fy(t))(-Fy'(t))$

The second order derivative being;
$D''(t)=2*(x-F_x(t))(-F''_x(t))+2*F'_x(t)^2+2*(y-F_y(t))(-F_y''(t))+2*F'_y(t)^2$

I should be able to use Newton's method of

$t_{n+1}=t_n-D'(t_n)/D''(t_n)$

to approximate the t with the minimum distance to $F_x(t)$ and $F_y(t)$, correct?

(Clamping on t as well from the range of $t\in\{0,1\}$)

答:
Beware that the Newton's method may fail in different ways:

it can find a root outside the usefult range of $t$, $[0,1]$ (you also need to check the distance at the endpoints of the curve).

it can find a local minimum which is not the global one (depending on the starting point).

The true challenge is to isolate the correct root.

This link https://computingandrecording.wordpress.com/2017/03/20/closest-point-to-a-cubic-spline/ suggests using the Sturm's sequence method to do this.

Title: Distance to a curve via Newton's Method


问:
A (not necessarily continuous) function on a compact metric space attaining its maximum.
I am studying for an exam and my study partners and I are having a dispute about my reasoning for $f$ being continuous by way of open and closed pullbacks (see below).  Please help me correct my thinking.  Here is the problem and my proposed solution:

Let $(K, d)$ be a compact metric space, and let $f: K \rightarrow \mathbb{R}$ be a function satisfying that for each $\alpha \in \mathbb{R}$ the set {$x \in K: f(x) \ge \alpha$} is a closed subset of $K$.  Show that $f$ attains a maximum value on $K$.

Proof: Notice that $A :=$ {$x \in K: f(x) \ge \alpha$} is precisely $f^{-1}[\alpha, \infty)$.  Since $[\alpha, \infty)$ is closed in $\mathbb{R}$ and $A$ is assumed to be closed in $K$, then it follows that $f$ is continuous on $A$.  On the other hand, $K-A = f^{-1}(-\infty, \alpha)$ is open in $K$ since $A$ is closed in $K$.  And since $(\alpha, \infty)$ is open in $\mathbb{R}$ and $K - A$ is open in $K$, then if follow that $f$ is continuous on $K - A$, hence $f$ is continuous on $K$.  Since $K$ is compact and $f$ is continuous, then $f(K)$ is compact in $\mathbb{R}$.  Compact sets in $\mathbb{R}$ are closed and bounded intervals.  Thus $\sup{f(K)} = \max{f(K)} = f(x_0)$ for some $x_0 \in K$.  Thus $f$ indeed attains its maximum value on $K$.  $\blacksquare$

答:
Nope, for continuity, you'd need to know that $f^{-1}(C)$ is closed for all closed sets $C$. You don't know that, you only know it for some closed sets $C$.

Hint for correct solution: Show that $f^{-1}(-\infty,\alpha)$ is open for any $\alpha$. Cover $X$ with sets of the form $f^{-1}(-\infty,\alpha)$. Then it must have  a finite sub-cover.


问:
Let $p$, $q$ be primes such that $q \equiv 2 \pmod{5}$ and $p = 4q+1$. Show that $100^{q} \not\equiv 1\pmod{p} $.
Let $p$, $q$ be two primes such that $q \equiv 2 \pmod{5}$ and $p = 4q+1$.

Show that $$100^{q} \not\equiv 1\pmod{p} $$

Here is one way that I tried to tackle this (and failed, obviously...):

Assume by contradiction that $100^{q} \equiv 1\pmod{p} $. By Fermat's little Theorem it follows that: $$100^{q} \equiv 100 \pmod{q}$$ This is where I got stuck. However, I did notice that:

$$p \equiv 1\pmod{q}$$
$$p \equiv -1\pmod{5}$$

But these congruences didn't help me either.

答:
If we negate the conclusion, we have
$$
100^q\equiv 1 (\textrm{ mod }p),$$
and this is equivalent to the existence $x$ in the following
$$
100\equiv x^4  (\textrm{ mod }p)$$
This reduces to $
10 \equiv x^2  (\textrm{ mod }p)$, or $
-10 \equiv x^2 (\textrm{ mod }p)$

Both of them cannot be true since
$$
\left(\frac{10}{p}\right)=\left(\frac{-10}{p}\right) = -1.$$


问:
Tennis match probability - is my logic incorrect?
A tennis match consists of sets, until one player wins three sets. Player $A$ has a two-thirds chance of winning. Player $B$ has a one-third chance of winning. If they play a match, what is the probability of $A$ winning?

My attempt:

The minimum games for $A$ victory is $3$, and the maximum is $5$ (both win $2$, then next round is decider)

I have probability = win in three + win in four + win in five

=$\left(\frac{2}{3}\right)^{3}+ {4\choose 1}\ \left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^{3}+{5\choose2}\left(\frac{1}{3}\right)^{2}\left(\frac{2}{3}\right)^{3}$

Which is more than $1$!

I don't understand why this is wrong though

答:
Your working: $\displaystyle \small \left(\frac{2}{3}\right)^{3}+ {4\choose 1}\ \cdot \frac{1}{3} \cdot \left(\frac{2}{3}\right)^{3}+{5\choose2}\left(\frac{1}{3}\right)^{2}\left(\frac{2}{3}\right)^{3}$

Please see the second term. You are allowing any $1$ of the $4$ sets to be won by player B. But if there are $4$ sets, player A  cannot win first three sets. Same logic for the next term.

So it should be,

$\displaystyle \small \left(\frac{2}{3}\right)^{3}+ {3\choose 1}\ \cdot \frac{1}{3} \cdot \left(\frac{2}{3}\right)^{3}+{4\choose2}\left(\frac{1}{3}\right)^{2}\left(\frac{2}{3}\right)^{3}$


问:
Percentage Problem
Trying to help a student with this, but I cannot see how I am doing it wrong. Here is the problem and the answer from the book.

Q. Tina invested $30,000$ in a stock. In the first year, the
stock increased in value by $10\%$. In the second year,
the stock decreased in value by $20\%$. What percent-
age gain is required in the third year for Tina’s stock
to return to its original value? (Round to the nearest
tenth of a percent.)

A. $13.6\%$

My work:
The first year as $30000*10\%=3000$,  $30000+3000=33000$

Second year as $33000*-20\%=-6600$, $33000-6600=26400$

Third year as $26400/30000=0.88,$ $1-0.88=.12=12\%$

By steps, please, what am I doing wrong?

Title: Percentage Problem

Body: You have $\frac{26400}{30000}$ but you should have $\frac{30000}{26400}$.


问:
Convergence for complex series tests
Suppose that
$∑a_nz^n$ has radius of convergence $R$ and let $C$ be the circle {$z ∈ C :
|z| = R$}. Prove or disprove the following (which may or may not be true):

(i) If
$∑ a_nz^
n$ converges at some point on $C$ then it converges everywhere on $C$.

(ii) If
$∑ a_nz^
n$ converges absolutely at some point on $C$ then it converges absolutely
everywhere on C.

(iii) If
$∑ a_nz^
n$ converges at every point on $C$, except possibly one, then it converges
absolutely everywhere on $C$.

Sorry since this question has already been asked, but I am thinking of solving it in the following way, for

i) it's false since we can put $z=1$ and it will diverge, and $z=-1$ and it will converge.

ii) I think it's true since the points on the radius all have the same absolute value, thus the absolute value of their series converge, thus they converge.

iii) I am not sure how to go with it

答:
i) You should provide an example, such as, for instance, $\sum_{n=1}^\infty\frac{z^n}n$, which converges when $z=-1$ and diverges when $z=1$.

ii) You are right.

iii) Again, take $\sum_{n=1}^\infty\frac{z^n}n$, which converges when $|z|=1$ and $z\ne1$ (by the Dirichlet test) but diverges when $z=1$. This shows that the assertion is false.

Title: Convergence for complex series tests


问:
Convergent successions is a closed set of bounded successions with the supremum distance
What I'm trying to prove is the following:
Be $l^{\infty} = \{x = \{x_n\} \subset \mathbb{R}: ||x||_{\infty}=sup_n |x_n| < \infty\}$ a metric space using $d(x,y) = ||x-y||_{\infty}$ for $x,y\in l^\infty$. Be $c$ the space of the convergent successions of real numbers. Then $c$ is a closed set of $l^\infty$.

My approach so far looks like this:

Be $\{z_k\}_k \subset c$ a convergent succession, with $\{z_{k,n}\}_n \subset \mathbb{R}$, $z_k \rightarrow z=\{z_n\}_n \subset \mathbb{R}$. Let's see that $z$ is convergent so $z\in c$: Be $\varepsilon > 0$.

Firstly, as $z_n \rightarrow z, \exists k_0 \in \mathbb{N}$ such that $\forall k \geq k_0, d(z_k, z)< \varepsilon / 2$, which, by definition of the distance, means that $\forall k \geq k_0, |z_{k,n} - z_n |< \varepsilon / 2, \forall n \in \mathbb{N} $.

Now, my approach would follow by using the inequality
$$
|z_n - a| \leq |z_n - z_{k,n}| + |z_{k,n} - a| < \varepsilon/2 + \varepsilon/2 = \varepsilon, \forall k\geq k_0, n\geq n_0
$$
where $n_0$ is such that $|z_{k,n} - a | < \varepsilon$ (as $z_{k,n}$ converges because it is from $c$).

The problem is that I don't know how (or is it even true) to prove that all the $z_{k,n}$ converge to the same $a$ for all $k$ (or at least $k$ big enough), and therefore this prove is invalid. My approach to prove this is by using that, as $z_k$ is convergent, it is also Cauchy, and, therefore, for two $k_a, k_b$ big enough, if $z_{k_a,n} \rightarrow a$, then so does $z_{k_b,n} \rightarrow a$, as
$d(z_{k_a,n}, z_{k_b,n}) < \varepsilon /4 \implies |z_{k_a,n} - z_{k_b,n}| < \varepsilon / 4, \forall n\in\mathbb{N}$, and therefore $|z_{k_b,n} - a| \leq |z_{k_b,n} - z_{k_a,n} + |z_{k_a,n} - a| < \varepsilon/4 + \varepsilon/4 = \varepsilon/2$. Is this approach correct? The main problem I see here is that  depending on the last $k_a$, $a$ could be different and therefore the prove is not solid as $z$ would converge to different values.

Thank you

答:
To avoid the probelm you are facing show that $(z_n)$ is  Cauchy sequence. That is enough to say that it is a convergent sequence.

Use the inequality $|z_n-z_m| \leq |z_{k,n}-z_{k,m}|+|z_{k,n}-z_n|+|z_m-z_{k,m}|$. Can you finish?


问:
Last step in proof of comparison theorem of etale and singular cohomology
I am trying to understand the argument in the very last sentence in the proof of Lemma 4.5 in SGA 4 Exposé XI, page 47 here:  https://www.normalesup.org/~forgogozo/SGA4/tomes/tome3.pdf

The argument that I am trying to understand is the following (english translation) "since $H^1(X_{ét}, \mathbb{Z}/n\mathbb{Z})\to H^1(X_{cl}, \mathbb{Z}/n\mathbb{Z})$ is bijective, we immediately obtain the vanishing in 4.5 for $q=1$".

Lemma 4.5 basically states the vanishing of $R^q \varepsilon_*\mathbb Z/n \mathbb Z$ for $q>0$, so what we are trying to deduce is $R^1 \varepsilon_*\mathbb Z/n \mathbb Z=0$.

My idea was to use the exact sequence of low degree of the Leray spectral sequence
\begin{align*}
  H^p(X_{ét},R^q\epsilon_{*}\mathbb Z/n \mathbb Z) \Longrightarrow H^{p+q}(X_{cl},\mathbb Z/n \mathbb Z)
 \end{align*}

induced by the morphism of sites $\epsilon \colon X_{cl} \to X_{ét}$. This yields
\begin{align}
0 \to H^1(X_{ét}, \epsilon_*\mathbb Z/n \mathbb Z) \to H^1(X_{cl}, \mathbb Z/n \mathbb Z) \to H^0(X_{ét}, R^1 \varepsilon_*\mathbb Z/n \mathbb Z) \to H^2(X_{ét}, \epsilon_*\mathbb Z/n \mathbb Z) \to H^2(X_{cl}, \mathbb Z/n \mathbb Z)
\end{align}
Now, knowing that the first map is an isomorphism doesn't seem to be enough to conclude that $H^0(X_{ét},R^1 \varepsilon_*\mathbb Z/n \mathbb Z)$ vanishes, so I am stuck here. Help with this argument, or an alternative argument, would be much appreciated.

答:
I think the intended argument is as follows: Since $H^1(X_{\mathrm{\acute{e}t}},\mathbb Z/n\mathbb Z)\rightarrow H^1(X_{\mathrm{cl}},\mathbb Z/n\mathbb Z)$ is bijective for all $X$ (under the assumptions of Théorème 4.3, as then both sides parametrise $\mathbb Z/n\mathbb Z$-principal étale coverings), it suffices to show:

Lemma. For all $\xi\in H^1(X_{\mathrm{\acute{e}t}},\mathbb Z/n\mathbb Z)$ and all geometric points $x$ of $X$ there exists an étale neighbourhood $X'\rightarrow X$ of $x$ such that $\xi$ vanishes in $H^1(X'_{\mathrm{\acute{e}t}},\mathbb Z/n\mathbb Z)$.

This is a completely general fact about cohomology; see [Stacks Project, Tag 01FW] for a reference. In your situation, this can also be seen geometrically: $\xi$ parametrises a $\mathbb Z/n\mathbb Z$-principal étale covering, and any such covering is étale-locally trivial.


问:
Diagonalizable Matrix $A$**$\in \mathcal{M_n}(\mathbb{R})$
Can someone please state the most known conditions for a Matrix $A$$\in \mathcal{M_n}(\mathbb{R})$ to be diagonalizable?

Without proof, just by shedding light on them please. In the means of being symmetric, or concerning eigenvalues, and so...

答:
If there are $n$ linearly independent eigenvectors, $$V1, V2, ..., Vn$$ then the matrix $P$ whose columns are these eigenvectors will satisfy $$ AP =PD $$ where D is the diagonal matrix of eigenvalues. We may solve for A or for D depending the application on hand.  $$P^{-1}AP=D
$$ or $$ A=PDP^{-1} $$


问:
How to integrate $\int_0^{\pi/2} \frac{\sin(x)}{\sin(x+\frac{\pi}{4})}{\rm d}x$ using substitution $x = \frac{\pi}{2} - y$?
I reached
$$
\int_0^{\pi/2}\frac{\sin(x)}{\cos(x) + \sin(x)}{\rm d}x
 = \int_0^{\pi/2}\frac{\cos(y)}{\cos(y) + \sin(y)}{\rm d}y
$$ by using the double angle formula and the substitution but not sure how the question wants me to "exploit the symmetry at the midpoint." Thanks in advance !

答:
Let$$I:=\int_0^{\pi/2}\frac{\sin x{\rm d}x}{\sin(x+\pi/4)}=\int_0^{\pi/2}\frac{\sqrt{2}\sin x{\rm d}x}{\sin x+\cos x}=\int_0^{\pi/2}\frac{\sqrt{2}\cos y{\rm d}y}{\sin y+\cos y}$$
so, averaging the last two expressions,$$I=\frac{1}{\sqrt{2}}\int_0^{\pi/2}{\rm d}x=\frac{\pi}{\sqrt{8}}.$$


答:
Another way is to realize that first to three is the same as best of five, ie. over five sets we must not lose more than two sets, so

$\dbinom50q^0p^5 + \dbinom51q^1p^4 + \dbinom53q^2p^3$

Title: Tennis match probability - is my logic incorrect?

Body:
A tennis match consists of sets, until one player wins three sets. Player $A$ has a two-thirds chance of winning. Player $B$ has a one-third chance of winning. If they play a match, what is the probability of $A$ winning?

My attempt:

The minimum games for $A$ victory is $3$, and the maximum is $5$ (both win $2$, then next round is decider)

I have probability = win in three + win in four + win in five

=$\left(\frac{2}{3}\right)^{3}+ {4\choose 1}\ \left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^{3}+{5\choose2}\left(\frac{1}{3}\right)^{2}\left(\frac{2}{3}\right)^{3}$

Which is more than $1$!

I don't understand why this is wrong though


问:
If $A$ is a deformation retract of $V$ then $H_n(X,A) \simeq H_n(X,V)$
I'd like to prove that if $A$ is a retraction of deformation of $V$, where $A \subset V \subset X$, then $H_n(X,V) \simeq H_n(X,A)$ using the five lemma. I thought I could reach the diagram from the exact sequence of pair of $(X,V)$ and $(X,A)$ in the following way. Consider

$$ \cdots \longrightarrow H_n(V) \longrightarrow H_n(X) \longrightarrow H_n(X,V) \longrightarrow H_{n-1}(V) \longrightarrow H_{n-1}(X) \longrightarrow \cdots$$

$$ \cdots \longrightarrow H_n(A) \longrightarrow H_n(X) \longrightarrow H_n(X,A) \longrightarrow H_{n-1}(A) \longrightarrow H_{n-1}(X) \longrightarrow \cdots$$

As far as concerns the vertical arrow I could use the isomorphism $i_* : H_n(A) \longrightarrow H_n(V)$ induced by the homotopic equivalence $i$. I'd like to conclude that $H_n(X,A) \simeq H_n(X,V)$ completing the diagram with the map induced by the identity of $X$, $id_* : H_n(X) \longrightarrow H_n(X)$ but I don't wheter the diagram commutes to correctly appy the five lemma, to exploit the naturality of the connection homomorphism $\partial : H_n(X,A) \longrightarrow H_{n-1}(A)$, any help would be appreciated.

答:
Consider the map of pairs $\phi : (X,A) \to (X,V), \phi(x) = x$. Its "absolute" part is the identity $id_X : X \to X$ and its "relative" part the inclusion $\phi_A : A \to V$ which is a homotopy equivalence in this case. Hence we get induced maps $(id_X)_* = id : H_n(X) \to H_n(X)$, $(\phi_A)_* : H_n(A) \to H_n(V)$ and $\phi_* : H_n(X,A) \to H_n(X,V)$, where $(\phi_A)_*$ is an isomorphism. These maps connect the two long exact sequences in your question. Each of the resulting squares commutes because the $H_n$ are functors and $\partial$ is natural for maps of pairs.

Now the five lemma applies.


问:
Units in $\mathbb{Q}[x]/(x^2-1)$
I know for integral domain $R$, units of $R[x]$ are units of $R$.

How about if we introduce a quotient ring? i.e., suppose I have an ideal $(x^2-1) \in \mathbb{Q}[x]$ and quotients by this ideal. Note since $x^2-1 = (x-1) (x+1) $ and $-1,1\in \mathbb{Q}$ $x^2-1$ is reducible so $(x^2-1)$ is not a maximal ideal.

I wonder what is units of $\mathbb{Q}[x]/(x^2-1)$.

Let $\bar{a} \in \mathbb{Q}[x]/(x^2-1)$ be units. then $\exists \bar b$ such that $\overline{ab}=1$

\begin{align}
&\bar{a} = a + (x^2-1), \quad \bar{b} = b+ (x^2-1) \\
&\Rightarrow \quad \overline{ab} = ab + (x^2-1) = (x^2-1)
\end{align}
then $ab\in(x^2-1)$, so $a$ can be either $c, c(x+1), c(x-1)$ where $c \in \mathbb{Q}$

Is this approach correct? How one can compute the units of $\mathbb{Q}[x]/(x^2-1)$ in a formal way?

答:
The "chinese remainder theorem" proves the following: If $I:=(x-1), J:=(x+1)$  it follows  $IJ=(x-1)(x+1)=(x^2-1)$ hence there is an isomorphism

$$\mathbb{Q}[x]/(x^2-1)\cong \mathbb{Q}[x]/I \oplus \mathbb{Q}[x]/J\cong R:=\mathbb{Q}\oplus \mathbb{Q}.$$

The units in $R$ are pairs $(a,b)$ with $a,b\in \mathbb{Q}^*$.

Question: "How one can compute the units of $\mathbb Q[x]/(x^2−1)$ in a formal way?"

Answer: There is a canonical map

$$u: \mathbb{Q}[x]/(x^2-1) \rightarrow R $$

defined by

$$u(f(x)):=(f(1),f(-1))$$

hence $f\in \mathbb{Q}[x]/(x^2-1)$ is a unit iff $f(1),f(-1)\in \mathbb{Q}^*$.

Example: In general if a polynomial $p(x)=p_1(x)\cdots p_l(x)\in B:=\mathbb{Q}[x]$ is a product of distinct irreducible polynomials $p_i(x)$, let $I_i:=(p_i(x))$ and $I:=I_1\cdots I_l$. It follows there is an isomorphism of rings

$$u: B/I\cong B/I_1\oplus \cdots \oplus B/I_l$$

with $u(f):=(\overline{f},.., \overline{f})\in B/I$. Hence

$\overline{f}\in B/I$ is a unit iff $\overline{f}\in B/I_i$ is a unit for all i. If $p_i(x):=(x-a_i)$ for all $i$ it follows $\overline{f}\in B/I$ is a unit iff $f(a_i)$ is a unit for all $i$.

Title: Units in $\mathbb{Q}[x]/(x^2-1)$


问:
Distinction between error estimator and error indicator
When solving differential equations numerically one can incur discretization error and one can construct a posteriori error estimates to approximate the true error. There is a distinction often made between "error estimators" and "error indicators". What is the difference between the two?</p>

答:
I will give an explanation applicable to adaptive mesh refinement (AMR).

An "error indicator" method uses the existing solution for a given mesh to estimate the error incurred by using that mesh. For example, to determine whether a particular region in the mesh needs to be refined, you can use the solutions at adjacent nodes to calculate the derivative (i.e., jump) at the region interfaces. If the jump exceeds some given criteria, then the mesh at those interfaces should be refined.

An "error estimator" method re-generates the solution using a finer mesh then estimates the error of the original mesh given the new refined solution. This method is more accurate than the "error indicator" method but it obviously involves more effort (computational work) to do.


问:
Homomorphism image of maximal torus
I’m asked to show given a surjective homomorphism between compact connected Lie groups: $G$ and $H$, then maximal torus in $G$ are mapped to maximal torus in $H$. Well, I’m not sure how can I  use the subjectivity to show that the torus we get by homomorphism is actually maximal. I doubt the maximal torus theorem comes into play, but I don’t see how, any suggestions?

答:
Hints:

Let $S$ be a compact connected Lie group, $K$ being a normal closed subgroup of $S$. If the factor group $S/K$ is an Abelian group, then there exists a connected closed Abelian subgroup $R$ of $S$ such that $S=KR$ and $|R\cap K|<\infty$.

Let $G$ and $H$ be compact connected groups Lie,
$T$ being a maximal torus of $G$,
$f:G\rightarrow H$ being a surjective homomorphism.
Since $f(T)$ is a connected Abelian subgroups of $H$,
there exists a maximal torus $T'$ of $H$ such that $f(T)<T'$.
Let $S=f^{-1}(T')$ and $K=Ker(f)$. Then $S/K\cong T'$ is an Abelian group.

Title: Homomorphism image of maximal torus


问:
Find the dimension in the subset
Find  the dimension of the subset of all polynomials {q(t)} of $P^3$ such that q(0) = 0 .

Thanks in advance

in my work, I think the equation of $P^3$ is like the $P_0$+$P_1q$ + $P_2q^2$+$P_3q^3$
as q(0) = 0, but I am not sure how is the matrix  in these polynomials
as the maximum dimension of $P^3$ is 4 I think.

答:
A polynomial of degree (at most) $3$ [which is what the notation $P^3$ usually refers to] has form $$q(t)=a+bt+ct^2+dt^3.$$ These form a vector space of dimension $4$ over the reals (or some other field such as complexes-- you didn't specify).

I think your attempt is too complicated. The subset you mention is given by $q(0)=0,$ i.e. by $a=0.$ So you're left with arbitrary $b,c,d$ and so dimension is $3.$

Title: Find the dimension in the subset


问:
Problem: Sum of absolute values of polynomial roots
Can you please give me some hints as to how I might approach this problem? Thanks!

Given the polynomial $f = 2X^3 - aX^2 - aX + 2, a \mathbb \in R$ and roots $x_1, x_2$ and $x_3,$ find $a$ such that $|x_1| + |x_2| + |x_3| = 3.$

Edit: We know $-1$ is one of the roots of that polynomial, regardless of the value of $a$. So, in essence, what we have to demonstrate is that $|x_2| + |x_3| = 2.$

答:
HINT

$|x_1| + |x_2| + |x_3| \geq 3 \sqrt[3]{|x_1||x_2||x_3|} = 3$

Hence, we have $|x_1| + |x_2| + |x_3| \geq 3$. Equality holds implies $|x_1| = |x_2| = |x_3| = 1$

We have $x_1 + x_2 + x_3$, $x_1x_2 + x_2x_3 + x_3x_2$ and $x_1 x_2 x_3$ to be real,  and further $x=-1$ satisfies the equation.

Hence $f(x) = 2 \left( x+1 \right) \left( x^2-\left(1 + \frac{a}{2} \right)x + 1 \right)$. We have $x_3 = -1$. And $|x_1| = 1$ and $|x_2| = 1$. Hence, $x_1 = e^{i \theta}$ and $x_2 = e^{i \phi}$.

$(x-x_1)(x-x_2) = \left( x^2-\left(1 + \frac{a}{2} \right)x + 1 \right)$

$x_1 x_2 = 1 \implies \phi = -\theta$

$x_1 + x_2 = 2 \cos(\theta) = 1 + \frac{a}{2}$

Hence, $\frac{a}{2} = 2 \cos(\theta) - 1 \implies a = 4 \cos(\theta) - 2$.

Hence, $a = 4 \cos(\theta) - 2$ and the roots are $-1,e^{i \theta},e^{-i \theta}$


问:
trigamma fuction
Explain me please why the equality $\ \displaystyle\psi_1(z)=\sum_{n=0}^\infty \frac 1{(z+n)^2}\ $ is right if we
define trigamma function like this : $\ \displaystyle\psi_1(z)=\frac{d^2}{dz^2}\ln\Gamma(z)$.</p>

答:
There is an easy way to come at the first equation if you note that :
$$\Gamma(z+1)=z\;\Gamma(z)$$
so that
$$\ln\Gamma(z+1)=\ln(z)+\ln\Gamma(z)$$
Differentiation gives
$$\psi(z+1)=\frac 1z+\psi(z)$$
Differentiating again :
$$\psi_1(z+1)=-\frac 1{z^2}+\psi_1(z)$$
so that
$$\psi_1(z)=\frac 1{z^2}+\psi_1(z+1)$$
$$\psi_1(z)=\frac 1{z^2}+\frac 1{(z+1)^2}+\psi_1(z+2)$$
$$\psi_1(z)=\frac 1{z^2}+\frac 1{(z+1)^2}+\frac 1{(z+2)^2}+\cdots$$

I hope that this will be a help too for your intuition !

Title: trigamma fuction


问:
Sequence of dependent variables
I have a question that seems easy but I couldn't solve it.

Let $X$, $Y$ and $N$ be random variables such that $X$ and $Y$ are dependent and $Y$ and $N$ are dependent.
I wonder if $X$ and $N$ has to be dependent or is it possible for them to be independent.
I think that $X$ and $N$ can be independent but couldn't find an example for this.

Would appreciate help. Thank You!

答:
If $X$ and $N$ are i.i.d. $N(0,1)$ and $Y=X+N$  then $X$ and $N$  are independent but $X, X+N$ and $Y, X+N$ are both dependent.

[Note that $EXY=EX(X+N)=EX^{2}+EX EN=1$ and $EX E(X+N)=0$. So $X$ and $Y$ are not independent].

Title: Sequence of dependent variables


问:
Can a closed region of the complex plane be star shaped?
I'm working with the same definition of a star domain as the one referred to in this post: Determine if the unit disc in the complex plan is star-shaped or not.  I read elsewhere in my notes, though, that a domain is an open subset of the complex numbers.  Does that mean that closed discs cannot be star domains, even though they are simply connected sets?  I'm new to complex analysis, and am a little confused by this.  Any insight would be appreciated.</p>

答: It is usual in Complex Analysis to deal only with functions whose domain is an open subset of $\Bbb C$ (that is, we only work with functions whose domain is an open set). So, if we define a domain as an open subset of $\Bbb C$, then yes, no closed disk is a domain, although it is star-shaped.


答: The use of "domain" in complex analysis is not the same as that in topology. The two concepts are independent, and a closed region can be a star domain.</p>

Title: Can a closed region of the complex plane be star shaped?

Body: I'm working with the same definition of a star domain as the one referred to in this post: Determine if the unit disc in the complex plan is star-shaped or not.  I read elsewhere in my notes, though, that a domain is an open subset of the complex numbers.  Does that mean that closed discs cannot be star domains, even though they are simply connected sets?  I'm new to complex analysis, and am a little confused by this.  Any insight would be appreciated.</p>


问:
True or False: Set theory
True or False:

$\forall y(\exists! x)(xy = y)$: False. Because in case $y = 0,$ there are indefinite number of $x$ that $xy = y.$

$\sim (\exists x)(\forall y)(x \le y)$: True: meaning: for all $x,$ there exists an y such that $x > y$ (I deleted the negation to make a new equivalent statement).

$\{\emptyset\} - \emptyset = \emptyset$: True ? I know that
$\emptyset \neq \{\emptyset\}$

$\emptyset \subseteq \{1, \emptyset\}$: True

Would you mind checking if my answers are correct? Thank you!

答:
Right, but you should say that there are numbers (infinitely many of them, in fact) $x$ such that $xy=y$.

Right, but for the wrong reasons: denying $x\leqslant y$ is the same thing as asserting that $x\color{red}>y$.

It's false: for any set $A$, $A\setminus\emptyset=A$. In particular $\{\emptyset\}\setminus\emptyset=\{\emptyset\}\ne\emptyset$. (Note: I use the notation $X\setminus Y$ instead of $X-Y$.)

It's true: the empty set is a subset of any set.

Title: True or False: Set theory


问:
Find the values of $k$ and $p$
When $kx^{3}+ px^{2}- x+ 3$ is divided by $x- 1,$ the remainder is $4.$

When $kx^{3}+ px^{2}- x+ 3$ is divided by $x- 2,$ the remainder is $21.$

Find the values of $k$ and $p.$

I've been stuck trying to solve this, so any help would be appreciated.

答:
The Remainder Theorem states that

For a polynomial $p(x)$, the remainder when it is divided by $x-a$ for some constant $a$, is equal to $f(a)$.

Can you take it from here?

Title: Find the values of $k$ and $p$


问:
Trouble understanding the definition of a parabola: Spivak Calculus Chapter 4-21b
Spivak's problem reads as follows:

Given the horizontal line $L$ defined by $g(x)=\gamma$ and a point $P=(\alpha, \beta)$ not on $L$ so that $\gamma \neq \beta$, show that the set of all points $(x,y)$ equidistant from $P$ and $L$ is the graph of a function of the form $f(x)=ax^2+bx+c$.

I interpreted this question as:

$W:=\{\langle x,y \rangle \in \mathbb R \times \mathbb R\  |\  \exists \langle x^*, \gamma \rangle \in L \Big [ \sqrt{(x-\alpha)^2+(y-\beta)^2}=\sqrt{(x-x^*)^2+(y-\gamma)^2}\Big ]\}$

After consulting Spivak's solution manual, it is clear Spivak wants the reader to only observe the points on the line $L$ that have the following relationship to a given ordered pair in $W$: $x=x^*$.

i.e. Spivak's set looks like this:

$W':=\{\langle x,y \rangle \in \mathbb R \times \mathbb R\  |\   \sqrt{(x-\alpha)^2+(y-\beta)^2}=\sqrt{(y-\gamma)^2}\}$

I've seen various definitions online which echo Spivak's above definition...and the point of confusion always seems to be the statement: "Equidistant From a Line (directrix)". Consider my quick sketches below:

Hopefully these sketches are sufficiently elucidating.

So what exactly does it formally mean to be "equidistant from a line?

答:
Distance from a line is defined as perpedicular distance to the line.

$$\sqrt{(x-\alpha)^2+(y-\beta)^2}=|y-\gamma| = \sqrt{(y-\gamma)^2}$$


问:
Converting SVG to drawing commands
Is there anyway I could convert SVG Cubic Bezier path to simple shapes?
I need this so that I can write a computer utility that converts a given SVG path to drawing commands.  For example:
Input is: M177.627,456.011 C 178.641,455.531 179.533,455.063 180.317,454.639
Output will be: drawCircle(177.627, 456.011, 45);
(signature here is: drawCircle(Xcentre, Ycentre, r);)

(Note that the above SVG path does not give a circle this is purely for illustration what I am after).

Simple shapes here are -

lines

circles

arcs

triangles

polygons

rectangles

Many Thanks,

答:
You can approximate the Bézier curve by a string of circular arcs. One easy approximation technique is using biarc curves (pairs of smoothly connected circular arcs).

If you googled the term “biarc approximation” or search this site, you’ll find lots of good material.

C++ code here.

Some graphics subsystems support Bézier curves directly, so you don’t need to do any conversion in order to draw them. Postscript and Windows Metafiles, for example.

Title: Converting SVG to drawing commands


问:
Functions of the form $f(x^2)=f^2(x)$
What are all the functions from $\mathbb{R}$ into $\mathbb{R}$ of the form $f(x^2)=f^2(x)$?

I have deduced the following properties :

Set $\{x^n: n \in \mathbb{N}\}$ belongs to the collection.

$f\equiv0$ and $f\equiv 1$ also belongs to the collection.

Such $f$ should satisfy $\{f(0),f(1)\} \subset \{0,1\}.$

What more can be said ?

答:
Then your $f(x)$ is anything that satisfies:

$f(0)=0$ or $1$

$f(1)=0$ or $1$ and $f(-1)=c(-1) f(1)$

for $x>1$, $f(x) = a(y-\lfloor y \rfloor)^{2^{\lfloor y \rfloor}}$ and $f(-x)=c(-x)f(x)$, where $y =  \log_2(\log_2(x))$

for $0 < x < 1$, $f(x) = b(z-\lfloor z \rfloor)^{2^{\lfloor z \rfloor}}$ and $f(-x)=c(-x) f(x)$, where $z =  \log_2(\log_2(1/x))$

for any three arbitrary functions $a:[0,1) \to \mathbb R_{\ge 0}$ and $b:[0,1) \to \mathbb R_{\ge 0}$ and $c:\mathbb R_{\lt 0} \to \{-1,+1\}$.

So you have choices for $f(0)$ and $f(1)$, plus the three arbitrary functions $a$ and $b$ and $c$.

Title: Functions of the form $f(x^2)=f^2(x)$


答:
Start from defining basic functions with this property.

$$f(x)=1$$
$$f(x)=x^n$$

Now if we represent $f(x)=x^{g(x)}$, we have $x^{2g(x)}=x^{2g(x^2)}$ from where $g(x)=g(x^2)$ which gives $g(x)=const.$ if continuous.

Notice that we cannot have a polynomial (not even as a Taylor series) with more than one term. Suppose it is, we would have that all coefficient if they exist that are not different from $0$ are $1$. There is no special reason to go any further from $ax+b$.

$$(ax+b)^2=ax^2+2abx+b^2=ax^2+b$$

From here either $b=1$ and then $a=0$ or $b=0$ and $a=1$. Higher exponents lead to a very similar conclusion.

Since it cannot have any solution as Taylor polynomial, we are very restricted over the form this function can have. As a continuous function the first family $x^n$ we have described is the only valid.

Notice, however, that $g(x)=g(x^2)$ hides a lot of local solutions, where you can simply agree to have any function from any two values $a$ and $a^2$ and then extend this in both directions. You would need to treat $a>1$ and $a<1$ separately.

Title: Functions of the form $f(x^2)=f^2(x)$

Body:
What are all the functions from $\mathbb{R}$ into $\mathbb{R}$ of the form $f(x^2)=f^2(x)$?

I have deduced the following properties :

Set $\{x^n: n \in \mathbb{N}\}$ belongs to the collection.

$f\equiv0$ and $f\equiv 1$ also belongs to the collection.

Such $f$ should satisfy $\{f(0),f(1)\} \subset \{0,1\}.$

What more can be said ?


问:
Uniform boundness question

I found the following statement in the internet and I have a conceptual question, please: Let $f_n$ be a sequence of functions such that $f_n\to f$ uniformly as $n\to\infty$ and each $f_n$ bounded. Show that $f_n$ is uniformly bounded, i.e. $\exists M>0 \ \forall x\in\mathbf{R} \ \forall n\in\mathbf{N}: |f_n(x)|\le M$.

So, the thing that I don't understand, is why the hypithesis that $f_n\to f$ uniformly is a necessary condition to get the uniform boundness? Why the pointwise convergence is not enough? I was thinking on some examples but I didn't find anything. If someone could provide some, I would really appreciate it. The key thing in this statement, I think, that with 2 conditions ($f_n\to f$ uniformly and $f_n$ bounded) we are sure that $f_n$ limit doesn't "explode", is it correct?

答: If $f_n(x)=n$ for $x>n$ and $0$ for $x \leq n$ then $f_n$ is bounded by $n$ for each $n$, $f_n \to 0$ point-wise but $f_n$ is not uniformly bounded.

Title: Uniform boundness question


问:
Radius of Convergence for series in $z^{3n}$ using Cauchy-Hadamard formula
If
$∑
a_nz
^n$ has radius of convergence R, use the formula $R = \left( \limsup_{n \to \infty} | a_n |
 ^{\frac{1}{n}} \right)^{-1} $
to
find the radius of convergence of

$∑
a_nz^{
3n}$

I tried to solve this question using the ratio test, I got an answer of $R^{\frac{1}{3}}$, however, I am not sure how to solve it using Cauchy-Hadamard formula.

Any hint will be appreciated.

答: For each $n\in\Bbb Z_+$, let$$b_n=\begin{cases}a_{n/3}&\text{ if }3\mid n\\0&\text{ otherwise.}\end{cases}$$Then $\sum_{n=0}^\infty a_nz^{3n}=\sum_{n=0}^\infty b_nz^n$ and, for each $n\in\Bbb N$,$$\sqrt[n]{|b_n|}=\begin{cases}\sqrt[n]{|a_{n/3}|}=\sqrt[3]{\sqrt[n/3]{|a_{n/3}|}}&\text{ if }3\mid n\\0&\text{ otherwise}\end{cases}$$and therefore$$\limsup\nolimits_n\sqrt[n]{|b_n|}=\sqrt[3]{\limsup\nolimits_n\sqrt[n]{|a_n|}}.$$


问:
Problem to find the matrix of linear transformation
Let $B = \left(\vec{e}_1, \vec{e}_2, \vec{e}_3\right)$ a direct orthonormal basis of $V^3$

and let $\vec{a}$ and $\vec{b}$ be two nonzero vectors of $V^3$ such that

$\vec{a}=a_{1}\vec{e}_1+a_{2}\vec{e}_2+a_{3}\vec{e}_2$

$\vec{b}=b_{1}\vec{e}_1+b_{2}\vec{e}_2+b_{3}\vec{e}_2$

Give the matrix associated with the following linear transformations

$T(\vec{u})=(\vec{a}\cdot\vec{u})\vec{b}$

$T(\vec{u})=\vec{a}\times\vec{u}$

Can anyone please give me an idea how to find these matrix?

I know the answers are

$$
        \begin{bmatrix}
        a_{1}b_{1} & a_{2}b_{1} & a_{3}b_{1} \\
        a_{1}b_{2} & a_{2}b_{2} & a_{3}b_{2} \\
        a_{1}b_{3} & a_{2}b_{3} & a_{3}b_{3} \\
        \end{bmatrix}
$$

$$
        \begin{bmatrix}
        0 & -a_{3} & a_{2} \\
        a_{3} & 0 & -a_{1} \\
        -a_{2} & a_{1} & 0 \\
        \end{bmatrix}
$$

but I have no idea what I need to do to get them

Thanks in advance

答: My suggestion would be to evaluate the transformation in $e_1$, $e_2$ and $e_3$. Also you must note that the matrix associated to a transformation in a given basis $\{e_1,e_2,e_3\}$ is $[T(e_1)T(e_2)T(e_3)]$ where each $T(e)$ is a column vector.</p>


问:
Equation of a graph VS formula of a graph
Simple question:

Why do refer to the equation of a straight line instead of the formula of a straight line?

For instance, formulae provide relationships between multiple variables, which is what something like $y=mx+c$ does. Equations, on the other hand, assert equality (which admittedly $y=mx+c$ does) but we tend to want them solved.

答:
One good reason to prefer the equation of a geometric object is if you go beyond
one linear equation in one variable. For example, an equation of a circle is
$\,(x-a)^2+(y-b)^2=r^2.\,$ Here, both variables $\,x\,$ and $\,y\,$ are treated
simultaneously. If written as a formula $\,y = \text{[something]}\,$ then there
are two values of $\,y\,$ for values of $\,x.\,$ A similar example is the equation
of a degenerate conic being two lines. The equation is something like
$\, 0 = (y-ax-b)(y-cx-d).\,$ Again, writing $\,y\,$ as a formula needs two values.


问:
Problem with system of equations $(x,y,z)$
System of equations:
$$\begin{cases}\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}=\frac{3}{xyz}\\xy+yz+zx=3\end{cases}$$
Solution:

$$\begin{cases}\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}=\frac{3}{xyz}\\xy+yz+zx=3\end{cases}\Rightarrow\begin{cases}x^2+y^2+z^2=3\\xy+yz+zx=3\end{cases}$$

$x^{2}+y^{2}+z^{2}=xy+yz+zx$

$(x+y+z)^{2}-2(xy+yz+zx)=xy+yz+zx$

$(x+y+z)^{2}-2*3=3$

$(x+y+z)^{2}=9$

What should I do next?

Body:
Hint:

$$(x-y)^2+(y-z)^2+(z-x)^2=?$$

Alternatively, $$x^2-x(y+z)+y^2+z^2-yz=0$$

As $x$ is real, the discriminant must be $\ge0$

But $$(y+z)^2-4(y^2+z^2-yz)=-3(y-z)^2\le0$$


问:
Using combinations to combinations for probability with dependent events
I am reading a problem about probabilities with dependent events.

There are $3$ people selected from $5$ freshmen, $2$ sophomores and
$4$ juniors. What is the probability that the first $2$ people
selected are freshmen and the third a junior?

I thought that this is solvable by using combinations.

I.e.

There are $\binom{11}{3}$ ways to select $3$ out of $11$ students (I hope that the notation is for combinations and not permutations as I get these 2 notations mixed).

This is $$\frac{11!}{3!\cdot 8!} = 165$$ ways

To choose $2$ freshman from $5$ is: $$\frac{5!}{2!\cdot 3!}$$
To choose $1$ junior from $4$ is: $$\frac{4!}{1!\cdot 3!}$$

So the probability we are looking for is:

$$\frac{\frac{5!}{2!\cdot 3!} \cdot \frac{4!}{1!\cdot 3!}}{165} = \frac{40}{165}=\frac{8}{33}$$

but the answer is wrong.

Is it not possible to use the combinations formula for calculating such a probability with $3$ dependent events?

答:
Your answer would be correct if the question were:

What is the probability that two of the people selected are freshman and one is a junior?

That would be a question about combinations - how many ways to choose 2 freshman out of 5 and one junior out of 4, over the number of ways to choose a group of 3 students from 11.

That is subtly different from the question which was asked. The question in your post says the first two people are freshman, and the third is a junior - i.e., the order of selection matters. So you should instead use permutations.


问:
Solutions of the functional equation $f(x+1)= xf(x)$
I have come across a question of functional equation
$$f(x+1)= xf(x).$$
My question is how can we solve these type of functions. I have tried to use substitutions like $x= x-1$ and some manipulations but could't find the function. I also have tried to use differentiation's basic definition. I am still unable to solve the functional equation. Please try to clear my query.

答:
The first idea is always to plug in special values for $x$ - but what is "special" may depend on the functional equation. Here, $x=0$ suggests itself as it leads to $f(1)=0$. After that, one readily finds by induction that $f(x)=0$ for all $x\in\Bbb N$. Incidentally, the recursive nature of the functional equation allows us to describe the quite large set of solutions:

Let $g\colon[0,1)\to\Bbb R$ be an arbitrary function and define $f\colon \Bbb R\to \Bbb R$ recursively as
$$f(x)=\begin{cases}g(x)&0\le x<1\\
(x-1)(x-2)\cdots(x-\lfloor x\rfloor)\cdot g(x-\lfloor x\rfloor)&x\ge1\\
\frac{g(x+\lceil -x\rceil)}{x(x+1)\cdots (x+\lceil -x\rceil)}&x<0\end{cases} $$
Then $f$ solves the functional equation, and evey solution of the functional equation is of this form.


答:
The $d$th forward difference operator applied to $\,x^d\,$ is defined by
$$ \Delta^dx^d :=\sum_{m=0}^{d}{(-1)}^{m}{d\choose m}(x+m)^{d}. \tag{1} $$
This is known to be equal to the constant function $\,d!\,$ This implies
that your summation expression is $\,d!-d!=0.$ The Wikipedia article
Finite difference has
some details but does not explicitly state the fact about $\,d!.$

Title: How to show $\sum_{m=0}^{d}(-1)^{m}{d\choose m}((x+m)^{d}-(x+1+m)^{d})=0$

Body: How to show $\sum_{m=0}^{d}{(-1)}^{m}{d\choose m}((x+m)^{d}-(x+1+m)^{d})=0$


问:
Elegant way to solve $n\log_2(n) \le 10^6$
I'm studying Tomas Cormen Algorithms book and solve tasks listed after each chapter.

I'm curious about task 1-1. that is right after Chapter #1.

The question is: what is the best way to solve: $n\lg(n) \le 10^6$, $n \in \mathbb Z$, $\lg(n) = \log_2(n)$; ?

The simplest but longest one is substitution. Are there some elegant ways to solve this? Thank you!

Some explanations: $n$ - is what I should calculate - total quantity of input elements, $10^6$ - time in microseconds - total algorithm running time. I should figure out nmax.

答:
If something cheesier than Newton's method is wanted, I recommend the fixed point iteration:

```<code>x := (10^6)/log_2(x)
</code>```

This is easily done with a calculator applet by storing constant (10^6) * log_10(2) and thereafter taking $\log_{10}(x)$, reciprocating, and then multiplying by the recalled constant.

Starting with obvious overestimate $x_0 = 10^6$, the fixed-point iterates alternate between under- and over-shooting:

```<code>x1 = 50,171.67
x2 = 64,042.69
x3 = 62,630.17
x4 = 62,756.63
x5 = 62,745.18
x6 = 62,746.21
x7 = 62,746.12
</code>```

Thus the largest possible integer is $n = 62746$, as Michael already stated.

Body:
If something cheesier than Newton's method is wanted, I recommend the fixed point iteration:

x := (10^6)/log_2(x)

This is easily done with a calculator applet by storing constant (10^6) * log_10(2) and thereafter taking $\log_{10}(x)$, reciprocating, and then multiplying by the recalled constant.

Starting with obvious overestimate $x_0 = 10^6$, the fixed-point iterates alternate between under- and over-shooting:

x1 = 50,171.67
x2 = 64,042.69
x3 = 62,630.17
x4 = 62,756.63
x5 = 62,745.18
x6 = 62,746.21
x7 = 62,746.12

Thus the largest possible integer is $n = 62746$, as Michael already stated.


问:
Probability of car starting first time

The Question

One a given day, Claude's car has an 80% chance to starting first time
and Andre's car has a 70% chance of the same, Given that one of the
cars has started first time, what is the chance that Andre's car
started the first time?

My Understanding

Event A: Claude's car starting first time

Event B: Andre's car starting first time

We want at least one car started first time, also event B occurs, so there are only these $2$ cases: AB and A'B. I calculated the probability of these $2$ separately and multiply them together, which is $0.8\times 0.7\ {+}\ 0.2 \times 0.7=0.7$, which makes sense since event A and B are independent events, the occurrence of A should not affect the probability of B. I have to try something else because this is not the answer.

Then I tried to calculate the probability of B', which is when Andre's car is not starting first time.
there's only one case for this to be happening while at least one car is starting first time. A B'.
the probability for this is $0.8 \times 0.3 =0.24$. then I subtracted this from $1$ to get $76\%$.
the answer for this is $\frac{35}{47}$

It's close but not really. Can someone help me with this? I can't think of more ways of solving it.

答:
I think simplest way is using a tree diagram

enter image description here

P(At least one of them starts)=0.8X0.7+0.2X0.7+0.3X0.8=0.94
P(Andre starts first)=0.8X0.7+0.2X0.7=0.7
P(Andre given started)=0.7/0.94=35/47
Hope this helps

Title: Probability of car starting first time


问:
Fixed point theorem with $\frac12 tx(t^2)=x(t)+2t^2$

Let $V=C[0,1]$ with the norm $\| \|_\infty$ and let $M=${$x\in V |\int_0^1|x(t)|dt\le 1$}.

a. Show that there exists a unique $x_\infty \in M$ that satisfies the equation $\frac12 tx(t^2)=x(t)+2t^2$.

b. If $x_0(t)=-1$ calculate successive approximations of $x_1$ and $x_2$.

c. Calculate $\| x_2-x_1\| _\infty$.

d. Using the previous results, give the upper bound of the value $x_\infty(\frac12)$.

I'm very confused about the fixed point theorem overall. Judging from the premise, I suppose that I need to show that the application is contractant. But how do I set the function in the first place? Should I isolate x(t)? And for the approximations we saw that $\|x_\infty -x_1\| \le \frac{c^n}{1-c} \|x_1 -x_0 \|$ and $\|x_\infty -x_n\| \le \frac{c}{1-c} \|x_n -x_{n-1} \|$. I replaced $x_0$ in the first inequality but that doesn't really give us anything, does it. For part c, same problem. I would use the second inequality, but how do you get $x_n$? And for the last part, I would use the second inequality as well, but would you then plug $\frac12$ in the equation at the start?

Any explanations of the fixed point theorem are welcome, I am very confused about how to apply it and what it means in general.

答:
Hint: The Banach Fixed Point Theorem says that if $(V, \| \cdot \|)$ is a Banach space (i.e a complete normed space) and $F :V \to V$ satisfies $\| F(x) - F(y) \| \leqslant C \| x - y\| $ with $0<C<1$, that is $F$ is a contraction, then $F$ has a unique fixed point.

For your question, $V = C[0,1]$ which is a Banach space with the norm $\| \cdot \|_\infty$. Your job is to find a specific $F : C[0,1] \to C[0,1]$ that is

a contraction;  and

if $F(x)(t)=x(t)$ then \begin{equation} \tag{1}\frac 1 2 t x(t^2) = x(t) + 2t^2. \label{eq:1}\end{equation}

In general, this could be very difficult, but in this case there is an obvious guess. Rearranging \eqref{eq:1} gives $$ x(t) = \frac 1 2 t x(t^2) - 2t^2,$$ so the obvious guess would be $F(x)(t) := \frac 1 2 t x(t^2) - 2t^2$. Now you need to show that:

If $x \in C[0,1]$ then $F(x) \in C[0,1]$. (Hence, $F$ is well-defined).

$F$ is a contraction.

If $x \in M$ then $F(x) \in M$. This implies that the fixed point will be in $M$.

This gets you to the end of (a). Are you able to do the rest?