Title: Expected value of pulling blue marble from sack on nth pull

Body:
I am doing a refresher for a work course and I'm a little stuck the question asks the following

You have a sack with 7 marbles inside 4 Blue and 3 Green

You draw all the marbles (without replacing them) and check their color.

If you let X represent the value of n if the first blue marble appears on the nth draw.

Obviously P(b) is 4/7 and P(g) = 3/7

My assumption was that:
4/7*(1+2+3+4+5+6+7) = (4/7)*1 + (4/7)*2 + (4/7)*3 ... (4/7)*7 = 16

But that seems entirely nonsensical value for X to take given that you can only pull 7 marbles from the sack.

I assume that is because the probability is obviously not static at 4/7.

If I look at the probability of the draws being green before being blue I end up with:
X=1 = 3/7
X=2 = 2/6
X=3 = 1/5

[(3/7)*1+ (2/6)*2 + (1/5)*3 ] = 1.67

But after this I'm a little stuck

Where am I going wrong?

Body:
$P(X=1)=\frac{4}{7}=\frac{20}{35}$

$P(X=2)=\frac{3}{7}\cdot \frac{4}{6}=\frac{10}{35}$

$P(X=3)=\frac{3}{7}\cdot \frac{2}{6} \cdot \frac{4}{5}=\frac{4}{35}   $

$P(X=4)=\frac{3}{7}\cdot \frac{2}{6} \cdot \frac{1}{5}\cdot \frac{4}{4}=\frac{1}{35}  $

Observe that $\Sigma_{i=1}^4 P(X=i)=1$ as it has to be...

Now, using the definition,

$$\mathbb{E}[X]=1\times\frac{20}{35}+2\times\frac{10}{35}+3\times\frac{4}{35}+4\times\frac{1}{35}=\frac{56}{35}=\frac{8}{5}$$


问:
How can I solve this limit with a definite integral inside?
This is the problem:

$$ \lim_{m \to 0}\left(\frac1m \int_{\sqrt\frac\pi 2}^{\sqrt\frac\pi 2+m}\sin(x^2)\,dx\right) $$

I don't have an idea how to solve it. I was thinking maybe it's an indeterminate form $ \infty \cdot 0$ and we can do something with l'hopital's rule. But I don't know what to do with the integral. What is the trick here?

答:
Another way to do it.

Let $x=t+\sqrt{\frac{\pi }{2}}$ to make
$$\int_{\sqrt\frac\pi 2}^{\sqrt\frac\pi 2+m}\sin(x^2)\,dx=\int_0^m \cos \left(\sqrt{2 \pi } t+t^2\right)\,dt$$ Now, By Taylor and binomial expansion
$$\cos \left(\sqrt{2 \pi } t+t^2\right)=1-\pi  t^2-\sqrt{2 \pi } t^3+O\left(t^4\right)$$
$$\int \cos \left(\sqrt{2 \pi } t+t^2\right)\,dt=t-\frac{\pi  t^3}{3}-\frac{1}{2} \sqrt{\frac{\pi }{2}} t^4+O\left(t^5\right)$$
$$\int_0^m \cos \left(\sqrt{2 \pi } t+t^2\right)\,dt\sim m-\frac{\pi  m^3}{3}+O\left(m^4\right)$$
$$\frac 1m\int_0^m \cos \left(\sqrt{2 \pi } t+t^2\right)\,dt\sim 1-\frac{\pi  m^2}{3}+O\left(m^3\right)$$

Just for the fun, computing the exact value of the integral for $m=0.1$ (far away from $0$), the expression would be $0.9889$ while the approximation gives $1-\frac{\pi }{300}=0.9895$.

ContentLicense: CC BY-SA 4.0


问:
Why does the Additive rule work as it does?
For non-mutually exclusive events, if there are probabilities for event A and event B, then I'm learning the additive theorem written as:

$P(A or B) = P(A)+ P(B) − P(AB)$

Why do we subtract P(AB)? The references I've read are terse. I gather it's because P(A) + P(B) accounts for P(AB) twice, but that doesn't make sense to me. Can anybody explain it?

答: We have $|A\cup B| = |A|+ |B|- |A\cap B|$ since the elements $x\in A\cap B$ are counted twice, in $|A|$ and in $|B|$, and so they need to be subtracted once in $|A\cap B|$.


问:
How to factor a four term polynomial without grouping?
$$2x^3 + 9x^2 +7x -6$$

This equation doesn't factor by grouping, and other than that I have no idea how to solve this problem. Will someone please help?

答:
Use the Rational Root Theorem to find possible candidates for roots. For example, $-2$ is a candidate, and it is indeed a zero of your polynomial (the polynomial is equal to zero when $x = -2$), hence $(x - (-2)) = (x + 2)$ is a factor.

Now use synthetic division on the polynomial, call it $p(x)$, to obtain the quotient when divided by $(x + 2)$.

You may simply be able to simply "eye up" the polynomial-quotient, this time with the resulting quadratic equation to $p(x)/(x+2).\;$ If the factorization of the quadratic is readily apparent, then you'll have three factors in all: $(x+ 2)$ and the factors of the quadratic.

Otherwise, use the coefficients of the quadratic and the Rational Root Theorem to find a second zero/root, $x = b$, and use polynomial division to divide the quadratic by the second factor $(x - b)$, to find the resulting quotient, which will be the "last" factor. If all had gone well, (and it should, in this case), you'll now havethree factors in all!

ContentLicense: CC BY-SA 3.0


问:
Solve a cubic polynomial?
I've been having trouble with this question:

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

I've tried methods such as Vieta's formula and simultaneous equations, assuming the roots are: $a$, $a$, $b$, but I am still unsuccessful.

Any help would be greatly appreciated.

答:
By Vieta's formula, $$\begin{cases}2a+b=\dfrac{24}5\\a^2+2ab=a(a+2b)=\dfrac95\\a^2b=-\dfrac{54}5 \end{cases}$$

Notice that $0$ is not a root of the original equation, so both $a,b$ are nonzero. Divide the second equation by the third equation to get $\dfrac{a+2b}{ab}=\dfrac2a
+\dfrac1b=-\dfrac1{6}$. Multiplying by the first equation, we have $$(2a+b)(\dfrac2a
+\dfrac1b)=5+2(\dfrac ba+\dfrac ab)=-\dfrac45 $$

Let $\dfrac ba=t$, then the above equation becomes $$5+2(t+\dfrac1t)=-\dfrac45\implies 10t^2+29t+10=(5t+2)(2t+5)=0$$

Hence $t=-\dfrac25$ or $-\dfrac52$. If $t=-\dfrac25, 2a+b=2a+at=\dfrac85a=\dfrac{24}5,$ so $a=3$. On the other hand, if $t=-\dfrac52, 2a+b=-\dfrac12a=\dfrac{24}5,$ so $a=-\dfrac{48}5$. But from the third equation we must have $b\lt 0$, and then $a,b$ are both $\lt 0$, so $2a+b\lt 0$ and then the first equation cannot hold. Therefore $a=3, b=-\dfrac{6}5$ are the roots of the original cubic.

Title: Solve a cubic polynomial?


问:
How to solve : $\,8^x=6x$
I am stuck on the following problem which one of my friends gave me:

Solve : $\,8^x=6x$.

MY ATTEMPTS:

We see that $$8^x=6x \implies 2^{3x}=6x.$$ Now I am not sure how to proceed further. Taking logarithm on both sides of the equation does not help much. Clearly, $x=\frac 1 3$ satisfies the given equation but I am not sure how to get it.

Can someone help? Thanks and regards to all.

答:
$8^x=6x\implies e^{x\log 8}=6x\implies -xe^{-x\log8}\log8=-\dfrac{\log8}6$

Recall the Lambert W-function is defined as the inverse of $ze^z$ in $\Bbb C$, so if we let $t=-x\log8$, then $te^t=-\dfrac{\log8}6\implies t=W(-\dfrac{\log8}6)$

Hence, $x=-\dfrac{W(\dfrac{\log8}6)}{\log8}$.

Title: How to solve : $\,8^x=6x$


问:
prove the identity tanA +cot A =2cosec2A
This question i tried doing tanA + cosA/sinA but then I cant continue the I tried the other side which I got 1/ 4sinAcosA and this doesn't look right I am stuck and I don't have a teacher to help m so if someone could help me that would mean so much.</p>

答:
Hint:
$$\tan A+\cot A=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}=\frac{\sin^2A+\cos^2A}{\sin A\cos A}$$

You can then use the Pythagorean identity in addition to multiplying the top and bottom by $2$.


答:
A quick way to determine if a polynomial like this one has any rational roots is by the rather easily proven theorem that for a polynomial with integer coefficients, a rational root $\frac{p}{q}$, in lowest terms, must satisfy that $p$ divides the free coefficient, and $q$ must divide the leading coefficient. This gives you a very short list of candidates for rational roots, and you can just try them one-by-one. If you find a root, then proceed to divide the polynomial and try to find more roots. If none of the rational candidates you found is a root (this is not the case for the polynomial you are considering now) then some other methods can be used.

There are general criteria for irreducibility as well as the (rather nasty, but workable) formulas for the solution of third and fourth degree polynomials with complex (hence real (hence integer)) coefficients. Again, you don't need any of that for your polynomial.

Title: How to factor a four term polynomial without grouping?

Body:
$$2x^3 + 9x^2 +7x -6$$

This equation doesn't factor by grouping, and other than that I have no idea how to solve this problem. Will someone please help?


Title: prove the identity tanA +cot A =2cosec2A

Body: This question i tried doing tanA + cosA/sinA but then I cant continue the I tried the other side which I got 1/ 4sinAcosA and this doesn't look right I am stuck and I don't have a teacher to help m so if someone could help me that would mean so much.</p>


问:
What does this definite integral theorem mean?
As I was solving a homework problem about definite integrals, I came across a theorem to help me solve the problem. Although I got the right answer to the problem, I do not really understand the theorem itself. I would be grateful if you provide me with a short explanation about what the theorem really means and when I could use it. Thank you!

Here is the theorem: picture of theorem

Theorem from image stated: If $f(x)$ is an odd function and continuous on $[-a,a]$, then

$$\int_{-a}^a f(x) \, dx = 0.$$

答:
I'm unsure what level of calculus you're at, but first, we can give some intuitive reasoning. If $f$ is an odd function, then $f(-x)=-f(x)$ for all $x$ in the domain of $f$ or equivalently, $f(x) = -f(-x)$. This simply means when $x < 0$, the function $f$ evaluates to the negative of $f$ values when $x > 0$. So, then the graph of an odd function will look something like the graph below:

$\hskip 4.5cm$

So, the graph of $f$ has a "negative symmetry," which I mean that the graph of $f$ for positive $x$-values are reflected across the $y$-axis and then negated (or multiplied by $-1$). Since, the definite integral is the signed area between the graph and the $x$-axis, then you can see that the areas on both sides of the $y$-axis are the same except that the definite integral from $-a$ to $0$ should be negative, and so the values of the definite integrals on both sides of the $y$-axis cancel out.

More analytically speaking, we can use $f(x) = -f(-x)$ to write

\begin{align}
\int_{-a}^a f(x) \, dx &= \int_{-a}^0 f(x) \, dx + \int_0^a f(x) \, dx \\
&= -\int_{-a}^0 f(-x) \, dx + \int_0^a f(x) \, dx.
\end{align}

Substituting $u = -x$ (and hence $-du = dx$) in the first integral on the RHS,

\begin{align}
\int_{-a}^a f(x) \, dx &= \int_a^0 f(u) \, du + \int_0^a f(x) \, dx \\
&= -\int_0^a f(u) \, du + \int_0^a f(x) \, dx \\ &= 0.
\end{align}


问:
Under what kind of set defintion, a, b will be constrained by a+b=c?
This is a soft question, but my mind has been struggling with it these days...(I'm reading Munkres' General Topology and doing Integer Programming with Gurobi)

Suppose $a, b, c \in A $ and $a+b=c$ holds.

When A is the set of real numbers (i.e., $A=\mathbb{R}$), then $a, b$ can be any real numbers if $c$'s value is not constrained.

When A is the set of integers (i.e., $A=\mathbb{Z}$), then $a, b$ can be any integers if $c$'s value is not constrained.

...etc

However, when $A =\{0,1\}$, then $a, b$ cannot be both 1 even if $c$ is not constrained.

So my question is, under what kind definition of $A$, there will certain constraints on $a, b$ even if $c$ is free.

答: As @DonThousand notes, such a constraint exists iff $A$ is not closed under addition, i.e. does not form a magma under $+$. @Divide1918 observes that $\{0,\,1\}$ forms a magma under modulo-$2$ addition, but not under "ordinary" addition.


问:
Big O notation Problems
I have read a lot of articles on Big O notation but never found a problem like this and I have no idea how to solve it.

$$3 \cdot O(n^3) = O(3 + n^3)$$

答:
I think the satatement is pretty intutive since both the LHS and RHS are dominated by the $n^{3}$, so both LHS and RHS are $O(n^{3})$. Anyhow, I tried proving it this way...

Starting with the following definition of Big-O: $f(x)=O(g(x))$ if there exists a positive real number $K$ and a real number $x_{0}$ such that $$\bigl|f(x)\bigr|\leq K \cdot g(x)$$ $\forall x>x_{0}$.

Now let $f_{1}(n)=3\cdot O(n^{3})$ and $f_{2}(n)=O(3+n^{3})$. By definition this means that $$f_{1}(n)\leq 3\cdot K_{1}\cdot n^{3}\leq K_{1}^{'}\cdot n^{3}$$ $\forall n> n_{0}$. So $\boxed{f_{1}(n)=3\cdot O(n^{3})=O(n^{3})}$ and $$f_{2}(n)\leq K_{2}\cdot (3+n^{3})=K_{2}\cdot 3+K_{2}\cdot n^{3}\leq K \cdot n^{3}$$ $\forall n>n_{0}$. That is $\boxed{f_{2}(n)=O(3+n^{3})=O(n^{3})}$. In conclusion we have that $f_{1}(n)=O(n^{3})=f_{2}(n)$, meaning in particular that $$3\cdot O(n^{3})=O(n^{3})=O(3+n^{3}).$$

Title: Big O notation Problems


问:
Relation between a family of a certain kind of function and linear differential equation
I was thinking of functions who have differential equations in which the function is 'factorable' from first order differential equation. Eg: Functions satisfying the equation:

$$ \frac{dy}{dx} = \eta(x) y(x)$$

Example functions:

$y=e^{ax}$, $ y= \frac{1}{1-x}$.. etc

Now, I notice that the solution to the differential equation is:

$$ y= e^{ \int \eta dx}$$

But here is the interesting thing, the above expression has the same form of the function which is used when we solve linear differential equation (see here) ... now what's the intuition which explains the integratnig factor of linear ode being function who's derivative has the factor of the original function?

答:
$$ \frac{dy}{dx} = \eta(x) y(x)$$
First, change of symbol $x$ for $t$ :
$$ \frac{dy}{dt} = \eta(t) y(t)$$
$$ \frac{dy}{dt} - \eta(t) y(t)=0$$
Second, let $\quad - \eta(t)=p(t)$
$$ \frac{dy}{dt} +p(t) y(t)=0$$
This is the equation from your cited document in the particular case $g(t)=0$:
$$ \frac{dy}{dt} +p(t) y(t)=g(t)$$
So it is not surprizing that you observe the relationship of solutions.


问:
Symmetric difference of a 3 subset of a finite set X
Let $A, B, C, D$ be subsets of a finite set $X$. Prove that $AΔB = CΔD$ if
$AΔC = BΔD$.

I have tried to applied some properties of symmetric difference but all to no avail according to my little knowledge about it. Please, kindly help me out!

答:
Proof

suppose $A, B, C, D$ be subsets of a finite set $X$.

we are to prove that $AΔB = CΔD$ if and only if $AΔC = BΔD$.

before we proceed we need some properties as being guided in comment session

$AΔB = BΔA$ $\longrightarrow$ Cummutative

$AΔ(BΔC)=(AΔB)ΔC$ $\longrightarrow$ Associative

$AΔA = \varnothing $

$AΔ\varnothing = A $

from $AΔB = CΔD$ multiply through by $C$, we have

$\implies AΔBΔC = CΔDΔC$

$\implies AΔBΔC = DΔCΔC = DΔ\varnothing = D$

$\implies AΔBΔC = D$, multiply through by $B$ we have

$\implies AΔBΔCΔB = DΔB = BΔD$

$\implies AΔCΔBΔB = BΔD$

$\implies AΔC = BΔD$


问:
A simple graph must have at least one pair of vertices whose degrees are equal.
I came across this proof while studying graph theory and although it instinctively makes sense, I have a problem with a part of it.

Proposition: A non-trivial simple graph $G$ must have at least one pair of vertices whose degrees are equal.

Proof:
Let $G$ be a graph with $n$ vertices. Because $G$ is a simple graph, then by definition it doesn't contain multi-edges or loops. Therefore, there appears to be $n$ possible degree values for any vertex, namely $0, ..., n-1$. However, there cannot be both a vertex of degree 0 and a vertex of degree $n - 1$, since the presence of a vertex of degree 0 implies that each of the remaining $n - 1$ vertices is adjacent to at most $n-2$ other vertices. Hence, the $n$ vertices of $G$ can realize at most $n − 1$ possible values for their degrees [This is the part where I have troubles]. Thus, the pigeonhole principle implies that at least two of the $n$ vertices have equal degree.

Isn't this proof based on assumption that there exists a vertex with degree $0$. What happens if that is not case? Does the proof still hold? And why can the $n$ vertices of the graph realize at most $n-1$ possible values? I don't understand that step.

答: There are $n$ vertices and there are $n$ possible degrees: $0,1,\ldots, n-1$. If no two vertices have same degree, then each possible degree must be attained. In particular, there must be one vertice $v$ of degree $0$ and one vertex $w$ of degree $n-1$. If $n>1$, this is a contradiction


答:
No, there is no assumption that the graph has a vertex of degree $0$: there is simply the observation that it is impossible for a graph on $n$ vertices to have both a vertex of degree $0$ and a vertex of degree $n-1$. If $G$ has a vertex $v$ of degree $n-1$, that vertex is adjacent to all of the other $n-1$ vertices of $G$, and therefore all of those vertices have degree at least $1$, because they all have an edge to $v$. Thus, $G$ cannot have both a vertex of degree $n-1$ and a vertex of degree $0$.

This means that if $G$ has a vertex of degree $n-1$, the only other possible degrees of its vertices are $1,2,\ldots,n-2$: degree $0$ is not possible. Thus, it can have at most the $n-1$ different degrees $1,\ldots,n-1$. And if it has a vertex of degree $0$, the only other possible degrees of its vertices are $1,2,\ldots,n-2$: degree $n-1$ is not possible. In this case it can have at most the $n-1$ degrees $0,\ldots,n-2$.

And of course if $G$ has neither a vertex of degree $0$ nor a vertex of degree $n-1$, the only possible degrees of its vertices are $1,\ldots,n-2$, so that it can have at most these $n-2$ degrees.

In every case, then, $G$ can have at most $n-1$ different degrees for its $n$ vertices, and the pigeonhole principle immediately tells us that two of the vertices must have the same degree.

Title: A simple graph must have at least one pair of vertices whose degrees are equal.

Body:
I came across this proof while studying graph theory and although it instinctively makes sense, I have a problem with a part of it.

Proposition: A non-trivial simple graph $G$ must have at least one pair of vertices whose degrees are equal.

Proof:
Let $G$ be a graph with $n$ vertices. Because $G$ is a simple graph, then by definition it doesn't contain multi-edges or loops. Therefore, there appears to be $n$ possible degree values for any vertex, namely $0, ..., n-1$. However, there cannot be both a vertex of degree 0 and a vertex of degree $n - 1$, since the presence of a vertex of degree 0 implies that each of the remaining $n - 1$ vertices is adjacent to at most $n-2$ other vertices. Hence, the $n$ vertices of $G$ can realize at most $n − 1$ possible values for their degrees [This is the part where I have troubles]. Thus, the pigeonhole principle implies that at least two of the $n$ vertices have equal degree.

Isn't this proof based on assumption that there exists a vertex with degree $0$. What happens if that is not case? Does the proof still hold? And why can the $n$ vertices of the graph realize at most $n-1$ possible values? I don't understand that step.


问:
Combinatorial problem - coloured balls
I have balls with $n$ colors, $n ≥ 3$, and I want to put them into boxes.

1/ Each box must have at last 2 colors.

2/ For every pair of colors and for every color in that pair there is a box that contains that color and doesn't contain the other one, i.e. $\forall \mathrm{color1} \, \forall \mathrm{color2} ≠ \mathrm{color1} \, \exists \mathrm{box}:(\mathrm{color1} \in \mathrm{box} \land \mathrm{color2} \notin \mathrm{box}) $

Take the product of the number of balls in a box, over all the boxes. Call this $f(n)$. What is the minimum value of $f(n)$?

It is easy to see $f(2n) ≤ [f(n)]^2$

For instance, for any n, here is a possible division into boxes:
$\{1,2\},\{2,3\},...,\{n-1,n\},\{1,n\}$, so $f(n) ≤ 2^n$.

However one can do better. For $2n$ one can divide into boxes as follows:
$\{1,...,n\},\{n+1,...,2n\},\{1,n+1\},\{2,n+2\},...,\{n,2n\}$

so $f(2n) ≤ n^2 \cdot 2^n$

And I can do (somewhat) better than this. Basically I'm looking for an upper bound if it's low, or a lower bound if it's high - an exact formula isn't really needed.

答:
Let's number the colors from $0$ to $mn - 1$.

We split the boxes into two categories:

The first category contains the following boxes:

$A_0:\{ 0, n, 2n,3n,...,(m-1)n \}$

$A_1:\{1,n+1,2n+1,...,(m-1)n+1\}$

$A_2:\{2,n+2,2n+2,3n+2,...,(m-1)n+2\}$

$\ldots$

$A_{n-1}:\{ n-1, 2n-1,3n-1,...,mn-1 \}$

The second category contains the boxes:

$B_0:\{0, 1,2,3,4,...,n-1\}$

$B_1:\{ n,n+1,n+2,...,2n-1 \}$

$\ldots$

$B_{m-1}:\{(m-1)n, (m-1)n+1,(m-1)n+2,...,mn-1\}$

If $n,m \ge 2$ then the first condition is satisfied.

Let's take two colors $c_1$ and $c_2$ whose Euclidian division by $n$ is $qn + r$ and $q'n + r'$ respectively.

The two colors are distinct so $q \neq q'$ or $r \neq r'$.

If $q \neq q'$ then $c_1 \in B_q$ and $c_2 \not\in B_q$.
If $r \neq r'$ then $c_1 \in A_r$ and $c_2 \not\in A_r$.

Either way, the second condition is satisfied.

The number of balls in each box in the first category is $m$ and there are $n$ boxes so the product is $m^n$.

The number of balls in each box in the second category is $n$ and there are $m$ boxes so the product is $n^m$.

As such, $f(mn) \le n^m \cdot m^n$ for all $m,n \ge 2$.

Title: Combinatorial problem - coloured balls


问:
How to prove that these both sequences have the limit equals to 3/2? And i want to know if my proof is correct?
[The image shows the problem and the proof]

Um the problem is my teacher said my proof is wrong but i guess is logical and true and i can't see where is the wrong so if cam someone enlightens me (i'm new to this but when i wrote the upper phrase a link appeared and the link opens the image so if the link didn't appear to you please tell me).
So this question is a part of an 3 parts exercice where this is the last question. The first question we are asked to prove that Un and Vn are positive. The 2nd one to prove that they are convergent and there limit exists. The 3rd one to prove that they have the same limit which is equal to 1.5.And when we arrived to the 3rd question the meeting with the professor was ended so he asked us to do it as a homework. So i tried to prove it by induction but when the prof saw it he said that it's not logical there is something missing, then we had a debate on it and finally he told me that he'll tell me where is the wrong on whatsapp after the end of the meeting to not waste time. But it have been 2 days and he didn't reply. So i've came to here to ask where is the wrong.
[1]: https://i.stack.imgur.com/RAakV.jpg

答:
You proved by induction both sequences satisfy the recursion $x_{n+1}=\tfrac{x_n}{3}+1$, which you wrote as $\tfrac{x_n+3}{3}$, although in your handwriting it looked more like $\tfrac{x_{n+3}}{3}$. Then you showed if $\ell:=\lim_{n\to\infty}x_n$ exists then $\ell=\tfrac32$, but the problem is you didn't show $\ell$ exists. The correct way to address that is to note if $x_n$ is a sequence of the above form and $\epsilon_n:=x_n-\tfrac32$ then$$\epsilon_{n+1}=x_{n+1}-\tfrac32=\tfrac13x_n-\tfrac12=\tfrac13\epsilon_n,$$so by induction $\epsilon_n=\tfrac{\epsilon_0}{3^n}$. Since this becomes arbitrarily small, $\lim_{n\to\infty}\epsilon_n=0$, which is equivalent to the desired result.

A more concise approach notes$$\begin{align}u_{n+1}+v_{n+1}&=u_n+v_n,\\u_{n+1}-v_{n+1}&=\frac{u_n-v_n}{3}\\\implies u_n+v_n&=3,\\u_n-v_n&=-3^{n}\\\implies u_n&=\frac{3-3^{-n}}{2},\\v_n&=\frac{3+3^{-n}}{2},\end{align}$$with the first $\implies$ a trivial induction.


问:
Show that $G=HN$ for $H$ subgroup and $N$ normal subgroup
I have a practice problem that states following:

Lad $G$ be a group and let N be a normal subgroup in $G$ with index $p$ which is a prime.

Let $H$ be a subgroup in $G$ which is not contained in $N$. Show that $G=HN$.

My answer:

I have for $x \in HN$ then clearly $x \in G$ and therefore $HN \subseteq G$

However for $x \in G$ I have $gNg^{-1} \in N$ and $gHg^{-1} \notin N$

But how do I continue from here to show $G=HN$?

答:
From $N\unlhd G$ and $H\le G$ it follows that $HN\le G$. Then:

\begin{alignat}{1}
p &= [G:N] \\
&= [G:HN][HN:N] \\
\tag 1
\end{alignat}

From the assumption $H\nsubseteq N$ it follows that $N\lneq HN$ (proper subgroup) or, equivalently, $[HN:N]\gneq 1$. Therefore, by $(1)$, we get $[G:HN]=1$, i.e. $G=HN$.


问:
Expected value and Variance of Xn in random walker
I have the following problem based on the random walker. I have no idea what steps I should take to solve this exercise. If you could solve it and explain each step to me it would be very helpful. Thank you!!!

A drunken person walk along a line, making steps sized 1. Each step is either to the
right or the left, and the probability to walk to the right is fixed: $P(right) = p$. Let $X_{n}$ be the position of the point after $n$ steps. Find $E(X_{n})$ and $V(X_{n})$

答:
We will assume each step is of length one. Denote $W_1, W_2, \cdots, W_n$ such that each $W_i$ states the direction of the person taken at time $i$, specifically
$$W_i=\begin{cases}1&\text{ probability }p,\\
-1&\text{ probability }1-p.\end{cases}$$
Therefore,
$$E[W_i]=p(1)+(1-p)(-1)=2p-1.$$
From here we know if $p>1/2$ then the person is expected to walk to the right, and vice versa. Moreover,
$$Var(W_i)=E[W_i^2]-E[W_i]^2=1-(2p-1)^2=4p(1-p).$$
Note that $W_i^2$ is always $1$.

Using the relation $X_n=\sum_{i=1}^n W_i$, we have
$$E[X_n]=E[W_1]+\cdots+E[W_n]=n(2p-1),$$
and since it is obvious each step is independent from other steps, we have $Cov(W_i,W_j)=0$ if $i\neq j$, hence
$$Var(X_n)=Var(W_1+\cdots+W_n)=Var(W_1)+\cdots+Var(W_n)=4np(1-p).$$


问:
Showing unique decomposition into parallel and orthogonal parts for any subspace
Given a general (possibly infinite dimensional) linear vector space $V$ with an inner product, how can you prove that for any subspace $S$, any vector v in $V$ can be uniquely expressed as

$$v = s + t$$

where $s \in S$  and $t$ is orthogonal to $S$?

答:
Set $d=\inf_{s\in S} d(v,s)$ to be the distance from $v$ to $S$. We show that finding the nearest point to $v$ lying within $S$ gives the solution.

Lemma 1 There exists $s\in S$ such that $\lVert v-s\rVert = d$ if we additionally assume $S$ complete.

Lemma 2 $t=v-s$ is orthogonal to $S$.

Lemma 3 This decomposition is unique.

Proof of 1 Take $d(s_n,v)\le d^2+\frac 1 n$. We show $s_n$ is Cauchy. Recall the parallelogram law $\lVert v_1+v_2\rVert^2+\lVert v_1-v_2\rVert^2=2(\lVert v_1\rVert^2+\lVert v_2\rVert^2)$. Set $v_1=v-s_n,v_2=v-s_m$. One finds $$\lVert s_m-s_n\rVert^2\le 2/m+2/n$$ and the result follows.

Proof of 2 Let $s'\in S$. $$\lVert v-(s+\lambda s')\rVert^2=\lVert v-s\rVert^2-2 \mathrm{Re}[\left<v-s,s'\right>]\lambda+\lVert s'\rVert^2 \lambda^2$$ but since $s$ minimized the length in question, considering small lambda the linear term must vanish. Hence $v-s$ is orthogonal to $S$.

Proof of 3 $v=s'+t'=s+t$ implies $(s'-s)+(t'-t)=0$. Taking inner products with each term shows both are zero.

As pointed out, for incomplete $S$ this fails. Just choose any situation like that of the finite sequences in $\ell^2$ where the natural 'orthogonal' terms form a series converging to a missing point in $S$.


问:
Finding $\lim_{n \to \infty}E\left(\frac{|\overline X_n|}{|\overline X_n| + 1}\right)^2$
Let $\{X_i\}$ be a sequence of i.i.d. random variables with mean $0$ and variance $\sigma^2 > 0$. Compute $$\lim_{n \to \infty}E\left(\frac{|\overline X_n|}{|\overline X_n| + 1}\right)^2$$

Please help, I've tried expansion and Markov's theorem. It's not working.

答: Divide the numerator and the denominator of the fraction by $n$. By SLLN you see that the ratio tends to $1$. Also, the fraction is bounded by $1$. Hence, by DCT we see that the limit is $1$.


问:
A consequence of quotient map from unit circle to itself
In Part II of General Topology, Munkres, Theorem 57.1 states that if $h:S^1\to S^1$ is continuous and antipode-preserving, then $h$ is not nulhomotopic.

There are some definitions I should clarify here: $S^1$ is the unit circle on the $xy$-plane, so we could parametrize $S^1$ as $(\cos \theta, \sin\theta)$, $0\leq \theta\leq 2\pi$. A map $f:S^1\to S^1$ is antipode-preserving if for $x\in S^1$ we have $f(-x)=-f(x)$, which means for every pair of antipodes $\{x, -x\}$, their images under $f$ are also antipodes $\{f(x), f(-x)\}=\{f(x),-f(x)\}$. Finally, $h$ is nulhomotopic means $h$ is path-homotopic to a constant map $e: S^1\to S^1$, by definition this means there is a continuous map $H:S^1\times [0,1]\to S^1$ such that $H(x, 0)=h(x), H(x, 1)=e(x)$. There are other details about this proof, but these should be enough for the question I would like to ask.

In the proof, by composition with some rotation map, we can further assume $h(b_0)=b_0$, where $b_0=(1,0)$. We define a map $q:S^1\to S^1$ by $q(x)=x^2$ (in the complex sense), hence
$$q(\cos \theta,\sin\theta)=(\cos2\theta, \sin2\theta).$$

It is easy to see $q$ is a quotient map, because it is a continuous closed surjective map. Note that $q$ maps each pair of antipode points to the same point, and the inverse of any point in $S^1$ under $q$ is a pair of antipode.

My question: The steps thereafter states that since $q$ is a quotient map, so $q\circ h$ induces a continuous map $k:S^1\to S^1$ such that $k\circ q=q\circ h$. I didn't quite get this point because I don't remember such result from the chapter of quotient map. My idea to get myself through this property is to let $k=q\circ h\circ q^{-1}$ and show that $k$ is well-defined. Indeed, for $x\in S^1$, $q^{-1}(x)$ is not a single point, but a pair of antipodes, then $h(q^{-1}(x))$ is also a pair of antipodes, then $q$ maps $h(q^{-1}(x))$ into a single point, thus $k$ is well-defined. I believe $k$ is continuous so I'm not troubling myself to show it, but is there a general way to see why $q$ being a quotient map will generate the continuous map $k$?

I have noted this question has been asked twice before, but the specific question is different from mine in this link and the answer to this link doesn't answer my question in more detail.

答: You have to apply Munkres' Theorem 22.2. Consider the map $g = p \circ h : S^1 \to S^1$. Each set $p^{-1}(z)$ consists of two antipodal points $w_1, w_2 = -w_1$ (these are the two complex square roots of $z$). But obviously $g$ is constant on all $p^{-1}(z)$ since $g(w_2) = h(w_2)^2 = h(-w_1)^2 = (-h(w_1))^2 = h(w_1)^2 = g(w_1)$.


问:
connected subspaces need to lie in one separated component
Let ($X,\tau)$ be a top space. Suppose $Y \subset X$ is a connected subspace given the subspace topology. If $A \cup B$ is a separation of the space $X$ then $Y$ lies entirely in $A$ or $B$.

I just want a verification:

Suppose other wise then $\exists a \in Y$ such that $a \in A$ and $\exists b \in Y$ such that $b \in B$.

I claim that

$$Y \cap X = Y \cap (A \cup B) = (Y \cap A) \cup (Y \cap B)$$

Forms a separation of $Y$.

As $Y$ inherits the subspace topology, the topology on $Y$ can be given by

$$\tau_Y = \{Y \cap U \mid U \in \tau\}$$

Then since $A,B \in \tau$ we have that $Y \cap A,Y \cap B \in \tau_Y$, i.e., are both open in $Y$.

Clearly both are nonempty by existence of $a,b$ up top.

Lastly for disjointness, suppose $\exists \alpha \in (Y \cap A) \cap (Y \cap B)$, then $\alpha \in A \cap B$ contradicting $A,B$ being disjoint (since they were assumed to form a separation of our parent space) and thus

$$(Y \cap A) \cap (Y \cap B) = \emptyset$$

hence

$$Y = (Y \cap A) \cup (Y \cap B)$$

forms a separation of $Y$ contradicting it being a connected subspace $\blacksquare$

答:
This community wiki solution is intended to clear the question from the unanswered queue.

Yes, your proof is correct.


问:
Test to determine if a point is inside a cardioid whose cusp is not at the origin
The Mandelbrot set is the set of all complex numbers $c$ that cause the function $z_{n+1} = z_n^2 + c$ to remain bounded within a circle of radius 2 when iterated from $z_0 = 0$. Plotted in the complex plane it includes a main cardioid, formed by rolling a circle of radius $0.25$ around another circle of the same radius centered on the origin with its cusp at $0.25 + 0i$; and a main bulb, also a circle of radius $0.25$, centered on the value $-1 + 0i$ and touching the main cardioid at the value $-0.75 + 0i$. These two regions are the only regions with well-defined boundaries.

When calculating the Mandelbrot set, in general the only way to be sure a given value of $c$ is in the set is to iterate it in the generating function an infinite number of times to see if it remains bounded. Because that is impossible in reality, a large integer is picked as a threshold value, and any points which do not exit the bounding circle within that number of iterations is declared to be part of the set. This means that every point inside the set requires the maximum number of iterations to be performed before moving on to the next value.

To save on calculation time, it helps to have an alternate way to determine if a given value is inside the set. This is easy for the largest circular region: if $|c + 1| ≤ 0.25$, $c$ is inside the disk and no further calculation is necessary.

Does a similar test exist for the main cardioid? I haven't been able to figure one out. My instinct is to check if the magnitude of $c$ is less than some function of the argument of $c$, but I can't find a function that works perfectly. The closest I've come is $\frac{1}{2}(\sin(\frac{φ}{2}))^{3/4} + \frac{1}{4}$, which is pretty close, but it isn't exact.

答:
The points in the main caridoid are those values of $c$ such that there is a fixed point $z_0$ with $|f_c'(z_0)|<1$. Now, the fixed points are

$$z_{\pm} = \frac{1\pm \sqrt{1-4c}}{2}$$
and $f'(z)=2z$ so I guess you need to check if

$$\left|1 +  \sqrt{1-4c}\right| \leq 1 \: \text{ or } \: \left|1 -  \sqrt{1-4c}\right| \leq 1.$$

It's not too hard to see that the first inequality is never satisfied so you're left with just the second condition to check.


问:
Limit of an $L^2$ sequence
We have functions $f_n\in L^2$ such that $f_n(x)$ tends to some $f(x)$ for almost all $x$, suppose that $||f_n||<M$. Does this mean that $f_n$ tend to $f$ weakly? In my previous question about $L^1$ I realized that the weak convergence fails; can one describe spaces for which this is true?</p>

答:
Yes.

By Fatou's lemma we have $f \in L^2$, so by replacing $f_n$ by $f_n - f$ we can assume that $f=0$.

Fix $g \in L^2$ and $\epsilon > 0$.  By a standard dominated convergence argument, there is a set $A$ of finite measure with $\int_{A^c} |g|^2 < \epsilon$.  There is also a $K$ so large that $\int_{|g| > K} |g|^2 < \epsilon$.  Let $B = A \cap \{|g| \le K\}$; then $\lvert\int_{B^c} f_n g \rvert \le M \sqrt{2 \epsilon}$ by Cauchy--Schwarz, for every $n$.

On the other hand, for each $n$ we have $\int_B |f_n g|^2 \le K^2 M$.  So $\{f_n g\}$ is an $L^2$-bounded sequence of functions on the finite measure space $B$, hence it is uniformly integrable on $B$.  We also have $f_n g \to 0$ almost everywhere, and hence they converge in measure on $B$; by the Vitali convergence theorem $f_n g \to 0$ in $L^1(B)$, so $\int_B f_n g \to 0$.  Letting $\epsilon \to 0$ completes the proof.

Edit: With respect to your other question (when can you say that an $L^1$-bounded, a.e. convergent sequence must converge weakly in $L^1$), the answer is: only in trivial cases.  Essentially, if the measure space is anything but a finite union of atoms, you can construct a sequence of nonnegative $f_n \in L^1$ with $f_n \to 0$ a.e. but $||f_n||_{L^1} = 1$, so that taking $g=1 \in L^\infty$, you have $\int f_n g = 1$ for all $n$.  (Idea: let $f_n = \mu(A_n)^{-1} 1_{A_n}$ where either $\mu(A_n) \uparrow \infty$, or $\mu(A_n) \downarrow 0$ fast (in particular try $\sum \mu(A_n) < \infty$)).

Title: Limit of an $L^2$ sequence


问:
The particular integral of a 2nd order differential equation
Consider the following 2nd order differential equation, $$\frac{\text{d}^2 z}{\text{d} t^2} + \omega^2 z = C \text{e}^{\text{i}\omega t}$$where $z$ is complex, and $C$ and $\omega$ are unknown real constants.

Now that I am about to find the particular integral. I tried $z=A\text{e}^{\text{i}\omega t}$, so $\frac{\text{d}^2 z}{\text{d} t^2}=-A\omega^2 \text{e}^{\text{i}\omega t}$, and the differential equation becomes $-A\omega^2 \text{e}^{\text{i}\omega t} + A \omega^2 \text{e}^{\text{i}\omega t} = C \text{e}^{\text{i}\omega t}$, which implies $C=0$. That is obviously nonsensical.

[Edit: I am sorry. I mixed some $z$s and $t$s at first.]

So, could someone please explain where I am wrong or what alternative method(s) should I use?

答:
Hint

The problem is easier if you switch variables to make
$$-\frac {t''}{[t']^3}+\omega^2 z=C e^{i \omega z}$$ Reduction of order $p=t'$ to get
$$-\frac {p'}{p^3}+\omega^2 z=C e^{i \omega z}\implies p=\pm \text{something}$$

Not sure about the remaining.


问:
Factor-critical graphs
$G$ is factor-critical $\Leftrightarrow$ $c_o(G-U) \leq |U|$, $\forall U \subseteq V(G)$ except when $U = \emptyset$ (where $c_o$ is the number of odd components - basically, the RHS means that $U = \emptyset$ violates Tutte's condition)

My attempt: ($\Rightarrow$) $G-v$ has a perfect matching $\forall v \in V(G)$. Consider $G' = G-v$ for some arbitrary $v$. Then $c_o(G'-W) = c_o(G - X - v) \leq |W| = |X+v|$, $\forall W \subseteq V(G')$. However, $|W| \geq 1$ since $W = X + v$ for some $v \in V(G)$, $X \subset V(G)$.

I feel like I'm just going in circles and not really getting anywhere, and I don't know how to begin proving the reverse direction.

答:
Your proof of the $\implies$ direction is correct.

For the converse, we must impose the condition that $G$ is connected--otherwise, the condition is satisfied by disjoint two factor-critical graphs (the union will not be factor critical).

Since Tutte's condition does not hold for $U = \varnothing$, $|G|$ is odd.  Remove a vertex $x$ and suppose $G - x$ has no matching.

Set $G - x = H$.  $|H|$ is even, so find a set for which $c_o(H - S) > |S|$.  It suffices to prove that $c_o(H - S) - |S| \ge 2$.  Hover below for the spoiler:

 To show this, it suffices to note that $c_o(H - S)$ and $|S|$ are both even or both odd, since, modulo $2$, they must sum to $|H| \equiv 0$.

Title: Factor-critical graphs


问:
Partial fractions of $\frac{x^4}{(x^2-1)^3}$ (is it possible to find coefficients without expanding?)
I want to decompose the fraction $\frac{x^4}{(x^2-1)^3}$ :
$$\frac{x^4}{(x^2-1)^3}=\frac A{x-1}+\frac B{(x-1)^2}+\frac C{(x-1)^3}+\frac D{x+1}+\frac E{(x+1)^2}+\frac F{(x+1)^3}$$

Because $f(x)=\frac{x^4}{(x^2-1)^3}$ is an even function we have $f(x)=f(-x)$. Hence we have $A=-D\;,\;B=E\;,\;C=-F$ :
$$\frac{x^4}{(x^2-1)^3}=\frac A{x-1}+\frac B{(x-1)^2}+\frac C{(x-1)^3}+\frac {-A}{x+1}+\frac B{(x+1)^2}+\frac {-C}{(x+1)^3}$$
Hence:

$$A(x-1)^2(x+1)^3+B(x-1)(x+1)^3+C(x+1)^3-A(x+1)^2(x-1)^3+B(x+1)(x-1)^3-C(x-1)^3=x^4$$

If we plug in $x=1$ we get $C=\frac18$. but I can't find $A$ and $B$ without expanding $(x-1)^2(x+1)^3\ldots$ is it possible to find them without expanding whole expression?

答:
Following my comments, we may try two simple values of $x$ to reduce computations. Here I believe $x=0$ and $x=2$ are simple enough.

Plugging in $x=0$ gives $$A-B+C-(-A)+(-B)-(-C)=0\implies A-B=-\frac 1 8$$
and plugging in $x=2$ gives $$A\cdot 3^3 - B\cdot 3^3 + C\cdot 3^3 - A\cdot 3^2 + B\cdot 3 - C = 2^4\implies A + \frac 5 3 B =\frac {17} {24}$$
which gives us $A=\frac 3 {16}, B=\frac 5 {16}$.


问:
Initial Conditions For Characteristic Projections
Given $u_t=(u_x)^2$ and an initial condition $u(x,0)=\begin{cases}
 1-x & \text{ if } x \in [0,1) \\
 0 & \text{ if } x \in [1,\infty)
\end{cases} $

Using Charpit's equations by defining $F(x,y,u,p=u_x,q=u_t)= u_t-(u_x)^2 = 0$, and parameterising characteristic projections by $(s,\tau)$, we obtain

$q=p^2 \implies F=q-p^2$ and so characteristic projections are defined by $ t'(s)=1,x'(s)=-2p$ and on the projections, $p$ and $q$ vary according to;$ \dot{p}=\dot{q}=0$
and the solution is given by $u'(s)=q-2p^2=p^2-2p^2=-p^2$

Now the problem is defining conditions on s=0, I have the solutions(given below), but I have no intuitive idea of how they are yielded, from the initial conditions given.

The aformentioned conditions are given by (on $s=0$) : $t=0,x=\tau,u=1-\tau,p=\mu, q=\mu^2$

答:
An obvious solution for the PDE $\quad u_t=u_x^2\quad$ is :
$$u(x,t)=\alpha x +\alpha^2 t+\beta$$
With the conditions the solution is :
$$u(x,t)=\begin{cases}
 -x+t+1 & \text{ if } x \in [0,1) \qquad\qquad \alpha=-1\quad;\quad \beta=1 \\
 0 & \text{ if } x \in [1,\infty) \qquad\qquad \alpha=0\quad;\quad \beta=0
\end{cases} $$
Or with the Heaviside step function
$$u(x,y)=(-x+t+1)\text{H}(1-x)$$


问:
En route to proving the periodicity of sine and cosine
This answer shows that $\cos(\kappa) = \sin(\kappa) = \frac{1}{\sqrt{2}}$ (but whilst $\cos$ and $\sin$ are yet to be defined, $c$ and $s$ are used).

The answer continues that if $0 < x < \kappa$, then $\frac{1}{\sqrt{2}} < x < 1$. Afterwards, it is claimed that $\frac{\kappa}{\sqrt{2}} < s(\kappa) < \kappa$ and $\frac{1}{\sqrt{2}} < \kappa < 1$.

Why do these bounds follow from a point of intersection?

答:
This does not directly answer the question but one can prove periodicity with another argument: for all $a\in{\mathbb R}$, one has
\begin{equation}
m:=\inf_{t\in[a, \infty)}|s(t)| = 0
\end{equation}
Indeed, if $b>a$, the mean value theorem applied to $c(x)$ tells us that for a certain number $t\in (a,b)$, one has
\begin{equation}
m\le |s(t)| = |c'(t)| = \left|\frac{c(b)-c(a)}{b-a}\right|\le \frac{2}{b-a}
\end{equation}
which tends to $0$ when $b\to\infty$.

A consequence of this is that $s$ cannot increase monotonically on $(0,\infty)$ otherwise we would have $0<s(a)\le m$ for any $a>0$ and this is a contradiction. Let $\alpha$ be the largest number such that $s$ increases on $(0, \alpha)$. We have $c(\alpha) = s'(\alpha) = 0$ and $s(\alpha)> 0$ and this implies $s(\alpha) = 1$.

The differential equation $y''+y=0$ now implies that $c(x) = s(x + \alpha)$ because both hand sides of this equality satisfy the same Cauchy problem at $x=0$. As $c$ is even, we conclude that $s(\alpha - x) = s(\alpha + x)$ and it follows that $s(x) = -s(x + 2\alpha)$ by the same Cauchy problem argument. Hence $s(x +  4\alpha) = -(-s(x)) = s(x)$ and $s$ is periodic with period $4\alpha$.


问:
Determining all triples $(a,b,c)$ of positive integers that are sides of a triangle inscribed in a circle of diameter $6.25$
An Olympiad Geometry question

Determine all triples $(a, b, c)$ of positive integers which are the lengths of the sides of a triangle inscribed in a circle of diameter $6.25$ units.

So, in this, basically diameter = 2R = $\frac{a}{sinA}$ and so on. Or we also have R = $\frac{abc}{4\Delta}$.

But I don't understand how this may help in this problem.

Any hint/ solution is appreciated,

Thanks!

答:
Observe that, since $6.25$ is the length of the longest chord(diameter) of the circle, $a$, $b$ and $c$ will belong to $\{1,2,3,4,5,6\}$.

Also, notice that, $25$ will divide $abc$. Then, at least $2$ of these terms will be equal to $5$($\Delta $ will be rational ).Let $a=5$ and $b=5$ without loss of generality.
Now, just plug in those values into the equation and you will get the value of $c$.


问:
Show that $(\mathbf{a}·\mathbf{a})(\mathbf{b}·\mathbf{b})-(\mathbf{a}·\mathbf{b})^2=(\mathbf{a}\times\mathbf{b})·(\mathbf{a}\times\mathbf{b})$?
I need to prove that, for two linearly independent vectors $\mathbf{a},\mathbf{b}\in\mathbb{R}^3$,

$$(\mathbf{a} · \mathbf{a}) (\mathbf{b} · \mathbf{b}) - (\mathbf{a} · \mathbf{b})(\mathbf{a} · \mathbf{b}) = (\mathbf{a} \times \mathbf{b})·(\mathbf{a} \times \mathbf{b})$$

Could someone give me a demonstration of this identity? Or a hint to prove it?

答: Observe that $(A\times B)\cdot (A\times B) = |A\times B|^2 = |A|^2|B|^2\sin^2\theta$, where $\theta$ is the angle between them. Now we have $\sin^2\theta = 1-\cos^2\theta$, and I will remind you that $A\cdot B = |A||B|\cos\theta$.


答:
The left hand side is the determinant of the Gramian matrix of $\mathbf a$ and $\mathbf b$:
$$
\det\begin{pmatrix}\mathbf a\cdot\mathbf a & \mathbf a\cdot\mathbf b \\
\mathbf b\cdot\mathbf a & \mathbf b\cdot\mathbf b\end{pmatrix}.
$$
In general, the determinant of a Gramian matrix is the square of the $n$-dimensional volume of a parallelotope, in this case the square of the area of the parallelogram spanned by $\mathbf a$ and $\mathbf b$.

The right hand side is $\|\mathbf a\times\mathbf b\|^2$, where $\|\mathbf a\times\mathbf b\|$ also is the area of the parallelogram spanned by $\mathbf a$ and $\mathbf b$, hence both sides are equal.

Title: Show that $(\mathbf{a}·\mathbf{a})(\mathbf{b}·\mathbf{b})-(\mathbf{a}·\mathbf{b})^2=(\mathbf{a}\times\mathbf{b})·(\mathbf{a}\times\mathbf{b})$?

Body:
I need to prove that, for two linearly independent vectors $\mathbf{a},\mathbf{b}\in\mathbb{R}^3$,

$$(\mathbf{a} · \mathbf{a}) (\mathbf{b} · \mathbf{b}) - (\mathbf{a} · \mathbf{b})(\mathbf{a} · \mathbf{b}) = (\mathbf{a} \times \mathbf{b})·(\mathbf{a} \times \mathbf{b})$$

Could someone give me a demonstration of this identity? Or a hint to prove it?


问:
Find a generator of the ideal generated by $3$ and $2-2\sqrt{-2}$ in ${\mathbb{Z}\left[\sqrt{-2}\right]}$
Exercise: Find a generator of the ideal generated by $3$ and $2-2\sqrt{-2}$ in ${\mathbb{Z}\left[\sqrt{-2}\right]}$.

I know that such a generator exists since $\mathbb{Z}\left[\sqrt{-2}\right]$ is a Euclidean domain and thus a PID, with a generator being given by an arbitrary element of $I$ with minimal norm, where $I$ denotes the ideal generated by $3$ and $2-2\sqrt{-2}$. Furthermore,  $3$ has norm $9$ and $2-2\sqrt{-2}$ has norm $12$.

How do I continue?

答: We want to find $\alpha\in \mathbb{Z}[\sqrt{-2}]$ such that $I = (3, 2 - 2\sqrt{-2}) = (\alpha)$. For such $\alpha$, there exists $\beta, \gamma \in \mathbb{Z}[\sqrt{-2}]$ such that $3 = \alpha\beta, 2 - 2\sqrt{-2} = \alpha \gamma$. If we observe norms of them, we have $9 = N(\alpha)N(\beta)$ and $12 = N(\alpha)N(\gamma)$, so $N(\alpha) = 1$ or $3$. When $N(\alpha) = 1$, $\alpha = \pm 1$ and one can show that $1 \not \in I$ by several ways (brute force actually works) Since there are only finitely many $\alpha \in \mathbb{Z}[\sqrt{-2}]$ with $N(\alpha)$ (and there are only two actually), and only one of them satisfies $\alpha | 3$ and $\alpha = 2 -2\sqrt{-2}$. Now we have $(3, 2 - 2\sqrt{-2}) \subseteq (\alpha)$, and you also need to show the opposite inclusion by seeking at $\beta_1, \beta_2$ with $\alpha = \beta_{1}\times 3 + \beta_{2} \times (2 - 2\sqrt{-2})$.


答:
As you observed, the right side is in resonance with the left side, the exponential factor is a root of the characteristic polynomial of the differential operator on the left. In these cases you have to add a monomial factor $t^m$ corresponding to the multiplicity of the characteristic root (in the non-resonance case $m=0$, as the factor is not a root, so also that case fits in).

So try again with
$$
z=At\mathrm{e}^{\mathrm{i}\omega t}.
$$

Title: The particular integral of a 2nd order differential equation

Body:
Consider the following 2nd order differential equation, $$\frac{\text{d}^2 z}{\text{d} t^2} + \omega^2 z = C \text{e}^{\text{i}\omega t}$$where $z$ is complex, and $C$ and $\omega$ are unknown real constants.

Now that I am about to find the particular integral. I tried $z=A\text{e}^{\text{i}\omega t}$, so $\frac{\text{d}^2 z}{\text{d} t^2}=-A\omega^2 \text{e}^{\text{i}\omega t}$, and the differential equation becomes $-A\omega^2 \text{e}^{\text{i}\omega t} + A \omega^2 \text{e}^{\text{i}\omega t} = C \text{e}^{\text{i}\omega t}$, which implies $C=0$. That is obviously nonsensical.

[Edit: I am sorry. I mixed some $z$s and $t$s at first.]

So, could someone please explain where I am wrong or what alternative method(s) should I use?


问:
$4^{2020} - 2020$ mod $48$

$4^{2020} - 2020$ mod $48$

My Approach:

I basically tried to separate $4^{2020}$ into multiple entities and tried to break them down but at one point it just wasn't computational.

Title: $4^{2020} - 2020$ mod $48$

Body:
Hint:
$4^{2018}\equiv1\pmod{3}$
implies
$4^{2020} \equiv 16\pmod {48}$


问:
Are these two curves birationally equivalent?
I'm trying to work out if the following two curves are birationally equivalent:

$$Y^2 = 2X^4 + 17X^2 + 12$$
$$2Y^2 = X^4 - 17$$

(I'm considering the above as the affine shorthands for the projective curves they represent)

I saw elsewhere online that I could use the following code in magma:

```<code>K<x,y>:=AffineSpace(Rationals(),2);
C1A:=Curve(K,2*x^4+17*x^2+12-y^2);
C2A:=Curve(K,x^4-17-2*y^2);
C1:=ProjectiveClosure(C1A);
C2:=ProjectiveClosure(C2A);
IsIsomorphic(C1,C2);
</code>```

However this gives me a runtime error.

Can anyone see an error with my code/approach above? Alternatively, if there's another way I can check if two curves like the above are birationally equivalent either by hand or by using Magma / Sage or something similar that would be really helpful.

答:
Smooth projective curves are birationally equivalent if and only if they are isomorphic (a rational map between smooth projective curves is automatically a morphism) - so we just want to find a smooth curve birationally equivalent to each $C_i$. The answer above is fine too, but I thought I'd continue your attempt at finding one with Magma.

I claim the curves are not birationally equivalent, so extending the field is fine (I will go to $L = \mathbb{Q}(\sqrt{-2}, \sqrt{31})$ ). I want to extend the base field simply to allow us to find some smooth points on $C_1$ and $C_2$ (PointSearch fails to find a smooth point up to height 10,000), at which point the intrinsic Magma function can form Elliptic Curves (it returns a smooth model $E_i$ over $L$, and a birational map from $C_i$ to $E_i$). Then we have

```<code>K<a> := QuadraticField(-2);
L<b> := PolynomialRing(K);
L<b> := ext<K | b^2 - 31>;

A2<x,y> := AffineSpace(L, 2);

C1A := Curve(A2, 2*x^4 + 17*x^2 + 12 - y^2);
C2A := Curve(A2, x^4 - 17 - 2*y^2);

C1 := ProjectiveClosure(C1A);
C2 := ProjectiveClosure(C2A);

assert Genus(C1) eq 1;
assert Genus(C2) eq 1;

P1 := C1![1,b,1];
P2 := C2![1,2*a,1];

assert IsNonsingular(P1);
assert IsNonsingular(P2);

E1 := EllipticCurve(C1, P1);
E2 := EllipticCurve(C2, P2);

assert jInvariant(E1) ne jInvariant(E2);

</code>```

which proves that $j(E_1) \neq j(E_2)$, so $E_1$ and $E_2$ are not isomorphic, and $C_1$ and $C_2$ are not birationally equivalent.

Body:
Smooth projective curves are birationally equivalent if and only if they are isomorphic (a rational map between smooth projective curves is automatically a morphism) - so we just want to find a smooth curve birationally equivalent to each $C_i$. The answer above is fine too, but I thought I'd continue your attempt at finding one with Magma.

I claim the curves are not birationally equivalent, so extending the field is fine (I will go to $L = \mathbb{Q}(\sqrt{-2}, \sqrt{31})$ ). I want to extend the base field simply to allow us to find some smooth points on $C_1$ and $C_2$ (PointSearch fails to find a smooth point up to height 10,000), at which point the intrinsic Magma function can form Elliptic Curves (it returns a smooth model $E_i$ over $L$, and a birational map from $C_i$ to $E_i$). Then we have

K<a> := QuadraticField(-2);
L<b> := PolynomialRing(K);
L<b> := ext<K | b^2 - 31>;

A2<x,y> := AffineSpace(L, 2);

C1A := Curve(A2, 2*x^4 + 17*x^2 + 12 - y^2);
C2A := Curve(A2, x^4 - 17 - 2*y^2);

C1 := ProjectiveClosure(C1A);
C2 := ProjectiveClosure(C2A);

assert Genus(C1) eq 1;
assert Genus(C2) eq 1;

P1 := C1![1,b,1];
P2 := C2![1,2*a,1];

assert IsNonsingular(P1);
assert IsNonsingular(P2);

E1 := EllipticCurve(C1, P1);
E2 := EllipticCurve(C2, P2);

assert jInvariant(E1) ne jInvariant(E2);

which proves that $j(E_1) \neq j(E_2)$, so $E_1$ and $E_2$ are not isomorphic, and $C_1$ and $C_2$ are not birationally equivalent.


问:
Prove that $β_0(G) ≥ δ(G)$ for any graph $G$.
Prove that $β_0(G) ≥ δ(G)$ for any graph $G$, where $δ(G)$ denotes the minimal vertex
degree in the graph and $β_0(G)$ is the minimum covering. Find a graph for which the equality holds.

A hint is given and I have tried to use follow it but, I couldn't get it. The is it below:

Hint: Consider a vertex $v$ of degree $0$. Prove that a covering of edges
incident to the vertex set $v \cup N(v)$ contains at least $0$ vertices.

Kindly help me out!

答: You can use induction on the number of vertices. For an empty graph the statement is trivial. Now let $G$ be a graph and $v$ be a minimum degree vertex. A vertex cover contains either $v$ or every neighbour of $v$ (why?). In the latter case it has at least $\delta(G)$ vertices. In the former case, deleting $v$ from the vertex cover gives a vertex cover of $G-v$. The minimum degree of $G-v$ is at least $\delta(G) - 1$, since the degree of every vertex apart from $v$ decreased by at most one. It follows by the induction hypothesis that a vertex cover of $G-v$ has size at least $\delta(G) - 1$, so after adding $v$ back in, it has size at least $\delta(G)$.


答:
Let $g\in\Bbb{Z}[\sqrt{-2}]$ be a generator of the ideal generated by $3$ and $2-2\sqrt{-2}$ in $\Bbb{Z}[\sqrt{-2}]$. Then
$$3=gx\qquad\text{ and }\qquad 2-2\sqrt{-2}=gy,$$
for some $x,y\in\Bbb{Z}[\sqrt{-2}]$. It follows that the norm of $g$ divides $3$ because
\begin{eqnarray*}
\mathcal{N}(g)\mathcal{N}(x)&=&\mathcal{N}(gx)=\mathcal{N}(3)&=&9,\\
\mathcal{N}(g)\mathcal{N}(y)&=&\mathcal{N}(gy)=\mathcal{N}(2-2\sqrt{-2})&=&12,
\end{eqnarray*}
and $\gcd(9,12)=3$. Because the norm is positive, either $\mathcal{N}(g)=1$ or $\mathcal{N}(g)=3$. As $g$ generates the smallest ideal containing $3$ and $2-2\sqrt{-2}$, we first check whether $\mathcal{N}(g)=3$ is possible; we find that if $g=a+b\sqrt{-2}$ then
$$3=\mathcal{N}(g)=\mathcal{N}(a+b\sqrt{-2})=a^2+2b^2,$$
from which it easily follows that $a=\pm1$ and $b=\pm1$. Multiplying by $-1$, a unit, if necessary we find that $g=1\pm\sqrt{-2}$. And indeed
$$3=(1+\sqrt{-2})(1-\sqrt{-2})\qquad\text{ and }\qquad 2-2\sqrt{-2}=2(1-\sqrt{-2}),$$
which shows that $g=1-\sqrt{-2}$ is a generator. And for good measure we can verify that
$$(1-\sqrt{-2})\cdot3+(-1)\cdot(2-2\sqrt{-2})=1-\sqrt{-2}=g,$$
which shows that indeed $g$ is contained in the ideal generated by $3$ and $2-2\sqrt{-2}$.

Body:
Exercise: Find a generator of the ideal generated by $3$ and $2-2\sqrt{-2}$ in ${\mathbb{Z}\left[\sqrt{-2}\right]}$.

I know that such a generator exists since $\mathbb{Z}\left[\sqrt{-2}\right]$ is a Euclidean domain and thus a PID, with a generator being given by an arbitrary element of $I$ with minimal norm, where $I$ denotes the ideal generated by $3$ and $2-2\sqrt{-2}$. Furthermore,  $3$ has norm $9$ and $2-2\sqrt{-2}$ has norm $12$.

How do I continue?


Title: $4^{2020} - 2020$ mod $48$

Body:
$4^{2020} - 2020$ mod $48$

My Approach:

I basically tried to separate $4^{2020}$ into multiple entities and tried to break them down but at one point it just wasn't computational.

Body:
Hint :

$4^{3}\pmod{48}\equiv4^{2}\pmod{48}$


问:
Order of an element in a group problem as follows
Let $(G,*)$ be a group, $a \in G$, and
$|a|=p$, where $p$ is a prime.

Prove that $|a^k|=p$ for all $1 \le k \lt p$.

Prove that for all $m \in \Bbb N$, either $a^m = e_G$ or $|a^m|=p$, where $e_G$ is the identity element in $G$.

attempt:

Suppose $|a^k| \ne p$ for all $1 \le k \lt p$. Then, $a^{pk} \ne e_G$. On the other hand, $|a|=p$ i.e. $a^p = e_G$. So, we have
\begin{equation*}
a^{pk} = (a^p)^k \ne e_G
\Leftrightarrow e_G^k \ne e_G
\end{equation*}
i.e. $e_G \ne e_G$, a contradiction.
Hence, $|a^k| = p$.

Alternatively, by using theorem:
Let $(G,*)$ be a group, $a \in G$, and
$|a| = n$. Then, for all $t \in \Bbb N$,
\begin{equation*}
|a^t| = \frac{n}{\gcd(t,n)}
\end{equation*}
by replacing $t=k$ and $n=p$, we have:
If $p \nmid k$, then we're done.
If $p \mid k$, then $|a^k| = \frac{p}{p} = 1$, a contradiction. Hence, we must have $p \nmid k$ i.e. proved.

Note that for all $m \in \Bbb N$,
\begin{equation*}
\gcd(m,p) =
\begin{cases}
p, \text{if $p \mid m$} \\
1, \text{if $p \nmid m$}
\end{cases}
\end{equation*}
Then, by theorem above, we have
\begin{equation*}
|a^m| = \frac{p}{\gcd(m,p)} =
\begin{cases}
p, \text{if $p \mid m$} \\
1, \text{if $p \nmid m$}
\end{cases}
\end{equation*}
Thus, either $|a^m| = 1 \Rightarrow a^m = e_G$ or $|a^m| = p$, as desired.

Does the above proof true? I still have a little bit confuse in proving 1. Any elementary or a simple approach?

答:
I think you might be having some misunderstanding about the statements. In your first proof for statement 1, you tried a proof by contradiction. To do so, you need to assume that the original statement is false. What does it mean for "$|a^k| = p$ for all $1 \leq k < p$" to be false? It means there exists a $k$ with $1 \leq k < p$ and $|a^k| \neq p$. It does not mean $|a^k| \neq p$ for all $k$. So your first proof is not correct.

In your second proof, you made a small mistake: it is not possible that $p|k$, because $k < p$. Thus, there is no "contradiction". You simply stop before your last sentence and the proof is concluded.

Your proof for the second statement is correct.


问:
Diagonalizability in $\mathbb{R}$ and $\mathbb{C}$
Give an example of a matrix $A\in M_{n\times n}(\mathbb{R})$ that is not diagonalizable, but A is diagonalizable viewed as a matrix over the field of complex numbers $\mathbb{C}.$</p>

答: You just need to write down a real matrix with distinct complex (non-real) eigenvalues.  Once you've found a $2 \times 2$ example, you can form an example of any size $n \times n$ for $n \ge 2$ by considering an appropriate block diagonal matrix.</p>


问:
A={1,2,3},B={1,2,2,3}. Is B⊆A? And is B⊂A?
I was thinking yes to the first question and no to the second. They are equal sets right even though B has duplicate elements.</p>

答:
$$A=B$$

So yes to the first and no to the second.

Don't worry about the duplicates. When we're talking about sets, only the existence matter and not how many times. Sets that counts how many times it appears are referred as multisets.


问:
A free group action with discrete orbits that is not a covering map
Suppose that a discrete group $G$ acts freely on a topological space $X$ such that each orbit is discrete for the subspace topology. Here $X$ is not necessarily Hausdorff.

Are there any easy examples such that the quotient map $X \to X/G$ is not a covering map?

I believe a counterexample is given without proof here, but I don't see why the quotient map fails to be a covering map in that case. Maybe there are easier counterexamples?

答:
I think both of the examples given at mathoverflow are actually covering maps.

I claimed in the comments that I had a somewhat discrete example but I transformed it to a manifold-based example. The space $X$ is the Möbius strip and $G = ℤ$. I haven't found a really nice way of describing the action but I'll do what I can.

We start by describing a free action of $ℝ$ on $X$. The quotient is $[0,1[$ and the fibers of the quotient $X↠[0,1[$ are shown on the drawing below.

Notice that the "cut" chosen above to draw the Möbius strip above is special. If we displace the cut, we see an angle appear in the fibers. Below are other drawings of two of the fibers. The one in red is the fiber above $0 ∈ [0,1[$.

The action of $ℝ$ is done by "following the fibers" in a chosen direction, but you need to slow down more and more as you approach the angle of the red line. Maybe it's a bit hard to see that it works on the drawings I gave. Here is a different way of drawing the Möbius strip and the two fibers. Imagine it has an infinite height.

On this drawing, the action of $ℝ$ is isometric on the fibers and easier to see (translate along the $y$ direction and teleport to the other side of the red line when you end up out of the drawing).

To get an action of a discrete group and discrete orbits, we restrict the above action to $ℤ⊆ℝ$. The quotient becomes $[0,1[ × S^1$ and the quotient map is not a covering above $(0,x)$ for any $x ∈ S^1$.

To give you an idea of the initial "discrete" example I had, here is a drawing of it. I don't explain more, it's very similar to what I said above. I built by trying to contradict the properness of the action in the simplest way possible.


问:
Complex-linear matrices
I'm reading Tapp's introduction to matrix groups.The book introduced complex-linear matrices and it gives this characterization of complex-linear matrix:

Let $B\in M_{2n}(\mathbb R)$, let $J$ be the matrix

$J = \left (\begin{array}{}
0 & -I \\
I  & 0
\end{array}\right )$,
where $I$ is the identity matrix ,  B is complex linear if and only if $BJ=JB$.

I understand why if $B$ is a complex-linear the $BJ=JB$ (by
using a commutative diagram) but i don't understand how i can prove
the other side (if $BJ=JB$ then $B$ is complex-linear).

Is it enough to say that if $B$ commutes with $i$ ($J$ is "the copy of $i$" in $M_{2n}(\mathbb R)$) then $B$ is linear complex ?
Do you have any hints to give me for the proof? Thank you all.

{Note}: Definition of complex linear matrix: B is a complex linear iff  B is in the image of a map  $\rho_n: M_n(\mathbb C) \to M_{2n}(\mathbb R)$ . This map associates to each matrix $(X + iY)$ a matrix  $\left (\begin{array}{}
X & -Y \\
Y & X
\end{array}\right)$ , $X,Y \in  M_n(\mathbb R)$

答:
More in line with Tapp's book, he notices (I use the same notations as Tapp) that it suffices to show that $F(iX)=iF(X)$, with $F=f_n^{-1}\circ R_{B}\circ f_{n}$. Also, $J_{2n}^{2}=-I$ so $BJ_{2n}=J_{2n}B$ implies $B=-J_{2n}BJ_{2n}$. Thus $F=f_n^{-1}\circ R_{-J_{2n}BJ_{2n}}\circ f_{n}=-f_n^{-1}\circ R_{J_{2n}} \circ R_B \circ R_{J_{2n}} \circ f_n$. Now, the point of $J_{2n}$ is that $ R_{J_{2n}} \circ f_n=f_n \circ R_{iI} $ so also, $f_n^{-1}\circ R_{J_{2n}} = R_{iI}\circ f_n^{-1} $. So, we get $F(iX)=-R_{iI}\circ f_n^{-1} \circ R_B\circ f_n \circ R_{iI} (iX)=-R_{iI}\circ f_n^{-1} \circ R_B\circ f_n (-X)=-R_{iI}\circ F(-X)$

Now, $F$ is $\mathbb{R}$-linear (as Tapp points out), so $F(iX)=R_{iI}(F(X))=iF(X)$

Title: Complex-linear matrices


问:
Application of Sylow to a group of order 60
Let $G$ be a group with $|G|=60=2^2 \cdot 3 \cdot 5$ and assume $G$ has a unique $3$-Sylow subgroup $H_3$. The claim is that then $G$ has also a unique $5$-Sylow subgroup.

This is the procedure to prove it:

i) Show that $H_3$ is normal in $G$. (Follows directly since all Sylow subgroups are conjugates).

ii) Show that $G/H_3$ has a unique $5$-Sylow $H_5$ (Follows from $|Syl_5(G)|\equiv 1 \, mod \, 5$ and $|Syl_5(G)|$ divides $4$).

Let $H$ be the inverse image of $H_5$ by the projection $ \pi :G\longrightarrow G/H_3$. Now the goal is to note how many Sylow-$5$ subgroups there are in $H$ and use this to prove the claim. I don't understand this last step at all.
Thank you for any clarifications.

答:
Let $H$ be its inverse image by the projection $\pi: G \rightarrow G/H_3$.

Whose inverse image? Is it the inverse image of $H_3$, or is it the inverse image of that unique 5-Sylow?


问:
$\int_{0}^{\infty } e^{-t}\cdot t^{3}\cdot \sin(t) dt$ using Laplace transform
I am currently on this question:

Use Laplace Transform to evaluate the following
integral:

$\int_{0}^{\infty } e^{-t}\cdot t^{3}\cdot \sin(t) dt$

What I did:

let the $\int_{0}^{\infty } e^{-t } dt$ = Limit $s\to s+1$

$t\sin(t) = (-1)^3  \frac{d^{3}}{ds^{3}} \frac{1}{s^{2}+1}$

However, my professor answered it differently, and it seems I'm either far off or both are correct.
I really have no idea what happens after the second line so If anyone could explain?

答:
Well, we are trying the following integral:

$$\mathscr{S}_\text{n}\left(\text{s},\omega\right):=\int_0^\infty x^\text{n}\sin\left(\omega x\right)\exp\left(-\text{s}x\right)\space\text{d}x\tag1$$

Using the definition of the Laplace transform, we can see that we can write:

$$\mathscr{S}_\text{n}\left(\text{s},\omega\right)=\mathscr{L}_x\left[x^\text{n}\sin\left(\omega x\right)\right]_{\left(\text{s}\right)}\tag2$$

Using the frequency-domain general derivative property of the Laplace transform, we can write:

$$\mathscr{S}_\text{n}\left(\text{s},\omega\right)=\left(-1\right)^\text{n}\cdot\frac{\partial^\text{n}}{\partial\text{s}^\text{n}}\left(\mathscr{L}_x\left[\sin\left(\omega x\right)\right]_{\left(\text{s}\right)}\right)\tag3$$

Using the table of selected Laplace transforms, we can see that:

$$\mathscr{S}_\text{n}\left(\text{s},\omega\right)=\left(-1\right)^\text{n}\cdot\frac{\partial^\text{n}}{\partial\text{s}^\text{n}}\left(\frac{\omega}{\text{s}^2+\omega^2}\right)=\omega\cdot\left(-1\right)^\text{n}\cdot\frac{\partial^\text{n}}{\partial\text{s}^\text{n}}\left(\frac{1}{\text{s}^2+\omega^2}\right)\tag4$$

Applying this to your integral, we get:

$$\mathscr{S}_3\left(1,1\right)=\lim_{\text{s}\to1}1\cdot\left(-1\right)^3\cdot\frac{\partial^3}{\partial\text{s}^3}\left(\frac{1}{\text{s}^2+1^2}\right)=0\tag5$$


问:
Volume of a set in phase space. How many dimensions?
Suppose I have a $6N$ dimensional space with points looking like this:

$$(r_x^{(1)},r_y^{(1)},r_z^{(1)}, p_x^{(1)}, p_y^{(1)}, p_z^{(1)},...,r_x^{(N)},r_y^{(N)},r_z^{(N)}, p_x^{(N)}, p_y^{(N)}, p_z^{(N)})$$

This is a phase space for a system of $N$ particles. In this space, I define a set by the following conditions:

$$\sum_{k=1}^{N} \left( \left[ p_{x}^{(k)} \right] ^2 +\left[ p_{y}^{(k)}  \right]^2 +\left[ p_{z}^{(k)} \right]^2 \right) =const>0 .$$

$$0\le r_{x}^{(k)} \le l ; \quad 0\le r_{y}^{(k)} \le l ; \quad 0\le r_{z}^{(k)} \le l ; \quad k=1,2,...,N; \quad l=const>0.$$

What is the dimension of this object? How to calculate its "volume"? I was thinking that the first equation specifies a sphere in $3N$ dimensions, which itself is a $3N-1$ dimensional object, but now I'm confused on how to treat the remaining conditions.

PS. Would it be helpful to first consider a similar problem, but with the first condition replaced by

$$0<\sum_{k=1}^{N} \left( \left[ p_{x}^{(k)} \right] ^2 +\left[ p_{y}^{(k)}  \right]^2 +\left[ p_{z}^{(k)} \right]^2 \right) \le const?$$

I've seen such procedures done in some textbooks but I don't really understand why not work with the problem directly.

Update: What would be the units of this set be? I'm assuming the $r$'s have units of length and $p$'s of momentum.

答:
Anything specifying a range on a continuous parameter does not typically reduce the dimensionality of the system. If there exist points 'in the middle of' all the ranges, they are locally unaware of the boundary, and hence locally the dimensionality is unaffected.

You have exactly one proper constraint on the motion, which is the total energy condition $\sum \mathbf p^2=E^2=\text{const.}$ There are $6N$ degrees of freedom. Hence your system is $6N-1$ dimensional.

The phase space volume is currently infinite since $r_y,r_z$ can take arbitrarily negative values. If you bound them below by zero then you have a $3N-1$ dimensional sphere of radius $E$ mutiplied by a $3N$ dimensional cube of side length $l$. The total phase space volume is the product of the volumes of these two objects, since they are independent. Note that by the volume of the sphere I mean the surface area in $3N$ dimensions, the intrinsic volume of the sphere as a $3N-1$ ddimensional manifold.

The units of the volume are $(\text{momentum})^{3N-1} (\text{length})^{3N}$


答:
The article on multisets on wiki is of interest:

In mathematics, a multiset (or bag, or mset) is a modification of the
concept of a set that, unlike a set, allows for multiple instances for
each of its elements. The positive integer number of instances given
for each element is called the multiplicity of that element in the
multiset. As a consequence, an infinite number of multisets exist
which contain only elements a and b, but vary in the multiplicities of
their elements:

The set {a, b} contains only elements a and b, each having multiplicity 1 when {a, b} is seen as a multiset.
In the multiset {a, a, b}, the element a has multiplicity 2, and b has multiplicity 1.
In the multiset {a, a, a, b, b, b}, a and b both have multiplicity 3.

These objects are all different, when viewed as multisets, although
they are the same set, since they all consist of the same elements. As
with sets, and in contrast to tuples, order does not matter in
discriminating multisets, so {a, a, b} and {a, b, a} denote the same
multiset

And later in the page :

Inclusion: A is included in B, denoted A ⊆ B, if

$\forall x\in U, m_A(x)\le m_B(x)$

Therefore seen as sets then $A=B$ and seens has multisets since $m_A(2)=1\le m_B(2)=2$ and other multiplicities are the same, then $A\subset B$ and $A\neq B$.

Title: A={1,2,3},B={1,2,2,3}. Is B⊆A? And is B⊂A?

Body: I was thinking yes to the first question and no to the second. They are equal sets right even though B has duplicate elements.</p>


问:
How can I describe a linear equation that becomes sinusoidal after a certain point?
For example, take $y = x$, but at $x \geq 5$, $y$ now equals $\sin x + 5$.
So for all $x$ values $\lt 5$, $y$ has linear relationship, but at $\geq 5$, $y$ now has sinusoidal relationship.
Is there a concise way to express this, perhaps a single equation?

答:
This question makes me think of a related but slightly different kind of function. A function which gradually/slowly changes behavior, instead of having this piecewise phenomenon.

The piecewise and discontinuous transition would be denoted as
$$ h(x) =
\begin{cases}
f(x)\quad\text{if}\quad x \leq x_0\\
g(x)\quad\text{if}\quad x > x_0\\
\end{cases}$$
A gradual shift from $f$ to $g$ can be written as $(1-t)\cdot f + t\cdot g$ with $t\in[0,1]$.
Therefore a function that gradually changes from $f$ after $x_0$ and becomes $g$ at $x_1 > x_0$ would be written
$$ h_{\mathrm{grad}}(x) =
\begin{cases}
f(x)\quad\text{if}\quad x \leq x_0\\
\\
\left(1 - \dfrac{x-x_0}{x_1-x_0}\right)\cdot f(x) + \left(\dfrac{x-x_0}{x_1-x_0}\right)\cdot g(x) \quad\text{if}\quad x_0 < x \leq x_1\\
\\
g(x)\quad\text{if}\quad x > x_1\\
\end{cases}$$
As an exemple, let us take $f(x)=x$ and $g(x) = \sin(x) + 5$ on the domain $[-5, 10]$.

The piecewise transition is pictured below:

The continuous/gradual transition is in green below:


问:
Does changing the size of a matrix, but not the number of elements, preserve rank?
Specifically I have a matrix $A \in \mathbb{C}^{n \times m}$ that I am reshaping (MATLAB) to a $\frac{n}{2} \times 2m $ matrix $A'$, where n is a multiple of 2.

Is the rank preserved? In limited testing in MATLAB it seems so, but I can't understand why. $\text{rank}\ A = m$, in my example if that matters.

Is this a change of Basis? I have seen this answer but it only concerns removing rows, not also adding them as columns. Does $A'$ still span a subspace (or the same space?) of $A'$?

I don't have a very deep knowledge of linear algebra so an intuitive answer, if it exists, would be welcome.

答: Yeah no this does not make sense at all. I was just looking at specific matrices that have this property and must've gotten confused..</p>


问:
Proving that an equation has exactly two solutions in the reals
Let $a_1,a_2,a_3,b_1,b_2,b_3\in \mathbb{R}$, $a_1,a_2,a_3>0$, $b_1<b_2<b_3$.

Prove that $\frac{a_1}{x-b_1}+\frac{a_2}{x-b_2}+\frac{a_3}{x-b_3} = 0$ has exactly two solutions in $\mathbb{R}$.

First, I showed that the solutions must be bounded in $[b_1,b_3]$.

Since there are 3 points that the function is undefined at, I couldn't use any theorem that requires the function to be continuous. I know that it's continuous between $(b_1,b_2)$ and $(b_2,b_3)$, so I was trying to use IVT, but it also didn't work.

Could you please give me a hint?

Body:
You need to use IVT in the two intervals $(b_1,b_2)$ and $(b_2, b_3)$ near the endpoints.

At $b_1^+$, $\frac{{a_1}}{x-b_1}+\frac{{a_2}}{x-b_2}+\frac{{a_3}}{x-b_3} \to \infty$.

That is because $\frac{{a_1}}{x-b_1}+\frac{{a_2}}{x-b_2}+\frac{{a_3}}{x-b_3}> \frac{{a_1}}{x-b_1}+\frac{{a_2}}{b-b_2}+\frac{{a_3}}{b-b_3} \to \infty$, where $b = \frac {b_1+b_2}2$.

Similarly at $b_2^-$, $\frac{{a_1}}{x-b_1}+\frac{{a_2}}{x-b_2}+\frac{{a_3}}{x-b_3} \to -\infty$.

By IVT, this shows that $\frac{{a_1}}{x-b_1}+\frac{{a_2}}{x-b_2}+\frac{{a_3}}{x-b_3} =0$ for some $x \in (b_1,b_2)$. The result for $(b_2, b_3)$ is similar.


问:
Recursive formula on Gram-Schmidt polynomials
Construct the sequence $\{p_n\}_{n\in \mathbb{N}_0}$ by applying the Gram-Schmidt procedure on $\{x^n\}_{n\in \mathbb{N}_0} \in L^2([0,1])$ whitout normalizing so that they are monic ($p_0=1$ , $p_1=x-1/2$ , $p_3=x^2-x+1/6$ , ...).

Prove that there exists sequences $a_n$, $b_n$ and $c_n$ such that $\forall n\ge 1$
$$p_{n+1} = \big(a_nx+b_n \big)p_n - c_np_{n-1}$$

Observation: we would expect $a_n=1$ for all $n$ since $p_{n+1}$ must be monic (this means that the coefficient of the largest exponent term is $1$).

Could you please help with this?

I was trying to prove it by induction but was not getting anywhere...

答:
I managed to come up with this argument

From the construction of our $p_i$ we can deduce that every polynomial of degree $m$ ($a_mx^m + \mbox{ ... } + a_1x + a_0$) is linear combination of the polynomials $p_0$, $p_1$, ... ,$p_m$, in fact $\forall k\in \{ 0$, $1$,... ,$m\}$
$$x^k= p_k(x) + \sum_{j=0}^{k-1}\frac{<x^k,p_j>}{<p_j,p_j>}p_j(x) ;$$
in particular $xp_m$ is linear combination of $p_0$, $p_1$, ... , $p_m$, $p_{m+1}$.

So, since $\langle p_j,p_i \rangle = 0$ when $i\not = j$, let $n$ e $m$ be two positive integers such that $m+1<n$, then
$$\langle xp_n, p_m \rangle = \int_0^1 xp_n(x) \cdot p_m(x)dx = \int_0^1 p_n \cdot xp_m dx = \langle p_n,xp_m \rangle = 0 ,$$
which implies that $xp_n$ is linear combination only of $p_{n+1}$, $p_n$ e $p_{n-1}$. Let $A_n$, $B_n$ and $C_n$ be 3 constants (depending on a fixed $n$) such that
$$xp_n = A_np_{n+1} + B_np_n + C_np_{n-1};$$
we can observe that $A_n \not = 0$, otherwise $xp_n$ would not have a term of degree $n+1$, so
$$p_{n+1} = \Big( \frac{1}{A_n}x - \frac{B_n}{A_n} \Big)p_n - \frac{C_n}{A_n}p_{n-1}.$$
This is exactly what we wanted since we can choose $a_n= \frac{1}{A_n}$, $b_n=-\frac{B_n}{A_n}$ and $c_n=\frac{C_n}{A_n}$.


问:
Prove that $\lim\limits_{x\to+\infty}f'(x)=0$

Give $f \in C^2(\mathbb{R})$. Prove that if
$\lim\limits_{x\to+\infty}f(x)=a \in \mathbb{R}$ and
$\lim\limits_{x\to+\infty}f''(x)=0$ then
$\lim\limits_{x\to+\infty}f'(x)=0$.

My attempt: Get $x_0 > 0$. By using the Taylor formula to degree 2 at $x_0$ we have
$$f(x)=f(x_0) + \frac{f'(x_0)}{1!}(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2 + \mathcal{O}((x-x_0)^2).$$
Then, if $x=x_0+\frac{1}{x_0}$ we have
$$f\left(x_0+\frac{1}{x_0}\right)=f(x_0)+\frac{f'(x_0)}{x_0} + \frac{f''(x_0)}{2x_0^2} + \mathcal{O}\left(\frac{1}{x_0^2}\right).$$
Let $x_0 \to +\infty$ we have
$$\lim\limits_{x\to+\infty}\frac{f'(x_0)}{x_0}=0.$$
So there are two case: $\lim\limits_{x_0\to+\infty}f'(x_0)=0$ or $\lim\limits_{x_0\to+\infty}f'(x_0)=+\infty$ ( $f'(x_0)$ goes to $+\infty$ slower than $x_0$).

Question: How can I show  $\lim\limits_{x_0\to+\infty}f'(x_0)=+\infty$ if my direction is correct? And if my direction is wrong, I hope you can hint the right way to me.

答: Try letting $x=x_0+1$ and using first degree Taylor polynomial with Lagrange remainder: $$f(x)=f(x_0)+f'(x_0)\cdot(x-x_0)+\dfrac{f''(c_{x,x_0})}{2}\cdot(x-x_0)^2, \ \ c_{x,x_0}\in(x,x_0)$$


答:
Obviously $x\ne b_1,b_2mb_3$ a simplification gives:
$$f(x)=a_1(x-b_2)(x-b_3)+a_2(x-b_1)(x-b_3)+a_3(x-b_1)(x-b_2)$$
So we get $$f(b_1)=a_3(b_1-b_2)(b_1-b_3)>0$$
$$f(b_2)=a_2(b_2-b_1)(b_1-b_3)<0$$
$$f(b_3)=a_3(b_3-b_1)(b_2-b_2)>0$$
By IVT $f(x)=0$ has two real roots one in $(b_1,b_2)$ and one in $(b_2,b_3)$. Effectively, the given eq. is a quadratic which has both rotts real.

Title: Proving that an equation has exactly two solutions in the reals

Body:
Let $a_1,a_2,a_3,b_1,b_2,b_3\in \mathbb{R}$, $a_1,a_2,a_3>0$, $b_1<b_2<b_3$.

Prove that $\frac{a_1}{x-b_1}+\frac{a_2}{x-b_2}+\frac{a_3}{x-b_3} = 0$ has exactly two solutions in $\mathbb{R}$.

First, I showed that the solutions must be bounded in $[b_1,b_3]$.

Since there are 3 points that the function is undefined at, I couldn't use any theorem that requires the function to be continuous. I know that it's continuous between $(b_1,b_2)$ and $(b_2,b_3)$, so I was trying to use IVT, but it also didn't work.

Could you please give me a hint?


问:
Condition for $k$ dimensional manifold
From Spivak's book chapter 5. A subset $M\subset \Bbb{R^n}$ is called a k-dimensional manifold if for every point $x\in M$ the following condition is satisfied : There is an open set $U$ containing $x$ and an open set $V\subset \Bbb{R^n}$ and a diffeomorphism $h : U \to V $ such that $$h(U\cap M)=V\cap(\Bbb{R^k}\times{0})=\{y\in V:y^{k+1}=\cdots y^n=0\}.$$

"In other words $U\cap M$ is "up to diffeomorphism" simply $\Bbb{R^k}\times {0}.$ From the figures in the book, it would seem $h(x)$ is always centered at the origin of the image $h(U\cap M).$ What confuses me most, is the term $\Bbb{R^k} \times \{0\}$. These are $k+1$ dimensional points. Does this definition by itself mean, that $h(U\cap M)$ is some sort of a plane where $h(x)$ lies in the origin? Because this would seem to be the case from given figures.

答:
First of all, Spivak's statement that $U\cap M$ is "up to diffeomorphism" simply $\Bbb{R^k}\times \{0\}$ is a bit too sloppy. At best $U\cap M$ is up to diffeomorphism an open subset of $\Bbb{R^k}\times \{0\}$, but even that formulation is misleading because it suggests that we have a concept for a map $U \cap M \to V \cap (\Bbb{R^k}\times \{0\})$ to be a diffeomorphism.

One can define such a concept, but in Spivak's sense a diffeomorphism is a certain  map between open subsets of $\mathbb R^n$ - and nothing else. What is meant is that $M$ can be "flattened" locally via diffeomorphisms on open subsets of the ambient $\mathbb R^n$. In fact, $\Bbb{R^k}\times \{0\}$ is a $k$-dimensional subspace of $\mathbb R^n$ and thus a special $k$-plane in $\mathbb R^n$ (note that arbitrary $k$-planes need not contain the origin). It is also important to observe that $h : U \cap M \to V \cap  (\Bbb{R^k}\times \{0\})$ is a homeomorphism, and since $\Bbb{R^k}\times \{0\}$ is canonically homeomorphic to $\mathbb R^k$ we see that $U \cap M$ is homeomorphic to an open subset of $\mathbb R^k$. This means that topologically $M$ looks locally like  $\mathbb R^k$.

The idea for calling $M \subset \mathbb R^n$ a $k$-manifold is that $M$ looks locally like  $\mathbb R^k$ and additionally must be "smooth" without corners, edges etc. For example the subset $M  = \{(x,y) \in \mathbb R^2 \mid x= 0 , y \ge 0 \text{ or } y = 0, x \ge 0\}$ is homeomorphic to the real line $\mathbb R$, but it has a corner at the origin: No diffeomorphisms $h : U \to V$ on an open neigborhood $U$ of $0$ can have the property that $h(U \cap M) \subset \mathbb R \times \{0\}$.

The above considerations show that a manifold $M$ can be described locally by a curvilinear smooth coordinate system $\xi_1,\ldots, \xi_n$ such that for $ \xi = (\xi_1,\ldots, \xi_n)$ we have  $ \xi \in M$ iff $\xi_i = 0$ for $i > k$. Examples for curvilinear smooth coordinate systems are polar coordinates in the plane and spherical coordinates for three-dimensional space.

In Spivaks' figures we have in fact $h(x) = 0$. However, this is not required in the definition, $h(x)$ can be any point of $\mathbb R \times \{0\}$. Nevertheless we may assume w.l.o.g. that $h(x) = 0$: The translations $t_a : \mathbb R^n \to \mathbb R^n, t_a(\xi) = \xi + a$, are diffemorphisms, and if we start with $h : U \to V$, then the diffeomorphism $\tilde h : U \stackrel{h}{\to} V  \stackrel{T_{-h(x)}}{\to} T_{-h(x)}(V) = \tilde V$ has the desired property.

Title: Condition for $k$ dimensional manifold


问:
Calculate limit using Maclaurin series
$$\lim_{x \to 0}\frac{1-(\cos{x})^{\sin{x}}}{x^3}$$

So what I have tried is defining three functions, $f(x)=\cos{x}, g(x) = \sin{x}, h(x) = x^3$

What I did next was do Maclaurin expansion for three functions up until the third order:

$f(x) = 1 - \frac{x^2}{2} + o(x^3)$

$g(x) = x - \frac{x^3}{3} + o(x^3)$

$h(x) = x^3 + o(x^3)$

So when I plug in these values I get:

$$\lim_{x \to 0}\frac{1-(1-\frac{x^2}{2}+o(x^3))^{x-\frac{x^3}{3} + o(x^3)}}{x^3}$$

So after plugging in $x=0$ I get $\frac{0}{0}$ and when I try to apply L'Hôpital's rule I find myself in a situation where I need to calculate the derivative of the denominator several times (which would take too long). I'm positive that I'm correct until this point (I checked on wolframalpha, the result is 1/2).

Does anyone know how to proceed from here? Do I need to calculate the derivative of the denominator several times or is there a smarter way?

答:
I propose an alternative: two steps as follows.

1 - Using MacLaurin Series

We just act on the term $$(\sin(x))^{\cos(x)}$$ by transforming it into an exponential:

$$(\sin(x))^{\cos(x)} = e^{\sin(x)\ln(\cos(x))}$$

Hence it's Taylor-MacLaurin series is $$1 + \sin(x)\ln(\cos(x)) + \ldots$$

Hence we have

$$\lim_{x\to 0} \dfrac{1 - (1 + \sin(x)\ln(\cos(x)))}{ x^3} = \lim_{x\to 0} \dfrac{-\color{red}{\sin(x)}\ln(\cos(x))}{\color{red}{x}\cdot x^2}$$

Notice that the red term goes to $1$ for it is a well known limit.

2 - De L'Hôpital

Now $$\lim_{x\to 0} \dfrac{\ln(\cos(x))}{x^2}$$ can be treated with De L'Hôpital twice, which will give us

$$\lim_{x\to 0} \dfrac{\cos(x)}{2\cos(x) - 2x\sin(x)} = \dfrac{1}{2}$$

Title: Calculate limit using Maclaurin series


问:
Conditional Probability Problem

An insurance company examines its pool of auto insurance customers and gathers the following information:

(i)  All customers insure at least 1 car

(ii)  64% of all customers insure more than one car

(iii)  20% of the customers insure a sports car

(iv)  Of those customers who insure more than one car, 15% insure a sports car.

What is the probability that a randomly selected customer insures exactly one car, and that car is not a sports car?

Let's use the following variable definitions:

O= owns 1 car, O' = owns more than 1 car

S= sports car, S' = Not sports car.

N() = Cardinal Number of a set

From statements (i)-(iii), we get the following:  $N(O') = 64, N(O) = 36, N(S) = 20$

From statement (iv):  $\Pr(S \mid O')=15$

We are asked to find $\Pr(S' \mid O)$

By definition:

$Pr(S' \mid O) = \cfrac{\Pr(S' \cap O)}{\Pr(O)}=\cfrac{\Pr(S' \cap O)}{N(O)}\tag{1}$

Pr()=N() since this is a uniform distribution--I interpret this when it says "randomly selected"

Next, I did:

$N(S)=N(S \cap O') + N(S \cap O)\tag{2}$

$0.2 = .64*.15 + .36 * x$

$x=0.28$, but we want 1-x because we want S' in $\Pr(S' \cap O)$ which equals 0.72.

See diagram below:

So plugging back into (1):

$Pr(S' \mid O) = \cfrac{0.71*.36}{.36}=.71$

but the answer is .26.

I mainly wanted to know why equation (2) is wrong.  I know that is the crux of the problem.  Why can't I use that equation in this case?

Any help is appreciated.  Thank you.

答:
This is problem number 10 from the SOA/CAS Exam P practice problems, which you can download  here

For the solution look here

Consider the following events about a randomly selected auto insurance customer:
A = Customer insures more than one car
B = customer insures a sport car

We want to find the probability of the complement of A intersecting the complement of B (exactly one car, non-sports). But $P\left(\bar{A} \cap \bar{B}\right) = 1 - P\left(A \cup B\right)$

And, by the Additive Law, $P\left(A \cup B\right) = P\left(A\right) + P\left(B\right) - P\left(A \cap B\right)$

By the Multiplicative law $P\left(A \cap B\right) = P\left(B|A\right)P\left(A\right) = 0.15 \times 0.64 = 0.096$

It follows that $P\left(A\cup B\right) = 0.64 + 0.20 - 0.096 = 0.744$ and $P\left(\bar{A} \cap \bar{B}\right) = 1 - 0.744 = 0.256$

Title: Conditional Probability Problem


答: Since $H_5$ is normal in $G/H_3$ the subgroup $H$ is normal in $G$. Furthermore $H$ has order $15$ and therefore has a unique Sylow $5$-subgroup $U$. Let $V$ be any Sylow $5$-subgroup of $G$. Since all Sylow $p$-subgroups are conjugate to each other there is some $g\in G$ with $V=gUg^{-1}$. Since $H$ is normal in $G$ we get $V\leq H$, hence $V$ is a Sylow $5$-subgroup of $H$ and therefore $V=U$, so $G$ has a unique Sylow $5$-subgroup.

Title: Application of Sylow to a group of order 60

Body:
Let $G$ be a group with $|G|=60=2^2 \cdot 3 \cdot 5$ and assume $G$ has a unique $3$-Sylow subgroup $H_3$. The claim is that then $G$ has also a unique $5$-Sylow subgroup.

This is the procedure to prove it:

i) Show that $H_3$ is normal in $G$. (Follows directly since all Sylow subgroups are conjugates).

ii) Show that $G/H_3$ has a unique $5$-Sylow $H_5$ (Follows from $|Syl_5(G)|\equiv 1 \, mod \, 5$ and $|Syl_5(G)|$ divides $4$).

Let $H$ be the inverse image of $H_5$ by the projection $ \pi :G\longrightarrow G/H_3$. Now the goal is to note how many Sylow-$5$ subgroups there are in $H$ and use this to prove the claim. I don't understand this last step at all.
Thank you for any clarifications.


问:
maybe crazy question on asymptotic notation
I see two time complexity.

$O ( m (\log m + \log n))$ and $O(m \log m \log n)$.

Is it correct that according to logarithm rules tell these are equals A) from asymptotic view and B) from math point of view,
to each other or I'm in a big mistake forever.

答:
Dont know exactly what you are asking for but I think the following is true:

Consider $f(x)=O\bigl(m(\log m+\log n)\bigr)$. Assuming that this is not some multivariable function with respect to $m$ and $n$ and instead that, for instance $m=K$ where $K$ is some constant. Then we have that $$f(x)=O\bigl(\underbrace{K\cdot \log K}_\text{const. C=O(1)}+K\cdot \log n)\bigr)=O\bigl(O(1)+K\cdot \log n)\bigr)=O(K\cdot \log n)=O(\log n)$$ $\forall n>n_{0}$. That is $$O\bigl(K\log K+K\log n)\bigr)=O(\log n).$$


问:
There are 100 towns in some country and each two towns are connected by a one-way road.
There are 100 towns in some country and each two towns are connected by a one-way
road.
we are to Prove that one can change the direction of at most one road so that after that
each town will be reachable from any other one.

please help with the solution, thanks.

答: The towns and roads represent a tournament, a complete graph with every edge given a direction. Every tournament has a directed Hamiltonian path; if the end of that path is not already pointing to its start, simply flipping the edge between start and end will form a Hamiltonian cycle, whereupon all towns can be reached from all other towns by traversing the cycle.


答:
Since $\lim\limits_{x\to+\infty}f(x)=a$ there is M such that for x > M we have $|f(x+1) - f(x)| < \epsilon$.
 And from the mean value theorem, for every such x, there is y = y(x) between x and x+1 for which $f(x+1) - f(x) = f^{'}(y)$. In words, for every $\epsilon$ there is M such that for x > M in each interval [x,x+1] there is a point y where first derivative is less than $\epsilon$.

On the other hand, mean value theorem when applied to the first derivative gives  $f^{'}(x+h) - f^{'}(x) = f^{''}(z)$ for some z between x and x+h.

Since second derivative tends to zero, we may assume that for x > M and any h > -1 we have
$|f^{'}(x+h) - f^{'}(x)| < \epsilon$.
If we now put for x value of y, where derivative is less than $\epsilon$ we see that
$|f^{'}(y+h)| < 2\epsilon$ If you now let h very from -1 to 1 we get that derivative is no bigger than $2*\epsilon$ in the interval y-1, y+1. But we proved above that such y exists in every interval [x,x+1] for sufficiently large x. That means that derivative of f is less than $2*\epsilon$ for x > M, that is it tends to zero.

Title: Prove that $\lim\limits_{x\to+\infty}f'(x)=0$

Body:
Give $f \in C^2(\mathbb{R})$. Prove that if
$\lim\limits_{x\to+\infty}f(x)=a \in \mathbb{R}$ and
$\lim\limits_{x\to+\infty}f''(x)=0$ then
$\lim\limits_{x\to+\infty}f'(x)=0$.

My attempt: Get $x_0 > 0$. By using the Taylor formula to degree 2 at $x_0$ we have
$$f(x)=f(x_0) + \frac{f'(x_0)}{1!}(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2 + \mathcal{O}((x-x_0)^2).$$
Then, if $x=x_0+\frac{1}{x_0}$ we have
$$f\left(x_0+\frac{1}{x_0}\right)=f(x_0)+\frac{f'(x_0)}{x_0} + \frac{f''(x_0)}{2x_0^2} + \mathcal{O}\left(\frac{1}{x_0^2}\right).$$
Let $x_0 \to +\infty$ we have
$$\lim\limits_{x\to+\infty}\frac{f'(x_0)}{x_0}=0.$$
So there are two case: $\lim\limits_{x_0\to+\infty}f'(x_0)=0$ or $\lim\limits_{x_0\to+\infty}f'(x_0)=+\infty$ ( $f'(x_0)$ goes to $+\infty$ slower than $x_0$).

Question: How can I show  $\lim\limits_{x_0\to+\infty}f'(x_0)=+\infty$ if my direction is correct? And if my direction is wrong, I hope you can hint the right way to me.


问:
Prove that limit doesn't exists at all
I have a function $f(x)=\frac{1}{x}+\sin(\frac{1}{x})$ which I have proved that $\lim_{x \to0^+}f(x)=\infty$, and I need to prove that $\lim_{x \to0^+}f'(x) \ne -\infty$.

I have got that $f'(x)=-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}$

I am trying to prove that:

$$\lim_{x \to 0^+}-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}$$

Does not equal to $-\infty$ (or to prove that it doesn't exists at all).

I have tried to assume by contradiction that
$$\lim_{x \to 0^+}-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}=-\infty$$

So there exists $\delta_{1}>0$ such that for every $0<x<\delta_{1}$ we say that $$f(x)<\square $$

We choose $\delta=\delta_{1}$, and let $0<x<\delta$ so:

$$f(x)=-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}=-\frac{1}{x^2}(1+\cos(\frac{1}{x}))\le -\frac{2}{x^2}<\frac{2}{\delta^2}=k$$

I do not know what to choose instead of $\square$, and if what I did is correct.

Appreciate your help!

I have also tried to think about the continuity of $f$, and saying that it is not bounded so it can't be continuous, so the limit doesn't exists.

Thanks a lot!

答:
Scrap the whole proof by contradiction thing, it's not necessary.

Hint: Find the values of $x$ such that $\frac{1}{x} = 2k\pi$, and $\frac{1}{x} = (2k+1)\pi,$ where $k \in \mathbb{Z}$.

Edit: In fact, as Calvin Khor pointed out in the comments to this answer, you only need one of the above two sequences to show that $\lim_{x \to0^+}f'(x) \ne -\infty$, which was the original question.

Further edit: I think OP doesn't properly understand what is meant by $lim_{x \to c^+}g(x) = -\infty$. I also needed a sanity check, which is why I posted this question.

If $g(x)$ is a real function, then a definition of $\lim_{x \to c^+}g(x) = -\infty$ is:

for each $\gamma \in \mathbb{R}, \exists \delta > 0 $ such that $x \in (c,c+\delta) \implies g(x) < \gamma$.

So an equivalent definition of $\ \neg\left(\lim_{x \to c^+}g(x) = -\infty\ \right) $ is:

there exists $\beta\in\mathbb{R}$ such that for every $\delta>0$, there exists $x'\in(c,c+\delta)$ with $g(x')\ge \beta$

If no such $\beta$ exists, then $\ \lim_{x \to c^+}g(x) = -\infty\ .$

Note that $g$ need not be continuous at $c$ in either of my definitions.

Title: Prove that limit doesn't exists at all


答:
Because there exists a subsequence converging to $0$:

if you pick $\{x_k\}$ to be such that $x_k=\frac{1}{\pi + 2\pi k}$, then
$$f^\prime (x_k) = -(\pi + 2\pi k)^2 \Big(1+cos(\pi + 2\pi k)\Big) = 0,$$
since
$$cos(\pi + 2\pi k)=-1 \mbox{, } \forall k;$$
Also the sequence of $x_k$ converges to $0$ as you can observe letting $k\to +\infty$.

Title: Prove that limit doesn't exists at all

Body:
I have a function $f(x)=\frac{1}{x}+\sin(\frac{1}{x})$ which I have proved that $\lim_{x \to0^+}f(x)=\infty$, and I need to prove that $\lim_{x \to0^+}f'(x) \ne -\infty$.

I have got that $f'(x)=-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}$

I am trying to prove that:

$$\lim_{x \to 0^+}-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}$$

Does not equal to $-\infty$ (or to prove that it doesn't exists at all).

I have tried to assume by contradiction that
$$\lim_{x \to 0^+}-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}=-\infty$$

So there exists $\delta_{1}>0$ such that for every $0<x<\delta_{1}$ we say that $$f(x)<\square $$

We choose $\delta=\delta_{1}$, and let $0<x<\delta$ so:

$$f(x)=-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}=-\frac{1}{x^2}(1+\cos(\frac{1}{x}))\le -\frac{2}{x^2}<\frac{2}{\delta^2}=k$$

I do not know what to choose instead of $\square$, and if what I did is correct.

Appreciate your help!

I have also tried to think about the continuity of $f$, and saying that it is not bounded so it can't be continuous, so the limit doesn't exists.

Thanks a lot!


问:
Existence of linear transformation when $T:V\rightarrow V$
Let V be a finite linear space and U is a subspace of V. Prove that there exists a linear transformation $T:V\rightarrow V$ when $ker(T)=U$ and $ker(T) ∩ Im(T) = \{0\}$.

What I though about that could be a solution for this problem, is to prove that a zero transformation $T(v)=0$ exists. But I am not sure about it.

答:
The zero transformation has $\text{ker}(T)=V$ which is in general larger than $U$.

To approach the problem, think about a decomposition $V=U\oplus W$.
(Or, equivalently, choose a basis of $U$ and extend it to a basis of $V$.)

Recall that is suffices to define a linear transformation on $U$ and $W$ (or on the basis vectors) to determine a linear transformation on $V$ uniquely.


问:
Infinite series simplification
Is there a clean way to simplify the following series:

$$
   0.5^2(1-2*0.5^2)+0.5^3(1-2*0.5^3)+\cdots +0.5^k(1-2*0.5^k)
$$
where k = 1, 2, ... ∞

Using R led to the convergence point = 1/3

```<code>n = 10000
x = rep(0,n-1)
for (i in 2:n){
  print(i)
  x[i]= (0.5^as.numeric(i))*(1-0.5^as.numeric(i)-0.5^as.numeric(i))
}
sum(x)
</code>```

Title: Infinite series simplification

OwnerDisplayName: Xi'an


问:
Mnemonic to remembering that inverse limits are limits and direct limits are colimits
Sometime ago I had trouble remembering that right adjoints preserve limits (and so left adjoints preserve colimits). But ever since R. Vakil suggested that we use RAPL as a mnemonic to Right Adjoint Preserve Limits, it stuck on my mind and I never forgot. However I still have trouble remembering that

Inverse limits (= projective limit) are special cases of limits and direct limits (= inductive limit) are special cases of colimits. Also, the arrow (in $\varinjlim$) in direct limits go to the right and the arrow in inverse limits go to the left.

I would like to know you do to remember it.

答:
A key example of projective limits is given by diagrams $(\mathbb{N}, \leq)^{\mathrm{op}} \to \mathscr{C}$. Dually, key examples of inductive limits are given by $(\mathbb{N}, \leq) \to \mathscr{C}$. This gives an indication for all of those nasty conventions:

Inductive limits might remember us of induction which might remember us of $\mathbb{N}$. Arguably, the most natural ordering associated with $\mathbb{N}$ gives a natural diagram $(\mathbb{N}, \leq) \to \mathscr{C}$.

Projective limits are just the dual. In key examples, "projective" stands for "projection". One of the most important examples for $(\mathbb{N}, \leq)^{\mathrm{op}} \to \mathscr{C}$ is given by $\mathbb{Z}_p$. The diagram is given by $\mathbb{Z}/p^{i+1} \mathbb{Z} \to \mathbb{Z}/p^i \mathbb{Z}$ which are projections! More generally, you may refer to the $I$-adic completion of a ring.

The conventions $\varprojlim$ and $\varinjlim$ are also indicated by those diagrams. Of course, this depends on conventions again, but it at least seems vaguely natural to write $(\mathbb{N}, \leq)$ as $$1 \to 2 \to 3 \to 4 \to 5 \to \dots. $$
Dually, we get $\varprojlim$.

In general, however, I wholeheartedly agree with Ittay Weiss. I really wish that more people would also just write $\lim_I$ and $\mathrm{colim}_I$ and also just say (categorical) limit and colimit.


问:
Help to prove: $1-\tan^2(A) = \cos(2A)\sec^2(A)$
I have to prove this identity and I keep getting stuck because I don’t know which side to work with and how to continue.
I rewrote $\tan^2 = \sin^2/\cos^2$ but then from there I got stuck.
Any hint? Thank you!

答:
$$1 - \tan^2(A) = 1 - \dfrac{\sin^2(A)}{\cos^2(A)} = \dfrac{\cos^2(A) - \sin^2(A)}{\cos^2(A)}$$

Now we use the trigonometric identity: $$\cos^2(A) - \sin^2(A) = \cos(2A)$$ to get

$$\dfrac{\cos(2A)}{\cos^2(A)}$$

But this is nothing but
$$\cos(2A)\sec^2(A)$$

Because of the definition of the secant: $$\sec(A) = \dfrac{1}{\cos(A)}$$


问:
Partition the remaining rectangle into equal parts.
There’s a rectangle. A small rectangle is cut from the bigger rectangle ( not necessarily from the center). How will you partition the original rectangle after removing the cut such that the remaining areas of both the partitions are the same?

Here I was thinking to partition it by extending the diagonal line of the cut part. Is it correct and Is it the best approach ?

答:
Observe that any line through the center of a rectangle bisects its area.

Then the line joining the centers of two rectangles partitions the remaining areas equally.


答:
Which diagonal would you want to cut along? It wouldn't give an answer, let alone be the best approach.

Of course, "best" does not mean much in this context, but cutting along the line connecting the centres of the two rectangles would work and be the closest thing to a "best" approach I can think of.

To elaborate, it would among all the lines cutting the remaining area in half, it would be the only one to also cut both rectangles in half, provided they do not share their centres.
Should that happen then any line passing through that point would work.

Title: Partition the remaining rectangle into equal parts.

Body:
There’s a rectangle. A small rectangle is cut from the bigger rectangle ( not necessarily from the center). How will you partition the original rectangle after removing the cut such that the remaining areas of both the partitions are the same?

Here I was thinking to partition it by extending the diagonal line of the cut part. Is it correct and Is it the best approach ?


问:
Can an artificial neural network with only one hidden layer suffice for all applications?
I have heard that only a single layer is needed for an artificial neural network to fit any possible function (input to output).

Is this true and where is this shown?

If this is true, then what is the advantage of having multiple layers?

答:
Search for the subject: How many hidden layers should I use?

http://www.faqs.org/faqs/ai-faq/neural-nets/part3/


答:
If $U=V$, just take
$$T(v):=0,\ \forall v\in V.$$
Then you have
$$\ker(T)=V,\quad T(V)=\{0\},\quad\ker(T)\cap T(V)=\{0\}.$$
If $U\ne V$ then $\dim U=k<n=\dim V$. Let $\mathcal{B}_U=\{v_1,\ldots,v_k\}$ be a basis for $U$.

You can complete it and obtain a basis for $V$ with the form $\mathcal{B}_V=\{v_1,\ldots,v_k,v_{k+1},\ldots,v_n\}$. Then define
$$V\ni v=a_1v_1+\cdots+a_nv_n\longrightarrow T(v):=a_{k+1}v_{k+1}+\cdots+a_nv_n\in V.$$
Clearly, so defined, $T$ is linear. Now, for any $u\in U$ you have $T(u)=0$ and for any other $v\in V$ is $T(v)\ne0$ (remember that the vectors in the basis are independent). So,
$$\ker(T)=U,\quad T(V)=\mathrm{span}\{v_{k+1},\cdots,v_n\},\quad\ker(T)\cap T(V)=\{0\}.$$

Title: Existence of linear transformation when $T:V\rightarrow V$

Body:
Let V be a finite linear space and U is a subspace of V. Prove that there exists a linear transformation $T:V\rightarrow V$ when $ker(T)=U$ and $ker(T) ∩ Im(T) = \{0\}$.

What I though about that could be a solution for this problem, is to prove that a zero transformation $T(v)=0$ exists. But I am not sure about it.