答:
$\phi:V\to V^\ast$

$\phi(v)\in V^\ast$ means that $\phi(v):V\to F$

$\phi(v)(w)$ is just the value of $\phi(v)$ at $w$.

Title: notation question (bilinear form)

Body:
So I have to proof the following:
for a given Isomorphism $\phi : V\rightarrow V^*$ where $V^*$ is the dual space of $V$ show that $s_{\phi}(v,w)=\phi(v)(w)$ defines a non degenerate bilinear form.

My question : Does $\phi(v)(w)$ denote the map from $v$ to a linear function $w$? (in this case i had serious trubles in showing linearity in the second argument,really confusing.

Or

maybe it just means $\phi(v)$ times $w$ where $w$ is the scalar value ( we get $w$ by applying  $v$ in the linear function it is mapped to)

I just started today with dual spaces and try my best with the notation , but i couldn't figure it out , please if you have any idea please help me with the notation , i will solve the problem on my own.


问:
Probability of meeting using integral
Here is a famous question- Two people agree to meet sometime between 9am and 10am. Each picks a time uniformly and waits for 15 minutes. What is the probability that they meet?

I know that this can be easily tackled by drawing rectangles. The problem eventually boils down to finding the probability of the event $|X-Y|<0.25$ where X and Y are uniform random variables. I would like to understand how to proceed through integration. Here is what I have
$$P(|X-Y|<0.25) = 2P(X-Y<0.25) = 2\int P(X<0.25+y)f_Y(y)dy $$
I am confused what would be the limit of the integral. I have used convolution and symmetry, and assumed that $X$ and $Y$ are independent. Are the assumptions correct? How to proceed? Thanks.

答:
I would like to understand how to proceed through integration

$$1-2\int_0^{\frac{3}{4}}\Bigg[\int_{x+\frac{1}{4}}^1dy\Bigg]dx=\frac{7}{16}$$

ContentLicense: CC BY-SA 4.0

Title: Probability of meeting using integral


问:
Understanding the definition of primitive recursion.
I'm a little (truthfully really) lost with the definition of primitive recursion. Here is the definition for primitive recursive functions that we have in our course (translated from German):

The class of primitive recursive functions is the smallest class of functions of $\mathbb{N}^l \to \mathbb{N}, l \geq 0$, which:

contains the following basic functions:

a. The constant function $f : \mathbb{N}^k \to \mathbb{N}, f(n_1, ..., n_k) = c$;

b. The projection $\pi_i^k : \mathbb{N}^k \to \mathbb{N}, \pi_i^k(n_1, ..., n_k) = n_i$;

c. The successor function $\text{succ}:\mathbb{N} \to \mathbb{N}, \text{succ}(n) = n+1$;

And is closed under the following operations:

a. Composition: If $f_1, ..., f_m : \mathbb{N}^k \to \mathbb{N}, g : \mathbb{N}^m \to \mathbb{N}$ are primitive recursive, then $g \circ (f_1, ..., f_m). \mathbb{N}^k \to \mathbb{N}$, $(n_1, ..., n_k) \mapsto g(f_1(n_1, ..., n_k), ..., f_m(n_1, ..., n_k))$ is also primitive recursive;

b. Primitive Recursive: If $g : \mathbb{N}^k \to \mathbb{N}, h : \mathbb{N}^{k+2} \to \mathbb{N}$ are primitive recursive then so if $f : \mathbb{N}^{k+1} \to \mathbb{N}$ with $f(0, n_1, ..., n_k) = g(n_1, ..., n_k)$ and $f(n + 1, n_1, ..., n_k) = h(n, f(n,n_1,...,n_k),n_1,...,n_k)$.

I don't understand what is happening in definition 2.b.

Why do $f,g,h$ have different domains?

What does $f(n + 1, n_1, ..., n_k) = h(n, f(n,n_1,...,n_k),n_1,...,n_k)$ do?

What is the goal with this method?

答:
I think in (2b) it will help to consider the case $k=1$. It's the simplest case where we really see all the parts of (2b) play nontrivial roles.

The simplest point is the role of $g$: this provides us with our "starting values." Intuitively when we define $f$ we're given $f(0, n)$ right off the bat (no recursion needed). This is what $g$ provides us.

The role of $h$ is a bit more complicated. Intuitively, $h$ tells us how to define $f(a,b)$ in terms of "previous values" of $f$. The details are a little subtle. In order to figure out what $f(a+1,b)$ should be, $h$ is allowed three pieces of information: what $a$ is, what $b$ is, and what the "previous value" $f(a,b)$ is. It doesn't have to use all this information, but that's what it's allowed in principle.

(If you like you could imagine $h$ being given $a+1$, $b$, and $f(a,b)$ instead - this may feel a bit more natural since it's "seeing" the current inputs. But since we can figure out $a$ from $a+1$ and vice versa, these both give the same ultimate definition.)

To see this all in action, consider the following definition of multiplication $f(a,b)=a\cdot b$ in terms of addition:

$f(0,b)=0$.

$f(a+1,b)=a+f(a,b)$.

The first point tells us that we're looking at $$g: n\mapsto 0$$ (all our starting values are $0$). The second point tells us that $h$ should take in the data $$a, f(a,b),\mbox{ and }b$$ and output $a+f(a,b)$. So this tells us that $$h(x,y,z)=x+y.$$

Note that the third coordinate isn't playing a role in $h$. Primitive recursion does allow the "next-step-provider" $h$ to see both inputs and the previous value, but we don't need to use that information. In most natural examples I think we don't in fact need that.

Finally, it may also help to go in the opposite direction: given a $g$ and $h$, try to compute the first few values of the resulting $f$. For example, try $$g(0,b)=b,\quad h(x,y,z)=xyz.$$ What is $f(2,3)$?


答:
I think, a simple example explains best what is happening: Let's take a look at the factorial function $f(n)=n!$. It is defined by

$f(0)=1$

$\forall n:f(n+1)=n\cdot f(n)$

Here, $h(n, f(n))=n\cdot f(n)$. So $h$ is the function that defines how you calculate the next value from the previous values.

The $f(n+1,n_1,\ldots,n_k)=h(n,f(n,n_1,\ldots,n_k),n_1,\ldots,n_k)$ just generalizes this concept to higher dimensions.

Title: Understanding the definition of primitive recursion.

Body:
I'm a little (truthfully really) lost with the definition of primitive recursion. Here is the definition for primitive recursive functions that we have in our course (translated from German):

The class of primitive recursive functions is the smallest class of functions of $\mathbb{N}^l \to \mathbb{N}, l \geq 0$, which:

contains the following basic functions:

a. The constant function $f : \mathbb{N}^k \to \mathbb{N}, f(n_1, ..., n_k) = c$;

b. The projection $\pi_i^k : \mathbb{N}^k \to \mathbb{N}, \pi_i^k(n_1, ..., n_k) = n_i$;

c. The successor function $\text{succ}:\mathbb{N} \to \mathbb{N}, \text{succ}(n) = n+1$;

And is closed under the following operations:

a. Composition: If $f_1, ..., f_m : \mathbb{N}^k \to \mathbb{N}, g : \mathbb{N}^m \to \mathbb{N}$ are primitive recursive, then $g \circ (f_1, ..., f_m). \mathbb{N}^k \to \mathbb{N}$, $(n_1, ..., n_k) \mapsto g(f_1(n_1, ..., n_k), ..., f_m(n_1, ..., n_k))$ is also primitive recursive;

b. Primitive Recursive: If $g : \mathbb{N}^k \to \mathbb{N}, h : \mathbb{N}^{k+2} \to \mathbb{N}$ are primitive recursive then so if $f : \mathbb{N}^{k+1} \to \mathbb{N}$ with $f(0, n_1, ..., n_k) = g(n_1, ..., n_k)$ and $f(n + 1, n_1, ..., n_k) = h(n, f(n,n_1,...,n_k),n_1,...,n_k)$.

I don't understand what is happening in definition 2.b.

Why do $f,g,h$ have different domains?

What does $f(n + 1, n_1, ..., n_k) = h(n, f(n,n_1,...,n_k),n_1,...,n_k)$ do?

What is the goal with this method?


问:
Graph Theory adjacency matrix
Is it possible for a graph to exist that meets these conditions?

For a Graph G, the adjacency matrix has all $1$’s in the first row and all $0$’s in the second row.

What I think: Such a graph cannot exist because if there were $1$’s in the first row, it means that v1 is connected to every single vertex in graph G. However, the second row being all $0$’s indicates that there is no vertex connected to v2, which is a contradiction?

Am I correct?

Also another question, with an adjacency matrix does a loop = 2 in the matrix?

答:
Also another question, with an adjacacny matrix does a loop = 2 in the matrix?

This is a matter of definition, and as such, we get to choose.  The obvious possibilities are $1$ or $2$.  So which is best?

In this case, there is a good reason for writing $1$ for a loop in the adjacency matrix.  Namely, provided we write $1$ for loops in the adjacency matrix,  if $A=(A[i,j])$ is the adjacency matrix of a graph, then $A^k[i,j]$ is the number of walks from $i$ to $j$ of length $k$.

However, for the vertex's degree, it can be better to count a loop as contributing $2$, otherwise the Handshaking Lemma doesn't work (or, at least, would need adjusting).

Title: Graph Theory adjacency matrix

ContentLicense: CC BY-SA 3.0


问:
Prove that $ o( g) =m$ iff is a multiplication of $ \frac{n}{m}$ disjoint cycles.
The problem:

Given a group $G$ or order $n$, and a Cayley embedding $\phi \ :\ G\to S_{n}$. Prove that some $g\in G$ is of order $m$ iff $\phi ( g)$ is a multiplication of $\frac{n}{m}$  disjoint cycles of length $m$.

I was able to prove that if $\phi ( g) \ $ is s a multiplication of $\displaystyle \frac{n}{m}$ disjoint cycles of order $m$, then the order of $g$ is $m$. This was fairly straightforward given the fact that:

The order of a multiplication of disjoint cycles is the $lcm$ of the length of the cycles.

In our case the $lcm$ is obviously $m$, so it was easy to prove from here that the order of $g$ is $m$.

The other direction, meaning if the order of $g$ is $m$, then $\phi ( g)$ is a multiplication of $\frac{n}{m}$ disjoint cycles of length $m$, I'm struggling to prove it.

I also want to admit that my understanding of Cayley embeddings, and Cayley's theorem in general, is very poor. What I know is simply that it's a homomorphism and also injective, not much more.

Any help?

答:
If $g$ has order $m$, then the orbit of $e$ under multiplication by $g$ is
$$e\mapsto g\mapsto g^2\mapsto g^3\mapsto\cdots\mapsto g^{m-1}\mapsto g^m=e.$$
That is, the cycle that contains $e$ is $(e,g,g^2,\ldots,g^{m-1})$.

Now, how about any other orbit/cycle? If we have $x\in G$, then the orbit of $x$ is given by $x\mapsto gx\mapsto g^2x\mapsto\cdots$ etc. What is the first repeat? If $g^ix = g^jx$, then $g^i=g^j$, so $i\equiv j\pmod{m}$. So the first repeat is $g^mx=x$. So the cycle/orbit of $x$ is $(x,gx,g^2x,\ldots,g^{m-1}x)$.


问:
How do I find the derivative of an integral
Find the derivative of $y$ with respect to $x$ if:

$$y=\int_0^{\ln(8x)}\sin (5e^t)\, dt$$
I don't understand how to approach this question as it has $\sin$ and exponentials combined.

答:
If you use the fundamental theorem of calculus, there is no need to do anything special with the "combination of sin and e". We can write $y$ in the form
$$
y = F(\ln(8x)) - F(0),
$$
where $F$ is a function for which $\frac{dF}{dx} = \sin(5e^x)$ (we don't have to say what the function $F$ "really is"). Using the chain rule, find $\frac{dy}{dx}$:
$$
\frac {dy}{dx} = \frac {d}{dx} [F(\ln(8x))] = \sin[5e^{[\ln(8x)]}] \cdot \frac{d}{dx} \ln(8x)
\\ = \sin[5 (8x)] \cdot \frac 1{x} = \frac{\sin(40x)}{x}.
$$


问:
Is its possible to modify the theorem like this let $P$ be a nonempty perfect set in $\mathbb{R}^K$ .Then $P$ is countable
I understand the Rudin theorem $2.43$ .

Suppose if we put uncountable instead of countable

i,e suppose $P$ is uncountable, and denote the points of $P$ by $\mathbf{x_1}, \mathbf{x_2}, \mathbf{x_3}, \ldots$

Then this will also  contradicts  $\bigcap_1^\infty K_n$ is nonempty
and   it will implies  $P$ is countable

My confusion : Is its possible to  modify the theorem like this  let $P$ be  a  nonempty perfect set in  $\mathbb{R}^K$ .Then $P$ is countable

Here is an outline of Rudin's proof(Theorem $2.43$):

Theorem 2.43 Let $P$ be a nonempty perfect set in $\mathbb{R}^k$. Then $P$ is uncountable.

Proof  Since $P$ has limit points, $P$ must be infinite. Suppose $P$ is countable, and denote the points of $P$ by $\mathbf{x_1}, \mathbf{x_2}, \mathbf{x_3}, \ldots$. We shall construct a sequence $\{V_{n}\}$ of neighborhoods as follows.

Let $V_1$ be any neighborhood of $\mathbf{x_1}$. If $V_1$ consists of all $y\in \mathbb{R}^k$ such that $|y−x_1|<r$, the closure $\overline{V_1}$ of $V_1$ is the set of all $y\in \mathbb{R}^k$ such that $|y−x_1|≤r$.

Suppose $V_n$ has been constructed, so that $V_n\cap P$ is not empty. Since every point of $P$ is a limit point of $P$, there is a neighborhood $V_{n+1}$ such that (i) $\overline{V_{n+1}} \subset V_n$, (ii) $x_n\notin \overline{V_{n+1}}$, (iii) $V_{n+1}\cap P$ is not empty. By (iii), $V_{n+1}$ satisfies our induction hypothesis, and the construction can proceed.

Put $K_n=\overline{V_n}\cap P$. Since $\overline{V_n}$ is closed and bounded, $\overline{V_n}$ is compact. Since $\mathbf{x_{n}}\notin K_{n+1}$, no point of $P$ lies in $\cap_1^\infty K_n$. Since $K_{n}\subset P$, this implies that $\cap_1^\infty K_n$ is empty. But each $K_n$ is nonempty, by (iii), and $K_n\supset K_{n+1}$, by (i); this contradicts the Corollary to Theorem 2.36.

答: If $P$ is uncountable you can't enumerate its points as ${\bf x_1, x_2,x_3,...}$ - the argument above requires that every point in $P$ is one of the ${\bf x_n}$s, and that means $P$ has to be countable.


问:
Which comes first?, the distribution $P_X$ or the measure $\text{Prob}$?
I'm currently re-learning probability through measure theory, and I came across these definitions:

A probability space is a measurable space $(\Omega, \mathcal{F})$ with a $\sigma$-additive measure $\text{Prob}\colon\mathcal{F}\to[0,+\infty)$ that obeys $\text{Prob}(\Omega) = 1$.

The distribution of a random variable $X\colon\Omega\to\mathbb{R}$ is the measure in $\mathbb{R}$ given by $P_X(A) = \text{Prob}(X\in A)$.

However, in practice, most probability calculations arise in the context of random variables. If we have e.g. $X \sim \text{Poisson}(\lambda)$, then we usually do computations like $\text{Prob}(X = k)$ by using the distribution of $X$ (and the fact that it has a density).

$$\text{Prob}(X=k) = p_X(k) = e^{-\lambda}\frac{\lambda^k}{k!}$$

My question is: what is $\text{Prob}$ here? (knowing that the sample space is $\Omega = \mathbb{N}_0$), it feels as if we are using the distributions of random variables to define $\text{Prob}$ instead of computing it. We almost never specify the sample space when we say things like $X\sim \text{Poisson}(\lambda)$, or any distribution for that matter. Is there a standard for choosing e.g. the counting measure in finite spaces and some other measure for the continuous case?

答:
In practice the underlying probability space is abstract, and has the features that we need to define the random variables that we would like to define. It is a separate mathematical task, which most users of probability theory do not concern themselves with, to show that a suitable probability space for a given situation actually exists. For example, it's not immediately obvious how to construct a probability space on which a sequence of independent continuously distributed variables can be defined, but it can be done, and in practice most people take this fact for granted.

Only very occasionally do we need to grapple with the possibility that no such space exists (for example, this derails any attempt to classically define white noise).


问:
Math school online?
As a hobbyist programmer with little to no math skills (no science skills either, thank you religious schools) I am finding math encroaching on my life more and more often. I have always laughed and said that I'm just not good at math because my brain doesn't work like that.

My younger brother challenged me recently claiming that it wasn't a lack of ability that kept higher math out of my hands but a lack of interest. Because my current realm of studies is really starting to delve into 3D modeling, programming for web, mobile and desktop, I feel that a stronger math skill set would benefit me immensely.

But I don't know where to start...

I'm wondering if there is a math version similar to: http://www.code.org/khan-academy

Some place where I can go and start challenging my brain, increasing my math as often as I have time? Some place online? I would also appreciate any tips for books or other resources that would be recommended for someone wanting to ignite an interest in math.

To give a background, my highest level of math in school was pre-algebra in 7th grade. After that the schools my parents sent me to didn't teach higher math (nor evolution) so when I hit college, I had to get a tutor to pass (with a C) the College Level Algebra... I barely did it and it was so long ago I don't remember anything I learned.

I hope this is the right place to ask this... I stumbled across math.stackexchange by accident. Whether this question gets answered or not I will start browsing around and seeing what all there is...

答:
I don't know anything about programming but you mentioned "3D" and to me that screams Linear Algebra. However, you only mention doing Gr.7 level math so you have lots of work to do. When growing up I learned a lot from the site Purple Math. It seems to have a lot more topics now then when I used to use it but I remember it being quite good.

Recently I have also come across Basic Mathematics by Serge Lange which seems to be a great book and prepares students for a first course in Calculus. I'm a fan of Lang's writing so I imagine (like most of his books) this one is very good too.

Here is a online lecture series for Linear Algebra by Strang: http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/

He uses his own text for the course.

Hope this helps in some way.

Title: Math school online?


问:
Is the matrix $A − 2I$ invertible?

Let $A$ be a $4 \times 4$ matrix with eigenvalues $1, 2, 3,$ and $4$. Is matrix $A − 2I_4$ invertible?

I tried to tackle this by constructing a matrix with eigenvalues $1, 2, 3,$ and $4$:
$$
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 2 & 0 & 0\\
0 & 0 & 3 & 0 \\
0 & 0 & 0 & 4 \\
\end{bmatrix}
\quad
$$
Now,

$$A − 2I_4 =
\begin{bmatrix}
-1 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 2 \\
\end{bmatrix}$$

Since the determinant of this matrix is $0$, it is not invertible. But this is a very specific case. How can I generalize it to all $4 \times 4$ matrices with eigenvalues $1, 2, 3,$ and $4$?

Any help is greatly appreciated!

答:
A scalar $\lambda$ is an eigenvalue of the $n\times n$ matrix $A$ if and only if there exists a nonzero vector $\mathbf{v}$ such that $A\mathbf{v}=\lambda\mathbf{v}$. That means that $(A-\lambda I_n)\mathbf{v}= A\mathbf{v}-\lambda\mathbf{v}=\mathbf{0}$.

That, in turn, means that the matrix $A-\lambda I_n$ has nontrivial nullspace.

So $\lambda$ is an eigenvalue of $A$ if and only if $A-\lambda I_n$ has nontrivial nullspace.

What do you know about the connection between the nullspace of a matrix and its invertibility?

Title: Is the matrix $A − 2I$ invertible?


问:
How can I see $e^z-1=ze^{z/2}\prod_{k=1}^{\infty}(1+\frac{z^2}{4\pi^2k^2})$?
I want to see why below equation holds.

$$
e^z-1=ze^{z/2}\prod_{k=1}^\infty \left(1+\frac{z^2}{4\pi^2k^2}\right)
$$

Just simply applying the Taylor series definition does not give a solution. What method can I use to see the statement is true?

答: The claim is $\sinh\frac{z}{2}=\frac{z}{2}\prod_k(1+\tfrac{z^2}{(2\pi k) ^2})$, or equivalently $\sin w=w\prod_k(1-\tfrac{w^2}{(k\pi)^2})$. This looks plausible given the roots of both sides and the expected small-$w$ asymptomatic behaviour, and the Weierstrass factorization theorem formalizes that intuition.


问:
Let $K \subseteq L \subseteq M$ be two field extensions then, $K \subseteq M$ is algebraic iff $K\subseteq L$ and $L\subseteq M$ are both algebraic.
Let $K\subseteq L\subseteq M$ be two field extensions then,
$K\subseteq M$ is algebraic, if and only if, $K\subseteq L$ and $L\subseteq M$ are both algebraic.

Please how can I prove the sufficient condition

答:
For one direction, suppose $K\subseteq M$ is an algebraic field extension. Then every element of $M$ satisfies a polynomial with coefficients in $K$; in particular, every element of $L\subseteq M$ does too, and so $K\subseteq L$ is an algebraic extension. Additionally, polynomials over $K$ can be considered as polynomials over $L$ via the inclusion $K\subseteq L$, so every element of $M$ hence satisfies a polynomial over $L$ and thus $L\subseteq M$ is an algebraic extension.

For the reverse direction, the main point is that, if $E\subseteq F$ is a field extension, then $f_1,\dots,f_n\in F$ are each algebraic over $E$ if and only if $[E(f_1,\dots,f_n):E]<\infty$. This can be proved by induction on $n$ and the tower law; I will leave it as an exercise for you.

Anyway, this fact makes the desired result easy to show. Suppose $K\subseteq L$ and $L\subseteq M$ are algebraic, and let $m\in M$. Then $m$ is algebraic over $L$, so there is a polynomial with coefficients in $L$ satisfied by $m$. Let $l_1,\dots,l_n$ be these coefficients. Since $L$ is algebraic over $K$, each $l_i$ is algebraic over $K$, so $[K(l_1,\dots,l_n):K]<\infty$ by the result mentioned above. Also, $m$ is algebraic over $K(l_1,\dots,l_n)$, and so, again by the result mentioned above, $$[K(l_1,\dots,l_n,m):K(l_1,\dots,l_n)]<\infty.$$ By the tower law, this means $[K(l_1,\dots,l_n,m):K]<\infty$, and so, by one final use of the result mentioned above, $m$ is algebraic over $K$.


问:
Finding the last 5 digits of $1! + 2\cdot 2! + \cdots + 24 \cdot 24!$
Find the last $5$ digits of $$1! + 2\cdot 2! + \cdots + 24 \cdot 24!$$

Body:
Hint:

$1! + 2\cdot 2! + \cdots + 24 \cdot 24! = (2-1)1! + (3-1) 2! + (4-1) 3! + ...+ (25-1)24!$

$ = 2! - 1! + 3! - 2! + 4! - 3! + ...+ 25! - 24! = 25! - 1!$


问:
Transformation of an equation with a fraction in the numerator of a fraction
Can someone explain the steps here -- how does it get the second part where it is equals to $20$? I was confused because of the fraction $1/2$ in the numerator.

I am studying some examples but I don't understand this one.

Body:
1) Given:

$$\dfrac{(1/2)x}{21}+ \dfrac {(1/2)x}{24} = 10$$

2) We have the same equation as:

$$\dfrac{0.5x}{21}+\dfrac{0.5x}{24} = 10$$

3) To get: $$\dfrac{x}{21}+\dfrac{x}{24} = 20$$

4) Simply multiply the entire equation in step 2) by 2:

$$2\left(\dfrac{0.5x}{21}+\dfrac{0.5x}{24} = 10\right)$$

5) We are left with:
$$\dfrac{x}{21}+\dfrac{x}{24} = 20$$


答:
Hint: since $5!=120$, all summand from $5\cdot 5!$ on have last digit $0$. Same goes for $10!=3628800$, which means that at least the last two digits of $10\cdot 10!$ (and all the following summands) are zeroes.

You can save a lot of time with other similar considerations. Take it from here.

Title: Finding the last 5 digits of $1! + 2\cdot 2! + \cdots + 24 \cdot 24!$

Body: Find the last $5$ digits of $$1! + 2\cdot 2! + \cdots + 24 \cdot 24!$$


问:
Show subset of $\ell^2$ is compact
How does one go about showing if the following subsets of $\ell^2$ are compact or not:

$$\{a\in \ell^2 : \|a\|_2=1\}$$

$$\{a\in \ell^2 : |a_k|\leq \frac{1}{k} \forall k\in\mathbb{N}\}$$

I thought that a subset being comapct meant that every sequence $\{x_n\}$ in the subset contains a subsequence that converges to an element in the metric space. But for example the first sequence when $a$ belongs to $\ell^2$ and it's norm is less than one isn't it 'trivial' that the sequence would converge to an element of $\ell^2$? For the second sequence I suppose that $k=0$ which would make the norm tend to infinity so that it's not compact in $\ell^2$?

答:
The first one is not compact since it contains the subset $A=\{\mathbf e_n \in \ell^2 : n\in\mathbb N \} $, where
$$\mathbf e_n(k) =\begin{cases} 1 \text{ if } n=k \\ 0 \text{ if } n\neq k \end{cases}$$
That subset $A$ does not have any limit point, because the distance between two (distinct) elements of $A$ is $\sqrt 2$. That is: there existe a $\varepsilon >0$ such that for every $n\in\mathbb N$, $d(x_n, n_{n+1})\geq \varepsilon$, which is the negation of a sequence being Cauchy. So that, any sequence in $A$ cannot be Cauchy, hence cannot converge.

Title: Show subset of $\ell^2$ is compact


答: We can prove by induction $\sum_{k=1}^nk!k=(n+1)!-1$, so the sum is $25!-1$. Since $2^{22}5^6|25!$, the last $6$ digits of $25!-1$, never mind $5$, are all $9$s.


问:
Setting up exponential equation
I want to devise a function $f(x)$ such that $f(4) = 160, f(16)=80, f(28)=40$ and $f(40)=20$, as well as all the other values in between $4-40$. As in, the numbers between $4-12$ have a proportional value between $160-80$. Numbers between $28-40$ have a proportional value between $40-20$ etc.

The values on the left go up linearly and the values on the right go down exponentially.

I hope this makes sense. But would appreciate any help. Thanks

答:
Since your $x$-values are spaced linearly and your $y$-values are spaced exponentially, we can find an exponential function based solely on the first two points and it will hold for all of them.  Let $f(x) = a(b)^x$ for unknown $a,b$.  Based on your first two points, we have
\begin{align}
160 &= a(b)^{4}\\
80 &= a(b)^{16}.
\end{align}
Dividing the first equation by the second lets us eliminate $a$:
$$2 = b^{-12} \implies b = 2^{-1/12}.$$

Plugging this back in to either equation (I'll use the second):
$$80 = a(2^{-1/12})^16) = a(2)^{-16/12} = a(2^{-4/3}) \implies a = 80(2)^{4/3}.$$

Putting it all together gives us $$f(x) = 80(2)^{4/3}(2^{-1/12})^{x},$$
which if we want we can rewrite as $$f(x) = 80(2)^{\frac{4}{3} - \frac{x}{12}} = 80(2)^{\frac{16-x}{12}}.$$

If for some reason you wanted to use base $e$ instead this would become $$\ f(x) = 80\left(2\right)^{\frac{4}{3}}e^{\ln\left(2^{-\frac{1}{12}}\right)x}.$$

Title: Setting up exponential equation


问:
Limit of the inverse of a sum
So I cant really wrap my head around this limit. It is from https://arxiv.org/abs/1209.2300. Let $N < \infty$, I want to show that:
$$
\lim_{t\rightarrow \infty}\sum_i^N\left(\sum_j^N e^{-tf(i,j)}\right)^{-1}=N
$$
knowing that $0\leq f(i,j)<\infty \quad \forall i,j$

I understand that $\lim_{t\rightarrow \infty} e^{-t} = 0$ but I have trouble justifying that $\lim_{t\rightarrow \infty}\left(\sum_i^N e^{-tf(i)}\right)^{-1}$ is not even undefined.

答:
Let $i\in[\![1,N]\!]$, then $f(i,j)>0$ for $j\neq i$ thus $\lim\limits_{t\rightarrow +\infty}e^{-tf(i,j)}=0$. We then have
$$ \sum_{j=1}^N e^{-tf(i,j)}=1+\sum_{j\neq i}e^{-tf(i,j)}\underset{t\rightarrow +\infty}{\longrightarrow}1 $$
Summing this for $i\in[\![1,N]\!]$ we get $$ \lim\limits_{t\rightarrow +\infty}\sum_{i=1}^N\left(\sum_{j=1}^N e^{-tf(i,j)}\right)^{-1}=N $$

Title: Limit of the inverse of a sum


问:
Contour Integral: $\int^{\infty}_{0}(1+z^n)^{-1}dz$
I'm working through Priestley's Complex Analysis (really good book by the way) and this Ex 20.2:

Evaluate $\int^{\infty}_{0}(1+z^n)^{-1}dz$ round a suitable sector of angle $\frac{2\pi}{n}$ for $n=1,2,3,...$

Can someone advise what the contour may be? If we use a sector that includes zero, surely we'll have to indent it to avoid the singularity (since it isn't a simple pole on account of the negative power).

Thanks.

答:
The contour $C$ is a wedge-shaped contour of angle $2 \pi/n$, as you state, with respect to the positive real axis.  This contour may be broken into 3 pieces:

$$\oint_C \frac{dz}{1+z^n} = \int_0^R \frac{dx}{1+x^n} + i R \int_0^{2 \pi/n} d\phi \, e^{i \phi} \frac{1}{1+R^n e^{i n \phi}} - e^{i 2 \pi/n} \int_0^R \frac{dx}{1+x^n} $$

The second integral vanishes in the limit as $R \to \infty$; in fact, it vanishes as $1/R^{n-1}$ in this limit.  The rest is equal to $i 2 \pi$ times the residue at the pole at $z=e^{i \pi/n}$; note that this pole is interior to $C$ and therefore no further deformation of $C$ is necessary.

The residue theorem then implies

$$\left ( 1-e^{i 2 \pi/n}\right) \int_0^{\infty} \frac{dx}{1+x^n} = \frac{i 2 \pi}{n e^{i \pi (n-1)/n}}$$

The final result is

$$\int_0^{\infty} \frac{dx}{1+x^n} = \frac{\pi}{n \sin{(\pi/n)}}$$


问:
Are the elements of a Lie algebra separated by its finite-dimensional representations?
Let $\mathbb{k}$ be some field and let $\mathfrak{g}$ be a $\mathbb{k}$-Lie algebra.

Does there exist for every nonzero element $x$ of $\mathfrak{g}$ a finite-dimensional representation of $\mathfrak{g}$ on which $x$ acts nonzero?

In other words, do the finite-dimensional representations of $\mathfrak{g}$ separate the elements of $\mathfrak{g}$?

If $\mathfrak{g}$ is finite-dimensional, then this is true by Ado’s theorem.

As pointed out in an answer to a similar question, the finite-dimensional representations of $\mathfrak{g}$ even separate the points in the universal enveloping algebra $\operatorname{U}(\mathfrak{g})$ if $\mathfrak{g}$ is finite-dimensional and $\mathbb{k}$ is of characteristic zero.
(This seems to be Theorem 2.5.7 in Dixmier’s Enveloping Algebras.)

Both of the above arguments show even stronger assertions, but also need some additional assumptions and quite a bit of work.

答:
There are many possible answers. <s>Here's one</s> Edit: Here are a few:

Consider the Lie algebra (over a fixed field of characteristic zero) with presentation
$$\mathfrak{g}=\langle x,y,z\mid [x,y]=y,\;[y,z]=z\rangle.$$

(1) It's easy to check that $f(z)=0$ for every finite-dimensional representation.

(2) However, $z\neq 0$. This is because this is by definition amalgam of two 2-dimensional Lie algebras $\langle x,y\mid [x,y]=y\rangle$ and $\langle y,z\mid [y,z]=z\rangle$ the common 1-dimensional subalgebra $Ky$, and it's known (see Encyclopedia of Math.) that subalgebras embed into their amalgam.

For (1), it's a simple consequence of the study of finite-dimensional representations of the 2-dimensional non-abelian Lie algebra $\langle x,y\mid [x,y]=y\rangle$, which we can assume to be over an algebraically closed field. Every such representation maps $y$ to a nilpotent matrix. Now consider a finite-dimensional representation of  $\mathfrak{g}$, mapping $x,y,z$ to $X,Y,Z$.. Using the first subalgebra, $Y$ is nilpotent. Also, we can make the second subalgebra act as upper triangular matrices, and $Z$ is nilpotent. So both $Y,Z$ are strictly upper triangular, and $[Y,Z]=Z$ forces $Z=0$.

Similarly we can deduce that the analogue of Higman group, the Lie algebra
$$\langle x_0,x_1,x_2,x_3\mid [x_{i-1},x_i]=x_i: i\in\mathbf{Z}/4\mathbf{Z}\rangle$$
has no non-trivial finite-dimensional representation. I guess one can elaborate using amalgams (but haven't checked details) that it's not trivial (hence infinite-dimensional).

$\DeclareMathOperator\h{\mathfrak{h}}$Here's now an example that is completely self-contained.

Consider the Lie algebra $\h$ with basis $u$, $(e_n)_{n\in\mathbf{Z}}$, law $[e_i,e_j]=(i-j)e_{i+j}$, $[u,e_i]=ie_i$, over a field $K$ of characteristic zero.

I claim that every finite-dimensional representation of $\h$ kills all $e_i$. Indeed, consider operators $U$, $E_n$ of a finite-dimensional vector space satisfying the same relations. Since $[U,E_n]=nE_n$, the $E_n$ are in distinct eigenspaces for $\mathrm{ad}(U)$, and hence the $KE_n$ generate their direct sum. Since the dimension is finite, there exists $n$ such that $E_n=0$. Then for $m\neq 2n$, $E_m=\frac{1}{2n-m}[E_n,E_{m-n}]=0$. In turn $E_{2n}=\frac{1}{2-2n}[E_1,E_{2n-1}]=0$, so $E_m=0$ for all $m\in\mathbf{Z}$.

Actually, in this example the subalgebra $\mathfrak{r}$ already has the property that every finite-dimensional representation is trivial, but using a slightly more elaborate argument, which however works in arbitrary characteristic $\neq 2$.

Let $W_n$ be the subspace generated by $\{E_k:k\ge n\}$, and $W_\infty=\bigcap_n W_n$, so $W_\infty=W_n$ for large enough $n$, say $n\ge n_0$. Then $[E_n,W_\infty]=W_{\infty}$ for all $n$.

Suppose by contradiction that $W_\infty\neq 0$. Choose $n\ge n_0$. Take a block-diagonal decomposition of $E_n$. Then the sum $M$ of characteristic subspaces for nonzero eigenvalues of $\mathrm{ad}(E_n)$ consists of those matrices in this block decomposition all of whose diagonal blocks are zero. The condition $[E_n,W_\infty]=W_\infty$ forces $W_\infty\subset M$. In particular, $E_n$ has this form. But by definition $E_n$ is block-diagonal. So $E_n=0$, and this works for all $n\ge n_0$.

So $W_{\infty}=0$, that is, $E_n=0$ for all large $n$. Similarly $E_{-n}=0$ for all large $n$. Using commutators we deduce that $E_n=\frac{1}{n+2q}[E_{n+q},E_{-q}]=0$ (choosing $q$ such that $n+2q\neq 0$ in $K$).


答:
So $Left = 4 + 12(t-1)$ or $-8 + 12t$.  And $Right = 160\cdot (\frac 12)^{t-1}$ or $320\cdot (0.5)^t$.

It's probably easiest to leave it as that $x(t) = Left = 12t -8$ while $y(t) = Right = 360\cdot (0.5)^{t}$.

But if you must have $Right(Left) = $some function in terms of $Left$ then:

Solve $t$ in terms of $Left$ and plug it into the formula for $Right$.

$Left = -8 + 12t$

$12t = Left +8$

$t = \frac {Left + 8}{12}$

$Right = 360\cdot (0.5)^{t}= 360\cdot (0.5)^{\frac {Left+8}{12}}$

If you like you can shift the offset:

$0.5^{\frac {Left+8}{12} } = 0.5^{\frac {Left}{12}+\frac 23}=[\frac 12]^{\frac 23}0.5^{\frac {Left}{12}}$ s

$Right = [\frac {360}{2^{\frac 23}}]\cdot (0.5)^{\frac {Left}{12}}$

Title: Setting up exponential equation

Body:
I want to devise a function $f(x)$ such that $f(4) = 160, f(16)=80, f(28)=40$ and $f(40)=20$, as well as all the other values in between $4-40$. As in, the numbers between $4-12$ have a proportional value between $160-80$. Numbers between $28-40$ have a proportional value between $40-20$ etc.

The values on the left go up linearly and the values on the right go down exponentially.

I hope this makes sense. But would appreciate any help. Thanks


问:
Is my explanation to this derivative problem correct?
Basically, I am trying to prove that number 13 relates to the graph of B, but I was wondering if my explanation was correct. Here is the problem and here is what I wrote.

From the interval of - ∞ to zero, the function is decreasing, so f ' < 0. The interval from zero to + ∞, the function is increasing, so f ' > 0. The graph that corresponds to this is B because the interval from - ∞ to zero, f ' is below zero on the y-axis and the interval from zero to + ∞, f ' is above zero on the y-axis.

Also, for this problem, is there a more efficient way of saying where the function is increasing or decreasing, rather than saying "in this interval, the function is increasing/decreasing," since there are 5 intervals and doesn't give the points on the axis?

,

答:
Hint: Your explanation is fine, but you could additionally be more specific.

Problem (13): The function $y^\prime$ is linear, so it's a function of the form $y^\prime = kx+d$. Since $(0,0)$ is a point of $y^\prime$ we have $d=0$. We also see the slope $k$ is positive. The function $y$ looks like a parabola and it is going through $0$. Which functions $y, y^\prime$ could be proper candidates?

Problem (15): We see periodical functions $y$ and $y^\prime$ with $y^\prime(0)=0$. The function $y^\prime$ is $y$ shifted to the left by a quarter of a period, i.e. shifted by $\frac{\pi}{2}$. Which trigonometric functions $y, y^\prime$ could be proper candidates?


问:
Show $E[(M_n-M_m)M_k]=0$

Let $M_n$ be an $(\mathcal{F}_n)$ - martingale on $\left(\Omega,\mathcal{F},P\right)$, show that
$$E[(M_n-M_m)M_k]=0,\qquad 0\leq k\leq m \leq n.$$

I want to know if this is correct:

By the multi-step property of a martingale we have that
$$E[M_n|\mathcal{F}_m]=M_m\implies E[M_n-M_m|\mathcal{F}_m]=0,\qquad 0\leq m\leq n.$$

Multiplying my $M_k$ gives
$$M_kE[M_n-M_m|\mathcal{F}_m]=0.$$
Since $\mathcal{F}_k\subseteq \mathcal{F}_m$, then
$$E[(M_n-M_m)M_k|\mathcal{F}_m]=0$$
and taking expectations gives
$$E[E[(M_n-M_m)M_k|\mathcal{F}_m]]=E[(M_n-M_m)M_k]=0.$$

Is this correct?

答:
It looks fine to me.  The only criticism I'd make is that your last line has the wrong order on the equalities, and
$$E[(M_n-M_m)M_k]=E[E[(M_n-M_m)M_k|\mathcal{F}_m]]=0$$
reflects your logical argument better (to someone who knows the subject, your intent is clear, but to someone learning it, it might look like you're begging the question).

On a sidenote, I'm assuming your martingale is $L^2$.  Otherwise, there's a problem when you first write $E[(M_n-M_m)M_k|\mathcal{F}_m]=0$

Title: Show $E[(M_n-M_m)M_k]=0$


问:
Parametric CDF estimation for exponential distribution
Suppose we have a random sample $(X_{1}, \dots,X_{n})$, where $X_{i}$ follows an exponential distribution with parameter $\lambda$. Hence, we have the following:

$$F(x) = 1 - \exp(-\lambda x)$$

$$\mathbb{E}(X_{i}) = \dfrac{1}{\lambda}$$

$$\text{Var}(X_{i}) = \dfrac{1}{\lambda^{2}}$$

I know that the MLE estimator is $\hat{\lambda} = \dfrac{n}{\sum_{i=1}^{n}X_{i}}$.

I am interested in estimating the CDF at the point $x = 1$, hence what I have is

$$\hat{F(1)} = 1 - \exp(- \hat{\lambda})$$

I want to discover if it is biased or not:

\begin{align*}
\mathbb{E}\left( \hat{F}(1) \right) &= 1 - \mathbb{E}(\exp(- \hat{\lambda}))
\end{align*}

Any clues how to proceed from here?

答:
I want to discover if it is biased or not:

The question is only to detect if the estimator is biased or not. They do not ask you to explicitly calculate the expectation of your estimator. Thus the solution is very easy: using Jensen's inequality you get

$$\mathbb{E}[\hat{F(1)}]=\mathbb{E}\Bigg[1-e^{-\frac{1}{\overline{X}_n}}\Bigg]\ne 1-e^{-\frac{1}{\mathbb{E}[\overline{X}_n]}}=1-e^{-\lambda}$$

Thus your estimator is biased

EDIT: I supposed it was easy also to calculate the needed expectation explicitly passing by Gamma function but I was wrong...the integral can be solved numerically... thus the solution (IMHO) is the one I showed


问:
Cramer-Rao Lower Bound
Assume that $X_1,X_2,\ldots,X_N\sim N(\mu,2^2)$ and $Y_1,Y_2,\ldots,Y_M\sim N(0,\sigma^2)$.

a)Find the Cramer-Rao Lower Bound (CRLB) for the variance of the unbiased estimators of $\mu$.

b)Find the CRLB for the variances of the unbiased estimators of $\mu^2$.

c)Is the MLE, $\hat{\mu}$, a uniformly minimum variance unbiased estimator (UMVUE) of $\mu$?

so for part a) I got $\dfrac{4}{n}$  and for part b) I got $\dfrac{\sigma^2}{n}$. Are these answers correct? Just want to know if I'm on the right track.

Lastly for part C, can anyone give me some guidance on where to start? Kind of lost haha

答:
So what have the $Y_i$ variables got to do with it?

Anyway, your answer to part (a) looks fine. Your answer to part (b) does not. You should be able to derive part (b) from part (a). In particular, if the CRLB on $\theta$ is say $CRLB(\theta)$, and if $g(\theta)$ is some differentiable function of $\theta$ that we are interested in estimating, then, subject to some regularity conditions, the CRLB on $g(\theta)$ is:

$$\left(\frac{\partial g(\theta) }{\partial \theta }\right)^2   CRLB(\theta) $$

In this case, $g(\theta)$ = $\mu^2$, and $\theta$ is $\mu$, so the rest is left to u.

And since he has now solved it :) (see comment below), one can also check one's work on such things with a computer algebra system. For this problem, part (b) would simply be:

$$\text{CRLB}=\frac{1}{n *\text{FisherInformation}\left[\mu ^2,f\right]}$$

where FisherInformation is a mathStatica funkeh monkeh, and $f$ is the $N(\mu,\sigma^2)$ pdf, and which returns, as output:

$\frac{4 \mu ^2 \sigma ^2}{n}$

Title: Cramer-Rao Lower Bound


答:
I'll give you a track, I may edit my answer later and give a complete solution, but for now here's a hint :

1- Prove that for any $ \left(n,z\right)\in\mathbb{N}^{*}\times\mathbb{C} $, we have : $$ \sinh{z}=n\sinh{\left(\frac{z}{n}\right)}\prod_{k=1}^{n-1}{\left(1+\frac{\sinh^{2}{\left(\frac{z}{2n}\right)}}{\sin^{2}{\left(\frac{k\pi}{2n}\right)}}\right)} $$

2- Prove that the previous product does in deed converge to the infinite product of the limit of what's inside, meaning : $$ \prod_{k=1}^{n-1}{\left(1+\frac{\sinh^{2}{\left(\frac{z}{2n}\right)}}{\sin^{2}{\left(\frac{k\pi}{2n}\right)}}\right)}\underset{n\to +\infty}{\longrightarrow}\prod_{k=1}^{+\infty}{\left(1+\frac{z^{2}}{k^{2}\pi^{2}}\right)} $$

3- Conclude.

Title: How can I see $e^z-1=ze^{z/2}\prod_{k=1}^{\infty}(1+\frac{z^2}{4\pi^2k^2})$?

Body:
I want to see why below equation holds.

$$
e^z-1=ze^{z/2}\prod_{k=1}^\infty \left(1+\frac{z^2}{4\pi^2k^2}\right)
$$

Just simply applying the Taylor series definition does not give a solution. What method can I use to see the statement is true?


问:
Probability bound involving two random variables
I am reading a proof of a theorem which seems to use the following result while going from one step to the next step.

Let $X$ and $Y$ be two positive random variables and let $\delta>0$, then $P(X+Y>\delta)\leq P(X>\delta/2)+P(Y>\delta/2)$.

Does this inequality hold in general?

答: Yes. When $X+Y>\delta$, at least one of them must be greater than $\delta/2$, since otherwise their sum would be at most $\delta$. Therefore, whenever $X+Y>\delta$, we must have $X>\delta/2$ or $Y>\delta/2$. Thus $P(X+Y>\delta) \leq P(X>\delta/2 \lor Y>\delta/2) \leq P(X>\delta/2)+P(Y>\delta/2)$. Note that this only applies when $\delta$, $X$, and $Y$ are all positive.


问:
Kernel of R-module homomorphism inclusion
Let $f:M \rightarrow N$ be an R-module homomorphism with $X$ being a submodule of $M$ and $Y$ being a submodule of $N$.

Prove that $Kerf \subseteq f^{-1}(Y)$.

I've previously proven that $f(X)$ is a submodule of $N$ and that $f^{-1}(Y)$ is a submodule of $M$, but I really have no idea how to approach this proof and I would need some help. Thanks!

答:
Tbh this is a very simple problem and if you're struggling with it that's a sign you need to rethink how you approach proofs, and/or give more thought to the problem before giving up and asking for help. Let me give you a rundown of what your thought process should be on a problem like this.

you are supposed to prove an inclusion of sets $A\subseteq B$. So you should always start by assuming $x\in A$, applying the definition of whatever $A$ is, and trying to deduce $x\in B$.

in our case we are supposed to prove $\ker(f)\subseteq f^{-1}(Y)$. So suppose $x\in\ker(f)$. Now you should apply the definition of the kernel: by definition, $x\in\ker(f)$ means $f(x)=0_N$.

we want to prove that $x\in f^{-1}(Y)$, which means we want to prove that $f(x)\in Y$. But we know that $f(x)=0_N$, so we want to know that $0_N\in Y$. But this is part of the definition of $Y$ being a submodule of $N$.


问:
What are some examples of numbering systems for which it is easy to generalize permutation polynomials?
I'm writing an assignment on permutation polynomials (PP). I've already explored quite a few generalizations and characterization of PPs over $\mathbb{Z}_p$ for $p$ prime (and finite fields in general) and $\mathbb{Z}$. I'm looking for another numbering system to analyze, ideally one where it is easy to derive a general form for any PP. This was was easily accomplishable for $\mathbb{Z}$ using basic properties of the evaluation map and for $\mathbb{Z}_p$ using Lagrange interpolation + related techniques. I would like this numbering system to be a non-finite field or any commutative ring with unity. If anyone has any suggestions, it would be much appreciated.

答:
The following springs to mind.

Quadratic permutation polynomials on $\Bbb{Z}_m$.

Let $m>1$ be any integer. Consider the quadratic polynomial function
$$
f:\Bbb{Z}_m\to\Bbb{Z}_m, x\mapsto ax+bx^2.
$$
Prove the following (it's relatively easy, ask if you need a hint).

Lemma. If $\gcd(a,m)=1$ and $b$ is divisible by every prime factor of $m$, then
$f$ is a permutation.

The reason I recommend this is that such permutation polynomials are used heavily in the LTE standard as turbo code interleavers (the version of the standard that was finalized in 2009, updates pending, and eventually this part is likely to become obsolete). In other words, unless my information is "dated", chances are your cell phone is calculating such permutations a few million times per second. The version of LTE I recall specified a range of values for $m$, each divisible by a relatively high power of two, and an optimized $(a,b)$ pair for each such $m$. The reasons for selecting such permutations are a bit technical, but I think this application is too cool to pass.

The idea was introduced in

J. Sun and O. Y. Takeshita, “Interleavers for turbo codes using permutation polynomials over integer rings,” IEEE Trans. Inf. Theory, vol.51, no. 1, pp. 101–119, Jan. 2005.

This is behind IEEE paywall, but hopefully your institute has access. Probably any reference to 3GPP minutes and/or specs I used back in the day are outdated. When I worked for at the time largish cellular player, I studied the inverse permutations a bit more intensely :-)


问:
Maximize a quadratic convex function with a symmetric structure
Let $S = \begin{pmatrix}
A & B \\
B & A
\end{pmatrix} \in \mathbb{R}^{2n \times 2n}$ be a positive semi-definite matrix with matrices $A, B \in \mathbb{R}^{n \times n}$ being symmetric.
Let $y_{1}, y_{2} \in \{0,1\}^{n\times 1}$ be two binary vectors.

The optimization problem is to maximize the following quadratic convex function with binary constraints:
$$
\text{maximize}_{y_{1}, y_{2} \in \{0,1\}^{n\times 1}}
\bigl(\begin{smallmatrix}
y_1 \\
y_2
\end{smallmatrix}\bigr)^{\mathrm{T}}
S
\bigl(\begin{smallmatrix}
y_1 \\
y_2
\end{smallmatrix}\bigr),
$$
i.e., selecting a principal submatrix to maximize the sum of entry values.

I wonder if it is possible to show that an optimal solution has $y_1 = y_2$? Since the problem is so symmetric, this seems a bit intuitive to me.

答: The answer is no. As a counterexample, with $n=1$: take $A = 1, B = -1$. The optimal solutions are $(1,0),(0,1)$.  A similar phenomenon occurs in general with $A = I, B = -I$.


问:
Every holomorphic function defined by power series with radius of convergence $1$ must have a point on $|z|=1$ beyond which $f$ cannot be extended.
I am sorry if the title of this post is confusing, feel free to edit it. My question is as follows:

I am working on an exercise stated as follows:

Let $\mathcal{E}$ be the set of holomorphic functions $f(z)$ defined as power series $f(z)=\sum_{n=1}^{\infty}a_{n}z^{n}$ with radius of convergence equal to $1$. Denoting $C$ to the unit circle, and for $f\in\mathcal{E}$, we define $$S(f):=\{z\in C:\ f\ \text{cannot be extended as a holomorphic function in a neighborhood of}\ z\}.$$

Show that $S(f)$ is closed and non-empty.

For $z_{0}\in C$, find an example such that $S(f)=\{z_{0}\}$.

I found a similar post here: Holomorphic function $f$ in $D$ which is not holomorphically extendible across any point in $\partial D$.

I tried to mimic the proof in the above post. As $\overline{\mathbb{D}}$ is sequentially compact, for any $z\in C$, there exists a sequence $\{z_{n}\}_{n=1}^{\infty}$ in $\mathbb{D}$ that converges to $z$. We know recall

[Weierstrass Theorem on Zeros of Holomorphic Functions] Assume $V$ is a proper open set of the Riemann Sphere $\mathbb{C}_{\infty}$, and let $A\subset V$ be a set that has no accumulation point in $V$. Then, there exists a function $f$ holomorphic on $V$ such that the set of zeros $Z(f)$ is exactly $A$, counting the multiplicities.

In our case, $V=\mathbb{D}$ and $A=\{z_{n}\}_{n=1}^{\infty}$. Note that the limit point of $A$ is on $\partial D$, so $A$ has no accumulation point in $\mathbb{D}$. Hence, by the theorem, there exists a holomorphic function $g$ in $\mathbb{D}$ such that $g(z_{n})=0$ for all $z_{n}\in A$.

But then I don't know how to proceed. The original post said that since the zeros are isolated, then we cannot extend beyond $z\in C$. I don't quite follow this argument.

Also, my proof seems backward... As saulspatz suggested, $f$ is given, and I need to show that there exists $z\in C$ such that $f$ cannot be analytically continued.

I don't quite know how to prove the set is closed.

I also don't really know how to find the example for $2$. I took a look at https://mathoverflow.net/questions/10831/example-of-continuous-function-that-is-analytic-on-the-interior-but-cannot-be-an, but did not find a proper example.

Thanks for any help!

Edit 1: Proof of Q1 part 1

Thanks to Conrad's suggestion, we show $C\setminus S(f)$ open. To show this, we show that for any $z\in C\setminus S(f)$, there exists $r=r(z)>0$ such that $D(z,r)\cap C\subset C\setminus S(f)$.

Indeed, let $z\in C\setminus S(f)$, i.e. $f$ can be analytically continued around $z$. Hence, there exists $r=r(z)>0$ such that there exists a holomorphic function $g$ on $D(z,r)$ such that $g=f$ on $D(z,r)\cap D$.

Let $z_{0}\in D(z,r)\cap C$, take $r_{0}>0$ such that $D(z_{0}, r_{0})\subsetneq D(z,r)$. Then, it is clear that $g$ is holomorphic on $D(z_{0}, r_{0})$ and $f=g$ on $D(z_{0}, r_{0})\cap\mathbb{D}\cap D(z,r)\cap\mathbb{D}$.

Hence, $f$ has an analytic continuation around $z_{0}$. This means that $z_{0}\in C\setminus S(f)\implies D(z,r)\cap C\subset C\setminus S(f)$.

We are done

Edit 2: Example for Q2.

Taking $z_{0}\in C$, i.e. $|z_{0}|=1$ and consider the series $$f(z):=\sum_{n=0}^{\infty}\Bigg(\dfrac{1}{z_{0}}\Bigg)^{n+1}z^{n}.$$ The radius of convergence is $$R:=\limsup_{n\rightarrow\infty}\Bigg|\dfrac{1}{z_{0}}\Bigg|^{\frac{n+1}{n}}=1.$$

So in $\mathbb{D}$, we have $$f(z)=\dfrac{1}{z_{0}}\sum_{n=0}^{\infty}\Bigg(\dfrac{z}{z_{0}}\Bigg)^{n}=\dfrac{1}{z_{0}-z}.$$

Note that the function $h(z):=\frac{1}{z_{0}-z}$ is holomorphic everywhere on $\overline{\mathbb{D}}$ except at $z_{0}$.

But $h(z)$ is not bounded around $z_{0}$ so by Riemann's theorem on removable singularity that $z_{0}$ is not removable.

Hence, we have $g$ analytic on $\overline{\mathbb{D}}\setminus\{z_{0}\}$ and $f=g$ on $\mathbb{D}$, and thus $f$ has an analytic continuation from $\mathbb{D}$ to $C$, except at the point $z_{0}$.

$S(f)=\{z_{0}\}$.

Edit 3:

Okay I proved that $S(f)$ is non-empty, see my own answer of the question below.

答:
Okay, for other proofs, please see the edit of my post. I am answering my own question to prove that $S(f)\neq\varnothing$.

Suppose $S(f)=\varnothing$, then all the points on $C$ can be analytically continued to for $f$. So for each $z\in C$, we have $D(z,r)$ for $f$ to have the analytic continuation.

Then, as $\overline{\mathbb{D}}$ is compact, the covering $\Big(\bigcup_{z\in C}D(z,r)\Big)\cup\mathbb{D}$ gives us finite sub cover $\{D(z_{i}, r_{i})\}_{i=1}^{N}$.

We define $\epsilon:=\min_{i}r_{i}$, and note that this finite sub-cover will cover the annulus $D(0,1+\epsilon)\setminus\mathbb{D}$.

Define $f_{i}$ to be the corresponding analytic continuation of $f$ to $D(z_{i}, r_{i})$. Then, for each $z$ in this annulus, we define $$g(z)=f_{i}(z)$$ if $z\in D(z_{i}, r_{i})$. This is well-defined since every point in the annulus is in at least one covering ball, and in the intersection of two balls, $f_{i}(z)=f=f_{j}(z)$ by the analytic continuation.

Now, define $h(z)=f(z)$ for $z\in\mathbb{D}$ and $h(z)=g(z)$ for $z$ in annulus.

It follows that $h(z)$ is holomorphic in $D(0,1+c)$ for some $c<\epsilon$. So it has a power expansion $\sum_{n=1}^{\infty}b_{n}z^{n}$, but this will also represent $f(z)=\sum_{n=1}^{\infty}a_{n}z^{n}$ inside $\mathbb{D}$, this forces $a_{n}=b_{n}$, so $f(z)$ has radius of convergence $1+c>1$, a contradiction.


问:
Least number of scalene triangles formed from $13$ points in a plane

Given $13$ points in a plane with no three on a line, prove that there are at least $130$ scalene triangles formed from the points.

I thought the highest number of non-scalene triangles with $13$ points would happen in a $13$ side polygon, so I tried to calculate this value, and I got $78$ so I subtracted it from the number of triangles formable by $13$ points, $13 \choose 3$ and then I got $208$. I know it's wrong but I can't think of anything else.

Any help would be greatly appreciated. Thanks!

答:
There are ${13 \choose 2}$ couple of points.

Each couple can be the endpoints of the base of at most $2$ isosceles triangles, otherwise with three or more triangles, the corresponding three or more vertices opposite to the base would be on a line (the perpendicular bisector of the base).
If one or more of the isosceles triangles are instead equilateral, we can choose a base with some criterion, e.g. the one formed with the two points nearest to the origin.

Therefore there are at most ${13 \choose 2} \cdot 2$ isosceles or equilateral triangles and at least ${13 \choose 3} - {13 \choose 2} \cdot 2 = 130$ scalene triangles. I believe there are actually much more than that though.

Instead of counting only at the base, we could count at all sides. In that case for each couple of points one can build from the corresponding side at most $6$ isosceles or equilateral triangles ($2$ triangles with the side as base and $4$ triangles with the side as leg), but in this way each triangle would be counted $3$ times, so we still have at least ${13 \choose 3} - {13 \choose 2} \cdot \frac{6}{3} = 130$ scalene triangles.


问:
A Possible Mistake in Lovasz's "Submodular functions and convexity"
The possible mistake appears in Lovász L. (1983) Submodular functions and convexity. In: Bachem A., Korte B., Grötschel M. (eds) Mathematical Programming The State of the Art. Springer, Berlin, Heidelberg. https://doi.org/10.1007/978-3-642-68874-4_10.

This a pdf of the paper and below is a screenshot where I believe there is a mistake.

Following the text, modular functions (2) are set functions $f : 2^E \rightarrow \mathbb{R}$ such that $f(X \cap Y) + f(X \cup Y) = f(X) + f(Y)$.

Lovasz claims that modular functions satsify the following property (3).

$f(X) = \sum_{x\in X} f(\{x\}) + f(\emptyset)$

I tried to prove it, but I think it's just false. Consider the following equality, which holds for modular functions.
$f(\{x\} \cup \{y\}) + f(\{x\} \cap \{y\}) = f(\{x\}) + f(\{y\})$

If $x \neq y$, this implies
$f(\{x, y\}) + f(\emptyset) = f(\{x\}) + f(\{y\})$ and then $f(\{x, y\}) = f(\{x\}) + f(\{y\}) - f(\emptyset)$

Then Lovasz can only be correct if $f(\emptyset) = 0$ for every modular function, which is not true. Am I missing something here or is this a genuine error in the paper?

答:
Yes I think this is a typo. The correct claim would be:

$$
f(X) = f(\emptyset) + \sum_{x\in X} (f(\{x\}) - f(\emptyset))
$$

It's common in the literature to assume $f(\emptyset)=0$ (sometimes referred to being normalized), so it's easy to see how that might have been overlooked.


问:
notation question (bilinear form)
So I have to proof the following:
for a given Isomorphism $\phi : V\rightarrow V^*$ where $V^*$ is the dual space of $V$ show that $s_{\phi}(v,w)=\phi(v)(w)$ defines a non degenerate bilinear form.

My question : Does $\phi(v)(w)$ denote the map from $v$ to a linear function $w$? (in this case i had serious trubles in showing linearity in the second argument,really confusing.

Or

maybe it just means $\phi(v)$ times $w$ where $w$ is the scalar value ( we get $w$ by applying  $v$ in the linear function it is mapped to)

I just started today with dual spaces and try my best with the notation , but i couldn't figure it out , please if you have any idea please help me with the notation , i will solve the problem on my own.

答:
This kind of notation occurs sometimes. In certain contexts you'll have a function from some set to a set of functions. In your case, $\phi : V \to V^{\ast}$ associates with each vector $v$ a function $\phi(v)$, so since the latter is a function $\phi(v) \in V^{\ast}$ it is just $\phi(v): V \to \mathbb{F}$ where $\mathbb{F}$ is the field, so it's perfectly good to write $\phi(v)(w)$ for the action of $\phi(v)$ on the vector $w$.

So in general, suppose $A$ is any set and $\mathcal{F}$ is a set of functions from some other set $B$ to some other set $C$, then if we define $\phi : A \to \mathcal{F}$ for each $a \in A$ we have $\phi(a) : B \to C$ so that the action on some $b \in B$ is just written as $\phi(a)(b)$. Note that some people prefer writing $(\phi(a))(b)$, however the first notation is more common.

Title: notation question (bilinear form)


问:
critical points of $f(x,y) = \frac{\sin(x)}{1+y^2}$
so I determined: $f'(x,y) = \left(\begin{array}{c} \frac{\cos\left(x\right)}{y^2+1}\\ -\frac{2\,y\,\sin\left(x\right)}{{\left(y^2+1\right)}^2} \end{array}\right)$

and: hess-f = $\left(\begin{array}{cc} -\frac{\sin\left(x\right)}{y^2+1} & -\frac{2\,y\,\cos\left(x\right)}{{\left(y^2+1\right)}^2}\\ -\frac{2\,y\,\cos\left(x\right)}{{\left(y^2+1\right)}^2} & \frac{8\,y^2\,\sin\left(x\right)}{{\left(y^2+1\right)}^3}-\frac{2\,\sin\left(x\right)}{{\left(y^2+1\right)}^2} \end{array}\right)$

My first problem: What $x$ and $y$ do I plug in the hessian-matrix? If I solve $f'(x,y)$ for $(0,0)$ I get:

$x = \pi \cdot (k+1/2)$ or $x = \pi \cdot k$

Now, do I have to substitute these values for $x$ and solve for $y$? Seems like a huge case-differentiation to me.

...There must be an easier way, right?

答:
Note that for:
$$f'(x,y) = (0,0) \iff \frac{cos(x)}{y^2+1} = 0 \quad \wedge \quad \frac{y\operatorname{sin}(x)}{(y^2+1)^2} = 0 $$
The first condition can only be achieved if $x\in \frac{\pi}{2}+\pi\mathbb{Z}$, and since cos and sin are linearly independent, then the only way they can both be 0 at the same time is for y to be equal to 0. Thus, the critical points are of the form $\lbrace \pi/2 + k\pi: k\in\mathbb{N}\rbrace\times\lbrace 0\rbrace$. Now you can just plug them in the Hessian


答:
Trying plugging in those values of $x$, you'll notice that the Hessian of $f$ doesn't change for any value of $k$.

You could consider studying the critical points of $f$ only in a $\{(x,y)\mid -\frac{\pi}2<x<\frac{\pi}2\}$ strip first, and deduce later that all the other critical points are just translated copies of these one.

Title: critical points of $f(x,y) = \frac{\sin(x)}{1+y^2}$

Body:
so I determined: $f'(x,y) = \left(\begin{array}{c} \frac{\cos\left(x\right)}{y^2+1}\\ -\frac{2\,y\,\sin\left(x\right)}{{\left(y^2+1\right)}^2} \end{array}\right)$

and: hess-f = $\left(\begin{array}{cc} -\frac{\sin\left(x\right)}{y^2+1} & -\frac{2\,y\,\cos\left(x\right)}{{\left(y^2+1\right)}^2}\\ -\frac{2\,y\,\cos\left(x\right)}{{\left(y^2+1\right)}^2} & \frac{8\,y^2\,\sin\left(x\right)}{{\left(y^2+1\right)}^3}-\frac{2\,\sin\left(x\right)}{{\left(y^2+1\right)}^2} \end{array}\right)$

My first problem: What $x$ and $y$ do I plug in the hessian-matrix? If I solve $f'(x,y)$ for $(0,0)$ I get:

$x = \pi \cdot (k+1/2)$ or $x = \pi \cdot k$

Now, do I have to substitute these values for $x$ and solve for $y$? Seems like a huge case-differentiation to me.

...There must be an easier way, right?


问:
Linear objective with quadratic equality constraint
I've been given a problem:

min $f(x) = c^Tx$

s.t.

$\sum_{i}x_i = 0$

$\sum_{i}x^2_i$ = 1

I've created the Lagrangian function

$L(x,\lambda) = c^Tx - \lambda_1\sum_{i}x_i - \lambda_2\left(\left(\sum_{i}x^2_i\right)-1\right)$

But I'm not sure how I can arrive at any particular solution given the $\nabla_x$ of the Lagrangian doesn't really help me.  Should I augment the function to be an unconstrained minimization by replacing the values of $x_i$ using the 2 equalities?  I'm not sure where to proceed.

答:
You derivate the Lagrangian with respect to each $x_i$ and $\lambda_1$ and $\lambda_2$ and put it equal to 0.
$$
\frac{\partial L}{\partial x_i}=c_i-\lambda_1-2x_i\lambda_2=0 \\
\frac{\partial L}{\partial \lambda_1}=-\sum_{i=1}^{n}x_i=0 \\
\frac{\partial L}{\partial \lambda_2}=-\sum_{i=1}^{n}x_i=0
$$
From the first equation, you get:
$$
x_i=\frac{c_i-\lambda_1}{2\lambda_2}
$$
If you fill it into the second equation, you get $\lambda_1$:
$$
\lambda_1=\frac{1}{n}\sum_{i=1}^{n}c_i
$$
If you fill it into the third equation, you get $\lambda_2$:
$$
\lambda_2 = \pm\frac{1}{2}\sqrt{\sum_{i=1}^{n}(c_i-\lambda_i)}
$$
So you have two solutions:
$$
x=\pm\frac{c-\frac{1}{n}\sum_i c_i}{||c-\frac{1}{n}\sum_i c_i||_2}
$$
And you can easily check that one is minimum and the other one is maximum. The minimum is this one:
$$
x=-\frac{c-\frac{1}{n}\sum_i c_i}{||c-\frac{1}{n}\sum_i c_i||_2}
$$


问:
How to prove $E[XY] = E[X^2]$ given $E[Y|X]=X$
Given $E[Y|X]=X$

I need to prove that $E[XY] = E[X^2]$

I'm not sure how to approach this. I would appreciate if someone could point to me what property of expectation I could use.

答:
As others have noted, $E[X]=E[Y]$ is not enough to conclude $E[XY]=E[X^2]$.

However, the original condition $E[Y \mid X=x] = x$ is enough. Note that this last condition implies $E[Y \mid X] = X$. So,
$$E[XY] = E[E[XY \mid X]] = E[X E[Y \mid X]] = E[X \cdot X] = E[X^2].$$


问:
Hamiltonian system ODE
I'm stuck in this exercise and I don't know how to proceed.

A system $$x'=f(x,y) \\ y'=g(x,y)$$

is a Hamiltonian if there is a function $H(x,y)$ such that $$f=H_y,\\ g=-H_x$$

The $H$ function is called Hamiltonian.

I need to prove that any conservative equation $x''=f(x)$ implies a Hamiltonian System, and I need to prove that the Hamiltonian function $H$ coincides to the total energy.

How I prove these two facts?

答:
Quick note, a Hamiltonian $H(x,y)$ for $x,y$ is a function such that
\begin{align}
\frac{dx}{dt} &= -\frac{\partial H}{\partial y}\\
\frac{dy}{dt} &= \frac{\partial H}{\partial x}\\
\end{align}
For the system $x''(t)=f(x)$ consider the function $y(t)=x'(t)$. It follows then that
\begin{align}
\frac{dx}{dt} &= y(t)\\
\frac{dy}{dt} &= x''(t)=f(x)\\
\end{align}
Thus your Hamiltonian function $H(x,x')$ must satisfy
\begin{align}
\frac{\partial H}{\partial y} &= -y\\
\frac{\partial H}{\partial x} &= f(x)
\end{align}
Well the function
$H(x,y)= -\frac{y^2}{2} + \int_0^x f(t)dt$ satisfies the PDE constraints above. Hope that helps!

Title: Hamiltonian system ODE


问:
Compositions of $n$ with largest part at most $m$
This is a problem from Stanley's Enumerative Combinatorics that I'm failing at a bit (lot):

Let $\bar{c}(m,n)$ denote the number of compositions of $n$ with largest part at most $m$.  Show that $$\sum_{n\geq 0}\bar{c}(m,n)x^n={{1-x}\over{1-2x+x^{m+1}}}$$

Some definitions:  A composition of $n$ is an $ordered$ list of positive integers that equals $n$.  If $\{a_1,...,a_k\}$ is one such composition, we say that the composition has $k$ $parts$.

I know it's pretty traditional to list a "what you've done so far" but I'm really about as blindly stuck as can be.

答:
This can be solved in two ways, both quite easy.

For the first, note that the given fraction simplifies due to the decomposition $1-2x+x^{m+1}=(1-x)(1-x-x^2-\cdots-x^m)$ to the rational function
$$
\frac1{1-x-x^2-\cdots-x^m}.
$$
The coefficients $a_i$ of the corresponding power series satisfy $a_0=1$ and the recurrence
$$
  a_i=a_{i-1}+a_{i-2}+\cdots+a_m \qquad\text{for $i>0$,}
$$
where  terms with negative index are taken to be$~0$. But taking $a_i=\bar c(n,i)$ this recurrence can be easily seen to be satisfied: a composition of $i>0$ has a final part$~k$ with $1\leq k\leq m$, and removing that part leaves a composition of $i-k$ into parts at most$~m$, each obtained exactly once, so $\bar c(n,i)=\sum_{k=1}^m\bar c(n,i-k)$.

The other way is to use the form of the fraction given in the question directly. Its denominator indicates that the recurrence
$
  a_i=2a_{i-1}-a_{i-(m+1)}
$
should be satisfied for sufficiently large$~i$. Here each composition of $i$ can be obtained in one of two ways from a composition of $i-1$: either adding a part $1$, or by increasing the last part by$~1$. However the latter method cannot be applied to all compositions of $i-1$ into parts at most $m$, since it may make the final part $m+1$. The number of such forbidden extensions equals $\bar c(n,i-(m+1))$ (each composition counted by that number produces one when a $m+1$ added at its end), so that number must be subtracted in the recurrence. With again the convention that recurrence terms with negative indices are taken to be $0$, the recurrence relation gives the correcct number for $i>1$, but the value$~2$ it would give for $i=1$ is one too much (since the composition of $0$ has no part to increase). This is reflected by the term "$-x$" in the numerator of the fraction $\frac{1-x}{1-2x+x^{m+1}}$.


问:
Use Doob’s convergence theorem to show almost sure convergence
Let $(X_i)_{i \in \mathbb{N}} \subset \mathcal{L}^2$ be independent and $(a_n)_{n \in \mathbb{N}} \subset \mathbb{R}_{\ge 0}$ an increasing an unbounded sequence such that

$$\sum_{i=1}^{\infty} \frac{\mathbb{V}[X_i]}{{a_i}^2} < \infty $$

I want to use Doob’s convergence theorem to show that

$$\lim_{n \to \infty} \frac{1}{a_n}\sum_{i=1}^{n}\big(X_i - \mathbb{E}[X_i]\big) = 0 \textit{ a.s.} $$

My problem is that while I can show that $R_n := \frac{1}{a_n}\sum_{i=1}^{n}\big(X_i - \mathbb{E}[X_i]\big)$ is a supermartingale and $\operatorname{sup}_{n \in \mathbb{N}} \mathbb{E}\big[R_n\big] < \infty$, Doob's convergence theorem doesn't give me any information about the limit itself.

答:
Assume without loss of generality that $X_i$ is centered. Let
$Y_i:=\sum_{k=1}^iX_k/a_k$ for $i\geqslant 1$ and $Y_0=0$. Then
$$
\frac{1}{a_n}\sum_{i=1}^nX_i=\frac{1}{a_n}\sum_{i=1}^na_i\left(Y_i-Y_{i-1}\right)
=Y_n+\frac{1}{a_n}\sum_{i=1}^{n-1}\left(a_i-a_{i+1}\right)Y_i.$$
From the assumption, $(Y_i)_{i\geqslant 1}$ is a martingale which is bounded in $\mathbb L^2$. Therefore, there exists a random variable $Y$ such that $Y_i\to Y$ almost surely. From the assumptions on $(a_i)$, we can see that $\frac{1}{a_n}\sum_{i=1}^{n-1}\left(a_i-a_{i+1}\right)Y_i\to -Y$ almost surely.


问:
Limit of $n$-th root of a polynomial as $x$ goes to infinity

Let $P(x)$ be a monic polynomial of degree $n$, find $\lim\limits_{x_\to \infty}([P(x)]^{\frac{1}{n}}-x)$.

I'm working through some practice problems for my analysis class (quiz tomorrow!) and am stumped by this one. Can someone help me with a hint or suggestion?

I tried rationalizing but was not able to get rid of the root. Taking the log was stymied by the subtraction. This was a problem in the chapter on derivatives, so I tried taking the derivative, and tried modifying it to make L'Hospital's rule work, but they did not make it any clearer.

I have a hunch that says a series expansion at $\frac{1}{X} = 0$ could help, but the root gets in my way every time. The problem:

答:
So $$P(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0.$$
For $x\gg0$, we have a good approximation
$$ P(x)\approx\left(x+\frac{a_{n-1}}n\right)^n$$
as in: The error is a polynomial of degree $\le n-2$.
Conclude that
$$ \sqrt[n]{P(x)}-x\approx \frac{a_{n-1}}n$$
(with error bounded by some $\frac Mx$).


问:
A sequence of order statistics from an iid sequence
Given a sequence of iid random variables $X_i$ (without loss of generality from $U(0,1)$), an integer $k \ge 1$ and some $p \in (0,1)$, construct the sequence of random vectors $Z^{(j)}$, $j=0,1,...$ in the following way. Let

$$Z^{(0)}=(X_{(1)},...,X_{(k)}),$$

where $X_{(l)}$ is the $l$-order statistic of sample $\{X_1,...,X_k\}$. Introduce notations

\begin{align}
Z^{(j)}&=(Z_{j,1},...,Z_{j,k}),\\
m_j&=\min(Z_{j-1,1},...,Z_{j-1,k},X_{k+j}),\\
M_j&=\max(Z_{j-1,1},...,Z_{j-1,k},X_{k+j})
\end{align}

Then

$$Z^{(j)}=(Y_{(1)},...,Y_{(k)})$$

where $Y_{(l)}$ is the $l$-order statistic of the following set which is

The set $\{Z_{j-1,1},...,Z_{j-1,k},X_{k+j}\}\backslash m_j$ with probability $p$

The set $\{Z_{j-1,1},...,Z_{j-1,k},X_{k+j}\}\backslash M_j$ with probability $1-p$

The decision between cases 1. and 2. is made independently from the $X_i$ (and hence from the $Z^{(i)}$).

The $Z^{(j)}$ are supported on the $k$-dimensional simplex $S_k = \{(x_1, \dots, x_k) \in \mathbb{R}^k \, | \, 0 \le x_1 \le x_2 \le \dots \le x_k \le 1 \}$.

It appears that the $Z^{(j)}$ converge in distribution. Is this known? Is anything known about the limiting distribution?

For the case $k=1$, the answer is the following. Denote the cdf of $Z^{(j)}$ by $F_j$.

The cdf of $\min(X_{n+1},Z^{(n)})$ (for $U(0,1)$ case) is

$$x+F_n(x)−xF_n(x)$$
and the cdf of $\max(X_{n+1},Z^{(n)})$ is

$$xF_n(x)$$.

Hence

\begin{align}
F_{n+1}(x)&=p(x+F_n(x)−xF_n(x))+(1−p)xF_n(x)\\
&=px+(p(1-x)+(1-p)x)F_n(x)
\end{align}

Since $p(1-x)+(1-p)x\in(0,1)$ we have that

$$\lim F_{n}(x)=\frac{px}{1-p(1-x)-(1-p)x}$$

I am looking for general results (case $k>1$) either for the limiting distribution of the whole vector $Z^{(j)}$ or of some of its components (marginal distributions).

答: I've asked this question in mathoverflow. It has one answer, which I do not find time to check, so I am giving the link as an answer here.</p>

OwnerDisplayName: Hans Engler

ContentLicense: CC BY-SA 2.5


问:
Math school online?
As a hobbyist programmer with little to no math skills (no science skills either, thank you religious schools) I am finding math encroaching on my life more and more often. I have always laughed and said that I'm just not good at math because my brain doesn't work like that.

My younger brother challenged me recently claiming that it wasn't a lack of ability that kept higher math out of my hands but a lack of interest. Because my current realm of studies is really starting to delve into 3D modeling, programming for web, mobile and desktop, I feel that a stronger math skill set would benefit me immensely.

But I don't know where to start...

I'm wondering if there is a math version similar to: http://www.code.org/khan-academy

Some place where I can go and start challenging my brain, increasing my math as often as I have time? Some place online? I would also appreciate any tips for books or other resources that would be recommended for someone wanting to ignite an interest in math.

To give a background, my highest level of math in school was pre-algebra in 7th grade. After that the schools my parents sent me to didn't teach higher math (nor evolution) so when I hit college, I had to get a tutor to pass (with a C) the College Level Algebra... I barely did it and it was so long ago I don't remember anything I learned.

I hope this is the right place to ask this... I stumbled across math.stackexchange by accident. Whether this question gets answered or not I will start browsing around and seeing what all there is...

答:
I will second Serge Lang's book "Basic Mathematics". It is definitely challenging though for those used to traditional high-school textbooks.

But as originally asked, AOPS has something called the "Alcumus" which you might find useful.

As mentioned on the website http://www.artofproblemsolving.com :

Art of Problem Solving's Alcumus offers students a customized learning
  experience, adjusting to student performance to deliver appropriate
  problems and lessons. Alcumus is specifically designed to provide
  high-performing students with challenging problems appropriate to
  their abilities.

```<code>-Over 9000 problems with solutions, including many taken from national math competitions
-An intelligent learning system that automatically provides problems of appropriate difficulty
-Detailed progress reports so students can track their performance
-Tools for teachers to monitor their students' progress
</code>```

Title: Math school online?

Body:
I will second Serge Lang's book "Basic Mathematics". It is definitely challenging though for those used to traditional high-school textbooks.

But as originally asked, AOPS has something called the "Alcumus" which you might find useful.

As mentioned on the website http://www.artofproblemsolving.com :

Art of Problem Solving's Alcumus offers students a customized learning
  experience, adjusting to student performance to deliver appropriate
  problems and lessons. Alcumus is specifically designed to provide
  high-performing students with challenging problems appropriate to
  their abilities.

-Over 9000 problems with solutions, including many taken from national math competitions
-An intelligent learning system that automatically provides problems of appropriate difficulty
-Detailed progress reports so students can track their performance
-Tools for teachers to monitor their students' progress


问:
Find all the cosets of $H$ in $G$ , where $G = \{ z\in \mathbb C \mid z \neq 0 \}$ , $H = \{ z \in\mathbb C \mid z^n = 1 , n\in\mathbb N \}$

Let, $G = \{z\in\mathbb C \mid z \neq 0\}$, $H = \{z\in\mathbb C \mid z^n = 1 , n\in\mathbb N\}$. Find the cosets of $H$ in $G$.

I have got the pictorial idea that how the cosets will look like concentric circle in complex plane, but I could not understant how to express those cosets and how to solve this problem in a rigorous way.

答:
Sketch of an idea:

$(1).\ $ Consider $\mathbb{Q}/\mathbb{Z}$ and note that for each equivalence class, there is exactly one representative $0\le q< 1.$

$(2).\ $ the map $[q]\to 1^q$ is an isomorphism: $: \mathbb{Q}/\mathbb{Z} \rightarrow H.$

$(3).\ $ If $[z] \in \mathbb{C}/(\mathbb{Q}/\mathbb{Z})$ then $z\sim w\Leftrightarrow zw^{-1}\in \mathbb{Q}/\mathbb{Z}.$ Now $z=re^{it}$ and $w=r'e^{is}$ then using  $(1)$ and $(2)$, we have that $z\sim w$ if and only if $zw^{-1}$ is a $p^{th}$ root of unity for some fixed rational $q=r/p.$ This means that $z$ and $w$ lie on the same circle and are such that $-s$ is the angle through which $z$ must be rotated to bring it to one of the $p$ roots of unity.


问:
What type of equation is this called?
I'm trying to solve it. Whatever it is, it's been out of my (very basic) understanding level thus far.  Hahaha I have sheets and sheets of paper trying to work it out. I dunno if it helps, but it's an optimization problem, I'm trying to work out for a piece of software I'm writing. The 'original' equation looks like this...

$\frac{2y}{2 + x} = n$

And I want to see if I'm able to feed in any particular number to N, and solve for the x and y. Like this...

$\frac{2y}{2 + x} = 4148$

At first I tried solving for X and Y like this.(I don't know how to show the steps, so I'm just going to show the 'results'.

$y = 2074 + 1037x$

$x = \frac{8296}{y -1}$

Then I thought I could replace one side with the other.

$x = \frac{8296}{2073 + 1037x}$

But I think that led down the road to madness. And I came here straight away. I think I'm okay. But could someone perhaps shine a bit of light upon my situation?

What I mean by this...is I keep getting different answers, which makes me think I'm really doing it wrong. Also, and maybe this doesn't matter, but x and y are a constant product formula themselves. So, x * y will always equal k(the invariant). I super appreciate any thoughts! Thanks =)

答:
It's unclear (to me at least) how you got from ${2y\over2+x}=4148$ to either $y=2074+1037x$ or $x={8296\over y-1}$, neither of which is correct, but here are simple algebraic steps that lead to correct equations:

$${2y\over2+x}=4148\implies{y\over2+x}=2074\implies y=2074(2+x)\implies y=4148+2074x$$

and

$${2y\over2+x}=4148\implies{y\over2+x}=2074\implies{2+x\over y}={1\over2074}\implies2+x={y\over2074}\\\implies x={y\over2074}-2$$

If you like, the equation for $x$ in terms of $y$ can be written as $x={y-4148\over2074}$.

Finally, in answer to the question in the title, because ${2y\over2+x}=n$ implies $y={n\over2}x+n$, it is, in essence, a disguised version of a linear equation. The name derives from the fact that the solution set, plotted in the $xy$-plane, consists of points on a line. (Technically the point $(-2,0)$ is not in the solution set since $2y/(2+x)$ is not defined when $2+x=0$. This is reflected in the use the implication symbol $\implies$, rather than the if-and-only-if symbol $\iff$.)

Title: What type of equation is this called?


问:
Sup and inf of a sequence
I have to find $\inf$ and $\sup$ of the given sequence:
$$ a_n := \left(1+ \frac{\cos(n\pi)}{n}\right)^n + \sin\left(\frac{n\pi}{2}\right)$$
First thought that come to my mind was to consider 4 cases dependinig on trigonometric functions. For $\sup$, 2 cases, when $\sin = 1$ or $\cos = 1$, and the same for $\inf$.

Is that correct or I'm completely wrong?

答:
Yes, that seems a good idea. Notice that $\cos(n\pi)$ takes values $\pm 1$, while $\sin(\frac{n\pi}{2})$ can also be $0$ (for $n$ even).

You can study the behaviour of the sequence splitting it in the four cases corresponding to $n\ (\operatorname{mod} 4)$.

Title: Sup and inf of a sequence


问:
Name for vector without sign/sense? (Just module and direction)
If vectors are the equivalence class of oriented segments, what is the name for the equivalence class of unoriented segments?

These "unsigned vectors" would have a module (or normal value) and a direction but no sense/sign.

An example of a use for such entities is the cross product: if we don't know the orientation/headiness of the space, we can "unify" the two possible answers as a single "unsigned vector".

答:
I would simply call this a line, which in a sense is a vector that points in both directions, and only unsigned direction matters. Whereas a vector space is a set of vectors, the set of all lines in a vector space is the corresponding projective space.

If you want to preserve magnitude as well, then it's a bit less clear-cut what to call it. You could say it's a line segment, though that is a more general term. Even a line segment through the origin could fail to have a well-defined magnitude. The magnitude could be the length, but you would really want it to be symmetric so that the length is the same in both directions. We are really looking at vectors that are identified by the equivalence relation $v\sim -v$, and these equivalence classes are the objects you seek. They can be represented by one or the other vector, but I wouldn't say they have a specific name.


答:
First of all, I am mystified why anyone would downvote the original query.  The OP clearly showed work.  Personally, I upvoted because of the shown work.

$\frac{2y}{2 + x} = n$

As others have indicated, this equation (by itself) can not be completely solved.  As Yves Daoust's answer indicates, if an additional equation is added, such as

$$xy = k ~: ~\text{with} ~k~ \text{a known value}$$

then, the equation can be completely solved.

Focusing only on the equation explicitly provided in the query:

$$\frac{2y}{2 + x} = n$$

The most that you can do is as follows:

$${2y} = n(2 + x) \implies y = \frac{n(2+x)}{2}.$$

What this partial solution signifies is that if $x$ is any known value, then $y$ can be computed.  An alternative explanation is that $y$ has been expressed as a function of $x$.

In a similar fashion a typical equation for a line might be

$$y = mx + b. \tag{1}$$

This equation would signify that a point $(x,y)$ is on the line if and only if equation (1) above is satisfied.

Addendum

Per OP's request.

Any chance you could walk through solving it if K is a known value? (which it is) Yves' answer seems to work for me, but I have no idea how to replicate it.

It is very difficult to add anything to Yves Daoust's answer.  Each step that he took was accurate, valid, and clear.  The best that I can do is walk you through it, one step at a time, providing explanation for why a specific step is taken, and (perhaps) extra detail on why the step is valid.

Assume that you are trying to solve the system:

$$\begin{cases}\dfrac{2y}{2+x}=n,\\xy=k,\end{cases}$$ where $n$ and $k$ are given.

Note: as indicated in Yves Daoust's answer, $(x = -2)$ is outlawed, because it would make the denominator in the first equation $0$.

The first question to ask is: what is your goal?

Do you want to know how to apply Yves Daoust's answer to a specific value for $n$ and $k$?

Do you want to be able to attack similar problems that will have somewhat different answers?

I will take these questions one at a time.  For the first question, as I understand it, you understand and agree with his work, and therefore accept that the answer is

$$y=\frac{n\pm\sqrt{n^2+2kn}}2.$$

Suppose that (for example) $[n = 10~$ and $~k = 15]$.

Then, the values for $y$ that satisfy the quadratic are

$$y ~=~ \frac{(10)\pm\sqrt{(10)^2+2(15)(10)}}{2}
~=~ \frac{10 \pm \sqrt{400}}{2} ~=~ \frac{10 \pm 20}{2}
~=~ (5 \pm 10).$$

First, the meaning of this result should be clarified.  By creating and solving a quadratic, the inference is that for any value of $(x,y)$ that solves the original equation, $y$ will have to be an element of $\{15,-5\}.$

This doesn't mean that each (or any) of the values of $y$ in $\{15, -5\}$ will necessarily generate an $(x,y)$ solution that satisfies both of the original equations.  What it does mean is that no other value for $y$ can work.

Therefore, what is required is to examine each candidate value of $y$, compute $x$ based on one of the two original equations, and then check to see whether that $(x,y)$ pairing satisfies the other original equation.

When $y = 15$ and $k = 15$, the second equation gives $x = 1$.  Checking the $(x,y)$ pair against the first equation asserts that

$$\frac{2[15]}{2+[1]}=10~: ~\text{this is true}.$$

Therefore, $y=15$ does lead to a satisfying solution.

When $y = (-5)$ and $k = 15$, the second equation gives $x = (-3)$.  Checking the $(x,y)$ pair against the first equation asserts that

$$\frac{2[-5]}{2+[-3]}=10~: ~\text{this is true}.$$

Therefore, the second value of $(y = -5)$ also leads to a solution: $(x,y) = (-3, -5).$  By all of the analysis so far, both solutions $(1,15)$ and $(-3,-5)$ are each solutions to both of the original equations.

Therefore, as far as the first question is concerned, you can use the above procedure against any given values of $n$ and $k$ to find all possible solutions (if any).  Note that any solution that results in $(x = -2)$ is outlawed, as previously mentioned.

Now, I will attack the second question.

First, start with the first original equation:

$$\dfrac{2y}{2+x}=n.$$

The first goal should be to clear the denominator, which is exactly what Yves Daoust did:

$$2y = 2n + nx \implies $$

$$2y-nx-2n=0. \tag{2}$$

Now, you have a major difficulty.  Once your intuition is developed, you will realize that you can not completely solve the equation without using both of the original equations (i.e. also using the second original equation).

However, the equation (2) above does not have an $(xy)$ term.  Let's be clear on what your goal is at this point.  $x$ and $y$ are variables, and $n$ and $k$ are known values.  Therefore, you want to solve for each of $x$ and $y$ in terms of $n$ and $k$.

What this means, at this point, is that equation (2) above must be altered so that the second (original) equation $(xy = k)$ can be applied.

There are only way two possible ways to do this:

multiply equation (2) by $x$, so that the first LHS term becomes $(2xy)$.

multiply equation (2) by $y$, so that the second LHS term becomes $(nxy)$.

Yves Daoust chose the second method, so let's follow his logic.  Equation (2) becomes

$$2y^2 - n(xy) - 2ny = 0. \tag{3}$$

At this point, you have succeeded in taking the only term in equation (2) that had a factor of $x$ and converting it into a factor of $(xy)$.  The motivation for this was because the second original equation is $(xy) = k$.

Now, the second original equation may be utilized, and equation (3) becomes

$$2y^2 - n(k) - 2ny = 0. \tag{4}$$

Examining equation (4) above, $x$ has been eliminated, so that the only variable is $y$.  Further, the first term in equation (4) is $(2y^2)$.  Therefore, you have a quadratic equation in $y$.  Therefore, the next step is to rearrange the terms to clarify the quadratic.

$$y^2(2) + y(-2n) + (-nk) = 0. \tag{5}$$

Since the quadratic equation $Ax^2 + Bx + C = 0$ is solved by $x = \frac{1}{2A} \times \left(-B \pm \sqrt{B^2 - 4AC}\right)$, equation (5) is solved by

$$y ~=~ \frac{1}{4} \times \left(2n \pm \sqrt{4n^2 + 8nk}\right) ~=~ \frac{1}{2} \times \left(n \pm \sqrt{n^2 + 2nk}\right). \tag{6}$$

Notice that this analysis exactly matches Yves Daoust's analysis.

The remaining steps are to consider each value of $y$ that satisfies the quadratic equation separately.  For each value of $y$, a value of $x$ will be generated by the equation $(xy = k).$  This will result in a candidate value for $(x,y)$ that may or may not satisfy the first original equation.  Your solution will then be each value of $(x,y)$ that satisfies both equations, where $x$ is not allowed to $= -2.$

If $y = \frac{1}{2} \times \left(n + \sqrt{n^2 + 2nk}\right),$ then

$$x ~=~ \frac{k}{y} ~=~ \frac{2k}{n + \sqrt{n^2 + 2nk}}.$$

Here, the denominator needs to be cleared.  This is done by multiplying the RHS fraction by
$\frac{n - \sqrt{n^2 + 2nk}}{n - \sqrt{n^2 + 2nk}}.$

This gives

$$x = 2k \times \frac{n - \sqrt{n^2 + 2nk}}{(-2nk)}
~=~ \frac{\sqrt{n^2 + 2nk} ~-~ n}{n}.$$

Now,

$$(x,y) ~=~ \left(\frac{\sqrt{n^2 + 2nk} ~-~ n}{n} ~,~
\frac{n + \sqrt{n^2 + 2nk}}{2} \right) \tag{7}$$

must be checked against

$$\dfrac{2y}{2+x}=n.$$

$$\frac{2y}{2+x} ~=~ \frac{n + \sqrt{n^2 + 2nk}}
{2 + \frac{\sqrt{n^2 + 2nk} ~-~ n}{n}}
~=~ \frac{n + \sqrt{n^2 + 2nk}}
{\frac{n + \sqrt{n^2 + 2nk}}{n}}
~=~ n.$$

Therefore, the candidate value identified in equation (7) satisfies both of the equations and is therefore a solution.

Now, the remaining candidate value must be checked.

If $y = \frac{1}{2} \times \left(n - \sqrt{n^2 + 2nk}\right),$ then

$$x ~=~ \frac{k}{y} ~=~ \frac{2k}{n - \sqrt{n^2 + 2nk}}.$$

Here, the denominator needs to be cleared.  This is done by multiplying the RHS fraction by
$\frac{n + \sqrt{n^2 + 2nk}}{n + \sqrt{n^2 + 2nk}}.$

This gives

$$x = 2k \times \frac{n + \sqrt{n^2 + 2nk}}{(-2nk)}
~=~ - \frac{n ~+~ \sqrt{n^2 + 2nk}}{n}.$$

Now,

$$(x,y) ~=~ \left(- \frac{n ~+~ \sqrt{n^2 + 2nk}}{n} ~,~
\frac{n - \sqrt{n^2 + 2nk}}{2} \right) \tag{8}$$

must be checked against

$$\dfrac{2y}{2+x}=n.$$

$$\frac{2y}{2+x} ~=~ \frac{n - \sqrt{n^2 + 2nk}}
{2 - \frac{n ~+~ \sqrt{n^2 + 2nk}}{n}}
~=~ \frac{n - \sqrt{n^2 + 2nk}}
{\frac{n - \sqrt{n^2 + 2nk}}{n}}
~=~ n.$$

Therefore, the candidate value identified in equation (8) also satisfies both of the equations and is therefore also a solution.

Note, that if this analyis (i.e. the Addendum's 2nd question) had been tackled before the Addendum's 1st question, then you could have reasoned as follows (re the specific example $n=10, k=15$).

From the above analysis, regardless of the values of $n$ and $k$ both of the solutions generated by $y = \frac{n \pm \sqrt{n^2 + 2nk}}{2}$ will result in an $(x,y)$ pair that satisfies both equations, assuming that the outlawed value of $(x = -2)$ is not generated.  Therefore, you could reason that the values of $(x,y) = (1,15)$ and $(x,y) = (-3,-5)$ must both satisfy both of the original equations.

Title: What type of equation is this called?

Body:
I'm trying to solve it. Whatever it is, it's been out of my (very basic) understanding level thus far.  Hahaha I have sheets and sheets of paper trying to work it out. I dunno if it helps, but it's an optimization problem, I'm trying to work out for a piece of software I'm writing. The 'original' equation looks like this...

$\frac{2y}{2 + x} = n$

And I want to see if I'm able to feed in any particular number to N, and solve for the x and y. Like this...

$\frac{2y}{2 + x} = 4148$

At first I tried solving for X and Y like this.(I don't know how to show the steps, so I'm just going to show the 'results'.

$y = 2074 + 1037x$

$x = \frac{8296}{y -1}$

Then I thought I could replace one side with the other.

$x = \frac{8296}{2073 + 1037x}$

But I think that led down the road to madness. And I came here straight away. I think I'm okay. But could someone perhaps shine a bit of light upon my situation?

What I mean by this...is I keep getting different answers, which makes me think I'm really doing it wrong. Also, and maybe this doesn't matter, but x and y are a constant product formula themselves. So, x * y will always equal k(the invariant). I super appreciate any thoughts! Thanks =)


问:
Complete theory with no atomic model
I'm stuck on the following problem. This is an old qualifying exam and a new homework question. So I need just a hint to think on it.

Let $T$ be complete theory in a countable language $\mathcal{L}$ with no atomic models. Let $\mathfrak{A}\vDash T$ be countable model. Prove that there is a countable model $\mathfrak{B}\vDash T$ such that neither $\mathfrak{A}$ nor $\mathfrak{B}$ can be elementarily embedded into the other.

One can deduce $T$ has no prime model, and the isolated types in $S_n(T)$ is not dense for some $n$. Also since $\mathfrak{A}$ is not atomic, it realizes a non-principal type $p$. So we can construct a model $\mathfrak{B}$ omitting $p$. However, this gives only that $\mathfrak{A}$ cannot be elementarily embedded into $\mathfrak{B}$. How can improve this idea? Or is there another way to prove this?

答:
Let $p$ be a non-isolated type which is omitted in $\mathfrak{A}$, and let $q$ be a non-isolated type which is realized in $\mathfrak{A}$. If we could find $\mathfrak{B}$ realizing $p$ and omitting $q$, we'd be done. Unfortunately, this strategy may not work directly: for some  pairs of non-isolated types $p$ and $q$, every model which realizes $p$ also realizes $q$.

So here's a different strategy. (I've edited this answer to provide a better hint that leads to a more direct solution than my previous answer.)

Let $\mathfrak{A}\models T$ be a countable model. Since $\mathfrak{A}$ is countable, there are only countably many non-isolated types realized in $\mathfrak{A}$, so we can apply the omitting types theorem to obtain a model $\mathfrak{B}$ omitting all of them. Since $T$ has no atomic models, $\mathfrak{B}$ is not atomic. Now show that $\mathfrak{B}$ omits some type realized in $\mathfrak{A}$ and $\mathfrak{B}$ realizes some type omitted in $\mathfrak{A}$.

Title: Complete theory with no atomic model


问:
Are eigenvectors of a matrix always contained in the nullspace?
If $v$ is a non-negative eigenvector of $A$, is it contained in the nullspace of A? I have an intuition that the eigenvectors for eigenvalue $\lambda$ are only in the null space of $A-\lambda I$, but I'm not sure how to prove this.

I tried justifying it by saying the eigenspace is the space of all vectors v satisfying $Av=\lambda v$. This is equivalent to $Av−\lambda v=0$ or $(A−\lambda I)v=0$. So the eigenspace of $\lambda$ is equal to the nullspace of $A−\lambda I$. If my intuition and justification are correct, then $v$ would not be in the nullspace of $A$ since it is non-negative.

Any help is greatly appreciated!

答: Given a linear transformation $A$, an eigenvector $v$ is in the nullspace of $A$ if and only if the associated eigenvalue is $0$. To see this fix some eigenvector $v$ with eigenvalue $\lambda$. If $\lambda \neq 0$, then by definition $v$ is not in the nullspace of $A$ since $Av = \lambda v \neq 0$. On the other hand, if $\lambda = 0$ then $Av = \lambda v = 0$ so clearly $v$ is in the nullspace.


答: The vector $v$ is in the nullspace of $A$ iff $Av = 0$. On the other hand, $v$ is in the eigenspace associated with $\lambda$ iff $Av = \lambda v$. Thus, it can only hold that $v$ is in the nullspace and in the eigenspace if $\lambda v = 0$. This only occurs if $\lambda = 0$ or $v = 0$.

Title: Are eigenvectors of a matrix always contained in the nullspace?

Body:
If $v$ is a non-negative eigenvector of $A$, is it contained in the nullspace of A? I have an intuition that the eigenvectors for eigenvalue $\lambda$ are only in the null space of $A-\lambda I$, but I'm not sure how to prove this.

I tried justifying it by saying the eigenspace is the space of all vectors v satisfying $Av=\lambda v$. This is equivalent to $Av−\lambda v=0$ or $(A−\lambda I)v=0$. So the eigenspace of $\lambda$ is equal to the nullspace of $A−\lambda I$. If my intuition and justification are correct, then $v$ would not be in the nullspace of $A$ since it is non-negative.

Any help is greatly appreciated!


问:
special case of Nagata's Lemma (Matsumura p.86)
Let $K$ be a field and $R$ a valuation ring of $K$ with maximal ideal $m_R$. Let $a \in R$ such that $1-a \in m_R$.

Statement: For any $s$ that is not a multiple of the characteristic of $R/m_R$, the element $(1+a+a^2+\cdots+a^{s-1})^{-1}$ is inside $R$.

How do we prove the above statement?

答: Reducing mod $m_R$, the element $1 + a + \cdots + a^{s-1}$ becomes $s$, which is invertible by assumption.  Since $R$ is a local ring, any element of $R$ which is invertible mod $R$ is invertible in $R$, thus $1 + a + \cdots + a^{s-1}$ is invertible in $R$.</p>


答:
As pointed out by @Matt Samuel, I would use the following definition for the line segment $l$ between two points $A$ and $B$:
$$l := \left\{ tA + (1-t)B \mid t \in [0, 1] \right\}$$

Hope that helps, please let me know if I can help further !

Title: Name for vector without sign/sense? (Just module and direction)

Body:
If vectors are the equivalence class of oriented segments, what is the name for the equivalence class of unoriented segments?

These "unsigned vectors" would have a module (or normal value) and a direction but no sense/sign.

An example of a use for such entities is the cross product: if we don't know the orientation/headiness of the space, we can "unify" the two possible answers as a single "unsigned vector".


答:
Being a monic polynomial, it looks like
$$p(x)=x^n+a_1x^{n-1}+\cdots +a_{n-1}x+a_n=x^n(1+\frac{a_1}{x}+\cdots +\frac{a_{n-1}}{x^{n-1}}+\frac{a_n}{x^n}),$$
so
$p(x)^{1/n}-x=x((1+\frac{a_1}{x}+\cdots +\frac{a_{n-1}}{x^{n-1}}+\frac{a_n}{x^n})^{1/n}-1),$
from there we see that we are in the problematic limit $0\cdot \infty.$ As, probably, you have seen before a way to approach this is using L'Hopital rule by changing the limit to be a quotient as
$$\lim _{x\rightarrow \infty}\frac{(1+\frac{a_1}{x}+\cdots +\frac{a_{n-1}}{x^{n-1}}+\frac{a_n}{x^n})^{1/n}-1}{1/x}=\lim _{x\rightarrow \infty}\frac{\frac{1}{n}(1+\frac{a_1}{x}+\cdots +\frac{a_{n-1}}{x^{n-1}}+\frac{a_n}{x^n})^{1/n-1}\cdot (\frac{-a_1}{x^2}-\frac{2\cdot a_2}{x^3}\cdots -\frac{n\cdot a_n}{x^{n+1}})}{-1/x^2},$$
where the second step is the derivative in the numerator and denominator.
Notice that now everything will cancel except the $a_1$ term and the $1/n$ giving you
$$\lim _{x\rightarrow \infty}p(x)^{1/n}-x=\frac{a_1}{n}.$$

Title: Limit of $n$-th root of a polynomial as $x$ goes to infinity

Body:
Let $P(x)$ be a monic polynomial of degree $n$, find $\lim\limits_{x_\to \infty}([P(x)]^{\frac{1}{n}}-x)$.

I'm working through some practice problems for my analysis class (quiz tomorrow!) and am stumped by this one. Can someone help me with a hint or suggestion?

I tried rationalizing but was not able to get rid of the root. Taking the log was stymied by the subtraction. This was a problem in the chapter on derivatives, so I tried taking the derivative, and tried modifying it to make L'Hospital's rule work, but they did not make it any clearer.

I have a hunch that says a series expansion at $\frac{1}{X} = 0$ could help, but the root gets in my way every time. The problem:


问:
Let $G$ a $k-$connected graph, then $L(G)$ is $(k-2)-$connected.
Let $G$ a $k-$connected graph, then $L(G)$ is $(k-2)-$connected.
Where $L(G)$ is the line graph of $G$.

I am very stuck in this problem. Try using Whitney's theorem for k-connected graphs.

答:
For any graph $H$ let $L(H)$ denote the line graph of $H$, and for each edge $e \in E(H)$ write the vertex in $L(H)$ that corresponds to $e$ as $u(e)$. First note the following:

(a) $L(H)$ is connected iff $H$ is connected

(b) $L(H \setminus \{e\}) = L(H) \setminus \{u(e)\}$, and $L(H \setminus S) = L(H) \setminus \{u(e); e \in S\}$ for any subset $S$ of $E(H)$.

(c) A $k$-connected graph $G$ is $k$-edge-connected.

So let $U$ be a set of $k-1$ vertices in $L(G)$. Then from (b) $L(G) \setminus U$ $=$ $L(G \setminus S)$ where $S =\{e; u(e) \in U\}$, and where $|S|=|U|$. Now,
if $G$ is $k$-connected then (c) implies that $G$ is $k$-edge-connected which
implies that $G \setminus S$ is connected. This in turn implies from (a) that
$L(G \setminus S)$ is connected. And so as $L(G \setminus S)=L(G) \setminus U$,
that $L(G \setminus U)$ is also connected.

So by definition $L(G)$ is $k$-connected, which implies that $L(G)$ is $(k-2)$-connected.


问:
Find the range and null space of T.
The inner product space with $\langle f(x),g(x)\rangle=\int_0^2 f(t)g(t)dt$ on $P_1(\mathbb{R})$. I need to prove that $T(f(x))=f(x)-f(1)$ is an orthogonal projection.

I know that I need to show $R(T)^{\bot}=N(T)$ and $N(T)^{\bot}=R(T)$. I'm stuck on actually solving R(T) and N(T) given T(f(x))=f(x)-f(1).

I started with the basis of $P_1(\mathbb{R})$, so $R(T)=span(\{T(1),T(x)\})$, but I'm not sure what $T(1)$ and $T(x)$ equals.

If it helps, in a class example we had that $T(1)$ and $T(x)$ equals $span(\{1,0\})$ when $T(f(x))=f(0)$

Additionally, I began to set up the null, and ran into a similar problem. I have $N(T)=\{a+bx\in P_1\mathbb(R)|T(ax+b)\}$ so far.

答: For each $f(x)\in P_1(\Bbb R)$, $T\bigl(f(x)\bigr)$ is a polynomial such that $T\bigl(f(1)\bigr)=0$. This suggests that $T$ is the orthogonal projection on$$\{p(x)\in P_1(\Bbb R)\mid p(1)=0\}=\{ax-a\mid a\in\Bbb R\}.\tag1$$On the other hand $\ker T$ is the set of all constant polynomials (since $p(x)\in\ker T\iff p(x)-p(1)=0$). And, if $k$ is a constant polynomial, then$$\langle k,ax-a\rangle=\int_0^2ka(x-1)\,\mathrm dx=0.$$So, $T$ is indeed the orthogonal projection on $(1)$.

Title: Find the range and null space of T.


问:
inner product space and fixed vector
My professor gave this question:

$V$ is an inner product space, $v$ and $w$ are fixed vectors in $V$. Let $T:V\to V$ be a linear map such that $T(u)=(u,v)w$. Find $T^*$.

Now I need to find the rank of $T$ and to find the matrix that represents $T$ by the standard basis of $\mathbb{C}^n$? So I found $T^*$ and according to my calculation it is $T^*(x)=\overline{(w,x)}v$.

Now assuming I am not wrong with $T^*$, how can I find rank$(T)$? I know that $$\text{rank}(T)=\text{rank}(TT^*)=\text{rank}(T*T)$$ but does this help me?

I do not know what $v$ and $w$ are and without that how can I  calculate what $T$ is doing the vectors of the basis in order to see how the matrix would look?

答: HINT: $(u, v)$ is always a scalar. You should be able to conclude on $Rank(T)$ from that, assuming that neither $v$ nor $w$ is the zero vector.</p>

Title: inner product space and fixed vector