问:
Help make Wonder Woman's box big.

Wonder Woman wants a box for her lasso. It is to be built from a rectangular piece of steel measuring 25 cm by 40 cm by cutting out a square from each corner and then bending up the sides. Find the size of the corner square which will produce a container that will hold the most.

Per suggestion by user Mark S, rewrote question to accept answer.

答:
Now the volume of a box is Length * Width * Height

I am thinking that the length of the box is $(40{\ \rm cm} - 2x)$, where $x$ is the side of one of the corner squares cut out of the piece of steel.  The width $= (25{\ \rm cm} - 2x)$ and the height $= x$.

This gives an expression $4x^3 - 130x^2 + 1000x$ for the volume, which has a max at $5$, thus the side of the squares cut out should be $5{\ \rm cm} \times 5{\ \rm cm}$.

Title: Help make Wonder Woman's box big.


问:
how can I calculate the sum of $i*2^i$ from $i=0$ to $\log n$?
What is $$\sum_{i=0}^{\log n}i\cdot 2^i\:?$$
I know it's $O(n\cdot\log n)$ but how do you calculate this sum?

ContentLicense: CC BY-SA 4.0

Body:
Define $f_{N}(t)$ such that
$$f_{N}(t) := \sum_{i=0}^{N} 2^{it}$$
this is a geometric sum which has the solution
$$f_{N}(t) = \frac{2^{Nt} - 1}{2^{t} - 1}$$
Now your sum is $\frac{1}{\ln 2}f_{\log n}'(1)$ so:
$$\sum_{i=0}^{\log n}i\cdot 2^i = \frac{1}{\ln 2}f_{\log n}'(1) = \frac{1}{\ln 2}\frac{d}{dt}(\frac{2^{(\log n)t} - 1}{2^{t} - 1})|_{t=1}$$


问:
Give a non-recursive definition for the language L
Consider the following recursive definition for $L ⊆ \{a, b\}^* : b ∈ L; ∀w ∈ L, bw, wa \text{ and } aw$ are in $L$. Give a non-recursive definition for L (using Kleene star or in English)

I'm trying to answer this problem, but I'm struggling finding the good answer using the Kleen star

Can someone explain me?

答:
The definition tells you that every word in $L$ can be formed by starting with $b$ and repeatedly adding an $a$ or a $b$ at the beginning or an $a$ at the end. For instance, the derivation

$$\color{red}b\overset{bw}\longrightarrow b\color{red}b\overset{aw}\longrightarrow ab\color{red}b\overset{bw}\longrightarrow bab\color{red}b\overset{wa}\longrightarrow bab\color{red}ba\overset{aw}\longrightarrow abab\color{red}ba\overset{wa}\longrightarrow abab\color{red}baa$$

shows a derivation of the word $ababbaa\in L$; above each arrow I’ve indicated which rule I’m using for that step. I’ve also colored the initial $\color{red}b$ red.

What kinds of strings can appear to the left of the initial $b$?

What kinds of strings can appear to the right of the initial $b$?

By looking at some examples and thinking about the rules for constructing members of $L$ you should be able to answer these two questions, and when you’ve done that, you’ve essentially described which words are in $L$. Once you can describe $L$ in words, we can help you describe $L$ symbolically with the aid of the Kleene star.


问:
Radon-Nikodym with a smaller $\sigma$-algebra
Let $(X,M, \mu)$ be a $\sigma$-finite measure. Let $M' \subset M$ be a smaller $\sigma$-algebra. Let $\mu' = \mu|_M$.

Show that there exists a unique $g \in L^1(\mu')$ such that $\int_E f d\mu = \int_E g d\mu'$ for all $E\in M'$ with $\|g\|_1 \leq \|f\|_1$.

My attempt:

I want to use Radon-Nikodym theorem here. But I'm not sure how to deal with the $\mu'$, since the version I learned requires both measures to be $\sigma$-finite.

Thank you for your help.

答:
The question is:

Let $(X,M, \mu)$ be a $\sigma$-finite measure space. Let $M' \subset M$ be a smaller $\sigma$-algebra. Let $\mu' = \mu|_{M'}$.

Show that for all $f \in L^1(\mu)$, there exists a unique $g \in L^1(\mu')$ such that $\int_E f d\mu = \int_E g d\mu'$ for all $E\in M'$ with $\|g\|_1 \leq \|f\|_1$.

This result, as stated, is false. Here is a counter-example.

Consider $(X,M, \mu)$ where $X=\Bbb R$, $M$ is the Borel $\sigma$-algebra and $\mu$ is the Lebesgue measure. We know that $(X,M, \mu)$ is a $\sigma$-finite measure space.

Now, let $M'=\{\emptyset , \Bbb R\}$. Then $M'$ is a $\sigma$-algebra and  $M' \subset M$. Let $\mu' = \mu|_{M'}$.

Let $f$ be any function in $L^1(\mu)$ such that $0< \int_{\Bbb R} f d\mu < +\infty$. Note that $g \in L^1(\mu')$ if and only if $g=0$.

So there is no $g \in L^1(\mu')$ such that $\int_{\Bbb R} f d\mu = \int_{\Bbb R} g d\mu'$

Remark: The key issue is that, even $\mu$ being $\sigma$-finite, $\mu'$ may not be $\sigma$-finite.

If we add to the question the additional hypothesis that $\mu'$ is $\sigma$-finite, then the result holds.

Just define on $M'$, the measure $\nu$ defined by, for all $E \in M'$, $\nu(E)= \int_E f d\mu$. Clearly, $\nu \ll \mu'$ and since now $\mu'$ is $\sigma$-finite, we can apply Radon-Nikodym theorem and conclude that  there exists a unique $g \in L^1(\mu')$ such that, for all $E\in M'$,
$$\int_E f d\mu = \nu(E)= \int_E g d\mu'$$

Then, let $\textrm{sign}(g) = \chi_{[g>0]} - \chi_{[g<0]}$. Since $g$ is $M'$-measurable function, so is $\textrm{sign}(g)$. So we have
\begin{align*}
\int_X |g| d\mu' & = \int_X  \textrm{sign}(g) g d\mu'=
 \int_X  \textrm{sign}(g)  d\nu =  \int_X  \textrm{sign}(g) f d\mu \leqslant
\int_X |f| d\mu
\end{align*}
So $\|g\|_1 \leq \|f\|_1$.


问:
What property allows me to integrate a gaussian function?
Whenever I integrate a gaussian function, I get to a step that makes me a little uncomfortable because I don't fully understand it. The only way I know of to analytically integrate the gaussian function is to multiply two of them together, like so...

$$\int_{- \infty}^{\infty} e^{-x^2}dx\int_{- \infty}^{\infty} e^{-y^2}dy = \int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{-x^2-y^2}\,dx\,dy$$

My questions is, what allows me to group two different integrands together into one integrand? When is this technique not allowed? Thanks in advance.

答: You can use this technique whenever the integrals converge.</p>

ContentLicense: CC BY-SA 3.0


问:
CGA Representation of an Ellipse
In conformal geometric algebra (CGA), I know the representation of a circle $s$ with center $c$ is:

$$
S = H(c) - \frac{1}{2}r^2n_{\infty}
$$

That $H(c)$ is the representation of $c$ (i.e., $n_0 + c + \frac{1}{2}c^2n_\infty$). Notice that $n_\infty$ is also denoted by $e_\infty$. Now my question is what is the representation of an ellipse with centers of $c_1$ and $c_2$ under these notations in CGA?

答:
CGA has no way to represent a non-circular ellipse.

CGA represents objects by their intersection with the paraboloid centered on the $n_\infty$ axis. Below is a diagram of 2D CGA I've drawn before to illustrate this.

Here, the thick blue point-pair and red circle are two 2D objects represented by CGA, while their 3D (ignoring $n_0$, which is just used to offset from the origin) representations are respectively the red plane and blue line.

The 3D subspaces are intersected with the white paraboloid and projected to the XY plane to obtain their 2D interpretation.

It should be straightforward to see that this system doesn't have enough degrees of freedom to represent arbitrary ellipses.

Title: CGA Representation of an Ellipse


问:
Construction of a Regular 5-gon
We were given the following true or false question in our Algebra exam

Q) The regular $5$-gon is not constructible by using a straight edge and compass.

The answer is false but we also had to provide an explanation. Followings was the explanation I gave

A regular $5$-gon is constructible since $5=2^{2}+1$ and for any prime $p$ if  $p=2^{r}+1$ for some $ r\in Z$ then a regular $p$-gon can be constructed.

The above explanation I believe is the statement of a theorem by Gauss. But my professor didn't find the answer satisfactory and wrote the words "Why ?"around the theorem. I am assuming because the theorem was only stated and not proved in our class. How else can we show that a regular $5$-gon is constructible by using compass and straight edge ?

答:
If you assume the side of the pentagon has length 1, then you can show that the length of a diagonal is the golden ratio $(1+\sqrt{5})/2$ by just drawing all the diagonals and comparing similar triangles.

Given a unit, construct a segment of length $2$ and then perpendicular to one endpoint, construct a segment of length $1$.  Draw the hypotenuse and you have segment of length $\sqrt{5}$.  Add one to that, then bisect and you have the golden ratio.  Use it and two segments of length $1$ to make a triangle.  Bisect the sides of length one and find the intersection of the bisectors.  This is the center of the pentagon.  Make a circle with this center and radius any vertex of the triangle. The three vertices of the triangle are also vertices of the pentagon.  Just copy those lengths about the circle.  Done.

Title: Construction of a Regular 5-gon


问:
Smallest example of two non-isomorphic 1-isomorphic groups
I just stumbled upon the definition of a map between to groups
$$\varphi:G \to H$$
being a 1-isomorphism. It means that it is bijective and satisfies
$$\varphi(g)^n=\varphi(g^n)$$
for all $g \in G$ and $n \in \mathbb Z$.

So my question is: What is the smallest example of two non-isomorphic 1-isomorphic groups $G$, $H$? Smallest here in terms of cardinality. For $|G|=|H|=16$ there are two such pairs. But maybe for even smaller cardinalities such pairs exist.

I don't want to know just the answer and proof by computer. I want to know if there is a simple argument why no smaller examples exist.

答:
It is clear that if $\phi$ is a $1$-isomorphism, then the order of $\phi(g)$ must be equal to the order of $g$: for otherwise, if the order of $\phi(g)$ were smaller, say $k$, then $\phi(g^k) = \phi(g)^k = e$ with $g^k\neq e$ (and $\phi(e)=e$ because $\phi(e) = \phi(e^2) = \phi(e)^2$). (If the order of $g$ were smaller, then apply the argument to $\phi^{-1}$.) Thus, the multiset of orders of elements of $G$ must be equal to the multiset for $H$.

(These are also called the "order sequences" of the groups, the ordered list of element orders. I think it's true that two finite groups are $1$-isomorphic if and only if they have identical order sequences, but I don't have time to work it out right now)

This makes it fairly straighforward to discuss the possibilities left.

Up to isomorphism, there is a unique group of order $p$ and of order $pq$ with $p\lt q$ both prime and $p\nmid q-1$; so none of those orders work. That discards $|G|=2$, $3$, $5$, $7$, $11$, $13$, and $15$.

For $p$ a prime, groups of order $p^2$ are abelian; the two abelian groups of order $p^2$ are $C_p\times C_p$ and $C_{p^2}$, and so you cannot establish the desired bijection unless the groups are isomorphic. That gets rid of $|G|=4$ and $9$.

For groups of order $pq$ with $p\mid q-1$, the (unique) abelian group is cyclic of order $pq$, and the nonabelian group has no elements of order $pq$, which gets rid of $|G|=6$, $10$, and $14$.

For groups of order $8$, the nonabelian groups $D_4$ and $Q_8$ have different number of elements of order $2$ (the first has four, the latter has one). The abelian groups are $C_8$, $C_2\times C_4$, and $C_2\times C_2\times C_2$. The first is the only one with elements of order $8$, so it cannot be part of such a pair; the second has exactly three elements of order $2$ (namely $(x,y^2)$, $(x,e)$, and $(e,y^2)$), so it cannot be part of a pair with $D_4$ or with $Q_8$; and the last has no elements of order $4$, so it cannot be part of a pair with the nonabelian ones or abelian ones, either.

That leaves only groups of order $12$. We can ignore $C_{12}$; the only remaining abelian one is $C_2\times C_6$, which has six elements of order $6$, no elements of order $4$, two elements of order $3$, and three elements of order $2$. The nonabelian groups are $D_{6}$ (which has two elements of order $6$ and six elements of order $2$, so it cannot be paired with the abelian groups), the dicyclic group $\mathrm{Dic}_{12}=\langle x,a\mid a^6=e, x^2=a^3, xax^{-1}=a^{-1}\rangle$, and $A_4$ (which has no elements of order $6$, hence cannot be paired with any other group of order $12$). The dicyclic group has elements of order $4$ (the Sylow $2$-subgroups are cyclic), so it cannot be paired with the dihedral group; and it cannot be paired with $C_2\times C_6$ because the latter lacks elements of order $4$.

Thus, there area also no examples in order $12$, and the smallest are in order $16$.

Relevant to this question are this previous question and this previous question


问:
Proof involving a simple inequality
I'm struggling to prove the following inequality (for $0 \leq a \leq c$ and $0 \leq b \leq c$)
$$|a-b| \leq c.$$
I see that it is true by drawing a number line and putting the numbers $a, b$ and $c$ there, but I have no idea how to prove it formally. It seems so trivial that it's tempting to write "obviously the following holds next to it." I tried playing around with triangle inequality but I got only this:
$$|a-b| \leq |a-c| + |b-c|=2c-a-b,$$
which is not helpful at all.

答:
What is $\color {red}{\text{max} |a-b| }$?

$$\begin{align}\color {gold}{\boxed {\color{black}{\text{max} |a-b| =|c-0|=|0-c|=c≤c.}}}\end{align}$$

Title: Proof involving a simple inequality


问:
Why does this converge to $\pi/4$?
The infinite series...

$\pi/4 = 1 - 1/3 + 1/5 - 1/7 ...$

...is very intriguing to me and seems like a crazy coincidence (its relationship to $\pi$).  Is it actually crazy or does it have an easy to explain, logical reasoning behind it that would make it seem not so magical?

答:
The infinite series $$\pi/4 = 1-1/3+1/5-1/7+ \ ...$$ can be established by finding the expression of Taylor series
\begin{equation}
f(x) = \sum_{k=0}^\infty \frac{1}{k!} f^{(k)}(a) (x-a)^k
\end{equation}
for $\arctan(x)$ for $x \in [-1,1]$ at $a = 0$ and applying the result for $x = 1$. The finite geometric sum formula
\begin{equation}
\sum_{k=0}^n q^k = \frac{1-q^{n+1}}{1-q}, \ \ q \in \mathbb{C}, \ q \neq 1
\end{equation}
is applied to find a Taylor-form series. The uniqueness of Taylor polynomial establishes the uniqueness of Taylor series. Note that we don't hence need to calculate all derivatives of $\arctan(x)$. We calculate
\begin{eqnarray}
\arctan(t) & = & \arctan(t) - \arctan(0) = \bigg\vert_0^t \arctan(x) = \int_0^t \frac{1}{1+x^2} dx \\
& = & \int_0^t \Big(\frac{1-(-x^2)^{n+1}}{1-(-x^2)} + \frac{(-x^2)^{n+1}}{1-(-x^2)} \Big) dx \\
& = & \int_0^t \frac{1-(-x^2)^{n+1}}{1-(-x^2)} dx + \int_0^t \frac{(-x^2)^{n+1}}{1-(-x^2)} dx \\
& = & \int_0^t \sum_{k=0}^n (-x^2)^k dx + \int_0^t \frac{(-x^2)^{n+1}}{1+x^2} dx \\
& = & \sum_{k=0}^n \int_0^t (-x^2)^k dx + \int_0^t \frac{(-x^2)^{n+1}}{1+x^2} dx \\
& = & \sum_{k=0}^n \int_0^t ((-1)x^2)^k dx + \int_0^t \frac{((-1)x^2)^{n+1}}{1+x^2} dx \\
& = & \sum_{k=0}^n \int_0^t (-1)^k (x^2)^k dx + \int_0^t \frac{(-1)^{n+1}(x^2)^{n+1}}{1+x^2} dx \\
& = & \sum_{k=0}^n \int_0^t (-1)^k x^{2k} dx + \int_0^t \frac{(-1)^{n+1}x^{2(n+1)}}{1+x^2} dx \\
& = & \sum_{k=0}^n (-1)^k \int_0^t x^{2k} dx + \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{1+x^2} dx \\
& = & \sum_{k=0}^n (-1)^k \bigg\vert_0^t \frac{x^{2k+1}}{2k+1} + \int_0^1 \frac{(-1)^{n+1}(tx)^{2n+2}}{1+(tx)^2} t dx \\
& = & \sum_{k=0}^n (-1)^k \frac{t^{2k+1}}{2k+1} + \int_0^1 \frac{(-1)^{n+1} t^{2n+2} x^{2n+2}}{1+(tx)^2} t dx \\
& = & \sum_{k=0}^n \frac{(-1)^k}{2k+1} t^{2k+1} + \int_0^1 \frac{(-1)^{n+1} t^{2n+3} x^{2n+2}}{1+(tx)^2} dx ,
\end{eqnarray}
where $t \in \mathbb{R}$ and $n \in \mathbb{N}$. Note that $-x^2 \neq 1$ for every $x \in \mathbb{R}$. Hence we can apply the finite geometric sum formula for every $x \in \mathbb{R}$, that allows us to calculate the Taylor polynomial for every $t \in \mathbb{R}$. Assume now $t \in [-1,1]$. To obtain the limit function we calculate
\begin{eqnarray}
\Bigg| \arctan(t) & - & \sum_{k=0}^n \frac{(-1)^k}{2k+1} t^{2k+1} \Bigg| = \Bigg| \int_0^1 \frac{(-1)^{n+1}t^{2n+3}x^{2n+2}}{1+(tx)^2} dx \Bigg| \\
& \leq & \int_0^1 \Bigg| \frac{(-1)^{n+1}t^{2n+3}x^{2n+2}}{1+(tx)^2} \Bigg| dx \\
& \leq & \int_0^1 \frac{|-1|^{n+1}|t|^{2n+3}|x|^{2n+2}}{|1+(tx)^2|} dx \\
& = & \int_0^1 \frac{1^{n+1} |t|^{2n+3} x^{2n+2}}{1+(tx)^2} dx \leq \int_0^1 \frac{|t|^{2n+3} x^{2n+2}}{1} dx \\
& = & \int_0^1 |t|^{2n+3} x^{2n+2} dx = |t|^{2n+3} \int_0^1 x^{2n+2} dx \\
& = & |t|^{2n+3} \bigg\vert_0^1 \frac{1}{2n+3} x^{2n+3} = \frac{|t|^{2n+3}}{2n+3} \leq \frac{1^{2n+3}}{2n+3} \\
& = & \frac{1}{2n+3} \rightarrow 0,
\end{eqnarray}
as $n \rightarrow \infty$. Hence
\begin{eqnarray}
\arctan(x) & = & \lim_{n \rightarrow \infty} \sum_{k=0}^n \frac{(-1)^k}{2k+1} x^{2k+1} = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} x^{2k+1}
\end{eqnarray}
for $x \in [-1,1]$. Now inserting $x = 1$ into the series expression of $\arctan(x)$ we obtain
\begin{eqnarray}
\pi/4 & = & \arctan(1) = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} 1^{2k+1} = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \\
& = & 1 - 1/3 + 1/5 - 1/7 + \ ... \ ,
\end{eqnarray}
that is the desired result. I hope that this was what you were searching for.</p>

ContentLicense: CC BY-SA 2.5

Title: Why does this converge to $\pi/4$?


问:
Determinant of a block matrix using row operations

Consider an $n \times n$ matrix of the form $$ A = \begin{pmatrix} M & P \\ 0 & N \end{pmatrix} $$ with $M \in \Bbb K^{r \times r}$, $P \in \Bbb K^{r \times s}$, and $0 \in \Bbb K^{s \times r}$ is the zero $s \times r$ matrix, $n = s+r$. Show that $$\det(A) = \det(M) \cdot \det(N)$$

I know that I have to approach this question by changing $M$ and $N$ into upper triangular matrix, but how do I do this? I am only allowed to use row operations. Thanks in advance.

Similar questions aren't showing the row operations method.

答: Let's agree to perform only row operations that multiply the determinant by $\pm 1$ and to record the product of the minus ones. If $|N|=0,$ then reduce the final $r$ rows until the last row is zero. Therefore, $|A|$ is also zero. Hence, $|N|$ is nonzero and we can reduce the final $r$ rows so that $N$ becomes a diagonal matrix with nonzero entries--which you can use to turn $P$ into a zero matrix! You can finish the proof.


答:
$\newcommand{LG}[1]{\left\lfloor \log{#1}\right\rfloor}$For $x\ne 1$, $$\sum_{k=0}^n k x^k=x\frac{d}{dx}\frac{x^{n+1}-1}{x-1}=x\frac{(n+1)x^n(x-1)-x^{n+1}+1}{(x-1)^2}=\\=\frac{nx^{n+2}-(n+1)x^{n+1}+x}{(x-1)^2}$$

Therefore $$\sum_{k=0}^{\LG n}k2^k=\frac{4\LG n2^{\LG n}-2\left(\LG n+1\right)2^{\LG n}+2}{(2-1)^2}=\\=2\LG n2^{\LG n}-2^{\LG n}+2$$

Now I presume that by $\log n$ you mean $\log_2n$, in which case we can see by $0\le\log n-\LG n\le 1$ that the quantity is in $\Theta(2^{\log n}\log n)=\Theta(n\log n)$.

If $\log n$ doesn't mean $\log_2n$, then the quantity is actually in $\Theta(n^{\log 2}\log n)$, and therefore not necessarily in $O(n\log n)$ (though it still is if you are working with $\log 2<1$).

Title: how can I calculate the sum of $i*2^i$ from $i=0$ to $\log n$?

Body:
What is $$\sum_{i=0}^{\log n}i\cdot 2^i\:?$$
I know it's $O(n\cdot\log n)$ but how do you calculate this sum?


问:
Integral transform acting on kernel with both arguments identified
This is purely a question sprung out of curiosity. If I have some $K(x,y)\in L^2(\mathbb{C}^2)$ and define the corresponding integral operator on $L^2(\mathbb{C})$

$$
L_K(f)=\int f(y)K(x,y)dy
$$

is there anything interesting that happens when we act this operator on its own kernel with both arguments identified? ie, if I define $K(x):=K(x,x)$, is there anything interesting about the action of $L_{K(x,y)}$ on $K(x)$?

答:
If $K$ is in $L^2({\mathbb C})$ then $K$ is actually not a function, but rather an equivalence class of square integrable functions modulo equality
a.e.  Since the diagonal
$$
  \Delta :=\{(x,x): x\in {\mathbb C}\}
  $$
has measure zero, you  may change the values of $K$ randomly on $\Delta $,  without changing the class of $K$  and, most
importantly,  without changing the integral operator.

I'd therefore say that your question is not a well posed question, as $K(x,x)$ can be anything you want.


问:
When addition distributes over multiplication
Everyone is familiar with distributivity of multiplication over addition of real numbers. The distributivity of two binary operations sometimes goes both ways (e.g. max and min, or for lattices in general.)

Out of curiosity, I looked at the set of real numbers for which addition distributes over multiplication. A simple computation shows that this set is

$$\{a,b,c\in \mathbb{R}\, |\, a+b+c=1\}.$$

Is this just a meaningless fluke, or is there reason to expect something like this?

答:
Addition may not be generally distributive over multiplication, but operators that are do exist.  For example the operator # defined by
a#b = exp(ln(a) * ln(b))
is distributive over multiplication:
a#(b*c) = (a#b) * (a#c)
(BTW, # so defined is also associative & commutative, and has a well-defined identity element and # inverses.  * and # can be manipulated to form a pretty variation on traditional calculus.)
Turning it around, the operator % defined by
a%b = ln(exp(a) + exp(b))
is distributed over by addition:
a+(b%c) = (a+b) % (a+c)</p>

OwnerDisplayName: Chester


问:
Complex Analysis bounding f(z)
Let $f(z)$ be an entire function such that $|f(z)|\le C|z|$ where $C$ is a positive constant. Show there exists a $D$ which is a complex constant such that $f(z)=Dz$.

I am trying to use a cauchy's estimate to show that $f''(z)=0$ but not sure how to go about this.

答:
Hint: Consider the function

$$ g(z) = \frac{f(z)}{z}$$

for $z\neq 0$. (How do you extend it to $0$?) What can you say about this function at infinity? What do you know about such functions?

Title: Complex Analysis bounding f(z)


问:
Whether the given series are convergent or divergent:
Whether the given series is absolutely convergent, conditionally convergent or divergent:

$a)\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{\cos nx}{n^2}:|(-1)^n\frac{\cos nx}{n^2}|\le\dfrac{1}{n^2}\implies$ Absolutely convergent.

$b)\displaystyle\sum_{n=1}^{\infty}(-1)^n\dfrac{n}{n+2}:(-1)^n\dfrac{n}{n+2}\to0\implies\dfrac{n}{n+2}\to0!\implies$ Divergent.

$c)\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{1}{2n+3}:u_n=\dfrac{1}{2n+3}\\\implies\dfrac{u_n}{u_{n+1}}=\dfrac{2n+3}{2n+5}=\dfrac{1+\dfrac{3}{2n}}{1+\dfrac{5}{2n}}=({1+\dfrac{3}{2n}})(1+\dfrac{5}{2n})^{-1}=({1+\dfrac{3}{2n}})(1-\dfrac{5}{2n}+\cdots)=1-\dfrac{1}{n}+O(\dfrac{1}{n^2})\\\implies \text{Not absolutely convergent. So conditionally convergent by Leibnitz test.}$

Am i right?

答: Looks good to me! Did you have any serious doubts, or are you just looking for confirmation? (If in serious doubt, why?) Assuming you mean that in $(b)$ that the limit does not approach $0$, therefore, divergent (fails absolute and conditional convergence) everything is good!</p>


问:
How can I calculate the average outcome / expected value from a list of probabilities?
I have a question regarding calculating the expected outcome from the list or probabilities.

Suppose each can play a max of 6 games. If you win the game you move to the next round. Below I have a sample of a team the probability of reaching the certain round, how can I calculate to get the average expected number of games won?

```<code>Number of Games    1      2      3        4       5       6
% Chance to win  0.97   0.92    0.83    0.679   0.545   0.384
</code>```

答:
The issue here is that you have the cumulative probabilities, and need the individual probabilities to calculate the expectation (using the standard method, at least).

If the probability of winning game 1 is $0.97$, that means this is the probability to win at least 1 game. The probability of winning exactly 1 game is the probability of winning at least 1, but not at least 2, i.e. $P(W=1)=0.97-0.92=0.05$. You can calculate the other probabilities similarly and use these to work out the expected number of games won in the usual manner.


问:
Gradient defined on level set
Suppose I have a continuously differentiable function $f:\mathbb{R}^N\rightarrow \mathbb{R}$ where the gradient is defined everywhere. Let $c$ be some constant in the range of $f$ and let $S=\{x \in \mathbb{R}^N: f(x)>c\}$. Assume $S$ is non-empty. Assume the gradient along the boundary of $S$, $\partial S$, is non-zero everywhere.

This is where I'm struggling a little with phrasing my question. If people have comments to help me fix this question's phrasing please tell me and I will change it ASAP. I have drawn the below picture to help show what I am asking.

I want to show that for any point, $p$, along the boundary of $S$, $\partial S$, if I draw a ball around $p$, the set $\partial S$ can only partition the ball into $2$ components.

On the left-hand side of the picture is an example where this condition can be seen to be the case for all $p$ on $\partial S$, whereas on the right-hand side it cannot be the case as the at the point $p$ labelled, the $\partial S$ divides the ball into 4 components.

Intuitively, I can explain why I think situations like on the right-hand side can't happen. The gradient $\nabla f$ must be normal to the level set and, in that example, I cannot draw a (non-zero) vector normal to $\partial S$ at $p$. However, I have no idea how to show this formally. Can anyone help me formalize my problem and solution?

答:
Your intuition that the gradient needs to be nonzero is good. These kinds of ideas are formalized usually by the implicit function theorem or inverse function theorem. Here  we will use the implicit function theorem, for which the idea is that "if $f$ is changing in a direction, then the solutions to $f=c$ locally are the graph of a function."

For a $C^1$ function $f: \mathbb{R}^N \to \mathbb{R}$ with $N > 1$, if the gradient at a point $p$ is nonzero, then some component is nonzero. Without loss of generality, the last component is nonzero. Then write $\mathbb{R}^N = \mathbb{R}^{N-1}\times \mathbb{R}$, with points written $(x, y)$ with $x \in \mathbb{R}^{N-1}$ and $y \in \mathbb{R}$. We have assumed that $\partial_y f(p) \neq 0$ and $f(p) = c$. Then the implicit function theorem tells us that there is an open ball $U \subseteq \mathbb{R}^{N-1}$ centered at $x_0$, where $(x_0, y_0) = p$, such that there is a unique $C^1$ function $g: U \to V \subseteq \mathbb{R}$ with $V$ a ball around $y_0$ such that $g(x_0) = y_0$ and $f(x, g(x)) = c$ for all $x \in U$
and the set in $U \times V$ where $f = c$ is the graph of $g$, i.e. $\{(x, g(x)): x \in U\} = \{(x,y) \in U \times V : f(x,y) = c\}$.

That is, locally, $g$ draws all the points where $f(x,y) = c$.
In particular this rules out your second picture, where points of the boundary are approaching from a different direction.
This nicely splits $U\times V$ in two, giving $A:=\{(x, y) \in U \times V : y > g(x)\}$ and $B:= \{(x, y) \in U \times V : y < g(x)\}$. At all the points in $A, B$ we know that
$f(x,y) \neq c$. It is not difficult to see that $A,B$ are open and path-connected (open because $g$ is continuous, path-connected by using a vertical-horizontal-vertical path.) Thus $f-c$ cannot have a sign change in $A$ or $B$, so we must have $f(A) > c$ and $f(B) < c$ or $f(A) < c$ and $f(B) > c$ pointwise, so indeed the boundary is locally given by $g$ and locally splits $f$ into the two open sets $A, B$ where $f > c$ (inside $S$) and $f< c$ (outside $S$).

Title: Gradient defined on level set


问:
The second derivative of the characteristic function and the integral of the random variable
I want to prove that if characteristic function $\bar\mu$ has a second derivative than $$\int_\mathbb{R}|x|^2\mu(dx) < +\infty.$$
I am thinking about how to apply characteristic functions, but I do not know how to do it. May somebody can help me? I would be grateful.

答:
Consider the second-order central difference of a function $f$:
$$
\delta_h^2[f](x)=f(x-2h)+f(x+2h)-2f(x).
$$
Then
$$
\delta_h^2[\bar{\mu}](t)=\int_{\mathbb{R}}e^{itx}(2i\sin xh)^2\, d\mu(x),
$$
and
$$
\left|\frac{\delta_h^2[\bar{\mu}](0)}{(2h)^2}\right|=\int_{\mathbb{R}}\left(\frac{\sin xh}{h}\right)^2\, d\mu(x).
$$
For each $N>0$,
$$
|\bar{\mu}''(0)|=\lim_{h\to 0}\left|\frac{\delta_h^2[\bar{\mu}](0)}{(2h)^2}\right|\ge \lim_{h\to 0}\int_{-N}^N\left(\frac{\sin xh}{h}\right)^2\, d\mu(x)=\int_{-N}^N x^2\,d\mu(x),
$$
which implies the result.


问:
How do I prove that all the eigenvalues are positive?

Show that the matrix

\begin{bmatrix}
2 & 0 & 0 & \cdots & 0 & 1\\
0 & 2^2 & 0 & \cdots & 0 & 1\\
0 & 0 & 2^3 & \cdots & 0 & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & 2^{n-1} & 1\\
1 & 1 & 1 & \cdots & 1 & a\\
\end{bmatrix}

only has positive eigenvalues if $a\ge 1.$

What should I use to prove this? Should I use the fact that the matrix is symmetric?

答:
$M$ is symmetric, which means that it will have positive eigenvalues if and only if it is positive definite. We can write this matrix as
$$
M = \pmatrix{D & x\\x^T & a},
$$
where
$$
D = \pmatrix{2\\ &\ddots \\&& 2^{n-1}}, \quad x = \pmatrix{1 \\ \vdots \\ 1}.
$$
It is clear that $D$ is positive definite since it is diagonal with positive diagonal entries. Thus, $M$ will be positive definite if and only if the Schur complement $M/D$ is positive definite. We compute
$$
M/D = a - x^TD^{-1}x = a - (2^{-1} + \cdots + 2^{-(n-1)}) = a - \left(1 - 2^{-n}\right).
$$
Thus, we find that the matrix $M$ is positive definite if and only if $a > (1 - 2^{-n})$.

We can therefore see that $M$ will be positive definite for all integers $n \geq 1$ if and only if
$$
a \geq \sup_{n \geq 1} (1 - 2^{-n}) = 1.
$$


答: Yes, though I don't know why you would choose to write $0!$ instead of $1$.</p>

Title: Whether the given series are convergent or divergent:

Body:
Whether the given series is absolutely convergent, conditionally convergent or divergent:

$a)\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{\cos nx}{n^2}:|(-1)^n\frac{\cos nx}{n^2}|\le\dfrac{1}{n^2}\implies$ Absolutely convergent.

$b)\displaystyle\sum_{n=1}^{\infty}(-1)^n\dfrac{n}{n+2}:(-1)^n\dfrac{n}{n+2}\to0\implies\dfrac{n}{n+2}\to0!\implies$ Divergent.

$c)\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{1}{2n+3}:u_n=\dfrac{1}{2n+3}\\\implies\dfrac{u_n}{u_{n+1}}=\dfrac{2n+3}{2n+5}=\dfrac{1+\dfrac{3}{2n}}{1+\dfrac{5}{2n}}=({1+\dfrac{3}{2n}})(1+\dfrac{5}{2n})^{-1}=({1+\dfrac{3}{2n}})(1-\dfrac{5}{2n}+\cdots)=1-\dfrac{1}{n}+O(\dfrac{1}{n^2})\\\implies \text{Not absolutely convergent. So conditionally convergent by Leibnitz test.}$

Am i right?


问:
Show that there exists a sequence with subsequences converging to every rational number.

The total question is:

(a) Give an example of a sequence with subsequences converging to $1, 2$, and $3$.

(b) Give an example of a sequence with subsequences converging to every integer.

(c) Show that there exists a sequence with subsequences converging to every rational
number.

My attempt

For (a) I got $(1,2,3,1,2,3,1,2,3,\dots)$

For (b), I got $(0,1,-1,2,-2,3,-3,4,-4,\dots)$

Both of these I think are right, but I have no idea how to start on (c), much less an example for (c). If someone could explain what I should be doing, I'd be so grateful. Thank you.

答:
Your example for (b) is wrong. There are no convergent subsequences of it.

Consider any countably infinite set $\{e_n : n \in \mathbb{N}\}$. Then consider the sequence $e_1, e_1, e_2, e_1, e_2, e_3, e_1, e_2, e_3, e_4, ...$

For all $n$, this sequence has a subsequence where every element is $e_n$.

Thus, for every countably infinite subset $E \subseteq \mathbb{R}$, we can find a sequence above where for each $e \in E$, there is a subsequence converging to $e$. In particular, we can do this for $\mathbb{Q}$ and $\mathbb{Z}$.


问:
If $(X, \mathcal O_X)$ is a scheme, what is $\mathcal O_X(U)$ where $U$ is any open subset (not necessarily affine open)?
If $(X, \mathcal O_X)$ is a scheme, what is $\mathcal O_X(U)$ where $U$ is any open subset (not necessarily affine open)?

Below is what I currently know.

Let $A$ be a ring.

Hartshorne defines the structure sheaf on $\operatorname{Spec}(A)$ as:

For an open subset $U \subset \operatorname{Spec}(A)$, we define $\mathcal O(U)$ to be the set of functions $s: U \to \coprod_{\mathfrak p \in U} A_{\mathfrak p}$, such that:

i.) $s(\mathfrak p) \in A_{\mathfrak p}$

ii.) for each $\mathfrak{p} \in U$, there is a neighborhood $V$ of $\mathfrak p$, contained in $U$, and elements $a, f \in A$ such that for each $\mathfrak q \in V$, we have $f \notin \mathfrak q$ and $s(\mathfrak q) = a/f \in A_{\mathfrak q}$.

Vakil first defines a sheaf of rings on the distinguished base $D(f) \subset \operatorname{Spec}(A)$:

Define $\mathcal O_{\operatorname{Spec} A}(D(f))$ to be the localization of $A$ at the multiplicative $\{g \in A \mid D(f) \subset D(g)\}$.

An exercise shows that $\mathcal O_{\operatorname{Spec} A}(D(f)) \cong A_f$, and that the stalk of at a point $\mathfrak p$ is $A_\mathfrak p$.

Since we have a sheaf on a base, we can extend this to all open $U \subset \operatorname{Spec} A$

Define $\mathcal O_{\operatorname{Spec} A}(U)$ to be the set of all $(s_{\mathfrak p})_{\mathfrak p \in U} \in \prod_{\mathfrak p \in U} A_\mathfrak p$ such that

i.) $s_\mathfrak p \in A_{\mathfrak p}$

ii.) for each $\mathfrak{p} \in U$, there is a neighborhood $D(f)$ of $\mathfrak p$, contained in $U$, and an element $a \in A$ such that for each $\mathfrak q \in D(f)$, we have $s_\mathfrak q = a/f^n \in A_{\mathfrak q}$.

I can see how the two definitions are related: given a function $s \in \mathcal O(U)$ as in Hartshorne, we get $(s(\mathfrak p))_{\mathfrak p \in U} \in \prod_{\mathfrak p \in U} A_\mathfrak p$ as in Vakil. The converse is similar.

Both authors define:

A scheme $(X,\mathcal O_X)$ is a ringed space such that any point of $X$ has an open neighborhood $U$ such that $(U, \mathcal O_X|_U)$ is isomorphic to an affine scheme.

I am confused about functions on a scheme $(X,\mathcal O_X)$.

It seems Hartshorne is explicitly defining what functions on an affine scheme should be in his definition, but not what functions on an arbitrary scheme should be. Why?

Vakil writes: Functions on an open subset $U$ of a locally ringed space have values at each point of $U$. The value at $p\in X$ of such a function lies in $\mathcal O_{X,p}/ \mathfrak m_p$ where $\mathfrak m_p$ is the maximal ideal of $\mathcal O_{X,p}$.

答:
Question: I am confused about functions on a scheme $(X,O_X)$. It seems Hartshorne is explicitly defining what functions on an affine scheme should be in his definition, but not what functions on an arbitrary scheme should be. Why?"

Answer: There are many discussions on this site and the MO site on the structure sheaf of a scheme and the relation with "sheaves of functions", real algebraic varieties etc. You find some information here:

Regular rational functions over real projective lines and projective spaces.

https://mathoverflow.net/questions/377922/building-algebraic-geometry-without-prime-ideals/378961#378961

Question: "Vakil writes: Functions on an open subset $U$ of a locally ringed space have values at each point of $U$. The value at $p∈X$ of such a function lies in $O_{X,p}/m_p$ where $m_p$ is the maximal ideal of $O_{X,p}$."

Answer: Locally if $U:=Spec(A)$, a section $s\in \mathcal{O}_U(U)\cong A$ is an element in the ring $A$, and given any prime ideal $\mathfrak{m}\in A$ you get a canonical "evaluation map"

$$ev: A \rightarrow A_{\mathfrak{m}}/\mathfrak{m}A_{\mathfrak{m}} :=\kappa(\mathfrak{m})$$

and $ev(s)$ is the "value" of the section $s$  at the point $\mathfrak{m}$. This is "unintuitive" since the residue field $\kappa(\mathfrak{m})$ changes with the prime ideal $\mathfrak{m}$, hence $s$ is not a function in the classical sense. There is however a way to construct for any Hilbert-Jacobson ring (scheme) $A$ a ringed topological space $Max(A)$ where you only considers maximal ideals in $A$. Such spaces are "more intuitive". One of the reasons for the introduction of prime ideals for "schemes" is to include nilpotent elements in geometry, and with the above construction of $Max(A)$ you get a locally ringed space with nilpotent elements in the structure sheaf. In the above links you find a discussion of this.

Example: Here:

Canonical bundle and skyscraper sheaf

I give an example to illustrate that a section $s$ of the structure sheaf of $S:=Spec(\mathbb{R}[x])$ does not give a function in the classical sense. This is because the (non-generic) points in $S$ have residue field $k:=\mathbb{R}$ or $K:=\mathbb{C}$. When a point has residue field $k$ you get a well defined function $s:S(k) \rightarrow k$. When the residue field is $K$ you have to make a choice. In the case of $S$ it is possible to make such a consistent choice.

For a real differentiable manifold $M$ and an open subset $U\subseteq M$ if you let $\mathcal{O}(U)$ be the set of differentiable real valued functions $f: U \rightarrow \mathbb{R}$ you get a sheaf of functions $\mathcal{O}$ on $M$ and the pair $(M, \mathcal{O})$ is a locally ringed space. In Hartshorne
he defines for any commutative unital ring $A$ a locally ringed space $(X, \mathcal{O}_X)$ where $X:=Spec(A)$ is the set of prime ideals in $A$ with the Zariski topology. A section $s \in \mathcal{O}_X(U)$ is not a function in the classical sense: There is no field $K$ with the property that you may interpret the ring $\mathcal{O}_X(U)$ as a "set of maps" $s: U \rightarrow K$ with multiplication and addition induced from the one on $K$.


问:
What property allows me to integrate a gaussian function?
Whenever I integrate a gaussian function, I get to a step that makes me a little uncomfortable because I don't fully understand it. The only way I know of to analytically integrate the gaussian function is to multiply two of them together, like so...

$$\int_{- \infty}^{\infty} e^{-x^2}dx\int_{- \infty}^{\infty} e^{-y^2}dy = \int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{-x^2-y^2}\,dx\,dy$$

My questions is, what allows me to group two different integrands together into one integrand? When is this technique not allowed? Thanks in advance.

答:
Two things are special here. Aside from the worry about improper integrals, you can express the double integral
$$\iint_R f(x,y)\, dA$$
as a product of single integrals when
(1) $R$ is a rectangle, and
(2) $f(x,y)=g(x)h(y)$ for some functions $g$ and $h$. To spare ourselves anomalies, let's assume all functions in question are continuous. </p>


问:
Prove that if there exists an infinite linearly independent subset E, of a vector space V, then V is infinite dimensional.
Def: A Hamel Basis for V is a linearly independent subset that spans V.

Def: If V has a finite Hamel Basis, F, then the dimension of V is equal to the number of elements in F, if not V is infinite dimensional.

Theorem: If V has a finite Hamel Basis then every Hamel Basis has the same number of elements.

So far I have supposed V has a finite Hamel Basis F, and E is an infinite linearly independent subset of V, I want to prove this does not hold by contradiction.

I have shown that if E spans V, F is not a finite Hamel Basis.

I think that it now suffices to show when E does not span V, F is once again not a finite Hamel Basis.

However, I am unsure of how to prove this, any help or alternative proofs would be greatly appreciated.

答: If the dimension of a vector space $V$ is $n < \infty$, then there are maximum $n$ linearly independent vectors. If there are infinitely many linearly indipendent vectors, then the dimension of $V$ is infinite.


答: If $E$ (infinite) and $F$ (finite) are both Hamel bases, then you can write the basis elements of $F$ using the basis elements of $E$. Each basis vector of $F$ by definition uses only finitely many elements of $E$, and since $F$ is finite, you can write all basis elements of $F$ using only finitely many elements of $E$. But $F$ is a basis, so any remaining basis element of $E$ that has not been used can be written in terms of elements of $F$, which can then be written in terms of finitely many elements of $E$, a contradiction, as the elements of $E$ are supposed to be linearly independent.

Title: Prove that if there exists an infinite linearly independent subset E, of a vector space V, then V is infinite dimensional.

Body:
Def: A Hamel Basis for V is a linearly independent subset that spans V.

Def: If V has a finite Hamel Basis, F, then the dimension of V is equal to the number of elements in F, if not V is infinite dimensional.

Theorem: If V has a finite Hamel Basis then every Hamel Basis has the same number of elements.

So far I have supposed V has a finite Hamel Basis F, and E is an infinite linearly independent subset of V, I want to prove this does not hold by contradiction.

I have shown that if E spans V, F is not a finite Hamel Basis.

I think that it now suffices to show when E does not span V, F is once again not a finite Hamel Basis.

However, I am unsure of how to prove this, any help or alternative proofs would be greatly appreciated.


答:
A cyclic permutation matrix takes the original (by congruence) into something more pleasant, I am calling it $H.$ . I took $a=1.$  The matrix I call $Q$ is upper triangular, all entries $\pm 1,$ with the main diagonal  and the first row are all $+1,$  the rest of the upper triangle is $-1$

$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$

Let's see, if $a > 1,$ we are simply adding my positive definite matrix $H$ below to a rank one semidefinite matrix $S,$ the single nonzero entry being $S_{11} = a-1.$

$$  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$

$$ P^T H P = D  $$
$$\left(
\begin{array}{rr}
1 & 0 \\
 - 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rr}
1 & 1 \\
1 & 2 \\
\end{array}
\right)
\left(
\begin{array}{rr}
1 &  - 1 \\
0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rr}
1 & 0 \\
0 & 1 \\
\end{array}
\right)
$$
$$ Q^T D Q = H  $$
$$\left(
\begin{array}{rr}
1 & 0 \\
1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rr}
1 & 0 \\
0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rr}
1 & 1 \\
0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rr}
1 & 1 \\
1 & 2 \\
\end{array}
\right)
$$

$$  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$

$$ P^T H P = D  $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
 - 1 & 1 & 0 \\
 - 2 & 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 1 & 1 \\
1 & 2 & 0 \\
1 & 0 & 4 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 &  - 1 &  - 2 \\
0 & 1 & 1 \\
0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 2 \\
\end{array}
\right)
$$
$$ Q^T D Q = H  $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
1 & 1 & 0 \\
1 &  - 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 2 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 1 & 1 \\
0 & 1 &  - 1 \\
0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrr}
1 & 1 & 1 \\
1 & 2 & 0 \\
1 & 0 & 4 \\
\end{array}
\right)
$$
$$  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$

$$ P^T H P = D  $$
$$\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
 - 1 & 1 & 0 & 0 \\
 - 2 & 1 & 1 & 0 \\
 - 4 & 2 & 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 1 & 1 & 1 \\
1 & 2 & 0 & 0 \\
1 & 0 & 4 & 0 \\
1 & 0 & 0 & 8 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 &  - 1 &  - 2 &  - 4 \\
0 & 1 & 1 & 2 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 4 \\
\end{array}
\right)
$$
$$ Q^T D Q = H  $$
$$\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
1 & 1 & 0 & 0 \\
1 &  - 1 & 1 & 0 \\
1 &  - 1 &  - 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 4 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 1 & 1 & 1 \\
0 & 1 &  - 1 &  - 1 \\
0 & 0 & 1 &  - 1 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrrr}
1 & 1 & 1 & 1 \\
1 & 2 & 0 & 0 \\
1 & 0 & 4 & 0 \\
1 & 0 & 0 & 8 \\
\end{array}
\right)
$$

$$  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$

$$ P^T H P = D  $$
$$\left(
\begin{array}{rrrrr}
1 & 0 & 0 & 0 & 0 \\
 - 1 & 1 & 0 & 0 & 0 \\
 - 2 & 1 & 1 & 0 & 0 \\
 - 4 & 2 & 1 & 1 & 0 \\
 - 8 & 4 & 2 & 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrrr}
1 & 1 & 1 & 1 & 1 \\
1 & 2 & 0 & 0 & 0 \\
1 & 0 & 4 & 0 & 0 \\
1 & 0 & 0 & 8 & 0 \\
1 & 0 & 0 & 0 & 16 \\
\end{array}
\right)
\left(
\begin{array}{rrrrr}
1 &  - 1 &  - 2 &  - 4 &  - 8 \\
0 & 1 & 1 & 2 & 4 \\
0 & 0 & 1 & 1 & 2 \\
0 & 0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrrrr}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 \\
0 & 0 & 0 & 4 & 0 \\
0 & 0 & 0 & 0 & 8 \\
\end{array}
\right)
$$
$$ Q^T D Q = H  $$
$$\left(
\begin{array}{rrrrr}
1 & 0 & 0 & 0 & 0 \\
1 & 1 & 0 & 0 & 0 \\
1 &  - 1 & 1 & 0 & 0 \\
1 &  - 1 &  - 1 & 1 & 0 \\
1 &  - 1 &  - 1 &  - 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrrr}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 \\
0 & 0 & 0 & 4 & 0 \\
0 & 0 & 0 & 0 & 8 \\
\end{array}
\right)
\left(
\begin{array}{rrrrr}
1 & 1 & 1 & 1 & 1 \\
0 & 1 &  - 1 &  - 1 &  - 1 \\
0 & 0 & 1 &  - 1 &  - 1 \\
0 & 0 & 0 & 1 &  - 1 \\
0 & 0 & 0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrrrr}
1 & 1 & 1 & 1 & 1 \\
1 & 2 & 0 & 0 & 0 \\
1 & 0 & 4 & 0 & 0 \\
1 & 0 & 0 & 8 & 0 \\
1 & 0 & 0 & 0 & 16 \\
\end{array}
\right)
$$

$$  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc  $$

$$ P^T H P = D  $$

$$\left(
\begin{array}{rrrrrr}
1 & 0 & 0 & 0 & 0 & 0 \\
 - 1 & 1 & 0 & 0 & 0 & 0 \\
 - 2 & 1 & 1 & 0 & 0 & 0 \\
 - 4 & 2 & 1 & 1 & 0 & 0 \\
 - 8 & 4 & 2 & 1 & 1 & 0 \\
 - 16 & 8 & 4 & 2 & 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrrrr}
1 & 1 & 1 & 1 & 1 & 1 \\
1 & 2 & 0 & 0 & 0 & 0 \\
1 & 0 & 4 & 0 & 0 & 0 \\
1 & 0 & 0 & 8 & 0 & 0 \\
1 & 0 & 0 & 0 & 16 & 0 \\
1 & 0 & 0 & 0 & 0 & 32 \\
\end{array}
\right)
\left(
\begin{array}{rrrrrr}
1 &  - 1 &  - 2 &  - 4 &  - 8 &  - 16 \\
0 & 1 & 1 & 2 & 4 & 8 \\
0 & 0 & 1 & 1 & 2 & 4 \\
0 & 0 & 0 & 1 & 1 & 2 \\
0 & 0 & 0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrrrrr}
1 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 & 0 \\
0 & 0 & 0 & 4 & 0 & 0 \\
0 & 0 & 0 & 0 & 8 & 0 \\
0 & 0 & 0 & 0 & 0 & 16 \\
\end{array}
\right)
$$
$$  $$

$$ Q^T D Q = H  $$

$$\left(
\begin{array}{rrrrrr}
1 & 0 & 0 & 0 & 0 & 0 \\
1 & 1 & 0 & 0 & 0 & 0 \\
1 &  - 1 & 1 & 0 & 0 & 0 \\
1 &  - 1 &  - 1 & 1 & 0 & 0 \\
1 &  - 1 &  - 1 &  - 1 & 1 & 0 \\
1 &  - 1 &  - 1 &  - 1 &  - 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrrrr}
1 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 & 0 \\
0 & 0 & 0 & 4 & 0 & 0 \\
0 & 0 & 0 & 0 & 8 & 0 \\
0 & 0 & 0 & 0 & 0 & 16 \\
\end{array}
\right)
\left(
\begin{array}{rrrrrr}
1 & 1 & 1 & 1 & 1 & 1 \\
0 & 1 &  - 1 &  - 1 &  - 1 &  - 1 \\
0 & 0 & 1 &  - 1 &  - 1 &  - 1 \\
0 & 0 & 0 & 1 &  - 1 &  - 1 \\
0 & 0 & 0 & 0 & 1 &  - 1 \\
0 & 0 & 0 & 0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrrrrr}
1 & 1 & 1 & 1 & 1 & 1 \\
1 & 2 & 0 & 0 & 0 & 0 \\
1 & 0 & 4 & 0 & 0 & 0 \\
1 & 0 & 0 & 8 & 0 & 0 \\
1 & 0 & 0 & 0 & 16 & 0 \\
1 & 0 & 0 & 0 & 0 & 32 \\
\end{array}
\right)
$$

$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$

Title: How do I prove that all the eigenvalues are positive?

Body:
Show that the matrix

\begin{bmatrix}
2 & 0 & 0 & \cdots & 0 & 1\\
0 & 2^2 & 0 & \cdots & 0 & 1\\
0 & 0 & 2^3 & \cdots & 0 & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & 2^{n-1} & 1\\
1 & 1 & 1 & \cdots & 1 & a\\
\end{bmatrix}

only has positive eigenvalues if $a\ge 1.$

What should I use to prove this? Should I use the fact that the matrix is symmetric?


问:
How to extract Parametric Equations from a Problem
How can I extract parametric equations from a scenario which is given in a problem like this:

A hitter's bat meets a ball about 3 ft above home plate. Suppose the horizontal component of the ball's initial velocity is 100 ft/s and the vertical component is 60 ft/s.

The equations that was shown in example were:

\begin{align*}
x&= 100t\\
y&=3+60t-16t^2
\end{align*}

答:
You need to understand what are the equations of motion for the ball. You can decompose the motion into two components - horizontal and vertical. In the horizontal direction you have no acceleration, so the velocity is constant (equal to the initial horizontal velocity). If you assume the initial $x$ coordinate to be $0$, $$x=x_0+v_xt$$ becomes $$x=0+100t=100t$$
In the vertical direction you start at $y_0=3$, with a vertical velocity upwards $v_y=60$. The acceleration is downwards $g=-32ft/s^2$. The equation of motion for constant acceleration is $$y=y_0+v_yt+\frac12gt^2=3+60t-16t^2$$


答:
You have
$$
\int_a^b \left( \int_c^d f(x)g(y)\,dx\right)\,dy.
$$
Look at the inside integral:
$$
\int_c^d f(x)g(y)\,dx.
$$
The thing you're integrating is $g(y)$ times something, and you have $x$ going from $c$ to $d$.  As $x$ goes from $c$ to $d$, notice that $g(y)$ DOES NOT CHANGE.  $g(y)$ does not depend on $x$.  And $g(y)$ is a FACTOR of the function you're integrating, i.e. what you're integrating is $g(y)$ times something.  Since this factor does not depend on $x$, it's a CONSTANT as a function of $x$, so it can be pulled out, getting
$$
g(y)\int_c^d f(x)\,dx.
$$
Now you have
$$
\int_a^b g(y)\left(\int_c^d f(x)\,dx\right) \,dy.
$$
Now notice that the expression in parentheses does not depend on $y$, and you're integrating with respect to $y$ as $y$ goes from $a$ to $b$.  In other words, as a function of $y$, that expression is a CONSTANT, so it can likewise be pulled out.  Then you're done.</p>

Title: What property allows me to integrate a gaussian function?

Body:
Whenever I integrate a gaussian function, I get to a step that makes me a little uncomfortable because I don't fully understand it. The only way I know of to analytically integrate the gaussian function is to multiply two of them together, like so...

$$\int_{- \infty}^{\infty} e^{-x^2}dx\int_{- \infty}^{\infty} e^{-y^2}dy = \int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{-x^2-y^2}\,dx\,dy$$

My questions is, what allows me to group two different integrands together into one integrand? When is this technique not allowed? Thanks in advance.


问:
How to evaluate integrals of type $\int\frac{\cos^2x}{\cos^2x+ 4\sin^2x}\,dx$
I think it is by substitution but I don't  have any clue on how to go further. Should I substitute $$\sin^2x=t~?$$

$$\int\frac{\cos^2x}{\cos^2x+ 4\sin^2x}\,dx$$

答:
Integrate as follows

\begin{align}
\int\frac{\cos^2x}{\cos^2x+ 4\sin^2x}\,dx
=\frac13 \int\left( -1+ \frac{4}{4-3\cos^2x}\right)dx\\
=-\frac13x +  \frac13\int  \frac{\sec^2x}{\frac14+\tan^2x}dx\\
=-\frac13x +  \frac23 \tan^{-1}(2\tan x)+C
\end{align}


问:
Prove we cannot make a homeomorphism from [0, 1] to ℝ by contradiction.
So start off contradiction proofs by assuming the opposite.
So we assume f is a homeomorphism from ℝ to [0, 1].

Since f is a surjection, there exists some  ∈ ℝ with $f() = 0$.
Let $x_1 =  – 1$ and $x_2 =  + 1$.

Since f is an injection, and since $x_1≠ x_2 ≠ $,we know that $f(x_1) ≠ f(x_2) ≠ 0$.

Let $M = min \{ f(x_1), f(x_2) \}$.

I know I have to use the intermediate value theorem at some point but I'm stuck at the next step.

答: Homeomorphisms are continuous. Continuous functions map compact sets to compact sets. $[0,1]$ is compact. $\mathbb{R}$ is not compact. Thus there is no continuous function from $[0,1]$ to $\mathbb{R}$ that has image all of $\mathbb{R}$.


答: You are allowed to do this due to  Fubini's theorem</p>


问:
Example of homeomorphisms with different supremum metric
I hope you can help me. I want to know if there are homeomorphisms $f,g:X\to X$, when $(X,d)$ is compact, such that
$$D(f,g)<D(f^{-1},g^{-1}),$$
where $D(a,b)=\sup_{x\in X}d(a(x),b(x))$.

Any suggestion is welcome :)

答: Yes. Let $X = [0,4]$. We shall specify piecewise linear homeomorphisms $f, g$ by their graphs in the square $[0,4]^2$. The graph of $f$ consists of the two line segments $F_1$ connecting $(0,0)$ and $(1,2)$ and $F_2$ connecting $(1,2)$ and $(4,4)$. The graph of $g$ consists of the two line segments $G_1$ connecting $(0,0)$ and $(1,3)$ and $G_2$ connecting $(1,3)$ and $(4,4)$. Drawing a picture is helpful. You can easily verify that the distance $D(f,g) = 1$ which is attained at $x=1$. The graphs of $f^{-1},g^{-1}$ are obtained by reflecting the graphs of $f,g$ at the diagonal $y = x$ (draw again a picture). You will see that $\lvert f^{-1}(3) - g^{-1}(3) \rvert > 1$, thus $D(f^{-1},g^{-1}) > 1 = D(f,g)$.


问:
If every prime ideal is a contracted ideal, does it imply that the induced map between spectrums is surjective?
Let $f: A \rightarrow B$ be a ring homomorphism between commutative rings with identity. Then there exists an induced map $f' : Spec(B) \rightarrow Spec(A)$. If $f'$ is surjective, then clearly every prime ideal of $A$ is a contracted ideal. Now my question is, is the converse true?</p>

答:
By atiyah-macdonald thm 3.15
If there is an homomorphism from A to B and p^{ec}=p, then p is a contraction of a prime ideal of B.

In your question p is contracted ideal, so p satisfies p^{ec}=p


问:
random sequence converges to a different value each time
I am simulating a random walk. Within each execution, the sequence converges to a fixed number. However, the number to which the sequence converges varies in different executions. Is this still called "the random variable converges?" If so, does it converge in probability / almost surely / in distribution? Thank you for your answer!</p>

答: In this situation, you are saying you have a sequence of random variables $(X_i)_i$ such that for every $\omega$ in your sample space you have $X_i(\omega)$ converges. In this case you would say that $X_i$ converges, or $X_i$ converges for every $\omega$. This kind of convergence is the strictest amongst the ones you mention. If $X_i$ converges for every $\omega$, then it also converges almost surely, and it also converges in probability, and it also converges in distribution.


问:
Finding point in 3D Space on line with fixed distance
I tried to look for similar questions but couldn't find something relating so here goes:

I have 3 points in 3D Space: Point A, Point B and Point O.
Point O is just the origin at [0, 0, 0].

I need to define a point C, which is between A and B, and exactly at distance d from O.

How I tried to approach this at first was to construct a direction from A to B, which can be used to define the line between A and B, then using the formula f = A + t * (A - B) to define the line. But then I got stuck and used the very ugly brute force method of looping over possible values of t until the distance between the point at said t was d from O, but I want to know if it can be done in one single simple mathematical equation to calculate C.

Kind regards and thanks in advance.

答:
If point $p(\lambda)$ in the segment $\lambda A+(1-\lambda)B,\ \ 0\le \lambda\le 1$ is located at a distance $d$ from $O$ then

$$
\|\lambda A+(1-\lambda)B\|^2= \|B+\lambda(A-B)\|^2 = \|B\|^2+2\lambda B\cdot (A-B)+\lambda^2\|A-B\|^2 = d^2
$$

or solving for $\lambda$

$$
\lambda^* = \frac{B\cdot(B-A)\pm\sqrt{(B\cdot(B-A))^2 -\|A-B\|^2(\|B\|^2-d^2) }}{\|A-B\|^2}
$$

Here if $\ 0\le \lambda^*\le 1$ then $C = \lambda^* A+(1-\lambda^*)B$


答:
Let's go through this line-by-line, and translate the written statements into mathematical statements.  To begin, let $x(t)$ and $y(t)$ be the horizontal and vertical positions, in feet, of the ball in relation to home plate at time $t$.  Then, we have the following:

A hitter's bat meets a ball about $3$ ft above home plate.

This tells us that that at time $t = 0$ the ball is at $(x,y) = (0,3)$, because it is at home plate ($x=0$), but $3$ ft off the ground ($y=3$).  This gives us $x(0) = 0$, and $y(0) = 3$.

The horizontal component of the ball's initial velocity is $100$ ft/s

This tells us that the $x'(0) = 100$, because $x'(t)$ is the horizontal component of the ball's velocity.

The vertical component (of the ball's initial velocity) is $60$ ft/s.

As before, this tells us that $y'(0) = 60$, because $y'(t)$ is the vertical component of the ball's velocity.

So, we've established the following information:
\begin{align*}
x(0) &=0\\
y(0) &= 3\\
x'(0) &= 100\\
y'(0) &= 60.
\end{align*}

Lastly, we need to remember an additional assumption about projectile motion, namely that acceleration is only due to gravity.  Remembering that $x''(t)$ and $y''(t)$ are the horizontal and vertical components of acceleration, respectively, this provides us with
\begin{align*}
x''(t) &= 0 \text{ ft/s$^2$}\\
y''(t) &= -32 \text{ ft/s$^2$}.
\end{align*}

So, working our way backwards, first for $x(t)$:
\begin{align*}
x''(t) = 0 &\implies x'(t) = c_1\\[5pt]
x'(0) = 100 &\implies c_1 = 100 \implies x'(t) = 100\\[5pt]
x'(t) = 100 &\implies x(t) = 100t + c_2\\[5pt]
x(0) = 0 &\implies c_2 = 0 \implies \boxed{x(t) = 100t}
\end{align*}

Next, for $y(t)$:
\begin{align*}
y''(t) = -32 &\implies y'(t) = -32t + c_3\\[5pt]
y'(0) = 60 &\implies c_3 = 60 \implies y'(t) = -32t + 60\\[5pt]
y'(t) = -32t + 60 &\implies y(t) = -16t^{2} + 60t + c_4\\[5pt]
y(0) = 3 &\implies c_4 =3 \implies \boxed{y(t) = -16t^2 + 60t + 3}
\end{align*}

Title: How to extract Parametric Equations from a Problem

Body:
How can I extract parametric equations from a scenario which is given in a problem like this:

A hitter's bat meets a ball about 3 ft above home plate. Suppose the horizontal component of the ball's initial velocity is 100 ft/s and the vertical component is 60 ft/s.

The equations that was shown in example were:

\begin{align*}
x&= 100t\\
y&=3+60t-16t^2
\end{align*}


问:
Notation Involving Sobolev Spaces I Cannot Find the Meaning Of
I've been reading some research papers in PDEs and I've run into a notation a couple of times that googling has not helped me define. It looks something like $W^{1,p}(0,T; L^2(\Omega))$. I know what each of these things mean individually.
$$L^p(\Omega)=\{u: \int_{\Omega}|u|^p<\infty\}$$
$$W^{1,p}=\{u:u\in L^p, Du\in L^p\}$$
and I understand that the $(0,T)$ mean we are talking about over the time (since it is usually talking about the heat equation or something similar).

I would understand if it were $W^{1,p}(0,T; \Omega)$ as I would guess that would be $$\{u:\int_{[0,T]}\int_{\Omega}|u|^p<\infty, \int_{[0,T]}\int_{\Omega}|Du|^p<\infty\}$$ (though I could be wrong). I just don't know what to do with the $L^2$ inside the $W^{1,p}.$ Is it a way to say that we're assuming $u\in L^2$? Is it saying we're integrating over $L^2(\Omega)$ somehow?

Any help with understanding what this notation means would be helpful.

答:
The solution of the PDE you are trying to solve is a function

$$u : [0,T] \times \Omega \to \mathbb{R}.$$

In other words, given a time and a spatial coordinate, you get a value. By currying, this is equivalent to a function $U$ from $[0,T]$ (the range) to functions from $\Omega$ to $\mathbb{R}$. From this point of view, for any time $t$, you have a function $U(t)(\cdot) = u(t, \cdot)$ on $\Omega$.

The notation $W^{1,p} (0,T; \mathbb{L}^2 (\Omega))$ is coherent with this second point of view: the indices $0$, $T$ correspond to the domain of $U$, while $\mathbb{L}^2 (\Omega)$ is its range. In other words, $U$ is seen as a function from $[0,T]$ taking its values in $\mathbb{L}^2 (\Omega)$. The derivative of $U$ is then the partial derivative of $u$ with respect to time.

To expand things a little in your example (a reference would be nice to check that I did not make a mistake here):

for (almost?) any $t \in [0,T]$, the function $u(t, \cdot)$ belongs to $\mathbb{L}^2 (\Omega)$:

$$\left( \int_\Omega |u(t,x)|^2 \ dx \right)^{\frac{1}{2}} < +\infty.$$

the function $U: t \mapsto u(t, \cdot)$ belongs to $\mathbb{L}^1 (0,T; \mathbb{L}^2 (\Omega))$:

$$\int_0^T\left( \int_\Omega |u(t,x)|^2 \ dx \right)^{\frac{1}{2}} \ dt< +\infty.$$

the function $U: t \mapsto u(t, \cdot)$ belongs to $W^{1,p} (0,T; \mathbb{L}^2 (\Omega))$:

$$\left( \int_0^T\left( \int_\Omega |u(t,x)|^2 \ dx \right)^{\frac{p}{2}} \ dt \right)^{\frac{1}{p}} < +\infty \\ \text{ and } \\ \left( \int_0^T\left( \int_\Omega |\partial_t u(t,x)|^2 \ dx \right)^{\frac{p}{2}} \ dt \right)^{\frac{1}{p}}< +\infty.$$


问:
Arbitrarily large $β < \kappa$ closed under a function $g : \kappa → \kappa$ where $\kappa$ is a regular unbounded cardinal
I'm reading through my axiomatic set theory lecture notes and came across the following result:

Suppose $\kappa$ is an uncountable regular cardinal. Let $g : \kappa → \kappa$ be any
function. Prove that for any $α < \kappa$, there exists $β < \kappa$, with $α ≤ β$, such that
$β$ is closed under $g$ (i.e. for all $γ < β$, $g(γ) < β$).

The proof given is the following:

We define recursively for
$n ∈ ω$, $α_0 = α$ and $α_{n+1} = \textrm{sup} g[α_n]$. Since $\kappa$ is a cardinal, if $α_n \in \kappa$ then
$\textrm{sup} g[α_n] \in \kappa$, so all $α_n \in \kappa$ (by induction). Since $\kappa$ is regular uncountable this
implies $β = \textrm{sup } α_n ∈ \kappa$. This $β$ works since if $δ \in β$ then $δ ∈ α_n$ for some $n ∈ ω$
and hence $g(δ) \in g[α_n] = α_{n+1} ⊆ β$.

I can understand how each step follows except where we conclude that since $\kappa$ is regular uncountable this implies $β = \textrm{sup } α_n \in \kappa$ - I don't see how this follows. The previous fact that if $α_n \in \kappa$ then
$\textrm{sup} g[α_n] \in \kappa$ clearly follows from regularity since if $α_n ∈ \kappa$ then $g|_{α_n}$ must be bounded in $\kappa$, but just because every $α_n$ is in $\kappa$ it doesn't follow that the supremum must be - presumably the supremum could be $\kappa$ itself.

Presumably uncountability comes into it somehow, but I can't work out how it does. If anyone could point me in the right direction I'd really appreciate it!

答: If $\sup_n\alpha_n$ were equal to $\kappa$, $\kappa$ would have countable cofinality: it would be the limit of the sequence $\langle\alpha_n:n\in\omega\rangle$. But $\kappa$ is regular and uncountable, so $\operatorname{cf}\kappa=\kappa>\omega$.


问:
Proof that the sum of exterior angles of a complex polygon or a non-planar polygon is greater than 2 Pi
In the context of writing a ray-tracing program, I must be able to determine if a polygon in three-dimensional space is a planar simple convex polygon. My conjecture is that if I take the sum of the exterior angles (absolute values, always in the range $[0, \pi]$) then this sum is equal to $2\pi$ if and only if it is a planar simple convex polygon.

I firmly believe this to be true, but I seek somewhat rigorous proof. I already managed to show that the sum equals $2\pi$ for convex polygons and that it exceeds $2\pi$ for concave ones. However, I do not have any ideas on how to continue to prove that complex and non-planar polygons have a sum of exterior angles greater than $2\pi$.

For complex planar polygons, I managed to show (w.l.o.g.) that the exterior angles which "turn to the right" minus those that "turn to the left" must be a $2\pi n$ for some natural $n$. From there it's trivial to show that if $n>1$ then the total must be greater than $2\pi$, but for $n=0$ and $n=1$ I do not know how to proceed.

For non-planar polygons, I thought about what would happen if we project it to 2-dimensions then somehow show that the projection has an exterior angle sum strictly between $2\pi$ and that of the non-planar version, but I once again cannot fathom how to show that. (Edit: this idea does not work since there are counterexamples of non-planar polygons whose projections onto two-dimensional space have a larger exterior angle sum.)

答:
The answer lies in Lemma 2.1 of Sullivan, Curves of Finite Total Curvature.

The paper refers to exterior angles as turning angles, and for a polygon $P$ designates the sum of turning angles (total curvature) as $TC(P)$

Here's a paraphrased version of the key lemma:

Lemma 2.1.  Suppose $P$ is a polygon in $\mathbb R^3$. If $P'$ is obtained from $P$ by deleting one vertex $v_n$ then $TC(P') ≤ TC(P)$.
We have equality here if $v_{n−1}v_n v_{n+1}$ are collinear in that
order, or if $v_{n−2}v_{n−1}v_{n}v_{n+1}v_{n+2}$ lie convexly in some
plane, but never otherwise.

Please refer to the paper for a proof, along with background and definitions.  It's a short simple argument, using a mapping that takes 3D polygons $P$ to spherical polygons $P_S$ on the unit sphere $\mathbb S^2$.  The total curvature of $P$ is the same as the boundary length of $P_S$.

Lemma 2.1 nicely takes care of both cases.

For a non-planar polygon $P$, removing vertices until only a triangle $T$ is left defines a sequence of polygons whose total curvature is strictly decreasing in at least one pair of the sequence (corresponding to where some $v_{n−2}v_{n−1}v_{n}v_{n+1}v_{n+2}$ is not planar).  Since $TC(T)=2\pi$ we conclude that  $TC(P)>2\pi$.

Similarly, for complex planar polygons $P$ there will be vertex
sequences $v_{n−2}v_{n−1}v_{n}v_{n+1}v_{n+2}$ that are not convex and again $TC(P)>2\pi$.


问:
Degenerate conic composed of two lines
In reading the text Multi-View Geometry for Computer vision, after the notion of a conic is introduced with its corresponding conic coefficient matrix, an example is given on page 32 stating

The conic $C= lm^T + ml^T$ is composed of two lines $l$ and $m$. Points on $l$ satisfy $l^T x = 0$, and are on the conic since $x^TCx = (x^T l)(m^Tx) + (x^Tm)(l^Tx) = 0$. Similarly, points satisfying $m^Tx = 0$ also satisfy $x^TCx=0$.

I'm not sure what this notation means. I'm guessing that $C$ is the sum of two outer products, since otherwise it would be a sum of scalars, and then $x^TCx$ would be a scalar multiple of the squared norm of $x$.

If it is indeed a sum of outer products, then is it generally true that for an outer product $l \bigotimes m^T$ that $x^T l \bigotimes m^T x = (x^T l)(m^Tx)$ ? (where the latter expression is a product of dot products)

答:
We are given column vectors
$$ x = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}, \quad
l = \begin{pmatrix} l_1 \\ l_2 \\ l_3 \end{pmatrix}, \quad
m = \begin{pmatrix} m_1 \\ m_2 \\ m_3 \end{pmatrix}.\tag{1} $$
Note that
$$ l\,m^T := \begin{pmatrix} l_1m_1 & l_1m_2 & l_1m_3 \\
l_2m_1 & l_2m_2 & l_2m_3 \\l_3m_1 & l_3m_2 & l_3m_3  \end{pmatrix}
\tag{2} $$ is a rank $1$ matrix.

The equation $$ l^T x := l_1x_1 + l_2x_2 + l_3x_3 = 0 \tag{3} $$
is the equation representing points $\,x\,$ on the line $\,l.\,$

The equation $$ m^T x := m_1x_1 + m_2x_2 + m_3x_3 = 0 \tag{4} $$
is the equation representing points $\,x\,$ on the line $\,m.\,$

Define the matrix
$$ C := lm^T + ml^T = \begin{pmatrix}
2l_1m_1 & l_1m_2+l_2m_1 & l_1m_3+l_3m_1 \\
l_1m_2+l_2m_1 & 2l_2m_2 & l_2m_3+l_3m_2 \\
l_1m_3+l_3m_1 & l_2m_3+l_3m_2 & 2l_3m_3
\end{pmatrix}. \tag{5} $$

This rank $2$ matrix represents a degenerate conic which
consists of the two lines $\,l\,$ and $\,m.\,$ The reason is
what you mentioned in your question. We have
$$ x^TCx = (x^Tl)(m^Tx) + (x^Tm)(l^Tx) = 2(l^Tx)(m^Tx)\tag{6} $$
since $\, l^Tx = x^Tl \,$ and $\, m^Tx = x^Tm \,$ are both
scalars. Their product is zero precisely when either of them
are zero and each factor is the equation of a line. Hence,
this proves the claim about $\,C\,$ representing two lines.

Title: Degenerate conic composed of two lines


问:
The maximum value of $p>0$ such that $(1+\frac{1}{n})^{n+p}<e$
Find the maximum value of $p>0$ such that $(1+\frac{1}{n})^{n+p}<e$ for all positive integer.

I don’t really know how to start this problem. I think an idea is to consider some function like $f(x) = \log_{1+\frac{1}{n}}{\frac{e}{(1+\frac{1}{n})^n}}$, $f(n)>p$, analyze it using derivatives and find its minimum, but I’m not sure. There might be some other ways. Can you help me?

答:
As will be shown below no such $p$ exists. However we indeed can find such $p>0$ that the inequality
$$
\left(1+\frac1n\right)^{n+p}\color{red}\le e
$$
is valid for any postive integer.

Preliminary we prove the following

Lemma 1:

$$\forall x\ge0:\quad\frac{x}{\sqrt{1+x}}\ge \log(1+x).\tag1$$

Proof:

Let $u(x)=\dfrac{x}{\sqrt{1+x}}-\log(1+x)$. Obviously  $u(0)=0$ and
$$
u'(x)=\frac1{(1+x)^{1/2}}-\frac x{2(1+x)^{3/2}}-\frac1{1+x}
=\frac{2+x-2\sqrt{1+x}}{2(1+x)^{3/2}}=\frac{(\sqrt{1+x}-1)^2}{2(1+x)^{3/2}}\ge0,
$$
so that claim $(1)$ is proved. Observe that $(1)$ is a strict inequality for all $x>0$.

Substituing in $(1)$ $t=\dfrac1x$ one obtains equivalent form:
$$
\forall t>0:\quad\frac{1}{\sqrt{t(1+t)}}> \log\left(1+\frac1t\right)
\iff \sqrt{t(1+t)}\,\log\left(1+\frac1t\right)<1.\tag2
$$

Consider now the following equivalent inequalities:
$$
\left(1+\frac1n\right)^{n+p}\le e\iff (n+p)\log\left(1+\frac1n\right)\le1
\iff p\le\frac1{\log\left(1+\frac1n\right)}-n.\tag3
$$
Let us prove that the function
$$
f(x)=\frac1{\log\left(1+\frac1x\right)}-x
$$
is increasing for $x>0$. Indeed
$$
f'(x)=\frac1{x(1+x)\left[\log\left(1+\frac1x\right)\right]^2}-1>0
$$
in view of $(2)$.

Since $f(n)$ is increasing the largest possible $p$ corresponds to $n=1$, i.e.
$$
p=\frac1{\log2}-1\approx0.442695.
$$


问:
Proving almost sure convergence
Assume the sequence of random variables $X_1, X_2, \cdots$ are IID with finite mean and finite variance. Define a random variable:

\begin{align}
Y_n = \frac{X_n}{n}
\end{align}
Show that $Y_n \to 0$ almost surely.

To converge to some value almost surely implies:
\begin{align}
\mathbb{P}\left( \lim_{n \to \infty}{Y_n} = 0\right) = 1
\end{align}
By the way $Y_n$ is defined, this is equivalent to:
\begin{align}
\mathbb{P}\left( \lim_{n\to \infty}{\frac{X_n}{n}} = 0\right) =1
\end{align}

But, the limit of $\frac{1}{n}$ as $n \to \infty$ is "obviously" $0$ and intuitively I would think that if the expectation of the numerator is finite then we can expect some finite number in the numerator. As $n$ runs off to $\infty$ it would just look like some finite number (it doesn't really matter what the number is so long as it is finite) being divided by a number growing larger and larger - approaching $\infty$. So, I would think the $\lim_{n\to \infty} \frac{X_n}{n}$ would behave similarly to $\lim_{n \to \infty} \frac{1}{n}$.

EDIT: but as pointed out in the comments, the limit of $X_n$ is not necessarily $0$ for all possible definitions of $X_n$.

Thank you.

答:
The result is still true even without the assumption of finite variance. Since $E|X|$ is finite, we have $E|X/\epsilon|$ is finite for all $\epsilon>0$. Using the layer cake formula, we get
$$
\infty>E|X/\epsilon |=\int_0^\infty P(|X/\epsilon|>t)\,dt\ge\sum_{n=0}^\infty P(|X/\epsilon|>n)=\sum_n P(|X/n|>\epsilon)
$$
Since the series $P(|X/n|>\epsilon)$ has a finite sum, the first Borel Cantelli lemma implies $P(|X_n/n|>\epsilon\text{ infinitely often})=0$. This holds for all $\epsilon>0$, which implies that $X_n/n\to 0$.

Title: Proving almost sure convergence


问:
How do I prove that all the eigenvalues are positive?

Show that the matrix

\begin{bmatrix}
2 & 0 & 0 & \cdots & 0 & 1\\
0 & 2^2 & 0 & \cdots & 0 & 1\\
0 & 0 & 2^3 & \cdots & 0 & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & 2^{n-1} & 1\\
1 & 1 & 1 & \cdots & 1 & a\\
\end{bmatrix}

only has positive eigenvalues if $a\ge 1.$

What should I use to prove this? Should I use the fact that the matrix is symmetric?

答:
I would personally endorse a caveman like approach to this.

Assume $\lambda$ is an eigenvalue and $\textbf{x} = (x_1, \dots x_n)$ is an eigenvector. Let's see what we get from the equation $Ax = \lambda x$ where $A$ is the matrix you've given.
We get the equations
\begin{align*}
2x_1 + x_n &= \lambda x_1  \ \ \  \ \ \ \Longrightarrow x_1 = \ \frac{x_n}{\lambda -2}\\
2^2x_2 + x_n &= \lambda x_2 \ \ \ \ \ \  \Longrightarrow x_2 = \frac{x_n}{\lambda -2^2}\\
& \ \  \vdots \ \\
2^{n-1}x_{n-1} + x_n &= \lambda x_{n-1} \ \ \Longrightarrow x_{n-1} = \frac{x_n}{\lambda -2^{n-1}} \\ \
\end{align*}

and lastly, we can use the above equations to obtain the following: $$ x_1  + \dots + ax_n  = \lambda x_n \ \  \Longrightarrow \frac{x_n}{(\lambda -2)} + \frac{x_n}{(\lambda -2^2)} + \dots \frac{x_n}{(\lambda -2^{n-1})} + a= \lambda x_n $$
The above equations imply that if $x_n = 0$ then $x_1 = \dots = x_{n-1} = 0$. So $\textbf{x}$ is the $0$ vector in which case x is not an eigenvector. So we may assume $x_n \neq 0$ and we can divide by $x_n$. We obtain

$$\frac{1}{(\lambda -2)} + \frac{1}{(\lambda -2^2)} + \dots \frac{1}{(\lambda -2^{n-1})} + a= \lambda   $$

If $a \geq 1$, then we have

$$\lambda - \frac{1}{(\lambda -2)} - \frac{1}{(\lambda -2^2)} - \dots  -\frac{1}{(\lambda -2^{n-1})}  = a\geq 1$$

from which we conclude

$$\lambda \geq 1+  \frac{1}{(\lambda -2)} + \frac{1}{(\lambda -2^2)} + \dots \frac{1}{(\lambda -2^{n-1})} $$

If $\lambda \leq 0$, this isn't possible. Geometric series shows the RHS is bounded below by $1/2^n$ if $\lambda \leq 0$.


问:
What can you say about groups if every homomorphism is trivial?
Suppose we have two groups $G,H$, and the premise is that every homomorphism from $G$ to $H$ is trivial.

This certainly can happen when either of $G,H$ is trivial, but is it possible that neither of $G,H$ are trivial?

答: By the First homomorphism theorem, this also happens if $G$ is simple and $|H|< |G|$. For example, $A_5\to S_4$.


问:
Finding the roots of a function applied twice
I saw this problem a few days ago and still haven't cracked it.
Therefore, I thought I might ask you all for help.

The graph of $f(x)$ has four roots on the interval $[-5,5]$. How many different roots does $f(f(x))$ have?
Here is a picture of the graph of $f(x)$:

We stumbled across this problem a week ago during math club and we were unable to solve it; our teacher didn't know the answer, either. Any help would thus be greatly appreciated!

答:
The roots of $f(x)$ are $$x=-4,-2,2,4$$, according to your graph, so for $f(f(x)) = 0$, we have $$f(x) = -4,-2,2,4$$ because $f(x)$ took the place of $x$. So, we need to examine when $f(x) = -4,-2,2,4$.

From the graph we see that $f(x)$ is never $-4$

From the graph we see that $f(x)=-2$ at $2$ values of $x$.

From the graph we see that $f(x) = 2$ at $4$ values of $x$

From the graph we see that $f(x) = 4$ at $2$ values of $x$

Thus, that makes $8$ values of $x$ total that $f(f(x)) = 0 $, so it has $8$ roots.


问:
Given $A \in \mathbb{R}^{n \times n}$ with spectral radius $\rho(A)<1$, is it possible for $||A||>1$ for Euclidean norm $|| \cdot ||$?
Given $A \in \mathbb{R}^{n \times n}$ with spectral radius $\rho(A)<1$, is it possible for $||A||>1$ for Euclidean norm $|| \cdot ||$?

I'm working on a proof where we are given the fact that $\rho(A)<1$, and in the case of $A$ being symmetric we know that $\rho(A)=||A||$ for the Euclidean norm. I am trying to make the proof more general, for when $A$ is not symmetric. All I would need to show is that, if $\rho(A) < 1$ then $||A|| \le 1$.

My attempt here is to use the singular value decomposition and the fact that the 2-norm is invariant under orthogonal transformations. Let the SVD of $A$ be
$$U^\intercal A V = D$$

where $D$ is diagonal with singular values along the diagonal. Then
$$||U^\intercal A V|| = ||A|| = ||D||$$
Since $\rho(A)<1$ then the singular values are also less than 1, then $||Dx|| < ||x||$, hence

$$\sup_{x \ne 0}\frac{||Dx||}{||x||} < 1$$

答:
Yes, it is possible. Here's a counterexample, let
$$
A = \begin{pmatrix} 0 & a \\ 0 & 0 \end{pmatrix}
$$
for some $a \in \mathbb{R}$. The eigenvalues of $A$ are $\{0\}$ thus $\rho(A) = 0$. However, for $x = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ we have $\|x \| = 1$ and $Ax = \begin{pmatrix} a\\0 \end{pmatrix}$. Hence $\|A\| \geq a^2$. As $a$ is arbitrary we can make the norm of $A$ arbitrarily large whilst keeping $\rho(A)  =0$.


答:
Notice that
$$ f(f(x)) = 0 $$
if and only if
$$ f(x) \in S=\{-4,-2,2,4\} , $$
since $S$ is the set of the roots of $f$.

So you should find how many times $f(x) = y$, with $y$ being a value in $S$. I suggest to draw four horizontal lines, respectively on $y=-4$ (which lies outside the graphics), $y=-2$, $y=2 and $y=4$.

You can notice that the line $y=-4$ has zero intersections with the graph of the function $f$; the line $y=-2$ has 2 intersections; the line $y=2$ has 4 intersections and the line $y=4$ has other 2 intersections$.

The equation
$$ f(f(x)) = 0 $$
has thus 8 solutions.

Title: Finding the roots of a function applied twice

Body:
I saw this problem a few days ago and still haven't cracked it.
Therefore, I thought I might ask you all for help.

The graph of $f(x)$ has four roots on the interval $[-5,5]$. How many different roots does $f(f(x))$ have?
Here is a picture of the graph of $f(x)$:

We stumbled across this problem a week ago during math club and we were unable to solve it; our teacher didn't know the answer, either. Any help would thus be greatly appreciated!


问:
Show that any open interval $(a,b) \subseteq \mathbb{R}$ is homeomorphic to $\mathbb{R}$
Show that any open interval $(a,b) \subseteq \mathbb{R}$ is homeomorphic to $\mathbb{R}$. I've seen similar questions on here, but they all use the fact that $(a,b)$ is homeomorphic to $(c,d)$, but can we prove this with without this fact, or without using the fact that the composition of homeomorphisms is homeomorphic?

答:
Well in fact the classical examples are all mapping $\mathbb R$ to either $(0,1)$ or $(-1,1)$.

For instance considering $\mathbb R\mapsto(-1,1)$ there is $f(x)=\begin{cases}\tanh(x)\\\frac 2{\pi}\arctan(x)\\\frac x{|x|+1}\end{cases}\quad$ to cite the most commonly used, and of course their reciprocals since all are continuous bijections $f^{-1}(x)=\begin{cases}\operatorname{argth}(x)\\\tan(\frac{\pi x}2)\\\frac x{1-|x|}\end{cases}$

A simple affine function $A(x)=\frac{b-a}2\,x+\frac{a+b}2$ and $A^{-1}(x)=\frac 2{b-a}\,x-\frac{a+b}{b-a}$ will scale and translate any $(a,b)$ to $(-1,1)$.

This is why composition of homeomorphisms is almost invariably invoked since $A(f(x))$ and $f^{-1}(A^{-1}(x))$ will do the job perfectly in mapping $\mathbb R$ and $(a,b)$. You can devise a direct mapping of course, but I guess the final form will resemble very much to this method.

Edit: 03/2022

To devise a mapping between $(a,b)$ and $\mathbb R$ you need a strictly monotonic function (let say increasing), and also since it should be continuous on $(a,b)$ then the limits in $a,b$ must be infinite (to cover all real values).
This is therefore calling for vertical asymptotes in $x=a$ and $x=b$.

You proposal $f(x)=\frac{x(a+b-2x)}{(a-x)(b-x)}$ verifies these two conditions as long as $a$ and $b$ are the same sign (else it doesn't work, not monotone), but the asymptotic behaviour is respected since $a,b$ are roots of the denominator.

In fact the additional $x$ on the numerator is the one causing problems, $f(x)=\frac{(a+b-2x)}{(a-x)(b-x)}$ work fine, this is because the root of numerator $x=\frac {a+b}2$ is between $a$ and $b$ and this make the discriminant of the numerator of the derivative negative, thus $f'$ of constant sign.


问:
Proof that a certain fraction is always an integer
Prove that $$\frac{(n+1)(n)^2(n-1)^2...(n-k+2)^2(n-k+1)}{(k+1)(k)^2(k-1)^2...(2)^2(1)}$$ is or is not an integer for $0\leq k \leq n$, where $k$ and $n$ are integer values.

This looks like $\frac{(n+1)!(n)!}{(n-k+1)(n-k)(k+1)!(k)!}$. The above statement is true for k=1 and k=2 as seen by observing modular residues. Not sure how to proceed.

答:
Transplanted answer, from duplicate query that was subsequently closed.  Actual (duplicate) query was:

For $n, k  \in \mathbb{N}$ such that $n \geq k, $ how do I show that

$$\prod_{\alpha = 1}^{k}\alpha(\alpha + 1) \quad \text{always divides} \quad \prod_{\alpha = n - k  + 1}^{n}\alpha(\alpha + 1) $$

Hint

Use Induction on Pascal's Triangle to demonstrate that $\binom{n}{k}$ is always an integer.

Then, demonstrate that the quotient equals $\binom{n}{k} \times \binom{n+1}{k+1}.$

Edit

I mistakenly thought that the RHS $\div$ LHS simplified to

$\binom{n}{k} \times \binom{n+1}{k+1}.$
This is wrong!

RHS $\div$ LHS $= \frac{1}{n-k+1} \times \binom{n}{k} \times \binom{n+1}{k+1}
~=~ \frac{1}{k+1} \times \binom{n}{k} \times \binom{n+1}{k}.$

Therefore, the problem reduces to either showing that

$(n-k+1)$ divides $\binom{n}{k} \times \binom{n+1}{k+1}$

or showing that

$(k+1)$ divides $\binom{n}{k} \times \binom{n+1}{k}.~~~$
This is what I will show.

Answer Repaired

For $x \in \mathbb{Z^+},$ and $p$ a prime,

let $V_p(x)$ denote the largest non-negative integer $y$ such that $p^y$ divides $x$.

Let $p$ be any prime factor of $(k+1)$, with
$V_p(k+1) = a$.

I am going to show that
$V_p\left[\binom{n}{k}\right] + V_p\left[\binom{n+1}{k}\right] \geq a.$

This will imply that $p^a$ divides
$\binom{n}{k} \times \binom{n+1}{k}.$

Since this demonstration will be applicable to any prime factor of $(k+1)$,

it will imply that
$(k+1)$ divides $\binom{n}{k} \times \binom{n+1}{k}.$

Note

By assumption, $~a \geq 1 \implies p~$ relatively prime to $~k \implies V_p(k) = 0$.

Let $b = V_p(n+2).$

$\underline{\text{Case 1 :}~ b \geq a}$

Since $p^a$ divides both $(n+2)$ and $(k+1)$,
it also divides $(n-k+1).$

Therefore, $V_p(n-k+1) \geq a.$

Further, since
$$\binom{n}{k-1} ~=~
\frac{n!}{(k-1)![(n-k+1)!]}
~~\text{is an integer}$$

$$V_p(n!) - V_p[(k-1)!]
- V_p[(n-k+1)!] \geq 0.$$

Since $$V_p(k) = 0, ~~V_p(k!) = V_p[(k-1)!].$$

Further,
$$V_p[(n-k+1)!] ~=~ V_p[(n-k)!] + V_p(n-k+1)
~\geq~ V_p[(n-k)!] + a.$$

Therefore,

$$V_p(n!) - Vp(k!) - V_p[(n-k)!] \geq a ~\implies~ $$

$$p^a \left|
\frac{n!}{(k!)[(n-k)!]} = \binom{n}{k}\right.. \tag1$$

$\underline{\text{Case 2 :}~ b < a}$

By assumption in Case 2, $V_p(n+2) = b.$

Further, since $p^b$ divides both $(n+2)$ and $(k+1)$,
it also divides $(n-k+1).$

Therefore, $V_p(n-k+1) \geq b.$

All of the corresponding analysis from Case 1 will apply here, and will lead to the conclusion that parallels equation (1) above:

$$p^b \left|\binom{n}{k}\right..$$

Therefore, it only remains to demonstrate that

$$p^{(a-b)} \left|\binom{n+1}{k}\right..$$

Since

$$\binom{n+2}{k+1} = \frac{(n+2)!}{[(k+1)!][(n-k+1)!]}
~~\text{is an integer}$$

$$V_p[(n+2)! - V_p[(k+1)! - V_p[(n-k+1)!] \geq 0.$$

Also,
$\displaystyle
V_p[(n+2)!] = V_p(n+2) + V_p[(n+1)!] = b + V_p[(n+1)!].$

Also, $\displaystyle
V_p[(k+1)!] = V_p(k+1) + V_p(k!) = a + V_p(k!).$

Therefore,

$$V_p[(n+1)!] ~+~ b ~-~ [V_p(k!) + a] ~-~ V_p[(n-k+1)!] ~\geq~ 0~ \implies $$

$$V_p[(n+1)!] - V_p(k!) - V_p[(n-k+1)!] \geq (a-b) \implies$$

$$p^{(a-b)} \left|
\frac{(n+1)!}{(k!)[(n-k+1)!]} = \binom{n+1}{k}\right..$$


问:
composition function with dense graph
Let $f\colon \Bbb R\to \Bbb R$ be a function  with dense graph .Let $g\colon \Bbb R \to \Bbb R$  be a function( Not constant).

Does $f\circ g$ have a dense graph? if not,   How about if $g$ is a homeomorphism? my guess is not but I could not have an example?

答:
Here are some examples to give an idea of what is possible.

if $f$ is a bijection and $g$ is its inverse then $f\circ g$ is the identity, so its graph is not dense

if $g$ is a homeomorphism then the graph of $f\circ g$ is the image of the graph of $f$ under the product map $g^{-1}\times\mathrm{Id}$, which is a homeomorphism, so $f\circ g$ is dense.

for many `common' continuous functions the result is again a dense graph, e.g., $g(x)=|x|$ takes the part of the graph of $f$ over $(0,\infty)$ and reflects it with respect to the $y$-axis, or $\sin x$ takes the part over $[-1,1]$ and stretches it to $[-\frac\pi2,\frac\pi2]$ and then alternately shifts it or its mirror image over the $x$-axis

if $g$ is constant on an interval then $f\circ g$ is constant over this interval and the graph is not dense

and so on

Title: composition function with dense graph


问:
Trace on $M_n(A)$ where $A$ is a tracial von Neumann algebra
Let $A$ be a tracial von Neumann algebra with trace $\tau$. If I'm not mistaken, this trace is unique. For $X=[x_{ij}]\in M_n(A)$, define
$$\tau_n(X):=\frac{1}{n}\sum_{i=1}^n\tau(x_{ii}).$$
I find this a very natural construction.

My questions are then:
Does this make $M_n(A)$ a tracial von Neumann algebra? If so, is it the unique one?

答:
You don't say what "trace" is, so it is a bit hard to discuss uniqueness.

With the mostly usualy meaning "state with the tracial property" (and this leaves out whether you want your trace to be normal), traces are not unique. For starters, in any commutative von Neumann algebra, any state is a trace. Even in non-commutative von Neumann algebras it is easy to find multiple traces. In fact, every time that the centre is not trivial, if there is one trace there are infinitely many others. For instance in $M_2(\mathbb C)\oplus \mathbb C$ given any $t\in[0,1]$ you have that
$$
\tau_t(A\oplus\lambda)=\tfrac12\Big[t\,\operatorname{Tr}(A)+(1-t)\lambda\Big]
$$
is a tracial state for any $t$.

Tracial states are unique on factors, because in such case any two projections are comparable and the factors is the norm-closure of the linear span of the projections.

Your $\tau_n$ is a tracial state if $\tau$ is. It's uniqueness depends on the uniqueness of $\tau$.

As a final comment, your construction is as natural as it can get, because what you are doing is define $\tau_n=\operatorname{tr}\otimes \tau$ on the algebra $M_n(\mathbb C)\otimes A$.


问:
Conditions for Which $\mathbb{F}[T]$ Is a Field
Given a linear operator $T$ on a finite-dimensional vector space $V$ over a field $\mathbb{F},$ consider the ring $\mathbb{F}[T]$ of polynomials in the linear operator $T.$ Under what conditions is $\mathbb{F}[T]$ a field? One sufficient condition with which I am already familiar is that $T$ maps no nonzero subspace of $V$ into itself.

Proof. Given an element $p(T) \in \mathbb{F}[T],$ let $v$ be a vector in $\ker p(T).$ Considering that $T$ commutes with $p(T),$ we have that $p(T)(T(v)) = T(p(T)(v)) = T(0) = 0,$ from which it follows that $T(\ker p(T)) \subseteq \ker p(T).$ Our assumption that $T$ maps no nonzero subspace of $V$ into itself implies that $\ker p(T) = \{0\},$ i.e., $p(T)$ is injective. Of course, $p(T)$ is a linear operator on $V,$ hence $p(T)$ is invertible. We conclude that every element of $\mathbb{F}[T]$ is invertible. Furthermore, $\mathbb{F}[T]$ is an integral domain since if $p(T)$ is nonzero and $p(T) q(T) = 0,$ then we must have that $q(T) = I q(T) = [p(T)]^{-1} p(T) q(T) = 0.$

Lately, I have been thinking about whether it is sufficient that $T$ is invertible. Using the evaluation map $\varphi_T : \mathbb{F}[x] \to \mathbb{F}[T]$ defined by $\varphi_T(p(x)) = p(T)$ in tandem with the fact that $\mathbb{F}[x]$ is a PID, it is clear that $\mathbb{F}[T] \cong \mathbb{F}[x]/(\mu_T(x)),$ where $\mu_T(x)$ is the minimal polynomial of $T.$ We have therefore that $\mathbb{F}[T]$ is a finite-dimensional vector space over $\mathbb{F}$ with a basis $\{I, T, T^2, \dots, T^{d-1} \},$ where $d = \deg \mu_T.$ We could conclude that $\mathbb{F}[T]$ is a field if we could show that $\mathbb{F}[T]$ is an integral domain since every integral domain that is a finite-dimensional vector space over a field is itself a field. But I am not entirely sure how to prove that $\mathbb{F}[T]$ is an integral domain in this case.

Generally, are there any other conditions than the one I initially stated for which $\mathbb{F}[T]$ is a field?

答:
Given a linear operator $T$ on a finite-dimensional vector space $V$ over a field $k,$ consider the ring $k[T]$ of polynomials in $T.$ Observe that there is a surjective ring homomorphism $\mathbf{ev}_T : k[x] \to k[T]$ defined by $\mathbf{ev}_T(p(x)) = p(T).$ Considering that $k[x]$ is a principal ideal domain (PID), we have that $k[T] \cong k[x]/(\mu_T(x)),$ where $\mu_T(x)$ is the minimal polynomial of $T.$

Question: Under what conditions is $k[T]$ a field?

Like I have mentioned in the original post and the subsequent comments, there are two sufficient conditions with which I am already familiar.

If $T$ maps no nonzero subspace of $V$ into itself, then $k[T]$ is a field.

Proof. Polynomials in $T$ commute with $T,$ hence $T$ maps $\ker(p(T))$ into $\ker(p(T))$ for all $p(T)$ in $k[T].$ Consequently, every nonzero element of $k[T]$ is invertible, hence $k[T]$ is a field. QED.

If $\mu_T(x)$ is irreducible, then $k[T]$ is a field.

Proof. If $\mu_T(x)$ is invertible, then it is prime, hence the ideal $(\mu_T(x))$ of $k[x]$ is maximal. QED.

One observation that I made above is that it is necessary but not sufficient that $T$ is invertible. For instance, the linear operator $T : k^2 \to k^2$ defined by $T(x, y) = (x, -y)$ is invertible; however, the linear operator $I + T$ satisfies $(I + T)(x, y) = (2x, 0)$ and hence is not invertible.

If anyone would like to suggest any additional sufficient conditions, please feel free!


问:
Can someone help me with this arc length integral?
I need to solve this arc length integral for a polar function:

$r=3\cos^4(\frac{\theta}{2})$

$\frac{dr}{d\theta}=-6\cos^3(\frac{\theta}{2})\sin(\frac{\theta}{2})$

So the formula for the arc length of a polar function is:

$\int_0^{2\pi} \sqrt{[\frac{dr}{d\theta}]^2+[r]^2} \,d\theta$

$\int_0^{2\pi} \sqrt{[-6\cos^3(\frac{\theta}{2})\sin(\frac{\theta}{2})]^2+[3\cos^4(\frac{\theta}{2})]^2} \,d\theta$

Body:
hint

Put $$t=\frac{\theta}{2}$$

it becomes

$$L=6\int_0^\pi |\cos^3(t)|\sqrt{4\sin^2(t)+\cos^2(t)}dt$$

$$=6\int_0^\pi |\cos(t)|(1-\sin^2(t))\sqrt{1+3\sin^2(t)}dt$$


问:
Prove that $A - B = A $ iff $A \cap B = \emptyset$
I have a lemma below:

$${A}-{B} = {A} \Longleftrightarrow A \cap B = \emptyset.$$

I start with the $\Longrightarrow$:

$$\text{Suppose}\ x \in {A}-{B},\text{then}\  x \in A\ \text{but}\  x \notin B.$$

$${A}-{B} = {A}\ \text{so}\ x \in A\ \text{but}\ x \notin B \equiv x \in A.$$

So the fact that $x \notin B$ does not matter at all. It's getting confused because this lemma is just proved to prove this theorem:

$$P(A-B) = P(A) \Longleftrightarrow A \cap B = \emptyset.$$

With the lemma, I can prove the theorem. Anyone has any advice on how to prove the lemma? Thank you!

答:
$A\cap B'=A\Longrightarrow$ $$A\cap B=(A\cap B')\cap B= A\cap (B'\cap B)=A\cap\emptyset=\emptyset $$
$A\cap B=\emptyset\Longrightarrow$
$$A\cap B'=(A\cap B')\cup(A\cap B)=A\cap(B'\cup B)=A$$


问:
Kalman filter initial prediction variance
I'm trying to learn about Kalman filters and I think that for the most part the explanations I found were sufficient (There is a great explanation on this forum). But I'm having trouble with the prediction variance and more specifically the initial value.

So from the literature I've read : the observation variance matrix R represents how accurate our measures are, and its concrete value would depend on the sensors we have.

But what about the initial variance of our prediction $P_0$ ? I don't understand what is the concrete meaning of this value. The only thing I found was the following :

It is often also natural to choose the stationary value of the
  variance as the initial value for the variance

Which if I understand correctly would mean that in some cases after a number of iterations, no matter the initial value, P(t) would converge towards a specific $P_{lim}$ and so we might as well use it as initial value ?

Is that what is meant ? And what if it doesn't converge ? Is it that the initial value for P is linked to how good we know our first guess for the state vector is ?

答:
But what about the initial variance of our prediction $P_0$ ? I don't understand what is the concrete meaning of this value.

It's how certain you are of the initial value of the state vector.  For instance, imagine that you're in the middle of a corn field that's 100 meters on a side, and you're trying to locate a bean.  You might model $x = \begin{bmatrix} x_{north} & x_{east} \end{bmatrix}^T$.  Let the middle point of the field be at $x = \begin{bmatrix}0&0\end{bmatrix}^T$.

You could decide that the bean must have a positional probability that's uniformly distributed across the field.  Since the Kalman filter is about Gaussian distributions and the uniform distribution isn't a good fit, it's not a bad idea to cheat a bit, and to assign an initial estimate of the position of the bean at $\hat x_0 = \begin{bmatrix}0&0\end{bmatrix}^T$, and the initial variance at $P_0 = \begin{bmatrix}2500 & 0 \\ 0 & 2500\end{bmatrix}^T$.  This madly over-represents the exterior of the field (because you know the bean isn't there) and somewhat under-represents the interior of the field -- but you do this because the Gaussian and uniform distributions are different, and you want to use a plain old Kalman filter.

Subsequent measures of the position of the bean will be noisy (it's a GPS-equipped bean?  You have bean-locating technology?  Who knows?), but each measurement will refine the bean's position, even if it moves (it's a jumping bean?).  At each point -- to the extent that your model is accurate -- your filter will tell you where the bean ought to be, and how accurately you should know it.

Note that if you're more interested in the end game than in the start, the "correct" $P_0$ is $P_0 = \begin{bmatrix}\frac{2500}{3} & 0 \\ 0 & \frac{2500}{3}\end{bmatrix}^T$.  This accurately captures the covariance of the bean, while ignoring a lot of important differences between Gaussian and uniform distributions.


问:
$\nabla f=0$ on an open set $U \subset C$ implies $f$ locally constant.
Suppose for some open subset $U \subset \mathbb{C}$ that $\partial_xf=\partial_yf=0$ for every $u \in U$. Show that $f$ is locally constant. I'm explicity told I can use the fact that if $h:\mathbb{R} \rightarrow \mathbb{R}$ with $h'(x)=0$ for $x \in [a,b]$ then $h$ is constant on $[a,b]$.

I for sure don't want a solution. I don't see the tension in this problem? What's stopping me from fixing two points $p_1=(x,y),p_2=(x',y') \in U$ and just applying the MVT in one dimension on both coordinates?

This is clearly wrong since it implies the stronger result that $f$ constant on $U$.

Can someone give an example where $\nabla f=0$ on a set but it is not constant?

答:
The issue is that $U$ may not be connected, so there will not neccesarily be a path from $(x,y)$ to $(x',y')$ inside $U$ (much less a path made of two straight lines!)

For a simple example, let $U = \{z \in \mathbb{C} : \lvert z \rvert < 1\} \cup \{z \in \mathbb{C} : \lvert 4 - z \rvert < 1\}$. This is a disjoint union of two open disks, and we can define a smooth function $f : U \to \mathbb{R}$ which sends one disk to $0$ and the other to $1$.

Instead, you should write a proof which directly shows that the definition of "locally constant" is satisfied: take a point $x \in U$, and produce an open neighborhood $x \in U_0 \subseteq U$ such that $f|_{U_0}$ is constant. If you pick $U_0$ correctly, you can use your path-connectedness argument!


问:
Reference request for 3-manifold
I am asking a soft question.

I am planning to learn $3$-manifold using the book "Geometry and topology of three-manifolds" by William Thurston. I want to know how much of Riemannian geometry, Algebraic topology, Smooth manifolds, Complex Geometry do I need to learn $3$- manifold. Also, I am looking for references of $3$-manifolds which contain exercises as I noticed that  William Thurston's contain does not contain exercises.

Also, it will be very much appreciatable if you also advise on book recommendations of $3$-manifolds.

Please advise me. Thanking in advance.

答: Hatcher's books on Algebraic topology and 3-manifolds are recommended. As is Sveta  Katok's little book on Fuchsian groups.</p>

Title: Reference request for 3-manifold


问:
The Coproduct of the Identity element
For a general Hopf algebra $H$, the coproduct of the identity element is
$$\Delta(1) = 1\otimes 1.$$
Now for a finite group $G$ and a field $k$, $k(G)$ forms a Hopf algebra with basis $\{\delta_{g}| g\in G\}$. The coproduct is given
$$\Delta(\delta_{g}) = \sum_{g_{1}g_{2} = g}\delta_{g_{1}}\otimes\delta_{g_{2}}.$$
The identity element $1$ is
$$1 = \sum_{g\in G}\delta_{g}.$$
I want to establish that $\Delta(1) = 1\otimes 1$. This is what I have done,

$$\Delta(1) = \Delta\left(\sum_{g\in G}\delta_{g}\right) = \sum_{g\in G}\Delta(\delta_{g}) =  \sum_{g\in G}\left(\sum_{g_{1}g_{2} = g}\delta_{g_{1}}\otimes\delta_{g_{2}}\right).$$ How do I proceed from here?

I am thinking I can use this change of indices but I am not too sure if it is correct
$$\sum_{g\in G}\sum_{g_1g_2 = G} = \sum_{g_2\in G}\sum_{g_1 = gg_2^{-1}}$$

答:
Observe $\delta_{g_1}\otimes\delta_{g_2}$ appears exactly once for every possible pair $(g_1,g_2)$, then we may factor:

$$ \sum_{g\in G} \Big(\sum_{g_1g_2=g} \delta_{g_1}\otimes\delta_{g_2}\Big)=\sum_{g_1\in G}\sum_{g_2\in G} \delta_{g_1}\otimes\delta_{g_2}=\Big(\sum_{g_1\in G}\delta_{g_1}\Big)\otimes\Big(\sum_{g_2\in G} \delta_{g_2}\Big)=1\otimes1. $$

Title: The Coproduct of the Identity element


问:
Whether the given series are convergent or divergent:
Whether the given series is absolutely convergent, conditionally convergent or divergent:

$a)\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{\cos nx}{n^2}:|(-1)^n\frac{\cos nx}{n^2}|\le\dfrac{1}{n^2}\implies$ Absolutely convergent.

$b)\displaystyle\sum_{n=1}^{\infty}(-1)^n\dfrac{n}{n+2}:(-1)^n\dfrac{n}{n+2}\to0\implies\dfrac{n}{n+2}\to0!\implies$ Divergent.

$c)\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{1}{2n+3}:u_n=\dfrac{1}{2n+3}\\\implies\dfrac{u_n}{u_{n+1}}=\dfrac{2n+3}{2n+5}=\dfrac{1+\dfrac{3}{2n}}{1+\dfrac{5}{2n}}=({1+\dfrac{3}{2n}})(1+\dfrac{5}{2n})^{-1}=({1+\dfrac{3}{2n}})(1-\dfrac{5}{2n}+\cdots)=1-\dfrac{1}{n}+O(\dfrac{1}{n^2})\\\implies \text{Not absolutely convergent. So conditionally convergent by Leibnitz test.}$

Am i right?

答:
Your solution to (c) is incomplete.  The Leibniz test is frequently used incorrectly by students.  It has three parts:

Terms alternate (usually easy)

Terms approach $0$ (fairly easy)

Absolute value of terms is monotone decreasing (!)

The third part is often hard to check, but is essential if you wish to apply the test.  Fortunately for you, in your problem it's not so bad.  $2n+5>2n+3$ so $\frac{1}{2n+3}>\frac{1}{2n+5}$ and you're done.