TITLE: Linear Independence easy question
QUESTION [6 upvotes]: I have these vectors $B = \{u, v, w\}$ with
$$u = (-1, 1, -1),\, v = (19, 10, -9),\, w = (-1, x, y)$$
And i want to prove that these vectors are linearly independent.
I have no problem to proove that three vectors without unknown variables are linearly indepedent but i have difficulties in this that i have two unknown variables $(x, y)$.
I find that $\det(B)$ is $-28x - 29y -1$ but i do not know how this helps.
The actual question is to prove that B forms a basis in R3
PS. New to Linear Algebra, dont shoot !

REPLY [7 votes]: Good question. In fact, the vectors are linearly independent whenever the determinant $-28x -29y -1$ is not equal to zero. So, the answer depends on $x$ and $y$.
How I know this: The determinant can be thought of as a function of $m$ vectors in $m$-dim'l space, in which case one of the properties that characterizes it is that if one of its inputs is a linear combination of some or all of the others, then the function will evaluate to zero. Also, the function will evaluate to zero only in this case. Hence, a set of $m$ vectors in an $m$-dim'l space will be linearly independent if and only if their determinant is nonzero. See this section of wiki article for more details, specifically where it says that properties 1, 7, and 8 completely characterize the determinant.
Back to your specific question. Set $$-28x - 29y - 1 = 0$$ and rewrite it as $$y = - \frac{28}{29}x  -\frac{1}{29}.$$ Your three vectors are linearly independent when $x,y$ lie away from this line, which you can easily plot now that it's written in slope-intercept form. And since you want to know when these vectors form a basis for $V := \mathbb{R}^3$, and since you know that any linearly independent set of 3 vectors in $V$ forms a basis, you know that these vectors form a basis when $x,y$ lie away from the line given above.
Hope that helps.


TITLE: Comparing a geometric definition of cross product to the "usual" one
QUESTION [1 upvotes]: Could you help me with my little problem?
Given this definition of cross product:
1) $a \times b$ is perpendicular to $a$ and $b$, whenever $ a,b$ are linearly independent
2) basis $a, \ b, \ a \times b$ is positively oriented
3) $\|a \times b\| = \|a\| \cdot \|b\| \cdot \sin( \angle(a, b))$
How can I prove that it's equivalent with the determinant definition of cross product (we put the basis vectors $e_1, e_2, e_3$ in the first row and the coordinates of $a$ and $b$ in the second and third rows)?

REPLY [1 votes]: I don't think I've completely understood your question but here is how I would answer it
By definition, if the basis vectors are $\{\vec{i},\vec{j},\vec{k} \}$, the cross product of $\vec{a}=(a_1,a_2,a_3)$ and $\vec{b}=(b_1,b_2,b_3)$ (expressed in those basis vectors) is
$$\vec{a}\times\vec{b}=\left| \begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \end{array}\right| $$
For the first one, just change the basis vectors to $\{\vec{e_1},\vec{e_2},\vec{e_3}\}$ such as the two vectors $\vec{a}$ and $\vec{b}$ are in the same plane given by the two vectors $\vec{e_1}$ and $\vec{e_2}$. Then they can be expressed as follows :
$$\vec{a}=(a'_1,a'_2,0) \quad \text{and} \quad \vec{b}=(b'_1,b'_2,0)$$
$$\vec{a}\times\vec{b}=\left| \begin{array}{ccc}
\vec{e_1} & \vec{e_2} & \vec{e_3} \\
a'_1 & a'_2 & 0 \\
b'_1 & b'_2 & 0 \end{array}\right| =
\left| \begin{array}{cc}
a'_1 & a'_2 \\
b'_1 & b'_2 \end{array}\right| \cdot \vec{e_3} $$
which is perpendicular to the plane and thus to $\vec{a}$ and $\vec{b}$.
For the second one, choose your basis vectors such as $\vec{a} = a''_1\cdot\vec{e_1}$ with $a''_1>0$. Then we have
$$\vec{a}=(a''_1,0,0) \quad \text{and} \quad \vec{b}=(b''_1,b''_2,0)$$
$$\left| \begin{array}{cc}
a''_1 & a_0 \\
b''_1 & b''_2 \end{array}\right|=a''_1\cdot b''_2$$
As $a''_1$ is by definition positive, only the sign of $b''_2$ matters... And if you make a quick drawing of the situation, you directly see that the sign of
$$\vec{a}\times\vec{b}=\left| \begin{array}{ccc}
\vec{e_1} & \vec{e_2} & \vec{e_3} \\
a''_1 & 0 & 0 \\
b''_1 & b''_2 & 0 \end{array}\right| =
\left| \begin{array}{cc}
a''_1 & 0 \\
b''_1 & b''_2 \end{array}\right| \cdot \vec{e_3} = a''_1 \cdot b''_2 \cdot \vec{e_3} $$
is positive if they are positively oriented, and negative if not. As $\vec{e_3}$ is positively oriented in $\{\vec{e_1},\vec{e_2},\vec{e_3}\}$, that's it.
For the third one I'm not really sure, but as $$|\vec{a}.\vec{b}|=||\vec{a}||\cdot||\vec{b}||\cdot|\cos(\phi)| = |a''_1\cdot b''_1|$$ and that $$||\vec{a}\times\vec{b}||=|a''_1\cdot b''_2| $$
I would do
$$\begin{align}
(|a''_1\cdot b''_1|)^2+(|a''_1\cdot b''_2|)^2 &= (a''_1)^2\cdot [(b''_1)^2+(b''_2)^2] \\
||\vec{a}||^2\cdot||\vec{b}||^2\cdot|\cos(\phi)|^2 + (|a''_1\cdot b''_2|)^2 &= ||\vec{a}||^2\cdot||\vec{b}||^2 \\
(|a''_1\cdot b''_2|)^2 &= ||\vec{a}||^2\cdot||\vec{b}||^2 - ||\vec{a}||^2\cdot||\vec{b}||^2\cdot|\cos(\phi)|^2 \\
||\vec{a}\times\vec{b}||^2&= ||\vec{a}||^2\cdot||\vec{b}||^2\cdot|\sin(\phi)|^2 \\
||\vec{a}\times\vec{b}||&= \pm ||\vec{a}||\cdot||\vec{b}||\cdot|\sin(\phi)|
\end{align}$$
Then as previously, the sign of $a''_1 \cdot b''_2$ show you the sign of the cross product. And justifies the $\sin\phi$
$$||\vec{a}\times\vec{b}||= ||\vec{a}||\cdot||\vec{b}||\cdot\sin(\phi)$$


TITLE: Determine local extrema of 2 functions
QUESTION [0 upvotes]: So I have 2 functions, $f(x)=cos(\frac{2}{x})$ and $g(x)=x^{\frac{1}{3}}$
For f(x), how many local extrema it has? It is very weird because near 0 it is many up and down. So do they belong to local maximum and minimum?
For g(x) in the interval (-3,2). does it have any local extrema? Because in this interval there is global maximum at g(2) and global minimum at g(-3).

REPLY [0 votes]: $f$ has infinitely many local maxima and infinitely many local minima as you have observed.
$g$ is strictly increasing in the given interval so it has no local maximum of minimum at any point of the open interval $(-3,2)$.


TITLE: Proving f(gh) = (fg)h
QUESTION [5 upvotes]: I am trying to prove that $f(gh)=(fg)h$ for $f,g,h \in R[\mathbb{N}^n]$. Context to the exercise:
We define the polynomial ring $R[X_1,...,X_n]$ in $n$ variables as
$$
R=[X_1,...,X_n] = R[\mathbb{N}^n] = \\\{ f:\mathbb{N}^n \to R | f(v)=0,|v| >> 0 \}
$$
where $v=(v_1,...,v_n)\in \mathbb{N}$ and $|v| = v_1+...+v_n.$
A polynomial $f \in R[X_1,...,X_n]$ is the same as a function $f:\mathbb{N}^n \to R$ that is non-zero for only finitely many $v \in \mathbb{N}^n$. We let $X^v \in R\left[\mathbb{N}^n\right]$ denote the function given by
$$
X^v(w)= \begin{cases}1 & \text { if } v=w, \\\\ 0 & \text { if } v \neq w .\end{cases}
$$
With this notation, every polynomial $f \in R\left[\mathbb{N}^n\right]$ can be written as a (finite) sum
$$
f=\sum_{v \in \mathbb{N}^n} a_v X^v,
$$
where $a_v \in R$ (an element $r \in R$ is identified with the function mapping the zero vector to $r$ and everything else to $0 \in R$ ). If $f, g \in R\left[\mathbb{N}^n\right]$ we define $f+g$ by $(f+g)(v)=f(v)+g(v)$ and $f g$ by the (finite) sum
$$
(f g)(v)=\sum_{v_1+v_2=v} f\left(v_1\right) g\left(v_2\right)
$$
where $v_1,v_2 \in \mathbb{N}^n$.

REPLY [1 votes]: Since it seems like you are learning university level algebra, this simpler but more theoretical proof could be instructive.
Note that the map $(fg)h - f(gh)$ is an $R$-module map
$$M: R[X_1, \ldots, X_n]^{\otimes 3} \to R[X_1, \ldots, X_n]$$
since it is obtained as compositions and subtractions of the multiplication, which is a $R$-module map. The key observation is that the triple tensor product is generated by tensor products of monomials $X^{\alpha}\otimes X^{\beta}\otimes X^{\gamma}$. If we manage to show that $M$ is zero in generators, we show associativity.
Product of monomials is particularly easy: you can show that the summation $X^{\alpha}X^{\beta}(\nu)$ is zero unless $\nu=\alpha+\beta$, and in this case the result is one, since only one summand is non-zero. We deduce that $X^{\alpha}X^{\beta} = X^{\alpha+\beta}$, which yields
$$M(X^{\alpha}\otimes X^{\beta}\otimes X^{\gamma}) = (X^{\alpha}X^{\beta})X^{\gamma} - X^{\alpha}(X^{\beta}X^{\gamma} ) = X^{(\alpha+\beta)+\gamma} - X^{\alpha+(\beta+\gamma)} = 0$$
Voilà!


TITLE: Using $\mathbb{E}(1/X)$ and $1/\mathbb{E}(X)$ to prove $X=Y$ a.s.?
QUESTION [0 upvotes]: Consider two r.v.s $X,Y\geq1$ defined on the probability space $(\Omega,\mathcal{F},\mathbb{P})$ which may not necessarily be integrable. Let $\mathbb{E}[X\mid \mathcal{G}]:=\lim_{n\to\infty}E[X\wedge n\mid\mathcal{G}]$ where $\mathcal{G}$ is any sub-$\sigma$-algebra of $\mathcal{F}$. I have $\mathbb{E}(X\mid Y)=Y$ and  $\mathbb{E}(Y\mid X)=X$. I need to prove the following: i) $\mathbb{E}\left(\frac{X}{Y}\right)=\mathbb{E}\left(\frac{Y}{X}\right)=1$; and ii) $X=Y$ a.s. by comparing $\mathbb{E}(1/X)$ and $1/\mathbb{E}(X)$. I also need to somehow use these to show that if instead $X,Y\geq-1$, $X=Y$ a.s.
I have read through the proofs outlined in here but the conditions for this case are not the same as the one here (since we cannot assume integrability anyway). What should I do in this case? I don’t really know to go about both cases. Thanks!

REPLY [1 votes]: Suppose that $\mathbb E\left[\frac{X}{Y}\right]=\mathbb E\left[\frac{Y}{X}\right]=1$.
Since $U=\frac{X}{Y}> 0$ is integrable and $x\rightarrow \frac{1}{x}$ is strictly convex on $]0,+\infty[$, Jensen inequality says that
\begin{align*}
\mathbb E\left[\frac{1}{U}\right]&\geq\frac{1}{\mathbb E[U]}
\end{align*}
with equality if and only if $U$ is almost surely constant (and so $U=\mathbb E[U]~a.s.$). By the previous inequality, we know that $\mathbb E[U]=1$ and $\mathbb E\left[\frac{1}{U}\right]=\mathbb E\left[\frac{Y}{X}\right]=1$ so we have equality. This means that $\frac{X}{Y}=\mathbb E[U]$ almost surely and so we indeed get $X=Y$ almost surely.

We can compute
\begin{align*}
\mathbb E\left[\frac{X}{Y}\middle| Y\right]
&= \lim_{n\to \infty} \mathbb E\left[\frac{X}{Y}\land n\middle| Y\right]\\
&= \frac{1}{Y}\lim_{n\to \infty} \mathbb E\left[X\land (Yn)\middle| Y\right]\\
&= \frac{1}{Y}\lim_{x\to \infty} \mathbb E\left[X\land x\middle| Y\right]\\
&= \frac{1}{Y}\mathbb E[X|Y]\\
&= 1
\end{align*}
almost surely. We change the limit from $n$ to $x$ by multiplying by $Y\geq 1$, observe that $X\land (Yn)$ and $X\land x$ will converge to the same limit almost surely by conditional dominated convergence theorem (actuelly they converge to $X$). However there may be some integrability conditions I missed for the conditional dominated convergence theorem.


TITLE: A boy is selected at random from among the children belonging to families with n children.
QUESTION [0 upvotes]: A boy is selected at random from among the children belonging to families with n children.

Question 1：What is the probability that the boy has k − 1 brothers?
The answer I got is (n-1)!/(2 ^(n-1)(k-1)!(n-k)!)

Question 2：If the boy has at least two sisters， what is the probability that the boy has k − 1 brothers now?

REPLY [1 votes]: We assume that children are born equally likely to be boys as girls.
Question 1.  All $2^n$ possible sequences of boys and girls are equally likely.  If we were to write out all such sequences, there would be $n2^n$ children, and $n2^{n-1}$ of them would be boys.  Any of them could be selected, with equal probability, for the purposes of this problem.
There would be, on the other hand, $\binom{n}{k}$ sequences with $k$ boys in that list, and thus $k\binom{n}{k}$ boys who could each say that they have $k-1$ brothers.  The probability we desire is therefore the ratio of these two quantities:
$$
\frac{k\binom{n}{k}}{n2^{n-1}}
$$
The answer you got, which can be written
$$
\frac{\binom{n-1}{k-1}}{2^{n-1}}
$$
is the probability that a randomly selected child can say that they have $k-1$ brothers.  But we didn't randomly select a child; we randomly selected a boy.  So the probabilities are different.
Question 2. We assume first that $n \geq k+2$.  The number of sequences with $k$ boys remains the same, then; all that need be adjusted is the denominator, which represents the possible range of boys.  From the total number of boys must be subtracted all boys in sequences with no girls at all, and all boys in sequences with exactly one girl.  There are $n\binom{n}{n} = n$ and $(n-1)\binom{n}{n-1} = n^2-n$ of those respectively, so the number of boys available for selection is now $n2^{n-1}-n^2$, and the new probability is
$$
\frac{k\binom{n}{k}}{n2^{n-1}-n^2}
$$
If $n < k+2$, the probability is $0$, as there are not enough children left to be boys, after the girls are accounted for.


TITLE: A ship is being pulled by two tugboats problems
QUESTION [1 upvotes]: A ship is being pulled by two tugboats. The larger tugboat exerts a force 25% greater
than the smaller tugboat and at an angle of 20 degrees west of north.
In what direction should the smaller tugboat pull in order for the ship to head
north?
So far, I assume the smaller force is $x$, so the The larger tugboat exerts a force is $1.25x
\tan(20)=1.25 x/m$, so $m=1.25x/tan20$. also, $x/(1.25x/\tan20)=tan\theta$ which $theta$ is the direction should the smaller tugboat pull in order for the ship to head north. $\theta=\arctan(\tan(20/1.25))$, so am I right of the track?

REPLY [2 votes]: Let the magnitude of the larger force be $1.25$. So the smaller one has magnitude  $1$.
The west component of the force exerted by the larger tug is $1.25 \sin(20^\circ)$. This must be balanced by the east component of the force exerted by the little tug, which is $\sin(\theta)$. I expect you can now find $\theta$.
Remark: The answer reached in the OP does not make physical sense, for it is clear that the little tug must head more than $20^\circ$ east of north to compensate for the extra power of the big tug.


TITLE: How to find the value $\sum_{n\geq2}^{\infty}(-1)^{n+1}\frac{n}{n^2-1}$ converges to?
QUESTION [1 upvotes]: How to find the value this sum converges to?$$\sum_{n\geq2}^{\infty}(-1)^{n+1}\frac{n}{n^2-1} $$
I've tried writing it like this
$$\sum_{n\geq2}^{\infty}(-1)^{n+1}·n·\Bigg(\frac{1/2}{n-1}-\frac{1/2}{n+1}\Bigg) $$
and writing a few terms, but they won't cancel and I ended up with no ideas, any hint?
I haven't learnt integration nor differentiation FYI.

REPLY [3 votes]: Try writing $$\begin{align}n\left({1\over n-1}-{1\over n+1}\right)&=
n\left({1\over n-1}-{1\over n} + {1\over n} -{1\over n+1}\right)\\
&=n\left({1\over n(n-1)}+{1\over n(n+1)}\right)\\&={1\over n-1}+{1\over n+1}\end{align}$$


TITLE: Is the morphism of affine varieties given by Noether Normalisation a closed map?
QUESTION [0 upvotes]: Let $X\subseteq \mathbb A^n(k)$ be an affine variety over the algebraically closed field $k$. Then by the geometric interpretation of Noether Normalisation Lemma there is a finite and surjective morphism $\varphi:X\to \mathbb A^m(k)$, where $m$ is the dimension of $X$ (see here for the process noether normalization theorem geometric meaning).
Is it true that the morphism $\varphi:X\to \mathbb A^m(k)$ is a closed map?

REPLY [2 votes]: Any finite morphism is proper (hence universally closed).


TITLE: Expression for $\|Ax\|_2^2$ in terms of rows of $A$
QUESTION [0 upvotes]: I am currently reading 1011.3027, Roman Vershynin's introduction to non-asymptotic random matrices. On page 24, there is an equation, (5.24), which reads
\begin{align*}
\|Ax\|_2^2 = \sum_{i=1}^n\langle A_i,x\rangle^2,
\end{align*}
where $A_i\in \mathbb{R}^n$ are the rows of the matrix $A\in \mathcal{M}_{r\times n}$ and $x\in \mathbb{R}^n$ is a vector (on the unit sphere).
I find myself believing this to be true, but I am not sure how to actually prove it. I tried playing around with an SVD and also decomposing $A$ as a sum of basic matrices but I didn't get anywhere useful.
Thus, I would appreciate a reference or proof of the above equation.
Thanks in advance.

REPLY [1 votes]: Hint: What is the $i$-th element of the vector $Ax$, using the definition of matrix multiplication?


TITLE: Connecting tangents of convex curves: at some point orthogonal?
QUESTION [2 upvotes]: Let $a(t)$ and $b(t)$ be two smooth, nested convex curves in the plane, $t\in[0,1]$:

Suppose the parametrization of $a()$ and $b()$ is such that $\dot{a}(t)$ is parallel to $\dot{b}(t)$:
at time $t$, the tangents at $a(t)$ and $b(t)$ are parallel. Let $ab(t) = b(t)-a(t)$ be the vector
from $a(t)$ to $b(t)$, and let $\theta(t)$ be the counterclockwise angle from $ab(t)$ to $\dot{a}(t)$.
I would like to claim that

$\theta(t) = \pi/2$ at least twice within $t\in[0,1]$.

My proof of this is inelegant, relying on the length $|ab(t)|$,
essentially showing that if, e.g, $\theta(t) < \pi/2$ for all $t$, then $|ab(1)| >|ab(0)|$
(contradicting $|ab(1)| = |ab(0)|$).
But I feel there might be a clever way to achieve this via the mean-value theorem that I am not
seeing.  Also, perhaps "twice" can be replaced by "four times," and perhaps
"nested" need not be assumed.  So I am seeking a cleaner proof that may yield further insights.
Thanks for contributing ideas or pointing me in the right direction!

REPLY [5 votes]: Take points $x, y\in b$ which minimize (correspondingly maximize) the function $\mathop{\rm dist}_a$.
Let $\bar x$ and $\bar y\in a$ be the closest points to $x$ and $y$ correspondingly.
Clearly $\bar x=a(t)$, $x=b(t)$ and $\bar y=a(\tau)$, $y=b(\tau)$ for some $t$ and $\tau$
and $\theta(t)=\theta(\tau)=\tfrac\pi2$.
If $a$ and $b$ are non-concentric circles then you get only these two values.


TITLE: How to prove that the Torus and a Ball have the same Cardinality
QUESTION [1 upvotes]: How to prove that the Torus and a Ball have the same Cardinality ?
The proof is using the Cantor Berenstein Theorem.
I now that they are subsets of $\mathbb{R}^{3}$ so I can write $\leq \aleph$ but I do not know how to prove $\geqslant  \aleph$.
Thanks

REPLY [1 votes]: Let's assume known the fact that $\Bbb R^m$ and $\Bbb R^n$ have the same cardinality for all pairs of positive integers $(m,n)$.
Then the following follows easily: Let $X\subset\Bbb R^n$ such that there exists a subset $A\subset X$ homeomorphic to an open subset of $\Bbb R^m$ for some $m\leq n$. Then $|X|=|\Bbb R^n|$.
Indeed, such an $A$ obviously contains an open segment $I$ (e.g. any radius of any open ball contained in $A$) and the chain of inclusions
$$
I\subset A\subset X\subset\Bbb R^n
$$
together with the fact that $|I|=|\Bbb R|=|\Bbb R^n|$ shows that $|X|=|\Bbb R^n|$.
It is clear that both the ball $B$ and the torus $T$ (no matter whether you take the solid shapes or just their surfaces) satisfy the property required for $X$, thus $|B|=|\Bbb R^3|=|T|$.


TITLE: why $f(y_{f_{\varphi}}) = \langle f, f_{\varphi}\rangle ?$.
QUESTION [1 upvotes]: Prove that  every Hilbert space is reflexive.
I found the answer here
I have some doubt in  given below  answer marked in red box

My thinking : I think this  sentence ($f_{\varphi} \in \mathcal H^*$. By the RRT, there exists a unique $y_{f_{\varphi}}$ such that for all $f \in \mathcal H^*$, we have $f(y_{f_{\varphi}}) = \langle f, f_{\varphi}\rangle$.)  is WRONG
why that given above sentence is wrong ?
Ans:  According to RRT  it should be like this $f_{\varphi}(y)=(y,y_{f_{\varphi}})$
My doubt :According to RRT :If $T$ is a bounded linear functional on a Hilbert space $H$ then
there exists some $g \in H$ such that for every $f \in H$ we have
$T(f) =\langle f, g \rangle $.
Here $\langle f, g \rangle $  contain $f$
If  $f(y_{f_{\varphi}}) = \langle f, f_{\varphi}\rangle$.then where $y_{f_{\varphi}}$ is gone
and $\langle f, f_{\varphi}\rangle$   doesn't contain $y_{f_{\varphi}}$
Therefore  $f(y_{f_{\varphi}}) \neq \langle f, f_{\varphi}\rangle$
So  im not getting why  $f(y_{f_{\varphi}}) = \langle f, f_{\varphi}\rangle ?$.

REPLY [0 votes]: Yes im right.
It should be  like this
Now, $f_{\varphi} \in \mathcal H^*$. By the RRT, there exists a unique $y_{f}$ such that for all $f \in \mathcal H^*$, we  have
$$\ f(y_{f_\varphi})=\langle\,y_{f_\varphi},y_f\rangle$$ for all $y_{f_\varphi} \in \mathcal H$
Using the RRT two times  we have  $||y_f||=||f_{\varphi}||$  and $||f||=||y_{f_\varphi}||$
Therefore $f(y_{f_{\varphi}}) = \langle f, f_{\varphi}\rangle$.


TITLE: Convert elliptic: $F\left(\sin^{-1}\left(\sqrt{\frac{R}{R+2}}\right)\biggr\rvert 1-\frac{4}{R^2}\right)$ to complete: $K\left(1-\frac{4}{R^2}\right)$
QUESTION [0 upvotes]: I explored the convolution of $\arcsin$ distributions SE. Here I found the following identity empirical (allot of trial, error and luck):

typo: $1-\frac{1}{R^2}$ should be $1-\frac{\color{red}{4}}{R^2}$. As pointed out by first answer.

$$F\left(\sin^{-1}\left(\sqrt{\frac{R}{R+2}}\right) \biggr\rvert 1-\frac{\color{red}{4}}{R^2}\right) \stackrel{?}{=}\frac{1}{2}\cdot K\left(1-\frac{\color{red}{4}}{R^2}\right) \tag{1}$$
With $F$ the incomplete and $K$ the complete elliptic integral and: $-2 \leq R \leq 2$. I only have numerical confirmation with WA online and Python (see: SE).
Looking at definitions:
$$F(\varphi \rvert k^2)=\int_{0}^{\varphi} \frac{ \ d \theta}{\sqrt{1-k^2 \sin^2(\theta)}}$$
$$K(k)=\int_{0}^{\frac{\pi}{2}}\frac{ \ d \theta}{\sqrt{1-k^2 \sin^2(\theta)}} $$
If the identity $(1)$ is correct how to interpret:
$$\sin^{-1}\left(\sqrt{\frac{R}{R+2}}\right) \neq \frac{\pi}{2}$$
Seems maybe related to: SE, this article is currently outside my math level.
Seeking a proof (comments: hints/tips to continue studying). Do not know where to start. Note that I am novice with elliptic integrals.

REPLY [0 votes]: The following is not a full answer but gives me more understanding exploring the function.
Incomplete elliptic integral:
$$F\left(\sin^{-1}\left(\sqrt{\frac{R}{R+2}}\right) \biggr\rvert 1-\frac{4}{R^2}\right) = \int_{0}^{\varphi}\frac{d \theta}{\sqrt{1-\left( 1- \frac{4}{R^2} \right) \sin^{2} \left(  \theta\right)}}$$
First I evaluated the range of: $\varphi=\sin^{-1}\left(\sqrt{\frac{R}{R+2}}\right)$. Maximum value $\varphi$:
$$\lim_{R \rightarrow \infty} \sin^{-1}\left(\sqrt{\frac{R}{R+2}}\right)=\lim_{R \rightarrow \infty} \sin^{-1}\left(\sqrt{\frac{1}{1+2/R}}\right)=\frac{\pi}{2}  $$
Plot:
$$Z=\frac{1}{\sqrt{1-\left( 1- \frac{4}{R^2} \right) \sin^{2} \left(  \theta\right)}}$$

Observations (from plot):

$F(a,b)$ from question is periodic with: $\pi$.
Integrating $\int Z \ d\theta$ gives $F(a,b)$ from $\varphi=0$ till $\varphi=\sin^{-1}\left(\sqrt{\frac{R}{R+2}}\right)$. $K(b)$ for $varphi=0$ till $\varphi=\pi/2$ giving: $F(a|b)=\frac{1}{2}K(b)$.


TITLE: Calculus implicit differentiation question
QUESTION [3 upvotes]: I stumbled upon this calculus implicit differential question: find
$\cfrac{du}{dy} $ of the function $ u = \sin(y^2+u)$.
The answer is $ \cfrac{2y\cos(y^2+u)}{1−\cos(y^2+u)} $.  I understand how to get the answer for the numerator, but how do we get the denominator part?
And can anyone point out the real intuitive difference between chain rule and implicit differentiation? I can't seem to get my head around them and when or where should I use them.
Thanks in advance!

REPLY [0 votes]: Starting with:
$ u = \sin(y^2+u)$
Perhaps more apparent and facile path (especially on an exam where an equivalent valid solution may suffice) would be just take the inverse of the  sine function for both sides. This results here in a rapid and easy separation of $u$ and $y$:
$ \arcsin(u) = y^2+u $
$ \arcsin(u) - u = y^2 $
Taking the differentials of both sides and applying the formula for differentiating the arc sine function:
$ du/\sqrt(1-u^2) - du = 2 y dy $
Whereupon it is evident:
$ \cfrac{du}{dy} = \cfrac{2y}{1/\sqrt(1-u^2)-1} $
which I claim as a likely equivalent and facile solution path. Note: the domain of the inverse sin function can be likely defined in a suitable fashion to accommodate real world conditions (see comments here).


TITLE: Binomial Distribution is Symmetric if and only if $p = 0.5$
QUESTION [2 upvotes]: I am trying to prove that whenever you have a binomial distribution $X \sim Bin(n, p)$, that it is symmetric whenever the probability is $p = 0.5$. I am not sure if there is a more precise way to prove it but my reasoning. I know the binomial distribution formula is
\begin{equation*}
P(X = k) = \dfrac{n!}{k!(n - k)!}p^k(1 - p)^{n - k}
\end{equation*}
My attempt to prove:
Case 1: If $p = 0.5$, then $X \sim Bin(n, 0.5)$ is symmetric.
When $p = 0.5$, then for any $k$, the binomial distribution becomes:
\begin{equation*}
P(X = k) = \dfrac{n!}{k!(n - k)!}(0.5)^k(1 - 0.5)^{n - k} = \dfrac{n!}{k!(n - k)!}(0.5)^n
\end{equation*}
The probability 0.5 in this case would mean that it is non-skewed to the left or right, and would be considered to be even probability. Otherwise, if $p < 0.5$, then the probability will be skewed to the left, or if $p > 0.5$, then the probability will be skewed to the right, in both which do not have any type of symmetry in the distribution. So to prevent any skewness, $p = 0.5$, and therefore, symmetric.
(Edit: Also, whenever $p = 0.5$, then we would have two mutually exclusive outcomes. For example, flipping a fair coin for $n$ amount of times)
Case 2: If $X \sim Bin(n, p)$ is symmetric, then $p = 0.5$.
I am not sure how to proceed with this case of the proof if it is possible to prove this. Also I am not sure if my first case is the right reasoning or if I am missing some details. Would appreciate some help.

REPLY [2 votes]: Assume $n>0,p\in(0,1)$ (the extreme cases are trivial since the distribution is then degenerate). You want to show
$$P(X=k)=P(X=n-k)\forall k\in \{0,...,n\}\iff p=0.5,$$
which follows since
$$P(X=k)={n \choose k} p^k (1-p)^{n-k}={n \choose n-k} p^k (1-p)^{n-k}$$
is equal to
$$P(X=n-k)={n \choose n-k} p^{n-k} (1-p)^k$$
for all $k$ if and only if
$$p^k (1-p)^{n-k} =p^{n-k} (1-p)^k\iff (1-p)^{n-2k}=p^{n-2k}\iff 1-p=p\iff p=0.5.$$


TITLE: Orthogonal projection on the Hilbert space .
QUESTION [10 upvotes]: I want to prove the following:

If $X$ is a Hilbert space and $Y$ is a closed subspace of $X$, then
  every $x\in X$ can be written as $x=y+z $ where $y\in Y$,
  $z \in Y^\perp$.
The projection (into $Y$) map $P:X\to Y$, given by $P(x)=y$ is linear,
  bounded, $P^2=P$, and $\langle x_1 , Px_2\rangle =\langle Px_1 , x_2\rangle$.

Here I have avoided subscripts (but the projection is always onto the $Y$ space):
Consider $x$ in $X$  , then there is a closest point to $x$ in $Y$ . Let us say that point as $Px$ , now we prove that $x-Px$ is orthogonal to $Y$ .   Choose  $y \in Y$ and $ |y|=1$
Now $|x-(Px+ y)|^2 =|x-Px|^2-2Re \alpha(x-Px,  y) + |\alpha|^2 y^2$   Let us choose $\alpha = (y, x-Px)$ then it becomes ,
$ := |x-Px|^2 - 2|(x-Px,  y) |^2 + |(y, x-Px)|^2 =|x-Px|^2 - |(x-Px, y) |^2 $
Which implies that distance of $x$ from $Px+y \in Y$ is less than the $|x-Px|$ , unless $|(x-Px, y) |^2 =0$ which gives us that $x$ and $x-Px$ are orthogonal .
Now let us see if $P : x \to Px$ is linear ,
Define $Qx =x-Px$ , We have already shown that $Qx$ is orthogonal to $Y$
Then $P(ax+by)+Q(ax+by) =ax+by =a(Px+Qx)+b(Py+Qy$, moving $P$ and $Q$ on two sides we get
$P(ax+by)-(aPx +bPy) = Q(ax+by) -(aQx+bQy)$, since the right side is in $Y$ and the left side is not in $Y$, both the side should be equal to $0$ ,
$P(ax+by)-(aPx +bPy)=0$ , hence we show that $P$ is linear .
And the boundedness follows because $|x|^2=|Px|^2+|Qx|^2$ , is that right ?
Am i right so far ? I am having a bit of difficulty in proving rest of the stuff.
Thanks for your help.

REPLY [11 votes]: The first part is often called the Orthogonal Decomposition Theorem and is found in just about any textbook on Hilbert spaces. Here (look at 3.6 and right below 3.9) is a readily available proof from the web.
For the second part, we can establish the following properties about $P$ rather quickly:

linear: Let $x_i=y_i+z_i$, where $x_i\in X$, $y_i\in Y$, $z_i\in Y^\perp$, and $\alpha,\beta$ be scalars. Then \begin{align}P(\alpha x_1+\beta x_2)&=P(\alpha(y_1+z_1)+\beta(y_2+z_2))\\&=P(\alpha y_1+\beta y_2+\alpha z_1+\beta z_2)=\alpha y_1+\beta y_2=\alpha P(x_1)+\beta P(x_2).\end{align}
bounded: Since $x=0$ is trivial, suppose $x\not=0$. Because the projection is orthogonal, the (generalized) Pythagorean Theorem says $\|x\|^2=\|y\|^2+\|z\|^2$, so $$\|Px\|^2=\|y\|^2=\|x\|^2-\|z\|^2\le \|x\|^2.$$ Therefore, $${\|Px\|^2\over \|x\|^2}\le 1 \implies \|P\|=\max_{x\not =0}{\|Px\|\over \|x\|}\le 1,$$ and hence $P$ is bounded.
idempotent: $P^2x=P(Px)=Py=y=Px$, so $P^2=P$.
self-adjoint: $$\langle Px_1,x_2\rangle=\langle y_1,y_2+z_2\rangle=\langle y_1,y_2\rangle+\langle y_1,z_2\rangle=\langle y_1,y_2\rangle+0=\langle y_1,y_2\rangle$$ and $$\langle x_1,Px_2\rangle=\langle y_1+z_1,y_2\rangle=\langle y_1,y_2\rangle+\langle z_1,y_2\rangle=\langle y_1,y_2\rangle+0=\langle y_1,y_2\rangle,$$ so $\langle Px_1,x_2\rangle=\langle x_1,Px_2\rangle$.


TITLE: Inequalities involving complex numbers
QUESTION [0 upvotes]: a) Prove that $|z^i|< e^\pi$ for all complex $z \neq 0$.
b) Prove that there is no constant $M > 0$ such that $|\cos z| < M$ for all complex $z$.
Where complex powers are defined as $z^w=e^{w Log\  z}$ and complex cosine is $\cos z=\frac{e^{iz}+e^{-iz}}2$.

REPLY [1 votes]: a) Since you define $z^i$ as $e^{i\log\lvert z\rvert-\operatorname{arg}z}$, where $\operatorname{arg}z$ is the only argument of $z$ from $[-\pi,\pi)$, then$$\lvert z^i\rvert=\lvert e^{i\log\lvert z\rvert-\operatorname{arg}z}\rvert=e^{-\operatorname{arg}z}<e^\pi.$$
b) Use the fact that $\cos(iz)=\cosh z$.


text:
\begin{document}

\mbox{}
\vspace{-1.1ex}
\title{Cohomology of toric line bundles via simplicial Alexander duality}
\author{Shin-Yao Jow}
\address{Department of Mathematics,
University of Pennsylvania,
Philadelphia, PA 19104}
\email{\texttt{jows@math.upenn.edu}}

\date{23 December 2010}

\begin{abstract}
We give a rigorous mathematical proof for the validity of the toric sheaf cohomology algorithm conjectured in the recent paper by R. Blumenhagen, B. Jurke, T. Rahn, and H. Roschy (\texttt{arXiv:1003.5217}). We actually prove not only the original algorithm but also a speed-up version of it. Our proof is independent from (in fact appeared earlier on the arXiv than) the proof by H. Roschy and T. Rahn (\texttt{arXiv:1006.2392}), and has several advantages such as being shorter and cleaner and can also settle the additional conjecture on ``Serre duality for Betti numbers'' which was raised but unresolved in \texttt{arXiv:1006.2392}.
\end{abstract}

\keywords{Toric variety; cohomology of line bundle; local cohomology; Alexander duality.}
\subjclass[2000]{14M25 (Primary); 13D45; 14Q99 (Secondary)}

\maketitle

\section{Introduction}

Recently in \cite{BJRR} a new algorithm for computing the cohomology groups of line bundles on a toric variety was conjectured, which the authors observed to be more efficient than previously known algorithms such as those described in \cite[\S 3.5]{Ful}, \cite{EMS}, or \cite[\S 9.1]{CLS}. However apart from having no proof, \cite{BJRR} also gave a somewhat vague treatment of a crucial step in the algorithm involving certain ``remnant cohomology''. The original goal of this article was to provide the first rigorous mathematical proof for the validity of the algorithm, as well as a precise procedure to compute the ``remnant cohomology''. A week after the first version of this article appeared on the arXiv, an independent and alternative proof was given in \cite{RR}, in the last section of which the authors further conjectured a ``Serre duality for Betti numbers'' that will speed up the algorithm even more. It turns out that this additional conjecture also follows quite easily from our original method of proof, possibly due to the fact that our language is more topological, which makes our proof cleaner and shorter than the more algebraic proof in \cite{RR}, so it is easier for us to have a clearer view. Thus in this updated version, the main theorem (Theorem~\ref{t:main}) is strengthened to incorporate this additional conjecture, and its immediate corollary (Corollary~\ref{c:algorithm}) is likewise improved to give the speed-up algorithm.

The basic framework we will use to explain the algorithm in \cite{BJRR} is the results in \cite[Section~2]{EMS} which connect toric sheaf cohomology with local cohomology over a polynomial ring (the Cox ring). Let $X$ be a $d$-dimensional toric variety over a field $\KK$ associated to a fan $\Delta$ in $N\cong\ZZ^d$. We denote by $\Delta(1)$ the set of
one-dimensional cones (i.e. rays) in $\Delta$, and assume that $\Delta(1)$ spans $N_{\RR}=N\otimes_{\ZZ}\RR$. The Cox ring of $X$ (\cite{Aud}, \cite{Mus}, \cite{Cox})  is the polynomial ring $S$ in the set of variables indexed by the rays in $\Delta$: \[
          S=\KK[x_{\rho}\mid \rho\in \Delta(1)]. \]
One can consider on $S$ the usual multigrading on a polynomial ring, which in this case is  the grading by $\ZZ^{\Delta(1)}$, the free abelian group with a basis indexed by the rays (of course in $S$ only the $\NN^{\Delta(1)}$-graded pieces are nonzero, but later we will also need to consider localizations of $S$ which invert some of the variables). There is also an important squarefree monomial ideal $B$ of $S$, called the irrelevant ideal, defined as follows. For each cone $\sigma$ in the fan $\Delta$, let $\sigma(1)=\{\rho\in\Delta(1) \mid \rho\subset\sigma\}$. Then \[
 B=\langle\prod_{\rho\notin\sigma(1)}x_\rho \mid \sigma\in\Delta \rangle. \]
The relations between $X$ and $(S,B)$ are very similar to those between a projective space and its homogeneous coordinate ring with the usual irrelevant ideal. For example it was shown in \cite{Cox} that $X$ is a quotient of the complement of the algebraic subset defined by $B$ in the affine space $\Spec S$. Moreover, under this quotient map, the image of the coordinate hyperplane $(x_\rho=0) \subset \Spec S$ is the torus-invariant prime divisor $D_\rho\subset X$ corresponding to the ray $\rho$. In view of this, it is natural to endow $S$ with another coarser grading by the class group of $X$: one simply sets the degree of a monomial $\prod_{\rho\in\Delta(1)}x_\rho^{a_\rho}$ to be the divisor class $[\sum_{\rho\in\Delta(1)} a_\rho D_\rho]$ in $\Cl(X)$. In the special case when $X$ is a projective space, this is simply the grading by the total degree. If $F$ is a
$\Cl(X)$-graded module over $S$, then an associated quasi-coherent sheaf $\widetilde{F}$ on $X$ can be constructed in a way similar to the projective space case, and it is also true that every coherent sheaf on $X$ is of the form $\widetilde{F}$ for some finitely generated $F$ (\cite[Proposition~3.3]{Cox} for simplicial $X$; \cite[Theorem~2.1]{EMS}). Using \v{C}ech cohomology, it was shown \cite[Proposition~2.3]{EMS} that for $i\ge 1$, there is an isomorphism of $\Cl(X)$-graded $S$-modules \[
 \bigoplus_{\alpha\in\Cl(X)} H^i\left(\widetilde{F(\alpha)}\right) \cong  H^{i+1}_B(F),\]
where $F(\alpha)$ is the shifted module $F(\alpha)_\beta=F_{\alpha+\beta}$, and $H^{i+1}_B(F)$ is the $(i+1)^{\text{th}}$ local cohomology of $F$ with support on $B$
\cite[Definition~13.1]{MS}. This isomorphism is the starting point for studying sheaf cohomology on a toric variety both in \cite{EMS} and in this paper. We will focus on the case $F=S$, since $\widetilde{S(\alpha)}\cong \OO_X(D)$ if $D$ is a divisor with divisor class $\alpha$, so $H^{i+1}_B(S)$ already encodes the $i^{\text{th}}$ sheaf cohomology of all line bundles on $X$. Following \cite{EMS}, we use the notation $H^i_\ast(\OO_X)$ to denote the left-hand side of the above isomorphism when $F=S$. So we have
\begin{equation} \label{e:1}
 H^i_\ast(\OO_X)=\bigoplus_{\alpha\in\Cl(X)} H^i\left(\widetilde{S(\alpha)}\right) \cong  H^{i+1}_B(S).
\end{equation}

The local cohomology $H^{i+1}_B(S)$ can be computed by a \v{C}ech complex whose terms involve only localizations of $S$ inverting some of the variables (\cite[Theorem~13.7]{MS}; see also Section~\ref{s:localcoh} of this paper). Consequently $H^{i+1}_B(S)$, and hence $H^i_\ast(\OO_X)$, enjoy the finer $\ZZ^{\Delta(1)}$-grading. It is known that each of the graded pieces $H^i_\ast(\OO_X)_p$, $p\in \ZZ^{\Delta(1)}$, depends only on which coordinates of $p$ are negative (\cite[Theorem~2.4]{EMS}; see also Proposition~\ref{p:3.1}). More precisely, if one defines \[
   \Neg(p)=\{ \rho\in\Delta(1) \mid \text{the $\rho$ coordinate of $p$ is negative}\}, \]
then $H^i_\ast(\OO_X)_p$ is canonically isomorphic to $H^i_\ast(\OO_X)_q$ for any $p,q\in \ZZ^{\Delta(1)}$ such that $\Neg(p)=\Neg(q)$. Hence it makes sense to define $H^i_\ast(\OO_X)_I$ for a subset $I\subset \Delta(1)$, by setting it to be $H^i_\ast(\OO_X)_p$ for any $p\in \ZZ^{\Delta(1)}$ such that $\Neg(p)=I$. One then wants to know which subsets $I$ give rise to nonzero graded pieces $H^i_\ast(\OO_X)_I$, and how these graded pieces can be computed. Our main theorem provides a simple combinatorial answer to this question when $X$ is a simplicial projective toric variety. To state it, it will be convenient to define two collections of subsets of $\Delta(1)$: first following the notation in \cite{BJRR} we define \[
 \SR=\Big\{\hat{J}\subset\Delta(1)\Bigm| \text{\parbox{9cm}{$\hat{J}$ does not span a cone in $\Delta$, but every proper subset of $\hat{J}$ spans a cone in $\Delta$.}}\Big\}. \]
The squarefree monomials of the form $\prod_{\rho\in\hat{J}} x_\rho$, $\hat{J}\in\SR$ are precisely the minimal generators of the so-called Stanley-Reisner ideal of $\Delta$. Then we define $\U$ to be the collection of all subsets of $\Delta(1)$ which can be expressed as a union $\hat{J_1} \cup \cdots \cup \hat{J_m}$ for some $\hat{J_1},\ldots,\hat{J_m}$ in $\SR$.

\begin{theorem} \label{t:main}
 Let $X$ be a simplicial projective toric variety of dimension $d$ associated to a fan $\Delta$. Let $I$ be a subset of $\Delta(1)$ and let $i\ge 1$ be a positive integer. Then
 \begin{enumerate}[\upshape (a)]
  \item
   $H^i_\ast(\OO_X)_I = 0$ unless $I\in \U$.
  \item
   Let $\hat{I}$ denote the complement of\/ $I$ in $\Delta(1)$. If\/ $i\ne d$, then $H^i_\ast(\OO_X)_I$ is naturally dual to $H^{d-i}_\ast(\OO_X)_{\hat{I}}$ (as $\KK$-vector spaces). Combined with part~(a) this implies in particular that $H^i_\ast(\OO_X)_I = 0$ unless both $I$ and $\hat{I}$ are in $\U$.
  \item
    If $I\in \U$, let $\hat{J_1},\ldots,\hat{J_m}$ be all of the elements in $\SR$ that are contained in $I$. Define $\Lambda_I$ to be the following abstract simplicial complex on the vertex set $\{1,\ldots,m\}$:  \[
    \Lambda_I=\bigg\{K\subset\{1,\ldots,m\} \biggm| \bigcup_{k\in K} \hat{J_k} \ne I \bigg\}. \]
   Then there is a natural isomorphism   \[
     H^i_\ast(\OO_X)_I \cong \widetilde{H}_{|I|-i-2}(\Lambda_I),  \]
   where the right-hand side is the reduced homology of $\Lambda_I$ (with coefficients in $\KK$).
 \end{enumerate}
\end{theorem}

We remark that part~(b) is the ``Serre duality for Betti numbers'' conjectured in the last section of \cite{RR}, while part~(c) is essentially the ``remnant'' cohomology $\mathcal{H}^i(\mathcal{Q})$ in \cite{BJRR}.\footnote{In our notations the ``remnant'' cohomology $\mathcal{H}^i(\mathcal{Q})$ is really the (reduced) relative homology of the full simplex on $\{1,\ldots,m\}$ modulo $\Lambda_I$, hence is essentially the same as the reduced homology of $\Lambda_I$ since the full simplex is contractible.}

 As an immediate corollary we obtain the following speed-up version of the algorithm in \cite{BJRR}:

\begin{corollary} \label{c:algorithm}
 Given a line bundle $L$ on $X$ and an integer $0<i<d$,\footnote{The cases $i=0$ or $d$ are known to be easy. When $i=0$ the cohomology can be computed by counting lattice points in a certain polytope: see for example \cite[p.66]{Ful}. The case $i=d$ can be reduced to the case $i=0$ by the usual Serre duality.} we have \[
h^i(X,L)=\sum_{I} \#\bigg\{p\in\ZZ^{\Delta(1)} \biggm| \Neg(p)=I \text{ and } \Big[\sum_{\rho\in\Delta(1)}p_\rho D_\rho\Big]=L\bigg\}\cdot \dim \widetilde{H}_{|I|-i-2}(\Lambda_I), \]
where the sum is over all\/ $I\subset\Delta(1)$ such that both $I$ and $\hat{I}$ are in $\U$. (In the original algorithm in \cite{BJRR} the sum is over all\/ $I$ in $\U$.)
\end{corollary}

We remark that from a theoretic point of view, the efficiency of this algorithm seems to come primarily from the vanishing in Theorem~\ref{t:main}~(a)(b), which greatly reduces the amount of homological computation one needs to perform. We refer the readers to the Introduction in \cite{RR} for more details on implementation of the algorithm and how it compares with other previously known algorithms.

The organization of the rest of this article is as follows: Section~\ref{s:topological} contains some topological preliminaries we will need in our proof, including our primary tool \emph{the simplicial Alexander duality} (Corollary~\ref{c:SAD}) and the topological source of the vanishing in Theorem~\ref{t:main}~(a) (Lemma~\ref{l:2.4}). In Section~\ref{s:localcoh} we review the method for computing the local cohomology $H^i_B(S)$ using the natural cellular resolution of $S/B$ supported on the moment polytope of $X$, as done in \cite[\S 3]{MS04}. After these preparations, we present our rather short proof for Theorem~\ref{t:main} in Section~\ref{s:proof}.

\section*{Acknowledgements}

The author would like to thank Ezra Miller for several useful comments on the first draft of this article.

\section{Combinatorial topological preliminaries} \label{s:topological}

In this section we collect some results from simplicial topology which will come into the proof of the main theorem, most notably the simplicial Alexander duality. Throughout this section $\Gamma$ will denote an (abstract) simplicial complex on a finite vertex set $V$, i.e. $\Gamma$ is a collection of subsets of $V$, such that if $\sigma\in\Gamma$ and $\tau\subset\sigma$ then $\tau\in\Gamma$. Each $\sigma\in\Gamma$ is called a simplex or a face of $\Gamma$. The underlying topological space of $\Gamma$ is denoted by $\|\Gamma\|$. Also recall that the link of a face $\sigma$ in $\Gamma$ is the subcomplex \[
  \link_\Gamma \sigma=\{\tau\in \Gamma \mid \tau\cup\sigma \in
  \Gamma\text{ and } \tau\cap\sigma=\varnothing\}. \]
Given any subset $\sigma\subset V$ (not necessarily a face of $\Gamma$), the notation $\hat{\sigma}$ will denote its complement $\hat{\sigma}=V\setminus\sigma$, and the notation $\Gamma_{\le\sigma}$ will denote the simplicial complex consisting of every face of $\Gamma$ that is contained in $\sigma$: \[
 \Gamma_{\le\sigma}=\{\tau\in\Gamma \mid \tau\subset\sigma \}. \]

\begin{definition}
 The \emph{Alexander dual} of $\Gamma$, denoted by $\Gamma^*$, is the following simplicial complex on the vertex set $V$: \[
 \Gamma^*=\{ \sigma\subset V \mid \hat{\sigma}\notin \Gamma \}. \]
\end{definition}

\begin{theorem}[Simplicial Alexander duality] \label{t:SAD}
For every integer $j$ there is an isomorphism \[
\widetilde{H}_j(\Gamma^*)\cong \widetilde{H}^{|V|-3-j}(\Gamma). \]
\end{theorem}

 A self-contained proof from first principles was given in \cite{BT}, and a proof using homological algebra can be found in \cite[Section~5.1]{MS}. The connection to the topological Alexander duality
$\widetilde{H}_j(S^{n-2}\setminus A)\cong \widetilde{H}^{n-3-j}(A)$ is that if $|V|=n$, then the simplicial complex consisting of all proper subsets of $V$ has an underlying space homeomorphic to $S^{n-2}$, and it contains the closed subcomplex $\Gamma$ whose underlying space plays the role of $A$. Furthermore it can be shown that $S^{n-2}\setminus \|\Gamma\|$ is homotopy equivalent to $\|\Gamma^*\|$: see Proposition~2.27 of the book \cite{BP}. What we will use in fact is the following equivalent version given in Proposition~2.29 of that same book:

\begin{corollary}[Simplicial Alexander duality---alternative version] \label{c:SAD}
 If $\sigma\in\Gamma^*$, then \[
  \widetilde{H}_j(\link_{\Gamma^*}\sigma)\cong \widetilde{H}^{|V|-3-j-|\sigma|}(\Gamma_{\le\hat{\sigma}}). \]
\end{corollary}

\begin{proof}
This follows from Theorem~\ref{t:SAD} because $\link_{\Gamma^*}\sigma$ is the Alexander dual of $\Gamma_{\le\hat{\sigma}}$ when both are viewed as simplicial complexes on the vertex set $\hat\sigma$. (On the other hand setting $\sigma=\varnothing$ recovers Theorem~\ref{t:SAD}, so the two statements are actually equivalent.)
\end{proof}

We will also need the following two simple lemmas:

\begin{lemma} \label{l:2.3}
 For any subset $\sigma\subset V$, it holds that $\|\Gamma_{\le\hat{\sigma}}\|$ is a deformation retract of $\|\Gamma\| \setminus \|\Gamma_{\le\sigma}\|$.
\end{lemma}

\begin{proof}
Let $\Gamma_{\sigma,\hat{\sigma}}$ be the set consisting of every face of $\Gamma$ which is neither contained in $\sigma$ nor in $\hat{\sigma}$. Each $\tau\in\Gamma_{\sigma,\hat{\sigma}}$ can be written as a proper disjoint union $\tau=\tau_{\sigma} \sqcup \tau_{\hat{\sigma}}$, where $\tau_{\sigma}=\tau\cap\sigma$ and $\tau_{\hat{\sigma}}=\tau\cap\hat{\sigma}$. Observing that \[
 \|\Gamma\| \setminus \|\Gamma_{\le\sigma}\|=\|\Gamma_{\le\hat{\sigma}}\| \cup
   \bigg( \bigcup_{\tau\in\Gamma_{\sigma,\hat{\sigma}}} \|\tau\|\setminus\|\tau_{\sigma}\|\bigg), \]
the lemma thus follows since each $\|\tau\|\setminus\|\tau_{\sigma}\|$ can be deformation retracted to $\|\tau_{\hat{\sigma}}\|$.
\end{proof}

\begin{lemma} \label{l:2.4}
Given $\sigma\in\Gamma$, let $\tau_1,\ldots,\tau_m\in\Gamma$ be the maximal faces of $\Gamma$ containing $\sigma$. If $\tau_1\cap\cdots\cap\tau_m\supsetneqq\sigma$, then
$\|\link_{\Gamma}\sigma\|$ is contractible.
\end{lemma}

\begin{proof}
The maximal faces of $\link_{\Gamma}\sigma$ are precisely $\tau_1\setminus\sigma,\ldots,\tau_m\setminus\sigma$, and by assumption they have a nonempty intersection. Hence the lemma follows.
\end{proof}

\section{Toric local cohomology} \label{s:localcoh}

Let $X$ be a $d$-dimensional simplicial projective toric variety associated to a fan $\Delta$ in $N\cong\ZZ^d$. Let $\Delta(i)$ denote the set of $i$-dimensional cones in $\Delta$. Recall from the Introduction that the Cox ring of $X$ is the polynomial ring $S=\KK[x_{\rho}\mid \rho\in \Delta(1)]$, and the irrelevant ideal is
$B=\langle\prod_{\rho\notin\sigma(1)}x_\rho \mid \sigma\in\Delta \rangle$. In this section we review the method for computing the local cohomology $H^i_B(S)$ using the natural cellular resolution of $S/B$ supported on the moment polytope of $X$. Our main reference is \cite{MS} (see also \cite[\S 3]{MS04} or \cite[Example~6.6]{Mil}).

A common way to compute local cohomology is to use the usual \v{C}ech complex \cite[Definition~13.5]{MS}. This is often not the most efficient complex one can use. In fact any free resolution $\mathcal{F}_\bullet$ of $S/B$ gives rise to a generalized \v{C}ech complex $\check{\mathcal{C}}_{\mathcal{F}}^\bullet$ \cite[Definition~13.28]{MS}, which can be used to compute $H^i_B(S)$ \cite[Theorem~13.31]{MS}. The usual \v{C}ech complex came from the Taylor resolution \cite[\S 4.3.2]{MS}, which tends not to be minimal. In our present toric situation, the best choice for $\mathcal{F}_\bullet$ is the natural cellular resolution supported on the moment polytope of $X$, which turns out to be minimal \cite[\S 4.3.6]{MS}. We will now briefly describe this resolution. The readers who are unfamiliar with the basic theory of cellular resolutions could consult \cite[\S 4.1]{MS} before going on.

First we need to describe the moment polytope of $X$. In the theory of projective toric varieties this is a well-known polytope in the dual space of $N$, but since we are only concerned about its combinatorial structure, it suffices to know that the set of all of its faces is in an inclusion-reversing bijection with $\Delta$. Hence we can identify the set of $i$-dimensional faces of the moment polytope with the set $\Delta(d-i)$, and label each face $\sigma$ with the monomial $\xx^{\hat{\sigma}}\overset{\text{def}}{=}\prod_{\rho\notin\sigma(1)}x_\rho$. The moment polytope together with this labeling fits the definition of a labeled cell complex \cite[Definition~4.2]{MS}, and   gives rise to the following cellular free resolution of $S/B$: \[
\mathcal{F}_\bullet \colon \ 0\longleftarrow S \longleftarrow \bigoplus_{\sigma\in\Delta(d)} S\,\xx^{\hat{\sigma}} \longleftarrow \bigoplus_{\sigma\in\Delta(d-1)} S\,\xx^{\hat{\sigma}} \longleftarrow \cdots \longleftarrow \bigoplus_{\sigma\in\Delta(0)} S\,\xx^{\hat{\sigma}} \longleftarrow 0, \]
where the homomorphism from a direct summand $S\,\xx^{\hat{\sigma}}$ of $\mathcal{F}_i$ to a direct summand $S\,\xx^{\hat{\tau}}$ of $\mathcal{F}_{i-1}$ is $0$ if $\tau\not\supset\sigma$, and is the inclusion map times $\pm 1$ if $\tau\supset\sigma$. The sign is determined by arbitrary but fixed orientations for the cones in $\Delta$.
The corresponding generalized \v{C}ech complex $\check{\mathcal{C}}_{\mathcal{F}}^\bullet$ is by definition \[
\check{\mathcal{C}}_{\mathcal{F}}^\bullet \colon \ 0\longrightarrow S \longrightarrow \bigoplus_{\sigma\in\Delta(d)} S\Big[\frac{1}{\xx^{\hat{\sigma}}}\Big] \longrightarrow \bigoplus_{\sigma\in\Delta(d-1)} S\Big[\frac{1}{\xx^{\hat{\sigma}}}\Big] \longrightarrow \cdots \longrightarrow   \bigoplus_{\sigma\in\Delta(0)} S\Big[\frac{1}{\xx^{\hat{\sigma}}}\Big] \longrightarrow 0, \]
where the homomorphism between two direct summands $S[\frac{1}{\xx^{\hat{\tau}}}]$ and $S[\frac{1}{\xx^{\hat{\sigma}}}]$ for each codimension one $\sigma\subset\tau$ is again the inclusion map times the sign determined by orientations.

By \cite[Theorem~13.31]{MS} we have $H^i_B(S)=H^i(\check{\mathcal{C}}_{\mathcal{F}}^\bullet)$. Since $\check{\mathcal{C}}_{\mathcal{F}}^\bullet$ is $\ZZ^{\Delta(1)}$-graded, so is $H^i_B(S)$. In fact given any $p\in\ZZ^{\Delta(1)}$ we can describe the graded piece $H^i_B(S)_p$ explicitly as follows. Let $I\subset\Delta(1)$ be the set of places where $p$ has negative coordinates, i.e. \[
   I=\Neg(p)\overset{\text{def}}{=}\{ \rho\in\Delta(1) \mid \text{the $\rho$ coordinate of $p$ is negative}\}. \]
Let $\hat{I}=\Delta(1)\setminus I$. Then the graded piece $S[\frac{1}{\xx^{\hat{\sigma}}}]_p$ is $0$ if $\sigma(1)\not\subset\hat{I}$, and is $\KK\xx^p$ if $\sigma(1)\subset\hat{I}$, where $\xx^p$ denotes the monomial whose exponent vector is $p$. Taking the graded piece of degree $p$ of every term in $\check{\mathcal{C}}_{\mathcal{F}}^\bullet$ and then computing the cohomology, one obtains the following description of $H^i_B(S)_p$, which is essentially the same as \cite[Theorem~2.7]{EMS} or \cite[Proposition~3.2]{MS04}:

\begin{proposition} \label{p:3.1}
Let $\PP$ be the abstract simplicial complex on the vertex set $\Delta(1)$ given by  $\PP=\{\sigma(1) \mid \sigma\in\Delta\}$. Then for every $p\in\ZZ^{\Delta(1)}$ and $i\ge 2$ there is an isomorphism \[
 H^i_B(S)_p\cong \widetilde{H}_{d-i}(\PP_{\le\hat{I}}),  \]
where $I=\Neg(p)$. In particular $H^i_B(S)_p$ only depends on $\Neg(p)$.
\end{proposition}

\section{Proof of Theorem} \label{s:proof}

We now present the proof of our main theorem.

\begin{proof}[Proof of Theorem~\ref{t:main}]
By \eqref{e:1} we have $H^i_\ast(\OO_X)_I \cong  H^{i+1}_B(S)_I$, and by Proposition~\ref{p:3.1} we have $H^{i+1}_B(S)_I\cong \widetilde{H}_{d-1-i}(\PP_{\le\hat{I}})$. Lemma~\ref{l:2.3} then implies that \[
\widetilde{H}_{d-1-i}(\PP_{\le\hat{I}})\cong \widetilde{H}_{d-1-i}(\|\PP\|\setminus\|\PP_{\le I}\|). \]
Since $\|\PP\|$ is homeomorphic to $S^{d-1}$, the topological Alexander duality gives \[ \widetilde{H}_{d-1-i}(\|\PP\|\setminus\|\PP_{\le I}\|)\cong \widetilde{H}^{i-1}(\PP_{\le I}). \]
Now there are two possible cases for $I$:
\begin{description}
\item[Case 1] $\hat{I}\notin\PP^*$. This is equivalent to $I\in\PP$, which implies that $\PP_{\le I}$ is the full simplex on $I$, so $\widetilde{H}^{i-1}(\PP_{\le I})=0$.
\item[Case 2] $\hat{I}\in\PP^*$. In this case we can apply Corollary~\ref{c:SAD} to get \[
  \widetilde{H}^{i-1}(\PP_{\le I})\cong \widetilde{H}_{|I|-2-i}(\link_{\PP^*}\hat{I}). \]
 Let $J_1,\ldots,J_m$ be the maximal faces of $\PP^*$ containing $\hat{I}$. There are two further subcases: either $J_1\cap\cdots\cap J_m\supsetneqq\hat{I}$ or $J_1\cap\cdots\cap J_m=\hat{I}$. If $J_1\cap\cdots\cap J_m\supsetneqq\hat{I}$ then $\widetilde{H}_{|I|-2-i}(\link_{\PP^*}\hat{I})=0$ by Lemma~\ref{l:2.4}. Hence $H^i_\ast(\OO_X)_I=0$ unless $J_1\cap\cdots\cap J_m=\hat{I}$, or equivalently $I=\hat{J_1}\cup\cdots\cup \hat{J_m}$. This proves part~(a) since \[
 \SR=\{\hat{J}\mid \text{$J$ is a maximal face of $\PP^*$}\}. \]
Now if $J_1\cap\cdots\cap J_m=\hat{I}$, then $J_1\setminus \hat{I}, \ldots,J_m\setminus \hat{I}$ are precisely the maximal faces of $\link_{\PP^*}\hat{I}$, and the nerve they form (in the sense of \cite{Gru}) is precisely $\Lambda_I$. Hence \[
 \widetilde{H}_{|I|-2-i}(\link_{\PP^*}\hat{I})\cong \widetilde{H}_{|I|-2-i}(\Lambda_I) \]
 by \cite[Theorem~10]{Gru}, which proves part~(c).
\end{description}
Finally to see part~(b), note that the series of isomorphisms in the beginning part of the proof showed that \[
 H^i_\ast(\OO_X)_I \cong \widetilde{H}_{d-1-i}(\PP_{\le\hat{I}}) \cong \widetilde{H}^{i-1}(\PP_{\le I}). \]
So $H^i_\ast(\OO_X)_I\cong \widetilde{H}^{i-1}(\PP_{\le I})$, and
$H^{d-i}_\ast(\OO_X)_{\hat{I}}\cong \widetilde{H}_{i-1}(\PP_{\le \hat{\hat{I}}})=\widetilde{H}_{i-1}(\PP_{\le I})$ if $i\ne d$, hence they are dual to each other.
\end{proof}

config: arxiv


TITLE: Is the induced map between etale cohomology groups independent of the compactification?
QUESTION [1 upvotes]: Let $k$ be an algebraically closed field, and $X$ be a smooth variety. For any compactification $i: X \hookrightarrow Y$ (so $X$ is a dense open subset of $Y$), consider the induced map
$i_!: H^i_{et,c}(X, \mathbb Z_l) \rightarrow H^i_{et,c}(Y,\mathbb Z_l)$ between compact supported etale cohomology groups.
Is $Ker(i_!)$ independent of the compactification (at least for $\mathbb Q_l$ coefficients )?

REPLY [2 votes]: The answer is yes, assuming $Y$ is smooth: this follows from the identification
$$ \operatorname{Ker}(H^n_c(X,\mathbf Q_\ell) \to H^n_c(Y,\mathbf Q_\ell)) = W_{n-1}H^n_c(X,\mathbf Q_\ell),$$
where $W$ denotes the weight filtration. Clearly any element of $W_{n-1}$ is in the kernel, since the image of the map is pure of weight $n$. For the reverse inclusion use the long exact sequence of the pair in compact support cohomology to see that the kernel is the image of $H^{n-1}_c(Y \setminus X,\mathbf Q_\ell)$, which is of course of weights $\leq n-1$.


TITLE: existence of finite irreducible reprentation of a nonunital $C^*$ algebra
QUESTION [0 upvotes]: Suppose $A$ is a non-unital $C^*$ algebra,can we conclude that there must exist a nonzero finite irreducible representation of $A$.

REPLY [0 votes]: We cannot guarantee the existence of such a representation.  Consider $K(H)$, for $H$ a separable infinite-dimensional Hilbert space.  This is non-unital and simple, so any representation of this has to be infinite-dimensional.


TITLE: Inverse Fourier transform of a sine function multiplied with a transfer function
QUESTION [0 upvotes]: I’ve got the following question:
The following relation holds:
$\hat{v}_F(\omega) = \hat{Y}(\omega)\hat{F}(\omega)$
With: $F(t)=\sum_k F_{k}\sin(\omega_k t - \phi_k)$
The transformation from frequency to time domain would be:
$v_F(t) = \sum_k F_{k}|\hat{Y}(\omega_k)| \sin(\omega_k t+\angle \hat{Y}(\omega_k) +\phi_k)$
How do you get to that solution?
This is how I got started:
$\hat{v}_F(\omega) = \hat{Y}(\omega)\hat{F}(\omega) \Leftrightarrow v_f(t)= y(t)*f(t)$
so, expressing the convolution:
$\int_0^\infty y(\tau)\sin(\omega_k(t-\tau)+\phi_k)d\tau$
$= \frac{1}{2i}\int y(\tau)[e^{i(\omega_k(t-\tau)+\phi_k}-e^{-i(\omega_k(t-\tau)+\phi_k}] =\frac{1}{2i} e^{i\omega_kt + \phi_k}\int y(\tau)e^{-i\omega_k \tau}d\tau- \frac{1}{2i} e^{-i\omega_kt + \phi_k}\int y(\tau) e^{+i\omega_k \tau}d\tau$
$= \frac{1}{2i} e^{i\omega_kt + \phi_k} \hat{Y}(i\omega) - \frac{1}{2i} e^{-i\omega_kt + \phi_k} \hat{Y}(-i\omega)$
and I do not see how to proceed.

REPLY [0 votes]: We have that $F(t)=\sum_k F_{k}\sin(\omega_k t - \phi_k)$ so
$$F(w) = \mathcal{F}(F(t)) = \sum_k F_{k} \mathcal{F}[\sin(\omega_k t - \phi_k)]$$
Note that
$$\sin(\omega_k t - \phi_k) = \cos(\phi_k) \sin (\omega_k t) - \sin(\phi_k)\cos(\omega_k t) $$
which means that
\begin{equation}
\tag{0}
\mathcal{F}[\sin(\omega_k t - \phi_k)] = \cos(\phi_k) \mathcal{F}[\sin (\omega_k t)] - \sin(\phi_k)\mathcal{F}[\cos(\omega_k t)]
\end{equation}
You could easily verify (using the definition of Fourier transform) that
\begin{align}
\mathcal{F}[\sin (\omega_k t)] &= j \pi [\delta(\omega + \omega_k) - \delta(\omega - \omega_k)]\\
\mathcal{F}[\cos (\omega_k t)] &=  \pi [\delta(\omega - \omega_k) + \delta(\omega + \omega_k)]
\end{align}
So
$$\mathcal{F}[\sin(\omega_k t - \phi_k)] =R_k(\omega) + j I_k(\omega)$$
where
$$ R_k(\omega) = - \sin(\phi_k)\pi[\delta(\omega - \omega_k) + \delta(\omega + \omega_k)]$$
and
$$ I_k(\omega) = \cos(\phi_k)\pi[\delta(\omega + \omega_k) - \delta(\omega - \omega_k)]$$
So
\begin{equation}
\tag{1}
\begin{split}
V_F(\omega) &= Y(\omega) F(\omega)\\
&= Y(\omega) \sum_k F_{k}(R_k(\omega) + jI(\omega) )\\
&=  \sum_k F_{k}(Y(\omega)R_k(\omega) + jY(\omega)I(\omega) )
\end{split}
\end{equation}
Notice that
$$Y(\omega)R_k(\omega)=- \sin(\phi_k)\pi[\delta(\omega - \omega_k)Y(\omega) + \delta(\omega + \omega_k)Y(\omega)]$$
$$Y(\omega)I_k(\omega)= \cos(\phi_k)\pi[\delta(\omega + \omega_k)Y(\omega) - \delta(\omega - \omega_k)Y(\omega)]$$
Notice that $\delta(\omega + \omega_k)Y(\omega)= Y(-\omega_k)\delta(\omega + \omega_k)$ and $\delta(\omega -\omega_k)Y(\omega)= Y(\omega_k)\delta(\omega -\omega_k)$ so
$$Y(\omega)R_k(\omega)=- \sin(\phi_k)\pi[Y(\omega_k)\delta(\omega -\omega_k) + Y(-\omega_k)\delta(\omega + \omega_k)]$$
$$Y(\omega)I_k(\omega)= \cos(\phi_k)\pi[Y(-\omega_k)\delta(\omega + \omega_k) - Y(\omega_k)\delta(\omega -\omega_k)]$$
Use exponential representation of your $Y(\omega_k) = \vert Y(\omega_k) \vert \exp(j \theta_k)$ where $\theta_k= \angle Y(\omega_k)$. If $y(t)$ is real, notice the following symmetry and anti-symmetry in frequency domain:
\begin{align}
\vert Y(\omega_k) \vert &= \vert Y(-\omega_k) \vert \\
\theta_k &= - \theta_{-k}
\end{align}
With this in mind, we get
\begin{equation}
\tag{2}
Y(\omega)R_k(\omega)=- \sin(\phi_k)\pi \vert Y(\omega_k) \vert[  \exp(j\theta_k)\delta(\omega -\omega_k) + \exp(-j\theta_k)\delta(\omega + \omega_k)]
\end{equation}
\begin{equation}
\tag{3}
Y(\omega)I_k(\omega)= \cos(\phi_k)\pi \vert Y(\omega_k) \vert[\exp(-j\theta_k)\delta(\omega + \omega_k) -\exp(j\theta_k)\delta(\omega -\omega_k)]
\end{equation}
Replacing (2) and (3) in (1) , and factoring common terms, we get
\begin{equation}
\begin{split}
V_F(\omega) &= -\sum_k F_{k} \vert Y(\omega_k)\vert \pi \big[ \sin \phi_k \exp(j\theta_k)+j\cos \phi_k \exp(j\theta_k) \big]\delta(\omega - \omega_k) \\
&+ \sum_k F_{k} \vert Y(\omega_k)\vert \pi \big[- \sin \phi_k \exp(-j\theta_k) + j\cos \phi_k \exp(-j\theta_k) \big]\delta(\omega + \omega_k) \\
\end{split}
\end{equation}
Using Euler's formula gives the terms in brackets as
\begin{equation}
\sin \phi_k \exp(j\theta_k)+j\cos \phi_k \exp(j\theta_k) =  \sin(\phi_k - \theta_k) + j\cos(\phi_k - \theta_k)
\end{equation}
and
\begin{equation}
- \sin \phi_k \exp(-j\theta_k) + j\cos \phi_k \exp(-j\theta_k) =  -\sin(\phi_k - \theta_k) + j\cos(\phi_k - \theta_k)
\end{equation}
Replacing we get
\begin{equation}
\begin{split}
V_F(\omega) &= -\sum_k F_{k} \vert Y(\omega_k)\vert \pi [\sin(\phi_k - \theta_k) + j\cos(\phi_k - \theta_k)]\delta(\omega - \omega_k) \\
&+ \sum_k F_{k} \vert Y(\omega_k)\vert \pi [-\sin(\phi_k - \theta_k) + j\cos(\phi_k - \theta_k)]\delta(\omega + \omega_k) \\
\end{split}
\end{equation}
Recollecting the real and imaginary terms,
\begin{equation}
\begin{split}
V_F(\omega) &= -j \sum_k F_{k} \vert Y(\omega_k)\vert \pi \cos(\phi_k - \theta_k)[\delta(\omega - \omega_k) -\delta(\omega + \omega_k) ]\\
&- \sum_k F_{k} \vert Y(\omega_k)\vert \pi \sin(\phi_k - \theta_k)[\delta(\omega - \omega_k) +\delta(\omega + \omega_k) ] \\
\end{split}
\end{equation}
That is
\begin{equation}
\begin{split}
V_F(\omega) &= -\sum_k F_{k} \vert Y(\omega_k)\vert \Big\lbrace j \pi \cos(\phi_k - \theta_k)[\delta(\omega - \omega_k) -\delta(\omega + \omega_k) ]\\
&+ \pi \sin(\phi_k - \theta_k)[\delta(\omega + \omega_k) +\delta(\omega - \omega_k) ] \Big\rbrace \\
\end{split}
\end{equation}
Note that the terms in brackets are in the same form of equation (0), hence you can now apply the inverse fourier transform and arrive at your formula.


TITLE: How does the relativistic Doppler effect model change if I want to include general relativistic effects also?
QUESTION [6 upvotes]: Derivations for the relativistic Doppler effect abound, which extends the description of the Doppler effect to include the effects of relativity in cases where the source or observer is moving at relativistic velocities. This is all pretty well-described in many sources that I have found.
But what if I would like to include the effects of general relativity instead? Assume the following scenario:

There is a stationary source on the earth emitting light at frequency $f$ (with that frequency measured on the reference frame of the source).
I launch a rocket that travels to orbit with some velocity profile $\mathbf{v}(t)$.
The rocket has a detector that continually observes the light source during its flight.
I would like to know the observed frequency of the light wave from the perspective of the rocket, as a function of time, $f'(t')$ (where $t'$ represents time in the rocket's reference frame).

I see three separate effects that would alter the observed frequency aboard the rocket:

The classical Doppler effect
Time dilation on the rocket due to special relativity
Time dilation on the rocket due to general relativity

All of these will change as a function of time as the rocket accelerates and as it moves within Earth's gravitational field. However, all analyses of the relativistic Doppler effect that I've seen only encompass the first two effects. Is there some model that includes all three?

REPLY [0 votes]: Definitions
Let's start with definitions since there are a lot of terms/symbols that are not commonly used by anyone outside the GPS timing community.  Most of these notes and comments are taken from Ashby [2003] and Zhang et al. [2006].

receiver $\equiv$ station receiving signal from spacecraft transmitter (subscript r)
satellite $\equiv$ object in orbit that is transmitting signal (subscript or superscript s)
LOS $\equiv$ line-of-sight
J2 = 1.0826300 x 10-3 $\equiv$ Earth's second zonal harmonic.
$P_{2}(x) = \tfrac{1}{2} \left( 3 x^{2} - 1 \right)$ $\equiv$ Legendre polynomial of degree two
ae = 6378.137 km $\equiv$ semi-major axis of the World Geodetic System ellipsoid
aorb $\equiv$ semi-major axis of satellite orbit
B $\equiv$ latitude of receiver
||r|| $\equiv$ distance from Earth center to receiver
GM = 3.986004418 x 1014 m3 s-2 $\equiv$ product of Earth's mass times Newtonian gravitational constant
$\Phi_{o}/c^{2}$ = 6.969290134 x 10-10 $\equiv$ potential of the geoid (see IERS reference PDF)
Ur $\equiv$ Newtonian gravitational potential (mass attraction only) at receiver location (see Equation 1 below)
Us $\equiv$ Newtonian gravitational potential (mass attraction only) at satellite location (okay treat Earth like point mass)
$\mathbf{v}_{r}$ $\equiv$ velocity of receiver relative to receiver-to-satellite LOS
$\mathbf{v}^{s}$ $\equiv$ velocity of satellite relative to receiver-to-satellite LOS
$\mathbf{n}_{r}^{s}$ $\equiv$ receiver-to-satellite LOS unit vector
$f_{o}$ $\equiv$ nominal signal frequency from satellite
$f_{r}$ $\equiv$ signal frequency at receiver from satellite

Doppler Effects
First we define the gravitational potential at the receiver, given by:
$$
U_{r} = - \frac{ G M }{ \lVert \mathbf{r} \rVert } \left[ 1 - \left( \frac{ a_{e} }{ \lVert \mathbf{r} \rVert  } \right)^{2} \ J_{2} \ P_{2}\left( \sin{B} \right) \right] \tag{1}
$$
Now we can define the frequency received from an orbiting satellite as:
$$
f_{r} = f_{o} \left[ 1 + \frac{ - U_{r} + v_{r}^{2}/2 + \Phi_{o} + 2 G M/a_{orb} + 2 \ U^{s} }{ c^{2} } \right] \ \left[ \frac{ 1 - \left( \mathbf{n}_{r}^{s} \cdot \mathbf{v}_{r} \right)/c }{ 1 - \left( \mathbf{n}_{r}^{s} \cdot \mathbf{v}^{s} \right)/c } \right] \tag{2}
$$
Technically, this is in a non-inertial reference frame and the $v_{r}^{2}/2$ must include centrifugal effects such that Equation 2 goes to:
$$
f_{r} = f_{o} \left[ 1 + \frac{ - U_{r} + \left(\omega^{2} \ \cos^{2}{B} \ \lVert \mathbf{v}_{r} \rVert^{2}\right)/2 + \Phi_{o} + 2 G M \left(a_{orb}^{-1} - \lVert \mathbf{r}^{s} \rVert^{-1} \right) }{ c^{2} } \right] \ \left[ \frac{ 1 - \left( \mathbf{n}_{r}^{s} \cdot \mathbf{v}_{r} \right)/c }{ 1 - \left( \mathbf{n}_{r}^{s} \cdot \mathbf{v}^{s} \right)/c } \right] \tag{3}
$$
where $\omega$ is the angular velocity of Earth's rotation.
Caveats
There are other errors that come into play like the non-vacuum index of refraction of the magnetosphere, ionosphere, and atmosphere between the satellite and receiver.  There are also errors/uncertainties in both the satellite and receiver clocks [e.g., Ashby, 2003; Zhang et al., 2006].
References

Ashby, N. "Relativity in the Global Positioning System," Living Rev. Relativity 6, 2003, Online article, https://link.springer.com/article/10.12942/lrr-2003-1. Accessed on 10/26/2018
Zhang, J., et al., "On the Relativistic Doppler Effect for Precise Velocity Determination using GPS," J. Geodesy 80, pp. 104--110, 2006.


TITLE: Prove a theorem on convergence of series
QUESTION [0 upvotes]: Let $a_n$ and $b_n$  be two sequences with $a_n >0 $ and $b_n>0$ for all $n\ge N$ and let $c_n=b_n-b_{n+1}a_{n+1}/a_n$. Prove that:
If there is a possitive constant $r$ such that $c_n \ge r \gt 0$ for all $n\ge N$,then $\sum a_n$ converges.
I tried :
$a_nb_n\gt a_{n+1}b_
{n+1}$. So, the sequence $a_nb_n$ is decreasing. Then for $n\ge N$,
$a_nb_n\le a_Nb_N$,
so $a_n\le a_Nb_N/b_n\lt a_Nb_N/r$, because $a_nb_n\gt a_nr$.
Each term of the sequence $a_n$ is smaller than the constant $a_Nb_N/r$ for $n\ge N$, but I dont know if this implies that the partial sums of $a_n$ are bounded and the series converges.

REPLY [5 votes]: Notice that the condition can be written as
$$ c_n a_n = b_n a_n - b_{n+1}a_{n+1} $$
and hence
$$ \sum_{n=1}^{N} a_n \leq \sum_{n=1}^{N} \frac{c_n}{r} a_n = \frac{b_{1}a_{1} - b_{N+1}a_{N+1}}{r} \leq \frac{b_{1}a_{1}}{r}. $$
So the partial sum of $(a_n)$ is monotone increasing and bounded above, hence converges.


TITLE: How to prove that the geometric distribution converges to the exponential distribution?
QUESTION [0 upvotes]: We have given $X_n$ to be a sequence of random variables s.t. $X_n$ has the geometric distribution with parameter $\frac{1}{n}$. I need to show that the distribution of $P_{X_n}$ converges to an exponential distribution. And I also need to find the parameter of this exponential distribution.
As a hint they gave us that we may us the formula for geometric series and also the definition of $e^x$ in terms of a limit.
I know that I need to show that $X_n$ converges to $X$ in distribution. By definition we need to show that for all continuous and bounded function $f:\Bbb{R}\rightarrow \Bbb{R}$ we have $$E(f(X_n))\rightarrow E(f(X))$$ so we only had this statement/definition until now therefore we need to work with it.
My idea was to using the definition of the integral by the Riemann sum. So I know that I should get something like $$\int_0^\infty f(x)\lambda e^{-\lambda }dx=\lim_{n\rightarrow \infty} \sum_{j=0}^\infty \frac{1}{n}f\left(\frac{j}{n}\right) \lambda e^{-\lambda \frac{j}{n}}$$ But now I wanted to take $\lambda =1$ because at the end I should get $$\sum_{k=1}^\infty f(k)\frac{1}{n} \left(1-\frac{1}{n}\right)^{k-1}$$But I somehow don't see how to proceed. Could maybe someone help me?

REPLY [1 votes]: Including the (necessary) normalization, we simply transform the sum into an integral:
$$\Bbb E\!\left[f\!\left(\frac{X_n}n\right)\right]=\sum_{k=1}^\infty\frac1n\left(1-\frac1n\right)^{k-1}f\!\left(\frac kn\right)
=\int_0^\infty\frac1n\left(1-\frac1n\right)^{\lfloor x\rfloor}f\!\left(\frac{\lfloor x\rfloor+1}n\right)\,\mathrm dx,$$
where $\lfloor\cdot\rfloor$ denotes integer part.
The change of variable $x=ny$ gives
$$\Bbb E[f(X_n)]=\int_0^\infty\left(1-\frac1n\right)^{\!\lfloor ny\rfloor}
f\!\left(\frac{\lfloor ny\rfloor+1}n\right)\mathrm dy.$$
The integrand tends to $\mathrm e^{-y}f(y)$ and is dominated by the integrable function $y\mapsto\mathrm e^{-y}\|f\|_\infty$ on $(0,\infty)$.


TITLE: Finding the homology of space comprised of $S^2$ and an intersecting torus.
QUESTION [1 upvotes]: I am reviewing old algebraic topology qualifying exams, and I need to compute the fundamental group and the homology of the following space.

First, a brief calculation of $\pi_1(X)$. Let $C$ be the sphere and let $D$ be the torus unioned with some small path connected open neighborhood of the intersection of $C$ and $D$. Note that $C\cap D$ (path-connected) will look like an oval on the sphere with two holes in it. It will deform rectract to $S^1\vee S^1$. Thus we have that $\pi_1(C)$ is trivial, $\pi_1(D)=\langle a,b| a b a^{-1} b^{-1}\rangle$ and $\pi_1(C\cap D)=\langle c,d| \hspace{3mm}\rangle$. We then have $(i_C)_*(c)=(i_C)_*(d)=0$ where $i_C:C\cap D\hookrightarrow C\cong S^2$ is the inclusion. $(i_D)_*(d)=(i_D)_*(d)=a\in \pi_1(T^2)$ where $i_D:C\cap D\hookrightarrow D\cong T^2$ is the inclusion. So SVK's theorem gives
$$ \pi_1(X)= \langle a, b | a b a^{-1} b^{-1}, a\rangle \cong \mathbb{Z}$$
Next, I need to calculate the homology of $X$. My first idea is to use the Mayer-Vietoris long exact sequence. I know that $H_0(X)$ is just $\mathbb{Z}$ (one path component) and that $H_1(X)=\pi_1^{ab}(X)\cong \mathbb{Z}$, but I would like to calculate $H_1(X)$ and $H_2(X)$ by hand for practice. Using $A\cong S^2$ and $B\cong T^2$ (perphaps with some small retractible neighborhoods so that $X=\textrm{int}(A)\cup \textrm{int} (B))$.

$(\Psi_1)_*$ will map $[a]=[(1,0)]\in \pi_1(T^2)$ to $[0]\in H_1(X)$ because that loop on the torus is homotopic to a constant map when it is seen as a loop in $S^2$. $(\Psi_1)_*$ will map $[b]=[(0,1)]\in \pi_1(T^2)$ to $[1]\in H_1(X)$. However, this analysis is based on knowledge of $\pi_1(X)$, which is isomorphic to $H_1(X)$, so this kind of reasoning seems contingent on the fact that I already know that $H_1(X)=\langle b| \hspace{3mm}\rangle$.  I think that that $(\Phi_1)_*$
will be neither injective nor surjective. It does seem to split and map one copy of $\mathbb{Z}$ to $0$ and the other copy of $\mathbb{Z}$ to the second summoned of $H_1(T^2)=\mathbb{Z}\oplus \mathbb{Z}$. $\textrm{Im}(\Phi_1)_*\cong\mathbb{Z}\cong\textrm{Ker} (\Psi_1)_*$ by exactness. So I'm unsure how to compute $H_1(X)$ without relying on the fundamental group.
for $n=2$ we have

It is immediate that $(\Phi_2)_*$ is injective. I know that given $[z]\in H_2(X)$ ($[z]$ is a cycle) , I can write $z=x+y$ with $x,y\in C_2(A)\oplus C_2(B)$, then $0=\partial([z])=\partial(z)+\partial(y)$ and $\delta_2([z])=[\partial(x)]=-[\partial(y)]$. I haven't decided on a cell decomposition for $X$, so I feel like I cannot proceed any further with the analysis of these maps. Is there a straight-forward way of proceeding with my long exact sequence, or would I be better off looking for a cell decomposition and computing the homology that way?

REPLY [3 votes]: As mentioned by JHF, $X$ is actually $S^1 \vee S^1 \vee S^2 \vee S^2 \vee S^2$. To see that, take a look at #1 here: https://www.math3ma.com/blog/clever-homotopy-equivalences, and contract what you called the oval, so that both halves of the torus become spheres with two identified points.
Your approach does lead to a result as well, but you have to be more careful. First of all, I don't think that $\pi_1(D)=\langle a,b| a b a^{-1} b^{-1}\rangle$. (Your definition of $D$ is recursive but I think I got it.) $D$ is more than a torus, and, accordingly, you seem to have lost one generator: "from the side" a torus looks like a big circle whereas $D$ looks like a wedge of two circles due to the component you've added.
Next, I don't think that to say what you've said about $(\Psi_1)_*$ is to use $\pi_1(X)$. One generator is mapped to zero and the image of the other generates a free subgroup - that's all you need to say here, no $\pi_1$ references. Then you can follow through with this calculation: first, by looking at what happens in $H_0,$ you can find the image of $\delta_1$ (as the kernel of $(\Phi_0)_*$). Then you can also find its kernel as the image of $(\Psi_1)_*$. Both will be $\mathbb{Z}$, which yields the right answer.
$H_2$ is even simpler. $(\Phi_2)_*$ is injective, of course, but you can say more: it's obviously zero. So $(\Psi_2)_*$ is injective. Thus you have a $\mathbb{Z}\oplus \mathbb{Z}$ subgroup in $H_2(X)$ and the quotient over it is the image of $\delta_2$, which is the kernel of $(\Phi_1)_*$, which is $\mathbb{Z}$. So, $H_2(X)=\mathbb{Z}^3.$ (Both times I'm using that extensions with the quotient a free group are trivial.)


TITLE: What's the notation for a 2 input 2 output function?
QUESTION [0 upvotes]: There are a couple of similar questions on this site but I don't think they answer the same issue (or I don't understand the answers enough to know if they do).
I have a function which takes X,Y coordinates and does some operations, let's say it multiplies the X coordinate by 2 and adds 3, multiplies the Y coordinate by 10, then outputs the new X',Y' coordinates. I'm a programmer and I need to translate my function into an equation suitable for publication.
What is the correct notation for this? I could describe f(x) and f(y) separately, but I keep thinking there must be a notation to do both in one line? There are lots of mathematical operations that work on coordinates...

REPLY [1 votes]: Based on your last comment, one possible way of writing your function is the following. Let $(x,y)$ denote a point in space. Because temperature and altitude are scalars, we can say that the temperature at $(x,y)$ is $T(x,y)$, and the altitude at $(x,y)$ is $A(x,y)$. Note that both $T$ and $A$ are functions that take two inputs $(x,y)$, and produce one output. Then, you can introduce a new function $F$ that also takes two inputs, but produces two outputs, i.e.,
$$
F(x,y) = \left(T(x,y), A(x,y) \right).
$$
Basically, every time you call $F$ with the inputs $(x,y)$, internally you call $T$ and $A$ with those same inputs and return their corresponding outputs.


TITLE: proofs of $n\sum_{d|n}\frac{(-1)^{d+1}}{d}=\sum_{d|n,d \text{ odd}}d$
QUESTION [2 upvotes]: I'm looking for other proofs of the identity :

For all integer $n\in \mathbb{N}^{+}$:
  $$n\sum_{d|n}\frac{(-1)^{d+1}}{d}=\sum_{d|n,d \text{ odd}}d $$
  where in the first sums is taken over all dividors of $n$, and the second over all odd divisors of $d$

I have this poof, let $n=2^k(2q+1)$ :
$$\begin{align}n\sum_{d|n}\frac{(-1)^{d+1}}{d}&=&&n\sum_{d|2q+1}\sum_{ i=0} ^k \frac{(-1)^{2^id+1}}{2^id} \end{align}$$
because every divisor of $n$ can be written uniquely as $2^id$ where $d|2q+1$ and $0\leq i \leq k$
$$\begin{align}n\sum_{d|n}\frac{(-1)^{d+1}}{d}&=&&n\sum_{d|2q+1}\left(\frac{1}{d} -\sum_{ i=1} ^k \frac{1}{2^id}\right)\\
&=&&\sum_{d|2q+1}\frac{n}{2^kd} \\
&=&&\sum_{d|2q+1} \frac{2q+1}{d}\\
&=&&\sum_{d|n,d \text{ odd}}d \end{align}$$
So i want to see if there is others proofs, mainly using generating functions because I think that they are powerful enough to solve this sorts of identities. Any hints, suggestions will be greatly appreciated.

REPLY [1 votes]: Following the suggestion from the comment we can prove algebraic equality as follows.
We have that
$$\sum_{n\ge 1} \frac{(-1)^{n+1}/n}{n^s}
= \sum_{n\ge 1} \frac{(-1)^{n+1}}{n^{s+1}}
= \left(1-\frac{2}{2^{s+1}}\right)\zeta(s+1)
= \left(1-\frac{1}{2^{s}}\right)\zeta(s+1).$$
Therefore
$$\sum_{n\ge 1} \left(\sum_{d|n} \frac{(-1)^{d+1}}{d}\right) \frac{1}{n^s} =
\zeta(s) \left(1-\frac{1}{2^{s}}\right)\zeta(s+1).$$
and the Dirichlet generating function of the LHS is
$$\zeta(s-1) \left(1-\frac{1}{2^{s-1}}\right)\zeta(s).$$
On the other hand we also have
$$\sum_{n\ge 1, \; n\;\mathrm{odd}} \frac{n}{n^s}
= \prod_{p,\; p\;\mathrm{odd}} \frac{1}{1-p^{-(s-1)}}
= \left(1-\frac{1}{2^{s-1}}\right)\zeta(s-1).$$
Therefore the Dirichlet generating function of the RHS is
$$\zeta(s) \left(1-\frac{1}{2^{s-1}}\right) \zeta(s-1)$$
and we have equality.
Remark. If we want to be rigorous about it, we have equality of the coefficients of the Dirichlet series in the intersection of the two half-planes of convergence because for the Dirichlet series $\Lambda(s) = \sum_{n\ge 1} \frac{\lambda_n}{n^s}$ we have
$$\lambda_n = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \Lambda(s) n^{s-1} ds.$$


TITLE: Sensitivity on A Farmer's Case
QUESTION [1 upvotes]: I have questions, from Bazaraa's book (Linear Programming).

I have calculated point a. This is the optimal table.

Where $x_1$ is wheat, $x_2$ is corn, dan $x_3$ is soybeans.
My question is, what the meaning of point b.?
Is that true if the point c. is adding a constraint $x_1\geq 120$?

REPLY [1 votes]: I think it is safe to say that questions (b) and (c) are separate, meaning that in (c) you are limited to the original labor supply. In (b), you are given the option to expand labor beyond the original 5,000 man-days, at a rate of $3/hour and assuming a work day is eight hours. The question asks you to decide whether or not to buy more labor (not how much to buy), based on sensitivity analysis.


TITLE: Can we arbitrarily add or remove single points from computations involving a continuous random variable?
QUESTION [0 upvotes]: Say I have a probability space $(\Omega, \Sigma, \mu)$ and some real-valued random variable $X$ such that the distribution of $X$ is continuous.
Say $\Sigma_{\mathbb{R}}$ is the typical Borel algebra on the reals. Is it always the case that for any non-empty $\sigma \in \Sigma_{\mathbb{R}}$ and for any $b \in \sigma$, we have that:
\begin{align*}
    \mu(\omega\ |\ X(\omega) \in \sigma) = \\
    \int_{a \in \sigma} \mu(\omega\ |\ X(\omega) \in da) = \\
    \mu(\omega\ |\ X(\omega) \in \left[\sigma - \{b\}\right]) = \\
    \int_{a \in \left[\sigma - \{b\}\right]} \mu(\omega\ |\ X(\omega) \in da)
\end{align*}
If this is not the case for all continuous distributions, does this hold if we assume that the distribution of $X$ is Gaussian?

REPLY [1 votes]: Yes.  For $X$ to be continuous means that $P(X=b)=0$ for each $b\in\mathbb R$.  The difference between your $P(X\in S)$ and your $P(X\in S\setminus \{ b\})$ is exactly $P(X\in\{b\})=P(X=b)$, which is by assumption zero.


TITLE: Supremum and infimum of $\{ \frac{|x|}{1+|x|}:x\in \mathbb{R} \}$ and $\{ x+x^{-1}:0<x \leq 2 \}$
QUESTION [1 upvotes]: I want to find out if these sets have a supremum and/or infimum:
$\{ \frac{|x|}{1+|x|}:x\in \mathbb{R} \}$
and
$\{ x+x^{-1}:0<x \leq 2 \}$
Regarding the first one I would say that all elements are non-negative and since $0 \in \mathbb{R}$ it follows that $0$ is minimum as well as infimum.
But is that enough?
Regarding the second one we can rewrite $x^{-1}$ as $\frac{1}{x}$ and since x can't be $0$ we could use $2$ and get $2,5$ as a supremum. Is that correct/and or enough reasoning?

REPLY [0 votes]: for $$A=\{ \frac{|x|}{1+|x|}:x\in \mathbb{R} \}$$
We have $\inf (A) = 0$ and $\sup (A) =1$
For $$B=  \{ x+x^{-1}:0<x \leq 2 \}$$
$\inf (B) = 2$ and $\sup (B)$ does not exist because it is not bounded above.


TITLE: Is chaos theory really a theory? Why not just call it non-linear dynamics?
QUESTION [6 upvotes]: This may just be semantics, but it's always confused me. What is the thesis of Chaos Theory? I have read an entire book about it, and as far as I can tell, its just a bunch of analytical techniques, but there is no underlying object "chaos" that is being characterized, its just a bunch of sophisticated tools for parsing nonlinear dynamical systems.
As a counterexample, Matrix Theory is about the behavior of vector spaces and operations on them, the theory of linear ODEs leads to elegant proofs of some properties. But what are some proofs/theorems about "chaos" -- the closest thing I can find is that phase-space trajectories have a positive Lyapunov exponent...there must be more than that!

REPLY [5 votes]: Nonlinear dynamical systems can be quite roughly divided in two huge realms: the realm of order (continuous time two dimensional systems) and the realm of chaos (dimension three and above for continuous time systems).
The former is mostly about 16'th Hilbert problem about the number of limit cycles on the plane and still is a huge area of research. The other one is about most "hyperbolicity" and how to classify the flows in higher dimensions.  So, despite the cynical comment of @ AndreNicolas, there is an excellent reason to talk about Chaos theory.


TITLE: Is it possible to use the principle of EHT to build a virtual gravitational wave detector to detect very long wavelength gravitational waves?
QUESTION [1 upvotes]: The press conference of the EHT is fascinating. While there are some features such as the asymmetry in the ring that is to be followed up by the team, I was actually wondering, whether it is possible to have multiple LIGO detectors scattered on the planet and in orbit in the future so as to create a virtual gravitational wave detector the size of our planet in order to detect and measure e.g. $10^{10}$ year period waves that are relevant to the primordial large scale structures of the universe?

REPLY [0 votes]: Very long baseline interferometry is based on measuring both the intensities and the phases of the incoming signals and then being able to resolve the position of the signals on the sky better thanks to comparing the phase differences. It does not change the wavelengths you observe on, all the nodes must be able on the same same wavelength even individually.
In this sense of the word, the LIGO/Virgo network is already using very-long-baseline interferometry. However, they have an extremely small number of nodes (3) and none of them provide directional information! A single-node network would thus ascribe a single pixel to the entire sky, and you get more and more pixels as you add more detectors. Once again, all the detectors must observe on the same frequencies ($\sim$ kHz) for LIGO/Virgo, interferometry will not be able to change that.
There are projects that want to extend to other frequencies/wavelength. The space-borne detector LISA is planned to observe $\sim$ mHz gravitational waves, and there might be some cosmologically interesting signals in that band (note that the LISA arm length, 25 million killometers, will be way larger than the Earth). However, the greatest promise for coherent old cosmological sources of gravitational-wave observations is the Pulsar Timing Array, that should be able to detect in the $\sim$ nHz regime. In a loose sense of the word, you can understand the Pulsar Timing Array and the way it will be used as a gigantic interferometer-type device.