text:
\section{Primitive of Root of p x + q over Root of a x + b}
Tags: Primitives involving Root of a x + b and Root of p x + q

\begin{theorem}
:$\displaystyle \int \sqrt {\frac {p x + q} {a x + b} } \rd x = \frac {\sqrt {\paren {a x + b} \paren {p x + q} } } a + \frac {a q - b p} {2 a} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }$
\end{theorem}

\begin{proof}
From [[Primitive of Power of p x + q over Root of a x + b|Primitive of $\dfrac {\paren {p x + q}^n} {\sqrt {a x + b} }$]]:
:$\displaystyle \int \frac {\paren {p x + q}^n} {\sqrt {a x + b} } \rd x = \frac {2 \paren {p x + q}^n \sqrt {a x + b} } {\paren {2 n + 1} a} + \frac {2 n \paren {a q - b p} } {\paren {2 n + 1} a} \int \frac {\paren {p x + q}^{n - 1} } {\sqrt {a x + b} } \rd x$
Putting $n = \dfrac 1 2$:
{{begin-eqn}}
{{eqn | l = \int \sqrt {\frac {p x + q} {a x + b} } \rd x
      | r = \frac {2 \paren {p x + q}^{1/2} \sqrt {a x + b} } {\paren {2 \cdot \frac 1 2 + 1} a} + \frac {2 \cdot \frac 1 2 \paren {a q - b p} } {\paren {2 \cdot \frac 1 2 + 1} a} \int \frac {\paren {p x + q}^{1/2 - 1} } {\sqrt {a x + b} } \rd x
      | c =
}}
{{eqn | r = \frac {\sqrt {\paren {a x + b} \paren {p x + q} } } a + \frac {a q - b p} {2 a} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }
      | c =
}}
{{end-eqn}}
{{qed}}
\end{proof}

config: wiki


TITLE: Is there a name for this kind of "Pascal's Triangle"?
QUESTION [14 upvotes]: I've been working on a problem and came across an interesting triangle that functions almost exactly like a Pascal Triangle. I'd like to see if I am able to find more about the properties of these triangles. I've found information on Pascal's simplexes, but that is not what this is.
Pascal's Triangle can be constructed by the fact that each cell is the sum of the $2$ above it. But I'm interested in the resulting triangles when you sum the above $3$ cells, $4$ cells, or more.
$n=3$:
$$
1\\
1\ \ 1\ \ 1\\
1\ \ 2\ \ 3\ \ 2\ \ 1\\
1\ \ 3\ \ 6\ \ 7\ \ 6\ \ 3\ \ 1\\
$$
$n=5$:
$$
1\\
1\ \ \ 1\ \ \ 1\ \ \ 1\ \ \ 1\\
1\ \ \ 2\ \ \ 3\ \ \ 4\ \ \ 5\ \ \ 4\ \ \ 3\ \ \ 2\ \ \ 1\\
1\ \ \ 3\ \ \ 6\ \ 10\ 15\ 18\ 19\ 18\ 15\ 10\ \ 6\ \ \ 3\ \ \ 1
$$
In particular what I'd like to know is how to generate an arbitrary row, like the $20$th row when $n=11$.

REPLY [6 votes]: Though I rather like the previous two answers, this configuration can also be thought of as a cellular automaton. Thus, your $n=3$ version can be computed in Mathematica like so:
radius = 1;
duration = 9;
ca = CellularAutomaton[{Total[##] &, {}, radius}, {{1}, 0}, duration];
Grid[ca /. 0 -> ""]

Simply increase radius and/or duration as desired.  We can generate your 20th row for $n=11$ and check it against the first, accepted answer as follows:
ca = First[CellularAutomaton[{Total[##] &, {}, 5}, {{1}, 0}, {{20}}]];
nom = CoefficientList[Expand[Sum[x^i, {i, 0, 10}]^20], x];
ca == nom
(* Out: True *)


TITLE: Point on the proof that $(C([0,1]),d_u)$ is complete.
QUESTION [1 upvotes]: The following is an excerpt from my notes on the example that $(C([0,1]),d_u)$ is a complete metric space.
I understand that to prove that $(C([0,1]),d_u)$ is complete, we need to show two things, namely that,

Every Cauchy sequence of elements in $C([0,1])$ converges to its limit in $C([0,1])$.
That these limits are also in $C([0,1])$, that is to say, that we need to show that the functions to which our sequence of functions converges are also continuous functions.

In moving to show the first of these points, I have the following in my notes:
"Let $(f_n)_{n=1}^\infty$ be a Cauchy sequence in our space $C([0,1])$, so that for every $\epsilon\gt0$ there is a $N$ so that,
$$d_u(f_n,f_m)=\sup_{x\in[0,1]}|f_n(x)-f_m(x)|\lt\epsilon$$
if $m,n\ge N$.
Here now is the point that I am not too sure I understand. In my notes it then goes on to say that,
"But in that case, for every single $x\in[0,1]$, $|f_n(x)-f_m(x)|\lt\epsilon$, which means that the sequence $(f_n(x))_{n=1}^\infty$ is a Cauchy sequence in $\mathbb R$ for each $x$."
How is it that we know that this is the case for every single $x\in [0,1]$? How is it that what is written above shows this fact?

REPLY [0 votes]: Well, for every $x\in [0, 1]$ you have
$$\left|f_n(x)-f_m(x)\right|\leq\sup\limits_{x\in [0, 1]}\left|f_n(x)-f_m(x)\right|$$
and that's just because the $\sup$ is the least upper bound for the $\left|f_n(x)-f_m(x)\right|$ ranging over all $x$. So if the $\sup$ is less than $\epsilon$, then so are all of the $\left|f_n(x)-f_m(x)\right|$.
It might be also helpful to note that, since $\left|f_n(x)-f_m(x)\right|$ is continuous, the $\sup$ is actually attained at some $x_0\in [0, 1]$.


TITLE: A remark about the Rellich-Kondrachov Compactness Theorem in Evans's PDE book
QUESTION [9 upvotes]: The following is the Rellich-Kondrachov Compactness Theorem in Evans's Partial Differential Equations

The author gives a remark as follows

I don't understand the first sentence in the Remark. Here are my questions:

How does one get $W^{1,p}(U)\Subset L^p(U)$ for $p=n$?

Why does "$p^*\to\infty$ as $p\to n$" imply the desired embedding?

REPLY [5 votes]: Remember $U$ is bounded, so $W^{1,n}(U) \subset W^{1,p}(U)$ for all $1 \leq p < n$ by Holder. Since $p^*\to \infty$ as $p\to n$, we can choose a fixed $p<n$ close enough to $n$ so that $p^*>n$. Then by the Rellich-Kondrachov compactness thoerem
$$W^{1,n}(U) \subset W^{1,p}(U) \subset\subset L^n(U).$$
Arguing along the same lines, we actually have $W^{1,n}(U) \subset\subset L^q(U)$ for all $1 \leq q < \infty$.


TITLE: problems about normal subgroups and the index
QUESTION [5 upvotes]: I'm stucked with these problems, could you help me?
I. Let $H$ and $N$ subgroups of a group $G$ such that $N \lhd G$ , $|N| < \infty$, $[G:H] < \infty$ and $([G:H], |N|) = 1$. Prove that $N$ is a normal subgroup of $H$.
Since $N$ is normal in $G$ then $gNg^{-1} = N$ for all $g\in G$ then in particular $hNh^{-1} = N$ for all $h \in H$ so the normality condition is easy, but I don't know how to show that $N < H$.
II. Let $H$ and $K$ be subgroups of a group $G$ such that the index of $H$ in $G$ and the index of $K$ in $G$ are finite. Prove that $[K:H \cap K] = [G:H]$ if and only if $G = HK = KH$.
For the sufficiency I used that $|G|=|HK|=|H||K||H \cap K|$ and $|G/K| = |G|/|K|$. And for the necessity I have no idea but I guess it has something to do with the second isomorphism theorem.

REPLY [2 votes]: Taking into account the comments:
1)  Already done by Derek holt. Note that $\,[NH:H]=[N:N\cap H]\,$ without assuming any normality of $\,H\,$: just define the following function between the sets of left cosets:
$$f:H^{\backslash NH}\to(N\cap H)^{\backslash N}\,\,,\,\,f(nH):=n(N\cap H)$$
Prove $\,f\,$ is a well-defined bijection (warning: we don't necessarily have group above so $\,f\,$ may not be a homomorphism).
2) (i) If $\, G= HK=KH\,$ , then define similarly as above :
$$f:H^{\backslash G}\to (H\cap K)^{\backslash K}\,\,,\,\,f(gH):=g(H\cap K)$$
$$(a)\;\;\text{Well-defined:}\;\;\;gH=xH\Longrightarrow g^{-1}x\in H . \,\text{Writing}\,\,g=kh\,\,,\,x=k'h'\,\,,\text{we get:}$$
$$h^{-1}k^{-1}k'h'=h_0\in H\Longrightarrow k^{-1}k'=hh_0h'^{-1}\in H\cap K\Longrightarrow k(H\cap K)=k'(H\cap K)\Longrightarrow $$
$$f(gH)=g(H\cap K)=k(H\cap K)=k'(H\cap K)=x(H\cap K)=f(xH)$$
and etc.


TITLE: What thoughts should I be having when proving the Euclidean Division Algorithm?
QUESTION [1 upvotes]: Given $m > 0 \mathbin{|} m ∊ ℕ$ and $n > 0 \mathbin{|} m ∊ ℕ$, let's say
$m \leftarrow 50$
$n \leftarrow 20$
Number Theory tells us that any positive integer can be written as a product of two other integers and remainder $r$, where $r=0$ or $r>0$, right?
How do I understand the following statements? (How can a number divide an equation?)

Any number that divides $\{m,n\}$ must divide
$m - qn = r$.
Any number that divides $\{n,r\}$ must divide
$qn + r = m$.

How do I conclude the following?
Therefore, the set of divisors that divide $\{m,n\}$ are the same as the set of divisors that divide $\{n,r\}$, in particular, the greatest divisor in both of those sets will be (must be) the same.
I ask this because I am pretty sure that the conclusion above is the core of the whole proof, yet I am not as mathematically talented as many others, because I am not at a point where I can confidently explain this to someone else.

Plugging in the numbers:

$50 - (2)20 = 10$ as $m - qn = r$
$(2)20 + 10 = 50$ as $qn + r = m$

REPLY [2 votes]: The argument is as follows:

We know there are integers $q$ and $r$ such that $m=qn+r$ and $0\le r<n$. If $r=0$ then $m=qn$ so the greatest divisor of $m$ and $n$ is $n$ and we are done. But if $r > 0$ we can proceed as follows.
Any number that divides both $m$ and $n$ will also divide $m-qn$. But $m-qn=r$. So any number that divides both $m$ and $n$ also divides $r$.
Any number that divides both $n$ and $r$ will also divide $qn+r$. But $qn+r=m$. So any number that divides both $n$ and $r$ also divides $m$.
Suppose $S$ is the set of numbers that divide both $m$ and $n$. Suppose $T$ is divide both $n$ and $r$. (2) shows that if $s \in S$ then $s \in T$, so $S \subset T$. (3) shows that if $t \in T$ then $t \in S$, so $T \subset S$. Since $S \subset T$ and $T \subset S$, the sets $S$ and $T$ must be the same set.
Since $S=T$, we can conclude that $\max(S) = \max(T)$. In other words, the greatest divisor of $m$ and $n$ is also the greatest divisor of $n$ and $r$. (Actually, there is one small step missing here - we must show that $S$ is not the empty set, so that it actually has a maximum. But we know for certain that $1$ divides any integer, so $1 \in S$, so $S$ is not empty.)
Now repeat the argument replacing $m$ and $n$ with $n$ and $r$. Since $r<n$ we are reducing the size of the numbers with each iteration. Eventually the process must finish at step (1) with $r=0$, and we have found the greatest divisor of the original $m$ and $n$.


TITLE: Diagonal $\Delta = \{x \times x : x \in X \}$ closed in $X \times X$ implies that $X$ is Hausdorff
QUESTION [3 upvotes]: I think I have solved a problem in Topology by Munkres, but there is a small detail that is bugging me. The problem is stated in this question's title. I will write down the proof and will highlight what is troubling me.
We prove by contradiction: Assume $X$ is not Hausdorff. Then there exist points $x,y$ where $x$ is different from $y$ such that no neighbourhoods $U$, $V$ about $x$ and $y$ respectively have trivial intersection. Now consider the point $(x,y)$ that is in the complement of $\Delta$. Now let $U \times V$ be any basis element that contains $(x,y)$ (such an element exists by definition of the product topology being generated by the basis $\mathcal{B}$ consisting of elements of the form $W \times Z$, where $W$ is open in $X$ and $Z$ is open in $Y$). Consider $(U \times V ) \cap \Delta$, which I claim to be $(U \cap X) \times (V \cap X)$.
By our choice of $x$ and $y$ there is $z \in U \cap V$, implying that the intersection $(U \times V ) \cap \Delta$ is not trivial.
Since $U \times V$ was any basis element containing $(x,y)$, this means that $(x,y) \in \overline{\Delta}$, which means that there exists a limit point of $\Delta$ that is not in it, contradicting $\Delta$ being closed.
The problem comes is in the way I have decomposed $\Delta$; the way I have put it seems I am saying that $\Delta$ is equal to $X \times X$, which is not the case. How  can I get round this?
Thanks.
Edit: Martin Sleziak has pointed out some mistakes, $(U \times V ) \cap \Delta$ should be $\{ (x,x) : x \in U \cap V\}$ and not as claimed.

REPLY [5 votes]: Your claim that $(U\times V)\cap \Delta=(U\cap X)\times(V\cap X)$ is incorrect. Since $U,V\subseteq X$, this is the same as claiming
$(U\times V)\cap \Delta=U\times V$.
If you use
$(U\times V)\cap \Delta = \{(x,x); x\in U\cap V\}$ instead, the rest of your proof should work fine, but there are still several minor details.

Minor nitpicks:

Wouldn't a direct proof (instead of using contradiction) more clear? I don't think you would have to change the proof much to do this. But perhaps this is a matter of taste.
If you're using proof by contradiction, you cannot choose $x$, $y$ arbitrarily, but you should choose $x,y\in X$, $x\ne y$, which witness that this space is not Hausdorff. (I.e., for any neighborhoods $U\ni x$, $V\ni y$ the intersection $U\cap V$ is non-empty.)

One more nitpick considering formatting:

It's not good to write two formulas after each other. If you write "$x,y$ $x \neq y$ such that..." this is quite difficult to read. You should separate such things at least with a little, e.g. "$x$, $y$; $x \neq y$ such that..."; in my opinion it is much better to separate them with text, e.g. "$x$, $y$ such that $x \neq y$ and..."

REPLY [4 votes]: A straightforward proof is a bit simpler:
If the diagonal is closed, then for $x\ne y$ in $X$, the point $(x,y)$ is not on the diagonal. So, there is an nhood of $(x,y)$ in $X\times X$ containing $(x,y)$ disjoint from the diagonal. But, then,  this nhood is of the form $U\times V$ where $U$ and $V$ are open disjoint sets containing $x$ and $y$ respectively.

Actually, contraposition would be best as you started in my opinion. If $X$ is not Hausdorf then there are $x$ and $y$ with $x\ne y$ such that for every open set $U$ containing $x$ and for every open set $V$ containing $y$ we have $U\cap V\ne\emptyset$ . Then, as you show, $(x,y)$ is in the closure of the diagonal. So, as $(x,y)$ is not on the diagonal but is in its closure, the diagonal is not closed.

The answer to your actual question is nicely explained in the other posts here. But I wanted to emphasise that an nhood $U\times V$ in $X\times X$ (which is as you describe) intersects the diagonal if and only if $U\cap V\ne\emptyset$. This follows from the relevant definitions is the key to the whole problem...

REPLY [3 votes]: The definition of Hausdorff is that for all distinct pairs $x,y \in X$ there exist disjoint open $U \ni x$ and $V \ni y$. Hence the negative is that there exist $x,y \in X$ each of whose open neighbourhoods $U \ni x$, $V \ni y$ have nonempty intersection. You seem to have said that this is the case for all $x,y \in X$, which isn't true.
Also your claim $(U \times V) \cap \Delta = (U \cap X) \times (V \cap X)$ seems to assume $\Delta = X \times X$, which could be the cause of your problems.

REPLY [3 votes]: You made the statement "Since $X$ is not Hausdorff, for any neighbourhoods $U$ and $V$ about $x,y$ respectively the intersection of $U$ and $V$ is not trivial." This actually has to only be true for a single $(x,y)$, not all $x$ and $y$.
It's probably best to not use proof by contradiction... If $\Delta$ is closed, then for any $(x,y) \notin \Delta$, there are open $U$ and $V$ such that $(x,y) \in U \times V$ and $U \times V$ is disjoint from $\Delta$. If you translate this correctly, you're done.


TITLE: If $\mathbb E_{\mathbb P} \vert f(X,Y)\vert <\infty$, is also $\mathbb E_{\mathbb P_1\times \mathbb P_2} \vert f(X,Y)\vert <\infty$?
QUESTION [5 upvotes]: Let $(E_1,\mathcal E_1)$ and $(E_2,\mathcal E_2)$ be two measurable spaces. Let $(E=E_1\times E_2,\mathcal E=\mathcal E_1\times \mathcal E_2)$. Let $\mathbb P$ a valid probability distribution on $(E,\mathcal E)$. Define the marginal distributions $\mathbb P_1(A)=\mathbb P (A\times E_2)$ and $\mathbb P_2(A)=\mathbb P (E_1\times A)$, and finally denote the product distribution by $\mathbb P_1\times \mathbb P_2$. Finally assume that we know
\begin{align*} \mathbb E_{\mathbb P} \vert f(X,Y)\vert <\infty.\end{align*}
My question is if it also holds that
\begin{align*} \mathbb E_{\mathbb P_1\times \mathbb P_2} \vert f(X,Y)\vert <\infty?\end{align*}
In particular, if it is false I hope to see a counterexample and I'd also be interested to know if it is true in the case $f(X,Y)=g(X)h(Y)$ for some functions $g$ and $h$.

REPLY [2 votes]: It is not true in general.
Let $E_1 = E_2 = \mathbb{R}$, and let $\mu$ be some probability measure on $\mathbb{R}$ with no first moment (e.g. Cauchy distribution).  Let $\mathbb{P}$ be the joint law of $(X, X)$ where $X \sim \mu$.  Finally let $f(x,y) = x-y$.  Under $\mathbb{P}$, we have $Y=X$ a.s., and therefore $f(X,Y)=0$ a.s., so $\mathbb{E}_{\mathbb{P}}|X-Y| = 0$.   But under $\mathbb{P}_1 \times \mathbb{P}_2$, we have $X,Y$ are iid $\mu$, and it follows that $\mathbb{E}_{\mathbb{P}_1 \times \mathbb{P}_2} |X-Y| = \infty$.  To see this, use Tonelli's theorem to write
$$\mathbb{E}_{\mathbb{P}_1 \times \mathbb{P}_2} = \int \mathbb{E}|X-y| \,\mathbb{P}_1(dy)$$ where the integrand is identically $+\infty$, using $|X-y| \ge |X|-|y|$.
It is not even true for $f(x,y) = g(x) h(y)$.  Let $W$ be a random variable with infinite first moment, and $Z$ a Bernoulli(1/2) random variable independent from $W$.  Let $\mathbb{P}$ be the joint law of $(WZ, W(1-Z))$, and set $f(x,y) = xy$, which is of the form $g(x) h(y)$.  Then
$$\mathbb{E}_{\mathbb{P}}|XY| = \mathbb{E} |W^2 Z (1-Z)| = 0$$ because $Z(1-Z)=0$ a.s.  But $WZ, W(1-Z)$ are identically distributed, so we have
$$\mathbb{E}_{\mathbb{P}_1 \times \mathbb{P}_2}|XY| = (\mathbb{E}|WZ|)^2 = (\mathbb{E}|W| \cdot \mathbb{E}|Z|)^2 = \frac{1}{4} (\mathbb{E}|W|)^2 = \infty.$$


TITLE: Perpendicular Projection Linear Transformations
QUESTION [0 upvotes]: If a linear transformation T has matrix A.
If T is a perpendicular projection onto the line y = -5x, then A has:
eigenvector [1, -5] (this is meant to be 2 rows, 1 column) with eigenvalue 1
how would I work out the second eigenvector and eigenvalue?
I thought that because it is a projection, the second eigenvalue would be 0, but does that mean the eigenvector is [0, 0]?

REPLY [1 votes]: Hint. You're correct about the first eigenvalue and eigenvector, as well as that the second eigenvalue is $0$. Now to find the eigenvector corresponding to the eigenvalue $\lambda=0$, we must find a nonzero vector $v$ such that $Av=\lambda v$, in other words $Av=0$. Thus we just need to find a nonzero vector that is taken to $0$ under the linear transformation. In alternative language, we want to find a vector that spans the nullspace of $A$. Can you continue?


TITLE: Is it true that for any square matrix $A$, if $A^4=\text{Id}$ then $A^2=-\text{Id}$ or $A^2=\text{Id}$?
QUESTION [1 upvotes]: I have tried some $A^2$ is not $-\text{Id}$ or $\text{Id}$ like $\pmatrix{1 & 0\\0 & -1}$, but I cannot find the corresponding matrix of $A$ in these cases.

REPLY [9 votes]: Permutation matrix to the rescue! We take the permutation $1 \mapsto 2 \mapsto 3 \mapsto 4 \mapsto 1$ and set entry $$a_{ij} = \begin{cases}1, &i\mapsto j\ \\ 0, &\text{otherwise}\end{cases}$$ to get the matrix
$$A = \begin{pmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\1&0&0&0\\\end{pmatrix}.$$
You can verify on WolframAlpha.


TITLE: If $f=g$ a.e. and $f$ and $g$ are continuous, then $f=g$ every where.
QUESTION [1 upvotes]: Let $f,g:\mathbb R\longrightarrow \mathbb R$ two continuous functions. Is it true that if $f=g$ a.e. then $f=g$ every where ?
I would say yes. My argument would be the following one : Let $$\Omega =\{x\mid f(x)\neq g(x)\}.$$
So let $x\in \Omega $. Is it true that there is a sequence in $\mathbb R\backslash \Omega $ that converge to $x$ ? If yes, the claim is proved. But if yes, How can I prove this claim ? If no, how can I do ?
Does this result can be adapted to an abstracted measurable space $(X,\mu)$ ?

REPLY [8 votes]: Since $f$ and $g$ are continuous, $\Omega$ is open, if it is not empty, its measure is not zero.


TITLE: Group homomorphism $\mathbb{Z}_p \longrightarrow\mathbb{F}_p$
QUESTION [0 upvotes]: Let $p$ be a prime number, denote by $\mathbb{Z}_p$ the ring of $p-$adic integers.
Can we define a group homomorphism $\mathbb{Z}_p \longrightarrow\mathbb{F}_p$?

REPLY [2 votes]: There is a homomorphism, called the reduction modulo $p$, given by
$$
\sum_{i\ge 0}a_ip^i\mapsto a_0 \mod p.
$$
This is a standard definition, which is used among other things to show that the invertible elements in
the ring $\mathbb{Z}_p$ consist of $\mathbb{Z}_p^{\times}=\{\sum_{i\ge 0}a_ip^i \mid a_0\neq 0\}$.


text:
\section{Generalized Lyubeznik numbers} \label{generalized}

The \emph{generalized Lyubeznik numbers} are a family of invariants associated to a local ring containing a field, which contains the Lyubeznik numbers.  These invariants can capture more subtle information about the ring than the (original) Lyubeznik numbers can.
Moreover, they can encode information about $F$-singularities in characteristic $p>0$, and have connections to $D$-module characteristic cycle multiplicities in characteristic zero.
Throughout our discussion, we use \cite{NuWiEqual} as our reference.
The generalized Lyubeznik numbers are defined as follows.

\begin{ThmDef}[Generalized Lyubeznik numbers] \label{GenDef}
Let $(R, \fm, K)$ be a local ring containing a field  and let $\widehat{R}$ be its completion at $\fm$.
Given a coefficient field  $L$ of $\widehat{R}$, there exists a surjection $\pi: S \to \widehat{R}$, where $S = K \llbracket x_1, \ldots, x_n \rrbracket $ for some $n \geq 1$, and such that $\pi(K) = L$.
For $1 \leq j \leq \ell$, fix  $i_j \in \NN$ and ideals $I_j \subseteq R$, and let $J_j = \pi^{-1} \left( I_j \widehat{R} \right) \subseteq S$.
The \emph{generalized Lyubeznik number of $R$ with respect to $L$, the $I_j$, and the $i_j$} is defined as
 $$\lambda^{i_\ell,\ldots, i_1}_{I_\ell,\ldots, I_1}(R;L):=\Length_{D(S,K)} H^{i_\ell}_{J_\ell}  \cdots  H^{i_2}_{J_2}H^{n-i_1}_{J_1}(S).$$
This number is finite and depends only on $R,$ $L$, the $I_j$, and the $i_j$ (\emph{i.e.}, it is independent of the choice of $S$ and of $\pi$).

When $\widehat{R}$ has a unique coefficient field (\emph{e.g}., when $K$ is a perfect field of characteristic $p>0$), or when the choice of $L$ is clear in the context, this invariant is denoted $\lambda^{i_\ell,\ldots, i_1}_{I_\ell,\ldots, I_1}(R)$.
\end{ThmDef}

In  the definition, we may take $I_1\subseteq \ldots \subseteq I_\ell$ without losing information:   For any $R$-module $M$ satisfying $H^0_I(M)=M$ for some ideal $I$ of $R$, then for every ideal $J$ of $S$, $H^i_{J}(M)=H^i_{I+J}(M)$.
Moreover, $\lambda^{i_\ell,\ldots, i_1}_{I_\ell,\ldots, I_1}(R, L)=\lambda^{i_\ell,\ldots, i_1}_{I_\ell,\ldots,I_2, 0}(R/I_1, L).$

We note that \`Alvarez Montaner has also given a generalization of the Lyubeznik numbers in the characteristic zero setting in \cite{AM-NumInv}; his invariants are defined in terms of $D$-module characteristic cycles.

{\edit{

In general,  to avoid the dependence on the choice of coefficient field of $\widehat{R}$ in the definition of generalized Lyubeznik numbers, one would need to answer the following question, asked by Lyubeznik.

\begin{question}[\cite{LyuSurveyLC}] \label{LyuQuestion}
Let $S$ be a complete regular local ring of equal characteristic.  For $1 \leq j \leq \ell$, fix $i_j \in \NN$ and ideals $J_j \subseteq S$.
Given any two coefficient fields $K$ and $L$ of $S$, is
$$\Length_{D(S,K)} H^{i_\ell}_{J_\ell}  \cdots
H^{i_2}_{J_2}H^{n-i_1}_{J_1}(S)=
\Length_{D(S,L)} H^{i_\ell}_{J_\ell}  \cdots
H^{i_2}_{J_2}H^{n-i_1}_{J_1}(S) \text{?}$$
\end{question}
\noindent To the best of our knowledge, this question is open even in the case that $s=1$.
}

As their nomenclature indicates, the generalized Lyubeznik numbers are, in fact, generalizations of the Lyubeznik numbers, as the following proposition makes explicit.
\begin{proposition}
If $(R, \fm, K)$ is a local ring containing a field, then $\lambda_{i,j} (R)=\lambda^{i,j}_{\fm,0} (R;L)$
for any coefficient field $L$ of $\widehat{R}.$
\end{proposition}

\begin{proof}[Sketch of proof]  Since the Bass numbers with respect to the maximal ideal are not affected by completion, $R$ is, without loss of generality, complete.

Consider a surjection $\pi: S:= K \llbracket x_1, \ldots, x_n \rrbracket  \surj R$, where $\pi(K)=L$.  If $\fn$ denotes the maximal ideal of $S$, then $H^i_{\fn} H^{n-j}_{I} (S)$ is injective by \cite[Corollary 3.6]{LyubeznikFinitenessLocalCohomologyModules}.
Thus, by  \cite[Lemma 1.4]{LyubeznikFinitenessLocalCohomologyModules}, $\dim_K \Hom(K, H^i_{\fn} H^{n-j}_I(S) ) = \dim_K \Ext^i_S(K, H^{n-j}_I(S))$, which, in turn, equals $\lambda_{i,j}(R).$  As $H^i_{\fn} H^{n-j}_{I} (S)$ is supported at $\fm$, it is isomorphic to $E_S(K)^{\oplus \ell}\cong E_S(L)^{\oplus \ell}$ for some $\ell\in\NN$.  Since $E_S(L)$ is a simple $D(S,L)$-module, we have that $\dim_K \Hom(K, H^i_{\fn} H^{n-j}_I(S) ) = \ell = \Length_{D(S,L)} H^i_{\fn} H^{n-j}_I(S) = \lambda_{\fm, 0}^{i,j}(R; L).$
\end{proof}

The proof that the generalized Lyubeznik numbers are well defined critically uses the key functor $G$ defined in Subsection \ref{Key Functor}; we now sketch this proof.

\begin{proof}[Sketch of proof of Theorem/Definition \ref{GenDef}]
{\edit{
First, note that the $D(S,K)$-module length  of the iterated local cohomology modules is finite by \cite{BjorkRingsDifferentialOperators,LyubeznikFinitenessLocalCohomologyModules,LyubeznikFModulesApplicationsToLocalCohomology}.
}}
Again, without loss of generality, we may assume that $R$ is complete.
Along with the surjection $\pi: S=K \llbracket x_1, \ldots, x_n \rrbracket  \surj R$,
fix $S' = K \llbracket y_1, \ldots, y_{n'} \rrbracket$ and
take a surjection $\pi' : S' \surj R$ satisfying $\pi'(K) = L$.
Set $I' = \Ker(\pi')$, and let $\fm'$ be the maximal ideal of $S'$.
 Let ${J_{j}}' = {(\pi')}^{-1}(I_j)$ for $1 \leq j \leq s$.

Let $S''=K \llbracket x_1,\ldots,x_n, y_1, \ldots, y_{n'} \rrbracket $ and
let $\pi''$ denote the surjection $\pi'': S''\surj R$ defined by
\begin{alignat*}{3}
&x_i \mapsto \pi(x_i) & \ \ & \text{for } 1\leq i\leq n, \text{ and} \\
&y_j \mapsto \pi'(y_j) & & \text{for } 1\leq j\leq n'.
\end{alignat*}
Let ${J_j}'' = (\pi'')^{-1}(I_j)$ for each $1 \leq j \leq s$.

Since $\pi$ is surjective, there exist $\sigma_j \in S$ such that $\pi(\sigma_j) = \pi''(y_j)$ for each $1 \leq j \leq n'$.
Let $\varphi: S'' \to S$ denote the map defined by $\varphi(x_i) = x_i$ for $1 \leq i \leq n$ and $\varphi(y_j) = \sigma_j$ for $1 \leq j \leq n'$, so that
$\varphi$ splits the inclusion $S \inj S''$.
Since for $1 \leq j \leq n'$, $y_1 - \sigma_1 \in \Ker \varphi$, and the map $S''/(y_1 - \sigma_1, \ldots, y_{n'}-\sigma_{n'}) \to S$ induced by $\varphi$ is an isomorphism, $\Ker(\varphi) = (y_1 - \sigma_1, \ldots, y_{n'}-\sigma_{n'}).$
Thus, ${J_j}'' = J_j + (y_1 - \sigma_1, \ldots, y_{n'}-\sigma_{n'})$.  Moreover, the elements $x_1, \ldots, x_n, y_1 - \sigma_1, \ldots, y_{n'}-\sigma_{n'}$ form a regular system of parameters for $S''$, by Properties \ref{FunctorProperties} {\edit{($6$) and ($2'$)}}

of the functor $G$ defined in Subsection \ref{Key Functor},
we obtain that
\[  \Length_{D(S,{K})} H^{i_\ell}_{J_\ell}  \cdots  H^{i_2}_{J_2}  H^{n-i_1}_{J_1}(S) = \Length_{D(S'',{K})} H^{i_\ell}_{J_\ell''}   \cdots H^{i_2}_{J_2''} H^{n'+n-i_1}_{J_1''}(S''),
\] which, by an analogous argument, equals $\Length_{D(S',{K})} H^{i_\ell}_{J_\ell'}  \cdots  H^{i_2}_{J_2'}  H^{n'-i_1}_{J_1'}(S')$ as well.
\end{proof}

Some vanishing properties of the generalized Lyubeznik numbers are as follows; \emph{cf.} Properties \ref{FirstProp}.

\begin{properties}
Given any local ring $(R, \fm, K)$ containing a field, ideals $I_1 \subseteq \ldots, \subseteq I_\ell$ of $R$, $i_j \in \NN$ for $1 \leq j \leq \ell$, and a coefficient field $L$ of $\widehat{R}$, the following hold.
\begin{enumerate}
\item $\lambda^{i_1}_{I_1}(R; L)\neq 0$ if $i_1=\dim(R/I_1)$.
\item  $\lambda^{i_2,i_1}_{I_2,I_1}(R; L)=0$ if $i_2>i_1$, and is nonzero if $i_1=\dim(R/I_1)$ and $i_2=\dim(R/I_1)-\dim(R/I_2)$.
\item  $\lambda^{i_\ell,\ldots, i_1}_{I_\ell,\ldots, I_1}(R; L)=0$ if either $i_1>\dim(R/I_1)$, or if $i_j>\dim(R/I_{j-1})$ and $2 \leq j\leq \ell$.
\end{enumerate}
\end{properties}

The generalized Lyubeznik numbers behave in the following way under finite field extensions:
if $K\subseteq K'$ is a finite field extension, $R=K \llbracket x_1,\ldots,x_n \rrbracket ,$ and $R'=K' \llbracket x_1,\ldots,x_n \rrbracket ,$ then for all ideals $I_j$ of $R$ and $i_j \in \NN$, $1 \leq j \leq \ell$, $\lambda^{i_\ell,\ldots, i_1}_{I_\ell,\ldots, I_1}(R)= \lambda^{i_\ell,\ldots, i_1}_{I_\ell R',\ldots, I_1 R'}(R').$

Although the Lyubeznik numbers cannot distinguish between complete intersection rings (see Properties \ref{FirstProp}), or between one-dimensional rings, the generalized Lyubeznik numbers can, as the following properties show.

\begin{properties} \label{exsGen} The following hold.
\begin{enumerate}
\item Fix a complete local ring $(R, \fm, K)$ of dimension one and a coefficient field $L$ of $R$, and let $P_1,\ldots P_t$ denote the minimal primes of $R$.  Then
$\lambda^{1}_{0} (R; L)=\lambda^{1}_{0} (R/P_1; L)+\ldots +\lambda^{1}_{0} (R/P_t ; L)+t-1.$
\item If $K$ is a field and $S=K \llbracket x_1,\ldots, x_n \rrbracket $,
let $f=f^{\alpha_1}_1\cdots f^{\alpha_t}_t$, where $f_1,\ldots ,f_t\in S$ are irreducible and each $\alpha_i \in \NN$. Then
$\lambda^{n-1}_{0} (S/f)\geq\lambda^{n-1}_{0} (S/f_1)+\ldots +\lambda^{n-1}_{0} (S/f_t)+t-1.$
\end{enumerate}
\end{properties}

Proposition \ref{exsGen} indicates that the generalized Lyubeznik numbers of the form $\lambda_0^j(R; L)$ are of specific interest.  One way they can be used is to define the following invariant of a local ring containing a field.

\begin{definition} [Lyubeznik characteristic]
Let $R$ be a $d$-dimensional local ring containing a field, and fix  a coefficient field $L$ of $\widehat{R}$.  The \emph{Lyubeznik characteristic of $R$ with respect to $L$} is defined as \[ \chi_{\lambda}(R; L) := \sum_{i=0}^d (-1)^i \lambda_{0}^i(R; L).\]

\end{definition}

As a consequence of the Mayer-Vietoris sequence for local cohomology (see Remark \ref{MVSequence}),

given ideals $I$ and $J$ of a local ring $R$ containing a field,
$\chi_{\lambda}(R/I; L) + \chi_{\lambda}(R/J;  L) = \chi_{\lambda}(R/(I+J); L) + \chi_{\lambda}(R/I\cap J; L)$
for any coefficient field $L$ of $\widehat{R}$.

Note that the Lyubeznik characteristic of a Stanley-Reinser ring is computed in Theorem \ref{LyuCharSR}.

\edit{
We now discuss some connections of certain generalized Lyubeznik numbers with singularities in  positive characteristic.
Suppose that $R$ is a ring of characteristic $p>0$.
In this case, we have the Frobenius endomorphism $F:R\to R$ defined by $F(r)= r^p$, as discussed in Section \ref{ProjSchemesSec}.
A variety of types of singularities can be defined via Frobenius (\emph{e.g.}, $F$-\emph{injective}, $F$-\emph{pure}, and $F$-\emph{regular} singularities).

If $R$ is reduced, let $R^{1/p^e}=\{r^{1/p^e}\mid r\in R\}$, the ring obtained by adjoining the $p^e$-th roots of elements of $R$.
We say that $R$ is \emph{$F$-finite} if $R^{1/p}$ is a finitely generated $R$-module.
Throughout this discussion, we assume that $R$ is reduced and $F$-finite.

We now recall some basic definitions from the theory of tight closure developed by Hochster and Huneke \cite{HochsterHunekeTC1,HochsterHunekeTightClosureAndStrongFRegularity};
we also refer to \cite{HunekeTightClosureBook} for a reference.
If $I$ is an ideal of $R$, the \emph{tight closure of $I$}, $I^*$, is the ideal of $R$
consisting of all elements $z\in R$ for which there exists some $c \in R$
that is not in any minimal prime of $R,$ such that
$$
cz^q \in I^{[q]} \hbox{ for all }q = p^e \gg 0.
$$

An ideal $I$ is \emph{tightly closed} if $I^* = I$.
A ring  is called  \emph{$F$-rational} if every parameter ideal is tightly closed, and is called \emph{weakly  $F$-regular} if every ideal of the ring is tightly closed.
A ring is called \emph{$F$-regular} if every localization of the ring is weakly $F$-regular.
It is not known whether every weakly $F$-regular ring is $F$-regular.
We point out that tight closure does not commute with localization in general \cite{Brenner-Monsky}.

The \emph{test ideal} of $R$ is defined by
$$
\tau(R)= \bigcap_{\substack{J \subseteq  R \text{ ideal} }} (J:J^*).
$$
The test ideal of $\tau(R)$ is an important object in understanding how ``far" a ring is from being weakly $F$-regular. In particular,
$\tau(R)=R$ if and only if $R$ is weakly $F$-regular.

If the $R$-module inclusion $R\hookrightarrow R^{1/p}$ splits, $R$ is called \emph{$F$-pure}. The $F$-purity of a ring simplifies computations for cohomology groups and implies vanishing properties of these groups \cite{HochsterRobertsFrobeniusLocalCohomology,LyubeznikCD}.
If for every $c\in R$ not contained in any minimal prime of $R$, there exists $e\geq 1$ such that the $R$-module map $R \to R^{1/p^e}$ defined by $1\mapsto c^{1/p^e}$ splits, then $R$ is called \emph{strongly $F$-regular} .
A local ring $(R,\mathfrak{m},K)$ is \emph{$F$-injective} if the action of Frobenius on the local cohomology module $H^i_{\mathfrak{m}}(R)$ is injective for every $i\in\NN$.
The relations among these properties are the following:
$$

\begin{Bmatrix}
F\hbox{-injective}\\
\hbox{rings}\\
\end{Bmatrix}
\supseteq
\begin{Bmatrix}
F\hbox{-pure}\\
\hbox{rings}\\
\end{Bmatrix}
\supseteq
\begin{Bmatrix}
\hbox{weakly}\\
F\hbox{-regular}\\
\hbox{rings}\\
\end{Bmatrix}
\supseteq
\begin{Bmatrix}
F\hbox{-regular}\\
\hbox{rings}\\
\end{Bmatrix}
\supseteq
\begin{Bmatrix}
\hbox{strongly}\\
F\hbox{-regular}\\
\hbox{rings}\\
\end{Bmatrix}
\supseteq
\begin{Bmatrix}
\hbox{regular}\\
\hbox{rings}\\
\end{Bmatrix}.
$$
}

The generalized Lyubeznik numbers of the form $\lambda_0^j(R)$ can
provide information on singularities in characteristic $p>0$.
In particular, results of Blickle indicate that they encode information on $F$-regularity and $F$-rationality.
\edit{ We present these results, as done in \cite{NuWiEqual}}.

\edit{
\begin{theorem} \label{BliThm} \cite{BlickleIntHomology}
Suppose that $(R, \fm, K)$ is a $d$-dimensional complete local domain of characteristic $p>0$.
If $R$ is $F$-injective and $\lambda_0^d(R;L)=1$, then $R$ is $F$-rational.
Moreover, if $K$ is perfect and $R$ is $F$-rational, then $\lambda_0^d(R)=1$.

\end{theorem}
}

{\edit{
\begin{example} Suppose that $K$ is a field, and let $R = K[X]$ be the polynomial ring over $K$ in the entries of an $r \times r$ square matrix $X$ of indeterminates, $r > 1$.
Let $\fm$ denote the homogeneous maximal ideal of $R$, and let $\det X$ denote the determinant of $X$.
When $K$ is a perfect field of characteristic $p>0$, $R_\fm/(\det X) R_\fm$ is strongly $F$-regular, so also $F$-rational, and $\lambda_0^{r^2-1} \left (R_\fm/(\det X) R_\fm\right)=1$ \cite{HochsterHunekeTightClosureOfParameterIdealsAndSplitting}.
However, when $K$ has characteristic zero, we have that  $\lambda_0^{r^2-1} \left(R_\fm/(\det X) R_\fm\right)\geq 2$ \cite{WaltherBS}.

\end{example}
}

Work in \cite{NunezPerezFJacobian} indicates that the generalized Lyubeznik numbers can measure how ``far" an $F$-pure hypersurface ring is from being $F$-regular.  Take an element $f$ of an $F$-finite regular local ring $R$ of characteristic $p>0$ such that $R/(f)$ is $F$-pure.
Let $\tau_1$ be the pullback of the test ideal of $R/(f)$ to $R$, and for $i>1$, and let $\tau_i$ denote the pullback of the test ideal of $R/\tau_{i-1}$ to $R.$ Vassilev showed that the chain $0 \subsetneq \tau_1 \subsetneq \tau_2 \subsetneq \cdots \subsetneq \tau_s = R$ is finite and each $R/\tau_i$ is $F$-pure \cite{Vassilev}.  In fact, $\lambda_0^{\dim\left(R/(f) \right)} \left(R/(f); L\right) \geq s$ for any coefficient field $L$ of $\widehat{R/(f)}$.
In particular, if $\lambda_0^{\dim\left(R/(f) \right)} \left(R/(f); L\right) =1$, then $R/(f)$ is $F$-regular,
which recovers the result of Blickle in this case (\emph{cf.} Theorem \ref{BliThm}).

{\edit{
\begin{question}
What further geometric and algebraic properties are captured by the generalized Lyubeznik numbers of a local ring
containing a field? In particular, do these numbers measure
singularities in characteristic zero?
\end{question}
}}

config: arxiv


TITLE: Reference request: English translation of Brieskorn 1970 paper
QUESTION [3 upvotes]: Is there any english (or french) translation of the following paper by Brieskorn (1970)?
Brieskorn, E., "Die Monodromie der Isolierten Singularitäten von Hyperflächen", Manuscripta Mathematica 2 (1970).
Thank you.

REPLY [3 votes]: Zentralblatt lists a translation in Russian [published in Matematika, Moskva 15, No.4, 130-160 (1971)], but that's all. I also checked the Brieskorn Anniversary Volume, which lists his complete works, and again only the Russian translation appears.

As suggested by @aginensky, here is how Google translates the introduction of the Brieskorn paper from German into English. I give the translation without any post processing, directly as it comes out of the Google translator. It seems like a viable road.

John Milnor has shown in [29] that can be topologically define a local
  Picard-Lefschetz monodromy for singularities of hypersurfaces and
  isolated that this monodromy up to a certain extent determines the
  topology of the singularity. On the other hand there is the
  Picard-Lefschetz monodromy of families of singular algebraic manifolds
  investigations of A. Borel, C.H. Clemens [5] Ph.A. Griffiths [12],
  [13], [14], A. Grothendieck [16], N.M. Katz [20], [21], A. Landman
  [22] and F. Pham [31], which the classical results of Picard [32] and
  Lefschetz [24] generalize. In particular, by Grothendieck and others
  for this purpose an algebraic theory was developed, the theory of
  Gauss-Manin connection. While this theory, however, families of
  manifolds without singularities presupposes one-parameter families of
  manifolds with singularities are examined in this paper, it is
  introduced a singular local Gauass-Manin connectivity for isolated
  singularities of hypersurfaces. This provides a purely algebraic
  calculation of the originally defined topologically local
  Picard-Lefschetz monodromy. The restriction to the special case of
  isolated singularities of hypersurfaces the theory is very simple and
  explicit.

and here is the French translation

John Milnor a montré dans [29] qui peut être définie topologiquement
  un monodromie locale Picard-Lefschetz de singularités des
  hypersurfaces isolé et que cette monodromie jusqu'à une certaine
  mesure détermine la topologie de la singularité. D'autre part, il ya
  la monodromie Picard-Lefschetz des familles de singulières collecteurs
  algébriques enquêtes de A. Borel, C.H. Clemens [5] Ph.A. Griffiths
  [12], [13], [14], A. Grothendieck [16], N.M. Katz [20], [21], A.
  Landman [22] et F. Pham [31], dont les résultats classiques de Picard
  [32] et Lefschetz [24] généraliser. En particulier, par Grothendieck
  et d'autres à cet effet une théorie algébrique a été développé, la
  théorie de la connexion de Gauss-Manin. Bien que cette théorie,
  cependant, les familles des collecteurs sans singularités suppose
  familles à un paramètre de collecteurs avec singularités sont
  examinées dans le présent document, il est introduit une connectivité
  locale singulière Gauass-Manin pour les singularités isolées de
  hypersurfaces. Cela fournit un calcul purement algébrique de la
  monodromie Picard-Lefschetz topologiquement locale défini à l'origine.
  La restriction au cas particulier des singularités isolées de
  hypersurfaces la théorie est très simple et explicite.


TITLE: Galois Group of $x^4 - x^2 - 3$
QUESTION [0 upvotes]: Find the Galois Group of $x^4 - x^2 - 3$

This is a qual question. I don't know how to find the splitting field of this polynomial.

REPLY [0 votes]: Considering $f(x) = x^4 - x^2 - 3 $ we must finds its roots. Well if you take $y = x^2$ it comes that
$$0=y^2 - y - 3  = \left(y - \frac{1}{2}\right)^2 - \frac{1}{4} - 3 = \left(y - \frac{1}{2}\right)^2 - \frac{13}{4} \implies y = \frac{1}{2} \pm \frac{\sqrt {13}}{2} $$
and from $x^2 = y$ you find all $4$ roots of $f$.


TITLE: eccentricity of locus of hyperbola
QUESTION [0 upvotes]: The locus of intersection of the lines $\sqrt 3 x-y-4\sqrt3 t=0$ and $\sqrt 3 tx+ty-4\sqrt 3=0$(where t is a parameter) is a hyperbola . we have to find its eccentricity .
I got the intersection point as  $x=(2+2t^2)/(t)$ and $y=(2\sqrt 3)/(2\sqrt 3 t^2 +t)$.
after that I got stuck .

REPLY [1 votes]: The family of lines is
$$\begin{cases}&I\;\;\sqrt3\,x-y=4\sqrt3\,t\,/\,\cdot(-t)\implies-\sqrt3\,tx+ty=-4\sqrt3t^2\\{}\\
&II\;\;\sqrt3\,tx+ty=4\sqrt3\end{cases}\stackrel{\text{sum up}\,-tI+II}\implies$$
$$\;2ty=4\sqrt3(1-t^2)\implies \color{red}{y=\frac{2\sqrt3(1-t^2)}t}\stackrel{\text{substitute in I}}\implies$$
$$\sqrt3\,x=4\sqrt3\,t+\frac{2\sqrt3(1-t^2)}t\implies$$
$$ \color{red}{x=4t+\frac{2(1-t^2)}t=\frac{2(t^2+1)}t}$$
so
$$\begin{cases}&x^2=\cfrac{4(t^2+1)^2}{t^2}\\{}\\
&y^2=\cfrac{12(t^2-1)^2}{t^2}\end{cases}\implies 3x^2-y^2=48\implies\cfrac{x^2}{4^2}-\cfrac{y^2}{(4\sqrt3)^2}=1$$
and thus the eccentricity is
$$e=\frac{\sqrt{a^2+b^2}}a=\frac{\sqrt{16+48}}{4}=\frac{\sqrt{64}}{4}=2$$


TITLE: Why does electric force depend on the medium?
QUESTION [14 upvotes]: Comparing the equations for the gravitational and electric force
$$\vec{F}_g=-\frac{Gm_1m_2}{r^2}$$
and
$$\vec{F}_e=\frac{Kq_1q_2}{r^2}$$
I noticed the only major difference between them is that the constant $K$ depends on the medium. Why does it?

REPLY [10 votes]: In electromagnetism there are both positive and negative charges. Hence the force due to electric charges can be attractive or repulsive. Gravity, when treated as a classical force field, can only be attractive, there are not two types of "gravitational charge".
What this means is that in electromagnetism, a given medium, may contain both positive and negative charges and these can be separated by the application of an electric field - a process called polarisation. The separation of the electric charges produces an electric field that when summed with the applied electric field leads to a different net electric field in the medium and thus a different force acting on a test charge in the medium.
Since there is only one type of "gravitational charge", this phenomenon does not occur when applying a gravitational field.


TITLE: Question about counting cards
QUESTION [0 upvotes]: A standard deck of cards contains $52$ cards divided into four suits:
the red suits, hearts and diamonds, and the black suits, clubs and spades. Each suit, in turn, is divided in 13 ordered ranks: ace through 10 followed by the face cards jack, queen, and king. The ace can act as either the lowest or highest card in the ranking.

What is the number of $5$-card hands in a $52$-card deck that contain two pairs (i.e., two pairs from different ranks, and a fifth card of a third rank)?

Answer: $\binom {13}{2}{\binom 42}^2 \binom {44}{1}$
Why is $\binom {13}{2}$ a factor in the answer?

REPLY [3 votes]: First you choose the two different ranks (one for each pair) so you get :
$$\begin{pmatrix}13\\2\end{pmatrix}\text{ choices.} $$
For each one of these choices you must choose two different suits for the first rank :
$$\begin{pmatrix}4\\2\end{pmatrix}\text{ choices.} $$
you must choose two different suits for the second rank :
$$\begin{pmatrix}4\\2\end{pmatrix}\text{ choices.} $$
And you must choose a last card which is among the remaining ranks :
$$(13-2)\times 4=44\text{ choices. } $$


TITLE: Special Relativity Explanation
QUESTION [1 upvotes]: I'm in a relatively basic Physics class in university, and was reading the section in my textbook about The Principle of Relativity.
It says:

Einstein's Theory of Relativity ... added the conjecture that the speed of a beam of light must be the same as measured by observers in different frames of reference in uniform motion with respect to each other.

I'd like to start off and say I certainly don't doubt that. People much more familiar with relativistic physics take this as fact so I see no reason to pretend I've found a loophole in the fabric of reality.
I'm just wondering... how? How can I be moving at 0.9c relative to the Cosmic Microwave Background (which my textbook says is the most reliable inertial reference frame) and watch light go past me, in the same direction, at 1.0c? Doesn't that mean the light is moving at 1.9c relative to the CMB?
Further, the textbook says that, in the case of two or more observers traveling at different speeds, the observers will see that the light has a different color. Is that evidence of a Doppler effect? Doesn't that imply that to see different colors observers would have to be observing the light moving at different velocities than each other?
I've just been puzzling about this for a while now. I'm hoping someone more learned than me can chime in and help clear things up, because so far the best explanation I can think of is localized time dilation changing just how long a "second" is in order to ensure that the photons always move at c, but that just seems like such a convenient answer.

REPLY [2 votes]: You're confused because you're holding two mutually contradicory postulates in mind. One is the one you mentioned that the speed of light is the same in every inertial reference frame. The other which you implicitly point out, is that the relation between speed measurements in two different inertial reference frames moving at relative speed v with respect to one another is given by adding or subtracting that speed.
Well, if both can not be right, and you are prepared to accept the first postulate, the second has to be wrong. And it indeed is.
What you're looking for is the Lorentz transformation.


TITLE: Find the closed-form for $\sum_{i=0}^n(-1)^i(\frac{1}{2})^i$
QUESTION [3 upvotes]: I start with simplifying:
$$\sum_{i=0}^n(-1)^i(\frac{1}{2})^i=\sum_{i=0}^n(-\frac{1}{2})^i$$
then:
$$S = 1 + (-\frac{1}{2}) + (-\frac{1}{2})^2 + ... +(-\frac{1}{2})^n$$
$$(-\frac{1}{2})S = (-\frac{1}{2}) + (-\frac{1}{2})^2 + ... +(-\frac{1}{2})^n+(-\frac{1}{2})^{n+1}$$
$$(-\frac{1}{2}-1)S = (-\frac{1}{2})^{n+1} - 1$$
am I on the right track?

REPLY [0 votes]: Here's another approach using the formula for finite geometric series
\begin{align*}
  \sum_{i=0}^nx^n=\frac{1-x^{n+1}}{1-x}\tag{1}
  \end{align*}

We obtain
  \begin{align*}
  \sum_{i=0}^n(-1)^i\left(\frac{1}{2}\right)^i&=\sum_{i=0}^n\left(-\frac{1}{2}\right)^i\\
&=\frac{1-\left(-\frac{1}{2}\right)^{n+1}}{1-\left(-\frac{1}{2}\right)}\\
&=\frac{2}{3}\left(1-\left(-\frac{1}{2}\right)^{n+1}\right)
  \end{align*}


text:
\begin{document}
\maketitle
\centerline{Columbia University

}

\begin{abstract}
Ramanujan recorded four reciprocity formulas for the Roger-Ramanujan continued fractions.  Two reciprocity formulas each are also associated with the Ramanujan--G\"ollnitz--Gordon continued fractions  and a  level-13 analog of the Roger-Ramanujan continued fractions. We show that all eight reciprocity formulas are related to a pair of quadratic equations. The solution to the first equation generalizes the golden ratio and is used to set the value of a coefficient in the second equation; and the solution to the second equation gives a pair of values for a continued fraction. We relate the coefficients of the quadratic equations to important formulas obtained by Ramanujan, examine a pattern in the relation between a continued fraction and its parameters, and use the reciprocity formulas to obtain close approximations for all values of the continued fractions. We highlight patterns in the expressions for certain explicit values of the Rogers-Ramanujan continued fractions by expressing them in terms of the golden ratio. We extend the analysis to reciprocity formulas for Ramanujan's cubic continued fraction and the Ramanujan-Selberg continued fraction.

\keywords{Roger-Ramanujan continued fractions \and  Ramanujan--G\"ollnitz--Gordon continued fractions \and Ramanujan's cubic continued fractions \and Ramanujan-Selberg continued fraction \and Reciprocal formulas}

\end{abstract}

\section{Introduction}
Ramanujan (\cite{rln}, \cite{Ram}) recorded four reciprocity formulas for the Rogers-Ramanujan continued fractions. Two other reciprocity formulas are associated with the Ramanujan--G\"ollnitz--Gordon continued fractions (Bagis \cite{Bag}), and another two with a level-13 analog of the continued fractions (Cooper and Ye \cite{CY}). We show that these eight formulas are obtained by choosing different values of a single parameter, $k$, in the quadratic equation $c^2-kc-1=0$, and then substituting the value of its root $c=(k+\sqrt{k^2+4})/2$ into the equation $(1-x_1x_2)/(x_1+x_2)=c$. The eight parameter values are  $k=\pm 1, \pm 2, \pm 3$ and $\pm 11$. Table \ref{tab1} shows the terms associated with $x_1$ and $x_2$ in a reciprocity formula for each value of $k$. In the table, $R(q)$ and $S(q)$ denote the Rogers-Ramanujan continued fractions; $V(q)$ and $-V(-q)$ denote the Ramanujan--G\"ollnitz--Gordon continued fractions; and $R'(q)$ and $S'(q)$ denote the level-13 analogs of the Rogers-Ramanujan continued fractions. The values of $c$ in the table are $\phi=(\sqrt{5}+1)/2$, $1/\phi=(\sqrt{5}-1)/2$, $\phi^5=(5\sqrt{5}+11)/2$, $1/\phi^5=(5\sqrt{5}-11)/2$, $\nu=\sqrt{2}+1$, $1/\nu=\sqrt{2}-1$, $\rho=(\sqrt{13}+3)/2$ and $1/\rho=(\sqrt{13}-3)/2$.

\begin{table}[htp]
\caption{Values of $k$ and $c$ associated with the Rogers-Ramanujan continued fractions, the Ramanujan--G\"ollnitz--Gordon continued fractions and their level-13 analog}
\begin{center}
\begin{tabular}{rrrrr}
\hline
$k$&$c$&$x_1$&$x_2$&$\alpha\beta$\\
\hline
$1$&$\phi$&$R\left(e^{-2\alpha}\right)$&$R(e^{-2\beta})$&$\pi^2$\\
$-1$&$1/\phi$&$S(e^{-\alpha})$&$S(e^{-\beta})$&$\pi^2$\\
$11$&$\phi^5$&$R^5\left(e^{-2\alpha}\right)$&$R^5\left(e^{-2\beta}\right)$&$\pi^2/5$\\
$-11$&$1/\phi^5$&$S^5(e^{-\alpha})$&$S^5(e^{-\beta})$&$\pi^2/5$\\
$2$&$\nu$&$V(e^{-\alpha})$&$V(e^{-\beta})$&$\pi^2$\\
$-2$&$1/\nu$&$-V(-e^{-\alpha})$&$-V(-e^{-\beta})$&$\pi^2$\\
$3$&$\rho$&$R'(e^{-2\alpha})$&$R'(e^{-2\beta})$&$\pi^2/13$\\
$-3$&$1/\rho$&$S'(e^{-\alpha})$&$S'(e^{-\beta})$&$\pi^2/13$\\
\hline
\end{tabular}
\end{center}
\label{tab1}
\end{table}
We use the two equations to examine the common properties of the continued fractions. If $x_1=x_2=x$, then $(1-x_1x_2)/(x_1+x_2)=c$ reduces to $(1/x)-x=c$. This form has been used to express important formulas for $x=R(q), R^5(q), V(q)$ and $R'(q)$.  If we set $x_1+x_2=2r$, then $(1-x_1x_2)/(x_1+x_2)=c$  can be written as the quadratic equation $x^2-2rx-(2rc-1)=0$, its two roots corresponding to the pairs of continued fractions shown in Table \ref{tab1}. The value of $r$ lies between $1/2c$ and $\sqrt{c^2+1}-c$. We examine the patterns in the relations between $r$, $x_1$, $x_2$, $k$ and $q$ when $x_1$ and $x_2$ represent the pairs of continued fractions in Table \ref{tab1}.  In each case, as $\alpha$ increases, $r$ first decreases and then increases, attaining a minimum value when $\alpha=\beta$. For example, consider $k=1$, $x_1=R\left(e^{-2\alpha}\right)$ and $x_2=R(e^{-2\beta})$. If $\alpha=0$, $R\left(e^{-2\alpha}\right)=1/\phi$ and  $R(e^{-2\beta})=0$. As $\alpha$ increases from zero to $\alpha=\beta=\pi$, $R\left(e^{-2\alpha}\right)$ decreases and $R(e^{-2\beta})$ increases to $\sqrt{\phi^2+1}-\phi$. The functions $R\left(e^{-2\alpha}\right)$ and $R(e^{-2\beta})$ change very slowly with $\alpha$ when $0<\alpha<\pi/5$. Thereafter, they change almost linearly until $\alpha=\beta=\pi$. Similar observations can be made about the continued fractions in the other pairs of reciprocity formulas. We highlight patterns in the expressions for certain explicit values of the Rogers-Ramanujan continued fractions by expressing them in terms of the golden ratio. Finally, we extend the present analysis to reciprocity formulas for Ramanujan's cubic continued fractions and the Ramanujan-Selberg continued fraction.

\section{Reciprocity formulas}
The Rogers-Ramanujan continued fractions are
$$R(q)={q^{1/5}\over 1+}{q\over 1+}{q^2\over 1+}{q^3\over 1+}\dots,$$
$$\qquad\qquad\qquad\quad\ \  =q^{1/5}\prod_{j=1}^\infty {(1-q^{5j-1})(1-q^{5j-4})
\over (1-q^{5j-2})(1-q^{5j-3})}$$
$$\qquad\  \ =q^{1/5}\prod_{j=1}^\infty (1-q^j)^{\left({j\over 5}\right)}$$
where $|q|<1$ and $\left({j\over p}\right)$ is the Legendre symbol; and
$$S(q)=-R(-q)={q^{-1/5}\over 1-}{q\over 1+}{q^2\over 1-}{q^3\over 1+}\dots.$$

\bigskip\noindent
Consider the equation
$${1-x_1x_2\over x_1+x_2}=c.\eqno(1)$$
We write it as
$x_1x_2+c(x_1+x_2)=1$, then add $c^2$ to both sides to obtain
$$(c+x_1)(c+x_2)=1+c^2.\eqno(2)$$
Let
$$c^2-kc-1=0.\eqno(3)$$
We substitute its root
$$c={k+\sqrt{k^2+4}\over 2}\eqno(4)$$
into equation (2) to obtain
$$\left({\sqrt{k^2+4}+k\over 2}+x_1\right)\left({\sqrt{k^2+4}+k\over 2}+x_2\right)=\sqrt{k^2+4}\left({\sqrt{k^2+4}+k\over 2}\right).\eqno(5)$$

\bigskip\noindent
1. Let $k=\pm 1$. Then equation (5) becomes
$$\quad\ \  \left(\phi+x_1\right)\left(\phi+x_2\right)=\left(\phi+{1\over \phi}\right)\phi,\eqno(6)$$
$$\left({1\over \phi}+x_1\right)\left({1\over \phi}+x_2\right)=\left(\phi+{1\over \phi}\right){1\over \phi},\eqno(7)$$
where $\phi=(\sqrt{5}+1)/2$, $1/\phi=(\sqrt{5}-1)/2$ and $\phi+(1/\phi)=\sqrt{5}$. Entry 39 in Chapter XVI of Ramanujan's second notebook \cite{Ram} records the pair of reciprocity formulas in equations (6) and (7), with $x_1=R\left(e^{-2\alpha}\right)$ and $x_2=R(e^{-2\beta})$ in equation (6), and $x_1=S(e^{-\alpha})$ and $x_2=S(e^{-\beta})$ in equation (7), where $\alpha\beta=\pi^2$.

\bigskip\noindent
2. Let $k=\pm 11$. Then equation (5) becomes
$$\quad \ \left(\phi^5+x_1\right)\left(\phi^5+x_2\right)=\left(\phi^5+{1\over \phi^5}\right)\phi^5,\eqno(8)$$
$$\left({1\over \phi^5}+x_1\right)\left({1\over \phi^5}+x_2\right)
=\left(\phi^5+{1\over \phi^5}\right){1\over \phi^5},\eqno(9)$$
where $\phi^5=(5\sqrt{5}+11)/2$, $1/\phi^5=(5\sqrt{5}-11)/2$ and $\phi^5+(1/\phi^5)=5\sqrt{5}$. Entry 3.2.9 in Ramanujan's lost notebook (\cite{rln}, p. 91) records the reciprocity formulas in equations (8) and (9), with $x_1=R^5\left(e^{-2\alpha}\right)$ and $x_2=R^5\left(e^{-2\beta}\right)$ in equation (8), and $x_1=S^5(e^{-\alpha})$ and $x_2=S^5(e^{-\beta})$ in equation (9), where $\alpha\beta=\pi^2/5$.

Since $\phi^5=5\phi+3$ and $1/\phi^5=5\phi-8$, equation (1) implies
$$\left({1-R\left(e^{-2\alpha}\right)R(e^{-2\beta})\over R\left(e^{-2\alpha}\right)+R(e^{-2\beta})}\right)^5=5\left({1-R\left(e^{-2\alpha}\right)R(e^{-2\beta})\over R\left(e^{-2\alpha}\right)+R(e^{-2\beta})}\right)+3={1-R^5(e^{-2\gamma})R^5(e^{-2\delta})\over R^5(e^{-2\gamma})+R^5(e^{-2\delta})}$$
and
$$\left({1-S^5(e^{-\alpha})S^5(e^{-\beta})\over S^5(e^{-\alpha})+S^5(e^{-\beta})}\right)^5=5\left({1-S(e^{-\alpha})S(e^{-\beta})\over S(e^{-\alpha})+S(e^{-\beta})}\right)-8={1-S^5(e^{-\gamma})S^5(e^{-\delta})\over S^5(e^{-\gamma})+S^5(e^{-\delta})}$$
where $\alpha\beta=\pi^2$ and $\gamma\delta=\pi^2/5$.

\bigskip\noindent
3. Let $k=\pm 2$. Then equation (5) becomes
$$\left(\nu+x_1\right)\left(\nu+x_2\right)=\left(\nu+{1\over \nu}\right)\nu,\eqno(10)$$
$$\left({1\over \nu}+x_1\right)\left({1\over \nu}+x_2\right)=\left(\nu+{1\over \nu}\right){1\over \nu},\eqno(11)$$
where $\nu=\sqrt{2}+1$, $1/\nu=\sqrt{2}-1$ and $\nu+(1/\nu)=2\sqrt{2}$. Ramanujan (\cite{Ram}, Chapter XIX, p. 225) recorded the following continued fraction, which is now known as the Ramanujan--G\"ollnitz--Gordon continued fraction (Chan and Huang \cite{CH}, Baruah and Saikia \cite{BS}):
$$V(q)={q^{1/2}\over 1+q+}{q^2\over 1+q^3+}{q^4\over 1+q^5+}{q^6\over 1+}\dots$$
$$\qquad\ \  =q^{1/2}\prod_{j=1}^\infty {(1-q^{8j-7})(1-q^{8j-1})\over (1-q^{8j-5})(1-q^{8j-3})}.$$
The reciprocity formulas in equations (10) and (11) are obtained when $x_1=V(e^{-\alpha})$ and $x_2=V(e^{-\beta})$ in equation (10), and $x_1=-V(-e^{-\alpha})$ and $x_2=-V(-e^{-\beta})$ in equation (11), where $\alpha\beta=\pi^2$ (Bagis \cite{Bag},  Corollary 3.1, p. 11).

\bigskip\noindent
4. Let $k=\pm 3$. Then equation (5) becomes
$$\quad\ \  \left(\rho+x_1\right)\left(\rho+x_2\right)=\left(\rho+{1\over \rho}\right)\rho,\eqno(12)$$
$$\left({1\over \rho}+x_1\right)\left({1\over \rho}+x_2\right)=\left(\rho+{1\over \rho}\right){1\over \rho},\eqno(13)$$
where $\rho=(\sqrt{13}+3)/2$, $1/\rho=(\sqrt{13}-3)/2$ and $\rho+(1/\rho)=\sqrt{13}$.
In his second notebook, Ramanujan recorded the following level-13 analog of $R(q)$:
$$R'(q)=q\prod_{j=1}^\infty {(1-q^{13j-12})(1-q^{13j-10})(1-q^{13j-9})(1-q^{13j-4})(1-q^{13j-3})(1-q^{13j-1})\over (1-q^{13j-11})(1-q^{13j-8})(1-q^{13j-7})(1-q^{13j-6})(1-q^{13j-5})(1-q^{13j-2})}$$
$$=q\prod_{j=1}^\infty (1-q^j)^{\left({j\over 13}\right)}.\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\ \ \ $$
Cooper and Ye \cite{CY} obtained the reciprocity formulas in equations (12) and (13) with $x_1=R'(e^{2\alpha})$ and $x_2=R'(e^{2\beta})$ in equation (12),  and $x_1=S'(e^{2\alpha})$ and $x_2=S'(e^{2\beta})$ in equation (13), where $S'(q)=-R'(-q)$ and $\alpha\beta=\pi^2/13$.

\bigskip\noindent
Andrews and Berndt \cite{rln} observed that the following are among the most fundamental and useful properties of the Rogers-Ramanujan continued fraction:
$$\quad\ \  {1\over R(q)}-R(q)=1+{f(-q^{1/5})\over q^{1/5}f(-q^{5})}=1+{1\over q^{1/5}}\prod_{j=1}^\infty {(1-q^{j/5})\over (1-q^{5j})}\eqno(14)$$
and
$${1\over R^5(q)}-R^5(q)=11+{f^6(-q)\over qf^6(-q^{5})}=11+{1\over q}\prod_{j=1}^\infty {(1-q^j)^6\over (1-q^{5j})^6},\eqno(15)$$
where
$$f(-q)=(q; q)_{\infty}=\prod_{j=1}^{\infty} (1-q^j)=\sum_{j=-\infty}^{\infty} (-1)^j q^{j(3j-1)/2}\ \ .$$
Ramanujan and subsequently others (e.g., Berndt and Chan \cite{BC}, Soon-Yi \cite{S-Y}) used equations (14) and (15) to evaluate $R(q)$ for specific values of $q$.

Ramanujan (\cite{Ram}, Chapter XIX, p. 225) recorded the following analogous expressions for $V(q)$:
$${1\over V(q)}-V(q)={\varphi(q^2)\over q^{1/2}\psi(q^4)}$$
and
$${1\over V(q)}+V(q)={\varphi(q)\over q^{1/2}\psi(q^4)},$$
where
$$\varphi(q)=\sum_{k=-\infty}^\infty q^{k^2}\qquad $$
and
$$\psi(q)=\sum_{k=0}^\infty q^{k(k+1)/2}\ \ .$$
The corresponding expression for $R'(q)$ is
$${1\over R'(q)}-R'(q)=3+{f^2(-q)\over qf^2(-13q)}=3+{1\over q}\prod_{j=1}^{\infty} {(1-q^j)^2\over (1-q^{13j})^2}.\eqno(16)$$

The left-hand side in each of these equations is obtained by setting $x_1=x_2=x$ in equation (1) and substituting $x=R(q)$, $x=R^5(q)$, $x=V(q)$ and $x=R'(q)$. In this case
$${1\over x}-x=2c=k+\sqrt{k^2+4},\eqno(17)$$
which gives
$$x=\sqrt{c^2+1}-c=\sqrt{{k(k+\sqrt{k^2+4})+4\over 2}}-{k+\sqrt{k^2+4}\over 2}.$$

\bigskip\noindent
1. If $x=R(q)$, then
$$2c-1=k+\sqrt{k^2+4}-1={f(-q^{1/5})\over q^{1/5}f(-q^{5})}\eqno(18)$$
and
$$k=c-{1\over c},\ \ c={1\over 2}\left(1+{f(-q^{1/5})\over q^{1/5}f(-q^{5})}\right).$$
The value $k=1$ is associated with $q=e^{-2\pi}$, which gives the following relations obtained by Ramanujan:
$${f(-e^{-2\pi/5})\over e^{-2\pi/5}f(-e^{-5\cdot 2\pi})}=2\phi-1=\sqrt{5}$$
and
$$2c=1+{f(-e^{-2\pi/5})\over e^{-2\pi/5}f(-e^{-5\cdot 2\pi})}=1+\sqrt{5}=2\phi.$$

\bigskip\noindent
2. If $x=R^5(q)$, then
$$k+\sqrt{k^2+4}-11={f^6(-q)\over qf^6(-q^{5})}\eqno(19)$$
and
$$k=c-{1\over c},\ \ c={1\over 2}\left(11+{f^6(-q)\over qf^6(-q^{5})}\right).\eqno(20)$$
The value $k=11$ is associated with $q=e^{-2\pi/\sqrt{5}}$, which gives
$${f^6(-e^{-2\pi/\sqrt{5}})\over e^{-2\pi/\sqrt{5}} f^6(-e^{-\sqrt{5}\cdot 2\pi})}=5\sqrt{5}$$
and
$$2c={1\over R^5(e^{-2\pi/\sqrt{5}})}-R^5(e^{-2\pi/\sqrt{5}})=11+\sqrt{11^2+4}=11+5\sqrt{5}=2\phi^5.$$

\bigskip\noindent
3. If $x=V(q)$, then
$$2c={1\over V(q)}-V(q)=k+\sqrt{k^2+4}={\varphi(q^2)\over q^{1/2}\psi(q^4)}\eqno(21)$$
and
$$k=c-{1\over c},\ \ c={\varphi(q^2)\over 2q^{1/2}\psi(q^4)}\ .\eqno(22)$$
The value $k=2$ is associated with $q=e^{-\pi}$, which gives
$$2c={1\over V(e^{-\pi})}-V(e^{-\pi})={\varphi(e^{-2\pi})\over e^{-\pi/2}\psi(e^{-4\pi})}=2\sqrt{2}+2=2\nu.$$

\bigskip\noindent
4. If $x=R'(q)$, then
$$2c={1\over R'(q)}-R'(q)=k+\sqrt{k^2+4}\eqno(23)$$
and
$$k+\sqrt{k^2+4}-3={1\over q}\prod_{j=1}^{\infty} {(1-q^j)^2\over (1-q^{13j})^2}\ .\eqno(24)$$
The value $k=3$ is associated with $q=e^{-2\pi/\sqrt{13}}$, in which case
$${1\over R'(e^{-2\pi/\sqrt{13}})}-R'(e^{-2\pi/\sqrt{13}})=2\rho.$$

\bigskip\noindent
Equation (3) implies that $c^n=a_n+b_nc$, where $a_1=0$, $b_1=1$, $a_n=b_{n-1}$ and  $b_n=(a_{n-1}+kb_{n-1})$, for all $n\ge 2$. As a result, the left-hand side of equation  (1) is a  linear function of its integer powers.  Thus
$$\left({1-x_1x_2\over x_1+x_2}\right)^n=a_n+b_n\left({1-x_1x_2\over x_1+x_2}\right),\ n\ge 2.\eqno(25)$$
Similarly, $1/c^{n}=A_n+B_nc$, where $A_1=-k$, $B_1=1$, $A_n=B_{n-1}-kA_{n-1}$ and $B_n=A_{n-1}$, for all $n\ge 2$.
 Thus
$$\left({x_1+x_2\over 1-x_1x_2}\right)^{n}=A_n+B_n\left({1-x_1x_2\over x_1+x_2}\right),\  n\ge 2.\eqno(26)$$
If $x_1=x_2=x$, then
$$\left({1-x^2\over 2x}\right)^n=a_n+b_n \left({1-x^2\over 2x}\right)$$
and
$$\left({2x\over 1-x^2}\right)^{n}=A_n+B_n\left({1-x^2\over 2x}\right)$$
For example, if $n=2$, then
$$\left({1\over x}-x\right)^2=2^2+2k\left({1\over x}-x\right),$$
which implies that
$${1\over x}-x=2k+{4\over 2k+}{4\over 2k+}{4\over 2k+}{4\over 2k+}\dots.$$
and
$${1\over x^2}+x^2=6+2k\left({1\over x}-x\right)$$
$$\qquad\qquad\qquad \ =6+2k(k+\sqrt{k^2+4}).$$
If $x=R(q)$, then
$${1\over R^2(q)}+R^2(q)=2+\left(1+{f(-q^{1/5})\over q^{1/5}f(-q^{5})}\right)^2.$$

\section{A second quadratic equation}
Watson \cite{W} showed that if $\alpha\beta=\pi^2$, then
$R\left(e^{-2\alpha}\right)$ is a root of the quadratic equation
$${1\over x}-1-x=\left({k_1k_1^{'}K_1^3\over k_2k_2^{'}K_2^3}\right)^{1\over 6}$$
and $R\left(e^{-2\beta}\right)$ is a root of the quadratic equation
$${1\over x}-1-x=\left({k_2k_2^{'}K_2^3\over k_1k_1^{'}K_1^3}\right)^{1\over 6},$$
where $k_1$ and $k_2$ are the moduli of elliptic functions whose quarter-periods satisfy the relations
$$\pi{K_1^{'}\over K_1}={\alpha\over 5},\  \ \pi {K_2^{'}\over K_2}=5\alpha,\ \
\pi{K_2\over K_2^{'}}={\beta\over 5},\  \ \pi {K_1\over K_1^{'}}=5\beta.$$

Equation (1) gives a different quadratic equation, the solution to which is a pair of values for the Rogers-Ramanujan continued fractions, the Ramanujan--G\"ollnitz--Gordon continued fractions or the level-13 analog. This equation is obtained as follows.

Let $x_1+x_2=2r$. Then equation (1) can be written as
$$2rc=1-x_1x_2=1-x_1(2r-x_1)=1-x_2(2r-x_2).$$
That is, $x_1$ and $x_2$ are the roots
$$x=r\pm \sqrt{r^2+2rc -1}\eqno(27)$$
of the quadratic equation
$$x^2-2rx-(2rc-1)=0.\eqno(28)$$
Observe that any quadratic equation $x^2+Bx+C=0$ can be expressed as equation (28) by setting $r=-B/2$ and $c=(C-1)/B$. Let $2rc\le 1$ so that both the roots in equation (27) are non-negative. If $r^2+2rc -1=0$, then $x_1=x_2=r=\sqrt{c^2+1}-c$.  Otherwise, the larger root has a value between $1/c$ and $\sqrt{c^2+1}-c$, and the smaller a value between $\sqrt{c^2+1}-c$ and zero.
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=0.75\textwidth]{fig1.pdf}\caption{Value of $2r=R\left(e^{-2\alpha}\right)+R(e^{-2\beta})$ as a function of $\alpha$, where $\alpha\beta=\pi^2$.}
\label{fig1}
\end{center}
\end{figure}
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=0.75\textwidth]{fig2.pdf}\caption{Values of  $R\left(e^{-2\alpha}\right)$ (blue line) and $R(e^{-2\beta})$ (yellow line) as a function of $\alpha$, where $\alpha\beta=\pi^2$.}
\label{fig2}
\end{center}
\end{figure}
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=0.75\textwidth]{rq_approx.pdf}\caption{Comparisons of $e^{-2\alpha/5}$ and $R\left(e^{-2\alpha}\right)$,  and $e^{-2\beta/5}$ and $R(e^{-2\beta})$, as functions of $\alpha$, where $\alpha\beta=\pi^2$.}
\label{fig3}
\end{center}
\end{figure}
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=0.75\textwidth]{rq_k.pdf}
\caption{$R\left(e^{-2\alpha}\right)$ as a function of $k$.}
\label{fig4}
\end{center}
\end{figure}

For example, consider $x_1=R\left(e^{-2\alpha}\right)$, $x_2=R(e^{-2\beta})$ and $2r=R\left(e^{-2\alpha}\right)+R(e^{-2\beta})$, where $\alpha\beta=\pi^2$. For each value of $r$, there are corresponding values of $\alpha$ and $\beta$ for which equation (28) has the roots $x_1=R\left(e^{-2\alpha}\right)$ and $x_2=R(e^{-2\beta})$. Figure 1 plots the values of $2r$ as a function of $\alpha$,  where $c=\phi$ and $\alpha\beta=\pi^2$. As the value of $\alpha$ increases from zero to $\pi$, the value of $2r$ decreases from $1/\phi$ to $2(\sqrt{\phi^2+1}-\phi)$. The minimum value of $2r$ is obtained when $\alpha=\beta=\pi$.

Figure 2 shows how the values of $R\left(e^{-2\alpha}\right)$ and $R(e^{-2\beta})$ change with $\alpha$.  As $\alpha$ increases from zero to $\pi$, $R\left(e^{-2\alpha}\right)$ decreases from $1/\phi$ to $\sqrt{\phi^2+1}-\phi$ and $R(e^{-2\beta})$ increases from zero to $\sqrt{\phi^2+1}-\phi$.  The horizontal line in Figure 2 passes through $r=R(e^{-2\pi})$, which corresponds to $\alpha=\beta=\pi$. It is sufficient to know the value of $R\left(e^{-2\alpha}\right)$ over $0<\alpha\le \pi$; the corresponding value of $R(e^{-2\beta})$ can be obtained by the value of  the symmetric function around $r=R(e^{-2\pi})$, shown by the yellow line in Figure 2. Observe that $R\left(e^{-2\alpha}\right)$ and $R(e^{-2\beta})$ change very little when $0<\alpha<\pi/5$. Thereafter, the functions change almost linearly until $\alpha=\pi$. Figures \ref{fig3} compares the values of $R\left(e^{-2\alpha}\right)$ and  $e^{-2\alpha/5}$, and $R(e^{-2\beta})$ and  $e^{-2\beta/5}$, over the interval $0<\alpha\le 4\pi$.  For $\alpha\ge \pi$, $R\left(e^{-2\alpha}\right)$ is very close to $e^{-2\alpha/5}$ and becomes increasingly closer as $\alpha$ increases; and for $0<\alpha\le \pi$, $R(e^{-2\beta})$ is very close to $e^{-2\beta/5}$ and becomes increasingly closer as $\alpha$ decreases. Using the reciprocity formula in equation (6), we can approximate  $R\left(e^{-2\alpha}\right)$ by
$$R\left(e^{-2\alpha}\right)\approx
\begin{cases}
{\left(1-\phi e^{-{2\pi^2/(5\alpha)}}\right)\Big/\left(\phi+e^{-{2\pi^2/(5\alpha)}}\right)}&{\rm if}\ 0<\alpha\le \pi,\\
e^{{-2\alpha/5}}&{\rm if}\ \alpha\ge \pi.\\
\end{cases}$$
The maximum difference between the values of $R\left(e^{-2\alpha}\right)$ and this approximate value is less than 0.000531 and occurs when $\alpha=\pi$. Figure \ref{fig4} shows the relation between $R\left(e^{-2\alpha}\right)$ and $k$.   Similar plots and results can be obtained about the terms in the other pairs of reciprocity formulas.

\subsection{Explicit expressions for $R(q)$}
Using equations (6) and (8), we can write the reciprocity formulas for $R\left(e^{-2\alpha}\right)$ and $R^5\left(e^{-2\alpha}\right)$ as
$$R\left(e^{-2\alpha}\right)=\left({\phi+{1\over \phi}\over \phi+R\left(e^{-2\beta}\right)}-1\right)\phi={1-\phi R\left(e^{-2\beta}\right)\over \phi+R\left(e^{-2\beta}\right)},\ \alpha\beta=\pi^2\eqno(29)\qquad\ \ $$
and
$$R^5\left(e^{-2\alpha}\right)=\left({\phi^5+{1\over \phi^5}\over \phi^5+R^5\left(e^{-2\beta}\right)}-1\right)\phi^5={1-\phi^5 R^5\left(e^{-2\beta}\right)\over \phi^5+R^5\left(e^{-2\beta}\right)},\
\alpha\beta={\pi^2\over 5}.\eqno(30)$$
Starting with a pair of values for $R\left(e^{-2\alpha}\right)$ and $R(e^{-2\beta})$, where $\alpha\beta=\pi^2$,  we can iterate between equations (29) and (30) to obtain a sequence of values for $R(e^{-2\cdot 5^n \alpha})$, $R(e^{-2\cdot 5^n \beta})$, $R(e^{-2\pi^2/(5^n \alpha)})$
and $R(e^{-2\pi^2/(5^n \beta)})$, for all $n\ge 1$. These values have notable patterns when expressed in terms of $\phi$. Below, we obtain these expressions for
$$R\left(e^{-2\pi(\sqrt{5})^n}\right)={1-\phi R\left(e^{-2\pi/(\sqrt{5})^n}\right)\over \phi+R\left(e^{-2\pi/(\sqrt{5})^n}\right)}$$
and
$$R^5\left(e^{-2\pi/(\sqrt{5})^{n+2}}\right)
={1-\phi^5 R^5\left(e^{-2\pi(\sqrt{5})^{n}}\right)\over \phi^5+R^5\left(e^{-2\pi(\sqrt{5})^{n}}\right)}.$$
Let
$$R\left(e^{-2\pi/(\sqrt{5})^n}\right)={A_n\over B_n}.$$
Let  $C_n=B_n-\phi A_n$ and $D_n=\phi + B_n$. Then
$$R\left(e^{-2\pi(\sqrt{5})^n}\right)={C_n\over D_n}$$
and
$$R^5\left(e^{-2\pi/(\sqrt{5})^{n+2}}\right)={D_n^5-\left(\phi C_n\right)^5\over \left(\phi D_n\right)^5+C_n^5}.$$

\bigskip\noindent
(1) Consider
$$R\left(e^{-2\pi}\right)={\sqrt{\phi^2+1}-\phi},$$
which is obtained by substituting $x_1=x_2=R(e^{-2\pi})$ in equation (6). Then
$$R^5\left(e^{-2\pi/5}\right)={{1-\phi^5 R^5\left(e^{-2\pi}\right)\over \phi^5+R^5\left(e^{-2\pi}\right)}}\qquad\qquad\qquad\qquad\quad$$
$$={{1-\phi^5 \left({\sqrt{\phi^2+1}-\phi}\right)^5\over \phi^5+\left({\sqrt{\phi^2+1}-\phi}\right)^5}}$$
and
$$R\left(e^{-2\cdot 5\pi}\right)={1-\phi R\left(e^{-2\pi/5}\right)\over \phi+R\left(e^{-2\pi/5}\right)}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad$$
$$\ \ \ ={\sqrt[5]{{\phi^5+\left({\sqrt{\phi^2+1}-\phi}\right)^5}}-\phi \sqrt[5]{1-\phi^5 ({\sqrt{\phi^2+1}-\phi})^5}\over \phi \sqrt[5]{{\phi^5+\left({\sqrt{\phi^2+1}-\phi}\right)^5}}+\sqrt[5]{1-\phi^5 ({\sqrt{\phi^2+1}-\phi})^5}}\ .$$

\bigskip\noindent
(2) Consider
$$R^5\left(e^{-2\pi/\sqrt{5}}\right)={{\sqrt{\phi^{10}+1}-\phi^5}},$$
which is obtained by substituting $x_1=x_2=R^5(e^{-2\pi/\sqrt{5}})$ in equation (8). Then
$$R\left(e^{-2\pi\sqrt{5}}\right)
={1-\phi R\left(e^{-2\pi/\sqrt{5}}\right)\over \phi+R\left(e^{-2\pi/\sqrt{5}}\right)}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$$
$$={1-\phi \sqrt[5]{{\sqrt{\phi^{10}+1}-\phi^5}}\over \phi +\sqrt[5]{{\sqrt{\phi^{10}+1}-\phi^5}}},\qquad\qquad\qquad\qquad\qquad$$
which Ramanujan expressed as follows in his second letter to Hardy (\cite{H}, p. 143):
$${1\over 1+}{e^{-2\pi\sqrt{5}}\over 1+}{e^{-4\pi\sqrt{5}}\over 1+ \dots}=\left[{\sqrt{5}\over 1+\sqrt[5]{5^{3/4}\left({\sqrt{5}-1\over 2}\right)^{5/2}-1}}-{\sqrt{5}+1\over 2}\right]e^{2\pi/\sqrt{5}}.$$
We use the expression for $R\left(e^{-2\pi\sqrt{5}}\right)$ to obtain
$$R^5\left(e^{-2\pi\over 5\sqrt{5}}\right)
={1-\phi^5 R^5\left(e^{-2\pi\sqrt{5}}\right)\over \phi^5+R^5\left(e^{-2\pi\sqrt{5}}\right)}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad$$
$$\qquad \qquad\quad ={{\left(\phi + \sqrt[5]{{\sqrt{\phi^{10}+1}-\phi^5}}\right)^5-\phi^5 \left(1-\phi \sqrt[5]{{\sqrt{\phi^{10}+1}-\phi^5}}\right)^5\over \phi^5\left(\phi + \sqrt[5]{{\sqrt{\phi^{10}+1}-\phi^5}}\right)^5+\left(1-\phi \sqrt[5]{{\sqrt{\phi^{10}+1}-\phi^5}}\right)^5}}$$
and then use this expression to obtain
$$R\left(e^{-2\cdot 5\sqrt{5}\pi}\right)
={1-\phi R\left(e^{-2\pi/(5\sqrt{5})}\right)\over \phi+R\left(e^{-2\pi/(5\sqrt{5})}\right)}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$$
$$\tiny ={\sqrt[5]{\phi^5 \left(\phi + \sqrt[5]{\sqrt{\phi^{10}+1}-\phi^5}\right)^5+
\left(1-\phi \sqrt[5]{\sqrt{\phi^{10}+1}-\phi^5}\right)^5}
-\phi \sqrt[5]{{\left(\phi + \sqrt[5]{\sqrt{\phi^{10}+1}-\phi^5}\right)^5-
\phi^5\left(1-\phi \sqrt[5]{\sqrt{\phi^{10}+1}-\phi^5}\right)^5}} \over
\phi\sqrt[5]{\phi^5 \left(\phi + \sqrt[5]{\sqrt{\phi^{10}+1}-\phi^5}\right)^5+
\left(1-\phi \sqrt[5]{\sqrt{\phi^{10}+1}-\phi^5}\right)^5}
+ \sqrt[5]{{\left(\phi + \sqrt[5]{\sqrt{\phi^{10}+1}-\phi^5}\right)^5-
\phi^5\left(1-\phi \sqrt[5]{\sqrt{\phi^{10}+1}-\phi^5}\right)^5}} }\ .$$

\bigskip\noindent
(3) Consider
 $$R(e^{-\pi})=\frac{-\sqrt{\sqrt{5}} \left(\sqrt{5}+5\right)+\sqrt{2 \sqrt{5}+5}+\sqrt{5 \left(2 \sqrt{5}+5\right)}}{2 \left(\sqrt{5}-\sqrt{2 \sqrt{5}+5}\right)},$$
 which can be expressed as follows in terms of $\phi$:
$$R(e^{-\pi})={\phi+{1\over \phi}-\phi\sqrt{\phi}\over \sqrt{\phi}-{1\over \phi}\sqrt{\phi+{1\over \phi}}}.$$
Starting with this expression, we can iterate between equations (29) and (30) to obtain the values of  $R(e^{-4\cdot 5^{n-1}\pi})$,  $R(e^{-4\pi/5^n})$ and $R(e^{-\pi/5^n})$, for $n\ge 1$. The first such values are
$$R(e^{-4\pi})={1-\phi R(e^{-\pi})\over \phi +R(e^{-\pi})}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$$
$$\ \ ={\left(\sqrt{\phi}-{1\over \phi}\sqrt{\phi + {1\over \phi}}\right) -\phi\left(\phi + {1\over \phi}-\phi\sqrt{\phi}\right)\over \phi+{1\over \phi}-\sqrt{\phi+{1\over \phi}}}\qquad\quad$$
$$R^5\left(e^{-4\pi/5}\right)={{1-\phi^5 R^5\left(e^{-\pi}\right)\over \phi^5+R^5\left(e^{-\pi}\right)}}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$$
$$={\left(\sqrt{\phi}-{1\over \phi}\sqrt{\phi + {1\over \phi}}\right)^5 -\phi^5\left(\phi + {1\over \phi}-\phi\sqrt{\phi}\right)^5\over\phi^5\left(\sqrt{\phi}-{1\over \phi}\sqrt{\phi + {1\over \phi}}\right)^5+\left(\phi + {1\over \phi}-\phi\sqrt{\phi}\right)^5}\ \ $$
and
$$R^5\left(e^{-\pi/5}\right)={{1-\phi^5 R^5\left(e^{-4\pi}\right)\over \phi^5+R^5\left(e^{-4\pi}\right)}}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$$
$$\qquad\qquad\qquad\qquad\  ={\left(\phi+{1\over \phi}-\sqrt{\phi+{1\over \phi}}\right)^5-\phi^5\left(\left(\sqrt{\phi}-{1\over \phi}\sqrt{\phi + {1\over \phi}}\right) -\phi\left(\phi + {1\over \phi}-\phi\sqrt{\phi}\right)\right)^5
\over
\phi^5 \left(\phi+{1\over \phi}-\sqrt{\phi+{1\over \phi}}\right)^5+\left(\left(\sqrt{\phi}-{1\over \phi}\sqrt{\phi + {1\over \phi}}\right) -\phi\left(\phi + {1\over \phi}-\phi\sqrt{\phi}\right)\right)^5}\ .$$

\section{Ramanujan's cubic continued fraction and the Ramanujan-Selberg continued fraction}
The preceding analysis extends to the following continued fraction, which appears in entry 3.3.1 of Ramanujan's lost notebook (\cite{rln}, p. 94) and is now known as Ramanujan's cubic continued fraction:
$$G(q)={q^{1/3}\over 1+}\  {q+q^2\over 1+}\ {q^2+q^4\over 1+}\ {q^4+q^6\over 1+}\dots$$
$$\qquad\qquad\qquad\quad\ \  =q^{1/3}\prod_{j=1}^\infty {(1-q^{6j-5})(1-q^{6j-1})
\over (1-q^{6j-3})^2},\ |q|<1.\quad\ \ \ $$

Consider the equation
$${2^k-x_1x_2\over x_1+x_2}=1,\eqno(31)$$
which can be re-written as
$$(1+x_1)(1+x_2)=1+2^k.\eqno(32)$$
Setting $k=\pm 1$ and $k=\pm 3$ in equation (32) yields the following four reciprocity formulas for Ramanujan's cubic continued fraction. Table \ref{tab2} shows the terms associated with $x_1$ and $x_2$ in a reciprocity formula for each value of $k$.
\begin{table}[htp]
\caption{Values of $k$ associated with Ramanujan's cubic continued fraction}
\begin{center}
\begin{tabular}{rrrr}
\hline
$k$&$x_1$&$x_2$&$\alpha\beta$\\
\hline
$1$&$1/G\left(-e^{-\alpha}\right)$&$1/G(-e^{-\beta})$&$\pi^2$\\
$-1$&$G(e^{-\alpha})$&$G(e^{-\beta})$&$2\pi^2$\\
$3$&$1/G^3\left(-e^{-\alpha}\right)$&$1/G^3(-e^{-\beta})$&$\pi^2/3$\\
$-3$&$G^3(e^{-\alpha})$&$G^3(e^{-\beta})$&$2\pi^2/3$\\
\hline
\end{tabular}
\end{center}
\label{tab2}
\end{table}

\bigskip\noindent
1. Let $k=1$. Adiga, Kim, Naika and Madhusudhan \cite{aknm} obtained the following reciprocity formula in which $x_1=1/G(-e^{-\alpha})$ and  $x_2=1/G(-e^{-\beta})$:
$$\left(1+{1\over G(-e^{-\alpha})}\right)\left(1+{1\over G(-e^{-\beta})}\right)=1+2,\ \ \alpha\beta=\pi^2.\eqno(33)$$

\bigskip\noindent
2. Let $k=-1$. In his lost notebook, Ramanujan (\cite{rln}, Theorem 3.3.2) recorded the following reciprocity formula in which $x_1=G(e^{-\alpha})$ and  $x_2=G(e^{-\beta})$:
$$\Big(1+G(e^{-\alpha})\Big)\Big(1+G(e^{-\beta})\Big)=1+{1\over 2},\ \ \alpha\beta=2\pi^2.\eqno(34)$$

\bigskip\noindent
3. Let $k=3$. Adiga, Kim, Naika and Madhusudhan \cite{aknm} obtained the following reciprocity formula in which $x_1=1/G^3(-e^{-\alpha})$ and $x_2=1/G^3(-e^{-\beta})$:
$$\left(1+{1\over G^3(-e^{-\alpha})}\right)\left(1+{1\over G^3(-e^{-\beta})}\right)=1+2^3,\ \ \alpha\beta={\pi^2\over 3}.\eqno(35)$$

\bigskip\noindent
4. Let $k=-3$. Cooper(\cite{coop}, Theorem 6.12, p. 373)  obtained the following reciprocity formula in which $x_1=G^3(e^{-\alpha})$ and  $x_2=G^3(e^{-\beta})$:
$$\Big(1+G^3(e^{-\alpha})\Big)\Big(1+G^3(e^{-\beta})\Big)=1+{1\over 2^3},\ \ \alpha\beta={2\pi^2\over 3}.\eqno(36)$$

\bigskip
Let $x_1+x_2=2r$, where $x_1$ and $x_2$ are a pair of values in Table \ref{tab2}.  For each value of $k$ in the table, $x_1$ and $x_2$ are the roots $r\pm \sqrt{r^2+2r-2^k}$ of the quadratic equation  $x^2-2rx-(2r-2^k)=0$. Let $r\le 2^{k-1}$ so that both the roots are non-negative. If one root is zero, the other obtains it's largest possible value of $2^k$. The two roots are equal if $r^2+2r - 2^k=0$, in which case  $x_1=x_2=r=\sqrt{2^k+1}-1$.  If the two roots are different, the larger has a value between $2^k$ and $\sqrt{2^k+1}-1$, and the smaller a value between $\sqrt{2^k+1}-1$ and zero. Starting with a pair of values for $G\left(e^{-\alpha}\right)$ and $G(e^{-\beta})$, we can use equations (33) and (35), and equations  (34) and (36), to iteratively obtain sequences of values for $G(q)$.

\bigskip
Finally, consider the following continued fraction, which Ramanujan recorded in his second notebook (\cite{Ram}, Chapter XIX, p. 225):
$$s(q)={q^{1/8}\over 1+}\  {q\over 1+q+}\ {q^2\over 1+q^2+}\ {q^3\over 1+q^3+}\dots, |q|<1.$$
 This continued fraction is now called the Ramanujan-Selberg continued fraction. Saikia (\cite{S}, Theorem 5.3, p. 52) obtained a reciprocity formula that can be expressed as
$$s^8(e^{-\alpha})+s^8(e^{-\beta})={1\over 16},\ \alpha\beta=\pi^2.$$
Thus, unlike the other preceding continued fractions, the Ramanujan-Selberg continued fraction is a solution to a linear equation, not a quadratic equation. Any positive number between zero and $1/16$ can be expressed as $x=s^8\left(e^{-\alpha}\right)$, and then ${1\over 16}-x$ can be expressed  as $s^8(e^{-\pi^2/\alpha})$.


TITLE: Is there a way to solve composite exponential equations precisely?
QUESTION [10 upvotes]: I have a, seemingly, trivial question. Find $x$ such that
$$
l = x^x,
$$
for some constant value $l \in \mathbb{R}^{>0}$ and $x \in \mathbb{R}^{>0}$.
Obviously, this equation has a unique solution that can aprroximated. Neverthelss, I do not see an obvious approach to solve this equation precisly, nor can I find one on this website or using google. Maybe I am only missing the appropriate terminology to express the question.
EDIT: I would also be fine with a good explanation why it is difficult or not possible.
EDIT 2: As discussed in the comments, the equation, of course, has no unique solution for $l, x \in \mathbb{R}^{>0}$ as stated by me above.

REPLY [1 votes]: It depends on what you mean by "precisely". Only rational numbers can be represented by a finite number of decimal points, so any real number can only be approximated by numerical methods, or represented by some mathematical expression such as "square root". $x \rightarrow x^x$ is continuous and increasing for $x>1$, so it's 1:1. So the cardinality of $l$ such that $x$ is rational is the same as the cardinality of the rationals, so "almost all" $l$ will result in irrational $x$.


TITLE: Evaluate $I=\iint_S (x^2+y^2)\, dS$ where $S$ is the surface of the cone $\sqrt{x^2+y^2}\leqslant z\leqslant 1$
QUESTION [0 upvotes]: I have to evaluate $I$
$$
I=\iint\limits_S (x^2+y^2)\ dS\ \ \text{where $S$ is the surface of the cone}\ \ \sqrt{x^2+y^2}\leqslant z\leqslant 1
$$

My attempt:
$$
\begin{aligned}
&I=\iint\limits_{S_1} (x^2+y^2)\ dS_1 + \iint\limits_{S_2} (x^2+y^2)\ dS_2,\ \ \text{where}\\
&S_1 \text{ is the side surface of the cone},\ \ S_2 \text{ is the base surface of the cone}
\end{aligned}
$$
The first integral is straigtforward:
$$
\begin{aligned}
&I_1=\iint\limits_{S_1} (x^2+y^2)\ dS_1=\iint\limits_{M_1}(x^2+y^2)\sqrt{(z'_x)^2+(z'_y)^2+1}\ dxdy,\ \ \text{where}\ z=\sqrt{x^2+y^2}\\
&I_1=\sqrt{2}\iint\limits_{M_1}(x^2+y^2)\ dxdy=\sqrt{2}\int\limits_0^{2\pi}d\phi\int\limits_0^1r^2\cdot r\ dr=\frac{\pi\sqrt{2}}{2}
\end{aligned}
$$
However, the answer I got for the second integral and the problem book answer don't match.
$$
\begin{aligned}
&\text{Since $z=1$, we have}\ \ I_2=\iint\limits_{S_2} (x^2+y^2)\ dS_2=\int\limits_\sigma(x^2+y^2)\ dl,\ \ \text{where}\ \ \sigma\ \ \text{is a circle}\ \ x^2+y^2=1\\
&I_2=\int\limits_\sigma(x^2+y^2)\sqrt{(x'_\phi)^2+(y'_\phi)^2}\ d\phi=\int\limits_0^{2\pi}(\cos^2\phi+\sin^2\phi)\cdot 1\ d\phi=2\pi
\end{aligned}
$$
So, my final answer is $I=I_1+I_2=\pi\left(\frac{\sqrt{2}}{2}+2\right)$.
However, the correct answer is $I=\pi\left(\frac{\sqrt{2}}{2}+\frac{1}{2}\right)$.
That means, I must have made a mistake while evaluating $I_2$, but I have no idea what is wrong. I would greatly appreciate it if someone pointed it out to me.

REPLY [1 votes]: Note the second integral is over the unit circle, not along the circumference,
$$I_2=\int\limits_{x^2+y^2<1}(x^2+y^2)dxdy
=\int_0^{2\pi}\int _0^1 r^2rdrd\phi=\frac\pi2
$$


TITLE: How did the Intersecting Storage Rings Actually Perform Tests
QUESTION [2 upvotes]: I am reading the book written in 1982 titled "Quarks, The Stuff of Matter" by Harald Fritzsch.  In it he writes about the ISR at Cern, that slammed protons together and in some of those collisions, 2 quarks bounced off each other at right angles to the protons.
I understand this part of it.  What I don't understand is how they detected anything.   None of the images of the ISR Interaction Areas show any detectors.
https://www.researchgate.net/figure/The-Intersecting-Storage-Rings-ISR-the-worlds-first-hadron-collider-photo-CERN_fig1_44218180
In this photo at almost the exact center is where the protons collided.  I do not see a large detector.  What am I missing?
Thanks

REPLY [2 votes]: The ISR was, in large part, built as an experiment in accelerator technology.  When it succeeded, CERN considered it a "facility for experiments". Unlike modern colliders, which are built along with detectors aimed at specific physics, the ISR was well along on construction while the physics was still being discussed.
Eventually, a number of detectors were built and given beam time.  Some sample pictures (from this retrospective):


TITLE: Characteristic function of Bernoulli distribution
QUESTION [0 upvotes]: How do I show that if $X$ is a Bernoulli-r.v. with parameter $p$, then
$$
\varphi_X(u)=E[e^{iuX}]=e^{iu0}(1-p)+e^{iu}p
$$
Without using the characteristic function of a binomial-r.v. with parameters n,p.

REPLY [2 votes]: We can evaluate the expected value $E[e^{iuX}]$ using the LOTUS. We have that
$$
E[e^{iuX}]=(1-p)e^{iu0}+pe^{iu}=(1-p)+pe^{iu}.
$$

REPLY [2 votes]: $Ee^{iuX }=Ee^{iuX }I_{\{X=0\}}+Ee^{iuX }I_{\{X=1\}}=Ee^{0 }I_{\{X=0\}}+Ee^{iu }I_{\{X=1\}}=P(X=0)+e^{iu}P(X=1)=(1-p)+e^{iu}p$.


text:
\begin{document}

\begin{frontmatter}

\title{Hyperviscosity-Based Stabilization for Radial Basis Function-Finite Difference (RBF-FD) Discretizations of Advection-Diffusion Equations}

\author[addr1]{Varun Shankar\corref{corresp}}
\address[addr1]{Department of Mathematics and School of Computing, University of Utah, UT, USA}
\ead{vshankar@math.utah.edu}
\cortext[corresp]{Corresponding Author}

\author[addr2]{Aaron L. Fogelson}
\address[addr2]{Departments of Mathematics and Bioengineering, University of Utah, UT, USA}
\ead{fogelson@math.utah.edu}

\begin{abstract}
We present a novel hyperviscosity formulation for stabilizing RBF-FD discretizations of the advection-diffusion equation. The amount of hyperviscosity is determined quasi-analytically for commonly-used explicit, implicit, and implicit-explicit (IMEX) time integrators by using a simple 1D semi-discrete Von Neumann analysis. The analysis is applied to an analytical model of spurious growth in RBF-FD solutions that uses auxiliary differential operators mimicking the undesirable properties of RBF-FD differentiation matrices. The resulting hyperviscosity formulation is a generalization of existing ones in the literature, but is free of any tuning parameters and can be computed efficiently. To further improve robustness, we introduce a simple new scaling law for polynomial-augmented RBF-FD that relates the degree of polyharmonic spline (PHS) RBFs to the degree of the appended polynomial. When used in a novel ghost node formulation in conjunction with the recently-developed overlapped RBF-FD method, the resulting method is robust and free of stagnation errors. We validate the high-order convergence rates of our method on 2D and 3D test cases over a wide range of Peclet numbers (1-1000). We then use our method to solve a 3D coupled problem motivated by models of platelet aggregation and coagulation, again demonstrating high-order convergence rates.
\end{abstract}
\begin{keyword}
Radial basis function; high-order method; hyperviscosity; meshfree; advection-diffusion.
\end{keyword}

\end{frontmatter}

\input{Intro}
\input{RBFReview}
\input{Stability}
\input{Methods}
\input{TimeStepping}
\input{Results}
\input{Discussion}

\section*{Acknowledgments}
This work was supported in part on NSF grants DMS-1160432 and DMS-1521748 and NIH grant 1R01HL120728.

\section*{References}
\bibliography{article_refs_mod}

\end{document}