text:
\section{Generic STGQs}

When an STGQ $(\mS^x,H)$ satisfies (*), or equivalently when $T = H/\mathbb{S}$ is abelian, one easily observes that
when $(\mF,\mF^*)$ is the associated Kantor-family, and $A, B$ are distinct members of $\mF$, that if $U \leq A$ and $V \leq B$, we have that
\begin{equation}
\langle U,V \rangle = H \ \ \longrightarrow \ \ U = A,\ \ V = B.
\end{equation}
(If at least one of $U, V$ is a proper subgroup, $\langle U\mathbb{S}/\mathbb{S}, V\mathbb{S}/\mathbb{S} \rangle$ is a proper
subgroup of $T$.) We will show that a generalization of this natural property suffices to classify the parameters of the generic case (that is, the
case where $T$ is not abelian), and to even classify to some extent. (So below, a {\em generic STGQ} is an STGQ for which $T$ is not abelian,
or equivalently, for which (*) is not satisfied. In particular, such STGQs cannot be square STGQs.)\\

More precisely, in a first instance our main objective is proving the following result. Before stating it, we need a definition. Let $(\mS^x,H)$ be as
above, and let $\Phi := \Phi(H)$ be the Frattini subgroup of $H$. Define a point-line geometry $\Gamma(\Phi)$ as follows.
Its lines are the $\Phi$-orbits on the lines incident with $x$ (where the trivial orbit $\{x\}$ is excluded); its points are the $\Phi$-orbits
in the affine point set of $\mS^x$. A point $u$ is incident with a line $V$ if at least one $\mS^x$-point of the orbit $V$ is collinear
with some point of the orbit $u$. If $u$ then is incident with $V$, it is easy to see that $V$ is surjectively projected on
$u$, so that for each $L \I x$, a $\Gamma(\Phi)$-point is incident with precisely one line which is a $\Phi$-orbit on $L$. So each $\Gamma(\Phi)$-point
is incident with precisely $t + 1$ $\Gamma(\Phi)$-lines.

\bt
\label{Frattsub}
Let $(\mS^x,H)$ be a generic STGQ of order $(s,t)$, with Kantor family $(\mF,\mF^*)$. Suppose that the projection lemma (cf. the next
subsection) holds.
Suppose also that at least one of the following properties is satisfied.
\begin{itemize}
\item[(a)]
$\Gamma(\Phi)$ is a dual partial linear space.
\item[(b)]
For each $A \ne B \in \mF$, we have that if $K$ is a maximal subgroup of $H$ which does not contain $A$, then
\begin{equation}
\langle A \cap K, B \rangle \ne H.
\end{equation}
(This is the generalization of the aforementioned natural property.)

\end{itemize}
Then the Frattini subgroup $\Phi(H)$ is the elation group of a subSTGQ of order $t$, so that $s = t^2$.
\et

The way to construct ideal subquadrangles goes as follows.
Let $(\mK,\mK^*)$ be a Kantor family of type $(s,t)$ in the group $G$, and put $\hF := \mK \cup \mK^*$.
A nontrivial subgroup $X$ of $G$ is an {\em $\hF$-factor} of $G$ if
\begin{quote}
$(U\cap X)(V \cap X) = X$ for all $U,V \in \hF$ satisfying $UV = G$.
\end{quote}

Define $\mK_X = \{U\cap X\vert U \in \mK\}$ and $\mK^*_X = \{U^*\cap X \vert U^* \in \mK^*\}$.
We say that $X$ is ``of type $(\sigma,\tau)$'' if $\vert X\vert = \sigma^2\tau$, $\vert A\cap X\vert = \sigma$ and $\vert A^*\cap X\vert = \sigma\tau$ for all
$A \in \mK$ (and in \cite{Hach} it is shown that such integers $\sigma,\tau$ always exist).

\bt[D.~Hachenberger \cite{Hach}]
\label{Ha1}
Let $X$ be an $\hF$-factor of type $(\sigma,\tau)$ in $G$. Then necessarily one of the following cases occurs:
\begin{itemize}
\item[$(a)$]
$\sigma = 1$, $\vert X\vert = \tau \leq t$ and $X$ is a subgroup of $\cap_{A \in \mK}A^*$;
\item[$(b)$]
$\sigma > 1$, $\tau = t$ and $(\mK_X,\mK^*_X)$ is a Kantor family in $X$ of type $(\sigma,\tau)$.
\end{itemize}
\et

If we are in case (b) of Theorem \ref{Ha1}, we call $X$ a {\em thick} $\hF$-factor.
An $\hF$-factor $X$ in $G$ is {\em normal} if $X$ is a normal subgroup of $G$.

\bt[D. Hachenberger \cite{Hach}]
\label{Ha2}
Let $G$ be a group of order $s^2t$ admitting a Kantor family $(\mK,\mK^*)$ of type $(s,t)$, with $s,t > 1$, and having a normal $\hF$-factor $X$ of type $(\sigma,\tau)$ with $\tau = t$.
Then one of the following cases occurs:
\begin{itemize}
\item[$(a)$]
$G$ is a group of prime power order;
\item[$(b)$]
$\sigma > 1$, $\vert G\vert$ has exactly two prime divisors, and $X$ is a Sylow subgroup of $G$ for one of these primes.
\end{itemize}
\et

Once we have ideal subSTGQs, we can use our theory of square STGQs to search for more structure in the generic case.\\

The reader might want to keep the following standard lemma in mind:

\bl
\label{Frattder}
Let $G$ be a finite $p$-group. Then $\Phi(H) = [H,H]G^p$, where $G^p := \langle g^p \vert g \in G\rangle$.
\el

\br
{\rm In the course of the series, we will show that ``most of the time'', $H$ has exponent $p$, so that
$\Phi(H) = [H,H]$. But when the exponent is $4$, it might happen that $\vert [H,H]\vert = t^3/2$ | in fact
this happens for the exotic elation groups of $H(3,q^2)$ with $q$ even.}
\er

\subsection{Projection Lemma}

Let $(\mS^x,H)$ be an STGQ of order $(s,t)$. Let $T = H/\mathbb{S}$ be the translation group of the net $\Pi_x$.
In this paragraph we wish to classify the extensions

\begin{equation}  \he \mapsto M \cap \mathbb{S} \mapsto  M \mapsto T \mapsto \he,     \end{equation}

with $M \leq H$. This will render us information about the Frattini subgroup of $H$.

The following lemma will be crucial (it is Theorem \ref{fix1}(iv) applied to STGQs with (*)).

\bl
\label{coll}
Suppose $\mS^x$ satisfies $(*)$.
If $s = t$ is odd, or $s \ne t$, any element $\alpha$ of $\cup_{A \in \hF}A^*$ maps some point of $\mP \setminus x^{\perp}$ to a collinear point.
\el
{\em Proof}.\quad
Suppose $\alpha \not\in \mathbb{S}$ does not have this property.
Applying Benson's lemma (while using $(*)$), we obtain

\begin{equation}  (t + 1)(s + 1)  + ts \equiv st + 1 \mod{s + t},    \end{equation}

so that $s + t$ divides $st$, contradiction. \eop \\

\bc
If $s = t$ is odd, or if $s \ne t$ and $T$ is abelian, the previous lemma applies,
\ec
{\em Proof}.\quad
By assumption, $T$ is abelian, so $(*)$ holds for $\mS^x$ by Theorem \ref{ab1}. \eop \\

(As we will later see, $(*)$ will often be satisfied when $T$ is not abelian.)\\

Suppose $\mS^x$ has the property that any element $\alpha$ of $\cup_{A \in \hF} A^*$ maps some point of $\mP \setminus x^{\perp}$ to a collinear point.
Suppose that $(*)$ holds.\\

Let $M$ be a subgroup as above, and note that without of loss of generality, we can suppose that $M$ is maximal in $H$. Let $(\hF,\hF^*)$ be the Kantor family
associated to $(\mS^x,H)$. Note that there must be some $A \in \hF$ not contained in $M$, since $\langle A \vert A \in \hF \rangle = H$.
Note that $[H : M] = p = [A^* : A^* \cap M]$, where $st$ is a power of the prime $p$.
Let $L \I x$ be such that $L \ne [A]$, so that there are $p$ $(A \cap M)$-orbits on $L \setminus \{x\}$. On the other hand, $M$ acts transitively on $L \setminus \{x\}$
since $M$ projects on $T$. Moreover, since $[A^*: A^* \cap M] = p$ and $\mathbb{S} \not\leq M$, $A^* \cap M$ also acts transitively
on $L \setminus \{x\}$.
Let $\beta \in M \cap A^*$ be such that it does not fix all $(A \cap M)$-orbits on $L \setminus \{x\}$. Then $\beta$ fixes some line $U \not\in x$ (since
it maps some affine point to a collinear point), and clearly $U \sim [A]$. Now $H_U$ is conjugate to $A$ by some element $\gamma$ of $H_{[L]}$, so
$\beta$ is conjugate to some element of $A$ with the same action on $L$. But as $M$ is a normal subgroup of $H$, $\beta \in M$, so that
$A \in M$, contradiction.

\bt[Projection Lemma]
\label{ext}
The only extension of $T$ in $H$ is $H$ itself, if one of the following properties is satisfied.
\begin{itemize}
\item[{\rm (i)}]
$ s = t$ is odd.
\item[{\rm (ii)}]
$s \ne t$, and $\mS^x$ satisfies $(*)$. \eop
\end{itemize}
\et

\br[Even case]
{\rm
When $s = t$ is even, Theorem \ref{ext} cannot work in general. For, let $\mS^x$ be a TGQ of order $2^h$, so that $x$ is a regular point.
Take any two distinct elements $A, B$ of the associated Kantor family. Then $AB = \langle A, B\rangle$, and $AB\mathbb{S}/\mathbb{S} = T$.
}
\er

 Let $M$ be maximal in $H$. By Theorem \ref{ext}, $M/(M \cap \mathbb{S})$ is an index $p$-group in $T$, so that $\mathbb{S} \leq M$.
 The following corollary is in some sense complementary to what we will encounter in the abelian factor section, later on. By $\Phi(H)$
we denote the Frattini subgroup of $H$.

 \bc[Structure of abelian $T$]
 Let $(\mS^x,H)$ satisfy one of $(i)$-$(ii)$ of the previous theorem.
 \begin{itemize}
 \item[{\rm (i)}]
 If $A, B \in \hF$ are distinct, $\langle A, B \rangle = H$. Whence $\mathbb{S} \leq \Phi(H)$.
 \item[{\rm (ii)}]
 If $T$ is abelian, $\mathbb{S} = [H,H]$.
 \item[{\rm (iii)}]
 $\Phi(H) = \mathbb{S} = [H,H]$ if and only if $T$ is elementary abelian (so in particular, if $s = t$).
 \item[{\rm (iv)}]
 If $T$ is abelian, every element of $\mathbb{S}$ is a commutator.
 \end{itemize}
 \ec
{\em Proof}.\quad
It is clear that $\langle A, B\rangle\mathbb{S}/\mathbb{S} = T$, so $\langle A, B\rangle = H$ by Theorem \ref{ext}. \\

Now let $T$ be abelian; then $[H,H] \leq \mathbb{S} = [H,H]$. Let $\gamma \in \mathbb{S}^\times$, and let $U \nI x$ be arbitrary.
Let $\xi$ be an element of $H \setminus \mathbb{S}$ mapping $U$ to $U^{\xi}$; then $\xi$ fixes some affine line $U' \ne U$ which meets
$\proj_xU$. Let $\beta \in H$ map $U'$ to $U$. Then $[\beta,\xi]$ maps $U$ to $U^\xi$,  while being  a symmetry. So every element of $\mathbb{S}$ is a
symmetry, and $[H,H] = \mathbb{S}$. Both groups coincide with $\Phi(H)$ if $T$ is elementary abelian. Vice versa, if $T$ is elementary abelian,
\begin{equation} \bigcap_{Y \leq T\ \mbox{maximal}}Y = \{\he\},  \end{equation}
so that $\Phi(H) = \mathbb{S}$.
\eop \\

 The following lemma follows in the same way as the Projection Lemma.

 \bt[Projection Lemma | general form]
\label{extg}
Suppose $(\mS^x,H)$ is an STGQ of order $(s,t)$, and let $(\hF,\hF^*)$ be the associated Kantor family.
Let one of the following properties be satisfied.
\begin{itemize}
\item[{\rm (i)}]
$ s = t$ is odd.
\item[{\rm (ii)}]
$s \ne t$, and $\mS^x$ satisfies $(*)$.
\end{itemize}
Let $N$ be a normal subgroup of $H$, $L \I x$, and $O$ an $N$-orbit in $L \setminus \{x\}$. If $O$ is also an $(A^*\cap N)$-orbit for $A^* \in \hF^*$,
then $O$ is an $A$-orbit.
\eop \\
\et

For a generic STGQ $(\mS^x,H)$, we say that {\em the projection lemma holds} if the following short exact sequence
\begin{equation}  \he \mapsto M \cap \mathbb{S} \mapsto  M \mapsto T \mapsto \he,     \end{equation}
with $M \leq H$, only has solutions for $M = H$.

\bl
If the projection lemma holds for the generic STGQ $(\mS^x,H)$, then $\mathbb{S} \leq \Phi(H)$.
\el
{\em Proof.}\quad
Let $M$ be any maximal subgroup of $H$, then
\begin{equation}
\frac{\vert H\vert}{\mathrm{char}(\mS^x)} = \vert M \vert = \vert M \cap \mathbb{S} \vert \times \vert M/(M \cap \mathbb{S}) \vert,
\end{equation}
so that the projection lemma implies that $\mathbb{S} \leq M$. \eop \\

\subsection{Generic STGQs with (b)}

In this subsection we suppose that $(\mS^x,H)$ is a generic STGQ of order $(s,t)$ with Kantor family $(\mF,\mF^*)$, so that $s \ne t$.
For the entire subsection, we assume to be in part (b) of the theorem.
We may suppose w.l.o.g. that $\mathbb{S}$ is contained in $\Phi = \Phi(H)$.
Let $\mathbb{S} \leq G \leq H$, and $A,B \in \hF$ with $A \ne B$.  We say that $G$ satisfies (F) ``at $A$ w.r.t. $B^*$'' if

\begin{equation}   G = (G \cap A)(G \cap B^*).    \end{equation}

Note that this property is symmetric in $A$ and $B$. If $G$ has (F) at $A$ w.r.t. any $B^* \in \hF^* \setminus \{A^*\}$, by definition $G$ satisfies (F) {\em at} $A$.\\

Let $A \in \hF$.
Define $\mR_A$ to be the set of maximal subgroups of $H$ that contain $A$; note that for $R \in \mR_A$, (F) is satisfied at $A$ w.r.t. any element of
$\hF^* \setminus \{A^*\}$. Let $\mK_A := \cap_{C \in \mR_A}C$. Then the reader verifies that $\mK_A$ also satisfies (F) at $A$ w.r.t. any element of $\hF^* \setminus \{A^*\}$.
(If $A \leq L \leq H$, $L$ has (F) at $A$ w.r.t. any such element.)

\bl
\label{FAB}
Let $A, B$ be different elements in $\hF$.  If $K$ is a maximal subgroup in $H$ not containing $B$, then there is a maximal subgroup $K'$ containing
$A$ for which $K' \cap B = K \cap B$. In particular, we have that $\mK_A \cap B \leq K \cap B \leq K$ for all maximal subgroups $K$. (So $\Phi \cap B = \mK_A \cap B$
if $A \ne B$ are elements in $\hF$, and whence for a fixed $B \in \hF$, $\mK_A \cap B$ is independent of the choice of $A \ne B$ in $\hF$.)
\el
{\em Proof}.\quad
As $\langle A,B\cap K\rangle \ne H$, there is a maximal subgroup $K'$ containing $\langle A,B\cap K\rangle$. Clearly, $K' \cap B = K \cap B$
(as otherwise $B \leq K'$, contradicting the projection lemma). \eop \\

Lemma \ref{FAB} implies that for any $C, D \ne C \in \hF$, and any $A \in \hF$ (where $A = C$ or $A = D$ are allowed),
\begin{equation}
A \cap \Phi = A \cap \mK_C \cap \mK_D.
\end{equation}
(Note that $\cap_{C \in \hF}C = \Phi$.) This is made slightly more explicit in the next lemma.

\bl
\label{FAB0}
Let $\mathbb{S} \leq L, M \leq H$, with $L$ and $M$ normal subgroups of $H$. Suppose that $L$ and $M$ have (F) at $A$ w.r.t. $B^*$ (and so also at $B$ w.r.t. $A^*$).
Then $L \cap M$ also has (F) at $A$ w.r.t. $B^*$.
\el
{\em Proof}.\quad
We have that
\begin{equation}
\vert L \cap M \vert = \frac{\vert L\vert \times \vert M\vert}{\vert LM\vert} = \frac{(\vert A\cap L\vert \times \vert L \cap B^*\vert) \times (\vert A \cap M\vert \times \vert B^* \cap M \vert)}{\vert LM\vert}.
\end{equation}
As $\vert L \cap M \cap A \vert = \frac{\vert L \cap A \vert \times \vert A \cap M\vert}{\vert (L \cap A)(A \cap M)\vert}$ and
$\vert L \cap M \cap B^* \vert = \frac{\vert L \cap B^* \vert \times \vert B^* \cap M\vert}{\vert (L \cap B^*)(B^* \cap M)\vert}$, it follows that
\begin{equation}
\label{eqpropf}
\vert L \cap M \vert =    (\vert L \cap M \cap A \vert \times \vert L \cap M \cap B^*\vert)\times\frac{\vert (L \cap A)(A \cap M)\vert\times\vert (L \cap B^*)(B^* \cap M)\vert}{\vert LM \vert}.
\end{equation}
Since $\vert L \cap M \vert \geq    \vert L \cap M \cap A \vert \times \vert L \cap M \cap B^*\vert$, we have that
\begin{equation}
\frac{\vert (L \cap A)(A \cap M)\vert\times\vert (L \cap B^*)(B^* \cap M)\vert}{\vert LM \vert} \geq 1.
\end{equation}
On the other hand, $(L \cap A)(A \cap M) \subseteq LM \cap A$ and $(L \cap B^*)(M \cap B^*) \subseteq LM \cap B^*$, while
$(LM \cap A)(LM \cap B^*) \subseteq LM$. It follows that
\begin{equation}
\frac{\vert (L \cap A)(A \cap M)\vert\times\vert (L \cap B^*)(B^* \cap M)\vert}{\vert LM \vert} =  1,
\end{equation}
so that
 $\vert L \cap M \vert =    \vert L \cap M \cap A \vert \times \vert L \cap M \cap B^*\vert$.
\eop

\bl
\label{FAB2}
Let $A, B$ ($\ne A$) $\in \hF$. Then $\mK_A \cap \mK_B$ satisfies $(F)$ at $A$ w.r.t. $B^*$. It follows that $\Phi = \mK_A \cap \mK_B$ for any choice of different $A, B \in \hF$.
\el

{\em Proof}.\quad
Consider $\mK_{A} \cap \mK_{B}$. Then applying Lemma \ref{FAB0} with $L = \mK_{A}$ and $M = \mK_B$, we conclude that $\mK_{A} \cap \mK_{B}$ has (F) at $A$ w.r.t. $B^*$.

Noting that $\Phi \leq \mK_A \cap \mK_B = (\mK_A \cap \mK_B \cap A)(\mK_A \cap \mK_B \cap B^*)$, the fact that  $\mK_A \cap \mK_B \cap A = \Phi \cap A \leq \Phi$ and
$\mK_A \cap \mK_B \cap B^* = \Phi \cap B^* \leq \Phi$, indeed implies that $\Phi = \mK_A \cap \mK_B$ for any choice of different $A, B \in \hF$.

\eop \\

\bc
Under the assumptions of this subsection, $\Phi = \Phi(H)$ is a normal $\hF$-factor in $H$.
\eop
\ec

\medskip
\subsection{Generic STGQs with (a)}

Now we suppose that $(\mS^x,H)$ is a generic STGQ of order $(s,t)$ with Kantor family $(\mF,\mF^*)$,
and  we assume to be in part (a) of the theorem. Let $A \ne B \in \mF$, and suppose that $abs = \phi$ with
$a \in A$, $b \in B$, $s \in \mathbb{S}$, and $\varphi \in \Phi$. (Note that such $a, b, s$ always exist.) Then $ab = \phi s^{-1} =: \varphi \in \Phi$.
As $ab \in \Phi$, $ab$ fixes
all points and lines of $\Gamma(\Phi)$. As $\Phi \unlhd H$, $a$ fixes all lines of $\Gamma(\Phi)$ of which the corresponding
orbits are subsets of $[B]$ (``$B$-lines''), and $b$ fixes all lines of $\Gamma(\Phi)$ of which the corresponding
orbits are subsets of $[A]$ (``$A$-lines''). As $\Gamma(\Phi)$ is a dual partial linear space, and as each $\Gamma(\Phi)$-point is
incident with one $A$-line and one $B$-line, it follows that $a$ and $b$ fix all $\Gamma(\Phi)$-points (and so all $\Gamma(\Phi)$-lines).
One immediately deduces that $a \in \Phi$ and $b \in \Phi$.  So for all $C \ne D \in \mF$, we have shown that
\begin{equation}
\Phi = (C \cap \Phi)(D^* \cap \Phi) = (D \cap \Phi)(C^* \cap \Phi).
\end{equation}

\bp
Under the assumptions of this subsection, $\Phi$ is a normal $\mF$-factor for $H$.
\eop \\
\ep

\subsection{Proof of Theorem \ref{Frattsub}}

 We have shown, under the assumptions of Theorem \ref{Frattsub}, that $\hF_{\Phi} := \{\Phi \cap A \vert A \in \hF\}$ defines a Kantor family of type $(s',t)$
in $\Phi$, with $s > s' > 1$, and $\hF^*_{\Phi} = \{\Phi \cap A^* \vert A \in \hF\}$. Also, since $x$ is regular, we have that $s'  = t = \sqrt{s}$ by Theorem \ref{2.2.2}.\\

config: arxiv


TITLE: Sub-Ads scale resolution of pluperfect tensor network
QUESTION [3 upvotes]: I am currently reading article "Bidirectional holographic codes and sub-AdS locality". [I]
This article presents tensor networks which are built from so called pluperfect tensors.
Authors claim that this kind of tensor network(tn) may "describe geometry at sub-Ads scale"
I am trying to show that we can define sub-Ads scale for this type of tn in case of const. time slice of $Ads_{3}$. (In my understanding it means that we can show that distance between two nodes in this tn can be smaller than $l_{Ads}$)
 I am using approach from article "Consistency Conditions for an AdS/MERA Correspondence"[II].
In this article authors compare lengths of two curves ($\gamma_{1,2}$) in Poincare Ads and in MERA correspondently. To calculate length of curves in MERA they introduce two length scales $L_{1}$ and $L_{2}$.  These distances will eventually be defined in terms of $l_{Ads}$.In this article it is also shown that no change of coordinates can make $L_{1}$ and $L_{2}$ be smaller than  $l_{Ads}$.

 In case of tensor network formed by pluperfect tensors for Poincare disc looks like(left image from [I], right one is toy model,I am currently working with)
To build analogy with MERA I am trying to introduce two length scales(I introduce only two because tn should describe 2-dim plane).
Let's define black lines to represent (vertical) distance between layers ($L_{2}$) and coloured ones - (horizontal) distance between nodes within layer ($L_{1}$). The problem is that this definition doesn't give us consistent way to define number of nodes as a function of layer. (For 1st we 5 nodes, for 2nd - 25, for 3d - 95). This problem becomes even worse if I try to use this approach for left picture.
  Is there any strict way to define such tensor network layer by layer?
May be there exists simpler way to prove that this pluperfect tensor network admits sub-Ads scale resolution. If there is one I would like to know about it as well.

REPLY [0 votes]: In order to answer this question one needs to follow different strategy.
 First let us consider tensor network formed by so called perfect tensors(PTN). We will investigate problem of sub-AdS scale resolution for it. This kind of TN is introduced in the article Toy models for the bulk/boundary correspondence. Once this problem is resolved the same problem for tensor network formed by pluperfect tensors may  be seen just as a generalization of above example.
In case of PTN one starts from constant time slice of $AdS_3$ given by Poincare disc. Now one needs to consider uniform tessellation of this disc. General idea is that Poincare disc in this case will be filled by polygons of equal area. There are many different ways to do so depending of particular form of polygon. Here is an example:

If we associate one perfect tensor to each polygon and connect them we will obtain following tensor network:

So now we can say that there is certain area associated to each tensor in tensor network. It may be shown that this area has to be of order of $ l^2_{AdS}$.This is why we can conclude that we are not allowed to have sub-AdS scale in our tensor network in aforementioned setup.
Obvious resolution of this "problem" will be to introduce non uniform tessellation of Poincare disc. However if we try to do so for PTN we will almost instantly arrive to following problem.  By increasing number of tensors in the bulk we increase number of uncontracted(red) bulk legs. Hence at some point dimension of bulk Hilbert space will become larger than dimension of boundary Hilbert space. Yet in this case our tensor network can't be used to mimic AdS/CFT anymore.
If instead of perfect tensors we consider pluperfect ones we will be able to avoid this problem. Because of certain properties of new setup one will be able to define so called "physical" bulk Hilbert space which will always have smaller dimension than boundary one. That is why nonuniform tessellation is allowed for pluperfect tensor networks. Hence one can have sub-AdS scale resolution in this new setup. Example of nonuniform tessellation may be seen in following image:


TITLE: An unknown pattern in $π^k$.
QUESTION [0 upvotes]: While trying to find a pattern in $π$, by taking the idea from $n$ such that the digits immediately after the decimal point of $\pi^n$ give $n$ again
By using wolfram alpha calculator first I am writing the examples of my invesgation
π3 = 31.00627668....
π5 = 306.0196847....
π9 = 29809.09933344....
π10 = 93648.0474760830....
π11 = 294204.01797389059....
π18 = 888582403.07126338067....
As we can see that in each example after a decimal place there is '0'. But I don't know any math programming language to calculate further more values.
Now the question is that is there are infinite values 'k' such that:
πk = .....abcdefhg.0wpywqx....      (in which after decimal place '0' will be there)
Any hint or solution is welcome. Thanks.

REPLY [2 votes]: Here's a histogram of the first digit (to the right of ".") values for $\pi^n$ for $1 \leq n \leq 500$:  Clearly the results are random and there is nothing special about the value $0$ in that digit position:

Here's the Mathematica code for generating the histogram:
piList = Table[N[π^n, 500], {n, 500}];
digitList = IntegerPart /@ (10 (FractionalPart /@ piList));
Histogram[digitList]


TITLE: Question about $2 \times 2$, determinant one matrices over a Euclidean domain
QUESTION [3 upvotes]: Let $R$ be a Euclidean domain. Let $a,b,c, d \in R$ with $ad - bc = 1_R = 1$. That is, the matrix:
$$ A = \left( \begin{array}{cc}
a & b  \\
c & d \\
\end{array} \right)  \in \operatorname{Mat}_{2}(R)$$
has determinant one.

Using elementary row and column operations, show that $A$ is equivalent to the identity matrix $I_2 \in \operatorname{Mat}_{2}(R).$

The elementary row/column operations are:

Interchanging two rows or columns,
Adding a multiple of one row or column to another,
Multiplying any row or column by a unit.

This does not seem too difficult at its core, and I have worked about several examples by setting $R = \mathbb{Z}$. However, I am having a difficult time crafting the argument for completely general $R$.
I am familiar with the concept of (and proof of) Smith normal form for matrices over a principal ideal domain. However, what I need to show here is more specific, since the particular diagonalization I seek is the identity matrix.
I do know that the equation $ad - bc = 1$ implies that $\gcd(a,b) = \gcd(c, d) = 1$. And, in an arbitrary Eulcidean domain, we do have the Euclidean algorithm available to us. So, my gut feeling is that this should be the strategy to make the proof work.
For instance, we invoke the Euclidean algorithm to get a sequence of equations:
$$a = q_0b + r_1,$$
 $$b = q_1r_1 + r_2,$$
 $$r_1 = q_2r_2 + r_3,$$
$$\vdots$$  $$r_{n-1} = q_n \gcd(a,b) + 0.$$
And we can use this sequence to start performing operations on $A$:
$$ A = \left( \begin{array}{cc}
a & b  \\
c & d \\
\end{array} \right)
=\left( \begin{array}{cc}
q_0b + r_1 & b  \\
c & d \\
\end{array} \right) \sim \left( \begin{array}{cc}
r_1 & b  \\
c - dq_0 & d \\
\end{array} \right) = \left( \begin{array}{cc}
r_1 & q_1r_1 + r_2   \\
c - dq_0 & d \\
\end{array} \right) \sim
\left( \begin{array}{cc}
r_1 & r_2  \\
c - dq_0 & d - cq_1 + dq_0q_1 \\
\end{array} \right). $$
So we have in a sense "simplified" the first row of $A$, and we could in practice continue this until the first row is of the form:
$$\left( \begin{array}{cc}
\gcd(a,b) & 0  \\
\ast & \ast \\
\end{array} \right) = \left( \begin{array}{cc}
1 & 0  \\
\ast & \ast \\
\end{array} \right). $$
Am I on the right track here? I am worried because the terms in the second row could become quite complicated.
Hints or solutions are greatly appreciated.

REPLY [1 votes]: Hint $\ $ Using row/col operations a la Euclid's algorithm you can reduce to a matrix where $\,a\mid b,c.\,$ Therefore $\,a\mid ad-bc = 1.\,$ Hence wlog $\,a = 1,\,$ so,  we can reduce $\,b,c\,$ to $\,0,\,$ so $\,d = 1.$


TITLE: Existence of $opt^c$-approximation of Dominating Set with $c < 1$?
QUESTION [10 upvotes]: Consider the Dominating Set problem in general graphs, and let $n$ be the number of vertices in a graph. A greedy approximation algorithm gives an approximation guarantee of factor $1 + \log n$, i.e. it's possible to find in polynomial-time a solution $S$ such that $|S| \leq (1 + \log n) opt$, where $opt$ is the size of a minimum dominating set. There are bounds showing that we cannot improve the dependency on $\log n$ much http://www.cs.duke.edu/courses/spring07/cps296.2/papers/p634-feige.pdf.
My question: is there an approximation algorithm which has a guarantee in terms of $opt$ instead of $n$? In graphs where $n$ is very large with respect to the optimum, a factor-$\log n$ approximation would be much worse than a factor $\log opt$ approximation. Is something like that known, or are there reasons why this cannot exist? I am happy with any polynomial-time algorithm which produces a solution $S$ such that $|S| \in O(opt^c)$ for some constant $c$.

REPLY [15 votes]: I think it is still open if Dominating Set or Hitting Set have a f(OPT) approximation for some (nontrivial) function f. This is should be a very difficult (and possible deep) question to answer. I consider it the most exciting question in parameterized approximation (along with the analogous question for Clique). You might want to have a look at my survey [1] that discusses this. Note that it is shown in the more recent paper [2] that "monotone circuit satisfiability for weft-2 circuits", a problem which is more general than Dominating Set, does not have f(OPT) approximation for any f.
[1] D. Marx. Parameterized complexity and approximation algorithms.
The Computer Journal, 51(1):60-78, 2008.
[2] D. Marx. Completely inapproximable monotone and antimonotone parameterized problems.
In Proceedings of the 25th Annual IEEE Conference on Computational Complexity, Cambridge, Massachusetts, 181-187, 2010.


TITLE: How do you calculate the probability of the difference between two normal distribution
QUESTION [1 upvotes]: If I have two normally distributed graphs, one with a mean of 70 and standard deviation of 3, and the other with a mean of 74 and standard deviation of 4.5, how would I go about calculating the probability of a random point from the first distribution being 2 larger than a random point on the second distribution?

REPLY [1 votes]: I will show how to work a similar problem, and let you make the required
changes for your own problem. I hope you can use my work for the alternative
problem as an outline for doing yours. Please check each step with formulas
and theorems in your text or notes.

Note: @DavidQuinn is correct, that the wording of
  your problem is faulty (unless it is intended to be trivial); to make sense it needs to say
  something like "larger by at least two." Also, you need to say that
  $X$ and $Y$ are independent random variables. Please look into that.

Alternative Problem: Suppose $X \sim \mathsf{Norm}(70, 3)$ and (independently)  $Y \sim \mathsf{Norm}(74, 4).$ [The second argument in my notation $\mathsf{Norm}$ is the standard deviation.]
Then find the distribution of $D = X - Y$ and $P(D > 2).$
Mean of the sum (or difference) of two random variables. A general formula states that $E(aX + bY) = aE(X) + bE(Y).$ In the
current problem $a = 1$ and $b = -1,$ so $E(X+Y) = 70-74 = -4.$
Variance of the sum (or difference) of two independent random variables. Another general formula states that for independent random variables
$X$ and $Y,$ one has $Var(aX + bY) = a^2Var(X) + b^2Var(Y).$ In the
current problem, this gives $Var(X - Y) = 3^2 + 4^2 = 25,$ so
$SD(X - Y) = \sqrt{25} = 5.$ Notice that the variances are added
even though one random variable is subtracted from the other.
Distribution of the sum of independent normal random variables. A third general result states that the sum (or difference) of two
independent normal random variables is is another normal random
variable with mean and SD given by the two relationships above.
Thus $D = X - Y \sim \mathsf{Norm}(-4, 5).$
The last step is to use software or printed standard normal tables to find $P(D > 2).$ [If you use standard normal tables, you will have to standardize $D.$
I assume you have done that before.] The answer to the current problem (from software) is $P(D > 2) = 0.1149.$ Because $Y$ has a larger mean than $X,$ it
is not surprise that this probability is considerably smaller than $1/2.$


TITLE: Primitive integral solutions to $x^2+y^3=z^2$
QUESTION [7 upvotes]: The Diophantine equation
$$x^2+y^3=z^2$$
has solutions $(\pm 1,2,\pm 3)$ and $(\pm 13,3,\pm 14)$.
Brown [Int. Math. Res. Not. IMRN 2012, no. 2, 423–436; MR2876388] states that "there are infinitely many parameterized solutions with $xyz \neq 0$ and $z^2 \neq 1$ when $n \leq 5$" to the equation
$$x^2 +y^3 = z^n,$$
but does not give a reference. I am looking for a reference that catalogs all primitive solutions.

REPLY [4 votes]: A warm-up:
$$ \left(\frac{s^n-t^n}2\right)^2 +\ (s\cdot t)^n\ =\
        \left(\frac{s^n+t^n}2\right)^2 $$
provides integer solutions for $\,\ s\equiv t \mod 2 $.
(Sorry, I couldn't help it).

REMARK   More generally, going in the abc direction:
$$ \left(\frac{s^m-t^n}2\right)^2 +\ s^m\cdot t^n\ =\
        \left(\frac{s^m+t^n}2\right)^2 $$
for $\ s\equiv t\equiv 1 \mod 2,\ $ and $\ \gcd(s\ t) = 1$.


TITLE: About list of discriminants of real quadratic fields with narrow class number 1?
QUESTION [0 upvotes]: I have a couple of questions regarding the list of discriminants of real quadratic fields with narrow class number 1.
The sequence A003655 in OEIS portraits a list of discriminants of real quadratic fields with narrow class number 1. In the sequence there is no indication that the list is complete. Q1: Is that the case? Q2: In any case, could you point to some relevant references about this list?
One further question following the first responses.

There are finitely many determinants $\Delta=4m>0$ with $m$ square-free and narrow class number 1. $\Delta=8$ is one such determinant, is it the only one? (references?).

Thanks in advance, and regards,
Guillermo

REPLY [3 votes]: The class number 1 problem was solved only for rather special positive fundamental discriminants. These are cases where the regulator of the underlying real quadratic field is automatically small, so that by Siegel's theorem the class number is automatically large (although non-effectively so). In particular, Biró compiled a full list of $d$'s with $h(d)=1$ when $d=n^2+4$ or $d=4n^2+1$ (see Acta Arith. 106 (2003), 85–104; Acta Arith. 107 (2003), 179–194). This result strikes as a real analogue of Gauss' original class number 1 problem (for negative fundamental discriminants).


TITLE: Why does Kramer's theorem only apply to systems with an odd number of fermions?
QUESTION [6 upvotes]: Kramer's theorem states that: In a time reversal symmetric system, where $T^2=-1$, all energy levels are (at least) doubly degenerate.
I claim to have proven this in a simple way.
We first establish that $|\Psi\rangle$ and $T|\Psi\rangle$ are different states. Let's assume the opposite, that the states are equivalent, then, $$T|\Psi\rangle=e^{i\phi}|\Psi\rangle.$$
Which implies that, $$T^2|\Psi\rangle=Te^{i\phi}|\Psi\rangle=e^{-i\phi}T|\Psi\rangle=e^{-i\phi+i\phi}|\Psi\rangle=|\Psi\rangle.$$
From which it follows that $T^2=1$, contrary to our requirement. Thus $|\Psi\rangle$ and $T|\Psi\rangle$ are unique states. Then trivially, $H|\Psi\rangle=E|\Psi\rangle$ and $$HT|\Psi\rangle=TH|\Psi\rangle=ET|\Psi\rangle.$$ So that both $|\Psi\rangle$ and $T|\Psi\rangle$ have the same energy! The system is doubly degenerate.
But nowehere was it assumed that the system must have an odd number of fermions. What am I missing here?

REPLY [6 votes]: You are assuming $T^2=-1$. You can have $T^2=1$ too, in which case it is perfectly fine to have $\psi$ and $T\psi$ be the same state, as you showed in your question. But it turns out the states with $T^2=-1$ are just the states with half-integer angular momentum, as I'll show.
$T$ anticommutes with angular momentum (since it should flip the sign). If I consider a particle at rest, and use a basis in terms of eigenstates of z-angular momentum $\psi_\sigma$, we have
$$T\psi_\sigma = \zeta_\sigma \psi_{-\sigma}$$
where $\zeta_\sigma$ is some phase factor that might depend on $\sigma$. Notice we see right away that we can't have $T\psi\propto\psi$ for a particle with half integer spin, since there is no $\sigma=0$ state. That is more or less what you wanted to see I think but let's go a little further.
If we apply the raising and lowering operator $J_\pm=J_x\pm iJ_y$
$$J_\pm T \psi_\sigma=\zeta_\sigma J_\pm \psi_{-\sigma}\propto \zeta_\sigma \psi_{-\sigma\pm 1}$$
where there's also that normalization factor with a square root you can look up. But we also have,
$$J_{\pm}T\psi_\sigma = -TJ_\mp \psi_\sigma \propto -\zeta_{\sigma\mp 1}\psi_{-\sigma\pm 1} $$
The normalization factor is the same in both cases so
$$\zeta_{\sigma}=-\zeta_{\sigma\mp 1}$$
So the sign flips each time you raise or lower z-angular momentum, so we can write (up to an overall phase we can absorb into our definition of $\psi$)
$$\zeta_\sigma = (-1)^{j-\sigma}$$
where $j$ is the total angular momenta. So we have
$$T^2\psi_\sigma=(-1)^{j+\sigma}(-1)^{j-\sigma}\psi_\sigma=(-1)^{2j}\psi_\sigma$$
So $T^2=\pm 1$ depending on whether $j$ is an integer or half integer.


TITLE: Does tightness imply exponential tightness?
QUESTION [2 upvotes]: I just learned about the notion of exponential tightness which is described as a generalisation of tightness. The first question that pops into my head is: So does tightness imply exponential tightness?
I haven't quickly found any answer for that, so I guess the answer must be obvious. Still, I'm heaving trouble verifying if it actually is true.
Let $X$ be a topological space equipped with its Borel-$\sigma$-algebra $\Sigma$ (the one generated by all open subsets). A series $(\mu_n)_n$ of probability measures on $(X,\Sigma)$ is called tight if for each $\epsilon>0$ there exists a compact subset $K_\epsilon\subset X$ (which is of course in $\Sigma$) such that $\mu_n(K_\epsilon^c)<\epsilon$ is true for all $n$ ($K_\epsilon^c$ means $X\setminus K_\epsilon$). $(\mu_n)_n$ is called exponentially tight on a scale $(\gamma_n)_n$ if for each $s>0$ there exists a compact subset $K_s\subset X$ such that $$\limsup_{n\rightarrow\infty} \frac{1}{\gamma_n} \log \mu_n(K_s^c) < -s.\tag{1}$$
I should add that a scale is a series $(\gamma_n)_n$, $\gamma_n\in(0,\infty)$, such that $\gamma_n\rightarrow \infty$ if $n\rightarrow \infty$.
Now I can put my question a bit more accurately: If $(\mu_n)_n$ is tight, is it also exponentially tight on every scale? Or can we at least impose some condition on the decay of the scale in order to conclude exponential tightness?
If I take $(\mu_n)$ to be tight and look at (1), I can see that we can choose $K_s$ such that $\mu_n(K_s^c)$ is bounded by an arbitrarily small $\epsilon>0$, so $\log \mu_n(K_s^c)$ gets (negatively) large and competes with the decay of $\frac{1}{\gamma_n}$ which by assumption converges to $0$.
Here I'm a little bit confused about how to proceed and conclude anything meaningful about the relation of tightness and exponential tightness. Can someone help me clear that up?

REPLY [1 votes]: No, tightness does, in general, not imply exponential tightness. Exponential tightness means that
$$\mu_n(K_s^c) \leq e^{-s \gamma_n} \qquad \text{for $n \gg 1$};$$
this is a much stronger assumption than tightness.
Example: Let $\mu$ be an arbitrary finite measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ and consider the sequence $\mu_n := \mu$, $n \in \mathbb{N}$. Then the sequence $(\mu_n)_{n \in \mathbb{N}}$ is tight (this follows e.g. from the continuity of the measure from below). On the other hand,
$$\lim_{n \to \infty} \frac{1}{\gamma_n} \underbrace{\mu_n(K^c)}_{\mu(K^c)} = 0$$
for any scale $(\gamma_n)_{n \in \mathbb{N}}$. Consequently, $(\mu_n)_{n \in \mathbb{N}}$ is not exponentially tight.


TITLE: For compact $A$, $\inf\{\varrho(y,x) : y \in A\}=\varrho(a,x)$
QUESTION [1 upvotes]: I need help with prooving that if non empty $A$ $\subset(X,\varrho)$ is compact, then:
$(\forall x \in X) (\exists a \in A)  \inf\{\varrho(y,x) : y \in A\}=\varrho(a,x) $
I found this solution:
$\varrho_{A}(x) =  \inf\{\varrho(y,x) : y \in A\}=\lim_{n} \varrho(y_{n},x)$ for some sequence $y_{n} \in A$.
We pick subsequence $y_{n_{k}}\rightarrow a$ and then:
$\varrho_{A}(x) = \lim_{n}\varrho(y_{n},x)=\lim_{n} \varrho(y_{n_{k}},x)=\varrho(a,x)$
I don't get this solution. Is there any other way to prove it?
Also, wouldn't it be enough if $A$ was closed, not compact? The way of proving that $\inf\{\varrho(a,b): a\in A, b\in B\}=\varrho(A,B)$ for compact $A$ and $B$ would be similar?
Thanks in advance!

REPLY [2 votes]: It seems the following.

$\varrho_{A}(x) = \lim_{n}\varrho(y_{n},x)=\lim_{n} \varrho(y_{n_{k}},x)=\varrho(a,x) $

This solution seems to be OK.

Is there any other way to prove it?

Yes. If a point $x\in X$ is fixed then $\varrho(x, a)$ is a continous function on a compact set $A$, therefore it attains its infimum at some point $a\in A$.

Also, wouldn't it be enough if $A$ was closed, not compact?

No. Let $(X,\varrho)$ be a $\Bbb Q$ endowed with the standard metric, $A=\{a\in\Bbb Q: a>\sqrt{2}\}$, and $x=0$. Then $A$ is a closed subset of the space $X$, but the distance from the point $x$ to a set $A$ is not realized by a point $a\in A$.

The way of prooving that $\inf\{\varrho(a,b): a\in A, b\in B\}=\varrho(A,B)$ for compact $A$ and $B$ would be similar?

Yes.


TITLE: What are these strange rays in the sky?
QUESTION [27 upvotes]: While driving, I noticed the sky had weird rays that appeared converge at some point on the horizon.  I took a few pictures in hopes I could figure out what they were.  There was a storm coming from the other direction if that's relevant information.

They don't show up quite as well on the pictures, but they were definitely visible in real-life so not an artifact from the camera or anything.

REPLY [24 votes]: Those are definitely anticrepuscular rays. The Sun was in the antipodes of the point of convergence of those rays in the sky, when you took that image. Here you have a similar observation:

These are formed just like crepuscular rays; clouds in the direction of the Sun project parallel shadows that cross the entire sky over your head into the opposite direction where the Sun was setting. You see them converging instead of being parallel because of the spherical projection of your spherical field of view (you can reinterpret this as the shadows are closer when they go over your head and are farther from you at the horizon so that rays that are parallel seem to converge at some point at infinity).


TITLE: Why are Markov chains better than naive algorithm for Random picking?
QUESTION [3 upvotes]: I have a question about the problem of coming up with a random distribution of non-overlapping coins in a given two dimensional space. I don't understand why using the random sequential adsorption technique (picking an initial random position for one coin, then do the same for the next; if it overlaps, keep trying until you find a non-overlapping position for it and so on) is not equivalent to using Markov chains to solve the same problem.
Thanks!

REPLY [3 votes]: The reason is that the successive picking process doesn't produce a uniform distribution on the possible center positions. Some positions are much more likely to be found by successive picking than others. For an obvious example, consider a near-packed configuration like oranges piled in a crate: in this case you will never make any such configuration by successive picks, you will just fail to fit that many oranges in the crate. But Monte-Carlo will eventually lead the oranges to settle into the right pattern, either by cooling and shrinking from a hot state in a big box, or by adiabatically raising the repulsion strength to infinity. The many-steps of the Monte-Carlo guaratee tha the distribution becomes uniform, because Markov chains converge to a unique stationary distribution generically.
To see specifically what's wrong with successive picks, go to one dimension, and consider putting two intevals of length 1 inside a region of length 3. If you plot the allowed positions of the two left most points you see that the distribution of the position of either one is far from uniform. Note that in this case, sequential picking will always succeed, but it will produce two intervals that are much closer than they have to be.

REPLY [1 votes]: This is a problem studied in Gravitational Waves (GW) data analysis for example, where you need to sample the surfaces of possible (GW) signals to obtain a finite subset of the signals to compare (see matched-filtering) to the output of your detector.
See for example this paper and references there.
The algorithms for non overlapping "coins" go under the name of "stochastic algorithms" in the GW literature, and they can be show to be more efficient than pure random sampling, but less efficient that regular lattice distributions (only true for the low dimensional case).
The point is that at least in 2 dimension it's possible to start from a stochastic distribution and start moving the "coins" around until they crystallize and reproduce an hexagonal lattice (which is know to be the optimal placement).
The subject is still open because: optimal placement are not know for higher dimensional spaces, the (hyper)-surface my be curved, the metric tensor might be difficult to obtain or difficult to diagonalize... etc.
So in general the problem is still open and the answer will change from case to case, but in 2d for a flat surface given in coordinates where the metric tensor is the identity the problem is basically solved.
Regarding your question in particular the efficiency of your algorithm can be estimated and it's worse then an hexagonal lattice. Since with a Markov chain you can get closer to a lattice the two algorithm have clearly different performances.
Notice that I don't know up to which dimension things have been actually tested.
"Weird" things can happen in higher dimensions: if you allow the coins (hyper-spheres) to overlap the random distribution becomes very efficient, but boundary effects become also determinant.


TITLE: Induction: show that $\sum\limits_{k=1}^n \frac{k}{2^k} = 2 - \frac{n+2}{2^n}$ for all n $\in Z_+$
QUESTION [0 upvotes]: So the question in my textbook is:
Show by induction that $\sum\limits_{k=1}^n \frac{k}{2^k} = 2 - \frac{n+2}{2^n}$ for all $n$ $\in Z_+$.
My attempt at a solution:
First of all $Z_+ = 1, 2, 3, 4, 5, 6, 7, 8 ...$
So I start to make sure that the base case is right... For that I choose $k = 1$
This gives us:
$$\sum\limits_{k=1}^1 \frac{1}{2^1} = 2 - \frac{1+2}{2^1} \rightarrow \frac{1}{2} = 2 - \frac{3}{2}$$
The base case does work.
The next step is to take make the inductive assumption that it is true for $n = k$. This gives us:
$$\frac{1}{2^1} + \frac{2}{2^2} + ... +  \frac{k}{2^k}  = 2 - \frac{k+2}{2^k}$$
The next step is to show that $n = k + 1$ holds:
$$\frac{1}{2^1} + \frac{2}{2^2} + ... +  \frac{k}{2^k} + \frac{k+1}{2^{k+1}}  = 2 - \frac{k+1+2}{2^{k+1}}$$
By using the induction hypothesis we have:
$$\frac{1}{2^1} + \frac{2}{2^2} + ... +  \frac{k}{2^k} + \frac{k+1}{2^{k+1}}  = 2 - \frac{k+2}{2^k} + \frac{k+1}{2^{k+1}}$$
The $RHS$ can be written as:
$$2 - \frac{k+2}{2^{k}} + \frac{k+1}{2^{k+1}}$$
How do I move on from here and show that:
$$2 - \frac{k+2}{2^{k}} + \frac{k+1}{2^{k+1}} = 2 - \frac{k+1+2}{2^{k+1}}$$
Thank you!

REPLY [1 votes]: It works in this way. Consider this (I will skip some passages to make the answer shorter, but they should be easy enough).
$$
S(x)=\sum_{k=1}^nkx^k
$$
we are interested in $S(1/2)$.
Now is easy to see that
$$
kx^k=x\frac{d}{dx}x^k
$$
so we can write
$$
S(x)=\sum_{k=1}^n x\frac{d}{dx}x^k=x \frac{d}{dx}\sum_{k=1}^n x^k
$$
but
$$
\sum_{k=1}^n x^k = \frac{x(x^n-1)}{x-1}
$$
so taking the derivative of this sum we get
$$
S(x)=x\frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2}
$$
Now we can calculate $S(1/2)$ and we get
$$
S(1/2) = \frac{1}{2}\left( 2\frac{n}{2^n} - 4\frac{n+1}{2^n} +4\right)
$$
and if you simplify this you get your result. I hope you can complete the missing passage. That is only algebra.


text:
\begin{document}
\begin{center}

{ \large\bf The Bailey transform and Hecke-Rogers identities for\\[5pt]
 the universal mock theta functions }
\end{center}

\vskip 5mm

\begin{center}
 {  Kathy Q. Ji}$^{1}$ \quad
and  \quad { Aviva X.H. Zhao}$^{2}$ \vskip 2mm

     Center for Combinatorics, LPMC-TJKLC\\[2pt]
   Nankai University, Tianjin 300071, P.R. China

   \vskip 2mm

 Email:  $^1$ji@nankai.edu.cn, \quad $^2$zhaoxiaohua@mail.nankai.edu.cn
\end{center}

\vskip 6mm \noindent {\bf Abstract.} Recently, Garvan obtained two-variable Hecke-Rogers identities for three universal mock theta functions $g_2(z;q),\,g_3(z;q),\,K(z;q)$ by using   basic hypergeometric functions,  and he proposed a problem of finding  direct proofs of these  identities by using Bailey pair technology. In this paper, we give proofs of Garvan's identities  by applying Bailey's transform with the conjugate Bailey pair of Warnaar   and three Bailey pairs deduced from two special cases of $_6\psi_6$ given by Slater. In particular, we obtain a compact form of two-variable Hecke-Rogers identity related to $g_3(z;q)$, which imply the corresponding identity given by Garvan. We also extend these two-variable Hecke-Rogers  identities into infinite families.

\noindent {\bf Keywords}: universal mock theta function, Hecke-Rogers identity, conjugate Bailey pair, Bailey pair,
Bailey transform

\noindent {\bf AMS Classifications}: 05A17, 11P83, 05A30

\section{Introduction}

This paper is concerned with two-variable Hecke-Rogers identities for the universal mock theta functions due to Garvan \cite{Garvan-2014}. Recall that universal mock theta functions are defined by  Gordon and McIntosh \cite{Gordon-2012} and Hickerson \cite{Hickerson-1988}  as
\begin{align*}
g_2(z;q)&=  \sum_{n=0}^{\infty}\frac{(-q;q)_nq^{\frac{1}{2}n(n+1)}}
  {(z;q)_{n+1}(z^{-1}q;q)_{n+1}},\\[5pt]
g_3(z;q)&=\sum_{n=0}^{\infty}
  \frac{q^{n(n+1)}}{(z;q)_{n+1}(z^{-1}q;q)_{n+1}},\\[5pt]
K(z;q) &=\sum_{n=0}^{\infty}\frac{(-1)^n(q;q^2)_nq^{n^2}}
  {(zq^2;q^2)_n(z^{-1}q^2;q^2)_n}.
\end{align*}
Here and throughout the paper,   we  adopt the standard notation on $q$-series \cite{Andrews-1976, Gasper-2004}.
\[
(a;q)_\infty =\prod_{n=0}^\infty (1-aq^n),
\]
\[ (a;q)_n=\frac{(a;q)_\infty}{(aq^n;q)_\infty},\\[3pt]
\]
\[(a_1,a_2,\ldots, a_k;q)_n=(a_1;q)_n(a_2;q)_n\cdots(a_k;q)_n,\]
and
\[(a_1,a_2,\ldots, a_k;q)_\infty=(a_1;q)_\infty(a_2;q)_\infty\cdots(a_k;q)_\infty.
\]

  Gordon and McIntosh \cite{Gordon-2012} and Hickerson \cite{Hickerson-1988}   noticed that all of Ramanujan's mock theta functions can be written in terms of  $g_2(z;q)$ and $g_3(z;q)$, that is why these two functions are called universal mock theta functions. The function $K(z;q)$ first appears in Ramanujan's lost notebook \cite{Andrews-1979} and is related by modular transformation to the universal mock theta function $g_2(z;q)$ in \cite{McIntosh-2012}.

In  \cite{Garvan-2014}, Garvan first
 obtained four Hecke-Rogers identities for these three universal mock theta functions, which  can be viewed as  two-variable generalizations of Hecke-Rogers identities.  These identities   can also be used to establish  Hecke-Rogers-type double sum representation for each of Ramanujan's mock theta functions.

More specifically,  Garvan \cite{Garvan-2014} considered the Dyson rank function $R(z;q)$ defined by
  \[R(z;q)=\sum_{n=0}^{\infty}
 \frac{q^{n^2}}{(zq;q)_n(z^{-1}q;q)_n},
  \]
  which has the following relation with $g_3(z;q)$.
   \[g_3(z;q)=-\frac{1}{z}+\frac{1 }{z(1-z)}R(z;q).\]
He obtained  the following Hecke-Rogers  identity  for $R(z;q)$.
\begin{align}\label{Garvan1}\nonumber
  &(zq, z^{-1}q,q;q)_{\infty}R(z;q)\\[3pt]\nonumber
  &=(zq, z^{-1}q,q;q)_{\infty}\sum_{n=0}^{\infty}
 \frac{q^{n^2}}{(zq;q)_n(z^{-1}q;q)_n}\\[3pt]
    &=\frac{1}{2}\sum_{n=0}^{\infty}\left(\sum_{j=0}^{[n/2]}(-1)^{n+j}(z^{n-3j}
  +z^{3j-n})q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n-j)}\right.\nonumber \\
  &\quad\hskip 2cm\left.+\sum_{j=1}^{[n/2]}(-1)^{n+j}(z^{n-3j+1}
  +z^{3j-n-1})q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n+j)}\right).
  \end{align}
  Putting $z=1$ in  \eqref{Garvan1},    we   see that  the Hecke-Rogers identity \cite{Hecke-1959,Rogers-1894} follows by noting  $R(1;q)=1/(q;q)_\infty$.
  \begin{equation}
  (q;q)_\infty^2=\sum_{n=0}^\infty \sum_{j=-[n/2]}^{[n/2]}(-1)^{n+j}q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n+j)}.
  \end{equation}
 Garvan \cite{Garvan-2014} also considered the overpartition rank function $H(z;q)$ defined by
 \begin{equation}H(z;q)=  \sum_{n=0}^{\infty}\frac{(-1;q)_nq^{\frac{1}{2}n(n+1)}}
  {(zq;q)_n(z^{-1}q;q)_n},
 \end{equation}
 which has the following relation with $g_2(z;q)$ \cite{McIntosh-2012}.
 \begin{equation}\label{r-h-g2}
 (1+z)H(z;q)=(1-z)+2z(1-z)g_2(z;q).
 \end{equation}
 The following Hecke-Rogers identity for $H(z;q)$ is given by Garvan \cite{Garvan-2014}.
  \begin{align}\nonumber
  &(1+z)(zq, z^{-1}q,q;q)_{\infty}
 H(z;q)\\[3pt]\nonumber
 &\hskip 0.5cm=(1+z)(zq, z^{-1}q,q;q)_{\infty}
  \sum_{n=0}^{\infty}\frac{(-1;q)_nq^{\frac{1}{2}n(n+1)}}
  {(zq;q)_n(z^{-1}q;q)_n}\\[3pt]
  &\hskip 0.5cm =\sum_{n=0}^{\infty}\sum_{\mid m\mid\leqslant[n/2]}
  (-1)^{n+m}(z^{n-2|m|+1}+z^{2|m|-n})
  q^{\frac{1}{2}(n^2-2m^2)+\frac{1}{2}n} \label{Garvan2}\\[3pt]
  &\hskip 0.5cm =\sum_{n=0}^{\infty}\sum_{\mid m\mid\leqslant[n/3]}
  (-1)^{n}(z^{n-4|m|+1}+z^{4|m|-n})
  q^{\frac{1}{2}(n^2-8m^2)+\frac{1}{2}n}. \label{Garvan3}
  \end{align}
  Letting $z=1$ in \eqref{Garvan2} and \eqref{Garvan3}, and using the fact that $H(1;q)=(q^2;q^2)_\infty/(q;q)^2_\infty$,   we  get  the following two Hecke-Rogers type identities firstly given by  Kac and Peterson \cite{Andrews-1984, Kac-Peterson-1980}.
  \begin{align}
  (q;q)_\infty(q^2;q^2)_\infty& =\sum_{n=0}^{\infty}\sum_{\mid m\mid\leqslant[n/2]}
  (-1)^{n+m}q^{\frac{1}{2}(n^2-2m^2)+\frac{1}{2}n}, \\[3pt]
(q;q)_\infty(q^2;q^2)_\infty&  =\sum_{n=0}^{\infty}\sum_{\mid m\mid\leqslant[n/3]}
  (-1)^{n} q^{\frac{1}{2}(n^2-8m^2)+\frac{1}{2}n}.
  \end{align}

  For $K(z;q)$, Garvan \cite{Garvan-2014} established the following identity.
  \begin{align}
  &(zq^2,z^{-1}q^2,q^2;q^2)_{\infty}
  K(z;q) \nonumber \\[3pt] \nonumber
  &\hskip 0.2cm =(zq^2,z^{-1}q^2,q^2;q^2)_{\infty}
  \sum_{n=0}^{\infty}\frac{(-1)^nq^{n^2}(q;q^2)_n}
  {(zq^2;q^2)_n(z^{-1}q^2;q^2)_n}\\[3pt]
  &\hskip 0.2cm =\sum_{n=0}^{\infty}\Big(\sum_{m=0}^n(-1)^nz^{m-n}
  q^{\frac{1}{2}(2n^2-m^2)+\frac{1}{2}(2n-m)}
  +\sum_{m=1}^n(-1)^nz^{n-m+1}q^{\frac{1}{2}(2n^2-m^2)
  +\frac{1}{2}(2n+m)}\Big). \label{Garvan4}
\end{align}
 Setting $z=1$ in \eqref{Garvan4}, and noting that  $K(1;q)=(q;q)_\infty/(q^2;q^2)^2_\infty$,  we   obtain  the following   Hecke-Rogers type identity of Bressoud \cite{Bressoud-1986}.
\begin{equation}
(q;q)_\infty(q^2;q^2)_\infty=\sum_{n=0}^\infty \sum_{m=-n}^n(-1)^n q^{\frac{1}{2}(2n^2-m^2)  +\frac{1}{2}(2n+m)}.
\end{equation}

In \cite{Garvan-2014}, Garvan showed that \eqref{Garvan2} and \eqref{Garvan3} are equivalent with the aid of a transform of Mine \cite{Milne-1994}. He then proved \eqref{Garvan1}, \eqref{Garvan2} and \eqref{Garvan4} are valid   by showing the coefficients of $z^k$ on both sides of these   identities are equal.    At the end of his paper, Garvan proposed a problem of finding  direct proofs of these three identities by the method of  Bailey pairs. In this paper, we will show these  identities hold by applying Bailey's transform with a conjugate Bailey pair  of Warnaar \cite{Warnaar-2003} and three Bailey pairs derived from three identities  of Slater \cite{Slater-1949}. In particular, we obtain a more compact form of Hecke-Rogers identity for $R(z;q)$.

\begin{thm} \label{Garvan-n1-th}  We have
\begin{align}\label{Garvan-n1}\nonumber
&(zq, z^{-1}q,q;q)_{\infty}R(z;q)\\[3pt]
  &=\sum_{n=0}^{\infty}\left(\sum_{j=0}^{[n/2]}(-1)^{n+j}z^{n-3j}
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n-j)}
  +\sum_{j=1}^{[n/2]}(-1)^{n+j}z^{n-3j+1}
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n+j)}\right)\\[5pt]
 &=\sum_{n=0}^{\infty}\sum_{0\leq j<n/3}(-1)^{n+j}(z^{n-3j}+z^{3j-n})
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n-j)}
 +\sum_{j=0}^{\infty}q^{3j^2+j}\nonumber\\[5pt]
 &\quad +\sum_{n=0}^{\infty}\sum_{1\leq j<(n+1)/3}(-1)^{n+j}(z^{n-3j+1}+z^{3j-n-1})
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n+j)}  -\sum_{j=1}^{\infty}q^{3j^2-j}.  \label{Garvan-n1-a}
\end{align}
\end{thm}
Note that Garvan's identity \eqref{Garvan1}  can  immediately follow from   \eqref{Garvan-n1}. To be specific, if we replace $z$ by $z^{-1}$ in \eqref{Garvan-n1}, we get
 \begin{align}  \label{tempt}
&(zq, z^{-1}q,q;q)_{\infty}R(z^{-1};q)\\ \nonumber
  &=\sum_{n=0}^{\infty}\left(\sum_{j=0}^{[n/2]}(-1)^{n+j}z^{-n+3j}
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n-j)}
  +\sum_{j=1}^{[n/2]}(-1)^{n+j}z^{-n+3j-1}
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n+j)}\right).
\end{align}
  Combining \eqref{Garvan-n1} and \eqref{tempt},  and noting    $R(z;q)=R(z^{-1};q)$, we get    Garvan's identity \eqref{Garvan1}.

Using the Bailey machinery, we obtain the following Hecke-Rogers identity for $g_2(z;q)$.

\begin{thm} \label{Garvan-n2-th}  We have
\begin{equation}
 \begin{split}\label{Garvan-n2}
  &(1-z)(zq, z^{-1}q,q;q)_{\infty}g_2(z;q)\\
    &=\sum_{n=0}^{\infty}\left(\sum_{m=0}^{[n/2]}
    (-1)^{m+n}z^{n-2m}q^{\frac{1}{2}(n^2-2m^2)+\frac{1}{2}n}
   +\sum_{m=1}^{[n/2]}
    (-1)^{m+n}z^{2m-n-1}q^{\frac{1}{2}(n^2-2m^2)+\frac{1}{2}n}\right).
 \end{split}
 \end{equation}
\end{thm}
When  substitute   \eqref{Garvan-n2} into the relation \eqref{r-h-g2}, and employ Jacobi's triple product identity,  we derive  Garvan's identity \eqref{Garvan2}.

More generally, we generalized  these two-variable Hecke-Rogers identities to infinite families. The next theorem embeds  \eqref{Garvan-n1-a} into an infinite family.

\begin{thm}\label{Garvan-n1-gen}
For $k\geq 2$,  we have
    \begin{align}\label{gener1}
&(zq, z^{-1}q,q;q)_{\infty}\sum_{n=0}^{\infty}
\frac{(q;q)_{2n}q^n}{(zq;q)_n(z^{-1}q;q)_n}\sum_{n_1,\cdots,n_{k-1}=0}^{\infty}
\frac{q^{N_1^2+\cdots+N_{k-2}^2+2N_{k-1}^2+N_1+\cdots+N_{k-2}}}
{(q;q)_{n-N_1}(q;q)_{n_1}\cdots(q;q)_{n_{k-2}}(q;q)_{2n_{k-1}}}\nonumber \\[3pt]
&=\sum_{n=0}^{\infty}\sum_{0\leq j<n/3}
(-1)^{n+j}(z^{n-3j}+z^{3j-n})q^{\frac{1}{2}(n^2-3j^2)+
\frac{1}{2}(n-j)+3(k-1)j(3j+1)}\nonumber  \\
&\quad+\sum_{n=0}^{\infty}\sum_{1\leq j
<(n+1)/3}(-1)^{n+j}(z^{n-3j+1}+z^{3j-n-1})q^{\frac{1}{2}(n^2-3j^2)+
\frac{1}{2}(n+j)+3(k-1)j(3j-1)}\nonumber \\
&\quad +\sum_{j=0}^{\infty}q^{(3k-2)j(3j+1)}
-\sum_{j=1}^{\infty}q^{(3k-2)j(3j-1)},
    \end{align}
  where  $N_j=n_j+n_{j+1}+\cdots+n_{k-1}$.
\end{thm}

We extends \eqref{Garvan-n2} into an infinite family.

\begin{thm}\label{gener2-thm}
For $k\geq 2$, we have
    \begin{align}\label{gener2}
&(zq, z^{-1}q,q;q)_{\infty}\sum_{n=0}^{\infty}
\frac{(q^2;q)_{2n}q^n}{(zq;q)_n(z^{-1}q;q)_{n+1}}\nonumber \\[2pt]
&\hskip 3cm \times\sum_{n_1,\cdots,n_{k-1}=0}^{\infty}
\frac{q^{N_1^2+\cdots+N_{k-2}^2+\frac{3}{2}N_{k-1}^2
+2(N_1+\cdots+N_{k-2})
+\frac{3}{2}N_{k-1}}(-q;q)_{n_{k-1}}}
{(q;q)_{n-N_1}(q;q)_{n_1}\cdots(q;q)_{n_{k-2}}
(q^2;q)_{2n_{k-1}}}\nonumber\\[2pt]
&=\sum_{n=0}^{\infty}\left(\sum_{m=0}^{[n/2]}
(-1)^{m+n}z^{n-2m}q^{\frac{1}{2}(n^2-2m^2)+\frac{1}{2}n
+4(k-1)(m^2+m)}\right. \nonumber\\
&\quad\hskip1cm\left.+\sum_{m=1}^{[n/2]}(-1)^{m+n}z^{2m-n-1}
q^{\frac{1}{2}(n^2-2m^2)+\frac{1}{2}n+4(k-1)(m^2-m)}\right),
    \end{align}
where   $N_j=n_j+n_{j+1}+\cdots+n_{k-1}$.
\end{thm}

Finally, we generalize \eqref{Garvan4} to an infinite family.

\begin{thm}\label{gener3-thm}
For $k\geq 2$, we have
\begin{align}\label{gener3}
&(zq^2, z^{-1}q^2,q^2;q^2)_{\infty}\sum_{n=0}^{\infty}
\frac{(q^2;q^2)_{2n}q^{2n}}{(zq^2;q^2)_n(z^{-1}q^2;q^2)_{n}}\nonumber \\[2pt]
&\hskip 4cm \times\sum_{n_1,\cdots,n_{k-1}=0}^{\infty}
\frac{(-1)^{n_{k-1}}q^{2N_1^2+\cdots+2N_{k-2}^2+3N_{k-1}^2
+2N_1+\cdots+2N_{k-2}}(q;q^2)_{n_{k-1}}}
{(q^2;q^2)_{n-N_1}(q^2;q^2)_{n_1}\cdots(q^2;q^2)_{n_{k-2}}
(q^2;q^2)_{2n_{k-1}}}\nonumber \\
&=\sum_{n=0}^{\infty}\left(\sum_{m=0}^n(-1)^{n}z^{m-n}
q^{\frac{1}{2}(2n^2-m^2)+\frac{1}{2}(2n-m)+2(k-1)(m^2+m)}\right. \nonumber \\
&\quad \hskip 1cm\left.+\sum_{m=1}^n(-1)^{n}z^{n-m+1}
q^{\frac{1}{2}(2n^2-m^2)+\frac{1}{2}(2n+m)+2(k-1)(m^2-m)}\right),
\end{align}
where  $N_j=n_j+n_{j+1}+\cdots+n_{k-1}$.
\end{thm}

\section{Conjugate Bailey pairs and Bailey pairs  }

In this section, we   present some conjugate Bailey pairs and Bailey pairs which are needed in the proof of Hecke-Rogers identities for the universal mock theta functions. Recall that a pair of sequences $(\delta_n, \gamma_n)$ is called a conjugate Bailey pair relative to $a$  if they satisfy
\begin{equation}
\gamma_n=\sum_{r=n}^\infty\frac{\delta_r}{(q;q)_{r-n}(aq;q)_{r+n}}.
\end{equation}
A pair of sequences $(\alpha_n, \beta_n)$ is called a Bailey pair relative to $a$ if they satisfy
\begin{equation}
\beta_n=\sum_{r=0}^n\frac{\alpha_r}{(q;q)_{n-r}(aq;q)_{n+r}}.
\end{equation}

From the definitions of Bailey pairs and conjugate Bailey pairs, one could easily obtain the following transform \cite{Bailey-1949}.

\begin{thm}[The Bailey transform] If $(\alpha_n,\beta_n)$ is a Bailey pair relative to $a$ and $(\gamma_n,\delta_n)$ is a conjugate Bailey pair relative to $a$. Then we have
\begin{equation*}
    \sum_{n=0}^{\infty}\alpha_{n}\gamma_{n}
    =\sum_{n=0}^{\infty}\beta_{n}\delta_{n}.
\end{equation*}

\end{thm}

 We first state a conjugate Bailey pair which is needed in the proof of all of Garvan's three identities.  The proof of this conjugate Bailey pair is mainly based on the following generalization
  of Jacobi's triple product identity given by Warnaar \cite{Warnaar-2003}. Andrews and Warnaar \cite{Andrews-Warnaar-2007} provided an alternative proof of this identity.
 \begin{equation}\label{Jacobi}
    1+\sum_{r=1}^{\infty}(-1)^rq^{r\choose 2}(a^r+b^r)
    =(a,b,q;q)_{\infty}
    \sum_{r=0}^{\infty}\frac{(ab/q;q)_{2r}q^r}{(q,a,b,ab;q)_r}.
 \end{equation}
 Setting $a=z$ and $b=z^{-1}q$ in \eqref{Jacobi}, we obtain  Jacobi's  triple product \cite[p.15]{Gasper-2004}
\begin{equation}\label{Jacobi-o}
1+\sum_{r=1}^{\infty}(-1)^rq^{r\choose 2}(z^r+z^{-r}q^r)
=(z,z^{-1}q,q;q)_{\infty}.
\end{equation}

 \begin{lem}\label{lem1}The pair of sequences $(\delta_n, \gamma_n)$ forms a conjugate Bailey pair relative to $ab$, where
  \begin{equation*}
 \delta_n=(aq,b,q;q)_{\infty}
   \frac{(ab;q)_{2n}q^n}{(aq;q)_{n}(b;q)_n},
 \end{equation*}
 and
 \begin{equation*}
    \gamma_n=\frac{(1-ab) q^{n}}{(1-abq^{2n})}\left(1+\sum_{r=1}^{\infty}
    (-1)^rq^{r \choose 2}((aq^{n+1})^r+(bq^n)^r)\right).
 \end{equation*}
 \end{lem}

 \pf
 By the definition of conjugate Bailey pair, we see that
 \begin{align*}
    \gamma_n&=\sum_{r=n}^{\infty}\frac{\delta_r}{(q;q)_{r-n}(abq;q)_{r+n}}\\[3pt]
    &=(aq,b,q;q)_{\infty}\sum_{r=n}^{\infty}
    \frac{(ab;q)_{2r}q^r}{(q;q)_{r-n}(abq;q)_{r+n}(aq;q)_{r}(b;q)_r}\\[3pt]
    &=(aq,b,q;q)_{\infty}\sum_{r=0}^{\infty}
    \frac{(ab;q)_{2n+2r}q^{r+n}}{(q;q)_{r}(abq;q)_{r+2n}(aq;q)_{r+n}(b;q)_{r+n}}\\[3pt]
    &=(aq,b,q;q)_{\infty}\frac{(ab;q)_{2n}q^n}
    {(abq;q)_{2n}(aq;q)_{n}(b;q)_{n}}\sum_{r=0}^{\infty}
    \frac{(abq^{2n};q)_{2r}q^{r}}{(q,abq^{2n+1},aq^{n+1},bq^n;q)_{r}}\\[3pt]
    &=\frac{(1-ab)q^n}{1-abq^{2n}}(aq^{n+1},bq^n,q;q)_\infty\sum_{r=0}^{\infty}
    \frac{(abq^{2n};q)_{2r}q^{r}}{(q,abq^{2n+1},aq^{n+1},bq^n;q)_{r}}.
    \end{align*}
 Using \eqref{Jacobi} with   $a$ replaced by $aq^{n+1}$ and $b$ by $bq^n$ in the summation of the above identity, we derive that
    \begin{align*}
     \gamma_n&=\frac{(1-ab)q^n}{1-abq^{2n}}(aq^{n+1},bq^n,q;q)_\infty \frac{1}{(aq^{n+1},bq^n,q;q)_\infty} \\[3pt]
    &\hskip 2cm \times \left(1+\sum_{r=1}^{\infty}
    (-1)^rq^{r \choose 2}((aq^{n+1})^r+(bq^n)^r)\right)\\[3pt]
    &=\frac{(1-ab)q^n}{1-abq^{2n}}\left(1+\sum_{r=1}^{\infty}
    (-1)^rq^{r \choose 2}((aq^{n+1})^r+(bq^n)^r)\right),
 \end{align*}
 as desired. This completes the proof.  \qed

To derive \eqref{Garvan-n1-a}, we also need  the following Bailey pair given by Warnnar \cite[p.375]{Warnaar-2003}, which can be established by a identity of Slater \cite[Eq.(3.4)]{Slater-1949}.

\begin{lem}\label{Garvan-BL-1}The pair of sequences $(\alpha_n,\beta_n)$ forms a Bailey pair relative to $q$, where
\begin{equation*}
\begin{split}
 \alpha_{3n}&=q^{(3n-2)n}\frac{1-q^{6n+1}}{1-q},  \quad \alpha_{3n+1}=0,
 \quad \alpha_{3n+2}=-q^{(3n+2)n}\frac{1-q^{6n+5}}{1-q}, \\[3pt]
 \beta_n&=\frac{q^{n(n-1)}}{(q;q)_{2n}}.
 \end{split}
\end{equation*}
\end{lem}

 The following Bailey pair is necessary in the proof of \eqref{Garvan-n2}. The proof of this Bailey pair is mainly based on a special case of $_6\psi_6$ given by Slater \cite[Eq.(4.2)]{Slater-1949}:
\begin{align}\label{Slater}
  &\sum_{r=-[n/2]}^{[n/2]}\frac{(1-aq^{4r})(q^{-n};q)_{2r}
  (d;q^2)_r(e;q^2)_r}{(1-a)(aq^{n+1};q)_{2r}(aq^2/d;q^2)_r(aq^2/e;q^2)_r}
 \left( \frac{a^{2}q^{2n+1}}{de}\right)^r \\\nonumber
  &\hskip1cm=\frac{(q^2/a,aq/d,aq/e,aq^2/de;q^2)_{\infty}}
  {(q,q^2/d,q^2/e,a^2q/de;q^2)_\infty}\cdot
  \frac{(q;q)_n(aq;q)_n(a^2q/de;q^2)_n}{(aq;q^2)_n(aq/d;q)_n(aq/e;q)_n}.
\end{align}

 \begin{lem}\label{Garvan-BL-2}
 The pair of sequences $(\alpha_n,\beta_n)$ forms a Bailey pair relative to $q^2$, where
 \begin{equation}\label{Bailey pair1}
    \begin{split}
     \alpha_{2n}&=(-1)^nq^{n^2-n}\frac{1-q^{4n+2}}{1-q^2},
     \ \ \ \alpha_{2n+1}=0,\\
     \beta_n&=\frac{(-q;q)_nq^{n\choose 2}}{(q^2;q)_{2n}}.
    \end{split}
 \end{equation}
 \end{lem}

\pf In \eqref{Slater}, let  $a=d=q^2$,  and
$e\rightarrow0$, noting  that $1/(q^2;q^2)_n=0$ when  $n<0$, we obtain
\[\sum_{r=0}^{[n/2]}\frac{(1-q^{4r+2})(q^{-n};q)_{2r}(-1)^rq^{2nr-r^2}}
{(1-q^2)(q^{n+3};q)_{2r}}=\frac{(q^3;q)_nq^{n\choose 2}}{(q^3;q^2)_n}.\]
Divide   both sides of the above identity  by $(q;q)_n(q^3;q)_n$   to get
\begin{equation*}
    \sum_{r=0}^{[n/2]}\frac{1}{(q;q)_{n-2r}(q^3;q)_{n+2r}}
    \frac{(1-q^{4r+2})(-1)^rq^{r^2-r}}{1-q^2}=\frac{(-q;q)_nq^{n\choose 2}}{(q^2;q)_{2n}},
\end{equation*}
which yields the desired Bailey pair. This completes the proof. \qed

To verify \eqref{Garvan4}, we also require the following Bailey pair.

 \begin{lem}\label{Garvan-BL-3}
 The pair of sequences $(\alpha_n,\beta_n)$ forms a Bailey pair relative to $q$ , where
 \begin{equation}\label{Bailey pair2}
    \begin{split}
      \alpha_n&=(-1)^n q^{\frac{n^2-3n}{4}}\frac{1-q^{2n+1}}{1-q},\\[3pt]
      \beta_n&=\frac{(-1)^nq^{\frac{n^2}{2}-n}(q^{\frac{1}{2}};q)_n}{(q;q)_{2n}}.
    \end{split}
 \end{equation}
 \end{lem}

 \pf In \eqref{Slater}, set $a=q,\,d=-q^{3/2}$
  and $e\rightarrow 0$ to get
  \begin{align*}
  \sum_{r=-[n/2]}^{[n/2]}\frac{(1-q^{4r+1})(q^{-n};q)_{2r}
  q^{2nr-\frac{1}{2}r-r^2}}
    {(1-q)(q^{n+2};q)_{2r}}=\frac{(q;q)_n(q^2;q)_n(-1)^nq^{\frac{n^2}{2}-n}}
    {(q^2;q^2)_n(-q^{\frac{1}{2}};q)_n}.
  \end{align*}
  Dividing  both sides of the above identity  by $(q;q)_n(q^2;q)_n$,  we obtain
  \begin{equation*}
    \sum_{r=-[n/2]}^{[n/2]}\frac{(1-q^{4r+1})q^{r^2-\frac{3}{2}r}}
    {(q;q)_{n+2r+1}(q;q)_{n-2r}}
    =\frac{(-1)^nq^{\frac{n^2}{2}-n}(q^{\frac{1}{2}};q)_n}{(q;q)_{2n}}.
  \end{equation*}
Simplifying the left hand side of the above identity yields
\begin{align*}
 &\sum_{r=-[n/2]}^{[n/2]}\frac{(1-q^{4r+1})q^{r^2-\frac{3}{2}r}}
    {(q;q)_{n+2r+1}(q;q)_{n-2r}}\\[5pt]
&\hskip 0.2cm= \frac{1}{(q^2;q)_n(q;q)_n}+\sum_{r=1}^{[n/2]}\frac{(1-q^{4r+1})
q^{r^2-\frac{3}{2}r}}
    {(q;q)_{n+2r+1}(q;q)_{n-2r}}+\sum_{r=-[n/2]}^{-1}\frac{(1-q^{4r+1})
    q^{r^2-\frac{3}{2}r}}{(q;q)_{n+2r+1}(q;q)_{n-2r}}\\[5pt]
    &=\frac{1}{(q^2;q)_n(q;q)_n}+\sum_{r=1}^{[n/2]}
    \frac{(1-q^{4r+1})q^{r^2-\frac{3}{2}r}}
    {(q;q)_{n+2r+1}(q;q)_{n-2r}}-\sum_{r=1}^{[n/2]}
    \frac{(1-q^{4r-1})q^{r^2-\frac{5}{2}r+1}}
    {(q;q)_{n-2r+1}(q;q)_{n+2r}}\\[5pt]
        &=\sum_{r=0}^n\frac{1}{(q;q)_{n-r}(q^2;q)_{n+r}}
    \frac{(1-q^{2r+1})(-1)^rq^{\frac{r^2-3r}{4}}}{1-q}.
\end{align*}
Hence, we have
\[\sum_{r=0}^n\frac{1}{(q;q)_{n-r}(q^2;q)_{n+r}}
    \frac{(1-q^{2r+1})(-1)^rq^{\frac{r^2-3r}{4}}}{1-q}=\frac{(-1)^nq^{\frac{n^2}{2}-n}(q^{\frac{1}{2}};q)_n}{(q;q)_{2n}},\]
which implies the desired Bailey pair. This completes the proof. \qed

To get an infinite family of Hecke-Rogers identities,
  we need the following limiting form of Bailey's lemma \cite{Andrews-1984-a, Paule-1985}.

\begin{thm}\label{BL-c-1}
Supposed   $(\alpha_n,\beta_n)$ is a Bailey pair relative to $a$. Then $(\alpha'_n,\beta'_n)$ is a new Bailey pair relative to $a$, where

\begin{equation*}\label{a}
\alpha_n' =a^n q^{n^2}\alpha_n
\end{equation*}
and
\begin{align*}\label{b}
\beta_n' &=\sum_{j=0}^{\infty}\frac{a^jq^{j^2}}
{(q;q)_{n-j}} \beta_j .
\end{align*}
\end{thm}
Iterating this lemma leads to the Bailey chain.
We now produce the following infinite sequence of Bailey pairs by iterating the Bailey pair in Lemma \ref{Garvan-BL-1} along this Bailey chain.

\begin{lem}\label{Garvan-BL-1-gen} For $k\geq 2$,   $(\alpha^{(k)}_n, \beta_n^{(k)}
)$ form   Bailey pairs relative to $q$, where
\begin{equation*}
  \begin{split}
    \alpha_{3n}^{(k)}&=q^{3(3k-2)n^2+(3k-5)n}
    \frac{(1-q^{6n+1})}{(1-q)}, \ \ \alpha_{3n+1}^{(k)}=0,\\[3pt]
     \alpha_{3n+2}^{(k)}&=-q^{3(3k-2)n^2+(15k-13)n+6(k-1)}
    \frac{(1-q^{6n+5})}{(1-q)},\\[3pt]
    \beta_n^{(k)}&=\sum_{n_1,\cdots,n_{k-1}=0}^{\infty}
\frac{q^{N_1^2+\cdots+N_{k-2}^2+2N_{k-1}^2+N_1+\cdots+N_{k-2}}}
{(q;q)_{n-N_1}(q;q)_{n_1}\cdots(q;q)_{n_{k-2}}(q;q)_{2n_{k-1}}},
  \end{split}
\end{equation*}
where  $N_j=n_j+n_{j+1}+\cdots+n_{k-1}$.
\end{lem}
\pf We proceed by induction on $k$. When $k=2$. Applying    Theorem  \ref{BL-c-1} to the Bailey pair in Lemma \ref{Garvan-BL-1},  we obtain a new Bailey pair $(\alpha^{(2)}_n,\beta^{(2)}_n)$  relative to $q$, where
\begin{align*}
    \alpha_{3n}^{(2)}&=q^{9n^2+3n}\cdot q^{(3n-2)n}\frac{1-q^{6n+1}}{1-q}
    =q^{12n^2+n}\frac{1-q^{6n+1}}{1-q}, \quad
      \alpha_{3n+1}^{(2)}=0,\\[5pt]
    \alpha_{3n+2}^{(2)}&=q^{(3n+2)(3n+3)}\cdot \left(-q^{(3n+2)n}\frac{1-q^{6n+5}}{1-q}\right)
    =-q^{12n^2+17n+6}\frac{1-q^{6n+5}}{1-q}
\end{align*}
and
\begin{align*}
    \beta_n^{(2)}=\sum_{j=0}^n\frac{q^{2j^2}}{(q;q)_{n-j}(q;q)_{2n}}
    =\sum_{j=0}^\infty\frac{q^{2j^2}}{(q;q)_{n-j}(q;q)_{2n}},
\end{align*}
as desired.

Assume that we have a Bailey pair $(\alpha^{(k-1)}_n,\beta^{(k-1)}_n)$ relative to $q$, where
\begin{equation*}
  \begin{split}
    \alpha_{3n}^{(k-1)}&=q^{3(3k-5)n^2+(3k-8)n}
    \frac{(1-q^{6n+1})}{(1-q)},  \ \ \alpha_{3n+1}^{(k-1)}=0,\\[3pt]
     \alpha_{3n+2}^{(k-1)}&=-q^{3(3k-5)n^2+(15k-28)n+6(k-2)}
\frac{(1-q^{6n+5})}{(1-q)},\\[3pt]
    \beta_n^{(k-1)}&=\sum_{n_1',\cdots,n_{k-2}'=0}^{\infty}
\frac{q^{N_1'^2+\cdots+N_{k-3}'^2+2N_{k-2}'^2+N_1'+\cdots+N_{k-3}'}}
{(q;q)_{n-N_1'}(q;q)_{n_1'}\cdots(q;q)_{n_{k-3}'}(q;q)_{2n_{k-2}'}},
  \end{split}
\end{equation*}
 where $N_j'=n_j'+n_{j+1}'+\cdots+n_{k-2}'$.

When we apply Theorem \ref{BL-c-1} to this Bailey pair, we  arrive at a new Bailey pair $(\alpha^{(k)}_n,\beta^{(k)}_n)$ relative to $q$, where for $\alpha^{(k)}_n$, we have
\begin{align*}
\alpha_{3n}^{(k)}&=q^{9n^2+3n}\alpha_{3n}^{(k-1)}\\[3pt]
&=q^{9n^2+3n+3(3k-5)n^2+(3k-8)n}\frac{(1-q^{6n+1})}{(1-q)}\\[3pt]
&=q^{3(3k-2)n^2+(3k-5)n}\frac{(1-q^{6n+1})}{(1-q)},\\[5pt]
\alpha_{3n+1}^{(k)}&=0, \\[5pt]
\alpha_{3n+2}^{(k)}&=q^{3(n+1)(3n+2)}\alpha_{3n+2}^{(k-1)}\\[3pt]
&=-q^{3(n+1)(3n+2)+3(3k-5)n^2+(15k-28)n+6(k-2)}
\frac{(1-q^{6n+5})}{(1-q)}\\[3pt]
&=-q^{3(3k-2)n^2+(15k-13)n+6(k-1)}\frac{(1-q^{6n+5})}{(1-q)}.
\end{align*}
For $\beta^{(k)}_n$, we have
\begin{align*}
    \beta_{n}^{(k)}&=\sum_{n_{k-1}'=0}^\infty\frac{q^{n_{k-1}'^2+n_{k-1}'}}
    {(q;q)_{n-n_{k-1}'}}\beta_{n'_{k-1}}^{(k-1)}\\
    &=\sum_{n_{k-1}'=0}^\infty\frac{q^{n_{k-1}'^2+n_{k-1}'}}{(q;q)_{n-n_{k-1}'}}
    \sum_{n_1',\cdots,n_{k-2}'=0}^{\infty}
\frac{q^{N_1'^2+\cdots+N_{k-3}'^2+2N_{k-2}'^2+N_1'+\cdots+N_{k-3}'}}
{(q;q)_{n_{k-1}'-N_1'}(q;q)_{n_1'}\cdots(q;q)_{n_{k-3}'}
(q;q)_{2n_{k-2}'}}.
\end{align*}
Recall that $N_j'=n_j'+n_{j+1}'+\cdots+n_{k-2}'$ for $1\leq j\leq k-2$. Let
\[n_1=n_{k-1}'-N_1',\  n_2=n_1',\  n_3=n_2',\ \ldots,\  n_{k-1}=n_{k-2}',\] and set $N_j=n_j+n_{j+1}+\cdots+n_{k-1}$ for $1\leq j\leq k-1$. Obviously, we have
\[N_2=N_1',\ N_3=N_2',\ \ldots,\ N_{k-1}=N_{k-2}',\ N_1=n_{k-1}',\]
and so
 \begin{align*}
   \beta_{n}^{(k)}&=\sum_{n_1,\cdots,n_{k-1}=0}^{\infty}
\frac{q^{N_1^2+\cdots+N_{k-2}^2+2N_{k-1}^2+N_1+\cdots+N_{k-2}}}
{(q;q)_{n-N_1}(q;q)_{n_1}\cdots(q;q)_{n_{k-2}}(q;q)_{2n_{k-1}}}.
 \end{align*}
This completes the proof. \qed

The following infinite sequence of Bailey pairs can be obtained from the Bailey pair in Lemma \ref{Garvan-BL-2} along the same line.

\begin{lem}\label{Garvan-BL-2-gen}For $k\geq 2$,   $(\alpha^{(k)}_n, \beta_n^{(k)}
)$ form   Bailey pairs relative to $q^2$, where
\begin{align*}
      \alpha_{2n}^{(k)}&=(-1)^nq^{(4k-3)n^2+(4k-5)n}\frac{1-q^{4n+2}}{1-q^2},
 \quad  \alpha_{2n+1}^{(k)}=0,\\[5pt]
       \beta_{n}^{(k)}&=\sum_{n_1,\cdots,n_{k-1}=0}^{\infty}
\frac{q^{N_1^2+\cdots+N_{k-2}^2+\frac{3}{2}N_{k-1}^2+2(N_1+\cdots+N_{k-2})
+\frac{3}{2}N_{k-1}}(-q;q)_{n_{k-1}}}
{(q;q)_{n-N_1}(q;q)_{n_1}\cdots(q;q)_{n_{k-2}}(q^2;q)_{2n_{k-1}}}.
\end{align*}
\end{lem}

Similarly, we obtain the following infinite sequence of Bailey pairs from the Bailey pair in Lemma \ref{Garvan-BL-3} which are used to establish the infinite family in Theorem \ref{gener3-thm}.

\begin{lem}\label{Garvan-BL-3-gen}For $k\geq 2$,   $(\alpha^{(k)}_n, \beta_n^{(k)}
)$ form   Bailey pairs relative to $q$, where
\begin{align*}
      \alpha_n^{(k)}&=(-1)^nq^{\frac{4k-3}{4}n^2+\frac{4k-7}{4}n}
   \frac{ 1-q^{2n+1}}{1-q},\\[5pt]
    \beta_n^{(k)}&=\sum_{n_1,\cdots,n_{k-1}=0}^{\infty}
\frac{(-1)^{n_{k-1}}q^{N_1^2+\cdots+N_{k-2}^2+\frac{3}{2}N_{k-1}^2
+N_1+\cdots+N_{k-2}}(q^{\frac{1}{2}};q)_{n_{k-1}}}
{(q;q)_{n-N_1}(q;q)_{n_1}\cdots(q;q)_{n_{k-2}}
(q;q)_{2n_{k-1}}}.
\end{align*}
\end{lem}

\section{A two-variable Hecke-Rogers identity for $R(z;q)$ and   generalization}

In this section, we first give a proof of \eqref{Garvan-n1-a} in Theorem \ref{Garvan-n1-th} by using the Bailey transform with conjugate Bailey pair in Lemma \ref{lem1} and Bailey pair in Lemma \ref{Garvan-BL-1}. We then show that \eqref{Garvan-n1} can be derived from \eqref{Garvan-n1-a} with the aid of  two identities given by Garvan \cite{Garvan-2014}. Finally, we give a proof of Theorem \ref{Garvan-n1-gen}.

\medskip

 \noindent{\it Proof of Theorem \ref{Garvan-n1-th}, Eq. \eqref{Garvan-n1-a}.}  Combining the Bailey
 transform with the Bailey pairs in Lemma \ref{Garvan-BL-1} and the conjugate Bailey pair in Lemma \ref{lem1} with $a=z$, and $b=z^{-1}q$, we obtain
\begin{align}
\sum_{n=0}^{\infty}\beta_n\delta_n&=(zq,z^{-1}q,q;q)_{\infty}
\sum_{n=0}^{\infty}\frac{q^{n^2}}{(zq;q)_n(z^{-1}q;q)_n}\nonumber\\
&=\sum_{j=0}^{\infty}\alpha_{3j}\gamma_{3j}+\sum_{j=1}^{\infty}\alpha_{3j-1}\gamma_{3j-1}\nonumber\\
&=\sum_{j=0}^{\infty}q^{j(3j-2)+3j}\left(1+\sum_{n=1}^{\infty}(-1)^nq^{n\choose 2}
((zq^{3j+1})^n+(z^{-1}q^{3j+1})^n)\right)\nonumber\\
&\quad -\sum_{j=1}^{\infty}q^{(j-1)(3j-1)+(3j-1)}\left(1+\sum_{n=1}^{\infty}(-1)^nq^{n\choose 2}
((zq^{3j})^n+(z^{-1}q^{3j})^n)\right)\nonumber
\end{align}
\begin{align}
&=\sum_{j=0}^{\infty}\sum_{n=1}^{\infty}(-1)^n(z^n+z^{-n})
 q^{{n+1\choose 2}+3j^2+j+3nj}+\sum_{j=0}^{\infty}q^{3j^2+j}\nonumber\\
 &\quad + \sum_{j=1}^{\infty}\sum_{n=1}^{\infty}(-1)^{n-1}(z^n+z^{-n})
 q^{{n\choose 2}+3j^2-j+3nj}-\sum_{j=1}^{\infty}q^{3j^2-j}\label{garvan}.
 \end{align}
Replace  $n$ by $n-3j$ in the first sum  of \eqref{garvan} and interchange  the order of  summation to get
\begin{align}\label{sum1}
&\sum_{j=0}^{\infty}\sum_{n=1}^{\infty}(-1)^n(z^n+z^{-n})
 q^{{n+1\choose 2}+3j^2+j+3nj}\nonumber \\[2pt]
 &\quad =\sum_{n=0}^{\infty}\sum_{0\leq j<n/3}(-1)^{n+j}(z^{n-3j}+z^{3j-n})
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n-j)}.
 \end{align}
Replace $n$ by $n-3j+1$ in
 the third sum of \eqref{garvan} and interchange the order of  summation to yield
 \begin{align}\label{sum2}
 &\sum_{j=1}^{\infty}\sum_{n=1}^{\infty}(-1)^{n-1}(z^n+z^{-n})
 q^{{n\choose 2}+3j^2-j+3nj}\nonumber\\[2pt]
 &=\sum_{n=0}^{\infty}\sum_{1\leq j<(n+1)/3}(-1)^{n+j}(z^{n-3j+1}+z^{3j-n-1})
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n+j)}.
 \end{align}
 Plugging  \eqref{sum1} and \eqref{sum2} into \eqref{garvan}, we get the desired identity \eqref{Garvan-n1-a}.\qed

We are now in position to show \eqref{Garvan-n1} in Theorem \ref{Garvan-n1-th} holds.

 \noindent{\it Proof of Theorem \ref{Garvan-n1-th}, Eq. \eqref{Garvan-n1}.}  Obviously, it suffices to show that
 \begin{align}
&\sum_{n=0}^{\infty}\sum_{0\leq j<n/3}(-1)^{n+j}(z^{n-3j}+z^{3j-n})
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n-j)}+\sum_{j=0}^{\infty}q^{3j^2+j}\label{Garvan-n1-pf}\\
 &\quad +\sum_{n=0}^{\infty}\sum_{1\leq j<(n+1)/3}(-1)^{n+j}(z^{n-3j+1}+z^{3j-n-1})
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n+j)} -\sum_{j=1}^{\infty}q^{3j^2-j} \nonumber\\[5pt]
&=\sum_{n=0}^{\infty}\left(\sum_{j=0}^{[n/2]}(-1)^{n+j}z^{n-3j}
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n-j)}
  +\sum_{j=1}^{[n/2]}(-1)^{n+j}z^{n-3j+1}
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n+j)}\right). \nonumber
 \end{align}

 To this end, we first show that
 \begin{align}\label{p1}
 &\sum_{n=0}^{\infty}\sum_{0\leq j<n/3}(-1)^{n+j}(z^{n-3j}+z^{3j-n})
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n-j)} +\sum_{j=0}^{\infty}q^{3j^2+j} \nonumber \\[3pt]
 & \quad =\sum_{n=0}^{\infty}\sum_{j=0}^{[n/2]}(-1)^{n+j}z^{n-3j}
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n-j)}.
 \end{align}

Using  the following identity given by Garvan \cite[Eq. (2.14)]{Garvan-2014},
\begin{align} \label{gposi-1}
&\sum_{n=0}^{\infty}\sum_{0\leq j<n/3}(-1)^{n+j} z^{n-3j}
   q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n-j)}\nonumber\\[3pt]
  & =\sum_{n=0}^{\infty}\sum_{n/3< j\leq n/2}(-1)^{n+j} z^{-n+3j}
   q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n-j)},
\end{align}
we get
\begin{align*}
 &\sum_{n=0}^{\infty}\sum_{0\leq j<n/3}(-1)^{n+j}(z^{n-3j}+z^{3j-n})
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n-j)} +\sum_{j=0}^{\infty}q^{3j^2+j}\\[3pt]
 &=\sum_{n=0}^{\infty}\sum_{0\leq j<n/3}(-1)^{n+j}z^{n-3j}
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n-j)}\\
 &\quad +\sum_{n=0}^{\infty}\sum_{n/3< j\leq n/2}(-1)^{n+j}z^{n-3j}
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n-j)}+\sum_{j=0}^{\infty}q^{3j^2+j}\\
 &=\sum_{n=0}^{\infty}\sum_{j=0}^{[n/2]}(-1)^{n+j}z^{n-3j}
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n-j)},
\end{align*}
as desired.

Similarly, applying the following identity of Garvan \cite[Eq. (2.15)]{Garvan-2014}
\begin{align*}
    &\sum_{n=0}^{\infty}\sum_{1\leq j<(n+1)/3}(-1)^{n+j} z^{n-3j+1}
q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n+j)}\nonumber\\[3pt]
  & \hskip 1cm  =\sum_{n=0}^{\infty}\sum_{(n+1)/3<j\leq n/2}(-1)^{n+j} z^{-n+3j-1}
  q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n+j)},
\end{align*}
we can show that
\begin{align}\label{p2}
   &\sum_{n=0}^{\infty}\sum_{0\leq j<n+1/3}(-1)^{n+j}(z^{n-3j+1}+z^{3j-n-1})
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n+j)}
 -\sum_{j=1}^{\infty}q^{3j^2-j}\nonumber\\
 &=\sum_{n=0}^\infty\sum_{j=1}^{[n/2]}(-1)^{n+j}z^{n-3j+1}
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n+j)}.
\end{align}
Combining \eqref{p1} and \eqref{p2},  we obtain
\eqref{Garvan-n1-pf}. Thus we show \eqref{Garvan-n1} holds.\qed

For the reminder of this section, we show \eqref{gener1} holds  by using the Bailey pair technology. \medskip

\noindent{\it Proof of Theorem \ref{Garvan-n1-gen}.} Applying the Bailey transform with the conjugate Bailey pair in Lemma \ref{lem1} by setting $a=z$ and $b=z^{-1}q$ and the Bailey pair in Lemma \ref{Garvan-BL-1-gen},  we get
\begin{align}
    \sum_{n=0}^{\infty}\beta_n^{(k)}\delta_n&=
   (zq,z^{-1}q,q;q)_{\infty}\sum_{n=0}^{\infty}
\frac{(q;q)_{2n}q^n}{(zq;q)_n(z^{-1}q;q)_n}\nonumber \\[2pt]
&\hskip 4cm \times \sum_{n_1,\cdots,n_{k-1}=0}^{\infty}
\frac{q^{N_1^2+\cdots+N_{k-2}^2+2N_{k-1}^2+N_1+\cdots+N_{k-2}}}
{(q;q)_{n-N_1}(q;q)_{n_1}\cdots(q;q)_{n_{k-2}}(q;q)_{2n_{k-1}}}\nonumber\\
&=\sum_{j=0}^{\infty}\alpha_{3j}^{(k)}\gamma_{3j}
+\sum_{j=1}^{\infty}\alpha_{3j-1}^{(k)}\gamma_{3j-1}\nonumber\\
&=\sum_{j=0}^{\infty}q^{(3k-2)j(3j+1)}
\left(1+\sum_{n=1}^{\infty}(-1)^nq^{n+1\choose 2}((zq^{3j})^n+(z^{-1}q^{3j})^n)\right)\nonumber\\
&\quad-\sum_{j=1}^{\infty}q^{(3k-2)j(3j-1)}
\left(1+\sum_{n=1}^{\infty}(-1)^nq^{n\choose 2}((zq^{3j})^n+(z^{-1}q^{3j})^n)\right)\nonumber\\
&=\sum_{j=0}^{\infty}q^{(3k-2)j(3j+1)}+\sum_{j=0}^{\infty}\sum_{n=1}^{\infty}
(-1)^n(z^n+z^{-n})q^{{n+1\choose2}+3jn+(3k-2)j(3j+1)}\nonumber\\\label{garvan2}
&\quad-\sum_{j=1}^{\infty}q^{(3k-2)j(3j-1)}+\sum_{j=1}^{\infty}\sum_{n=1}^{\infty}
(-1)^{n-1}(z^n+z^{-n})q^{{n\choose2}+3jn+(3k-2)j(3j-1)}.
\end{align}
Replace  $n$ by $n-3j$ in the second sum  of \eqref{garvan2} and reverse   the order of summation to get
\begin{align}\label{sum3}
&\sum_{j=0}^{\infty}\sum_{n=1}^{\infty}(-1)^n(z^n+z^{-n})
 q^{{n+1\choose 2}+3nj+(3k-2)(3j^2+j)}\nonumber \\
 &=\sum_{n=0}^{\infty}\sum_{0\leq j<n/3}(-1)^{n+j}(z^{n-3j}+z^{3j-n})
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n-j)+3(k-1)j(3j+1)}.
 \end{align}
Replace $n$ by $n-3j+1$ in
 the fourth sum of \eqref{garvan2} and interchange the order of summation. This gives
 \begin{align}\label{sum4}
 &\sum_{j=1}^{\infty}\sum_{n=1}^{\infty}(-1)^{n-1}(z^n+z^{-n})
 q^{{n\choose 2}+3nj+(3k-2)(3j^2-j)}\nonumber\\[2pt]
 &=\sum_{n=0}^{\infty}\sum_{1\leq j<(n+1)/3}(-1)^{n+j}(z^{n-3j+1}+z^{3j-n-1})
 q^{\frac{1}{2}(n^2-3j^2)+\frac{1}{2}(n+j)+3(k-1)j(3j-1)}.
 \end{align}
 Substituting \eqref{sum3} and \eqref{sum4} in \eqref{garvan2}, we are led to the desired identity \eqref{gener1}.\qed

\section{A two-variable Hecke-Rogers identity for $H(z;q)$ and generalization}

In this section  we first use the Bailey transform to show Theorem \ref{Garvan-n2-th}. We then derive Garvan's identity   \eqref{Garvan2} from  Theorem \ref{Garvan-n2-th}. Last, we give a proof of  Theorem \ref{gener2-thm}.

\medskip

\noindent{\it Proof of Theorem \ref{Garvan-n2-th}.}
Apply the Bailey transform with the conjugate Bailey pair in Lemma \ref{lem1} by setting $a=z^{-1}q$ and $b=zq$ and the Bailey pair in Lemma \ref{Garvan-BL-2} to yield
\begin{align}
\sum_{n\geq 0}\beta_n\delta_n &=(zq, z^{-1}q,q;q)_{\infty}
 \sum_{n=0}^{\infty}\frac{(-q;q)_nq^{n+1\choose 2}}{(z^{-1}q;q)_{n+1}(zq;q)_{n}}\nonumber \\
 &=(1-z)(zq, z^{-1}q,q;q)_{\infty}
 g_2(z;q)\nonumber \\
 &=\sum_{m=0}^{\infty}\alpha_{2m}\gamma_{2m} \nonumber\\
 &=\sum_{m=0}^{\infty}(-1)^mq^{m^2+m}\left(1+\sum_{n=1}^{\infty}
 (-1)^nq^{n+1\choose 2}((z^{-1}q^{2m+1})^n+(zq^{2m})^n)\right) \nonumber\\
 &=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}
(-1)^{m+n}z^{n}q^{{n+1\choose 2}+2mn+m^2+m}\nonumber\\\label{h1}
&\quad\ \ +\sum_{m=0}^{\infty}
\sum_{n=1}^{\infty}(-1)^{m+n}z^{-n}q^{{n+1\choose 2}+(2m+1)n+m^2+m}.
\end{align}
Shift  $n$ to $n-2m$ in the first sum of \eqref{h1} and interchange the order of summation. This gives
\begin{equation}\label{sum-h1}
  \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}
(-1)^{m+n}z^{n}q^{{n+1\choose 2}+2mn+m^2+m}
= \sum_{n=0}^{\infty}\sum_{m=0}^{[n/2]}
    (-1)^{m+n}z^{n-2m}q^{\frac{1}{2}(n^2-2m^2)+\frac{1}{2}n}.
\end{equation}
Shift  $n$ to $n-2m-1$ in the second sum of \eqref{h1},  and then replace $m$ by $m-1$ ,  change the order of  summation. This summation becomes
\begin{equation}\label{sum-h2}
  \sum_{m=0}^{\infty}
\sum_{n=1}^{\infty}(-1)^{m+n}z^{-n}q^{{n+1\choose 2}+(2m+1)n+m^2+m}
= \sum_{n=0}^{\infty}\sum_{m=1}^{[n/2]}
    (-1)^{m+n}z^{2m-n-1}q^{\frac{1}{2}(n^2-2m^2)+\frac{1}{2}n}.
\end{equation}
Plugging  \eqref{sum-h1} and \eqref{sum-h2} into \eqref{h1}, we get the desired identity \eqref{Garvan-n2}. Thus we complete the proof of Theorem \ref{Garvan-n2-th}.\qed

Now it's an easy matter to prove \eqref{Garvan2}.

\noindent{\it Proof of Garvan's identity \eqref{Garvan2}.} By the  relation \eqref{r-h-g2}, we see that
 \begin{align}\label{r-h-g2-pf}
 &(1+z)(zq, z^{-1}q,q;q)_{\infty}H(z;q)\nonumber \\[3pt]
 &=(1-z)(zq, z^{-1}q,q;q)_{\infty}+2z(1-z)(zq, z^{-1}q,q;q)_{\infty}g_2(z;q).
 \end{align}
Invoking    Jacobi's  triple product identity \eqref{Jacobi-o}
 and  Theorem \ref{Garvan-n2-th} in \eqref{r-h-g2-pf}, we find that
\begin{align*}
&(1+z)(zq, z^{-1}q,q;q)_{\infty}H(z;q)\\
&=\sum_{n=0}^{\infty}(-1)^nz^{-n}q^{n+1\choose2}+\sum_{n=1}^\infty
(-1)^nz^{n}q^{n\choose2}+2\sum_{n=0}^{\infty}\sum_{m=0}^{[n/2]}
    (-1)^{m+n}z^{n-2m+1}q^{\frac{1}{2}(n^2-2m^2)+\frac{1}{2}n}\\
   &\quad+2\sum_{n=0}^{\infty}\sum_{m=1}^{[n/2]}
    (-1)^{m+n}z^{2m-n}q^{\frac{1}{2}(n^2-2m^2)+\frac{1}{2}n}.
\end{align*}
Replacing $n$ by $n+1$ in the second term, and subtracting the terms with $m=0$ in the
third sum, we  obtain
\begin{align*}
  &(1+z)(zq, z^{-1}q,q;q)_{\infty}H(z;q)\\
  &= \sum_{n=0}^{\infty}(-1)^nz^{-n}q^{n+1\choose2}-\sum_{n=0}^\infty
(-1)^nz^{n+1}q^{n+1\choose2}+2\sum_{n=0}^{\infty}(-1)^nz^{n+1}q^{n+1\choose2}\\ &\quad+2\sum_{n=0}^{\infty}\sum_{m=1}^{[n/2]}
    (-1)^{m+n}(z^{n-2m+1}+z^{2m-n})q^{\frac{1}{2}(n^2-2m^2)+\frac{1}{2}n}\\
  &=\sum_{n=0}^{\infty}(-1)^n(z^{n+1}+z^{-n})q^{n+1\choose2} +2\sum_{n=0}^{\infty}\sum_{m=1}^{[n/2]}
    (-1)^{m+n}(z^{n-2m+1}+z^{2m-n})q^{\frac{1}{2}(n^2-2m^2)+\frac{1}{2}n}\\
    &=\sum_{n=0}^{\infty}\sum_{\mid m\mid\leqslant[n/2]}
  (-1)^{n+m}(z^{n-2|m|+1}+z^{2|m|-n})
  q^{\frac{1}{2}(n^2-2m^2)+\frac{1}{2}n}.
\end{align*}
Thus we obtain Garvan's identity  \eqref{Garvan2}.\qed

We proceed to prove Theorem \ref{gener2-thm}.

\noindent{\it Proof of Theorem \ref{gener2-thm}.} Substituting the Bailey pair
 in lemma \ref{Garvan-BL-2} with the conjugate Bailey pair in Lemma \ref{lem1} with $a=z^{-1}q$, and $b=zq$ into the Bailey transform, we obtain
\begin{align}
  \sum_{n=0}^{\infty}\beta_n^{(k)}\delta_n&=(zq, z^{-1}q,q;q)_{\infty}\sum_{n=0}^{\infty}
\frac{(q^2;q)_{2n}q^n}{(zq;q)_n(z^{-1}q;q)_{n+1}}\nonumber\\
&\quad\ \ \ \times\sum_{n_1,\cdots,n_{k-1}=0}^{\infty}
\frac{q^{N_1^2+\cdots+N_{k-2}^2+\frac{3}{2}N_{k-1}^2+2(N_1+\cdots+N_{k-2})
+\frac{3}{2}N_{k-1}}(-q;q)_{n_{k-1}}}
{(q;q)_{n-N_1}(q;q)_{n_1}\cdots(q;q)_{n_{k-2}}(q^2;q)_{2n_{k-1}}}\nonumber\\
&=\sum_{m=0}^{\infty}\alpha_{2m}^{(k)}\gamma_{2m}\nonumber\\
&=\sum_{m=0}^{\infty}(-1)^mq^{(4k-3)(m^2+m)}\left(1+\sum_{n=1}^{\infty}
(-1)^nq^{n+1\choose 2}((z^{-1}q^{2m+1})^n+(zq^{2m})^n)\right)\nonumber
\end{align}
\begin{align}
&\hskip -2cm =\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}
(-1)^{m+n}z^{n}q^{{n+1\choose 2}+2mn+(4k-3)(m^2+m)}\nonumber\\\label{h2}
&\hskip -2cm \quad\ \ +\sum_{m=0}^{\infty}
\sum_{n=1}^{\infty}(-1)^{m+n}z^{-n}q^{{n+1\choose 2}+(2m+1)n+(4k-3)(m^2+m)}.
\end{align}
Replacing $n$ by $n-2m$ in the first term of \eqref{h2}  and interchanging the order of summation,
 we find that
\begin{align}\label{sum-h3}
& \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}
(-1)^{m+n}z^{n}q^{{n+1\choose 2}+2mn+(4k-3)(m^2+m)} \nonumber\\
&\quad = \sum_{n=0}^{\infty}\sum_{m=0}^{[n/2]}
(-1)^{m+n}z^{n-2m}q^{\frac{1}{2}(n^2-2m^2)+\frac{1}{2}n+4(k-1)(m^2+m)}.
\end{align}
Replace $n$ by $n-2m-1$ in the second sum of \eqref{h2}, and then shift $m$ to $m-1$,  reverse the order of summation to yield
\begin{align}\label{sum-h4}
 &  \sum_{m=0}^{\infty}\sum_{n=1}^{\infty}(-1)^{m+n}z^{-n}q^{{n+1\choose 2}+(2m+1)n+(4k-3)(m^2+m)}\nonumber \\
 &\quad  =\sum_{n=0}^\infty\sum_{m=1}^{[n/2]}(-1)^{m+n}z^{2m-n-1}
q^{\frac{1}{2}(n^2-2m^2)+\frac{1}{2}n+4(k-1)(m^2-m)}.
\end{align}
Substituting \eqref{sum-h3} and \eqref{sum-h4} into \eqref{h2}, we get the desired identity \eqref{gener2}. Thus, we show Theorem \ref{gener2-thm}.\qed

\section{A two-variable Hecke-Rogers identity for $K(z;q)$ and generalization}

In this section, we give a proof of Garvan's identity \eqref{Garvan4} by the method of Bailey pair. We then give a generalization  \eqref{gener3} of this identity.

\medskip

\noindent{\it Proof of Garvan's identity \eqref{Garvan4}.}
Applying the Bailey transform with the conjugate Bailey pair in Lemma
\ref{lem1} by setting $a=z$ and $b=z^{-1}q$ and the Bailey pair in Lemma \ref{Garvan-BL-3}, we find that
\begin{align}
   \sum_{n=0}^{\infty}{\beta_n\delta_n}
   &=(zq,z^{-1}q,q;q)_{\infty}
   \sum_{n=0}^{\infty}\frac{(-1)^nq^{n^2/2}(q^{1/2};q)_n}
   {(zq;q)_n(z^{-1}q;q)_n}\nonumber\\
  &= \sum_{n=0}^{\infty}{\alpha_n\gamma_n}\nonumber\\
    &=\sum_{m=0}^{\infty}(-1)^mq^{(m^2+m)/4}\left(1+\sum_{n=1}^{\infty}
    (-1)^nq^{n\choose 2}((zq^{m+1})^n+(z^{-1}q^{m+1})^n)\right)\nonumber
     \end{align}
\begin{align}\label{k1}
    &\hskip -4cm =\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}(-1)^{m+n}z^{-n}
    q^{{n+1\choose 2}+\frac{m^2+m}{4}+mn}\nonumber \\[2pt]
    &\hskip -4cm \quad +\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}
    (-1)^{m+n}z^nq^{{n+1\choose 2}+\frac{m^2+m}{4}+mn}.
    \end{align}
Shifting $n$ to $n-m$ in the first sum of \eqref{k1}, and interchanging the order of summation,  we  get
    \begin{equation}\label{sum-k1}
      \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}(-1)^{m+n}z^{-n}
    q^{{n+1\choose 2}+\frac{m^2+m}{4}+mn}
    = \sum_{n=0}^{\infty}\sum_{m=0}^n(-1)^nz^{m-n}
  q^{\frac{1}{4}(2n^2-m^2)+\frac{1}{4}(2n-m)}.
    \end{equation}
   For the second sum of \eqref{k1},  we first shift  $n$ to $n-m$,   then    replace $m$ by $m-1$, and  reverse the order of summation, thus  the second sum of \eqref{k1} becomes
    \begin{equation}\label{sum-k2}
        \sum_{m=0}^{\infty}\sum_{n=1}^{\infty}
    (-1)^{m+n}z^nq^{{n+1\choose 2}+\frac{m^2+m}{4}+mn}
    =\sum_{n=0}^{\infty}\sum_{m=1}^n(-1)^nz^{n-m+1}
    q^{\frac{1}{4}(2n^2-m^2)
  +\frac{1}{4}(2n+m)}.
    \end{equation}
Substituting  \eqref{sum-k1} and \eqref{sum-k2} into \eqref{k1} and replacing $q$ by $q^2$,  we  recover     \eqref{Garvan4}. This completes the proof. \qed

Using the similar argument, we could show \eqref{gener3} is valid.
\medskip

\noindent{\it Proof of Theorem \ref{gener3-thm}.}  Applying the Bailey transform with Bailey pair in Lemma \ref{Garvan-BL-3-gen} and the conjugate Bailey pair in Lemma \ref{lem1} with $a=z$, and $b=z^{-1}q$,   we have
\begin{align}
\sum_{n=0}^{\infty}\beta_n^{(k)}\delta_n
&=(zq, z^{-1}q,q;q)_{\infty}\sum_{n=0}^{\infty}
\frac{(q;q)_{2n}q^n}{(zq;q)_n(z^{-1}q;q)_{n}}\nonumber\\
&\quad \times\sum_{n_1,\cdots,n_{k-1}=0}^{\infty}
\frac{(-1)^{n_{k-1}}q^{N_1^2+\cdots+N_{k-2}^2+\frac{3}{2}N_{k-1}^2
+N_1+\cdots+N_{k-2}}(q^{\frac{1}{2}};q)_{n_{k-1}}}
{(q;q)_{n-N_1}(q;q)_{n_1}\cdots(q;q)_{n_{k-2}}
(q;q)_{2n_{k-1}}}\nonumber\\
&=\sum_{n=0}^{\infty}\alpha_n^{(k)}\gamma_n\nonumber\\
&=\sum_{m=0}^{\infty}(-1)^mq^{\frac{4k-3}{4}(m^2+m)}
\left(1+\sum_{n=1}^{\infty}(-1)^nq^{n+1\choose 2}
((zq^m)^n+(z^{-1}q^m)^n)\right)\nonumber\\
&=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}(-1)^{m+n}z^{-n}
q^{{n+1\choose 2}+mn+\frac{4k-3}{4}(m^2+m)}\nonumber\\\label{k2}
&\quad+\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}(-1)^{m+n}z^{n}
q^{{n+1\choose 2}+mn+\frac{4k-3}{4}(m^2+m)}.
\end{align}
Replace $n$ by $n-m$ in the first sum of \eqref{k2} and interchange the order of summation to get
\begin{align}\label{sum-k3}
 &\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}(-1)^{m+n}z^{-n}
q^{{n+1\choose 2}+mn+\frac{4k-3}{4}(m^2+m)}\nonumber \\[2pt]
&\quad = \sum_{n=0}^{\infty}\sum_{m=0}^n(-1)^{n}z^{m-n}
q^{\frac{1}{4}(2n^2-m^2)+\frac{1}{4}(2n-m)+(k-1)(m^2+m)}.
\end{align}
Shifting $n$ to $n-m$ in the second sum of \eqref{k2},   replacing  $m$ by $m-1$,  and then reversing the order of summation, this gives
\begin{align}\label{sum-k4}
&\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}(-1)^{m+n}z^{n}
q^{{n+1\choose 2}+mn+\frac{4k-3}{4}(m^2+m)}\nonumber \\[2pt]
&\quad =\sum_{n=0}^{\infty}\sum_{m=1}^n(-1)^{n}z^{n-m+1}
q^{\frac{1}{4}(2n^2-m^2)+\frac{1}{4}(2n+m)+(k-1)(m^2-m)}.
\end{align}
Hence  \eqref{gener3} is obtained  upon substituting \eqref{sum-k3} and \eqref{sum-k4} in \eqref{k2} and replacing $q$ by $q^2$. This completes the proof. \qed

\noindent{\bf Acknowledgments.} This work was supported by the 973 Project, the PCSIRT Project of the Ministry of Education and the National Science Foundation of China.


TITLE: Image of a Prime ideals in quotient ring
QUESTION [2 upvotes]: I am having confusion regarding the relationship between the prime ideals of a ring and a certain quotient of a ring.

Let $A$ be a commutative ring with identity, and $a$ an ideal of $A$. Clearing there is a one-to-one order preserving correspondence between the ideals of $A/a$ and the ideals of $A$ containing $a$, but how does this relate to prime ideals in the respective rings? If $p$ is a prime ideal in $A$, under what conditions is its image in $A/a$ also prime?

REPLY [3 votes]: If $p$ contains $a$, then $p/a$ should be prime in $A/a$, since $(A/a)/(p/a)\cong A/p$, and so both are integral domains.
If $p$ does not contain $a$, then this need not hold: for example, let $A=\mathbb Z$, let $a=(2)$, $p=(3)$. Then $A/a=\mathbb Z/2\mathbb Z$, and the image of $(3)$ is the entirety of the quotient ring, which is not prime.
This counterexample is not ideal (no pun intended) since only a triviality prevents the ideal from being prime. If someone can think of a counterexample where the image is a nonprime proper ideal, that would be best.


TITLE: Why don't stars re-emit the photons they absorb, thus restoring a continuous emission spectrum?
QUESTION [3 upvotes]: If you shine white light through a gas, electrons can absorb sufficiently energetic photons to reach higher excited states. This produces gaps in the spectrum and it's how Helium was discovered. So goes the story.
But an excited electron now exists above a gap in a lower shell. It should fall back down, re-emitting a photon of the same characteristic frequency.
So why doesn't this plug the gaps in the spectrum?

REPLY [1 votes]: When the electron falls to its ground state, it will emit the photon in a random direction.This means that the photon might not travel in the direction as the rest of the white light.
In the absorption spectrum of the sun, there will be some photons of the characteristic frequency of helium, but significantly less than the rest of the spectrum, meaning a gap is created.


TITLE: Does the following function have a name and what properties does it have?
QUESTION [3 upvotes]: (Apologies for the vague way the question is stated.)
We define the function like this:
(1) $f(x) = x$ for $0\leq x \leq a$ (for some $a \geq 15 \in \mathbb{R}$) $\land$ $f(x) = a$ if $ x > a$ . For $x<0$, it doesn't really matter what $f(x)$ looks like.
Has someone studied this type of function? Does it have a 'nicer' definition than the one stated above? Any references on the subject? It is not absolutely necessary for the function to have all of the above characteristics. If you know a function that is similar to it, please let me know.
Motivation:
I'm looking for a function that assigns a constant to 'infinite values' of x and does not 'deform' finite values (too much). So the function should satisfy $f(\zeta(1))=a$ for some finite value of $a \in \mathbb{R}$ and $f(3)=3$, for example. It doesn't really matter whether $f(b)=b$ or any multiple of b or that $f(x)$ is some finite polynomial, as long as it's 'easy' to find the value of $f(b)$.
Thanks,
Max Muller

REPLY [2 votes]: f(x)=min(x,a)
We can rewrite this in terms of the absolute value function as (x+a)/2-|x-a|/2, the average minus half the difference. If you prefer this can be seen as getting the graph of the absolute value function, and manipulating it to fit the function you want. This is possible since they are both formed of two straight lines meeting at a point.


TITLE: Multivariable Function Optimization with Semidefinite Hessian Matrix
QUESTION [0 upvotes]: I'm writing to ask for support in carrying out an exercise about the optimization of a multivariable function.
The function is: $f(x,y,z)=x^2+y^2z+z^2-2x$
Clearly the domain is $\mathbb{R}^3$, and $f$ is infinitely differentiable on it.
Well... I started to calculate critical points. It turned out to be only one of them, which is: $P=(1,0,0)$.
Then I determined the Hessian Matrix: $Hf(P=(1,0,0))=\begin{pmatrix}2
 & 0 & 0 \\ 0
 &0  &0 \\ 0
 & 0 & 2
\end{pmatrix}$
It's associated to a positive semidefinite quadratic form, so the point $P=(1,0,0)$ must be a local minimum or a saddle point, but we can't conclude anything more.
I thought to calculate the sign of the increment $\Delta f=f(x,y,z)-f(P)=x^2+y^2z+z^2-2x-(-1)=(x-1)^2+y^2z+z^2$, in a neighbourhood of $P$.
Firstly I translated the axes (with the substitution $u=x-1$), and then I tried to use spherical coordinates. I guess $P=(1,0,0)$ is a local minimum, but I'm not sure. I would be grateful if someone could show me how to rigorously prove the nature of this critical point.
Thank you so much for the support.

REPLY [1 votes]: Hint:
Define the curve $\mu(t) = (1,0,|t|^{3/2})$. Now, investigate the composite functions $f \circ \mu$ near $t=0$.