text:
100+ Best Tombstone Quotes: Exclusive Selection
January 12, 2021 April 3, 2020 by Tahsincan
Tombstone is a 1993 American Western film directed by George P. Cosmatos, written by Kevin Jarre (who was also the original director, but was replaced early in production), and starring Kurt Russell and Val Kilmer, with Sam Elliott, Bill Paxton, Powers Boothe, Michael Biehn, and Dana Delany in supporting roles, as well as narration by Robert Mitchum. Profoundly inspirational tombstone quotes will brighten up your day and make you feel ready to take on anything.
Famous Tombstone Quotes
Live so that when the final summons comes you will leave something more behind you than an epitaph on a tombstone or an obituary in a newspaper. — Billy Sunday
James Joyce buried himself in his great work. _Finnegan's Wake_ is his monument and his tombstone. A dead end. — Edward Abbey
We had some errors, and it dropped the girls' morale, then they lost confidence in themselves. It's our one bad game for the week. We are looking forward to playing Tombstone on Thursday. If we win against Tombstone, we'll go to state. — Rand Baker
I used to want the words 'She tried' on my tombstone. Now I want 'She did it. — Katherine Dunham
The Greatest Blues Singer in the World Will Never Stop Singing. — Bessie Smith
I used to have nightmares that they would put 'He played Ted' on my tombstone. — Keanu Reeves
Death looses its sting when the epitaph drafted for the tombstone of the departed even by those with whom he crossed swords in his lifetime is filled with encomiums. — Uche Ohia
I never understood why when you died, you didn't just vanish, everything could just keep going on the way it was only you just wouldn't be there. I always thought I'd like my own tombstone to be blank. No epitaph, and no name. Well, actually, I'd like it to say 'figment. — Andy Warhol
Some people hate funerals. I find them comforting. They hit the pause button on life and remind us that it has an end. Every eulogy reminds me to deepen my dash, that place on the tombstone between our birth and our death. — Regina Brett
I did not want my tombstone to read, 'She kept a really clean house. — Ann Richards
I always remember an epitaph which is in the cemetery at Tombstone, Arizona. It says: 'Here lies Jack Williams. He done his damnedest.' I think that is the greatest epitaph a man can have – When he gives everything that is in him to do the job he has before him. That is all you can ask of him and that is what I have tried to do. — Harry S Truman
I have tried my best to give the nation everything I had in me. There are probably a million people who could have done the job better than I did it, but I had the job and I always quote an epitaph on a tombstone in a cemetery in Tombstone, Arizona: Here lies Jack Williams. He done his damndest. — Harry S Truman
How meagre one's life becomes when it is reduced to its basic facts. And the last, most complete reduction is on one's tombstone: a name, two dates. — Bill Vaughan
On my tombstone, I want written: 'He never did 'Love Boat! — Orson Welles
Nothing in Moderation We all loved him. — Ernie Kovacs
I want my tombstone to read: If this is a joke, I don't get it. — David Brenner
I don't want my tombstone to say actress. I want it to say human being. — Laurie Holden
Steel True, Blade Straight. — Arthur Conan Doyle
If you're going to sit on someone's tombstone, you might as well know something about them, right? — Nicholas Sparks
If I thought you weren't my friend, I just don't think I could bear it. — Doc Holliday
That's the one for my tombstone… Here lies David Byrne. Why the big suit? — David Byrne
…there occurred to me the simple epitaph which, when I am no more, I intend to have inscribed on my tombstone. It was this: He was a man who acted from the best motives. There is one born every minute. — P. G. Wodehouse
I've had such a satisfying life professionally and personally. I hope my tombstone says, Never boring. — Alan Dershowitz
But I am greedy for life. I do too much of everything all the time. Suddenly one day my heart will fail. The Iron Crab will get me as it got my father. But I am not afraid of The Crab. At least I shall have died from an honourable disease. Perhaps they will put on my tombstone. 'This Man Died from Living Too Much'. — Ian Fleming
My closest friend, who died not long ago, is buried near Marx's grave in Highgate cemetery, so I see the gaggle of admirers laying roses at the foot of his tombstone regularly. I have never been tempted to leave flowers there myself. Great theories, shame about the practice. Marx did many things. But inventing class was not one of them. — Alastair Campbell
I've always felt that, when I looked at my tombstone, it shouldn't say, 'Mehmet Oz banged out 10,000 open-heart operations.' I've probably done 5,000. Am I any better at it than 10,000? He shook his head. It's just a different number on the tombstone. — Mehmet Oz
They didn't understand what they were doing. I'm afraid that will be on the tombstone of the human race. — Michael Crichton
A genius of comedy His talent brought joy and Laughter to all the world. — Oliver Hardy
Earth has scarcely an acre that does not remind us of actions that have long preceded our own, and its clustering tombstones loom up like reefs of the eternal shore, to show us where so many human barks have struck and gone down. — Edwin Hubbel Chapin
The act of writing is for me often nothing more than the secret or conscious desire to carve words on a tombstone: to the memory of a town forever vanished, to the memory of a childhood in exile, to the memory of all those I loved and who, before I could tell them I loved them, went away. — Elie Wiesel
I told my kids I just want three words on my tombstone, if I have one. I'll probably be cremated. One is woman. I'm very comfortable in that role. I've loved being a woman, I've loved being a mother, I've loved being a grandmother. I want three words: Woman, Atheist, Anarchist. That's me. — Madalyn Murray O'Hair
Novelist-Citizen of Two Countries Interpreter of his Generation on both Sides of the Sea. — Henry James
The other day, the way people [do] who are approaching their 80th birthday, I was thinking about all the last business – funerals and where do you want to be buried – and I thought if anything were to be inscribed on my tombstone, I said let it be that. — Frederick Buechner
A book has to be easy to open and you don't have to be a bodybuilder to lift it. I like books I can read in bed. Those big tombstones would kill me. — Karl Lagerfeld
Everybody Loves Somebody Sometime. — Dean Martin
I want one word on my tombstone – dancer. — Agnes de Mille
He made us laugh, he took my pain away. I love you, Lauretta. — Marty Feldman
I conceive disgust at these impertinent and misbecoming familiarities inscribed upon your ordinary tombstone. — Charles Lamb
The chances that your tombstone will read 'Killed by Asteroid' are about the same as they'd be for 'Killed in Airplane Crash. — Neil deGrasse Tyson
Some people collect paperweights, or pre-Columbian figures, or old masters, or young mistresses, or tombstone rubbings, or five-minute recipes, or any of a thousand other things … My own collection is sunrises; and I find that they have their advantages. Sunrises are usually handsome, they can't possibly be dusted, and they take only a little room, so long as it has a window to see them from. — Peg Bracken
The tombstone is about the only thing that can stand upright and lie on its face at the same time. — Mary Wilson Little
Leaning her silly, beautiful, drunken head on my shoulder, she said, Oh, Esther, I don't want to be a feminist. I don't enjoy it. It's no fun. I know, I said. I don't either. People think you decide to be a radical, for God's sake, like deciding to be a librarian or a ship's chandler. You make up your mind, you commit yourself (sounds like a mental hospital, doesn't it?). I said Don't worry, we could be buried together and have engraved on our tombstone the awful truth, which some day somebody will understand: WE WUZ PUSHED. — Joanna Russ
If the Confederacy fails, there should be written on its tombstone: Died of a Theory. — Jefferson Davis
A Gentle Man and a Gentleman. — Jack Dempsey
Your next SMS will probably be around longer, and remain more legible, than your tombstone. For, unlike your tombstone or even your mortal coil, your texts may be worth something. — Douglas Rushkoff
On my tombstone, I really hope that someday they will write: He was true but partial. — Ken Wilber
I must say, some are not very beautifully made. They're coffee-table books for people who drink alcohol. I have nothing against coffee-table books as long as they are well done. They must not look like gravestones on a table. Sometimes they are too big, they come in boxes and things like this. No, a book has to be easy to open and you don't have to be a bodybuilder to lift it. I like books I can read in bed. Those big tombstones would kill me. — Karl Lagerfeld
(Palm Pictures founder) Chris Blackwell said to me 6 or 7 years ago that DVD doesn?t have to be a tombstone for creative content. He said (DVD) is such a great format that people will make content solely for DVD, and not just cheap franchise movies, but things that will extend the audio and visual arts. — Richard Browne
The city itself swung slowly toward us silent as a dream. No sign of life but puffs of steam from skyscraper chimneys, the motion of the traffic. The mighty towers stood like tombstones in a graveyard, leaning against the sky and waiting for — for what? Someday we'll know. — Edward Abbey
My uncle Sammy was an angry man. He had printed on his tombstone: What are you looking at? — Margaret Smith
I am a worker, a tombstone mason, anxious to pleace averyburies and jully glad when Christmas comes his once ayear. — James Joyce
We had a good day. Tombstone and Safford were missing some of their top guns, but I'm really pleased with what our team did. That's the important thing. — Adam Estrada
When I am dead and buried, on my tombstone I would like to have it written, 'I have arrived.' Because when you feel that you have arrived, you are dead. — Yul Brynner
Cemeteries in Bohemia are like gardens. The graves are covered with grass and colourful flowers. Modest tombstones are lost in the greenery. When the sun goes down, the cemetery sparkles with tiny candles… no matter how brutal life becomes, peace always reigns in the cemetery. Even in wartime, even in Hitler's time, even in Stalin's time. — Milan Kundera
Bateman becoming an icon made me resentful. It will be on my tombstone that I created that character. And I resented that character haunting me. — Christian Bale
Having read the inscriptions upon the tombstones of the great and little cemeteries, Wang Peng advised the Emperor to kill all the living and resurrect the dead. — Paul Eldridge
If I take the wings of the morning and dwell in the uttermost parts of the sea. — Anne Morrow Lindbergh
The Stone the Builders Rejected. — Jack London
I conceive disgust at those impertinent and misbecoming familiarities, inscribed upon your ordinary tombstones. Every dead man must take upon himself to be lecturing me with his odious truism, that such as he now is, I must shortly be. Not so shortly, friend, perhaps, as thou imaginest. In the meantime I am alive. I move about. I am worth twenty of thee. Know thy betters! — Charles Lamb
I will not be back after these messages. — Merv Griffin
Who's afraid of the big, bad buildings? Everyone, because there are so many things about gigantism that we just don't know. The gamble of triumph or tragedy at this scale — and ultimately it is a gamble — demands an extraordinary payoff. The trade center towers could be the start of a new skyscraper age or the biggest tombstones in the world. — Ada Louise Huxtable
On my tombstone just write, 'The sorest loser that ever lived. — Earl Weaver
A star on earth – a star in heaven. — Karen Carpenter
I never met a man I didn't like until I met Will Rogers. — Mort Sahl
I know what I'm having 'em put on my tombstone: 'I have nothing more to say'. — Ted Turner
I hope that on my tombstone it says 'Born 1933, died 2043.' I hope that's my legacy. — Quincy Jones
Left love behind many years ago. Now it rests under a cross in the cemetery in Tombstone. — Franco Nero
Having 'Oscar winner' on your tombstone is a great thing. — George Clooney
You might be a redneck if your favorite hunting dog has a bigger tombstone than your grandfather. — Jeff Foxworthy
The burden of suffering seems a tombstone hung about our necks, while in reality it is only the weight which is necessary to keep down the diver while he is hunting for pearls. — Jean Paul
Good Night Sweet Prince and a flight of angels sing to thy rest. — Douglas Fairbanks
Hear the bells ringing they're singing Christ is risen from the dead! The angel up on the tombstone said, He is risen just as he said! — Keith Green
Fame – a few words upon a tombstone, and the truth of those not to be depended on. — Christian Nestell Bovee
Nonsense, I have not yet begun to defile myself. — Doc Holliday
Many years ago I sent an old, beloved jacket to a cleaner, the Sycamore Cleaners. It was a leather jacket covered in Guinness and blood and marmalade, one of those jobs… and it came back with a little note pinned to it, and on the note it said, 'It distresses us to return work which is not perfect.' So that will do for me. That can go on my tombstone. — Peter O'Toole
In testimony of their Respect For The Patriot of incorruptible Integrity, The Soldier of approved Valour The Statesman of consummate Wisdom; Whose Talents and Virtues will be admired By Grateful Posterity Long after this Marble shall have mouldered into Dust. — Alexander Hamilton
Here lies one whose name was writ in water. — John Keats
I shall be better satisfied if the same can be said of me as was said of the prophet of old, That I walked in the fear of the Lord, and begat sons and daughters [Genesis 5:22], than if it were inscribed on my tombstone that I governed the councils or commanded the arms of the whole continent of America. — Benjamin Rush
When I go, if there's a tombstone it will say, She doesn't give in. She doesn't give up. And she never takes no for an answer. — Doris Roberts
Actually I like the idea of being a Renaissance hack. If tombstones were still in style, I would want to have the two words chiseled right under my name. — Dennis Flanagan
The one thing I would hope would go on my tombstone is, 'I made my parents proud. — Roger Goodell
I work to please myself. I'm still not sure if movies are an art form. And if they're not, then let them inscribe on my tombstone what they could about any craftsman who loves his job: Here lies Vincente Minnelli. He died of hard work. — Vincente Minnelli
The old dead trees are the most fascinating – the countless trees lying in the gullies and up the hills that fell perhaps a century ago, pulling up their roots from the earth as they toppled. The great upheavals left rocks in their huge tentacles and, as they slowly rot, the trunks are home to populations of creatures, from goannas to wild pigs. As grey as tombstones in a cemetery they lie there, having outlasted generations of farmers, as they'll outlast me. In their own way they are as beautiful, more beautiful, than living trees. — Phillip Adams
Satire does not look pretty upon a tombstone. — Charles Lamb
The Doc Holliday of legend is a gambler and gunman who appears out of nowhere in 1881, arriving in Tombstone with a bad reputation and a hooker named Big Nose Kate. — Mary Doria Russell
Surely with as good reason as had Archimedes to have the cylinder, cone and sphere engraved on his tombstone might our distinguished countrymen leave testamentary directions for the cubic eikosiheptagram to be engraved on theirs. Spirit of the Universe! wither are we drifting, and when, where, and how is all this to end? — James Joseph Sylvester
I did not want my tombstone to read, 'She kept a really clean house.' I think I'd like them to remember me by saying, 'She opened government to everyone. — Ann Richards
She did it the hard way. — Bette Davis
What Muddy Waters did for us is what we should do for others. It's the old thing, what you want written on your tombstone as a musician; HE PASSED IT ON. — Keith Richards
Sleep after toyle, port after stormie seas, Ease after warre, death after life, does greatly please. — Joseph Conrad
At Rest An American Soldier And Defender of the Constitution. — Jefferson Davis
I always thought I'd like my own tombstone to be blank. No epitaph, and no name. Well, actually, I'd like it to say 'figment. — Andy Warhol
I loved Stand By Me. I loved Tombstone. Give men a little credit. Maybe we'll be able to humanize women more and see more of their depth as people. — Evan Rachel Wood
Another thing I don't want on my tombstone, Shane said. You have others? Claire asked. He held up one finger. I thought it wasn't loaded, Shane said. Second finger. Hand me a match so I can check the gas tank. Third finger. Killed over ice cream. Basically, any death that requires me to be stupid first. — Rachel Caine
I had A Lover's Quarrel With The World Robert Lee Frost (Old Bennington Cemetery, Bennington, Vermont) Our Darling Eva We Love You. — Eva Gabor
I don't want a tombstone. You could carve on it 'She never actually wanted a tombstone. — Ali Smith
I now believe when I'm dead and buried my tombstone will read, 'I'm not entirely sure the band's over. — Nick Mason
I cannot feel my legs from the waist down any longer. But who cares? I look good and that's all that matters. And when I die of hypothermia for wearing formal shorts in winter, tell them to put that on my tombstone. — Eliza Coupe
The most touching epitaph I ever encountered was on the tombstone of the printer of Edinburgh. It said simply: He kept down the cost and set the type right. — Gregory Nunn
We live to love you more each day. — Jayne Mansfield
82+ Best Family Time Quotes: Exclusive Selection
54+ Best Garden Quotes: Exclusive Selection
225+ EXCLUSIVE Famous Quotes That Will Leave You Speechless
68+ Best Grandson Quotes That Will Warm Your Heart
100+ Best Machine Gun Kelly Quotes: Exclusive Selection
100+ Best Trauma Quotes: Exclusive Selection


text:
Metallica, 'The Unforgiven' – Official Music Video
Chuck Armstrong Published: November 19, 1991
Making its worldwide debut on Nov. 19, 1991, 'The Unforgiven' is the third music video released by Metallica. Like 'One,' there are multiple versions of this music video: the standard video that clocks in at six minutes and 24 seconds, and the "theatrical" version that runs more than 11 and a half minutes long.
The black and white video was filmed by Matt Mahurin in Los Angeles, and like the videos released before it, it follows along the theme of the lyrics while also showing clips of the band performing the song. The theatrical version of 'The Unforgiven' includes the same music video as the original version, but it also features several minutes of scenes prior to the start of the song.
Nearly five years after shooting this video, Mahurin would return to help Metallica with two more music videos. In addition to Metallica, he has worked with the likes of Alice in Chains, Bush, Hole and Soundgarden.
'The Unforgiven' is one of four songs on the band's 1991 self-titled album (aka The Black Album) that credits lead guitarist Kirk Hammett as a co-writer.
Filed Under: Metallica (Black Album), The Unforgiven
Categories: Metallica Videos, Official Metallica Music Videos

redpajama_set_name: RedPajamaCommonCrawl


text:
\section{Introduction}\label{IntroSec}

\subsection{Identities and Virasoro Characters}\label{AGSec}

The $q$-series identities of And\-rews--Gordon \cite{andrews74,gordon}
take the form:
\begin{equation}\label{AG}
\sum_{N_1 \ge \cdots \ge N_{k-1} \ge 0}
\frac{ q^{ N_{1}^2 + \cdots + N_{k-1}^2 + N_{i} + \cdots + N_{k-1} }}
     {(q)_{N_{1}-N_{2}} \cdots (q)_{N_{k-2}-N_{k-1}} (q)_{N_{k-1}}}
=\prod_{\begin{subarray}{c}
  n=1\\n\not\equiv0, \pm i \, (\mod 2k+1) \end{subarray}}^{\infty}
\frac{1}{1-q^n},
\end{equation}
where $|q|<1$ and, as usual,
$(q)_0=1$ and $(q)_n=\prod_{i=1}^{n}(1-q^i)$ for $n>0$.
Here $k\ge2$ and $1\le i\le k$.
The $k=2$ cases are the famous Rogers-Ramanujan identities
\cite{rogers,rogers-ramanujan}.

In the past twenty years, it was recognised that \eqref{AG} is
an identity for the (normalised) character $\chi_{1, i}^{2, 2k+1}$ of the
minimal model $M(2, 2k+1)_2$ of the Virasoro algebra.
In fact, the left side of \eqref{AG} has a combinatorial
interpretation in terms of particles which are forbidden to overlap.
It is for this reason that the left side of \eqref{AG} is
termed a {\em fermionic} expression.

The Virasoro minimal models $M(p,p')_2$ are labelled by
coprime integers $p$ and $p'$ for which $1<p<p'$.%
\footnote{$M(p,p')_2$ is often denoted $\mathcal M(p,p')$ or
$\mathcal M^{p,p'}$.}
They contain irreducible modules labelled by $r$ and $s$
with $1\le r<p$ and $1\le s<p'$.
In \cite{feigin-fuchs,rocha-caridi}, the characters of these
modules were calculated to be
$\hat\chi^{p, p'}_{r, s}=
q^{\Delta^{p, p'}_{r, s}} \chi^{p, p'}_{r, s}$,
where the normalised character $\chi^{p, p'}_{r, s}$ is given by:
\begin{equation}\label{bosonic-equation}
\chi^{p, p'}_{r, s}=
{\frac{1}{(q)_\infty}}
\sum_{\lambda=-\infty}^\infty
(
q^{\lambda^2pp'+\lambda(p'r-ps)}
-
q^{(\lambda p+r)(\lambda p'+s)}
),
\end{equation}
with $(q)_{\infty}=\prod_{i=1}^\infty (1-q^{i})$,
and the {\em conformal dimension} $\Delta^{p, p'}_{r, s}$ is given by:
\begin{equation}\label{conformal-equation}
\Delta^{p, p'}_{r, s}=
\frac{(p'r-ps)^2-(p'-p)^2}{4pp'}.
\end{equation}
This expression \eqref{bosonic-equation} for $\chi^{p, p'}_{r, s}$
is of a different nature to those on either side of \eqref{AG}.
It is known as a {\em bosonic} expression.
For later purposes, it will be useful to note that
$\chi^{p,p'}_{r,s}=\chi^{p,p'}_{p-r,p'-s}$ and that
$\chi^{p,p'}_{r,s}|_{q=0}=1$.

Following the recognition that \eqref{AG} is an identity for
$\chi^{2,2k+1}_{1,i}$, it was natural to seek
fermionic expressions for other Virasoro characters.
Gradually, as described in \cite{bms,welsh},
beginning with the pioneering work of the Stony Brook group \cite{kkmm},
an increasingly wide range of characters were tackled in a series of works,
culminating in \cite{welsh} giving fermionic expressions for all
minimal model Virasoro characters.
In the simplest cases, such as the subset of the $p=3$
cases of $\chi^{p,p'}_{r,s}$
tackled by \cite{andrews84,foda-quano}, the fermionic
expressions are strikingly similar to that in \eqref{AG},
with merely the coefficients of some of the parameters changed.
However, expressions for other characters \cite{kkmm,berkovich-mccoy},
most notably the unitary characters $\chi^{p,p+1}_{r,s}$,
necessitated the summand to have one or more Gaussian polynomial
factors, where as usual, the Gaussian polynomial $\qbinom{P}{N}$
is defined by:
\begin{equation}
\qbinom{P}{N}=
\left\{
  \begin{array}{cl}
    \displaystyle
    \frac{(q)_P}{(q)_N(q)_{P-N}}& \text{if $0\le N\le P$;}\\[3mm]
    0& \text{otherwise.}
  \end{array} \right.
\end{equation}
Such a multi-sum expression is known as a {\em fundamental fermionic form}.
In \cite{welsh}, the fermionic expression for $\chi^{p,p'}_{r,s}$
is generally a sum over a number of fundamental fermionic forms.

Of course, to obtain an identity which is an analogue of \eqref{AG},
it is necessary to find an expression for the character which is
a product similar in nature to that on the right.
In fact, as explained in \cite{christe}, $\chi^{p,p'}_{r,s}$
is a product of terms of the form $(1-q^n)^{-1}$ if and only
if $p=2r$, $p'=2s$, $p=3r$ or $p'=3s$:
\begin{equation} \label{ProdEq1}
\chi^{2r, p'}_{r, s}=
\hskip-1mm
\prod_{\begin{subarray}{c}
  n=1\\n\not\equiv0,\pm rs\,(\mod rp')\end{subarray}}^{\infty}
\hskip-2mm
\frac{1}{1-q^n},
\qquad
\chi^{p, 2s}_{r, s}=
\hskip-1mm
\prod_{\begin{subarray}{c}
  n=1\\n\not\equiv0,\pm rs\,(\mod sp)\phantom{'}\end{subarray}}^{\infty}
\hskip-2mm
\frac{1}{1 - q^n},
\end{equation}
\begin{equation}
\label{ProdEq3}
\chi^{3r, p'}_{r, s}=
\hskip-1mm
\prod_{\begin{subarray}{c}
  n=1\\n\not\equiv0,\pm rs\,(\mod 2rp')\\n\not\equiv\pm2r(p'-s)\,(\mod 4rp')\\
       \end{subarray}}^{\infty}
\hskip-3mm\frac{1}{1-q^n},
\qquad
\hskip-1mm
\chi^{p, 3s}_{r, s}=
\prod_{\begin{subarray}{c}
  n=1\\n\not\equiv0,\pm rs\,(\mod 2sp)\\n\not\equiv\pm2s(p-r)\,(\mod 4sp)\\
       \end{subarray}}^{\infty}
\hskip-3mm\frac{1}{1-q^n}.
\end{equation}
These expressions are easily derived by applying Jacobi's triple
product identity \cite[eq.\ (II.28)]{gasper-rahman}
in the cases $p=2r$ and $p'=2s$, and
Watson's quintuple product identity \cite[ex.\ 5.6]{gasper-rahman}
in the cases $p=3r$ and $p'=3s$, to the expression \eqref{bosonic-equation}.
On identifying one of these product expressions with a
fermionic expression for the same $\chi^{p,p'}_{r,s}$,
we obtain an identity which we refer to as a $M(p,p')_2$-identity.

\subsection{Combinations of Virasoro characters}

In \cite{bytfri}, the sum and difference of certain pairs of
characters were considered.
Firstly, use of Watson's quintuple product identity
and \eqref{bosonic-equation} leads to the following
expression (\cite[eqn.~(2.26)]{bytfri}):
\begin{equation}\label{Composep3Eq1}
\chi^{3,p'}_{1,s}\pm q^{\frac{p'}4-\frac s2}\chi^{3,p'}_{2,s}=
 \frac{(\mp q^{\frac{p'}4-\frac s2},\mp q^{\frac{p'}4+\frac s2},
                     q^{\frac{p'}2};q^{\frac{p'}2})_\infty
       (q^{s},q^{p'-s};q^{p'})_\infty}
         {(q)_\infty},
\end{equation}
for $p'\not\equiv0\,(\mod3)$ and $1\le s<p'$,
where
$(a_1,a_2,\ldots,a_t;z)_\infty=
\prod_{j=1}^t (a_j;z)_\infty$
with
$(a;z)_\infty=\prod_{i=0}^\infty (1-az^i)$ as usual.
It will be useful to note the following alternative form of this
expression when $p'\ne2s$:
\begin{equation}\label{Composep3Eq2}
\chi^{3,p'}_{1,s}\pm q^{\frac{p'}4-\frac s2}\chi^{3,p'}_{2,s}=
\frac{
  (q^{s},q^{\frac{p'}2-s},q^{\frac{p'}2};q^{\frac{p'}2})_\infty}
 {(q)_\infty
    (\pm q^{\frac{p'}4-\frac s2},\pm q^{\frac{p'}4+\frac s2};
                                         q^{\frac{p'}2})_\infty}.
\end{equation}

Secondly, use of Jacobi's triple product identity
and \eqref{bosonic-equation} leads to the following
expression (\cite[eqn.~(2.22)]{bytfri}):
\begin{equation}\label{Composep4Eq2}
\chi^{4,p'}_{1,s}\pm q^{\frac{p'}2-s}\chi^{4,p'}_{3,s}=
\frac{(q^{s},\mp q^{\frac{p'}2-s},\mp q^{\frac{p'}2};
                                    \mp q^{\frac{p'}2})_\infty}
         {(q)_\infty},
\end{equation}
for $p'\not\equiv0\,(\mod2)$ and $1\le s<p'$.
This may also be written:
\begin{equation}\label{Composep4Eq1}
\chi^{4,p'}_{1,s}\pm q^{\frac{p'}2-s}\chi^{4,p'}_{3,s}=
\frac{(\mp q^{\frac{p'}2-s},\mp q^{\frac{p'}2},\mp q^{\frac{p'}2+s},
      q^{s},q^{p'-s},q^{p'};q^{p'})_\infty}
         {(q)_\infty}.
\end{equation}

In what follows, we derive fermionic expressions for the \lq$+$\rq\
cases of the above character combinations.
This leads to analogues of the
Andrews-Gordon identities

In recognition of its origin in a character combination,
we refer to the identity for each sum
$\chi^{3,p'}_{1,s}+ q^{\frac{p'}4-\frac s2}\chi^{3,p'}_{2,s}$
of characters, as an $M(3,p')^+_2$-identity.
Similarly, the identity for each sum
$\chi^{4,p'}_{1,s}+ q^{\frac{p'}2-s}\chi^{4,p'}_{3,s}$
of characters, is referred to as an $M(4,p')^+_2$-identity.

\subsection{\boldmath $M(3,7)_3$-identities}

For $n>2$, the $W_n$ algebra \cite{zamo,fatlyk} is a generalisation
of the Virasoro algebra $W_2$. In \cite{asw}, Andrews, Schilling and
Warnaar applied a modified Bailey transform to obtain identities
(referred to in \cite{asw} as Rogers-Ramanujan type identities)
for three (of the possible four) characters of the minimal model
$M(3, 7)_3$ of $W_3$.\footnote{One of these identities
is also proved in \cite{warnaar} using Hall-Littlewood polynomials.}
In Section \ref{ASWSec} below,
we list these identities together with a conjectured identity for
the fourth character.
Notably, the summand on the fermionic side of each of these identities
contains a Gaussian polynomial.

In attempting to understand these $M(3,7)_3$-identities, we noticed that,
since the $W_n$ minimal model $M(p,p')_n$ has central charge
\begin{equation}\label{Ccharge}
c^{p,p'}_n = (n-1)\left(1 - \frac{n(n+1)(p' - p)^2}{p p'}\right),
\end{equation}
the $W_3$ minimal model $M(3, 7)_3$ has the same central
charge as that of the $W_2$ minimal model $M(3, 14)_2$.
This indicated that the characters of the $M(3, 7)_3$ theory should
be expressible as linear combinations of those of $M(3, 14)_2$.
In fact, the required combinations are precisely those of
\eqref{Composep3Eq1} for $s\in\{1,3,5\}$,
together with the single character $\chi^{3,14}_{1,7}$.
In each of the first three cases, the known fermionic expressions for
the characters $\chi^{3,14}_{1,s}$ and $\chi^{3,14}_{2,s}$
each comprises a single fundamental fermionic form \cite{bms,welsh},
and sum conveniently to yield a single fundamental fermionic form.
In the latter case, the known fermionic expression
is a sum of two fundamental fermionic forms \cite{welsh}.
However, they sum conveniently to yield a single fundamental fermionic form
in a manner that is similar to the other cases.

\subsection{\boldmath $M(p,p')^+_2$-identities}

Since the above means of combining $W_2$ characters extends readily to
many $p=3$ and $p=4$ cases,
the statement and proof of the $M(3,14)^+_2$-identities
have been subsumed into the general case.
The general $M(3,p')^+_2$-identities and $M(4,p')^+_2$-identities are
stated in Sections \ref{Resp3Sec} and \ref{Resp4Sec} respectively.
Their proofs are given in Section \ref{ProofSec}.

The $M(3,p')^+_2$-identities that we have obtained were conjectured
by Bytsko \cite{bytsko}. Our $M(4,p')^+_2$-identities are (to the
best of our knowledge) new.

As explained above, the $M(3,14)^+_2$-identities are also identities
for the $M(3,7)_3$ characters. In order to compare our
$M(3,14)^+_2$-identities with
the identities of \cite{asw} listed in Section \ref{ASWSec},
we write out the former explicitly in Section \ref{NewSec}.
We see that we now have two different fermionic expressions
for each of the four characters.
In each case, the two fermionic expressions are not simply
related to one another.

\section{Andrews-Gordon type identities}\label{ResSec}

In this section, we list our results.
In each identity, if $s>k$ a sum of the form
$(N_s+N_{s+1}+\cdots+N_k)$ is to be taken as $0$.
Each identity is proved in Section \ref{ProofSec}
using the procedure discussed above.

\subsection{\boldmath $M(3,p')^+_2$-{} and $M(3,p')_2$-identities}
\label{Resp3Sec}

Theorems \ref{FermCase3Cor} and \ref{FermCase1Cor} below give
$M(3,p')^+_2$-identities.
Theorems \ref{FermCase4Cor} and \ref{FermCase2Cor} below give
$M(3,p')_2$-identities.
The $M(3,p')^+_2$-identities originate from the sum of two
Virasoro characters, one that has $r=1$ and one that has
$r=2$, as explained above.
The $M(3,p')_2$-identities originate from one Virasoro character
only, as can be seen from the proofs in Section \ref{ProofSec}.

In Theorems \ref{FermCase3Cor} and \ref{FermCase1Cor}, the identities
are indexed by the integers $g$ and $s$. In these two cases, $p'$ is
related to $g$ by $p'=3g+1$ and $p'=3g+2$ respectively.
In Theorems \ref{FermCase4Cor} and \ref{FermCase2Cor}, the identities
are indexed by the integer $h$ only. From $h$, one can determine both
$p'$ and $s$, as explained in Section \ref{ProofSec}.

\begin{Theorem}\label{FermCase3Cor}
If $g\ge1$ and $1\le s\le g+1$, then:
\begin{equation}
\boxed{
\begin{aligned}
&\sum_{\begin{subarray}{c}
           N_1\ge\cdots\ge N_{g-1}\ge0\\[0.5mm]
           M\ge0
        \end{subarray}}
\hskip-4mm
\frac{q^{\sum_{j=1}^{g-1} N_j(N_j+M)
+\frac {g+1}4 M^2 +\frac {g-s}2 M
+(N_{s}+N_{s+1}+\cdots N_{g-1})}}
{(q)_{N_{1}-N_{2}} \cdots (q)_{N_{g-2}-N_{g-1}} (q)_{N_{g-1}} (q)_M}\\
&\hskip17mm
=\frac{(-q^{\frac{3g+1}4-\frac s2},-q^{\frac{3g+1}4+\frac s2},
                              q^{\frac{3g+1}2};q^{\frac{3g+1}2})_\infty
       (q^s,q^{3g+1-s};q^{3g+1})_\infty}
     {(q)_\infty}.
\end{aligned}
}
\end{equation}
\end{Theorem}

\begin{Theorem}\label{FermCase4Cor}
Let $h\ge1$. Then:
\begin{equation}\label{FermCor4}
\boxed{
\begin{aligned}
&\sum_{\begin{subarray}{c}
           N_1\ge\cdots\ge N_{2h}\ge0\\[0.5mm]
           M\ge0
        \end{subarray}}
\hskip-4mm
\frac{q^{\sum_{j=1}^{2h} N_j(N_j+M)
+\frac{h+1}2 M(M+1)-M
+(N_h+N_{h+1}+\cdots+N_{2h})}}
{(q)_{N_{1}-N_{2}} \cdots (q)_{N_{2h-1}-N_{2h}} (q)_{N_{2h}} (q)_M}\\
&\hskip55mm
=\frac{(q^{3h+2};q^{3h+2})_\infty}{(q)_\infty}.
\end{aligned}
}
\end{equation}
\end{Theorem}

\begin{Theorem}\label{FermCase1Cor}
If $g\ge1$ and $1\le s\le g+1$, then:
\begin{equation}
\boxed{
\begin{aligned}
&\sum_{\begin{subarray}{c}
           N_1\ge\cdots\ge N_{g-1}\ge0\\[0.5mm]
           M\ge0
        \end{subarray}}
\hskip-4mm
\frac{q^{\sum_{j=1}^{g-1} N_j(N_j+M)
+\frac g4 M^2 +\frac {g-s+1}2 M
+(N_{s}+N_{s+1}+\cdots+N_{g-1})}}
{(q)_{N_{1}-N_{2}} \cdots (q)_{N_{g-2}-N_{g-1}} (q)_{N_{g-1}} (q)_M}\\
&\hskip16mm
=\frac{(-q^{\frac{3g}4-\frac{s-1}2},-q^{\frac{3g}4+\frac{s+1}2},
                              q^{\frac{3g}2+1};q^{\frac{3g}2+1})_\infty
       (q^s,q^{3g+2-s};q^{3g+2})_\infty}
     {(q)_\infty}.
\end{aligned}
}
\end{equation}
\end{Theorem}

\begin{Theorem}\label{FermCase2Cor}
Let $h\ge1$. Then:
\begin{equation}
\boxed{
\begin{aligned}
&\sum_{\begin{subarray}{c}
           N_1\ge\cdots\ge N_{2h-1}\ge0\\[0.5mm]
           M\ge0
        \end{subarray}}
\hskip-4mm
\frac{q^{\sum_{j=1}^{2h-1} N_j(N_j+M)
+\frac h2 M(M+1)
+(N_h+N_{h+1}+\cdots+N_{2h-1})}}
{(q)_{N_{1}-N_{2}} \cdots (q)_{N_{2h-2}-N_{2h-1}} (q)_{N_{2h-1}} (q)_M}\\
&\hskip55mm
=\frac{(q^{3h+1};q^{3h+1})_\infty}{(q)_\infty}.
\end{aligned}
}
\end{equation}
\end{Theorem}

We note that, in view of the equality between the right sides
of \eqref{Composep3Eq1} and \eqref{Composep3Eq2},
in the cases in which both $p'$ and $p'/2-s$ are even, the identities
do not involve fractional powers of $q$, with the right side
readily written as a product of terms of the form
$(1-q^{n})^{-1}$ for various $n\in\field{Z}_{>0}$.
Similarly, in the cases in which $p'$ even and $p'/2-s$ is odd,
the right side is readily written as a product of terms of the form
$(1-q^{\frac n2})^{-1}$ for various $n\in\field{Z}_{>0}$,
and the identities
may be naturally viewed as identities in the indeterminate $q^{1/2}$,

When $p'$ is odd, the identities may be naturally viewed as identities
in the indeterminate $q^{1/4}$, with in particular, the
product expression able to be readily written as a product of terms of
the forms $(1-q^{\frac n4})^{-1}$ and $(1+q^{\frac m2})^{-1}$
for various $n,m\in\field{Z}_{>0}$.

The identities of Theorems \ref{FermCase4Cor} and \ref{FermCase2Cor},
which are indexed by the integer $h$,
do not involve fractional powers of $q$.
They are $M(3,6h+4)_2$-{} and $M(3,6h+2)_2$-identities respectively.

\subsection{\boldmath $M(4,p')^+_2$-identities}\label{Resp4Sec}

Theorems \ref{FermCase5Cor} and \ref{FermCase6Cor} below
give $M(4,4g+1)^+_2$-identities,
and Theorems \ref{FermCase7Cor} and \ref{FermCase8Cor} below
give $M(4,4g+3)^+_2$-identities.

\begin{Theorem}\label{FermCase5Cor}
Let $g\ge1$ and $1\le s\le g+1$. Then:
\begin{equation}\label{FermCase5CorEq}
\boxed{
\begin{aligned}
&\sum_{\raisebox{0mm}{$
      \begin{subarray}{c}
           N_1\ge\cdots\ge N_{g-1}\ge0\\[0.5mm]
           M\ge0
        \end{subarray}$}}
\frac{q^{\sum_{j=1}^{g-1} N_j(N_j+2m)
+(g+\frac12)M^2+(g-s)M
+(N_s+N_{s+1}+\cdots+N_{g-1})}}
{(q)_{N_{1}-N_{2}} \cdots (q)_{N_{g-2}-N_{g-1}} (q)_{N_{g-1}}
(q^{\frac12};q)_M(q^2;q^2)_M}\\
&\hskip15mm
=\frac{(-q^{2g-s+\frac12},-q^{2g+\frac12},-q^{2g+s+\frac12},
      q^{s},q^{4g+1-s},q^{4g+1};q^{4g+1})_\infty}
         {(q)_\infty}.
\end{aligned}
}
\end{equation}
\end{Theorem}

\begin{Theorem}\label{FermCase6Cor}
Let $g\ge1$. Then:
\begin{equation}\label{FermCase6CorEq}
\boxed{
\begin{aligned}
&\sum_{\raisebox{0mm}{$
      \begin{subarray}{c}
           N_1\ge\cdots\ge N_{g-1}\ge0\\[0.5mm]
           M\ge0
        \end{subarray}$}}
\frac{q^{\sum_{j=1}^{g-1} N_j(N_j+2M+1) + (g+\frac12)M^2+gM}}
{(q)_{N_{1}-N_{2}} \cdots (q)_{N_{g-2}-N_{g-1}} (q)_{N_{g-1}}
(q^{\frac12};q)_{M+1}(q^2;q^2)_M}\\
&\hskip30mm
=\frac{(-q^{\frac12},-q^{2g+\frac12},-q^{4g+\frac12},
      q^{2g},q^{2g+1},q^{4g+1};q^{4g+1})_\infty}
         {(q)_\infty}.
\end{aligned}
}
\end{equation}
\end{Theorem}

\begin{Theorem}\label{FermCase7Cor}
Let $g\ge1$ and $1\le s\le g+1$. Then:
\begin{equation}
\boxed{
\begin{aligned}
&\sum_{\raisebox{0mm}{$
      \begin{subarray}{c}
           N_1\ge\cdots\ge N_{g-1}\ge0\\[0.5mm]
           M\ge0
        \end{subarray}$}}
\frac{q^{\sum_{j=1}^{g-1} N_j(N_j+2M)
+M(Mg+g+1-s)
+(N_s+N_{s+1}+\cdots+N_{g-1})}}
{(q)_{N_{1}-N_{2}} \cdots (q)_{N_{g-2}-N_{g-1}} (q)_{N_{g-1}}
(q^{\frac12};q)_{M}(q^2;q^2)_M}\\
&\hskip15mm
=\frac{(-q^{2g-s+\frac32},-q^{2g+\frac32},-q^{2g+s+\frac32},
      q^{s},q^{4g+3-s},q^{4g+3};q^{4g+3})_\infty}
         {(q)_\infty}.
\end{aligned}
}
\end{equation}
\end{Theorem}

\begin{Theorem}\label{FermCase8Cor}
Let $g\ge1$. Then:
\begin{equation}
\boxed{
\begin{aligned}
&\sum_{\raisebox{0mm}{$
      \begin{subarray}{c}
           N_1\ge\cdots\ge N_{g-1}\ge0\\[0.5mm]
           M\ge0
        \end{subarray}$}}
\frac{q^{\sum_{j=1}^{g-1} N_j(N_j+2M+1) + gM(M+1)}}
{(q)_{N_{1}-N_{2}} \cdots (q)_{N_{g-2}-N_{g-1}} (q)_{N_{g-1}}
(q^{\frac12};q)_{M+1}(q^2;q^2)_M}\\
&\hskip25mm
=\frac{(-q^{\frac12},-q^{2g+\frac32},-q^{4g+\frac52},
      q^{2g+1},q^{2g+2},q^{4g+3};q^{4g+3})_\infty}
         {(q)_\infty}.
\end{aligned}
}
\end{equation}
\end{Theorem}

In each case above, the identities may be viewed as identities in the
indeterminate $q^{1/2}$, with the product expression readily
written as a product of terms of the forms
$(1-q^{\frac n2})^{-1}$ and $(1+q^{m})^{-1}$
for various $n,m\in\field{Z}_{>0}$.

\subsection{\boldmath $M(3,7)_3$ Andrews-Gordon identities}\label{NewSec}

The $g=4$ cases of Theorem \ref{FermCase1Cor} for
$s\in\{1,3,5\}$ and the $h=2$ case of Theorem \ref{FermCase2Cor},
now yield identities for the four $M(3,7)_3$ characters.
Written out in full, these are:

\begin{subequations}
\begin{multline}
\sum_{n_1, n_2, n_3, n_4 \geq 0}
\frac{
q^{
(n_1+n_2+n_3)^2 + (n_2+n_3)^2 + {n_3}^2 + {n_4}^2
+ (n_1 + 2 n_2 + 3 n_3) n_4
+ (n_1 + 2 n_2 + 3 n_3)
+ 2 n_4
}
}
{(q)_{n_1} (q)_{n_2} (q)_{n_3} (q)_{n_4}}\\
=
\prod_{n=1}^{\infty}
\frac{1}{(1-q^{7n-2})(1-q^{7n-3})^2(1-q^{7n-4})^2(1-q^{7n-5})},
\end{multline}

\begin{multline}
\sum_{n_1, n_2, n_3, n_4 \geq 0}
\frac{
q^{
(n_1+n_2+n_3)^2 + (n_2+n_3)^2 + n_3^2 + n_4^2
+ (n_1 + 2 n_2 + 3 n_3) n_4 + n_3 + n_4
}}
{
(q)_{n_1} (q)_{n_2} (q)_{n_3} (q)_{n_4}}
\\
=
\prod_{n=1}^{\infty}
\frac{1}{(1-q^{7n-1})(1-q^{7n-2})^2 (1-q^{7n-5})^2
(1-q^{7n-6})},
\end{multline}

\begin{multline}
\sum_{n_1, n_2, n_3, n_4 \geq 0}
\frac{q^{
(n_1+n_2+n_3)^2 + (n_2+n_3)^2 + n_3^2 + n_4^2
+ (n_1 + 2 n_2 + 3 n_3) n_4
}}{
(q)_{n_1} (q)_{n_2} (q)_{n_3} (q)_{n_4}
}\\
=
\prod_{n=1}^{\infty}
\frac{1}
{(1-q^{7n-1})^2(1-q^{7n-3})(1-q^{7n-4})(1-q^{7n-6})^2},
\end{multline}

\begin{multline}
\sum_{n_1, n_2, n_3, n_4 \geq 0}
\frac{q^{
(n_1+n_2+n_3)^2 + (n_2+n_3)^2 + n_3^2 + n_4^2
+ (n_1 + 2 n_2 + 3 n_3) n_4 + (n_2 + 2 n_3) + n_4
}}{
(q)_{n_1} (q)_{n_2} (q)_{n_3} (q)_{n_4}
}\\
=
\prod_{n=1}^{\infty}
\frac{1}
{(1-q^{7n-1})(1-q^{7n-2})(1-q^{7n-3})(1-q^{7n-4})(1-q^{7n-5})(1-q^{7n-6})}.
\end{multline}
\end{subequations}

\subsection{\boldmath $M(3,7)_3$ ASW identities}\label{ASWSec}

In \cite{asw}, Andrews, Schilling and Warnaar obtained
$q$-series identities for three characters of
the $M(3, 7)_3$ minimal model.
These identities differ from those we have derived above.
For comparison purposes,
we list these identities here, together with a conjectured identity
\eqref{ConjASW} for the fourth $M(3, 7)_3$ character:
\begin{subequations}
\begin{multline}
\sum_{n_1, n_2 \geq 0}
\frac{q^{n_1^2 - n_1 n_2 + n_2^2 + n_1 + n_2}}{(q)_{n_1}}
\qbinom{2n_1}{n_2} \\
=
\prod_{n=1}^{\infty}
\frac{1}{(1-q^{7n-2})(1-q^{7n-3})^2(1-q^{7n-4})^2(1-q^{7n-5})},
\end{multline}

\begin{multline}\label{ConjASW}
\sum_{n_1, n_2 \geq 0}
\frac{q^{n_1^2 - n_1 n_2 +n_2^2 + n_2}}{(q)_{n_1}}
\qbinom{2 n_1 + 1}{n_2} \\
=
\prod_{n=1}^{\infty}
\frac{1}{(1-q^{7n-1})(1-q^{7n-2})^2 (1-q^{7n-5})^2 (1-q^{7n-6})},
\end{multline}

\begin{multline}
\sum_{n_1, n_2 \geq 0}
\frac{q^{n_1^2 -n_1 n_2 + n_2^2}}{(q)_{n_1}}
\qbinom{2n_1}{n_2} \\
=
\prod_{n=1}^{\infty}
\frac{1}{(1-q^{7n-1})^2(1-q^{7n-3})(1-q^{7n-4})(1-q^{7n-6})^2},
\end{multline}

\begin{multline}
\sum_{n_1,n_2\geq 0}
\frac{q^{n_1^2-n_1 n_2+n_2^2+n_1}}{(q)_{n_1}}\qbinom{2n_1+1}{n_2}
=
\sum_{n_1,n_2\geq 0}
\frac{q^{n_1^2-n_1 n_2+n_2^2+n_2}}{(q)_{n_1}}\qbinom{2n_1}{n_2}\\
=
\prod_{n=1}^{\infty}\frac{1}{(1-q^{7n-1})(1-q^{7n-2})(1-q^{7n-3})
(1-q^{7n-4})(1-q^{7n-5})(1-q^{7n-6})}.
\end{multline}
\end{subequations}

\subsection{Special cases}\label{Intp3Sec}
Here we note that the simplest cases of the results in Section \ref{ResSec}
yield known identities of Rogers-Ramanujan type.

The $g=1$ cases of Theorem \ref{FermCase3Cor} yield the
$z=q^{\frac12}$ and $z=1$ specialisations of Euler's formula
\cite[II.2]{gasper-rahman}.

The $g=1$ cases of Theorem \ref{FermCase1Cor} yield the
following identities after substituting $q\rightarrow q^4$:
\begin{subequations}
\begin{equation}\label{RogersEq1}
\sum_{m=0}^\infty
\frac{q^{m(m+2)}}{(q^4;q^4)_m}=
\frac{(-q^3,-q^7,q^{10};q^{10})_\infty (q^4,q^{16};q^{20})_\infty}
     {(q^4;q^4)_\infty},
\end{equation}
\begin{equation}\label{RogersEq2}
\sum_{m=0}^\infty
\frac{q^{m^2}}{(q^4;q^4)_m}=
\frac{(-q,-q^9,q^{10};q^{10})_\infty (q^8,q^{12};q^{20})_\infty}
     {(q^4;q^4)_\infty}.
\end{equation}
\end{subequations}
These identities are equivalent to identities of Rogers \cite{rogers}.
They also appear in \cite[eqns.\ (16) \&\ (20)]{slater}.

The $g=1$ cases of Theorems \ref{FermCase5Cor} and \ref{FermCase6Cor}
yield the following identities after substituting $q\rightarrow q^2$:
\begin{subequations}
\begin{equation}\label{RogersEq3}
\sum_{m=0}^\infty
\frac{q^{3m^2}}{(q;q^2)_m(q^4;q^4)_m}=
\frac{(-q^3,-q^5,-q^7;q^{10})_\infty}
     {(q^4,q^6;q^{10})_\infty},
\end{equation}

\begin{equation}\label{RogersEq4}
\sum_{m=0}^\infty
\frac{q^{3m^2+2m}}{(q;q^2)_{m+1}(q^4;q^4)_m}=
\sum_{m=0}^\infty
\frac{q^{3m^2-2m}}{(q;q^2)_m(q^4;q^4)_m}=
\frac{(-q,-q^5,-q^9;q^{10})_\infty}
     {(q^2,q^8;q^{10})_\infty}.
\end{equation}
\end{subequations}
In an alternative form, \eqref{RogersEq3} and the second identity
in \eqref{RogersEq4} are due to Rogers \cite{rogers}.
They also appear in \cite[eqns.\ (19) \&\ (15)]{slater}.
Bailey \cite{bailey47} also derives \eqref{RogersEq3} and
the two identities \eqref{RogersEq4}.

{}From Theorems \ref{FermCase7Cor} and \ref{FermCase8Cor}, we obtain:
\begin{subequations}
\begin{equation}\label{SelbergEq1}
\sum_{m=0}^\infty
\frac{q^{2m(m+1)}}{(q;q^2)_m(q^4;q^4)_m}=
\frac{(-q^5,-q^7,-q^9;q^{14})_\infty}
     {(q^4,q^6,q^8,q^{10};q^{14})_\infty},
\end{equation}

\begin{equation}\label{SelbergEq2}
\sum_{m=0}^\infty
\frac{q^{2m^2}}{(q;q^2)_m(q^4;q^4)_m}=
\frac{(-q^3,-q^7,-q^{11};q^{14})_\infty}
     {(q^2,q^6,q^8,q^{12};q^{14})_\infty},
\end{equation}

\begin{equation}\label{SelbergEq3}
\sum_{m=0}^\infty
\frac{q^{2m(m+1)}}{(q;q^2)_{m+1}(q^4;q^4)_m}=
\frac{(-q,-q^7,-q^{13};q^{14})_\infty}
     {(q^2,q^4,q^{10},q^{12};q^{14})_\infty}.
\end{equation}
\end{subequations}
These identities are alternative forms of the Rogers-Selberg identities
which are originally due to Rogers \cite{rogers}.
They also appear in \cite[eqns.\ (32), (33) \&\ (31)]{slater}.

\section{Proofs of fermionic combinations}\label{ProofSec}

In this section, we prove the expressions of Section \ref{ResSec}.
In each case, the first step is to extract fermionic expressions for
characters $\chi^{3,p'}_{1,s}$ or $\chi^{4,p'}_{1,s}$ from \cite{welsh}.
Some of these expressions can also be found in \cite{berkovich-mccoy,bms}.
Having obtained the fermionic forms, the required proofs
result upon combining them appropriately and then using
either \eqref{Composep3Eq1} and \eqref{Composep4Eq2}.

In each of the cases below, we use the notation of \cite{welsh}.

\subsection{The $(p,p')=(3,3g+1)$ case}\label{p3SecA}

The continued fraction of $p'/p=(3g+1)/3$ is $[g,3]$.
Then, from \cite[\S1.3]{welsh}, $n=1$, $t=g+1$ and $t_1=g-1$.
The set ${\mathcal T}$ of Takahashi lengths is given by
${\mathcal T}=(1,2,\ldots,g,g+1)$.
The set $\tilde{\mathcal T}$ of truncated Takahashi lengths is
given by $\tilde{\mathcal T}=(1)$.
{}From \cite[\S1.11]{welsh},
$\boldsymbol{B}$ is the $(g-1)\times (g-1)$ matrix
$\boldsymbol{B}=\{B_{j\ell}\}_{1\le j,\ell\le g-1}$
with entries $B_{j\ell}=\min\{j,\ell\}$, and
$\overline{\boldsymbol{C}}$ is the $1\times 1$ matrix $(2)$.
The matrix $\overline{\boldsymbol{C}}^*$ is the $1\times 1$ matrix $(-1)$.

Throughout this subsection, we take
$N_j=n_j+n_{j+1}+\cdots+n_{g-1}$ for $1\le j\le g-1$,
and the $(g+1)$-dimensional vectors $\boldsymbol{e}_j$ are defined by
$\boldsymbol{e}_j=(\delta_{1j},\delta_{2j},\ldots,\delta_{g+1,j})$
for $0\le j\le g+2$.

\begin{Lemma}\label{FermCase3}
Let $g\ge1$ and $1\le s\le g+1$. Then:
\begin{subequations}
\begin{equation}\label{FermEx3a}
\chi^{3,3g+1}_{1,s}=
\hskip-5mm
\sum_{\begin{subarray}{c}
           n_1,n_2,\ldots,n_{g-1}\in\field{Z}_{\ge0}\\
           m\in\field{Z}_{\ge0}\\
           m\equiv0\,(\mod2)
        \end{subarray}}
\hskip-8mm
\frac{q^{\sum_{j=1}^{g-1} N_j(N_j+m)
+\frac{g+1}4 m^2 +\frac {g-s}2 m
+(N_s+N_{s+1}+\cdots+N_{g-1})}}
{(q)_{n_1}(q)_{n_2}\cdots (q)_{n_{g-1}}(q)_m}
\end{equation}
and
\begin{equation}\label{FermEx3b}
\chi^{3,3g+1}_{2,s}=
q^{-\frac{3g}4+\frac s2-\frac14}
\hskip-5mm
\sum_{\begin{subarray}{c}
           n_1,n_2,\ldots,n_{g-1}\in\field{Z}_{\ge0}\\
           m\in\field{Z}_{\ge0}\\
           m\equiv1\,(\mod2)
        \end{subarray}}
\hskip-8mm
\frac{q^{\sum_{j=1}^{g-1} N_j(N_j+m)
+\frac{g+1}4 m^2 +\frac {g-s}2 m
+(N_s+N_{s+1}+\cdots+N_{g-1})}}
{(q)_{n_1}(q)_{n_2}\cdots (q)_{n_{g-1}}(q)_m}.
\end{equation}
\end{subequations}
Each term $(N_s+N_{s+1}+\cdots+N_{g-1})$ is to be taken as $0$
whenever $s\ge g$.
\end{Lemma}

\medskip\noindent {\it Proof: }
Since with $1\le s\le g+1$, we have $s\in{\mathcal T}$,
and for $r=1$, we have $r\in\tilde{\mathcal T}$, the fermionic form
\cite[(1.16)]{welsh} for $\chi^{p,p'}_{r,s}$ comprises a single
fundamental form \cite[(1.17)]{welsh}.
The summation in \cite[(1.17)]{welsh} is then over integers
$n_1,n_2,\ldots,n_{g-1}$ and $m$
(this last parameter is named $m_{g}$ in \cite[(1.17)]{welsh}),
where the parity of $m$ is restricted.

{}From \cite[\S1.7]{welsh},
we obtain:
\begin{equation*}
\boldsymbol{u}^L=\begin{cases}
\boldsymbol{e}_{g}&\text{if }s=g+1;\\
0&\text{if }s=g;\\
\boldsymbol{e}_{s-1}-\boldsymbol{e}_{g-1}&\text{if }s\le g-1.
\end{cases}
\end{equation*}
{}From \cite[\S1.9]{welsh}, we then get
$\overline{\boldu}^L_\flat=(1)$ if $s=g+1$ and
$\overline{\boldu}^L_\flat=(0)$ if $s\le g$.
On the other hand, $\boldsymbol{u}^R=\boldsymbol{e}_g$ and
$\overline{\boldu}^R_\sharp=(0)$.
For \cite[(1.18)]{welsh}, we then get
$\tilde\boldsymbol{n}=(\tilde n_1,\tilde n_2,\ldots,\tilde n_{g-1})$
where, if $s\in\{g,g+1\}$ then $\tilde n_{g-1}=n_{g-1}+\frac12m$
and $\tilde n_j=n_j$ for $1\le j<g-1$,
and if $1\le s<g$ then $\tilde n_{g-1}=n_{g-1}+\frac12m+\frac12$,
$\tilde n_{s-1}=n_{s-1}-\frac12$
and $\tilde n_j=n_j$ for $1\le j<s-1$ and $s\le j<g-1$.
The expression \cite[(1.17)]{welsh} then yields equation \eqref{FermEx3a}
up to an overall factor, where $m$ is summed over even integers
because here $\overline{\boldu}=(0)$ leads to $\overline{\boldsymbol{Q}}(\boldsymbol{u})=(0)$.
The overall factor may be obtained after calculating $\gamma$ as in
\cite[\S1.10]{welsh}.
However, we can bypass this calculation by noting that
the smallest value of the exponent in the numerator of the summand in
equation \eqref{FermEx1a} occurs when $n_1=n_2=\cdots=n_{2g-1}=m=0$.
Since $\chi^{p,p'}_{r,s}|_{q=0}=1$,
it follows that the required factor is simply $1$.

For the $\chi^{3,p'}_{2,s}$ case, we obtain $\boldsymbol{u}^L$ as above.
Using $r=2$ instead of $r=1$ results in
$\boldsymbol{u}^R=\boldsymbol{e}_g+\boldsymbol{e}_{g+1}$.
This yields $\overline{\boldu}^R_\sharp=(0)$ as above.
However here, $m\equiv1\,(\mod2)$ in the summation
because $\overline{\boldu}=(1)$ leads to $\overline{\boldsymbol{Q}}(\boldsymbol{u})\equiv(-1)$.
Thus, apart from the overall factor,
the fermionic form \cite[(1.17)]{welsh} for
$\chi^{3,p'}_{2,s}$ differs from the $r=1$ case above only
in that $m$ is summed over odd integers.
The smallest value of the exponent in the numerator of the summand
in equation \eqref{FermEx3b} occurs when
$n_1=n_2=\cdots=n_{2g-1}=0$ and $m=1$.
It follows that the required factor is
$q^{-\frac{3g}4+\frac s2-\frac14}$.
\hfill $\Box$ \medskip

\Proofof{Theorem \ref{FermCase3Cor}}
Combining Lemma \ref{FermCase3} with equation \eqref{Composep3Eq1}
immediately gives the required result.
\hfill $\Box$ \medskip

\begin{Lemma}\label{FermCase4}
Let $h\ge1$ and set $s=3h+2$. Then:
\begin{equation}\label{FermEx4}
\chi^{3,2s}_{1,s}=
\hskip-5mm
\sum_{\begin{subarray}{c}
           n_1,n_2,\ldots,n_{2h}\in\field{Z}_{\ge0}\\
           m\in\field{Z}_{\ge0}
        \end{subarray}}
\hskip-8mm
\frac{q^{\sum_{j=1}^{2h} N_j(N_j+m)
+\frac{h+1}2 m(m+1)-m
+(N_h+N_{h+1}+\cdots+N_{2h})}}
{(q)_{n_1}(q)_{n_2}\cdots (q)_{n_{2h}}(q)_m},
\end{equation}
where $N_j=n_j+n_{j+1}+\cdots+n_{2h}$.
\end{Lemma}

\medskip\noindent {\it Proof: }
Here $p'=6h+4$, and so the above data applies for $g=2h+1$.
Since $s\not\in{\mathcal T}$, the fermionic expression \cite[(1.16)]{welsh} is
a sum over more than one fundamental fermionic form.
In fact, the Takahashi tree for $s$ (\cite[\S1.6]{welsh}) has
precisely two leaf nodes. Since the truncated Takahashi tree for $r=1$
has one leaf node, the sum \cite[(1.16)]{welsh} is over two terms.

For the leaf node $a_{00}$ of the Takahashi tree for $s$,
we obtain (\cite[\S1.7]{welsh}) $d=2$,
$\Delta_1=1$, $\sigma_1=g$, $\tau_1=g+2$,
$\Delta_2=-1$, $\tau_2=g-2$ and $\sigma_2=(g-3)/2$.
This leads to
$\boldsymbol{u}^L=\boldsymbol{e}_{(g-3)/2}-\boldsymbol{e}_{g-2}+\boldsymbol{e}_{g}+\boldsymbol{e}_{g+1}$.
Thereupon, $\overline{\boldu}^L_\flat=(1)$.
On the other hand $r=1$ leads to
$\boldsymbol{u}^R=\boldsymbol{e}_{g}$ and $\overline{\boldu}^R_\sharp=(0)$.
For this case, the calculation described in (\cite[\S1.10]{welsh})
yields the constant term $\gamma(\chi^L,\chi^R)=-(g-1)/2$.
The fundamental fermionic form \cite[(1.17)]{welsh} for
this case is then given by precisely the right side of equation
\eqref{FermEx4}, with the summation restricted to odd integers $m$
because $\overline{\boldu}=(1)$ here, which yields $\overline{\boldsymbol{Q}}(\boldsymbol{u})\equiv(-1)$.

For the leaf node $a_{10}$ of the Takahashi tree for $s$,
we obtain $d=2$,
$\Delta_1=-1$, $\sigma_1=g-1$, $\tau_1=g+2$,
$\Delta_2=1$, $\tau_2=g-2$ and $\sigma_2=(g-3)/2$.
This leads to $\boldsymbol{u}^L=\boldsymbol{e}_{(g-3)/2}-\boldsymbol{e}_{g-2}+\boldsymbol{e}_{g}$.
Thereupon, $\overline{\boldu}^L_\flat=(1)$.
Again we then obtain $\gamma(\chi^L,\chi^R)=-(g-1)/2$.
The fundamental fermionic form for this case
is then given by precisely the right side of equation \eqref{FermEx4},
with the summation restricted to even integers $m$ because
$\overline{\boldu}=(0)$ here, which yields $\overline{\boldsymbol{Q}}(\boldsymbol{u})=(0)$.
The expression \eqref{FermEx4} follows.
\hfill $\Box$ \medskip

\Proofof{Theorem \ref{FermCase4Cor}}
Combining Lemma \ref{FermCase4} with the $p'=2s$ case of
equation \eqref{Composep3Eq1}, and cancelling a factor of 2 from
each side, yields the required result.
\hfill $\Box$ \medskip

\subsection{The $(p,p')=(3,3g+2)$ case}\label{p3SecB}

The continued fraction of $p'/p=(3g+2)/3$ is $[g,1,2]$.
Then, from \cite[\S1.3]{welsh}, $n=2$, $t=g+1$, $t_1=g-1$ and $t_2=g$.
The set ${\mathcal T}$ of Takahashi lengths is given by
${\mathcal T}=(1,2,\ldots,g+1)$.
The set $\tilde{\mathcal T}$ of truncated Takahashi lengths is
given by $\tilde{\mathcal T}=(1)$.
{}From \cite[\S1.11]{welsh},
$\boldsymbol{B}$ is the $(g-1)\times(g-1)$ matrix
$\boldsymbol{B}=\{B_{j\ell}\}_{1\le j,\ell\le g-1}$
with entries $B_{j\ell}=\min\{j,\ell\}$, and
$\overline{\boldsymbol{C}}$ is the $1\times 1$ matrix $(1)$.
The matrix $\overline{\boldsymbol{C}}^*$ is the $1\times 1$ matrix $(-1)$.

Throughout this subsection, we take
$N_j=n_j+n_{j+1}+\cdots+n_{g-1}$ for $1\le j\le g-1$,
and the $(g+1)$-dimensional vectors $\boldsymbol{e}_j$ are defined by
$\boldsymbol{e}_j=(\delta_{1j},\delta_{2j},\ldots,\delta_{g+1,j})$
for $0\le j\le g+2$.

\begin{Lemma}\label{FermCase1}
Let $g\ge1$ and $1\le s\le g+1$. Then:
\begin{subequations}
\begin{equation}\label{FermEx1a}
\chi^{3,3g+2}_{1,s}=
\hskip-5mm
\sum_{\begin{subarray}{c}
           n_1,n_2,\ldots,n_{g-1}\in\field{Z}_{\ge0}\\
           m\in\field{Z}_{\ge0}\\
           m\equiv0\,(\mod2)
        \end{subarray}}
\hskip-8mm
\frac{q^{\sum_{j=1}^{g-1} N_j(N_j+m)
+\frac g4 m^2 +\frac{g-s+1}2 m
+(N_s+N_{s+1}+\cdots+N_{g-1})}}
{(q)_{n_1}(q)_{n_2}\cdots (q)_{n_{g-1}}(q)_m}
\end{equation}
and
\begin{equation}\label{FermEx1b}
\chi^{3,3g+2}_{2,s}=
q^{-\frac{3g}4+\frac s2-\frac12}
\hskip-5mm
\sum_{\begin{subarray}{c}
           n_1,n_2,\ldots,n_{g-1}\in\field{Z}_{\ge0}\\
           m\in\field{Z}_{\ge0}\\
           m\equiv1\,(\mod2)
        \end{subarray}}
\hskip-8mm
\frac{q^{\sum_{j=1}^{g-1} N_j(N_j+m)
+\frac g4 m^2+\frac{g-s+1}2 m
+(N_s+N_{s+1}+\cdots+N_{g-1})}}
{(q)_{n_1}(q)_{n_2}\cdots (q)_{n_{g-1}}(q)_m}.
\end{equation}
\end{subequations}
Each term $(N_s+N_{s+1}+\cdots+N_{g-1})$ is to be taken as $0$
whenever $s\ge g$.
\end{Lemma}

\medskip\noindent {\it Proof: }
Since the expressions here are obtained in a manner very similar to those of
Lemma \ref{FermCase3}, we do not give the details.
We only note that with $1\le s\le g+1$ here,
$s\in{\mathcal T}$ and then:
\begin{equation*}
\boldsymbol{u}^L=\begin{cases}
0&\text{if }s=g+1;\\
-\boldsymbol{e}_{g}&\text{if }s=g;\\
\boldsymbol{e}_{s-1}-\boldsymbol{e}_{g-1}-\boldsymbol{e}_g&\text{if }s\le g-1.
\end{cases}
\end{equation*}
\hfill $\Box$ \medskip

\Proofof{Theorem \ref{FermCase1Cor}}
Combining Lemma \ref{FermCase1} with equation \eqref{Composep3Eq1}
immediately yields the required result.
\hfill $\Box$ \medskip

\begin{Lemma}\label{FermCase2}
Let $h\ge1$ and $s=3h+1$. Then:
\begin{equation}\label{FermEx2}
\chi^{3,2s}_{1,s}=
\hskip-5mm
\sum_{\begin{subarray}{c}
           n_1,n_2,\ldots,n_{2h-1}\in\field{Z}_{\ge0}\\
           m\in\field{Z}_{\ge0}
        \end{subarray}}
\hskip-8mm
\frac{q^{\sum_{j=1}^{2h-1} N_j(N_j+m)
+\frac h2 m(m+1)
+(N_h+N_{h+1}+\cdots+N_{2h-1})}}
{(q)_{n_1}(q)_{n_2}\cdots (q)_{n_{2h-1}}(q)_m},
\end{equation}
where $N_j=n_j+n_{j+1}+\cdots+n_{2h-1}$.
\end{Lemma}

\medskip\noindent {\it Proof: }
Since the expression here is obtained in a manner very similar to that of
Lemma \ref{FermCase4}, we do not give the details.
We only note that, after setting $g=2h$,
the leaf node $a_{00}$ of the Takahashi tree for $s$
yields $d=2$,
$\Delta_1=1$, $\sigma_1=g$, $\tau_1=g+2$,
$\Delta_2=-1$, $\tau_2=g-1$ and $\sigma_2=g/2-1$,
which leads to
$\boldsymbol{u}^L=\boldsymbol{e}_{g/2-1}-\boldsymbol{e}_{g-1}+\boldsymbol{e}_{g+1}$,
$\overline{\boldu}^L_\flat=(0)$ and
$\gamma(\chi^L,\chi^R)=-g/2$;
and the leaf node $a_{10}$ yields $d=2$,
$\Delta_1=-1$, $\sigma_1=g+2$, $\tau_1=g+2$,
$\Delta_2=1$, $\tau_2=g-1$ and $\sigma_2=g/2-1$,
which leads to $\boldsymbol{u}^L=\boldsymbol{e}_{g/2-1}-\boldsymbol{e}_{g-1}$,
$\overline{\boldu}^L_\flat=(0)$ and
$\gamma(\chi^L,\chi^R)=-g/2$.
\hfill $\Box$ \medskip

\Proofof{Theorem \ref{FermCase2Cor}}
Combining Lemma \ref{FermCase2} with the $p'=2s$ case of
equation \eqref{Composep3Eq1}, and cancelling a factor of 2 from
each side, yields the required result.
\hfill $\Box$ \medskip

\subsection{The $(p,p')=(4,4g+1)$ case}\label{p4SecA}

In this and the following subsection, we concentrate on the
cases where $p=4$.
In these cases, we make use of the following simple consequences
of the $q$-binomial theorem \cite[II.4]{gasper-rahman}:
\begin{subequations}
\begin{align}
\label{CauchyEq1}
&\sum_{k=0}^\infty q^{\frac12 k^2}\qbinom Pk
=(-q^{\frac12};q)_P,\\
\label{CauchyEq2}
&\sum_{k=0}^\infty q^{\frac12 k^2-Pk}\qbinom Pk
=q^{-\frac12 P^2} (-q^{\frac12};q)_P.
\end{align}
\end{subequations}

The continued fraction of $p'/p=(4g+1)/4$ is $[g,4]$.
Then, from \cite[\S1.3]{welsh}, $n=1$, $t=g+2$ and $t_1=g-1$.
The set ${\mathcal T}$ of Takahashi lengths is given by
${\mathcal T}=(1,2,\ldots,g,g+1,2g+1)$.
The set $\tilde{\mathcal T}$ of truncated Takahashi lengths is
given by $\tilde{\mathcal T}=(1,2)$.
{}From \cite[\S1.11]{welsh},
$\boldsymbol{B}$ is the $(g-1)\times(g-1)$ matrix
$\boldsymbol{B}=\{B_{j\ell}\}_{1\le j,\ell\le g-1}$
with entries $B_{j\ell}=\min\{j,\ell\}$, and
$\overline{\boldsymbol{C}}$ is the $2\times 2$ matrix
$\left(\begin{smallmatrix}2&-1\\-1&2\end{smallmatrix}\right)$.
The matrix $\overline{\boldsymbol{C}}^*$ is the $2\times 2$
matrix $\left(\begin{smallmatrix}-1&2\\0&-1\end{smallmatrix}\right)$.

Throughout this subsection, we take
$N_j=n_j+n_{j+1}+\cdots+n_{g-1}$ for $1\le j\le g-1$,
and the $(g+2)$-dimensional vectors $\boldsymbol{e}_j$ are defined by
$\boldsymbol{e}_j=(\delta_{1j},\delta_{2j},\ldots,\delta_{g+2,j})$
for $0\le j\le g+3$.

\begin{Lemma}\label{FermCase5}
Let $g\ge1$ and $1\le s\le g+1$. Then:
\begin{subequations}
\begin{multline}\label{FermEx5a}
\chi^{4,4g+1}_{1,s}=
\sum_{\raisebox{0mm}{$
      \begin{subarray}{c}
           n_1,n_2,\ldots,n_{g-1}\in\field{Z}_{\ge0}\\
           m_1,m_2\in\field{Z}_{\ge0}\\
           m_1,m_2\equiv0\,(\mod2)
        \end{subarray}$}}
\hskip-8mm \\
\frac{q^{\sum_{j=1}^{g-1} N_j(N_j+m_1)
+\frac{g+1}4 m_1^2+\frac12 m_2^2-\frac12m_1m_2+\frac12(g-s)m_1
+(N_s+N_{s+1}+\cdots+N_{g-1})}}
{(q)_{n_1}(q)_{n_2}\cdots (q)_{n_{g-1}}(q)_{m_1}}
\qbinom{\frac12 m_1}{m_2}
\end{multline}
and
\begin{multline}\label{FermEx5b}
\chi^{4,4g+1}_{3,s}=
q^{s-2g-\frac12}
\hskip-5mm
\sum_{\raisebox{0mm}{$
      \begin{subarray}{c}
           n_1,n_2,\ldots,n_{g-1}\in\field{Z}_{\ge0}\\
           m_1,m_2\in\field{Z}_{\ge0}\\
           m_1\equiv0\,(\mod2)\\
           m_2\equiv1\,(\mod2)
        \end{subarray}$}}
\hskip-8mm \\
\frac{q^{\sum_{j=1}^{g-1} N_j(N_j+m_1)
+\frac{g+1}4 m_1^2+\frac12 m_2^2-\frac12m_1m_2+\frac12(g-s)m_1
+(N_s+N_{s+1}+\cdots+N_{g-1})}}
{(q)_{n_1}(q)_{n_2}\cdots (q)_{n_{g-1}}(q)_{m_1}}
\qbinom{\frac12 m_1}{m_2}.
\end{multline}
\end{subequations}
Each term $(N_s+N_{s+1}+\cdots+N_{g-1})$ is to be taken as $0$
whenever $s\ge g$.
\end{Lemma}

\medskip\noindent {\it Proof: }
Since with $s\le g+1$ and $r=1$, we have $s\in{\mathcal T}$ and $r\in\tilde{\mathcal T}$,
the fermionic expression
\cite[(1.16)]{welsh} for $\chi^{p,p'}_{r,s}$ comprises a single
fundamental fermionic form \cite[(1.17)]{welsh}.
The summation in \cite[(1.17)]{welsh} is then over integers
$n_1,n_2,\ldots,n_{g-1}$ and $m_1$ and $m_2$
(these latter two parameters are named $m_{g}$ and $m_{g+1}$
in \cite[(1.17)]{welsh}), with the parity of $m_1$ and $m_2$ are restricted.

{}From \cite[\S1.7]{welsh},
we obtain:
\begin{equation*}
\boldsymbol{u}^L=\begin{cases}
\boldsymbol{e}_g&\text{if }s=g+1;\\
0&\text{if }s=g;\\
\boldsymbol{e}_{s-1}-\boldsymbol{e}_{g-1}&\text{if }s<g.
\end{cases}
\end{equation*}
{}From \cite[\S1.9]{welsh}, we then get
$\overline{\boldu}^L_\flat=(1,0)$ if $s=g+1$ and
$\overline{\boldu}^L_\flat=(0,0)$ if $s\le g$.
On the other hand, $\boldsymbol{u}^R=\boldsymbol{e}_g$ and
$\overline{\boldu}^R_\sharp=(0,0)$.
For \cite[(1.18)]{welsh}, we then get
$\tilde\boldsymbol{n}=(\tilde n_1,\tilde n_2,\ldots,\tilde n_{g-1})$
where,
if $s\in\{g,g+1\}$ then $\tilde n_{g-1}=n_{g-1}+\frac12m_1$
and $\tilde n_j=n_j$ for $1\le j<g-1$,
and if $1\le s<g$ then $\tilde n_{g-1}=n_{g-1}+\frac12m_1+\frac12$,
$\tilde n_{s-1}=n_{s-1}-\frac12$,
and $\tilde n_j=n_j$ for $1\le j<s-1$ and $s\le j<g-1$.
The expression \cite[(1.17)]{welsh} then yields equation
\eqref{FermEx5a} up to an overall factor, where $m_1$ and $m_2$
are each summed over even integers
because here $\overline{\boldu}=(0,0)$ leads to $\overline{\boldsymbol{Q}}(\boldsymbol{u})=(0,0)$.
The overall factor may be obtained by noting that the smallest
value of the exponent in the numerator of the summand in equation
\eqref{FermEx5a} occurs when $n_1=n_2=\cdots=n_{g-1}=m_1=m_2=0$.
Since $\chi^{p,p'}_{r,s}|_{q=0}=1$,
it follows that the required factor is simply $1$.

For the second expression, we obtain $\boldsymbol{u}^L$ as above. Using
$r=3$ instead of $r=1$ results in $\boldsymbol{u}^R=\boldsymbol{e}_{g}+\boldsymbol{e}_{g+2}$.
This yields $\overline{\boldu}^R_\sharp=(0,0)$ as above. However here,
$m_1\equiv0\,(\mod2)$ and $m_2\equiv1\,(\mod2)$ in the summation
because $\overline{\boldu}=(0,1)$ leads to $\overline{\boldsymbol{Q}}(\boldsymbol{u})\equiv(-2,-1)$.
Thus, apart from the overall factor, the sum form
\cite[(1.17)]{welsh} for $\chi^{4,4g+1}_{3,s}$ differs from the
$r=1$ case above only in that $m_2$ is summed over odd integers.
The smallest value of the exponent in the numerator of the summand
in equation \eqref{FermEx5b} occurs when
$n_1=n_2=\cdots=n_{2g-1}=0$ and $m_1=2$ and $m_2=1$.
It follows that the required factor is $q^{s-2g-\frac12}$.
\hfill $\Box$ \medskip

\Proofof{Theorem \ref{FermCase5Cor}}
The expressions \eqref{FermEx5a} and \eqref{FermEx5b} of Lemma \ref{FermCase5}
may be combined to yield a sum over all $m_2\in\field{Z}_{\ge0}$.
On performing this summation using \eqref{CauchyEq2} with $P=\frac12m_1$,
we obtain:
\begin{equation*}
\begin{aligned}
&\chi^{4,4g+1}_{1,s} + q^{2g+\frac12-s} \chi^{4,4g+1}_{3,s}\\
&\hskip5mm
=
\sum_{\raisebox{0mm}{$
      \begin{subarray}{c}
           n_1,n_2,\ldots,n_{g-1}\in\field{Z}_{\ge0}\\
           m_1\in\field{Z}_{\ge0}\\
           m_1\equiv0\,(\mod2)
        \end{subarray}$}}
\hskip-7mm
\frac{q^{\sum_{j=1}^{g-1} N_j(N_j+m_1)
+\frac{2g+1}8 m_1^2+\frac12(g-s)m_1
+(N_s+\cdots+N_{g-1})} (-q^{\frac12};q)_{\frac12m_1} }
{(q)_{n_1}(q)_{n_2}\cdots (q)_{n_{g-1}}(q)_{m_1}}.
\end{aligned}
\end{equation*}
On setting $m_1=2M$ and noting that
$(q)_{2M}=(q;q^2)_M(q^2;q^2)_M
=(-q^{\frac12};q)_M$ $(q^{\frac12};q)_M(q^2;q^2)_M$,
we obtain the left side of \eqref{FermCase5CorEq}.
The right side follows from \eqref{Composep4Eq1}.
\hfill $\Box$ \medskip

\begin{Lemma}\label{FermCase6}
Let $g\ge1$. Then:
\begin{subequations}
\begin{equation}\label{FermEx6a}
\chi^{4,4g+1}_{1,2g+1}=
q^{-\frac{g+1}4}
\hskip-10mm
\sum_{\raisebox{-6.5mm}{$
      \begin{subarray}{c}
           n_1,n_2,\ldots,n_{g-1}\in\field{Z}_{\ge0}\\
           m_1,m_2\in\field{Z}_{\ge0}\\
           m_1\equiv1\,(\mod2)\\
           m_2\equiv0\,(\mod2)
        \end{subarray}$}}
\hskip-10mm
\frac{q^{\sum_{j=1}^{g-1} N_j(N_j+m_1)
+\frac{g+1}4 m_1^2+\frac12 m_2^2-\frac12m_1m_2-\frac12m_2}}
{(q)_{n_1}(q)_{n_2}\cdots (q)_{n_{g-1}}(q)_{m_1}}
\qbinom{\frac12(m_1+1)}{m_2}
\end{equation}
and
\begin{equation}\label{FermEx6b}
\chi^{4,4g+1}_{3,2g+1}=
q^{-\frac{g-1}4}
\hskip-10mm
\sum_{\raisebox{-5mm}{$
      \begin{subarray}{c}
           n_1,n_2,\ldots,n_{g-1}\in\field{Z}_{\ge0}\\
           m_1,m_2\in\field{Z}_{\ge0}\\
           m_1,m_2\equiv1\,(\mod2)
        \end{subarray}$}}
\hskip-10mm
\frac{q^{\sum_{j=1}^{g-1} N_j(N_j+m_1)
+\frac{g+1}4 m_1^2+\frac12 m_2^2-\frac12m_1m_2-\frac12m_2}}
{(q)_{n_1}(q)_{n_2}\cdots (q)_{n_{g-1}}(q)_{m_1}}
\qbinom{\frac12(m_1+1)}{m_2}.\!\!\!\!
\end{equation}
\end{subequations}
\end{Lemma}

\medskip\noindent {\it Proof: }
Here $s=2g+1$ so that $s\in\mathcal T$.
Thereupon, $\boldsymbol{u}^L=\boldsymbol{e}_{g+1}$, ${\overline{\boldu}}^L_\flat=(0,1)$
and $\tilde\boldsymbol{n}=(n_1,n_2,\ldots,n_{g-2},n_{g-1}+\frac12m_1)$.
The proof now follows the lines of Lemma \ref{FermCase5},
noting that for $r=1$, we obtain $\boldsymbol{u}=(1,0)$ whereupon
${\overline{\boldsymbol{Q}}}(\boldsymbol{u})=(1,0)$;
and for $r=3$, we obtain $\boldsymbol{u}=(1,1)$ whereupon
${\overline{\boldsymbol{Q}}}(\boldsymbol{u})=(1,1)$.
\hfill $\Box$ \medskip

\Proofof{Theorem \ref{FermCase6Cor}}
After noting that $\chi^{4,4g+1}_{1,2g+1}=\chi^{4,4g+1}_{3,2g}$
and $\chi^{4,4g+1}_{3,2g+1}=\chi^{4,4g+1}_{1,2g}$,
expressions \eqref{FermEx6a} and \eqref{FermEx6b} of Lemma \ref{FermCase6}
may be combined to yield a sum over all $m_2\in\field{Z}_{\ge0}$.
On performing this summation using \eqref{CauchyEq2} with $P=\frac12(m_1+1)$,
we obtain:

\begin{equation*}
\begin{aligned}
&\chi^{4,4g+1}_{1,2g} + q^{\frac12} \chi^{4,4g+1}_{3,2g}\\
&\hskip10mm
=
q^{-\frac{2g-1}8}
\hskip-5mm
\sum_{\raisebox{-5mm}{$
      \begin{subarray}{c}
           n_1,n_2,\ldots,n_{g-1}\in\field{Z}_{\ge0}\\
           m_1\in\field{Z}_{\ge0}\\
           m_1\equiv1\,(\mod2)
        \end{subarray}$}}
\hskip-5mm
\frac{q^{\sum_{j=1}^{g-1} N_j(N_j+m_1)
+\frac{2g+1}8 m_1^2-\frac14m_1} (-q^{\frac12};q)_{\frac12(m_1+1)} }
{(q)_{n_1}(q)_{n_2}\cdots (q)_{n_{g-1}}(q)_{m_1}}.
\end{aligned}
\end{equation*}
On setting $m_1=2M+1$ and noting
$(q)_{2M+1}=(q;q^2)_{M+1}(q^2;q^2)_M
=(-q^{\frac12};q)_{M+1}$ $(q^{\frac12};q)_{M+1}(q^2;q^2)_M$,
we obtain the left side of \eqref{FermCase6CorEq}.
The right side follows from \eqref{Composep4Eq1}.
\hfill $\Box$ \medskip

\subsection{The $(p,p')=(4,4g+3)$ case}\label{p4SecB}

The continued fraction of $p'/p=(4g+3)/4$ is $[g,1,3]$.
Then, from \cite[\S1.3]{welsh}, $n=1$, $t=g+2$, $t_1=g-1$ and $t_2=g$.
The set ${\mathcal T}$ of Takahashi lengths is given by
${\mathcal T}=(1,2,\ldots,g,g+1,2g+1)$.
The set $\tilde{\mathcal T}$ of truncated Takahashi lengths is
given by $\tilde{\mathcal T}=(1,2)$.
{}From \cite[\S1.11]{welsh},
$\boldsymbol{B}$ is the $(g-1)\times(g-1)$ matrix
$\boldsymbol{B}=\{B_{j\ell}\}_{1\le j,\ell\le g-1}$
with entries $B_{j\ell}=\min\{j,\ell\}$, and
$\overline{\boldsymbol{C}}$ is the $2\times 2$ matrix
$\left(\begin{smallmatrix}1&1\\-1&2\end{smallmatrix}\right)$.
The matrix $\overline{\boldsymbol{C}}^*$ is the $2\times 2$
matrix $\left(\begin{smallmatrix}-1&2\\0&-1\end{smallmatrix}\right)$.

Throughout this subsection, we take
$N_j=n_j+n_{j+1}+\cdots+n_{g-1}$ for $1\le j\le g-1$,
and the $(g+2)$-dimensional vectors $\boldsymbol{e}_j$ are defined by
$\boldsymbol{e}_j=(\delta_{1j},\delta_{2j},\ldots,\delta_{g+2,j})$
for $0\le j\le g+3$.

\begin{Lemma}\label{FermCase7}
Let $g\ge1$ and $1\le s\le g+1$. Then:
\begin{subequations}
\begin{multline}\label{FermEx7a}
\chi^{4,4g+3}_{1,s}=
\hskip-5mm
\sum_{\raisebox{-1mm}{$
      \begin{subarray}{c}
           n_1,n_2,\ldots,n_{g-1}\in\field{Z}_{\ge0}\\
           m_1,m_2\in\field{Z}_{\ge0}\\
           m_1,m_2\equiv0\,(\mod2)
        \end{subarray}$}}
\hskip-8mm\\
\frac{q^{\sum_{j=1}^{g-1} N_j(N_j+m_1)
+\frac g4 m_1^2+\frac12 m_2^2+\frac12(g+1-s)m_1
+(N_s+N_{s+1}+\cdots+N_{g-1})}}
{(q)_{n_1}(q)_{n_2}\cdots (q)_{n_{g-1}}(q)_{m_1}}
\qbinom{\frac12 m_1}{m_2}
\end{multline}
and
\begin{multline}\label{FermEx7b}
\chi^{4,4g+3}_{3,s}=
q^{s-2g-\frac32}
\hskip-5mm
\sum_{\raisebox{-1mm}{$
      \begin{subarray}{c}
           n_1,n_2,\ldots,n_{g-1}\in\field{Z}_{\ge0}\\
           m_1,m_2\in\field{Z}_{\ge0}\\
           m_1\equiv0\,(\mod2)\\
           m_2\equiv1\,(\mod2)
        \end{subarray}$}}
\hskip-8mm\\
\frac{q^{\sum_{j=1}^{g-1} N_j(N_j+m_1)
+\frac g4 m_1^2+\frac12 m_2^2+\frac12(g+1-s)m_1
+(N_s+N_{s+1}+\cdots+N_{g-1})}}
{(q)_{n_1}(q)_{n_2}\cdots (q)_{n_{g-1}}(q)_{m_1}}
\qbinom{\frac12 m_1}{m_2}.
\end{multline}
\end{subequations}
Each term $(N_s+N_{s+1}+\cdots+N_{g-1})$ is to be taken as $0$
whenever $s\ge g$.
\end{Lemma}

\medskip\noindent {\it Proof: }
Since with $s\le g+1$ and $r=1$, we have $s\in{\mathcal T}$ and $r\in\tilde{\mathcal T}$,
the fermionic expression
\cite[(1.16)]{welsh} for $\chi^{p,p'}_{r,s}$ comprises a single
fundamental fermionic form \cite[(1.17)]{welsh}.
The summation in \cite[(1.17)]{welsh} is then over integers
$n_1,n_2,\ldots,n_{g-1}$ and $m_1$ and $m_2$
(these latter two parameters are named $m_{g}$ and $m_{g+1}$
in \cite[(1.17)]{welsh}), with the parity of $m_1$ and $m_2$ are restricted.

{}From \cite[\S1.7]{welsh},
we obtain:
\begin{equation*}
\boldsymbol{u}^L=\begin{cases}
0&\text{if }s=g+1;\\
\boldsymbol{e}_g&\text{if }s=g;\\
\boldsymbol{e}_{s-1}-\boldsymbol{e}_{g-1}-\boldsymbol{e}_g&\text{if }s<g.
\end{cases}
\end{equation*}
{}From \cite[\S1.9]{welsh}, we then get
$\overline{\boldu}^L_\flat=(0,0)$ if $s=g+1$ and
$\overline{\boldu}^L_\flat=(-1,0)$ if $s\le g$.
On the other hand, $\boldsymbol{u}^R=0$ and
$\overline{\boldu}^R_\sharp=(0,0)$.
For \cite[(1.18)]{welsh}, we then get
$\tilde\boldsymbol{n}=(\tilde n_1,\tilde n_2,\ldots,\tilde n_{g-1})$
where,
if $s\in\{g,g+1\}$ then $\tilde n_{g-1}=n_{g-1}+\frac12m_1$
and $\tilde n_j=n_j$ for $1\le j<g-1$,
and if $1\le s<g$ then $\tilde n_{g-1}=n_{g-1}+\frac12m_1+\frac12$,
$\tilde n_{s-1}=n_{s-1}-\frac12$,
and $\tilde n_j=n_j$ for $1\le j<s-1$ and $s\le j<g-1$.
The expression \cite[(1.17)]{welsh} then yields equation
\eqref{FermEx7a} up to an overall factor, where $m_1$ and $m_2$
are each summed over even integers because here $\overline{\boldu}=(0,0)$
leads to $\overline{\boldsymbol{Q}}(\boldsymbol{u})=(0,0)$.
The overall factor may be obtained by noting that the smallest
value of the exponent in the numerator of the summand in equation
\eqref{FermEx7a} occurs when $n_1=n_2=\cdots=n_{g-1}=m_1=m_2=0$.
Since $\chi^{p,p'}_{r,s}|_{q=0}=1$,
it follows that the required factor is simply $1$.

For the second expression, we obtain $\boldsymbol{u}^L$ as above.
Using $r=3$ instead of $r=1$ results in
$\boldsymbol{u}^R=\boldsymbol{e}_{g+2}$.
This yields $\overline{\boldu}^R_\sharp=(0,0)$, as above.
However here, $m_1\equiv0\,(\mod2)$ and $m_2\equiv1\,(\mod2)$ in the summation
because $\overline{\boldu}=(0,1)$ leads to $\overline{\boldsymbol{Q}}(\boldsymbol{u})\equiv(-2,-1)$.
Thus, apart from the overall factor,
the fermionic form
for $\chi^{4,4g+1}_{3,s}$ differs from the $r=1$ case above only
in that $m_2$ is summed over odd integers.
The smallest value of the exponent in the numerator of the summand
in equation \eqref{FermEx7b} occurs when
$n_1=n_2=\cdots=n_{2g-1}=0$ and $m_1=2$ and $m_2=1$.
It follows that the required factor is $q^{s-2g-\frac32}$.
\hfill $\Box$ \medskip

\Proofof{Theorem \ref{FermCase7Cor}}
The required result is obtained from Lemma \ref{FermCase7}
and \eqref{CauchyEq2} along the lines used to prove
Theorem \ref{FermCase5Cor}.
\hfill $\Box$ \medskip

\begin{Lemma}\label{FermCase8}
Let $g\ge1$. Then:
\begin{subequations}
\begin{equation}\label{FermEx8a}
\chi^{4,4g+3}_{1,2g+1}=
q^{-\frac{g}4}
\hskip-10mm
\sum_{\raisebox{-6.5mm}{$
      \begin{subarray}{c}
           n_1,n_2,\ldots,n_{g-1}\in\field{Z}_{\ge0}\\
           m_1,m_2\in\field{Z}_{\ge0}\\
           m_1\equiv1\,(\mod2)\\
           m_2\equiv0\,(\mod2)
        \end{subarray}$}}
\hskip-3mm
\frac{q^{\sum_{j=1}^{g-1} N_j(N_j+m_1)
+\frac g4 m_1^2+\frac12 m_2^2}}
{(q)_{n_1}(q)_{n_2}\cdots (q)_{n_{g-1}}(q)_{m_1}}
\qbinom{\frac12(m_1+1)}{m_2}
\end{equation}
and
\begin{equation}\label{FermEx8b}
\chi^{4,4g+3}_{3,2g+1}=
q^{-\frac{g+2}4}
\hskip-10mm
\sum_{\raisebox{-5mm}{$
      \begin{subarray}{c}
           n_1,n_2,\ldots,n_{g-1}\in\field{Z}_{\ge0}\\
           m_1,m_2\in\field{Z}_{\ge0}\\
           m_1,m_2\equiv1\,(\mod2)
        \end{subarray}$}}
\hskip-3mm
\frac{q^{\sum_{j=1}^{g-1} N_j(N_j+m_1)
+\frac g4 m_1^2+\frac12 m_2^2}}
{(q)_{n_1}(q)_{n_2}\cdots (q)_{n_{g-1}}(q)_{m_1}}
\qbinom{\frac12(m_1+1)}{m_2}.
\end{equation}
\end{subequations}
\end{Lemma}

\medskip\noindent {\it Proof: }
Here $s=2g+1$ so that $s\in\mathcal T$.
Thereupon, $\boldsymbol{u}^L=\boldsymbol{e}_{g+1}$, ${\overline{\boldu}}^L_\flat=(0,0)$
and $\tilde\boldsymbol{n}=(n_1,n_2,\ldots,n_{g-2},n_{g-1}+\frac12m_1)$.
The proof now follows the lines of the proof of Lemma \ref{FermCase7},
noting that for $r=1$, we obtain $\boldsymbol{u}=(1,0)$ whereupon
${\overline{\boldsymbol{Q}}}(\boldsymbol{u})=(1,0)$;
and for $r=3$, we obtain $\boldsymbol{u}=(1,1)$ whereupon
${\overline{\boldsymbol{Q}}}(\boldsymbol{u})=(1,1)$.
\hfill $\Box$ \medskip

\Proofof{Theorem \ref{FermCase8Cor}}
The required result is obtained from Lemma \ref{FermCase8}
and \eqref{CauchyEq2} along the lines used to prove
Theorem \ref{FermCase6Cor}.
\hfill $\Box$ \medskip

\section{Discussion}

In this paper, for $p=3$ and $p=4$, we have proved fermionic-product
identities for sums $\chi^{p, p'}_{1, s} + q^{\Delta} \chi^{p, p'}_{p-1,s}$
of characters for particular choices of $\Delta$.

For $p>5$, Bytsko and Fring \cite{bytfri} showed that the corresponding
sums of characters do not have a product form.
Furthermore, a detailed investigation of the fermionic expressions
of \cite{welsh}
(which is not reported in this work) shows that it is also not possible
to directly sum such expressions to obtain a single fermionic form.
Thus these $p>5$ cases do not lead to analogues of the Andrews-Gordon
identities.

In \cite{bytfri}, product forms were also derived for
the difference of two characters when $p=3$, $p=4$ and $p=6$.
In the $p=3$ cases in which $p'$ and $p'/2-s$ are both even, these
lead to new identities similar to our $M(p, p')^+_2$-identities with
a sign factor $(-1)^M$ incorporated into the summands.
In the other $p=3$ cases and all of the $p=4$ cases, the identities
so obtained are equivalent to our $M(p, p')^+_2$-identities.

\section*{Acknowledgements}

This work started when two of us (BF and OF) were visiting Kyoto
University, and Tokyo University, Japan. We thank our colleagues
there for their hospitality, and Ole Warnaar for discussions.

BF is partially supported by grant number RFBR 05-01-01007, and
OF by the Australian Research Council (ARC).

redpajama_set_name: RedPajamaArXiv


text:
Q: Image is missing required "src" property. Make sure you pass "src" in props to the `next/image` component. İ got really stuck this time Image is missing required "src" property. Make sure you pass "src" in props to the next/image component. İ got really stuck this time.  i've checked everywhere but didnt fixed it
import Link from "next/link";
import Image from "next/image";
import { Flex, Box, Text, Button } from "@chakra-ui/react";

const Banner = ({
  purpose,
  imageUrl,
  buttonText,
  linkName,
  title1,
  title2,
  desc1,
  desc2,

}) =>

   (
    <Flex flexWrap="wrap" justifyContent="center" alignItems="center" m="10">
      <Image src={imageUrl} width="500" height="300" alt="banner"/>
      <Box p="5">
        <Text color="gray.500" fontSize="sm" fontWeight="medium">
          {purpose}
        </Text>
        <Text fontSize="3xl" fontWeight="bold">
          {title1}
 {title2}
        </Text>
        <Text
          fontSize="lg"
          paddingTop="3"
          paddingBottom="3"
          color="gray.700"
          fontWeight="medium"
        >
          {desc1}

          {desc2}
        </Text>
        <Button fontSize="xl" bg="blue.300" color="white">
          <Link href={linkName}>{buttonText}</Link>
        </Button>
      </Box>
    </Flex>
  );

export default function Home() {
  return (

Emlak Evim

      <Banner
        purpose="Ev Kirala"
        title1="Kiralık Evler"
        title2="Herkes için"
        desc1="Daireleri keşfedin"
        desc2="ve daha fazlası"
        buttonText="Kiralık Evleri keşfet"
        linkName="/search?purpose=for-rent"
        imageUrl="https://bayut-production.s3.eu-central-1.amazonaws.com/image/145426814/33973352624c48628e41f2ec460faba4"
      />
      <Banner purpose={"kiralıkk"} />
      <Banner
        purpose="Ev Satın Al"
        title1="Bul ve Satın Al"
        title2="Hayalinizdeki Evinizi Bulun"
        desc1="Daireleri, Villaları, Yazlıkları keşfedin"
        desc2="ve daha fazlası"
        buttonText="Satılık Evleri keşfet"
        linkName="/search?purpose=for-sale"
        imageUrl="https://bayut-production.s3.eu-central-1.amazonaws.com/image/110993385/6a070e8e1bae4f7d8c1429bc303d2008"
      />
      <Banner purpose={"kiralıkk"} />

  );
}

// next config file.
module.exports = {
  reactStrictMode: true,
  images:{
    domains:['bayut-production.s3.eu-central-1.amazonaws.com']
},
}

A: You try to render <Banner purpose={"kiralıkk"} /> without imageUrl prop given, so <Image/> gets this prop undefined

redpajama_set_name: RedPajamaStackExchange


text:
Eileen DeLucia to Present at Benchmark Litigation's Women in Litigation Forum
New York—Holwell Shuster & Goldberg LLP counsel Eileen DeLucia will present at Benchmark Litigation's New York Women in Litigation Forum, taking place virtually Nov. 17-18.
Eileen will participate in a panel titled "Finding your voice as a woman in law: Credibility, confidence, and resilience," where she and her panelists will share various perspectives on how women can best position themselves for success, including techniques to ensure your voice is heard and respected, conveying confidence in your daily role, and the importance of active listening and critical thinking.
The panel will take place at 9:10 a.m. on Nov. 18 and will include the VP and assistant general counsel for Pfizer; the VP, associate general counsel, and global head of litigation for Conduent; the associate general counsel for litigation at AIG Investments; and the assistant general counsel for litigation and compliance for Dr. Reddy's Laboratories.
Recognized as a "Woman Worth Watching" by Profiles in Diversity Journal, Eileen has worked on some of the highest-profile and most groundbreaking litigations touching the financial markets, including one that recovered $2.4 billion from the Lehman Brothers estate. Eileen and her colleagues have also brought more than 20 RMBS putback actions, which are putting to rest massive liabilities associated with the 2008 housing collapse.
In addition to her active litigation practice, Eileen dedicates significant time to pro bono work, most recently working with the Reporters Committee for Freedom of the Press to file a brief on behalf of a range of news organizations and publishing entities in support of singer Kesha in her ongoing dispute with Dr. Luke involving First Amendment issues that touch on the #MeToo movement.
Additional information about Benchmark's seventh annual Women in Litigation Forum can be found here.


text:
Case Study – Yale Art University
The primary objective for this report was to choose a website to redesign and improve the user interface and customer experience by suggesting various possible solutions to those issues that were found through the research aspect of the report. I decided to choose Yale University of Art (www.art.yale.edu) to base off my case study report on, as for my initial thoughts of the website, I felt it heavily lacked on the professionalism aspect and interface and it was fairly a struggle to navigate throughout the content of the pages.
START DATE: Jan 28, 2020
END DATE: Mar 3, 2020
SOFTWARE(S): InVision, Adobe XD, Uxpressia.com, Adobe Photoshop,
OVERVIEW OF YALE UNIVERSITY
Yale University is a private Ivy League research university in New Haven, Connecticut as it is known to be one of the most prestigious institutions in United States. As their reputation can be at its highest, it does not indicate that their website is UI UX oriented as I discovered that their Yale University of Arts website seemed very unorganized and unprofessional. Keeping in mind of the key aspects to a well built user based website, I decided to partake how i would design their site that would encourage more on its functionality & interface elements.
WEBSITE PREVIEW
PAIN POINTS OF THE SITE
4 major pain points / issues of the site that stand out and are deemed necessary to improve:
Site looks very unprofessional and substandard due to half of the screen being displayed with unnecessary animation in the back. The images that are presented are so poorly taken and everything looks so messy and out of place.
Information on the programs heavily lacks on elaborating the entrance requirements, tuition cost or any other significant details. Lacks in visual presentation like videos or images of the program so the page looks vapid and dull.
No search bar which makes it difficult and a hassle for the user to find the information the they are seeking. The concept of any staff or
student from Yale University having access to the site so they can edit it the way they want to present their creativity is unique however, its not the best choice as the result is not very appealing.
The site does not mention any opportunities or potential jobs that will be provided to the post graduates after the program so users who have a strong desire for a job after graduating wouldn't bother to check the programs out.
JOURNEY MAP (DRAFT)
On the right side, it displays the first version, essentially the draft of a journey map I designed. The journey map basically details on what a typical user would go through when they first enter the website. The primary audience I depicted for the Yale University website is students who are looking for a prestigious institution that have a strong passion for digital or creative arts based around technology.
JOURNEY MAP (FINAL VERSION / REVISED)
On the right side, I have my final version, essentially the revised version of the journey map. I had numerous people from my peers to evaluate my journey map and help me come up ideas to improve further. Some of the aspects I revised include, addition of touch points, elaborated further on the pain points at a perspective as a high school student for this particular persona, addition of channels & quotes, and also added a new stage after apply stage.
PAPER PROTOTYPE
This is a paper prototype, essentially a sketch that I drew out of what my ideal interface of the application would become. I tried to implement a lot of interactivity that is more mobile friendly with more visuals that will be eye catching.
These are the low fidelity prototype stages where I designed these wireframes on Adobe XD. From the left, I have the home page that displays the general information with various areas that most students would particularly be looking for on the website, and to its right, I have the about page where I display all the necessary information about Yale University and the Arts program it offers etc. Right beside, I have the long listed programs page where it displays every general program the university has to offer along with a small description and view course section that will showcase every course in that particular program area. On to the very right, I have the navigation menu that will toggle when clicked on the left hamburger menu icon.
STYLE TILE
Yale University brands itself using the yellow and blue colors to polish their aesthetic of the website. So I decided to go with the flow and came up with different shades or blue and yellow that can possibly be used for my high fidelity prototypes and colors for the website. I kept the fonts very simple and a common type which is the Myriad Pro.
HIGH FIDELITY PROTOTYPES (DRAFT)
Now moving on to the high fidelity prototype portion of this project, this is essentially how my ideal graphics would look for the mobile application of the Yale University of Art would look like. As you can see, I heavily incorporated the yellow and blue colors all around the page to emphasize their identity and tried not to change the site information as much compared to the original website.
HIGH FIDELITY PROTOTYPE (FINAL VERSION / REVISED)
This is the revised version of the high fidelity prototype that I designed and went through the same exact process as the journey map which was having my peers / classmates to help me find the issues and changes that needed to be made to improve its interface. From the revision process, I changed quite a lot of things from the original as it may not seem to different at first glance but it includes, lowering the opacity of the blue box where the text is labeled "#1 Ranked Art School in the United States" in the home page (far left image), increasing the size of the buttons and text, and images overall to increase the area where the user would interact with, and adding supporting images behind headers for the page titles and programs. There were definitely more changes that were made however, these factors were the dominant ones that I agreed with which in result turned far better and more user friendly than the original.
Link to inVision


text:
United StatesFeb 9th 2008 edition
For the bin
But before that, George Bush's fiscal plan will set off an epic fight
APART from its record-breaking size—over $3 trillion, for the first time ever—the most memorable novelty of George Bush's budget proposal was the method he used to submit it to Congress. On February 4th Mr Bush held up a tablet PC and showed off his 2009 e-budget. "It saves paper, it saves trees, it saves money. I think it's the first budget submitted electronically," he noted. All the easier for the Democrats to drag it to the recycling bin.
One click and it's gone
Since the plump document aims to keep Mr Bush's policies in place long past his departure from the Oval Office, it seems tailor-made to antagonise the Democratic Congress. The president's request for defence is up 8%, or $36 billion, from last year. He is insistent, in the face of strenuous Democratic opposition, that his 2001 and 2003 tax cuts should be made permanent, which would cost nearly a trillion dollars over the next five years.
To cut spending, Mr Bush slashes a range of government-run domestic programmes, including a scheme that trains paediatric specialists. His plan, which features a short-term stimulus package and generous assumptions about the economy's strength, envisions a deficit of $407 billion (2.7% of GDP) in fiscal 2009. By 2012 the budget is meant to be in surplus.
That is unlikely. Mr Bush sets aside $70 billion to finance the war on terror next year, but the cost in past years has been more than double that. The document overstates likely tax revenue by assuming that Congress will not extend relief from the Alternative Minimum Tax, even though it is sure to do so. Given these two factors, the Centre on Budget and Policy Priorities, a Washington think-tank, calculates that under the plan the deficit will be $118 billion in 2012.
Some of the programmes Mr Bush wants to cut, including certain farm subsidies, deserve cutting. More important, his blueprint is a step towards reforming America's ballooning entitlement schemes. Spending on the federal health programmes, Medicare and Medicaid, has increased some 2.5% faster than GDP per person over the past 40 years. If current policies were to persist, these health outlays would go from 4.6% of GDP to 12% in 2050 and 19% in 2082—about the same proportion as total federal spending is now.
Mr Bush's solution—slashing payments to health-care providers—is insufficient and unsustainable, and might even force some patients into more expensive government programmes. But at least he attempts to address the government's long-term fiscal insolvency, which is rather more than the Democrats are likely to do this year.
Regardless of its merits, Mr Bush's plan may well lead to "world war three", as one pundit put it: a budget fight made all the more bitter by the election that is coming. In last year's budget row, Mr Bush and his allies all but refused to compromise with the leadership in Congress. This year the Democrats look just as wary of giving the president anything he wants. But in this particular war Congress has time on its side. It can always wait until Mr Bush has retired to pass a budget whose priorities are more to its liking.
This article appeared in the United States section of the print edition under the headline "For the bin"
More from United States
Labour constraints
Staffing shortages in America are a glimpse into its future
Handing out money
America's experiments with guaranteed-income schemes show promise
Changing gun culture
Gun-ownership in America is diversifying, because of safety fears
Flags and free speech
The Supreme Court looks askance at Boston's refusal to fly a Christian flag
Voting rights and wrongs
Joe Biden's voting rights push is futile
Even as the pandemic ebbs, the pool of potential workers may be permanently smaller than once assumed
Could the pilot programmes be scaled up?
Concerns over safety lead more women and minorities to arm themselves
A pole dance over the First Amendment
He doesn't have the votes—and his proposals are misguided


A spare HMI lamp for ProDaylights that reduces UV by 98% for safety and ideal color rendering. This halogen bulb comes with a hot restrike function.
The HMI Lamp HR UV-C is available in three variants – 200W, 400W and 800W.


text:
Bali, Jyoti S, KLE Technological University
balog, florica, Polytechnic University of Timisoara, Romania
Bama, Sathya, Sri Krishna College of Technology, Coimbatore
Bamber, Sukhvinder Singh, Punjab University SSG Regional Centre, Punjab, India
Bandhyapadhya, Parimalendu, North Bengal University
Bandyopadhyay, Prof. Samir K, University of Calcutta, Kolkata
Bandyopadhyay, Samir k, University of Calcutta
Banerjee, Debanjan, NIT Durgapur, 713209, West Bengal, India
Banerjee, Dr. P.K., Professor, ETCE, Jadavpur University Kolkata-700032
Banerjee, Durga, Institution,Rathtala,P.O.-Ranaghat,District-Nadia,Pin Code-741201,West Bengal
Banerjee, Sreeparna, School of Computer Science & Engineering, West Bengal University of Technology, Kolkata
Banica, Logica, University of Pitesti, Faculty of Economic Sciences
Banica, Logica, University of Pitesti
Bansal, Alka, Adesh Institute of Engineering and Technology,Faridkot
Bansal, Avinash, GNIT Mullana
Bansal, Ginni, Department of Information Technology, CEC Landran
Bansal, Jyoti, Associate Prof. & Dean Academic Affairs, B.F.C.E.T., Bathinda
Bansal, Prof.(Dr.) Sanjeev, Amity School of Engineering and Technology Amity University, Sec-125, NOIDA, (U.P.)
Bansiya, Akhilesh, Research Scholars Computer Science & Engineering Samrat Ashok Technological Institute, Vidisha (MP)
Barkouti, Wahid, Science Faculty of Tunis 1060 Tunisia
Baronak, Ivan, Slovak University of Technology BratislavaIlkovičova 3, Bratislava
Barras, Thomas, SECA GROUP
Baskar, S. S., Research Scholar,St.Joseph's college, Trichy
Baskaran, R, Anna University, Chennai.
326 - 350 of 2643 items << < 9 10 11 12 13 14 15 16 17 18 > >>


What do alarm systems and auto tires have in common? This may sound like a bizarre ask to the amateur ear, but lend me yours and I will make sense of it.
The aim I cite car tires is that they are a basic that most all of us have some encounter purchasing, at least once in a while.
Even all the same we need good tires to keep our cars confidently on the road, when they are bearing out we often put the acquire off for a few weeks. If we are patient, a Saturday will in due course come along that we have no plans for, and we force ourselves to go to our local tire store.
Maybe it's the one with the big giant marshmallow looking thing on the roof, or the shop down the road who seems to be generous away free franks, and has a racing car crew helping them, that gets us to stop.
If you are jumping the gun and think I'm correlating putting off the acquire of a must such as car tires with a must such as an alarm approach you would be wrong. The first basis you would be wrong is as most associates don't see an alarm as a necessity, but as a luxury. Only after they have been burglarized, do they see an add-on of an alarm arrangement as a little they need as a replacement for of want.
Do you know why they appreciate after the fact that they need one? Not just since they are upset that a big name took the electronics. Not just since the gun that they painstaking the only alarm arrangement they would ever need is missing, and god only knows whose hands it will end up in. Not even for the reason that they are freaked out calculating that some stranger was rifling all the way through the underwear drawer. It is since this is the first time most ancestors start to think about how this awkward event could have effected the lives of their loved ones and themselves, if they had been home or walked in on a break-in in progress.
What if my wife came home for lunch and walked in on them?
What if I came home from work and had my own gun cutting at me when I came in?
What if we were fast asleep while they entered, and how would that alarm the family?
After asking ourselves these types of questions is when an alarm approach becomes a necessity.
Assuming you do know that you need and want a approach as part of your anticipatory defense plan, and you do have researching one on your to do list, let's care about how else an alarm arrangement might have a little in collective with auto tires.
How often does the salesperson, take the time to defend belongings that are so crucial to your assessment such as Load Range, Tread Wear, Rubber Composites, Tread Patterns and the Elements, Sidewall Strength, Speed Ratings, etc. If they did, you would not only know much more than you before now do about tires, but you would most apt adjust your choices and the cost with life- safety, as you should. You will also most expected buy your tires at that store the next time you need them, no be of importance what they are generous away down the street.
Would you deem that alarm systems are often sold the same way tires are. " We have, this one or that one, which do you prefer?" When building an chief conclusion such as how a collateral arrangement will fit into your lifestyle, delight take the time to appreciate all of your options. The existing differences in attribute of all the plans you are making an allowance for for asset is astounding, and will make a big discrepancy in how happy you are with your choices. If you can't find a celebrity to defend all of these belongings to you, then you have not found the memorable "perfect tire store"
Someone Spying? So are the Teddy Bear and the Conserved PlantSee the cuddly teddy bear over on the shelf? Well, it sees you too. "Right here in the camera, at the back the left eye.
What is going on with these ancestors who feel they can walk onto your assets or in front of your home and make off with your child? If those parents had installed a covert camera in their home, the control could use the recordings from the camera and have a quick lead to find the child and come again them safely. Being equipped for the absurd act of your child being abducted by installing clandestine cameras is much advance than wishing later you had installed cameras!You can order a concealed camera and you can certainly ensconce it inside the week!! Buried cameras will make our offspring safer.
As the damage and death toll from deadly Tempest Katrina becomes more apparent, it is a grim reminder that we live in precarious times.Every year, millions of colonize about the world lose their lives or have them dangerously disrupted for the reason that of floods, tsunamis, wildfires, earthquakes, blizzards and hurricanes.
If you are about to get hit by a Storm are you prepared? Do you have an adequate amount of plywood to board up your windows? Often the local clump companies will be sold out. In fact if you wait till the last detailed Home Depot will be out too.
Smoke & Heat Detectors- Smoke detectors are obtainable that will edge with your defense system. The smoke detectors that are by now in your home can't be tied into your confidence system.
Have you found manually disgusted with the crime lately. You know much of it is gang and drug associated yet cannot appreciate how it has infiltrated your quiet town or how it has develop into such a menus in your city? Have you a short time ago been a victim of a crime, vandalism or petty theft.
Is Big Brother Inspection You? Or Are You Inspection Big Brother?
Isn't equipment a breathtaking thing? I accepted wisdom this as I customary the text to my cellular phone phone illuminating me that my home confidence camera had detected movement. A big cheese or a touch was where they shouldn't be at the rear of my house and I was 2000 miles away on escape - yet I still knew they were there.
Passive Infrared Beckon Detectors- These detectors are also known as PIR detectors. The equipment they exploit is "passive infrared".
This years fire spell has previously ongoing and so far the score is viewing the fire armed forces in the lead with numerous large fires under their belts in California, Nevada and Arizona. So far in 2005 we have not had any huge or "Mega Fires" yet, cross your fingers that our luck continues.
It's one of those belongings we don't give a back up brain wave about. Many of us move into a house and never even alter the amount that's on the house.
Generally when it rains the water runs off the house and yards into the gutter and into pipes or storm drains which feed it into the river, course or ocean. This normal accost is knotty in that many of the countries regions are in awful drought.
Alarm Contacts- Call switches are attractive switches used to keep doors and windows for the most part. There are three main styles that are used by most alarm companies.
It is a known fact that marketing is made to be a magnet for our eyes. Marketers work very hard to plant visual memories in your mind.
Over the last 22 years that I have been complicated in the collateral alarm industry, I have seen quite an evolution. Changes have not only taken place in what type of gear is available, but also in how the consumer buys and use's the equipment.
The alarm charge is the brain of your collateral system. It is typically located in an area that is out of the way like a basement, attic, closet or office.
Stand Alone Keyless Approach - How Does It Work?
Security and convenience are as a rule carefully balanced. If a bit is acquire it isn't opportune and vice versa.
Buying the right garage door introduction has not never been more simpler.Whether you want to buy one for suburban use or the business-related use read the tips below and buy the garage door opening which ideally suits you .
You need to know how to guard yourself, since your confidence area or any consultant may not be truly aware of these threats and hence not dependable. Since September 11th, a add up to of businesses entered the defense field with barely to no data of the trade itself; only of its profit potential.
The best way to turn your home into a fort is to cautiously and purposefully believe all your options. First you need to make a arrangement of your assets and look at an aerial diagram of your home.
Choosing the Right Interrupt SprayThere are so many types of interleave spray on the promote today that it has befit mystifying for you to conclude which one to get. There are three basic substance compounds that collect yourself these cynical sprays.


Hayanga, H., Barnett, D., Shallow, N., et al. (2017). Anesthesiologists and Disaster Medicine: A Needs Assessment for Education and Training and Reported Willingness to Respond. (Abstract only.) Anesthesia and Analgesia. 124(5):1662-1669.
A web-based survey of anesthesiologists was conducted to assess self-reported perceptions of knowledge and skills, as well as attitudes and beliefs regarding education and training, employee development, professional obligation, safety, psychological readiness, efficacy, personal preparedness, and willingness to respond during a natural disaster, pandemic flu, and a radiological event. Few respondents indicated that they think they receive sufficient education for disaster response, and most think that their hospitals should provide this training to them. The authors recommend that additional training be provided, and support for staff to meet family obligations be put in place to encourage providers to come to work during disasters.