This document comes with a set of sample web pages that show ways to integrate BA Server functionality and content into your web application. This document describes the techniques used by the various samples, so we recommend you deploy the samples on your BA Server when following this document.
The samples are packaged into a file named pentaho-sdk<version>.zip accessible for download in your Customer Support Portal. The readme file contains the information for deployment.
Copy the integration-examples directory to your biserver-ee/tomcat/webapps directory.
Paste them into the .../biserver-ee/pentaho-solutions/system directory.
The samples are deployed and can be accessed by opening a web browser and navigating to http://localhost:8080/integration-examples.
If you want to access the OEM tools, browse to http://localhost:8080/pentaho/conten...web/index.html.
If you want to access the example-visualization, look under the Charts menu in Analyzer. It will be listed as Example KPI.


This Aston Martin DB6 had been locked away for over 30 years. We are now performing a complete nut & bolt rebuild to the owner's extremely high standards.
Copyright © 2007-2019 Hightone Restorations Ltd. All rights reserved.


The sneak peeks are rolling in from our attendees from the California Film Workshop and it's so fun seeing their work! We partnered with Lucy Munoz Photography, Trynh Photo, and Siren Floral Co. to create this seaside engagement shoot along the coast in Malibu. We know there are so many more beautiful images to come, but in the meantime enjoy these lovely photos and a quick video taken by C & C photography!
It's going to be an anniversary to remember!
We have had the privilege of helping plan weddings for three of Craig and Marilyn's kids and we were so excited when they approached us about their 40th anniversary party at their new property near Julian! Cowboy boots served as our jump off point for inspiration as they knew they wanted to have fun night filled with line dancing and their loved ones! Take a peek at some of our key elements for their upcoming day!


text:
@interface HKPDateFormatter (HKPInspectableDateFormatter)

@property (nonatomic, strong, readonly) NSMutableDictionary *localeDictionary;

+ (NSDictionary *)sharedInstances;
+ (instancetype)sharedInstance;

@end

redpajama_set_name: RedPajamaGithub


This is a very exciting year for Bramham International Horse Trials, with record entries and two action packed days of cross country, competition will be fierce amongst riders looking to shine for Olympic selection.
Look out for the Suregrow fertiliser Owl Hole at fence eight on the cross country course. This is an exciting, new and very challenging combination fence, comprising of a three fence maze of tight angles and a central hole to be jumped through.
During the event we will be keeping you updated on our sponsored riders – Caroline Powell riding the famous Lenamore in the CIC3* and Thornfield Jones in the CCI3*, also Ruth Edge riding the striking black gelding Elite Opposition in the CIC3*.


The affected hdd has been replaced and all services has resumed back to normal. Do report to us if you face any issue after this short maitnenance.
One of the raid disk has gone degraded and need urgent replacement to ensure that the raid unit running at very optimum level.
During the maintenance timeframe, all services on cinnamon server will not be available such as web, mysql, email, ftp.


In the past several years New Yorkers have witnessed their city change before their eyes as new buildings have sprung up throughout the boroughs and helped the city recover from financial uncertainty. Perhaps the most visually striking aspect of this new New York is the abundance of newly-built towers in Manhattan, Brooklyn and Queens that are typically luxury condominiums. Skyscrapers are nothing new to the city, but gargantuan residences have irritated an audible portion of New Yorkers resistant to such developments. In particular, the developments along West 57th Street, which seem to be competing for height records, are the target of criticism.The freshly completed condo/hotel building, One57 at 157 West 57th Street, is the latest mega-tower in Manhattan and is the first of many to come.
Developed by Extell and designed by the renowned Christian de Portzamparc, One57 was started in 2009 and now stands at 75 stories, about 1004 feet. At this staggering height, One57 is currently the tallest residential building in New York and even the western hemisphere. Yet these records came at a price: you may remember how, during Hurricane Sandy, a crane on the budding tower collapsed and teetered above the city, forcing nearby residents to be evacuated. If that weren't enough, a new crane was erected on the tower back in May, forcing more local evacuations (for safety reasons) and causing further frustration and controversy.
Justin Davidson of The New Yorker has recently attacked One57, as well as its future counterparts, for what it represents and portends. Davidson poignantly comments how One57 "is a luxury object for people who see the city as their private snow globe. Tall and clunky, preening yet graceless, the tower recapitulates an assortment of commercially proven stylistic gestures from New York's recent past." While the views from One57 are indeed astounding, Davidson doubts that the multi-million-dollar apartments could possibly be worth their price no matter how good the views, high the ceilings or "how unusably glossy the kitchen[s]" are. Despite the ludicrously high asking prices, most of the units of One57 have already been filled and it is only a matter of time before residents arrive with moving trucks.
The intended buyers of One57 units and units in similar towers are not surprisingly among the wealthiest in the world and this will by no means change as development continues. The money these developments as well as their buyers bring into the city is also obvious and has helped the city recover from the worst financial disaster since the Great Depression. Yet Davidson's piece, although admittedly highly-opinionated, brings up the question of what's to come in New York as these developments continue. Successful luxury condo towers will inevitably breed more such developments in the city but such trends benefit the elite minority and largely fail to consider the rest of New York sitting below these ivory towers.


text:
Blog, Braehead, Business, Events, Glasgow, Paisley, Paisley Events, Paisley Photographs, People from Paisley, Press Release, Renfrew, Renfrewshire, renfrewshire events
Students in the frame for photo prize
[vc_row][vc_column][vc_column_text]THE boss at intu Braehead has put three students in the frame as being top photographers.
The trio of snappers doing the photography course at West College Scotland were chosen by the mall's general manager as prize winners in a competition, which has seen dozens of photographs being put on display in the busy mall.
First place went to George Plu for his photograph of snow-covered hills at Newton Stewart, in Dumfries and Galloway; and runners-up were Susan Soutar, 48, with a photograph of a sunset in the Isle of Skye and 24-year-old Kirsty Thompson who photographed Royal Exchange Square, in Glasgow.
Head of Photography at West College Scotland, Robert Tabor said: "Students work on lots of different photographic projects and for so many people to see their work is a real boost to them.
"Recognition for their efforts is going to build their confidence as photographers and we can't thank intu Braehead enough for displaying our students' photographs."
The photographic competition winner received a £100 gift card and £50 gift cards were presented to the runners-up.
Gary Turnbull said: "The standard of photographs from the students is extremely high and the display of their pictures in the lower mall – which is changed every month – has attracted a lot of attention from shoppers passing by.
"The public should get the chance to see a display of the fine work being done by the photography students at West College Scotland and we're delighted to make this happen."
Photo winners pic 1: From left George Plu, Kirsty Thompson, Susan Soutar , with their winning photographs and Gary Turnbull, from intu Braehead.
Photo winners pic 2: From left George Plu, Kirsty Thompson, Gary Turnbull and Susan Soutar.[/vc_column_text][/vc_column][/vc_row]
https://i2.wp.com/www.paisley.org.uk/wp-content/uploads/2015/02/photo-winners-1.jpg?fit=1500%2C1046&ssl=1 1046 1500 Brian McGuire https://www.paisley.org.uk/wp-content/uploads/2015/09/paisley-logo-trans.png Brian McGuire2015-02-09 15:44:502015-02-09 15:44:50Students in the frame for photo prize
Six of the best go for national bar awards Edward has 'em mall on their feet with Sinatra song parody

redpajama_set_name: RedPajamaCommonCrawl


This page lists the song chart entries of Nirvana.A list of the hit albums by Nirvana is also available. The songs are listed with the most widely successful first.
An American rock band that was formed in 1987. The band consisted of Kurt Cobain, Krist Novoselic and a succession of drummers, the longest lasting being Dave Grohl. Despite the fact that the band ceased to exist in 1994 (with the death of Cobain) it continues to have a major influence.
There was also a British band (formed in 1965) called by exactly the same name. The song Pentecost Hotel was, of course by this Nirvana rather than the more famous one.
Need to take Pentecost Hotel off, that was a 1960s band that shared the same band name as the 90s band, but not the same band!
You are of course correct. It was the only minor hit for a British band of the 1960s.
Since we keep a complete record of all chart entries we can't just remove the song. So we have added a note to the top of the page that explains the situation.


text:
package alluxio.cli.fs.command;

import alluxio.AlluxioURI;
import alluxio.Constants;
import alluxio.cli.CommandUtils;
import alluxio.client.file.FileInStream;
import alluxio.client.file.FileSystemContext;
import alluxio.client.file.URIStatus;
import alluxio.exception.AlluxioException;
import alluxio.exception.ExceptionMessage;
import alluxio.exception.status.InvalidArgumentException;
import alluxio.util.FormatUtils;

import com.google.common.base.Preconditions;
import org.apache.commons.cli.CommandLine;
import org.apache.commons.cli.Option;
import org.apache.commons.cli.Options;

import java.io.IOException;

import javax.annotation.concurrent.ThreadSafe;

/**
 * Prints the file's last n bytes (by default, 1KB) to the console.
 */
@ThreadSafe
public final class TailCommand extends AbstractFileSystemCommand {
  private static final Option BYTES_OPTION = Option.builder("c")
      .required(false)
      .numberOfArgs(1)
      .desc("number of bytes (e.g., 1024, 4KB)")
      .build();

  /**
   * @param fsContext the filesystem of Alluxio
   */
  public TailCommand(FileSystemContext fsContext) {
    super(fsContext);
  }

  @Override
  public String getCommandName() {
    return "tail";
  }

  @Override
  protected void runPlainPath(AlluxioURI path, CommandLine cl)
      throws AlluxioException, IOException {
    URIStatus status = mFileSystem.getStatus(path);
    int numOfBytes = Constants.KB;
    if (cl.hasOption('c')) {
      numOfBytes = (int) FormatUtils.parseSpaceSize(cl.getOptionValue('c'));
      Preconditions.checkArgument(numOfBytes > 0, "specified bytes must be > 0");
    }

    if (status.isFolder()) {
      throw new IOException(ExceptionMessage.PATH_MUST_BE_FILE.getMessage(path));
    }
    try (FileInStream is = mFileSystem.openFile(path)) {
      byte[] buf = new byte[numOfBytes];
      long bytesToRead;
      if (status.getLength() > numOfBytes) {
        bytesToRead = numOfBytes;
      } else {
        bytesToRead = status.getLength();
      }
      is.skip(status.getLength() - bytesToRead);
      int read = is.read(buf);
      if (read != -1) {
        System.out.write(buf, 0, read);
      }
    }
  }

  @Override
  public int run(CommandLine cl) throws AlluxioException, IOException {
    String[] args = cl.getArgs();
    AlluxioURI path = new AlluxioURI(args[0]);
    runWildCardCmd(path, cl);
    return 0;
  }

  @Override
  public String getUsage() {
    return "tail [-c <bytes>] <path>";
  }

  @Override
  public String getDescription() {
    return "Prints the file's last n bytes (by default, 1KB) to the console.";
  }

  @Override
  public void validateArgs(CommandLine cl) throws InvalidArgumentException {
    CommandUtils.checkNumOfArgsEquals(this, cl, 1);
  }

  @Override
  public Options getOptions() {
    return new Options().addOption(BYTES_OPTION);
  }
}


text:
VIRGINIA ATTORNEY GENERAL MARK HERRING TERMINATES CONCEALED CARRY RECIPROCITY AGREEMENTS WITH 25 STATES THAT DON'T MEET COMMONWEALTH'S SAFETY STANDARDS
FOR IMMEDIATE RELEASE: December 22, 2015
MEDIA CONTACT: apatrick@csgv.org, Cell: (828) 712-7603
Coalition to Stop Gun Violence applauds AG's action to enforce Virginia law and prohibit dangerous out-of-state visitors from carrying loaded firearms
Richmond, VA—Today, Virginia Attorney General Mark Herring announced that the Commonwealth of Virginia will be terminating concealed carry reciprocity agreements with 25 states that fail to meet the standards Virginia has set to ensure the safety of its citizens.
Specifically, the Virginia statute governing reciprocity requires that: (1) The issuing state must provide instantaneous verification of the validity of permits, and; (2) The issuing state's requirements and qualifications must be sufficient to prevent the issuance of a permit to an individual who would be denied a permit in the Commonwealth of Virginia. This latter group includes stalkers; those convicted of violent misdemeanor offenses (i.e., assault/battery, weapons charges, sex offenses, etc.); habitual drunkards; juveniles convicted of offenses that would have been felonies had they been adults; and persons whose mentally illness causes them to be a potential danger to self or others.
For years, Virginia has maintained concealed carry reciprocity agreements with states that fail to meet the basic standards outlined in Virginia code. This oversight has allowed potentially violent individuals to carry loaded, concealed guns in public spaces across the Commonwealth.
"Virginia's lack of enforcement of its own standards has been irresponsible and dangerous," said Coalition to Stop Gun Violence (CSGV) executive director Josh Horwitz. "While the Commonwealth has plenty of room to improve their own standards for issuing concealed carry permits, the actions taken today by the Attorney General are a good first step toward making Virginia a safer place for its citizens and visitors alike."
"We express our sincere thanks to Attorney General Herring for putting the safety of our families above the financial interests of the gun lobby," said Lori Haas, CSGV's Virginia State Director. "As a mother and survivor of gun violence, it gives me some peace of mind knowing that dangerous individuals from other states will no longer be allowed to legally carry concealed weapons on the streets of Virginia."
The Coalition to Stop Gun Violence (CSGV) seeks to secure freedom from gun violence through research, strategic engagement and effective policy advocacy. CSGV can be found on Facebook and Twitter.


Woke up at 4:45am this morning to catch the "Super Blue Blood Moon". This is a combination of a full moon that's at its closest point to earth (super), the second full moon of the month (blue) and a total lunar eclipse (blood). It's the first time we've had one of these in 152 years!
This photo is a composition of seven photos as unfortunately, there aren't six moons. The moons are unretouched, laid over a photo of the downtown San Diego skyline and convention center.


text:
CA Technologies Issues CA Smart Government Challenge to U.S. Developers: Help the Federal Government Enhance its "Citizen Experience" Using Open Source Software
Developers To Compete For Prize Money and Technical and Business Support From the CA Accelerator, CA's In-House Startup Incubator
LAS VEGAS--(BUSINESS WIRE)--CA WORLD '17 — CA Technologies (NASDAQ:CA) today announced the CA Smart Government Challenge, a new innovation competition that invites U.S.-based developers to submit their most innovative ideas for improving the "citizen experience" of the U.S. federal government through open source software.
CA will award up to five finalists a cash prize of $5,000 and the chance to vie for a grand prize of $75,000 and access to the CA Accelerator Boot Camp, an immersive course that has never before been open to developers outside of CA.
Launched internally in 2016, the CA Accelerator operates like a venture capital fund within the company. It provides financial and incubator support to CA employees who have ideas for new software offerings that are relevant to CA's customers, which include some of the most prestigious companies and government agencies in the world. To date, the CA Accelerator has funded 14 teams, 11 of which are still in business today. Two notable examples are Waffle.io, which provides project management powered by GitHub, and CA Jarvis, an advanced analytics platform, which has been embedded in several products across the company's portfolio.
With the CA Smart Government Challenge, CA will put this powerful engine of innovation to work in a new area: the "citizen experience" of the U.S. federal government.
"Governments around the world increasingly are looking for ways to improve the citizen experience through software," CA Technologies CEO Mike Gregoire said in unveiling the CA Smart Government Challenge today at CA World '17, CA's annual customer conference in Las Vegas. "Whether it's in the form of smart, sensor-driven trash pickup in Washington, D.C., or the 'whole-government API' that has been created in Singapore – both developed, incidentally, with help from CA – governments are embracing the opportunity to use software to deliver experiences to the citizens they serve that match modern customer experiences delivered through the private sector. We are excited to help bring fresh ideas that can support the modernization of government IT and complement the good work underway in many parts of the federal government."
Developers participating in the CA Smart Government Challenge will engage in the following process:
Submit their idea in the form of a one-page Lean Canvas style business plan.
A panel of judges led by CA Accelerator founder and CA Chief Technology Officer Otto Berkes will select five finalists to receive a $5,000 prize and a consultation with members of the CA Accelerator team, who will provide suggestions on how to move the finalist's idea forward.
After six months of incubation, finalists will present their progress to the CA Accelerator for consideration for the $75,000 grand prize.
The grand prize winner will travel to participate in the CA Accelerator boot camp session, with travel expenses covered by CA.
The judging panel will select the best software idea based on criteria that include:
Citizen value: impact on the experience of interacting with the federal government
Impact on federal government efficiency
Market opportunity, including potential for growth and applicability to other government agencies or governments
Creativity and differentiation from other submissions
Sophistication: use of modern agile/devops development techniques
Likelihood of long-term success.
"Until today, the CA Accelerator's work has been focused on developing new technologies that can enhance our portfolio of enterprise IT products," said Berkes. "We are excited now to apply the Accelerator model, which combines the learn-fast creativity of a VC-backed startup with the significant resources of a major company like CA, to the customer experience of the federal government. The result could be a new company, a free app, some solution that hasn't even been conceived of yet, or simply a better understanding of the complexity of government services. That's the beauty of true innovation driven by a pure desire to solve a problem that matters – we don't know what the end result will be. And as with all projects in the Accelerator, we're excited to see where this journey takes us."
Developers interested in learning more can register on the CA Technologies Smart Government Challenge website to receive application materials that will be made available by December 1. Applications will be accepted during the month of December and the finalists will be selected and announced in February 2018.
For more information on all news being announced at CA World '17, please visit the CA World '17 Newsroom. To access replays of the CA World '17 keynotes, please click here.
About CA Technologies
CA Technologies (NASDAQ:CA) creates software that fuels transformation for companies and enables them to seize the opportunities of the application economy. Software is at the heart of every business in every industry. From planning, to development, to management and security, CA is working with companies worldwide to change the way we live, transact, and communicate – across mobile, private and public cloud, distributed and mainframe environments. Learn more at www.ca.com.
Follow CA Technologies
Copyright © 2017 CA Inc. All Rights Reserved. All trademarks, trade names, service marks, and logos referenced herein belong to their respective companies.
Rita O'Brien, +1-631-342-6687
rita.obrien@ca.com
CA Technologies announced the CA Smart Government Challenge, a new innovation competition for U.S.-based developers.
#volunteer
#CATogetherInAction


Coloring Pages Info. Tsum Disney Colouring Pages Coloring Printable With Regard To Info Plan 1. Exclusive Colouring Pages Cupcake Chic Shopkins Season 2 Coloring Within Info Prepare 7. Page A Colorier Loup Print Cute Wolf Adult Mandala Grown Up Coloring For Pages Info Decor 10. Coloring Pages Info Miraculous Ladybug Printable Book Throughout Inspirations 8. Miraculous Coloring Pages Unique Ladybug And Cat Noir Intended For Info Prepare 18. Print Frozen Elsa Disney Princess Christmas Coloring Pages Inside Info Decorations 4. Shopkins Coloriage T L Chargement Gratuit Coloriages Imprimer With Regard To Coloring Pages Info Designs 16.
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text:
\section{Introduction}

According to the Standard Model of Cosmology, supported by a large amount of
observational
data, during its evolution the universe has experienced two phases of accelerated
expansion, the inflationary one at early times and the current one at late times
\cite{SN,Planck:2015xua, Array:2015xqh, Hinshaw:2012aka,LSS,Jain:2003tba}.
In order to explain these accelerated phases one should introduce small deviations to
the standard physics paradigm, and there are two main directions that one can follow. The
first is to maintain general relativity as the gravitational theory and alter the content
of the universe by introducing novel, exotic forms of matter fields, either as inflaton
field(s) \cite{Inflation} or as dark energy sector \cite{Peebles:2002gy,Cai:2009zp}.
The second way is to modify the gravitational sector, constructing a theory that
possesses
general relativity as a particular limit, but with additional degrees of freedom that can
drive an accelerating expansion \cite{R-D-M}.

Although most of the works in modified gravity start from the usual gravitational
description based on curvature and modify the Einstein-Hilbert action, with the
simplest example being the $F(R)$ scenario \cite{F-R}, one could equally well construct
gravitational modifications starting from the torsion-based description of gravity. In
particular, one could start from the Teleparallel Equivalent of General
Relativity (TEGR) \cite{ein28,Hayashi79,Pereira.book,Maluf:2013gaa}, in which
the gravitational Lagrangian is the torsion scalar $T$, and build various extensions,
with
the simplest example being the $F(T)$ theory \cite{Bengochea:2008gz,Linder:2010py} (for a
review see \cite{Cai:2015emx}). Interestingly enough, although TEGR is completely
equivalent with general relativity at the level of equations, $F(T)$ does not coincide
with $F(R)$ gravity, and thus its cosmological application has led to novel features,
either at inflationary stage \cite{F-T-Inf} or at the late-time accelerated epoch
\cite{Chen:2010va,Geng:2011aj}.

On the other hand, unimodular gravity \cite{U-G} is an interesting gravitational theory,
which can be considered as a specific case of general relativity. In particular, while in
standard general relativity the origin of the cosmological constant is not well
understood
\cite{C-C}, in unimodular gravity it arises as the trace-free part of the gravitational
field equations, as long as the determinant of the metric is fixed to a number or a
function. The great theoretical advantage of this procedure is that since the trace-free
part of the field equations is not related to the vacuum expectation value of any matter
field, one can fix its value without facing the cosmological constant problem. Hence, one
can use unimodular gravity in order to describe the inflationary regime
\cite{Cho:2014taa}
and the late-time cosmic acceleration \cite{UG-LC}. Additionally, one can start from
unimodular gravity in order to construct extensions, such as unimodular $F(R)$ gravity
\cite{Nojiri:2015sfd} and unimodular $F(T)$ gravity \cite{Nassur:2016yhc}, which prove to
have interesting cosmological implications.

In the present work we are interesting in investigating inflationary cosmology in the
framework of unimodular $F(T)$ gravity. Specifically, we study the Lagrange multiplier
method to represent the action of unimodular $F(T)$ gravity and we build its
reconstruction procedure. In addition, we extract the
observables of the inflationary regime,
namely the spectral index of the curvature perturbations and its running, the tensor
spectral index, and the tensor-to-scalar ratio,
 showing that they
are in very satisfactory agreement with observations. Finally, we study the instability
of the de Sitter solution, by investigating its perturbations, showing that a graceful
exit can always be realized for specific parametric regions, which is an advantage of
inflation in unimodular $F(T)$ gravity.

The plan of the work is the following.
In Sec. \ref{unimodulars} we formulate unimodular $F(T)$ gravity using Lagrange
multipliers and we present the reconstruction procedure. In Sec. \ref{Inflationarys} we
investigate the inflationary realization in the context of unimodular $F(T)$ gravity, in
the case of de Sitter, power-law and Starobinsky inflation, as well as for a specific
$F(T)$ form, extracting the inflationary observables and comparing them with the Planck
data. In Sec. \ref{grexit} we discuss the graceful exit from inflation in the scenario at
hand. Finally, Sec. \ref{Conclusions} is devoted to the conclusions.

\section{Unimodular $F(T)$ gravity}
\label{unimodulars}

In this section we will present unimodular $F(T)$ gravity and then we will formulate it
using Lagrange multipliers. Finally, we will provide the method for reconstructing
unimodular $F(T)$ gravity in the general case.

\subsection{Teleparallel and $F(T)$ gravity}

In teleparallel formulation of gravitation one uses the vierbeins $e^\mu_A$ as dynamical
variables, which at each point $x^{\mu}$ of a generic manifold form an orthonormal base
for the tangent space. The metric is then given as
\begin{equation}
g_{\mu\nu}=\eta_{A B} e^A_\mu
e^B_\nu,
\label{metricvierbein}
\end{equation}
 where greek indice span the coordinate space while latin indices span the
tangent space). Furthermore, one uses the curvatureless
Weitzenb{\"{o}}ck connection  \cite{Pereira.book}
$\overset{\mathbf{w}}{\Gamma}^\lambda_{\nu\mu}\equiv e^\lambda_A\:\partial_\mu e^A_\nu$,
instead of the standard torsionless Levi-Civita one, and thus the gravitational field is
described not by the curvature tensor but by the torsion one, which reads as
\begin{equation}
T^\rho_{\verb| |\mu\nu} \equiv e^\rho_A
\left( \partial_\mu e^A_\nu - \partial_\nu e^A_\mu \right).
\end{equation}
The Lagrangian of such a theory is just the torsion scalar $T$, which is constructed by
contractions of the torsion tensor, namely \cite{Maluf:2013gaa}
\begin{equation}
\label{Tdefscalar}
T\equiv\frac{1}{4}
T^{\rho \mu \nu}
T_{\rho \mu \nu}
+\frac{1}{2}T^{\rho \mu \nu }T_{\nu \mu\rho}
-T_{\rho \mu}{}^{\rho }T^{\nu\mu}{}_{\nu}.
\end{equation}
Since in $F(R)$ gravity one extends the Einstein-Hilbert Lagrangian, namely the Ricci
scalar $R$ to an arbitrary function $F(R)$, one can follow a similar procedure
in the context of teleparallel gravity, i.e. generalize $T$ to $F(T)$ obtaining the
$F(T)$ gravitational modification \cite{Bengochea:2008gz,Linder:2010py,Cai:2015emx}:
\begin{equation}
  S= \int d^4x e \left[
\frac{F(T)}{2{\kappa}^2}
\right],
\label{actiontotfT}
\end{equation}
 where
$e= \det \left(e^A_\mu \right)=\sqrt{-g}$ and $\kappa^2 =8\pi G=M_p^{-1}$ is the
gravitational constant, with $M_p$ the Planck mass.

The equations of motion for $F(T)$ gravity arise by variation of the total action
$S+ S_{\mathrm{M}}$, where $ S_{\mathrm{M}}$ is the matter action, in terms of the
vierbeins, and they write as
\begin{eqnarray}
\label{eomsgeneral}
&&\!\!\!\!\!\!\!e^{-1}\partial_{\mu}(ee_A^{\rho}S_{\rho}{}^{\mu\nu})\frac{d
F(T)}{dT}
 +
e_A^{\rho}S_{\rho}{}^{\mu\nu}\partial_{\mu}({T})\frac{d^2 F(T)}{dT^2} \ \ \ \ \  \
\ \
\ \ \nonumber\\
&&\!
-\frac{d
F(T)}{dT}  e_{A}^{\lambda}T^{\rho}{}_{\mu\lambda}S_{\rho}{}^{\nu\mu}+\frac{1}{4} e_ { A
} ^ {\nu}F(T) = \frac{{\kappa}^2}{2} e_{A}^{\rho}\,{T^{(\mathrm{M})}}_\rho^{\verb| |\nu},
\end{eqnarray}
where the ``super-potential'' tensor $S_\rho^{\verb| |\mu\nu} = \frac{1}{2}
\left(K^{\mu\nu}_{\verb|  |\rho}+\delta^\mu_\rho
  T^{\alpha \nu}_{\verb|  |\alpha}-\delta^\nu_\rho
 T^{\alpha \mu}_{\verb|  |\alpha}\right)$ is defined in terms of the co-torsion tensor
$K^{\mu\nu}_{\verb| |\rho}=-\frac{1}{2}\left(T^{\mu\nu}_{\verb|  |\rho} - T^{\nu
\mu}_{\verb|  |\rho} - T_\rho^{\verb| |\mu\nu}\right)$. Additionally, ${
T^{({\mathrm{M}})}}_\rho^{\verb| |\nu}$ denotes the energy-momentum tensor
corresponding to $S_{\mathrm{M}}$. We mention that in the case where $F(T)=T$ one obtains
teleparallel equivalent of general relativity, in which case equations
(\ref{eomsgeneral}) coincide with the field equations of the latter.

\subsection{Unimodular conditions}

Let us now present briefly the basic idea of unimodular gravity. In such a construction
the determinant $g$ of the metric  $g^{\mu\nu}$ is imposed to be a constant value,
namely the metric components are constrained in order for $\sqrt{-g}$ to be fixed.
Without
loss of generality, one can set $\sqrt{-g} = 1$ \cite{Nojiri:2015sfd}. In the case of
cosmological applications one considers a flat
Friedmann-Lema\^{i}tre-Robertson-Walker (FLRW) space-time
with metric
\begin{equation}
ds^2 = dt^2 - a^2(t)
\left[ \left(dx^1\right)^2 + \left(dx^2\right)^2 +
\left(dx^3 \right)^2 \right]\,,
\label{eq:2.5}
\end{equation}
which arises from the vierbein
\begin{equation}
e_{\mu}^A={\rm diag}(1,a(t),a(t),a(t))
\label{eq:2.5vierbein},
\end{equation}
where $a(t)$ is the scale factor and $t$ is the cosmic time. Introducing a new time
variable $\tau$  through
\begin{equation}
d\tau \equiv a^3 (t) dt,
\label{dtdatu}
\end{equation}
the FLRW metric is rewritten as
\begin{equation}
ds^2 = a^{-6} (\tau) d\tau^2
- a^2(\tau) \left[ \left(dx^1\right)^2 + \left(dx^2\right)^2 +
\left(dx^3 \right)^2 \right]\,,
\label{eq:2.6}
\end{equation}
with $a(\tau) \equiv a(t(\tau))$.
In this case we have $g_{\mu \nu}= \mathrm{diag} (a^{-6}(\tau), -a^2(\tau),
-a^2(\tau), -
a^2(\tau))$ and the vierbein components are given by
\begin{equation}
e^A_\mu = \mathrm{diag}(a^{-3}(\tau), a(\tau), a(\tau), a(\tau))
\label{eq:2.6vierbein2}.
\end{equation}
 Hence, we can easily verify the satisfaction of the unimodular gravity  constraint,
namely that  $|e|= \det \left(e^A_\mu \right)=\sqrt{-g} =1$.

\subsection{Lagrange multiplier formulation of unimodular $F(T)$ gravity}

In this subsection we will present the Lagrange multiplier formulation of unimodular
$F(T)$ gravity, following the corresponding procedure developed for $F(R)$ gravity in
\cite{Nojiri:2015sfd}. In particular, we will use the Lagrange multiplier
method \cite{L-M-M} in the framework of $F(T)$ gravity in order to ensure that the
unimodular condition is satisfied. Introducing the Lagrange multiplier $\lambda$,
the action of unimodular $F(T)$ gravity with matter can be written
as \cite{Nassur:2016yhc}
\begin{equation}
S = \int d^4x \left\{ |e| \left[ \frac{F(T)}{2{\kappa}^2}
-\lambda \right] + \lambda \right\}
+{S}_{\mathrm{M}} \,,
\label{eq:2.15}
\end{equation}
and in the following we set $2{\kappa}^2 = 1$ for simplicity.

By varying the action in Eq.~(\ref{eq:2.15}) with respect to
the vierbein we acquire \cite{Nassur:2016yhc}
\begin{eqnarray}
\label{eq:2.16}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
e^{-1}\partial_{\mu}(ee_A^{\rho}S_{\rho}{}^{\mu\nu})\frac{d
F(T)}{dT}
 +
e_A^{\rho}S_{\rho}{}^{\mu\nu}\partial_{\mu}({T})\frac{d^2 F(T)}{dT^2}
\nonumber\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
-\frac{d
F(T)}{dT}  e_{A}^{\lambda}T^{\rho}{}_{\mu\lambda}S_{\rho}{}^{\nu\mu}\!+\!\frac{1}{4} e_ {
A
} ^ {\nu}\left[F(T)\! - \!\lambda \right] \!=\! \frac{1}{4}
e_{A}^{\rho}\,{T^{(\mathrm{M})}}_\rho^{\verb| |\nu}\!.
\end{eqnarray}
In the following we consider the matter energy-momentum tensor
${T^{(\mathrm{M})}}_\rho^{\verb| |\nu}$ to correspond to a perfect fluid, namely
  ${T^{(\mathrm{M})}}_\rho^{\verb| |\nu} =
\mathrm{diag} (\rho_{\mathrm{M}}, -P_{\mathrm{M}}, -P_{\mathrm{M}}, -P_{\mathrm{M}})$,
where $\rho_{\mathrm{M}}$ and $P_{\mathrm{M}}$ are the energy density
and pressure respectively.

In the case of FRLW geometry of (\ref{eq:2.5vierbein}) or (\ref{eq:2.6vierbein2}), the
torsion scalar $T$ defined in (\ref{Tdefscalar}) becomes
\begin{eqnarray}
T=-6 H(t)^2=-6 a^6 (\tau) \mathcal{H}(\tau)^2,
\label{Tscalartau}
\end{eqnarray}
where
$\mathcal{H}(\tau) \equiv \frac{1}{a(\tau)}\frac{d a(\tau)}{d\tau}$ is the new function
that plays the role of the
Hubble parameter $H(t) \equiv \dot{a}(t)/a(t)$ (with dots denoting derivatives with
respect to $t$). Hence, in this case the general field equations
(\ref{eq:2.16}) give rise to the two Friedmann equations as
\begin{equation}
12a^6(\tau)\mathcal{H}^2 \frac{d F(T)}{dT}
+ \left[ F(T) - \lambda \right] -\rho_{\mathrm{M}} =0\,,
\label{eq:2.17}
\end{equation}
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
-48a^{12}(\tau)\mathcal{H}^2
\left(3\mathcal{H}^2 + \frac{d\mathcal{H}}{d\tau}\right)
\frac{d^2 F(T)}{dT^2}+ \left[ F(T) - \lambda \right]
\nonumber\\
&&\ \ \
+4a^6(\tau) \left( 6\mathcal{H}^2
+ \frac{d\mathcal{H}}{d\tau}\right)\frac{d F(T)}{dT}
+ P_{\mathrm{M}} =0 \,.
\label{eq:2.18}
\end{eqnarray}
Eliminating the term $\left( F(T) - \lambda \right)$ between (\ref{eq:2.17}) and
(\ref{eq:2.18}), and using the relation (\ref{Tscalartau})
we acquire
\begin{equation}
\!
4a^6(\tau) \!
\left(3\mathcal{H}^2 + \frac{d\mathcal{H}}{d\tau}\right) \!
\left[2T\frac{d^2 F(T)}{dT^2}\!+\!\frac{d F(T)}{dT}\right]
+ \rho_{\mathrm{M}} + P_{\mathrm{M}} = 0\,.
\label{eq:2.20}
\end{equation}
Finally, the system of equations closes by considering the  continuity equation for the
matter fluid, namely
\begin{equation}
\frac{d\rho_{\mathrm{M}}}{d\tau} + 3\mathcal{H}
\left(\rho_{\mathrm{M}} + P_{\mathrm{M}}\right) = 0\,.
\label{eq:2.19}
\end{equation}

\subsection{Reconstruction of unimodular $F(T)$ gravity}
\label{lagmultform}

In this subsection we will present the method of reconstructing the $F(T)$ form that
generates a given scale-factor evolution. We start by differentiating (\ref{Tscalartau})
in order to obtain the useful relation
\begin{equation}
\frac{dT}{d\tau} = -12a^6(\tau)\mathcal{H}
\left(3\mathcal{H}^2 + d\mathcal{H}/d\tau \right),
\label{dTdtau}
\end{equation}
and inserting it into Eq. (\ref{eq:2.20}) we acquire
\begin{equation}
-\frac{1}{3\mathcal{H}}
\left[2T\frac{d}{d\tau}
\left(\frac{d F(T)}{dT}\right)
+ \frac{dF(T)}{d\tau}
\right]
+ \rho_{\mathrm{M}} + P_{\mathrm{M}} = 0\,.
\label{eq:2.23}
\end{equation}

Let us now consider a specific scale factor. For simplicity we choose the general
power-law form
 \begin{equation}
 a(t) = a_* \left( \frac{t}{t_*} \right)^p,
 \label{apowergener}
 \end{equation}
 where $a_*$ is a value of $a$ at time $t_*$ and
$p $ is a constant, but the reconstruction procedure can be applied in a general
$a(t)$ too. In this case relation (\ref{dtdatu}) leads to
\begin{eqnarray}
\tau \Eqn{=} \frac{a_*^3\,t_* }{3p+1} \left(\frac{t}{t_*}\right)^{3p+1}\,,
\label{tautrelation}
\end{eqnarray}
and in terms of the new time variable $\tau$ the above scale factor reads as
\begin{eqnarray}
a(\tau) \Eqn{=} \left(\frac{\tau}{\tau_*}\right)^q \,,
\label{eq:2.1200}
\end{eqnarray}
with
\begin{eqnarray}
q \Eqn{\equiv} \frac{p}{3p+1} \,,
\nonumber\\
\tau_* \Eqn{\equiv} \frac{t_*}{a_*^{1/p} \left(3p+1\right)} \,.
\label{eq:2.14}
\end{eqnarray}
In this case $H=p/t$ and   $\mathcal{H} = q/\tau$, and hence (\ref{Tscalartau}) leads to
\begin{equation}
T= -6\left(\frac{q}{\tau_*}\right)^2
\left(\frac{\tau}{\tau_*}\right)^{2\left(3q-1\right)}.
\label{Ttauaux}
\end{equation}
 Thus, we can easily find that
$d/dT = -\left\{\tau_*^3/\left[12q^2\left(3q-1\right)\right]\right\}
\left(\tau/\tau_*\right)^{-3\left(2q-1\right)} d/d\tau$.
 Additionally, if the matter perfect fluid has a constant equation-of-state parameter
$w=P_{\mathrm{M}}/\rho_{\mathrm{M}}$, then the continuity equation (\ref{eq:2.19}) gives
$\rho_{\mathrm{M}} = \rho_{\mathrm{M}*} \left(\tau/\tau_*\right)^{-3q\left(1+w\right)}$,
where $\rho_{\mathrm{M}*}$ is the value of $\rho_{\mathrm{M}}$
at $\tau = \tau_*$. Inserting these into Eq. (\ref{eq:2.23})  we obtain
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\frac{d^2F(\tau)}{d\tau^2} + \frac{\left(2-3q\right)}{\tau} \frac{dF(\tau)}{d\tau}
\nonumber\\
&&
-\frac{3q\left(
3q-1\right)\left(1+w\right)\rho_{\mathrm{M}*}}{\tau_*^{-3q\left(1+w\right)}}
\tau^{-3q\left(1+w\right)-2} = 0\,,
\label{eq:2.24}
\end{eqnarray}
which is a differential equation in terms of $\tau$. The general solution of
(\ref{eq:2.24}) reads as
\begin{equation}
\!\!
F(\tau) = c_1 \tau^{3q-1}
-
\frac{\left(3q-1\right)\rho_{\mathrm{M}*}}{\left[3q\left(2+w\right)-1\right]
}\left(\frac{ \tau}{\tau_*}\right)^{-3q\left(1+w\right)} +c_2 ,
\label{eq:2.25}
\end{equation}
where $c_1$ and $c_2$ are integration constants. Therefore, using (\ref{Ttauaux}) we can
express this solution in terms of $T$ as
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
F(T) = c_1 \frac{\tau_*^{3q}}{ \sqrt{6}q} \sqrt{-T} +c_2\nonumber\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!
- \frac{\left(3q-1\right)\left(6q^2\right)
^{\frac{3q\left(1+w\right)}{2\left(3q-1\right)}}
\tau_*^{-\frac{3q\left(1+w\right)}{3q-1}}
\rho_{\mathrm{M}*}}{\left[3q\left(2+w\right)-1\right]}
(-T)^{-\frac{3q\left(1+w\right)}{2\left(3q-1\right)}}  .
\label{eq:2.26}
\end{eqnarray}
Hence, we have reconstructed the $F(T)$ form that generates the power-law scale factor
evolution (\ref{eq:2.1200}). Finally, for completeness we give the expression for the
Lagrange multiplier too. In particular, inserting (\ref{eq:2.26})  into
Eq.~(\ref{eq:2.17}) we find that
\begin{equation}
\lambda (\tau) = -2 \rho_{\mathrm{M}*}
\left(\frac{\tau}{\tau_*}\right)^{-3q\left(1+w\right)} + c_2
\,,
\label{eq:2.27}
\end{equation}
which using (\ref{Ttauaux}) leads to
\begin{equation}
\lambda (T) =
-2\rho_{\mathrm{M}*} \left(6q^2\right)^{\frac{3q\left(1+w\right)}{2\left(3q-1\right)}}
\tau_*^{-\frac{3q\left(1+w\right)}{3q-1}}
(-T)^{-\frac{3q\left(1+w\right)}{2\left(3q-1\right)}} +c_2 \,.
\label{eq:2.28}
\end{equation}
Lastly, note that in the vacuum case, i.e. when $\rho_{\mathrm{M}} =0$ and
$P_{\mathrm{M}} = 0$, we find that
\begin{equation}
F(T) = c_1 \sqrt{-T} + c_2 \,,
\label{eq:2.21}
\end{equation}
while
\begin{equation}
\lambda = c_2 \,.
\label{eq:2.22}
\end{equation}
Thus, the Lagrange multiplier becomes constant and the $F(T)$ form can be reconstructed
without using the expression of the scale factor. However, we mention that in this simple
case the theory becomes trivial, since the Lagrangian, in the absence of the constant
term, becomes a total derivative \cite{Basilakos:2013rua}.

\section{Inflationary Cosmology}
\label{Inflationarys}

In the previous section we presented unimodular $F(T)$ gravity and we analyzed the
procedure with which one can reconstruct the specific $F(T)$ form that can generate a
given scale-factor evolution. Hence, in the present section we will apply these in
inflationary cosmology, in order to investigate inflation realization in unimodular
$F(T)$
gravity. Additionally, we will extract various inflation-related observables, such as
the scalar and tensor spectral indices, the running spectral index, and the
tensor-to-scalar ratio, and we will compare them with observational data.

\subsection{Slow-roll parameters and inflationary observables}

In every inflationary scenario one needs to calculate the values of various
inflation-related observables, such as the scalar spectral index of the curvature
perturbations $n_\mathrm{s}$, the running $\alpha_\mathrm{s} \equiv d n_\mathrm{s}/d
\ln k$ of the spectral index
$n_\mathrm{s}$, where $k$ is the absolute value of the wave number $\Vec{k}$,
the tensor spectral
index $n_\mathrm{T}$ and the tensor-to-scalar ratio $r$, since these quantities are
determined very accurately by observational data, and thus confrontation can constrain of
exclude the scenarios at hand. In principle, the calculation of the above observables
requires a detailed and lengthy perturbation analysis. However, one can bypass this
procedure by transforming the given scenario to the Einstein frame, where all the
inflation information is encoded in the (effective) scalar potential $V(\phi)$, defining
the slow-roll parameters $\epsilon$, $\eta$ and $\xi$ in terms of this potential and its
derivatives as \cite{IM-PDP,Martin:2013tda}
\begin{eqnarray}
\label{epsV}
\epsilon&\equiv&\frac{M_p^2}{2}\left(\frac{1}{V}\frac{ dV}{ d\phi}\right)^2\,
,~~\,
\,\\
\eta &\equiv& \frac{M_p^2}{V}\frac{ d^2V}{ d\phi^2}\,
,
\label{etaV}\\
\xi^2&\equiv&\frac{M_p^4}{V^2}\frac{ dV}{ d\phi}\frac{
 d^3V}{d\phi^3}\, ,
\label{xiV}
\end{eqnarray}
(inflation ends when $\epsilon=1$),
and then using the approximate expressions for the observables in terms of
these potential-related slow-roll parameters \cite{Martin:2013tda}:
\begin{eqnarray}
 r &\approx&16\epsilon ,
 \label{eps111}\\
 n_\mathrm{s} &\approx& 1-6\epsilon+2\eta  ,
\label{eps2222} \\
\alpha_\mathrm{s} &\approx& 16\epsilon\eta-24\epsilon^2-2\xi^2 ,
\label{eps333}
\\
 n_\mathrm{T} &\approx& -2\epsilon .
\label{eps444}
\end{eqnarray}

However, the above procedure cannot be applied in modified gravity scenarios where the
conformal transformation to the Einstein frame is absent, since in this case one cannot
define a scalar potential and then the potential-related slow-roll parameters. In such
scenarios one should instead introduce the ``Hubble slow-roll'' parameters $\epsilon_n$
(with $n$ positive integer), defined as \cite{Martin:2013tda}
\begin{eqnarray}
\epsilon_{n+1}\equiv \frac{d\ln |\epsilon_n|}{dN},
\end{eqnarray}
with $\epsilon_0\equiv H_{ini}/H$ and $N\equiv\ln(a/a_{ini})$  the e-folding number,
and where $a_{ini}$ is the scale factor at the beginning of inflation and $H_{ini}$ the
corresponding Hubble parameter (inflation ends when $\epsilon_1=1$). In terms of the
first three $\epsilon_n$, which are straightforwardly extracted to be
\begin{eqnarray}
\label{epsVbb}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\epsilon_1\equiv-\frac{\dot{H}}{H^2},
\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\epsilon_2 \equiv  \frac{\ddot{H}}{H\dot{H}}-\frac{2\dot{H}}{H^2},
\label{etaVbb}\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\epsilon_3 \equiv
\left(\ddot{H}H-2\dot{H}^2\right)^{-1}
\nonumber\\
&&\!\!\!\!
\cdot\!\left[\frac{H\dot{H}\dddot{H}-\ddot{H}(\dot{H}
^2+H\ddot{H}) } { H\dot { H } }-\frac{2\dot{H}}{H^2}(H\ddot{H}-2\dot{H}^2)
\right]\!,
\label{xiVbb}
\end{eqnarray}
the inflationary observables write as \cite{Martin:2013tda}
 \begin{eqnarray}
 r &\approx&16\epsilon_1 ,
 \label{eps111bb}\\
 n_\mathrm{s} &\approx& 1-2\epsilon_1-2\epsilon_2  ,
\label{eps2222bb} \\
\alpha_\mathrm{s} &\approx& -2 \epsilon_1\epsilon_2-\epsilon_2\epsilon_3  ,
\label{eps333bb}\\
 n_\mathrm{T} &\approx& -2\epsilon_1  .
\label{eps444bb}
\end{eqnarray}
Obviously, in cases where both the potential-related slow-roll parameters and the Hubble
slow-roll parameters can be defined, the final expressions for the observables $r$,
$ n_\mathrm{s}$, $\alpha_\mathrm{s}$ and $ n_\mathrm{T}$ coincide.

In the present work we are interested in investigating inflationary cosmology in the
framework of unimodular $F(T)$ gravity. Similarly to usual  $F(T)$ gravity, and in
contrast with $F(R)$ gravity, in this case there is not a conformal transformation to the
Einstein frame \cite{Cai:2015emx,Yang:2010ji}, where one could find the effective scalar
potential and then use it in order to calculate  the potential-related slow-roll
parameters. Hence, in order to calculate the inflationary observables we must use the
Hubble slow-roll parameters (\ref{epsVbb})-(\ref{xiVbb}), and then insert them
into expressions (\ref{eps111bb})-(\ref{eps444bb}).

Finally, let us stress here that the above procedure pre-assumes that there is only one
extra (comparing to general relativity) degree of freedom that drives inflation, and which
will affect the scalar and tensor perturbations. However, in general, $F(T)$ gravity
possesses three extra degrees of freedom, corresponding to one massive vector field or to
one massless
vector field and one scalar  \cite{Li:2011rn}. Nevertheless, as it was shown in
\cite{Li:2011wu}, in the case of FLRW geometry, and for viable $F(T)$
scenarios, only one extra degree of freedom arises at the background and linear
perturbation levels. Hence, as a first approximation, the observables can be calculated
with the aforementioned procedure, through the use of Hubble slow-roll parameters.

\subsection{de Sitter and power-law inflation}

Let us first provide the kinematical expressions for inflation realization. Without loss
of generality we focus on two basic inflationary scale-factor evolutions, namely the de
Sitter and the power-law ones. The investigation of arbitrary evolutions is
straightforward.

For the de Sitter inflation the scale factor has the well-known exponential form
\begin{eqnarray}
a(t) = e^{ H_\mathrm{inf} t },
\label{deSittat}
\end{eqnarray}
where
$H_\mathrm{inf}$ is the constant Hubble parameter at the inflationary stage.
In this case,  following the procedure of the previous section, and in particular
relations (\ref{dtdatu}) and (\ref{eq:2.6}), we find
\begin{eqnarray}
\tau \Eqn{=} \frac{1}{3H_\mathrm{inf}}
e ^{ 3H_\mathrm{inf} t}\,,
\label{eq:2.7}\\
a (\tau) \Eqn{=} \left( H_\mathrm{inf} \tau \right)^{1/3}\,.
\label{eq:2.8}
\end{eqnarray}
Similarly, for the power-law inflation the scale factor has the form (\ref{apowergener}),
namely
 \begin{equation}
 a(t) = a_* \left( \frac{t}{t_*} \right)^p,
 \label{apowergener2}
 \end{equation}
 where $a_*$ is the value of $a$ at time $t_*$ and
$p$ is a constant. In this case, and as was analyzed in the previous section, in
expressions (\ref{eq:2.1200}) and (\ref{eq:2.14}), we obtain
\begin{eqnarray}
a(\tau) \Eqn{=} \left(\frac{\tau}{\tau_*}\right)^q \,,
\label{eq:2.1200bb}
\end{eqnarray}
with
$q = p/(3p+1) $ and $
\tau_* = t_*/[a_*^{1/p} (3p+1)] $. Note that in the limit $p \to \infty$ and $t_* \to
\infty$ with  $ p/t_*=H_*=const.$, the power-law expansion (\ref{apowergener2}) gives
the de Sitter expansion (\ref{deSittat}), with $H_\mathrm{inf}=3 H_*$. Hence, we can
study both cases in a simultaneous way, using (\ref{eq:2.1200bb}), and thus the
unimodular metric (\ref{eq:2.6}) becomes
\begin{equation}  \!
ds^2 = \left( \frac{\tau}{\tau_*} \right)^{-6q}\!\! d\tau^2
- \left( \frac{\tau}{\tau_*} \right)^{2q}\!\!
\left[ \left(dx^1\right)^2 + \left(dx^2\right)^2 +
\left(dx^3 \right)^2 \right].
\label{eq:2.11}
\end{equation}
In summary, for  $q=1/3$ we re-obtain the de Sitter expansion, while for  $1/4 < q <
1/3$, i.e. for $p>1$,  the power-law inflation is realized. Additionally, note that for
$p <0$, i.e. for $q<0$ or $q>1/3$, we acquire $\dot{H}=-p/t^2 >0$, which corresponds to
the realization of super-inflation. Finally, for $0 < p \leq 1$, i.e. for $0 < q \leq
1/4$, accelerated expansion, and thus inflation, is not realized.

Let us now investigate the observables in the case of power-law inflation.
As it was shown in \cite{Bamba:2014daa}, it is more convenient to use the e-folding
number
$N$ as the independent variable
(for related considerations, see~\cite{Bamba:2014wda, Mukhanov:2014uwa}). Hence, for every
function
$g$ we have that
$\dot{g}=g'(N) H(N)$, where primes denote derivatives with respect to $N$. Thus, the
Hubble slow-roll parameters (\ref{epsVbb})-(\ref{xiVbb}) can be re-expressed as
\begin{eqnarray}
\label{epsVbbxx}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\epsilon_1(N)\equiv-\frac{H'(N)}{H(N)},
\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\epsilon_2(N) \equiv  \frac{H''(N)}{H'(N)}-\frac{H'(N)}{H(N)},
\label{etaVbbxx}\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\epsilon_3(N) \equiv
\left[\frac{H(N)H'(N)}{H''(N)H(N)-H'(N)^2}\right]
\nonumber\\
&&\!\!\!\!
\cdot\left[ \frac{H'''(N)}{H'(N)}-
\frac{H''(N)^2}{H'(N)^2}-\frac{H''(N)}{H(N)}
+\frac{H'(N)^2}{H(N)^2}
\right].
\label{xiVbbxx}
\end{eqnarray}

Let us now consider the power-law inflation of the form (\ref{apowergener2}) or,
expressed
in unimodular terms, of (\ref{eq:2.1200bb}) and (\ref{eq:2.11}). For this case we obtain
\begin{eqnarray}
&&\epsilon_1=\frac{1}{p},
\\
&&\epsilon_2=0,
\\
&&\epsilon_3=\frac{1}{p},
\end{eqnarray}
and thus (\ref{eps111bb})-(\ref{eps444bb}) give
 \begin{eqnarray}
 &&r \approx\frac{16}{p},
 \label{eps111bbcd}\\
&& n_\mathrm{s} \approx 1-\frac{2}{p} ,
\label{eps2222bbcd} \\
&&\alpha_\mathrm{s} \approx 0 ,
\label{eps333bbcd}\\
 &&n_\mathrm{T} \approx -\frac{2}{p},
\label{eps444bbcd}
\end{eqnarray}
where we remind that $p=q/(1-3q)$.
Finally, eliminating $p$ between (\ref{eps111bbcd}),(\ref{eps2222bbcd}) we obtain
\begin{eqnarray}
 r=8(1-n_\mathrm{s}).
\end{eqnarray}
Hence, if we take $p=100$, i.e. for $q=0.332$, we acquire $r\approx0.16$
$n_\mathrm{s} \approx 0.98$,
$\alpha_\mathrm{s}=0$, and $n_\mathrm{T} \approx -0.02$, which is in good agreement with
the Planck results \cite{Planck:2015xua}. However, one can see that the power-law
inflation is a quite simple scenario and thus the expressions for the observables,
although in agreement with observations, are simple and one does not have large
parametric
freedom to change their values. Hence, in the next subsection we examine a more
complicated scenario, with improved phenomenological behavior.

\subsection{Starobinsky inflation}

Let us consider a simple but efficient model, namely the $R^2$ or
Starobinsky inflation.
In curvature modified gravity, and in particular in $R^2$ inflation, the action is given
by
$S = \int d^4 x \sqrt{-g} \left\{ R + \left[1/\left(6M^2\right)\right] R^2 \right\}$,
where $M$ is
a constant with mass dimension.
In such a scenario the Hubble parameter can be described
as \cite{Starobinsky:1980te, BC-OO}
\begin{equation}
H = H_* - \frac{M^2}{6} \left(t-t_*\right) \,,
\label{eq:3.4}
\end{equation}
where $H_*$ is the value of the Hubble function at $t=t_*$.
Straightforwardly, the scale factor reads as
$a(t) = a_* \exp \left[H_* \left(t-t_*\right) - \left(M^2/12\right) \left(t-t_*\right)^2
\right]$. For $t \ll t_*$, that is at the early stages of inflation,
we approximately acquire
$a(t) = a_* \exp \left\{ \left[H_* + \left(M^2/6\right) t_* \right]t
-\left[H_* + \left(M^2/12\right) t_* \right]t_* \right\}$.
Thus, using the expressions extracted in Section \ref{unimodulars}, and in particular
(\ref{dtdatu}), in the framework of unimodular gravity and the new time variable we have
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!
\tau = \bar{\tau}
+ \frac{a_*^3}{3H_* + \left(M^2/2\right) t_*} \exp \left[
\left(3H_* + \frac{M^2}{2} t_* \right) t \right.\nonumber\\
&&\left. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \  \ \  \ \  \ \  \ \  \ \  \ \
-\left(3H_* + \frac{M^2}{4} t_* \right) t_*
\right],
\label{eq:3.5}\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!
a (\tau) = \left( 3H_* + \frac{M^2}{2} t_*
\right)^{1/3}
\left( \tau-\bar{\tau} \right)^{1/3},
\label{eq:3.6}
\end{eqnarray}
where $\bar{\tau}$ is a constant. From (\ref{eq:3.6}) we obtain $\mathcal{H}
= 1/\left[3\left( \tau-\bar{\tau} \right)\right]$, and therefore, using
(\ref{Tscalartau}) for the torsion scalar
we acquire $T = -6\left[ H_* + \left(M^2/6\right) t_* \right]^2$. Thus, we deduce that
in the early stage of inflation $T$ is approximately constant.

The e-folding number $N$ at the inflationary stage is defined as
\begin{equation}
N \equiv \ln \left( \frac{a_\mathrm{f}}{a_\mathrm{i}} \right)
= - \int_{t_\mathrm{f}}^{t_\mathrm{i}} H (\tilde{t}) d\tilde{t} \,,
\label{eq:3.3}
\end{equation}
where $a_\mathrm{i} = a(t=t_\mathrm{i})$ is the value of the scale factor $a$ at the
beginning of inflation $t_\mathrm{i}$, and $a_\mathrm{f} = a(t=t_\mathrm{f})$ is its
value  at the end
 of inflation  $t_\mathrm{f}$. Inserting (\ref{eq:3.4}) into
(\ref{eq:3.3}) we acquire
\begin{equation}
N = -\left(H_* + \frac{M^2}{6} t_* \right) \left(t_\mathrm{i}-t_\mathrm{f}\right)
+ \frac{M^2}{12} \left(t_\mathrm{i}^2-t_\mathrm{f}^2\right),
\label{eq:3.8}
\end{equation}
which for positive $t_\mathrm{f}$ can be inverted to express $t_\mathrm{f}$ in terms of
$N$, namely
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!
t_\mathrm{f}=6\frac{H_*}{M^2}+t_*+M^{-2}\Big\{
\left(6 H_*+M^2 t_*\right)^2
\nonumber\\
&&\
+M^2\left\{t_\mathrm{i}\left[
M^2\left(t_\mathrm{i}-2t_*\right)-12 H_*\right]-12N\right\}
\Big\}^{\frac{1}{2}}.
\label{tNrel}
\end{eqnarray}

We can now use (\ref{eq:3.4}) in order to
calculate the Hubble slow-roll parameter $\epsilon_1$ from (\ref{epsVbb}), namely
\begin{eqnarray}
\label{epsVbb200}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\epsilon_1=\frac{6M^2}{\left[6H_*-M^2\left(t_\mathrm{f}-t_*\right)\right]^2},
\end{eqnarray}
and thus eliminating $t_\mathrm{f}$ in favor of $N$ using (\ref{tNrel}) we obtain
{\small{
\begin{equation}
\label{epsVbb2002}
\!\!\!\!\!\!
\epsilon_1(N)\!=\!\frac{6M^2}{\left(6 H_*\!+\!M^2
t_*\right)^2\!+\!M^2\left\{t_\mathrm{i}\left[
M^2\left(t_\mathrm{i}-2t_*\right)\!-\!12 H_*\right]\!-\!12N\right\}}.
\end{equation}}}
Similarly, from (\ref{etaVbb}),(\ref{xiVbb}) we obtain
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\epsilon_2(N) =2\epsilon_1(N),
\label{etaVbb2}\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\epsilon_3(N) = 2\epsilon_1(N).
\label{xiVbb2}
\end{eqnarray}
Inserting these into
(\ref{eps111bb})-(\ref{eps444bb}) we find
 \begin{eqnarray}
 r(N) &=&16\, \epsilon_1 (N),
 \label{eps111bb3}\\
 n_\mathrm{s}(N)&=& 1-6\, \epsilon_1 (N) ,
\label{eps2222bb3} \\
\alpha_\mathrm{s}(N) &=& -8\, \epsilon_1^2(N)  ,
\label{eps333bb3}\\
 n_\mathrm{T}(N) &=& -2\, \epsilon_1(N).
\label{eps444bb3}
\end{eqnarray}%
These expressions for the inflationary observables can be in a very good agreement with
observations \cite{Planck:2015xua}. For instance, taking  $t_\mathrm{i} = 1/H_*$, $t_* =
3/H_*$,  $H_*/M = 0.04$, and for e-folding number $N=50$, we find  $r\approx0.049$
$n_\mathrm{s} \approx 0.981$,
$\alpha_\mathrm{s}=-7.78 \times 10^{-5}$, and $n_\mathrm{T} \approx -0.0062$, while for
 $N=60$, we find  $r\approx0.053$
$n_\mathrm{s} \approx 0.98$,
$\alpha_\mathrm{s}=-8.85 \times 10^{-5}$, and $n_\mathrm{T} \approx -0.0067$.

We can eliminate the complicated function $\epsilon_1(N)$ between (\ref{eps111bb3}) and
(\ref{eps2222bb3}), and between (\ref{eps111bb3}) and (\ref{eps333bb3}),   obtaining
respectively
\begin{equation}
\label{rnsrel}
r=\frac{8}{3}(1- n_\mathrm{s}),
\end{equation}
and
\begin{equation}
\label{asnsrel}
\alpha_\mathrm{s}=-\frac{2}{9}(1- n_\mathrm{s})^2,
\end{equation}
which prove to be very useful, since they allow us to compare the predictions of our
scenario with the observational data. In particular, the Planck
results \cite{Planck:2015xua}
suggest that $n_{\mathrm{s}} = 0.968 \pm 0.006\, (68\%\,\mathrm{CL})$,
$r< 0.11\, (95\%\,\mathrm{CL})$,
and $\alpha_\mathrm{s} = -0.003 \pm 0.007\, (68\%\,\mathrm{CL})$.
The combined analysis of the BICEP2 and Keck Array data with
the Planck data shows $r< 0.07\, (95\%\,\mathrm{CL})$ \cite{Array:2015xqh}. As we
can see, using (\ref{rnsrel}) and (\ref{asnsrel}),  for $n_{\mathrm{s}} \approx 0.97$ we
obtain $r\approx 0.08$ and
$\alpha_\mathrm{s} \approx -0.0002 $, which reveals a very good agreement with
observations.

\subsection{ A specific model: $F(T) =  -T + \alpha_1 (-T)^n+\Lambda$ and $\lambda (T)
= \alpha_2+\alpha_3 (-T)^m$  }
\label{specificmodel}

Let us close the investigation of inflationary cosmology in the framework of unimodular
$F(T)$ gravity, by examining a specific model. As we showed in subsection
\ref{lagmultform}, in unimodular $F(T)$ gravity inflation may arise from power-law
forms of the $F(T)$ and Lagrange multiplier $\lambda (T)$ functions. Hence, as a specific
example we choose
\begin{eqnarray}
F(T) \Eqn{=}  -T + \alpha_1 (-T)^n+\Lambda \,,
\label{eq:4.3} \\
\lambda (T) \Eqn{=}\alpha_2+\alpha_3 (-T)^m \,,
\label{eq:4.4}
\end{eqnarray}
where $\alpha_1$, $\alpha_2$, $\alpha_3$, $\Lambda$ and
$n,m$ are constants (the minus sign in front of $T$ is chosen for convenience, since in
the usual conventions of $f(T)$ formulation, in FRW geometry $T=-6H^2<0$). In the
following we will focus on the case where $n=2$, $m=1/2$. Inserting the above  ansatzes
into the first Friedmann equation (\ref{eq:2.17}), using the relation $d\tau = a^3(\tau)
dt$, and considering also the matter sector to correspond to radiation, i.e
$\rho_{\mathrm{M}}=\rho_{r0}/a(t)^4$, we obtain the following differential equation:
\begin{equation}
 \Lambda-\alpha_2-6H(t)^2-\sqrt{6}\alpha_3 H(t)+\frac{\rho_{r0}}{a(t)^4} -108
\alpha_1 H(t)^4=0,
\label{adiffeq}
\end{equation}
with $H(t)\equiv\dot{a}(t)/a(t)$ and  $\rho_{r0}$ a constant. Choosing small values for
$\alpha_1$ (such that
$\alpha_1 H(t)^2\ll 1$) and for $\rho_{r0}$ (such that $\rho_{r0}\ll
4\alpha_2-\alpha_3^2-4\Lambda$) we can extract the solution as
\begin{equation}
H(t)\approx\frac{\beta_2  }{\beta_3  +2e^{\beta_1t}},
\label{Hgensol}
\end{equation}
where the new constants are defined as
\begin{eqnarray}
&&
\!\!\!\!\!\!\!
\beta_1=\sqrt{\frac{2}{3}}\left(\alpha_3-   \sqrt{
\alpha_3^2+4\Lambda-4\alpha_2}
 \right)
\nonumber\\
&&
\!\!\!\!\!\!\!
\beta_2=
\frac{\rho_{r0}^{-1}}{2\sqrt{6}}\, \Big(\sqrt{\alpha_3^2+4\Lambda-4\alpha_2}-\alpha_3
\Big)\nonumber\\
&& \ \ \ \
\cdot
\exp
\Big\{
4\Big[
(\alpha_3^2\!+\!4\Lambda\!-\!4\alpha_2)\Big(4\alpha_2\!+\!\alpha_3\sqrt{
\alpha_3^2\!+\!4\Lambda\!-\!4\alpha_2}
\Big)
\nonumber\\
&& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,
-\alpha_3^2-4\Lambda
\Big]C_1
\Big\}
\nonumber\\
&&
\!\!\!\!\!\!\!
\beta_3=
\rho_{r0}^{-1}  \exp\Big\{4\sqrt{\alpha_3^2+4\Lambda-4\alpha_2}
\Big[\alpha_3(1-4\alpha^2+4\Lambda)
\nonumber\\
&& \ \ \ \ \ \ \ \ \  \
\ \ \ \ \ +\alpha_3^2
+(4\alpha_2-1)\sqrt{
\alpha_3^2\!+\!4\Lambda\!-\!4\alpha_2}
\Big] C_1\Big\},
\label{const1}
\end{eqnarray}
with $C_1$ is an integration constant.

Let us now use the solution (\ref{Hgensol}) in order to calculate the inflationary
observables. We start by calculating the e-folding number $N$ defined in (\ref{eq:3.3}),
obtaining
\begin{equation}
N=\frac{\beta_2}{\beta_3 \beta_1}
\left[\beta_1(t_i-t_f)+
\ln
\left(\frac{\beta_3 +2 e^{\beta_1t_f }    }{\beta_3
 +2 e^{\beta_1t_i }  }\right)\right],
\label{eq:3.3gensol}
\end{equation}
where $t_\mathrm{i}$ is the beginning of inflation and  $t_\mathrm{f}$  its end. Relation
(\ref{eq:3.3gensol}) can be easily inverted to give $t_i(N)$. Next, we use
(\ref{Hgensol})  in oder to
calculate the Hubble slow-roll parameters   from
(\ref{epsVbb}),(\ref{etaVbb}),(\ref{xiVbb}), acquiring
\begin{eqnarray}
\label{epsVbb200gen}
&&
\epsilon_1=\frac{2 \beta_1 e^{\beta_1t_i } }{\beta_2}
\\
&&\epsilon_2=
\frac{ \beta_1 \left(\beta_3+2 e^{\beta_1t_i } \right)}{\beta_2}
\\
&&\epsilon_3=\frac{2 \beta_1 e^{\beta_1t_i } }{\beta_2}.
\end{eqnarray}
Inserting these into
(\ref{eps111bb})-(\ref{eps444bb}) we find
 \begin{eqnarray}
 &&\!\!\!\!\!\!\!  \!\!\!\!\!\!
 r =
 \frac{32 \beta_1 e^{\beta_1t_i } }{\beta_2},
 \label{eps111bb3gen}\\
 &&\!\!\!\!\!\!\!  \!\!\!\!\!\!
 n_\mathrm{s}= 1-\frac{ 2 \beta_1 \left(\beta_3+4 e^{\beta_1t_i } \right)}{\beta_2}
  ,
\label{eps2222bb3gen} \\
 &&\!\!\!\!\!\!\!  \!\!\!\!\!\!
 \alpha_\mathrm{s} =
 -\frac{ 8 \beta_1^2  e^{\beta_1t_i }
 \left(\beta_3+2 e^{\beta_1t_i }
\right)}{\beta_2^2},
\label{eps333bb3gen}\\
 &&\!\!\!\!\!\!\!  \!\!\!\!\!\!
 n_\mathrm{T} =- \frac{4 \beta_1 e^{\beta_1t_i } }{\beta_2}.
\label{eps444bb3gen}
\end{eqnarray}

Hence, we can insert (\ref{eq:3.3gensol}) into the above relations in order to eliminate
$t_i$ in favor of $N$, obtaining
 \begin{eqnarray}
 &&\!\!\!\!\!\!\!  \!\!\!\!\!\!\!\!\!\!\!\!
 r(N) =
 \frac{32 \beta_3 \beta_1 e^{\beta_1 \left(\frac{\beta_3}{\beta_2}N +t_f    \right)
 } }{\beta_2\left[\beta_3-2\left(
e^ {\frac{ \beta_1 \beta_3 } {\beta_2 }N}
-1
\right)e^{\beta_1 t_f}
\right]},
 \label{eps111bb3genN}\\
 &&\!\!\!\!\!\!\!  \!\!\!\!\!\!\!\!\!\!\!\!
 n_\mathrm{s}(N)
 =
 1-
 \frac{2 \beta_3   \beta_1
 \left[  \beta_3+2
   e^{\beta_1 t_f}
\left(  1+      e^ {\frac{ \beta_1 \beta_3 } {\beta_2 }N}    \right)
\right]}
{\beta_2\left[\beta_3-2\left(
e^ {\frac{ \beta_1 \beta_3 } {\beta_2 }N}
-1
\right)e^{\beta_1 t_f}
\right]}
  ,
\label{eps2222bb3genN} \\
 &&\!\!\!\!\!\!\!  \!\!\!\!\!\!\!\!\!\!\!\!
 \alpha_\mathrm{s}(N) =
 -
 \frac{8 \beta_3^2   \beta_1^2
 \left( \beta_3 +   2  e^{\beta_1 t_f}   \right)
    e^ { \beta_1 \left(
    \frac{\beta_3}{\beta_2}N +t_f
    \right)      }}
{\beta_2^2
\left[\beta_3-2\left(
e^ {\frac{ \beta_1 \beta_3 } {\beta_2 }N}
-1
\right)e^{\beta_1 t_f}
\right]},
\label{eps333bb3genN}\\
 &&\!\!\!\!\!\!\!  \!\!\!\!\!\!\!\!\!\!\!\!
 n_\mathrm{T}(N) =
 -
 \frac{4 \beta_3 \beta_1 e^{\beta_1 \left(\frac{\beta_3}{\beta_2}N +t_f    \right)
} }{\beta_2\left[\beta_3-2\left(
e^ {\frac{ \beta_1 \beta_3 } {\beta_2 }N}
-1
\right)e^{\beta_1 t_f}
\right]}
 .
\label{eps444bb3genN}
\end{eqnarray}
Finally, we can insert (\ref{eps2222bb3genN}) into (\ref{eps111bb3genN}) and
(\ref{eps333bb3genN}),(\ref{eps444bb3genN}),
in order to  eliminate $N$ and
$t_f$ in favor of $ n_\mathrm{s}$, resulting to
  \begin{eqnarray}
 &&\!\!\!\!\!\!\!  \!\!\!\!\!\!\!\!\!
 r =4(1   -n_\mathrm{s} )
 -\frac{8 \beta_3  \beta_1 }{\beta_2}
 ,
 \label{eps111bb3genNb00}\\
 &&\!\!\!\!\!\!\!  \!\!\!\!\!\!\!\!\!
 \alpha_\mathrm{s} =-
 \frac{1}{4}
  ( n_\mathrm{s}-1)^2+   \frac{  \beta_3^2  \beta_1^2    }{ \beta_2^2}
,
\label{eps333bb3genNb00}\\
 &&\!\!\!\!\!\!\!  \!\!\!\!\!\!\!\!\!
 n_\mathrm{T} = -\frac{1}{2}(1   -n_\mathrm{s} )
 +\frac{ \beta_3  \beta_1 }{\beta_2},
\label{eps444bb3genNb00}
\end{eqnarray}
or, restoring the original parameters using (\ref{const1}), we obtain
{\small{
  \begin{eqnarray}
 &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
 r =4(1   -n_\mathrm{s} )\nonumber\\
 &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!  \!\!\!\!\!\!\!\!
 +32\exp\Bigg\{\!
 \frac{4\left[\alpha_3^3\!
 +\!4\alpha_3\Lambda\!+\!4\alpha_2\left(
\sqrt{\alpha_3^2\!+\!4\Lambda\!-\!4\alpha_2}\!-\!\alpha_3
 \right)
 \right]\!C_1}
 {\sqrt{\alpha_3^2+4\Lambda-4\alpha_2}}\!
\Bigg\} \!
 ,
 \label{eps111bb3genNb}\\
 &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
 \alpha_\mathrm{s} =  -
 \frac{1}{4}
  ( n_\mathrm{s}-1)^2\nonumber\\
  &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!  \!\!\!\!\!\!\!\!
  +
 16\exp\Bigg\{\!
 \frac{4\left[\alpha_3^3\!
 +\!4\alpha_3\Lambda\!+\!4\alpha_2\left(
\sqrt{\alpha_3^2\!+\!4\Lambda\!-\!4\alpha_2}\!-\!\alpha_3
 \right)
 \right]\!C_1}
 {\sqrt{\alpha_3^2+4\Lambda-4\alpha_2}}\!
\Bigg\}  \!
,
\label{eps333bb3genNb}\\
 &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
 n_\mathrm{T} = -\frac{1}{2}(1   -n_\mathrm{s} )\nonumber\\
 &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!  \!\!\!\!\!\!\!\!
-4\exp\Bigg\{\!
 \frac{4\left[\alpha_3^3\!
 +\!4\alpha_3\Lambda\!+\!4\alpha_2\left(
\sqrt{\alpha_3^2\!+\!4\Lambda\!-\!4\alpha_2}\!-\!\alpha_3
 \right)
 \right]\!C_1}
 {\sqrt{\alpha_3^2+4\Lambda-4\alpha_2}}\!
\Bigg\} .
\label{eps444bb3genNb}
\end{eqnarray}}}

Relations (\ref{eps111bb3genNb})-(\ref{eps444bb3genNb}) prove to be very useful, since
they allow us to compare the
predictions of our scenario with the observational data. In particular, in
Fig.~\ref{rns} we use (\ref{eps111bb3genNb}) and we present the estimated
tensor-to-scalar ratio of the specific scenario
(\ref{eq:4.3})-(\ref{eq:4.4}) of
inflation in unimodular $F(T)$ gravity, for two cases, on top of
the
1$\sigma$ and
2$\sigma$ contours of
the Planck 2013 results \cite{Planck:2013jfk} as well as of the  Planck 2015 results
\cite{Ade:2015xua}. Additionally, in Fig. \ref{asns} we use (\ref{eps333bb3genNb}) and we
show the predictions of
our scenario for the running  spectral index $\alpha_\mathrm{s}$  on top of the
1$\sigma$ and 2$\sigma$ contours of the Planck 2013
results \cite{Planck:2013jfk} as well as of the  Planck 2015 results
\cite{Ade:2015xua}. The  agreement with observations is inside the 2$\sigma$ region for
the tensor-to-scalar ratio and it is very satisfactory for the running  spectral index.
We mention that the agreement with observations is obtained through the unimodular
construction, since taking $\lambda$ to zero, i.e. for $\alpha_2,\alpha_3\rightarrow0$,
one arrives to unacceptable deviations.
\begin{figure}[ht]
\centering
\includegraphics[scale=.50]{rnsn.eps}
\caption{{\it{ 1$\sigma$ (magenta) and 2$\sigma$ (light magenta) contours for Planck 2015
results ($TT+lowP+lensing+BAO+JLA+H_0$)  \cite{Ade:2015xua}, and 1$\sigma$ (grey) and
2$\sigma$ (light grey) contours for Planck 2013 results  ($Planck+WP+BAO$)
\cite{Planck:2013jfk} (note that the  1$\sigma$ region of Planck 2013 results  is behind
the Planck 2015 results, hence we mark its boundary by a dotted curve), on
$n_{\mathrm{s}}-r$ plane.
Additionally, we present the predictions of the specific scenario
(\ref{eq:4.3})-(\ref{eq:4.4}) of
inflation in unimodular $F(T)$ gravity, with  $n=2$, $m=1/2$, and for  $\alpha_2=0.5$,
$\alpha_3=2$,  $\Lambda=0.1$, $C_1=-1$ (black - solid curve),
and for  $\alpha_2=1$,
$\alpha_3=0.5$,  $\Lambda=1$, $C_1=-0.5$    (blue - dashed
curve), in units where $2{\kappa}^2 = 1$.
}}}
\label{rns}
\end{figure}
\begin{figure}[ht]
\centering
\includegraphics[scale=.47]{asnsn.eps}
\caption{{\it{ 1$\sigma$ (magenta) and 2$\sigma$ (light magenta) contours for Planck 2015
results ($TT,TE,EE+lowP$)  \cite{Ade:2015xua}, and 1$\sigma$ (grey) and
2$\sigma$ (light grey) contours for   Planck 2013 results  ($\Lambda
CDM+running+tensors$)
\cite{Planck:2013jfk}, on $n_{\mathrm{s}}-\alpha_\mathrm{s}$ plane. Additionally, we
present the predictions of the specific scenario
(\ref{eq:4.3})-(\ref{eq:4.4}) of
inflation in unimodular $F(T)$ gravity,  with  $n=2$, $m=1/2$, and for  $\alpha_2=0.5$,
$\alpha_3=2$,  $\Lambda=0.1$, $C_1=-1$ (black - solid curve),
and for  $\alpha_2=1$,
$\alpha_3=0.5$,  $\Lambda=1$, $C_1=-0.5$    (blue - dashed
curve), in units where $2{\kappa}^2 = 1$. The two curves are indistinguishable in
the resolution scale of the figure.   }}}
\label{asns}
\end{figure}

\section{Graceful exit from the inflationary stage}
\label{grexit}

In this section we are interested in investigating the graceful exit from inflation
in unimodular $F(T)$ gravity. In general, if the de Sitter solution describing the
inflationary stage is unstable, then the graceful exit from
inflation can be realized. When the de Sitter inflation occurs
the Hubble parameter is described by
$H = H_\mathrm{inf}$, where $H_\mathrm{inf}$ is a positive constant.
In order to examine the instability of the de Sitter solution, in general one  considers
the perturbations of
the Hubble parameter, which can be expressed as
\begin{eqnarray}
H \Eqn{=} H_\mathrm{inf} + H_\mathrm{inf} \delta(t) \,,
\label{eq:4.1} \\
\delta(t) \Eqn{\equiv} \delta_0 e^{\beta t} \,,
\label{eq:4.2}
\end{eqnarray}
with $\delta_0$ and $\beta$ constants. In   (\ref{eq:4.1})
 the   term $H_\mathrm{inf} \delta(t)$ describes
 the perturbations from the de Sitter solution, i.e.
the constant part $H_\mathrm{inf}$ of the Hubble parameter, and thus we assume
$\left| \delta(t) \right| \ll 1$. Note that in this way we can quantify
the instability of the de Sitter solution.
If we obtain a positive solution of $\beta$ then the value of $\left| \delta(t) \right|$
increases in time during inflation, and thus
the de Sitter solution is unstable.  Consequently, the inflationary stage
ends successfully and the universe can enter the reheating stage and its standard thermal
history.

Let us focus on the specific scenario analyzed in subsection \ref{specificmodel}. In this
case, the solution for the Hubble function (\ref{Hgensol}), in the $\left| \delta(t)
\right| \ll 1$ regime, can take the form  (\ref{eq:4.1}) with $
H_\mathrm{inf}=\beta_2/\beta_3$, $\delta_0=-2/\beta_3$ and $\beta=\beta_1$. Hence, one can
easily deduce that for $\beta_1>0$ we have a graceful exit from the inflationary regime.

We close this section by making some comments on the difference   between the
present scenario of inflation in unimodular $F(T)$ gravity and
$R^2$ inflation (Starobinsky inflation), focusing on the graceful exit.
In particular, in $R^2$ inflation, for the flat FLRW geometry,
by differentiating an equation derived from
the gravitational field equations in the absence of matter \cite{R-D-M}, in terms of $t$,
one finds
\begin{equation}
\dddot{R} + 3\dot{H}\dot{R} +3H\ddot{R} +M^2 \dot{R} =0 \,,
\label{eq:4.8}
\end{equation}
where $R = 6\left(H^2 +\dot{H}\right)$.
Inserting (\ref{eq:4.1}), with $\delta(t) = e^{ \tilde{\beta} t}$,
into Eq.
~(\ref{eq:4.8}) and keeping first-order terms in $\delta(t)$, we eventually  acquire
\begin{equation}
  \tilde{\beta}
\left[ \tilde{\beta}^3 + 10H_\mathrm{inf} \tilde{\beta}^2 +
\left( 24H_\mathrm{inf}^2 + M^2 \right) \tilde{\beta} +4H_\mathrm{inf} M^2  \right]   =
0.
\label{eq:4.9}
\end{equation}
It is seen from Eq.~(\ref{eq:4.9}) that even if there   exists
a real solution for $\tilde{\beta}$, it would be a non-positive solution, which
 implies that through the perturbative analysis  the instability of the de Sitter solution
does not appear. In summary,   the de Sitter solution for $R^2$ inflation would be more
stable than that for inflation in unimodular $F(T)$ gravity. This difference between
$R^2$ inflation and  inflation in unimodular $F(T)$ gravity acts as an advantage for the
latter.

\section{Conclusions}
\label{Conclusions}

In the present paper, we have investigated the inflationary realization in the context of
unimodular $F(T)$ gravity. The action of the theory is based on the $F(T)$ modification
of teleparallel gravity, in which one imposes the unimodular condition through the
use of Lagrange multipliers. Hence, we have developed the general reconstruction
procedure of the $F(T)$ form that can give rise to a given scale-factor evolution.

Having presented the general machinery, we have applied it in the inflationary regime. In
particular, we have extracted the Hubble slow-roll parameters that allow us to calculate
various inflation-related observables, such as the scalar spectral index and its running,
the tensor-to-scalar ratio, and the tensor spectral index. Then, we examined the
particular cases of de Sitter and power-law inflation, of Starobinsky inflation, as well
as inflation in a specific model of unimodular $F(T)$ gravity. As we showed, in all cases
the predictions of our scenarios are in a very good agreement with observational data
from Planck probe.

Apart from the very satisfactory agreement with observations, the scenario of inflation
in unimodular $F(T)$ gravity has the additional advantage that it always allows for a
graceful exit for specific regions of the model parameters, as it can be seen by
examining the instability of the de Sitter phase. This is in contrast with inflationary
realizations in curvature-based modified gravity, such as the Starobinsky inflation, where
a graceful exit is not guaranteed. The above features make inflation in unimodular $F(T)$
gravity a successful candidate for the description of the early universe.

\section*{Acknowledgments}

This work was partially supported by the JSPS Grant-in-Aid for
Young Scientists (B) \# 25800136 and
the research-funds provided by Fukushima University (K.B.), and
MINECO (Spain) project FIS2013-44881 (S.D.O.).

redpajama_set_name: RedPajamaArXiv


The ISK-SODEX in Istanbul is a perfect market place for the HVAC Industries and their suppliers. ZUWAs turkish partner Kulez has displayed his product portfolio as well for the first time the flushing station DESCAL from ZUWA at the ISK-SODEX 2016.
Serkan Kulez (CEO at Kulez) said he is very pleased with the outcome of this trade show and the products of ZUWA.


I remember one particular footnote took me over a day to find, and I am not proud of it. I pushed through it, though, and I am glad that I did. Despite the occasional difficulties, reading this book was a truly unique experience. The structure of the book and how the words are laid across the pages add to the stories being told. There were chapters where I felt lost, confused, turned around, alone, and overwhelmed. That may not seem encouraging to the reader debating whether to pick this up, but what is great about this book is how every detail is meticulously designed to evoke these feelings. Highly recommend.
After finishing this book, every book afterwards seems a little off-putting, as if the lack of scattered typography and braille is a sin against man.
wow, this is a good book!


It was some random night in 1978, when my sister came home in what you could call an agitated state. She'd just seen the movie Halloween and was so excited, she had to tell my Mom and I all about it. Lesley described, in graphic detail, everything she'd seen onscreen. She then went to the living room and picked out the theme-song on our piano.
I was in elementary school at the time and was so freaked out by Lesley's detailed story, I couldn't sleep for a week. I didn't SEE Halloween until it played on HBO much later, but it was somehow even more horrible (meaning AWESOME) than I'd imagined.


Sponsorship of events in the City of Subiaco for its destination brand, See Subiaco, is focussed on shining a spotlight on Subiaco as a vibrant and diverse entertainment, event and leisure centre. It aims to assist producers in the staging of events throughout Subiaco, highlight the benefits of living, working and playing in Subiaco and contribute to the city's vibrancy and diversity.
The program has two annual sponsorship rounds to a maximum value of $15 000 per event and a year-round quick fire funding for events and festivals for between $1000 to $2000.
The success of an application is determined by its merits against the funding priorities and in competition with other submissions.
Please see the Event Sponsorship Guideline for further information before applying.
Ready to apply? Complete the See Subiaco Event Sponsorship form.


text:
India, All India
Envoy invite to 'Khalistani' triggers Justin Trudeau visit row
AGE CORRESPONDENT WITH AGENCY INPUTS
Published : Feb 23, 2018, 6:09 am IST
Updated : Feb 23, 2018, 6:09 am IST
Sources said that Atwal's name had been removed from the blacklist by the Union home ministry.
Jaspal Atwal, a convicted Khalistani terrorist, who was active in the banned International Sikh Youth Federation, posed with Canadian Prime Minister Justin Trudeau's wife Sophie Trudeau at an event in Mumbai on February 20. (Photo: ANI | Twitter)
New Delhi: Canadian Prime Minister Justin Trudeau's India visit plunged into yet another controversy on Thursday over a dinner invitation to convicted a Khalistani terrorist, Jaspal Singh Atwal, by the Canadian high commissioner to India.
Though the Canadian high commission swiftly withdrew the invitation, the episode has embarrassed both the Canadian and Indian governments as it raises questions about how Mr Atwal, a Canadian national, managed to get an Indian visa and on the guest list for a dinner reception being hosted for Mr Trudeau in New Delhi on Thursday.
While the Canadian high commission said it has "rescinded Mr Atwal's invitation," and refused to comment on the matter further as it concerns "the PM's security," India's external affairs ministry said it is trying to "ascertain" how the pro-Khalistan terrorist had entered India.
Mr Atwal, a Sikh separatist active in the banned International Sikh Youth Federation, was convicted for trying to kill the then Punjab minister Malkiat Singh Sidhu in Vancouver in 1986.
Sidhu was shot at twice on a road and survived, but was later assassinated in India.
A judge called it "an act of terrorism" and sentenced Mr Atwal and three others to 20 years in prison. Sources said that Mr Atwal's name had been removed from the blacklist by the Union home ministry.
Reacting to the controversy, Mr Trudeau told reporters, "Obviously we take this situation extremely seriously. The individual in question never should have received an invitation and, as soon as we found out, we rescinded the invitation immediately... The member of Parliament who included this individual has, and will, assume full responsibility for his actions."
The Canadian PMO also said in a statement that Mr Atwal was not part of Mr Trudeau's official delegation.
When asked how Mr Atwal got a visa, external affairs ministry spokesperson Raveesh Kumar said the government is "ascertaining" the details. "We are checking to see how it happened... We are trying to find out how he managed to come in (into India)... whether it was through an OCI card or e-visa," he said.
When asked whether action could be taken now against Mr Atwal, Mr Kumar said law enforcement agencies would have to be consulted on this. He described the Canadian high commission's earlier invitation to Mr Atwal as an "oversight".
The invite cancellation comes at a time when Mr Trudeau's alleged pro-Khalistan approach is being criticised by many in India, including Punjab chief minister Amarinder Singh, who raised the issue with the Canadian PM during a meeting. On his part, Mr Trudeau assured Mr Singh that his country does not support separatism in India or elsewhere.
Top Indian government sources told this newspaper that Mr Atwal — who was originally on an Indian government blacklist for his terror activities — had been removed from the blacklist by the Union home ministry as part of an "outreach" to moderate elements in the pro-Khalistan lobby overseas who were seen to have given up their radical activities.
Review of the blacklist is always done in consultation with the Punjab government and central security agencies. And after the government removes names from the blacklist, they can move freely around the world and even return to India.
Reacting to the controversy, Mr Atwal told Canada's Postmedia that he did not plan to attend the dinner as he was in Mumbai on business, the Vancouver Sun reported. He also said it was unfair to raise his criminal conviction for shooting a visiting Punjabi Cabinet minister in Vancouver in 1986 given how long ago the crime occurred.
Tags: justin trudeau, amarinder singh, jaspal singh atwal
Location: India, Delhi, New Delhi
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The Master Pyramorphix, also known as the Mastermorphix, is a puzzle similar to the Pyramorphix but with one more layer.
The Mastermorphix (Master Pyramorphix) has two seperate versions. There is the one color version, and the four color version. This guide is written for the four color version.
I can't really ever find guides for the mastermorphix, I have seen a few forum posts talking about ideas. This is from the ideas, and the method that works for me.
The method used to solve this 4 sided puzzle is loosely based on the Petrus method. This puzzle doesn't have many people doing it, or solvers with any posted times. I see some people say this is one of the most difficult 3x3x3 shape mods. I am not doubtful about that, but I believe this puzzle can be achieved to sub-30, and maybe people have already done this. I have not seen any times below that though.
This puzzle is simply a shape mod of the 3x3x3 similar to Mirror Blocks, and many other puzzles. You can see in the diagram to the right where the corners and centers for this puzzle are located in the solved state.
This puzzle has 2 big differences to most 3x3x3 shape mods.
It's centers can become rotated; although this occurs on a normal 3x3x3 this doesn't affect you because these are solid colors. Unless of course you are solving a picture cube.
One corner can be rotated, unlike any 3x3x3, just one corner can be rotated while the rest of the cube is currently in the solved position. (This is only required for the actual corners of this puzzle.
This actually occurs because unseen by you another "corner" is actually rotated. the small triangular wedges can be rotated, and this is not visible. This means when 2 or more corners are rotated, it is very likely this will fall on one of the corners that has three colors. This means you will need to fix one or two of these at the end of every solve.
The following guide was written from scratch.
o Corners: The three colored actual corners of this puzzle are also corners.
o Centers: The two colored diamond pieces on the edge of this puzzle.
o Edges: The solid colored "wedge" pieces.
You need to first orient 3 center pieces in the correct orientation (looking at this from the top going clockwise your centers may be :yellow-blue—blue-red—red-yellow: remember, the centers of this puzzle have 2 colors on them.
Now you have to begin to build your 2x2x2 block, you have to insert 2 edges so that they match. These edges (obviously) will be between the two centers on the puzzle. so for example between the yellow-blue and blue-red centers we would place a blue edge. Make sure you keep the proper orientation of the centers.
Now that you have 2 edges correctly placed you do one f2l placement. Find the corner for the slot, match it with the missing edge and place it into the slot.
You should now have completed a 2x2x2 block, you will hold the point of this to the bottom of the puzzle pointing towards the floor or table.
* You can insert the other 2 edges by repeatedly inserting f2l pairs with the missing corner.
* There three edges of the same color, and any of these can be used in the pairs.
I am not going to go into detail for this, once you get the 2nd edge added you can place this into the appropriate position. This should be done intuitively to learn your own tricks for this part.
After you place 2 of these along the 2x2x2 block you will probably have to insert another f2l pair. Once again this should be obvious of where it is, and should be a very simple fix.
Once this has been finished one "edge" of the puzzle will be unsolved, the rest of the cube will be in the correct position. See center rotation if required.
The above will take practice.
* Did you wind up with an inversed triangle around your face, you just can't figure out how to get it in position?
* Yes this step requires a lot of practice, but once you get the hang of it it's easy.
For this step think about the cube as a 3x3x3, you can see this by rotating the puzzle slightly so you have a 3x3 on the top of the puzzle! This is determined by correct orientation of edges as well, the corners are easy to orient!
if the edges are oriented they will be in the apropriate position.
There are some great youtube videos for this, I'm sure you can find them.
Use the above to determine if you have the bar case; or the crooked bar case.
The last layer of this puzzle is pretty much a permutation problem, more so than an orientation problem!
* Rotate one center 180degrees ( [R U R' U]x6 ) - this will keep all other pieces oriented.
Enjoy this puzzle, and good luck.
This page was last edited on 13 August 2014, at 17:47.


text:
Munster signs on until 2019
Wed 11 May 2016, 11:25 AM
Storm fans have received some positive news on Wednesday with young fullback Cameron Munster signing a two-year extension on his current deal.
The new deal takes the 21-year-old through to 2019 with Melbourne, and it's a club that Munster has made his new home.
"It's a good feeling, I've signed for another two years with the club and I'm excited to see where it all goes. I'm happy enough to be with the boys again and I love Melbourne," Munster said.
"We had an option for next year in their favour and they came to me with the two-year extension and I was pretty excited for them to want to keep me for the next couple of years, so I guess I was never going to say no.
"I'm pretty happy staying here, the Storm were the first NRL team that gave me an opportunity and I'm very grateful for that so I was more then happy to stay down here."
With Storm stars Billy Slater and Cameron Smith locked up until the end of 2017 and 2018 respectively, Munster hopes to be able to follow in their footsteps and have a long career in Melbourne after a little advice from his mother.
"I was talking to my mum about it the other day and she always telling me to be like Billy and Cameron and be a one club player, so I guess she's given me a bit of advice, so hopefully I can be a one-club man."
Munster has been a mainstay in an ever-changing backline this season for a Storm team that has been superb defensively, holding their opposition to nil in two consecutive games for the first time in their history.
"I think all the boys know their role in defence, everyone knows Melbourne is a good defensive side that's what we pride ourselves on, so I think the boys are working hard for each other, and it's working for us in the last couple of weeks."
The Storm will travel to Brisbane this weekend for a match against the Cowboys at Suncorp Stadium as part of a double header on Saturday night.
Munster will come up against another in-form fullback in Lachlan Coote, but said the Storm would need to shut down more than just Coote if they're going to get the points.
"He's been playing really well and he's one of the best form fullbacks at the moment, so full credit to him for playing really well. We've got a lot of things to do to get the win, contain [Johnathan] Thurston, contain Coote and [Michael] Morgan too," Munster said.
"It'll be a tough game against the Cowboys, they won the premiership last year, and it'll be a big task for us."


What is the minimum space is required(Filesystem) for Control-M/Agent to submit a job?
But considering number of jobs and retention, this size could be not enough.
Thanks for the details. In our environment, we use Control-M V7 and the dedicated file system GB for the application server.
But depends, as Fyot says, on how many jobs you execute on that agent and how many days you retain sysouts and logs on it (history).
Set agent parameters in manner to have allarms on space left and avoid disk space full incidents.
The 60 MB minimal space requirement to store the job output (sysout) generated during the job execution.


text:
Flybe paying Government's Airport Departure Tax for passengers
Leading regional airline highlights unfairness of Air Passenger Duty by reducing one way fares by the equivalent amount on 116 routes
Flybe, Europe's largest regional airline, will pay the Government's Air Passenger Duty departure tax on behalf of customers making new bookings online between Friday 5th and Tuesday 9th April across 116 routes* effective for travel between 27th October 2013 and 25th October 2014. This limited time 'tax rebate' effectively lowers one way fares by £13, and by £26 on return UK domestic fares.
The move follows last month's Budget where the Chancellor failed to cut the UK's air taxes, leaving them the highest in Europe. By giving passengers an equivalent 'tax rebate', Flybe is highlighting just how much APD adds to the cost of flying in the UK, indicating how punitive the tax is not only to those travelling for leisure but also for those conducting business so necessary to the country's economic success and development. Flybe will be presenting a petition to the Chancellor of the Exchequer calling upon him to introduce a fair deal for UK domestic aviation and is asking customers to show their support by signing the petition at www.flybe.com/taxrebate
It is just another move the airline is taking in continuing to raise public awareness of the punitive APD departure tax that:
Is a £13 tax that must be paid by all passengers departing a UK airport on a domestic or European flight
Hits those travelling on a return flight within the UK particularly hard as they have to pay APD not just once, but twice = £26 per passenger. In reality, this means that a passenger travelling on a return ticket between Manchester and Southampton (total 366 miles) not only has to pay £13 APD when departing Manchester, but also another £13 APD when departing on the return part of their journey from Southampton. This adds an extra £26 to the total cost of their return ticket.Compare this to those passengers travelling abroad with a return ticket from Manchester to Cyprus (total 4,030 miles) who only need to pay £13, the return journey being exempt from UK APD.
Has increased by a massive 160% since 2007 when it was just £5 per passenger per flight
Is the highest of its kind in Europe
Is a direct charge collected by airlines on behalf of the Government.
Andrew Strong, Managing Director Flybe UK comments: "The entire aviation industry was disappointed that, in the face of compelling evidence, the Chancellor did nothing in last month's budget to lower UK APD which remains the highest in Europe. The Government penalises domestic passengers who have to pay double the tax, while those flying abroad pay just once because APD is a departure tax from a UK airport and that can't be fair.
"Many UK travellers prefer to fly rather than use any other means of transport and in many cases, often have no access to convenient or affordable options. This is why we are hoping that our limited time 'tax rebate' will not only help people to save money but also raise awareness of just how much this 'double dip' tax is costing them.
"Flybe operates more domestic flights than any other airline in the UK and its services help drive the economies of the UK regions, On behalf of our passengers, we vow to continue putting pressure on the Government to re-consider this punitive and unfair tax."
APD Petition
Flybe is proud to operate more domestic flights than any other airline. Our services help drive the economies of the UK regions at a time when they need all the help they can get...
However, passengers departing UK airports continue to pay the highest levels of tax in Europe and we'd like your help to tell the government that this is not right. Air Passenger Duty (APD) – a tax charged on all passengers departing a UK airport – was first introduced in 1994 with a £5 rate for flights within the UK/EU and £10 elsewhere. Since then, it has seen an increase of 160% since 2007, with a one-way domestic journey now being taxed at £13 per passenger.
The tax is particularly punitive for UK domestic passengers as they have to pay on both legs of their journey because APD is a departure tax. What that means is someone travelling between say, Belfast and Manchester, Edinburgh and Southampton or Exeter and Newcastle get stung twice by the taxman - £26 in government tax.
We don't think that's fair and that's why Flybe, will be presenting a petition to the Chancellor of the Exchequer calling upon him to introduce a fair deal for UK domestic aviation. If you think the same as us, you can show your support by signing the petition below.
"We the undersigned call upon the Chancellor to bring to an end the outrageous and punitive practice of double taxation for UK domestic air travel. Air Passenger Duty on domestic flights has risen from £5 in 2007 to £13 and hits UK regional passengers hardest because they pay both ways on a return journey. That's bad for passengers, bad for jobs, constrains growth and puts our hard-pressed regional economies at a disadvantage at a time when they need all the support they can get."
British citizen Yes
Why this matters to me


Karisma Kapoor shows us how to work pastels in your festive wardrobe in this Kiran Uttam Ghosh outfit, which she paired with Karan Johar for Tyaani jewellery.
Parineeti Chopra makes for a vibrant picture in an Anita Dongre creation.
The actor opts for festive blues in these Arpita Mehta separates.
Sonakshi Sinha goes for an elegant blue in an Anita Dongre ensemble.
Take cues from Kalki Koechlin on how to play with deep-hued velvet this festive season. She wears Noura separates.
Kajol was out and about for Durga Puja celebrations wearing a yellow Anavila sari.
Shruti Haasan monotone Anavila sari is the perfect day-to-night outfit.
Malaika Arora styled her Satya Paul sari in an unconventional manner.
Kriti Sanon proved that green is the new black.
Sonakshi Sinha also made a strong case for green in this Anita Dongre ensemble.
Tanishaa Mukerji experimented with different shades of pink this festive season.
Shilpa Shetty Kundra added a cheerful shade to her festive style with this Sukriti & Aakriti ensemble.
Sushmita Sen looked legant in a Sashi Vangapalli outfit.
Katrina Kaif chose to wear millennial pink for the festivities.
Jhanvi Kapoor paired her Manish Malhotra ensemble with drop earrings and signature side-swept waves.
Priyanka Chopra made a strong case for millennial pink in an Abhinav Mishra outfit.
Sara Ali Khan also chose a bright shade of pink. She paired her Abu Jani Sandeep Khosla outfit with golden heels and star-shaped earrings.
The Khan family came out in full force in metallic for the festivities. Azad looks adorable in that tangerine outfit.
Deepika Padukone chose a golden tissue Sabyasachi sari for the occasion.
Sonakshi Sinha looks gorgeous in this white Seema Khan ensemble.
Amrita Arora looked her festive best in Ashley Rebello.
Malaika Arora Khan also chose a Seema Khan outfit for Ganpati celebrations.
Jacqueline Fernandez's cheery Arpita Mehta outfit is one of our faves.
Karisma Kapoor went for an Ananika Khanna ensemble.
Iulia Vantur looked lovely in these soft pastel shades.
Yellow seemed to be a favourite with our leading ladies. Vidya Balan opted for a deep yellow Punit Balana sari.
Swara Bhasker's white Sukriti and Aakriti sari made for an interesting change.
Divya Khosla Kumar went for a pristine Anjul Bhandari outfit.
We would love to copy Aditi Rao Hydari's Anita Dongre look.
Disha Patani teamed her Arpita Mehta outfit with interesting gold jewellery.
Aishwara Rai Bachchan and Aaradhya Bachchan adorned Abu Jani Sandeep Khosla for the festivities.
Anushka is our fave girl-next-door. Here's why!


text:
const signAnswer = ans => {
  if (ans === null) {
    return null;
  } else {
    return 1;
  }
}

const sumNotCoercion = (a,b) => {
  if (a === null && b === null) {
    return null;
  } else {
    return a + b;
  }
}

export function countAnswered(list) {
  return list.skip(1).map(q => q.get('answer'))
             .reduce((red,val) => sumNotCoercion(red,signAnswer(val)),
                     signAnswer(list.first().get('answer')));
}

export function countQuestions(list) {
  return list.size;
}

export function onAnswerQuestion(actions, option, questionIndex) {
  actions.answer(questionIndex, option);
}


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