text
stringlengths 14
5.77M
| meta
dict | __index_level_0__
int64 0
9.97k
⌀ |
|---|---|---|
\section{Introduction}
The first type I X-ray bursting sources were discovered with SAS 3 \citep{Lewin1976} and OSO-8 \citep{Swank1977}.
\cite{Woosley1975} and \cite{Maraschi1976} independently discussed the origin of the phenomenon: type
I X-ray bursts are explained by thermonuclear flashes of the material accreting from the companion star on the surface
of the neutron star. All X-ray sources showing type I bursts are low mass X-ray binaries (LMXBs).
The proprieties and the theory of the X-ray bursts are discussed in the review of \cite{Lewin1995}.
The X-ray burster KS 1741-293 was firstly reported by \cite{Zand1991} as one of the two new transient sources near
the Galactic Centre (GC) detected during the observations performed with the X-ray wide-field camera TTM on board the
Kvant module of the Mir space station. KS 1741-293 was detected,
on 3 consecutive days in the energy range 5.7$-$27.2 keV, during which it exhibited two type I X-ray bursts.
KS 1741-293 may be identified with either MXB 1743-29 and MXB 1742-29, two bursting sources detected in 1976 with SAS-3.
The single peak burst profile excludes the identification of KS 1741-293 with MXB 1743-29 \citep{Zand1991}. Therefore KS 1741-293 and MXB 1742-29
are likely to be the same source. In the BeppoSAX era (1996-2002), KS 1741-293 was detected, together with a large sample of
galactic sources, during the Wide Field Camera (WFC) monitoring of the GC region \citep{Zand2004} at
a peak flux of the order of 30 mCrab in the 2-28 keV energy range.
From the Medium Energy Concentrator Spectrometer (MECS, on board BeppoSAX) observations, \cite{Sidoli1999}
report a 2-10 keV luminosity of the source $<10^{35}$ erg s$^{-1}$ and 10$^{36}$ erg s$^{-1}$ (corrected for absorption)
on September 1997 and March 1998, respectively, assuming a distance of 8.5 kpc.
KS 1741-293 was also detected by ASCA during 107 pointing observations of a $5 \times 5$ deg$^2$
region around the GC showing an apparent variability by a factor of 50, while on the contrary no burst
has been found \citep{Sakano2002}. No hard X-ray detection has been
reported by the first gamma-ray imager SIGMA on board the GRANAT satellite and indeed the source is not in the
hard X-Ray SIGMA catalogue, covering the 40-100 keV range \citep{Revnivstev2004}.
KS 1741-293 is listed in the BATSE/CGRO instrument deep sample as one of the 179 sources
monitored along the CGRO operative life \citep{Harmon2004} even though it is not a firm detection.
KS 1741-293 is reported in the third IBIS catalogue \citep{Bird2007} at a significance level of 67 sigma with a flux of
($5.2 \pm 0.1$) mCrab in the 20-40 keV band. Recently an X-ray burst has been reported from KS 1741-293 with IBIS/ISGRI
in the 15-25 keV band by \cite{Che2006}. It occurred on March 30, 2004.
A search for optical, infrared and radio counterparts was made by \cite{Chere1994} without finding a firm candidate. A Chandra source
inside all the KS 1741-293 high energy error circles has been proposed by \cite{Marti2007} as a possible counterpart. These authors
discuss also a possible association with a non-thermal radio nebula that could be the supernova remnant produced
by the accretion induced collapse in the binary system. However this association is still under debate due to
the estimated age (about 500 years) of the SNR, which is very low for a LMXB.
In this work we show that, within February 2003 and May 2005, KS 1741-293 has been clearly detected in the
hard X energy band ($> 20$ keV) with the IBIS imager during two visibility periods, while during the other observations it
appears to be in a quiescent status. Using the combined data from the X-ray monitor JEM-X and the IBIS hard X telescope,
we obtain in the second period of visibility the wide band X-ray spectrum from about 5 keV to 100 keV. We show for the first time
the soft component simultaneously with the hard tail for this source.
We report also the detection of two bursts with JEM-X and their temporal and spectral properties.
In section \S2 we show the observations and the data analysis tools. In section \S3 we present the data analysis results from:
flux monitoring, spectral analysis and X-ray burst analysis. Finally the conclusions are summarised in section \S4.
\section{Observations and data analysis}
\begin{table*}
\begin{center}
\caption{KS1741-293 IBIS observations. The visibility periods 1, 2, 3 include all the public data and
the core program data. Periods 4 and 5 include the core program data only. The source significance and average
flux are reported in the 20-40 keV energy band, the upper limits, in the same energy band, are
estimated at 3 sigma of significance level.}
\vspace{1em}
\renewcommand{\arraystretch}{1.2}
\begin{tabular}{lrccccc}
\hline
Period & Rev. & Start (MJD) & End (MJD) & Exp. (ks) & Significance & Average Flux (mCrab)\\
\hline
1 & 46-63 & 52698 & 52792 & 306 & 28 $\sigma$ & $9.4 \pm 0.3$ \\
2 & 103-120 & 52871 & 52921 & 1333 & not visible & $<$ 0.5 \\
3 & 164-185 & 53052 & 53115 & 921 & 62 $\sigma$ & $11.1 \pm 0.2$ \\
4 & 229-249 & 53246 & 53306 & 159 & not visible & $<$ 1.4 \\
5 & 291-307 & 53431 & 53479 & 43 & not visible & $<$ 3 \\
\hline \\
\end{tabular}
\label{tab:table1}
\end{center}
\end{table*}
\begin{table*}
\begin{center}
\caption{KS 1741-293 JEM-X observations. The source significance and the average flux are reported in the
4-15 keV (Sigma$_I$, Flux$_I$) and 15-30 keV energy bands (Sigma$_{II}$, Flux$_{II}$)}
\vspace{1em}
\renewcommand{\arraystretch}{1.2}
\begin{tabular}{lrccccccc}
\hline
Period & Rev. & Start (MJD) & End (MJD) & Exp. (ks) & Sigma$_I$ & Flux$_I$ (mCrab) & Sigma$_{II}$ & Flux$_{II}$ (mCrab)\\
\hline
1 & 46-63 & 52698 & 52792 & 66 & not visible & $<$ 1.5 & not visible & $<$ 2\\
2 & 103-120 & 52871 & 52921 & 413 & not visible & $<$ 0.5 & not visible & $<$ 1\\
3a & 164-169 & 53052 & 53069 & 39 & not visible & $<$ 2 & not visible & $<$ 2\\
3b & 170-185 & 53070 & 53115 & 164 & 27 & $ 11.6 \pm 0.5 $ & 16 & $20.2 \pm 1.5$ \\
4 & 229-249 & 53246 & 53306 & 77 & not visible & $<$ 2 & not visible & $<$ 1.7 \\
5 & 291-307 & 53431 & 53479 & 16 & not visible & $<$ 2.5 & not visible & $<$ 3\\
\hline \\
\end{tabular}
\label{tab:table1b}
\end{center}
\end{table*}
The X- and gamma-ray observatory INTEGRAL was launched on October 17, 2002 by the Russian PROTON launcher. The
satellite orbits around the Earth in three days, along a highly eccentric orbit and the observing time
is optimised by this choice. The wide-field Gamma-ray imaging and wide-band spectral capabilities of INTEGRAL
coupled with the Core Program strategy \citep{Winkler2003}, are a powerful tool to investigate deeply
the high energy behaviour of X-ray bursters as firstly reported by \cite{Bazzano2004}.
The scientific instruments on board are the hard X-ray and gamma-ray imager IBIS \citep{Ubertini2003}
covering the energy band 20 keV$-$10 MeV, the gamma-ray spectrometer SPI \citep{Vedrenne2003}, that works in the same energy
band of IBIS but is devoted to fine spectroscopy, the X-ray monitor JEM-X (3$-$35 keV) \citep{Lund2003} and the
optical camera OMC \citep{Mas2003}. The angular resolution of SPI is not good enough to disentangle KS 1741-293 from the nearby sources
in a crowded region as it is the Galactic Center. In the optical band the X-ray sources located in the galactic center region are generally
obscured. Thus, SPI and OMC data are not useful for our purposes and we did not analyse them.
We have used the public data from revolution 46 (2003-02-28) to revolution 185 (2004-04-19), and the Core
Program data from the revolution 46 to the revolution 307 (2005-05-19). We have selected the data from
all pointings in which KS 1741-293 is in the Fully Coded Field of View (FCFV, equal to a square of $9 \times 9$ degrees and
$4.8$ degrees diameter for IBIS and JEM-X respectively), where the instrument sensitivity has the best value.
The data reduction was carried out with the release 5.1 off-line scientific analysis
(OSA) software \citep{Courvoisier2003} for IBIS and the last 6.0 release for
JEM-X. The reason of this choice is that OSA 6.0 has been released recently and
includes relevant updates in the analysis methods of JEM-X but not in IBIS.
In the OSA environment, the standard analysis pipeline requires to create firstly a set of data (observation group) and
then to run the single tasks. Therefore we have created an observation group for each KS 1741-293 visibility
period and we have then performed the data reduction in two steps, obtaining firstly from the raw data the single
pointings images and after a mosaic image combining all single images.
During the visibility periods when the source is detectable, the IBIS mean spectrum has been obtained by the single pointing spectra.
Because the quiescent emission of the source is below the JEM-X sensitivity for each individual pointing, the JEM-X mean spectrum
has been extracted from the mosaic image. The JEM-X bursts analysis has been performed selecting the appropriate good time intervals.
For the spectral analysis, we have used the standard fitting tool Xspec, version 11.3.1.
\section{Scientific results}
\subsection{Flux monitoring}
\begin{figure}
\centering
\includegraphics[width=0.9\linewidth]{fig1.ps}
\caption{IBIS/ISGRI mosaic, including all the data during the visibility period 3 (see Table 1),
in the 20 - 40 keV energy band. The burster KS 1741-293 has been detected during this period at
62 sigma significance level.
\label{fig:image_ibis}}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=1.0\linewidth, height=4cm]{fig2a.ps}
\includegraphics[width=1.0\linewidth, height=4cm]{fig2b.ps}
\includegraphics[width=1.0\linewidth, height=4cm]{fig2c.ps}
\caption{From top to bottom: IBIS/ISGRI KS 1741-293 20-40 keV, 40-60 keV light curves and
hardness ratio during the visibility period 3. Each point in the plots represents the average flux
within a single INTEGRAL orbital period. The time is expressed in the
INTEGRAL Julian Date (IJD = MJD - 51544.0). The log of the observation is quoted in table 1.}
\label{fig:time_ibis}
\end{figure}
Table \ref{tab:table1} shows the log of our IBIS observations.
KS 1741-293 has been clearly detected in the visibility periods labelled as 1 and 3,
with a 20-40 keV significance of 28 sigma, and 62 sigma respectively. The average flux during these periods does
not show any meaningful variation, ranging from ($9.4 \pm 0.3$) mCrab during period 1 to ($11.1 \pm 0.2$) mCrab for period 3.
For this last period the statistics are good enough to allow a study of the source's temporal behaviour.
When the source is in a quiescent status, i.e. the flux is below the instrument sensitivity, we report the 20-40 keV flux with upper limits
estimated at 3 sigma of significance level.
The image mosaic during the period 3 is shown in figure \ref{fig:image_ibis},
where the sources are labelled according with \cite{Liu2001}. KS 1741-293 and SLX 1744-299 are associated with MXB 1741-29 and
MXB 1743-29 respectively, detected by SAS-3 in 1976.
Note that at that time 1A 1742-294 (21 arcminutes off our source) was not resolved by SAS-3.
On the contrary thanks to the good IBIS angular resolution (12 arcmin) and
to the long exposure, we are able to obtain a good separation between these two sources.
In figure \ref{fig:time_ibis} we show, during period 3, the light curves in the energy band 20-40 keV, 40-60 keV and
the hardness ratio, here defined as the ratio (count(40-60 keV)/count(20-40 keV)) between the high energy to low energy band.
To increase the statistics of the single IBIS pointings, we
have grouped all the IBIS pointing data associated to one satellite revolution.
The bin width used for the flux integration is different from one revolution to another, ranging from about 30 min (one IBIS pointing) to 30 hours.
This explains the differences in the error bars. The light curve in the 20-40 keV shows a variability of about a factor of four, smoothly
increasing from 0.8 counts/s to 2.7 counts/s in 30 days. Moreover we find an indication of spectral softening at IJD equal to about 1550.
The JEM-X observations are reported in the table \ref{tab:table1b}. JEM-X was operated with JMX2 until revolution 170 and since then
it is operating mainly with JMX1. Our source is detected in the 4-15 keV and 15-30 keV energy bands at a significance level of
27 and 16 respectively in period 3 from March 6 (revolution 170, IJD= 1526) to April 20 2004 (revolution 185, IJD = 1571),
while is not detected during periods 1 and 3a (table 2). The lack of JEM-X detection
during these periods can be explained by both the less exposure and by the source flux variability
on temporal scales of the order of few days as we observe during period 3.
The JEM-X exposures during periods 1 and 3a are a factor of 2.5 and 4.2 lower than in period 3b.
Assuming a constant flux during all the observations and taking into account the less exposure,
we would expect a signal at 17 and 13 sigma in the period 1 and 3a, respectively. However, the
flux is not constant. In fact, the 20-40 keV IBIS light curve during the period 3 (figure 2, top plot)
shows that flux can varies more than a factor of two, reaching its maximum (26 mCrab) during
period 3b, that is when we have the JEM-X detection.
The mean 20-40 keV flux in period 3b (18 mCrab) is about two times the one during
period 1 (9.4 mCrab). Thus, we would expect from JEM-X in period 1 a signal that is a factor of
two less than the above calculations (17 sigma), that is of about 9 sigma.
This is a marginal significance level for a detection with a coded mask telescope. The same
arguments can be applied to the lack of detection in period 3a. Moreover,
during period 1 and 3a the signal could be even less significant since we have indication of
spectral softening during period 3b (figure 2, bottom plot) when the source is detected by JEM-X.
\subsection{Spectral analysis}
\begin{figure}
\centering
\includegraphics[width=6cm,height=8cm,angle=-90]{fig3.ps}
\caption{IBIS/ISGRI and JEM-X KS 1741-293 average spectrum obtained in the 3th visibility period.The solid line is the
best fit model assuming a multi blackbody plus a Comptonized model.}
\label{fig:spectrum_ibis}
\end{figure}
\begin{table*}
\begin{center}
\caption{Best fit parameters from a simultaneous IBIS JEM-X spectral analysis during period 3 assuming a multicolor disk blackbody
component plus a cut-off power law (cutoffpl in Xspec) or a Comptonized model (CompTT in Xspec)
with a plan geometry. $T_{in}$ is the temperature at
inner disk radius, $E_{cut}$ is the cut-off energy, $T_0$ is the input soft photon (Wien) temperature, $kT_e$ and $\tau$ are the plasma
temperature and optical depth, respectively. The errors are at 68 \% significance level.
}
\vspace{1em}
\renewcommand{\arraystretch}{1.2}
\begin{tabular}[h]{cccccccccc}
\hline
\hline
Model & $N_H$& $T_{in}$ & $\Gamma$ & $E_{cut}$ & $T_0$ & $kT_e$ & $\tau$ & Flux (2-100 keV) & $\chi^2_{\nu}(\nu)$ \\
& $10^{22}$cm$^{-2}$ & keV & & keV & keV & keV & & $10^{-10}$erg cm$^{-2}$ $s^{-1}$ & \\
\hline
const*wabs*(diskbb+cutoffpl) & $35\pm10$ & $1.0\pm0.2$ & $2.0\pm0.2$ & $90^{+40}_{-20}$ & - & - & - & 7.7 & 1.1(6) \\
\hline
const*wabs*(diskbb+compTT) & $31^{+12}_{-8}$ & $1.1\pm0.2$ & - & - & 0.1 & $28^{+16}_{-4}$ & $1.0\pm0.5$ & 7.7 & 1.0(6) \\
\hline
\end{tabular}
\label{tab:table2}
\end{center}
\end{table*}
The IBIS/ISGRI average spectrum has been extracted in the 20-150 keV energy range during the two periods corresponding to source detection, namely
period 1 and period 3 (see table \ref{tab:table1}). We fitted the extracted spectra with a simple power law, a cut-off power law and a Comptonized
model with Xspec (v.11.3.2). During Period 1 we find that all these models provide a good agreement with the data, with a marginal evidence of
improvement by using the cut-off power law or a Comptonized model (the probability of a chance improvement in $\chi^2$
is $\sim7\%$ using the F-test).
During period 3, the increased statistics due to the longer exposure than for period 1, enable us to confidently exclude a simple power law
model (with reduced $\chi^2_{\nu}=3.3$ with 5 degrees of freedom). We obtain a comparable good fit good fit with a cut-off power law ($\chi^2_{\nu}(\nu)=0.5(4)$) and with a
Comptonized model ($\chi^2_{\nu}(\nu)=0.5(4)$).
We find no evidence of spectral variability between Period 1 and Period 3 in the 20-150 keV energy range.
In order to constrain the model parameters, we extended our spectral analysis to low energies of 4 keV thanks to the simultaneous
detection with JEM-X during Period 3. Analysis at low energies of this source have already been performed by \cite{Sidoli1999}
with BeppoSAX/MECS observations taken on the 31st of March 1998. From their analysis, the 2-10 keV spectrum was equally well
modelled by a simple power law, a thermal bremmstrahlung and a black body.
Interestingly, in the two former cases, high photoelectric absorption was required by the data, with an equivalent hydrogen column density $N_H$ of
a few $10^{23}$cm$^{-2}$.
We then performed a simultaneous analysis of the JEM-X and IBIS/ISGRI spectra on the basis of the previous results. A constant normalisation factor
has been introduced in the fitting models in order to take into account the systematic errors in the knowledge of the absolute JEM-X and IBIS/ISGRI inter-calibration.
We find that, assuming a cut-off power-law model at high energies, none of the previous models used to fit the low energies (Sidoli et al. 1999) can
fit the JEM-X data. On the contrary, a multicolor disk blackbody component (\rm diskbb in \rm xspec), typically invoked to model the emission from an accretion
disk, provides a good fit to the data. The available statistics do not allow us to disentangle among a cut-off power-law or a Comptonized model at high
energies, thus we obtained similar results also assuming a Comptonized model. At the same time, we were able to confidently exclude a simple power
law model (with $\chi^2(\nu)=4.8(7)$) rather than a cut-off power-law or a Comptonized model. The best fit model parameters from these analysis are summarised in Table 2.
We also compared the 2-10 keV flux obtained on March 1998 from the BeppoSAX/MECS observations (Sidoli et al. 1999) with our estimate from the best fit models
in the same energy range. We find a 2-10 keV flux of $2.4\times10^{-10}$ cm$^{-2}$ s$^{-1}$ that is consistent with the previous measure.
Assuming the distance of 8.5 Kpc for this source as quoted by \cite{Sidoli1999}, we computed the source luminosity from the flux measured during the period 3 in
the 20 - 100 keV energy band as 1.7 $\times$ $10^{36}$ erg s$^{-1}$.
\subsection{X-ray bursts}
\begin{figure}
\centering
\includegraphics[width=0.9\linewidth]{fig4.ps}
\caption{JEM-X KS 1741-293 image in the 4 - 15 keV energy band during the bursts occurred in the revolution 53 (March 22 2003) with superimposed
the IBIS positions with a 0.6 arcmin 90 \% error circle radius. The JEM-X error circle radius with the same confidence level
(not plotted in this picture) is equal to 3 arcmin. The 1A 1742-294 source is not detected due to a very short (about 20 seconds) integration time.}
\label{fig:image_jemx}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=0.74\linewidth, angle=-90]{fig5a.ps}
\includegraphics[width=0.74\linewidth, angle=-90]{fig5b.ps}
\caption{Light curves of the two bursts detected during the revolution 53 (upper plot) and 63 (bottom plot) in the 4-15 keV band. The zero time corresponds
to 1176.359 and 1207.412 IJD for the revolution 53 and 63 respectively.}
\label{fig:time_jemx}
\end{figure}
\begin{figure}
\centering
\includegraphics[height=6cm]{fig6.eps}
\caption{JEM-X KS 1741-293 spectrum obtained during the bursts detected in the revolutions 53 and 63. The solid lines are the best fit
assuming a blackbody model.}
\label{fig:spectrum_jemx}
\end{figure}
Two X-ray bursts from KS 1741 has been detected in the 4-15 keV band by the JEM-X telescope during the
integral revolution 53 (March 22 2003) and 63 (April 22 2003). On the instrument image of the first burst
(see fig. \ref{fig:image_jemx}) we have superimposed the IBIS position as it is reported with good accuracy
(0.6 arcmin 90\% error circle) in the survey of \cite{Bird2007}. Note that the JEM-X position accuracy (3 arcmin 90\% error circle)
is lower then the IBIS ones due to shorter exposure time. As the JEM-X position of KS 1741-293, obtained during the X-ray burst,
is in agreement with IBIS, the burst is firmly associated to our source.
The one-second resolved light curves obtained from JEM-X data exhibit two X-ray bursts in the energy range 4 - 15 keV,
one in revolution 53 (science window 58) and the other in revolution 63 (science window 92).
Both bursts lasted around 20 s, with a maximum flux of 1.3 and 2.1 Crab respectively.
The burst morphology (fig. \ref{fig:time_jemx}) confirms previous observations with a single peaked time profile unlike
double peaked time profile reported for MXB 1743-29 \citep{Zand1991}.
The JEM-X spectra during the bursts have been extracted selecting the Good Time Intervals (GTI) 1176.359-1176.360 and 1207.4116-1297.4120 IJD
for the $1^{st}$ and $2^{nd}$ bursts respectively. Both spectra have been fitted with a black body model (Fig. 6), with
a temperature kT$_{bb}$ equal to $2.1^{+0.3}_{-0.3}$ keV for the first burst and $2.5^{+0.5}_{-0.4}$ keV
for the second one at 90 \% of confidence level. The 4-20 keV fluxes are
3.6 $\times$ $10^{-9}$ erg cm$^{-2}$ s$^{-1}$ and 1.1 $\times$ $10^{-8}$ erg cm$^{-2}$ s$^{-1}$, respectively. Assuming a distance of 8.5 kpc, the radius
of the first and the second bursts are $3.9^{+1.1}_{-1.0}$ km and $4.9^{+2.1}_{-1.5}$ km respectively.
The burst emission is not detected by IBIS. This lack of high energy ($>$ 20 keV) detection can be
explained by the soft spectrum of the source emission during the burst activity and by the short exposure.
Indeed, taking into account the IBIS sensitivity, we estimate a 2 sigma flux upper limit of 140 mCrab in
the 20-40 keV energy band for a 20 seconds exposure (i.e. the burst time interval). Extrapolating the JEM-X
burst spectrum we obtain in this band a flux of 9 mCrab and 7 mCrab for the first and the second burst
respectively. These values are significantly below the IBIS 20 seconds upper limit.
\section{Conclusions}
Despite KS 1741-293 being reported for the first time many years ago \citep{Zand1991}, this source was
poorly studied, in particular at high energies.
This is mainly due to the source faintness, both in the soft and hard X-ray energy range,
and to its position in the Galactic Centre crowed region, requiring good angular resolution.
The IBIS angular resolution, good sensitivity, large field of view and long exposure in the region of the
Galactic Center are then appropriate to fulfil this task. In particular the IBIS angular resolution ($12'$) allowed us to clearly resolve
KS 1741-293 from other X-ray sources in the field of view, especially from the nearest source 1A 1742-294.
Using Open Time and Core Programme data, we have monitored, with a mCrab sensitivity in the 20-40 keV band, the hard X-ray emission
for a time period of more then two years. We have obtained a clear IBIS detection only during the visibility periods
52698-52792 MJD and 53052-53115 MJD, showing that hard X-ray emission from this source is not persistent.
The measured orbital periods for LMXBs range from a fraction of hour to tens of hours (see \cite{White1995}).
During the $3^{rd}$ visibility period the source shows a smooth flux variation by a factor of four, without evidence
of periodicity in the light curve on a time scale of 60 days. During the first visibility period there is no evidence
of source flux variation.
We have obtained for the first time a wide band (from 5 to 100 keV) spectrum using the simultaneous JEM-X and IBIS
data. The spectrum is fitted with a two component model. While the blackbody soft component could originate from the surface
of an accretion disk, the neutron star surface, or both, the hard tail, fitted by a cut-off power law, or by a Comptonized
model \citep{Titarchuk1994}, is due to a Comptonization of soft photons in a hot plasma around the neutron star.
We have detected two type I X-ray burst with JEM-X, at a position clearly consistent with the IBIS KS 1741-293 detection.
The temporal analysis with JEM-X confirms the single peak bursts of KS 1741-293, as firstly reported by \cite{Zand1991}.
The spectral proprieties of the hard tailed LMXBs are discussed in \cite{Disalvo2002}. KS 1741-293 can be included in this
class of sources.
\section*{Acknowledgements}
Authors are grateful to the anonymous referee for the useful comments.
Authors thanks M. Federici (IASF/Rome) for the continuous effort to update the INTEGRAL archive in Rome.
This work has been supported by Italian Space Agency by the grant I/R/046/04.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 6,537
|
package com.google.enterprise.cloudsearch.sdk;
import static com.google.common.base.Preconditions.checkNotNull;
import static org.hamcrest.core.StringContains.containsString;
import static org.junit.Assert.assertEquals;
import static org.junit.Assert.assertFalse;
import static org.junit.Assert.assertNull;
import static org.junit.Assert.assertThat;
import static org.junit.Assert.assertTrue;
import static org.mockito.Mockito.spy;
import static org.mockito.Mockito.verify;
import com.google.api.client.googleapis.json.GoogleJsonResponseException;
import com.google.api.client.googleapis.services.json.AbstractGoogleJsonClient;
import com.google.api.client.googleapis.services.json.AbstractGoogleJsonClientRequest;
import com.google.api.client.googleapis.testing.auth.oauth2.MockGoogleCredential;
import com.google.api.client.http.HttpRequestInitializer;
import com.google.api.client.http.HttpTransport;
import com.google.api.client.json.GenericJson;
import com.google.api.client.json.Json;
import com.google.api.client.json.JsonFactory;
import com.google.api.client.json.jackson2.JacksonFactory;
import com.google.api.client.testing.http.MockHttpTransport;
import com.google.api.client.testing.http.MockLowLevelHttpRequest;
import com.google.api.client.testing.http.MockLowLevelHttpResponse;
import com.google.api.client.util.Key;
import com.google.common.collect.ImmutableList;
import com.google.common.collect.ImmutableSet;
import java.io.IOException;
import java.security.GeneralSecurityException;
import java.util.Collection;
import java.util.Collections;
import java.util.Iterator;
import java.util.List;
import java.util.Set;
import java.util.concurrent.atomic.AtomicBoolean;
import java.util.concurrent.atomic.AtomicInteger;
import org.junit.Rule;
import org.junit.Test;
import org.junit.rules.ExpectedException;
/** Unit tests for validating functionality provided by {@link BaseApiService} */
public class BaseApiServiceTest {
private static final JsonFactory JSON_FACTORY = JacksonFactory.getDefaultInstance();
@Rule public ExpectedException thrown = ExpectedException.none();
private static class TestingHttpTransport extends MockHttpTransport {
private final Iterator<MockLowLevelHttpRequest> requests;
private TestingHttpTransport(Collection<MockLowLevelHttpRequest> requests) {
this.requests = checkNotNull(requests).iterator();
}
@Override
public MockLowLevelHttpRequest buildRequest(String method, String url) throws IOException {
if (!requests.hasNext()) {
throw new UnsupportedOperationException("Unexpected request for URL " + url);
}
return requests.next();
}
}
private static class TestApi extends AbstractGoogleJsonClient {
protected TestApi(Builder builder) {
super(builder);
}
private static class Builder extends AbstractGoogleJsonClient.Builder {
protected Builder(
HttpTransport transport,
JsonFactory jsonFactory,
HttpRequestInitializer httpRequestInitializer) {
super(
transport,
jsonFactory,
"https://www.googleapis.com/",
"mock/v1/",
httpRequestInitializer,
false);
}
@Override
public TestApi build() {
return new TestApi(this);
}
}
}
private static class TestRequest extends AbstractGoogleJsonClientRequest<TestItem> {
protected TestRequest(TestApi client) {
super(client, "GET", "/mock/v1", null, TestItem.class);
}
}
/** Public for json parsing */
public static class TestItem extends GenericJson {
@Key private Boolean flag;
@Key private Integer number;
@Key private String text;
@Key private List<String> collection;
@Key private TestItem child;
}
private static class TestApiService extends BaseApiService<TestApi> {
private final AtomicBoolean isRunning = new AtomicBoolean();
private TestApiService(Builder builder) {
super(builder);
}
private static class Builder extends BaseApiService.AbstractBuilder<Builder, TestApi> {
@Override
public Builder getThis() {
return this;
}
@Override
public Set<String> getApiScopes() {
return ImmutableSet.of("https://www.googleapis.com/mock.read");
}
@Override
public TestApi.Builder getServiceBuilder(
HttpTransport transport,
JsonFactory jsonFactory,
HttpRequestInitializer requestInitializer) {
return new TestApi.Builder(transport, jsonFactory, requestInitializer);
}
@Override
public TestApiService build() throws GeneralSecurityException, IOException {
setupServiceAndCredentials();
return new TestApiService(this);
}
}
@Override
protected void startUp() throws Exception {
isRunning.compareAndSet(false, true);
}
@Override
protected void shutDown() throws Exception {
isRunning.compareAndSet(true, false);
}
private TestItem get() throws IOException {
return executeRequest(new TestRequest(service), StatsManager.getComponent("TestApi"), true);
}
}
@Test
public void testRequestWithInitializePrimitiveTypes() throws Exception {
HttpTransport transport =
new TestingHttpTransport(ImmutableList.of(buildRequest(200, new GenericJson())));
TestApiService apiService =
new TestApiService.Builder()
.setTransport(transport)
.setCredentialFactory(scopes -> new MockGoogleCredential.Builder().build())
.build();
validateEmptyItem(apiService.get());
}
@Test
public void testRequestWithNonEmptyResponse() throws Exception {
TestItem expected = new TestItem();
expected.number = 10;
expected.collection = ImmutableList.of("item1");
expected.flag = true;
expected.child = new TestItem();
expected.text = "golden";
HttpTransport transport =
new TestingHttpTransport(ImmutableList.of(buildRequest(200, expected)));
TestApiService apiService =
new TestApiService.Builder()
.setTransport(transport)
.setCredentialFactory(scopes -> new MockGoogleCredential.Builder().build())
.build();
TestItem result = apiService.get();
assertTrue(result.flag);
assertEquals(expected.number, result.number);
assertEquals(expected.text, result.text);
assertEquals(expected.collection, result.collection);
validateEmptyItem(result.child);
}
@Test
public void testRequestWithNonRetryableError() throws Exception {
HttpTransport transport =
new TestingHttpTransport(ImmutableList.of(buildRequest(400, new GenericJson())));
TestApiService apiService =
new TestApiService.Builder()
.setTransport(transport)
.setCredentialFactory(scopes -> new MockGoogleCredential.Builder().build())
.build();
thrown.expect(GoogleJsonResponseException.class);
apiService.get();
}
@Test
public void testRequestWithRetryableError() throws Exception {
TestingHttpTransport transport =
new TestingHttpTransport(
ImmutableList.of(
buildRequest(503, new GenericJson()), buildRequest(200, new GenericJson())));
TestApiService apiService =
new TestApiService.Builder()
.setTransport(transport)
.setCredentialFactory(scopes -> new MockGoogleCredential.Builder().build())
.build();
validateEmptyItem(apiService.get());
assertFalse(transport.requests.hasNext());
}
@Test
public void testRequestWithRetryableErrorFor502() throws Exception {
TestingHttpTransport transport =
new TestingHttpTransport(
ImmutableList.of(
buildRequest(502, new GenericJson()), buildRequest(200, new GenericJson())));
TestApiService apiService =
new TestApiService.Builder()
.setTransport(transport)
.setCredentialFactory(scopes -> new MockGoogleCredential.Builder().build())
.build();
validateEmptyItem(apiService.get());
assertFalse(transport.requests.hasNext());
}
@Test
public void testRequestWithTimeouts() throws Exception {
AtomicInteger connectionTimeoutHolder = new AtomicInteger();
AtomicInteger readTimeoutHolder = new AtomicInteger();
MockLowLevelHttpRequest request =
new MockLowLevelHttpRequest("https://www.googleapis.com/mock/v1") {
@Override
public MockLowLevelHttpResponse execute() throws IOException {
MockLowLevelHttpResponse response = new MockLowLevelHttpResponse();
response
.setStatusCode(200)
.setContentType(Json.MEDIA_TYPE)
.setContent(JSON_FACTORY.toString(new GenericJson()));
return response;
}
@Override
public void setTimeout(int connectTimeout, int readTimeout) throws IOException {
connectionTimeoutHolder.set(connectTimeout);
readTimeoutHolder.set(readTimeout);
}
};
HttpTransport transport =
new TestingHttpTransport(ImmutableList.of(request));
TestApiService apiService =
new TestApiService.Builder()
.setTransport(transport)
.setCredentialFactory(scopes -> new MockGoogleCredential.Builder().build())
.setRequestTimeout(10, 15)
.build();
validateEmptyItem(apiService.get());
assertEquals(10000, connectionTimeoutHolder.get());
assertEquals(15000, readTimeoutHolder.get());
}
@Test
public void testRequestWithNonDefaultRootUrl() throws Exception {
MockLowLevelHttpRequest request = buildRequest(200, new GenericJson());
TestingHttpTransport transport = spy(new TestingHttpTransport(ImmutableList.of(request)));
TestApiService apiService =
new TestApiService.Builder()
.setTransport(transport)
.setCredentialFactory(scopes -> new MockGoogleCredential.Builder().build())
.setRootUrl("https://staging.googleapis.com")
.build();
validateEmptyItem(apiService.get());
verify(transport).buildRequest("GET", "https://staging.googleapis.com/mock/v1");
}
@Test
public void testApiServiceWithPrebuiltClient() throws Exception {
MockLowLevelHttpRequest request = buildRequest(200, new GenericJson());
HttpTransport transport = new TestingHttpTransport(ImmutableList.of(request));
TestApi apiClient =
(TestApi)
new TestApi.Builder(transport, JSON_FACTORY, null)
.setApplicationName("bring your own client")
.build();
TestApiService apiService =
new TestApiService.Builder()
.setService(apiClient)
.setTransport(transport)
.setCredentialFactory(scopes -> new MockGoogleCredential.Builder().build())
.build();
validateEmptyItem(apiService.get());
assertThat(
request.getHeaderValues("user-agent").get(0), containsString("bring your own client"));
}
@Test
public void testIOExceptionRetry() throws Exception {
MockLowLevelHttpRequest request =
new MockLowLevelHttpRequest("https://www.googleapis.com/mock/v1") {
@Override
public MockLowLevelHttpResponse execute() throws IOException {
throw new IOException("something is wrong");
}
};
MockLowLevelHttpRequest retryRequest = buildRequest(200, new GenericJson());
HttpTransport transport = new TestingHttpTransport(ImmutableList.of(request, retryRequest));
TestApiService apiService =
new TestApiService.Builder()
.setTransport(transport)
.setCredentialFactory(scopes -> new MockGoogleCredential.Builder().build())
.build();
validateEmptyItem(apiService.get());
}
private static MockLowLevelHttpRequest buildRequest(int responseCode, GenericJson apiResponse) {
return new MockLowLevelHttpRequest("https://www.googleapis.com/mock/v1") {
@Override
public MockLowLevelHttpResponse execute() throws IOException {
MockLowLevelHttpResponse response = new MockLowLevelHttpResponse();
response
.setStatusCode(responseCode)
.setContentType(Json.MEDIA_TYPE)
.setContent(JSON_FACTORY.toString(apiResponse));
return response;
}
};
}
private static void validateEmptyItem(TestItem result) {
assertFalse(result.flag);
assertEquals(Integer.valueOf(0), result.number);
assertNull(result.text);
assertNull(result.child);
assertEquals(Collections.emptyList(), result.collection);
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 316
|
{"url":"https:\/\/kar.kent.ac.uk\/4651\/","text":"Skip to main content\n\n# The relative trace ideal and the depth of modular rings of invariants\n\nFleischmann, Peter, Shank, R. James (2003) The relative trace ideal and the depth of modular rings of invariants. Archiv der Mathematik, 80 (4). pp. 347-353. ISSN 1420-8938. (doi:10.1007\/s00013-003-0794-0) (The full text of this publication is not currently available from this repository. You may be able to access a copy if URLs are provided) (KAR id:4651)\n\n The full text of this publication is not currently available from this repository. You may be able to access a copy if URLs are provided. Official URLhttps:\/\/doi.org\/10.1007\/s00013-003-0794-0\n\n## Abstract\n\nWe prove that for a modular representation, the depth of the ring of invariants is the sum of the dimension of the fixed point space of the p-Sylow subgroup and the grade of the relative trace ideal. We also determine which of the Dickson invariants lie in the radical of the relative trace ideal and we describe how to use the Dickson invariants to compute the grade of the relative trace ideal.\n\nItem Type: Article 10.1007\/s00013-003-0794-0 Q Science > QA Mathematics (inc Computing science) Divisions > Division of Computing, Engineering and Mathematical Sciences > School of Mathematics, Statistics and Actuarial Science James Shank 05 Sep 2008 10:58 UTC 16 Nov 2021 09:42 UTC https:\/\/kar.kent.ac.uk\/id\/eprint\/4651 (The current URI for this page, for reference purposes) https:\/\/orcid.org\/0000-0002-3317-4088\n\u2022 Depositors only (login required):","date":"2022-06-28 16:40:20","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8184275031089783, \"perplexity\": 1510.0480706912304}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": false}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-27\/segments\/1656103556871.29\/warc\/CC-MAIN-20220628142305-20220628172305-00120.warc.gz\"}"}
| null | null |
{"url":"https:\/\/stats.stackexchange.com\/questions\/331073\/how-can-i-teach-someone-sampling-from-a-given-distribution-is-hard","text":"# How Can I teach someone \u201csampling from a given distribution\u201d is hard?\n\nFor many people I know, they do not think sampling from a given distribution is a hard problem in general. For example, many software provide functions do to sample from normal distribution or uniform distribution.\n\nHow can I teach to other person that, in general, sampling is hard (in terms of time complexity), which is why we have many algorithms to do sample, such as McMC? Is there any good intuitive example of to show sampling is a non-trivial problem?\n\n\u2022 You'll have to bring up the example, that's the only way to show that something's difficult. Bring up a case where there is not analytical expression, and you have to do MCMC, for instance. \u2013\u00a0Aksakal Feb 28 '18 at 18:50\n\u2022 @Aksakal thanks for the comment. Do you have couple of examples to show sampling is not trivial? I am thinking about some examples from rejection sampling, that trying to say, if we sample for P(Y|X) is hard, if we conditional on something almost never happen. \u2013\u00a0Haitao Du Feb 28 '18 at 18:55\n\u2022 In \"deep learning\" by Goodfellow, Bengio & Courvile there is a nice observation in the context of generative models that is relevent to sampling too. They say that if your target is a set of cats & dogs, you should play on to levels. First, your items should represent well a cat (or a dog). Other than that, you should generate\/sample both cats and dogs, as in the target. Producing only excellent cats won't be enough. I think that this example can help someone understand that there are more constraints to the problem than it seems first. \u2013\u00a0DaL Mar 1 '18 at 7:53\n\nHow about the following distribution on variables $x_i \\in \\texttt{ascii}$:\n\n$P(x_i ... x_k) = 0$ if $x_1x_2\\ldots x_k$ is not a valid program in C (or your favorite formal language)\n\n$P(x_i ... x_k) \\propto w^{-1}$ otherwise, where $w$ is the number of characters $x_i$ which are not spaces.\n\nI think this is a reasonable example because\n\n1. It's fairly interpretable: Sample random C programs inversely proportional to their (non whitespace) length.\n\n2. There is no trivial solution -- simply sampling valid programs is nontrivial. Rejection sampling is also hard, since most of the mass lies with in a small number of programs. Computing the partition function is also intractable for moderate values of $k$.\n\n3. An MCMC approach seems feasible in this case. There are many reasonable proposal distributions. Although the mixing time might be huge.\n\n\u2022 I liked your example very much (+1), however, I cannot think about a MCMc method on this, could you tell me more about it? \u2013\u00a0Haitao Du Feb 28 '18 at 21:09\n\u2022 Yes, we can use metropolis-hastings where we sample from the proposal distribution by randomly picking one $j \\in 1 \\ldots k$ and then replacing $x_j$ with a new character at random. This proposal distribution is symmetric. Start off the chain with a simple hello world program. \u2013\u00a0shimao Feb 28 '18 at 21:27","date":"2019-10-19 12:25:43","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6654700636863708, \"perplexity\": 494.616270653177}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-43\/segments\/1570986693979.65\/warc\/CC-MAIN-20191019114429-20191019141929-00049.warc.gz\"}"}
| null | null |
Free rpg games?Or maybe 3d games?
You can find different types of games on Gamehag! Visir our site now, and join our magical world!
Also get for free Grand Theft Auto V (GTA V) | Fifa 17 | Fallout 4!
Steam (at least lvl 5), Twitter, Instagram, Facebook, Youtube, Discord and Gamehag.
can i get a key pls?
I got a duplicate, thanks a lot cancerhag!
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 1,068
|
{"url":"https:\/\/math.stackexchange.com\/questions\/2558538\/understanding-a-proof-of-the-uniform-lln\/2558604","text":"# Understanding a proof of the Uniform LLN\n\nI'm trying to understand this proof of the uniform law of large numbers. The proof I mean is on page 2, the theorem titled \"ULLN1\".\n\nFirst of all, I don't understand the first inequality:\n\n$$\\max_k \\sup_\\theta\\left[ \\frac1n\\sum g(X_i, \\theta)-Eg(X_i,\\theta) \\right] \\leq \\max_k\\left[\\frac1n\\sum\\sup_\\theta g(X_i, \\theta)-E\\inf_\\theta g(X_i,\\theta)\\right]$$\n\nIs the idea to forget about the $\\max$, and bound each $\\sup$ separately? It seems like we're just saying $\\sup (a-b)\\leq \\sup a - \\inf b$, is that basically it?\n\nSecondly, two lines down, I get that we're applying the standard LLN, but how do we pull the $o_P(1)$ out of the $\\max$?\n\nFinally, I'm not sure I really understand the overall structure of the proof. We basically want to prove $f_n\\to f$ uniformly, but instead of trying to bound $\\sup|f_n-f|$, it seems like we're separately bounding $\\sup(f_n-f)$ and $\\inf(f_n-f)$, is that right? Can the proof not be rewritten in a more standard way where we just bound an absolute value directly?\n\n\u2022 What does \"this proof\" refer to in the first line? Page 2 of what? Did you forget to stick in a URL? My understanding of the \"ULLN\" is that it is the LLN for vectors in the Banach space $C(\\Theta)$: see math.stackexchange.com\/questions\/2469152. Dec 9, 2017 at 14:23\n\u2022 @kimchilover Yes, I forgot the link. Dec 9, 2017 at 14:35\n\nA key step is covering the space of functions with a collection of finitely many compact sets of functions that catch most of the probability mass. Recall the Arzel\u00e0\u2013Ascoli theorem, which describes relatively compact subsets of continuous functions: they have to bounded, and they have to be equicontinuous. That is, they cannot wiggle too much, oscillate too much. The mysterious sup - inf terms in the $$\\max_k$$ expression you ask about has to do with bounding the oscillation of the summand functions over certain $$\\delta$$-balls covering $$\\Theta$$. The $$\\max$$ picks the worst such $$\\delta$$-ball, and guess what, it's tame enough, after all.","date":"2022-05-25 10:57:57","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 5, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8925185203552246, \"perplexity\": 349.1391561133315}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-21\/segments\/1652662584398.89\/warc\/CC-MAIN-20220525085552-20220525115552-00170.warc.gz\"}"}
| null | null |
As an alternative to the laminar flow box of the FBS series "SuSi", this version is offered without a sliding door and perforated metal floor. As a result, work can be performed directly on the existing table.
A well thought out air circulation ensures that the air flow happens from the inside to the outside and no contamination through outside air occurs on the inside of the box.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 2,165
|
Bzowski:
Baltazar Bzowski
Abraham Bzowski
Stanisław Bzowski (Aleksander Kostka-Napierski)
Kazimierz Janota-Bzowski
Władysław Roman Janota Bzowski
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 1,199
|
Q: EFD-Reinf - Versão do lote inválida. Deve ser utilizada a versão 1.04.00 Estou gerando o registro R-1000 para teste de envio no ambiente de produção restrita do EFD-Reinf, e já obtive ajuda aqui no SOpt mesmo:
Como consumir WebService do EFD-Reinf no C#? (Envio Eventos).
Consegui autenticar e fazer o consumo do serviço com sucesso, porém o retorno do XML enviado diz que a versão está inválida:
Versão do lote inválida. Deve ser utilizada a versão 1.04.00.
Sendo que aponto para a nova:
http://www.reinf.esocial.gov.br/schemas/envioLoteEventos/v1_04_00
Já pesquisei aqui no SOpt, inclusive no post:
C# EFD-REINF 1.03.02 - Assinatura do evento inválida. Assinatura Digital do documento XML é inválida, mas não encontrei a causa do problema.
Segue o XML que estou enviando, sem assinatura:
<?xml version="1.0" encoding="utf-8"?>
<soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/" xmlns:sped="http://sped.fazenda.gov.br/">
<soap:Header />
<soap:Body>
<REINF xmlns="http://www.reinf.esocial.gov.br/schemas/envioLoteEventos/v1_04_00">
<loteEventos>
<evento id="ID1">
<Reinf xmlns="http://www.reinf.esocial.gov.br/schemas/evtInfoContribuinte/v1_04_00">
<evtInfoContri id="20181121110000000000000001">
<ideEvento>
<tpAmb>2</tpAmb>
<procEmi>1</procEmi>
<verProc>1.0</verProc>
</ideEvento>
<ideContri>
<tpInsc>1</tpInsc>
<nrInsc>01628604</nrInsc>
</ideContri>
<infoContri>
<inclusao>
<idePeriodo>
<iniValid>2018-01</iniValid>
</idePeriodo>
<infoCadastro>
<classTrib>99</classTrib>
<indEscrituracao>1</indEscrituracao>
<indDesoneracao>0</indDesoneracao>
<indAcordoIsenMulta>0</indAcordoIsenMulta>
<indSitPJ>0</indSitPJ>
<contato>
<nmCtt>XXXX</nmCtt>
<cpfCtt>XXX</cpfCtt>
<foneFixo>(011)XXXX6427</foneFixo>
<email>XXXXX</email>
</contato>
</infoCadastro>
</inclusao>
</infoContri>
</evtInfoContri>
</Reinf>
</evento>
</loteEventos>
</REINF>
</soap:Body>
</soap:Envelope>
O XML com a assinatura:
<soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/" xmlns:sped="http://sped.fazenda.gov.br/">
<soap:Header />
<soap:Body>
<REINF xmlns="http://www.reinf.esocial.gov.br/schemas/envioLoteEventos/v1_04_00">
<loteEventos>
<evento id="ID1">
<Reinf xmlns="http://www.reinf.esocial.gov.br/schemas/evtInfoContribuinte/v1_04_00">
<evtInfoContri id="20181121110000000000000001">
<ideEvento>
<tpAmb>2</tpAmb>
<procEmi>1</procEmi>
<verProc>1.0</verProc>
</ideEvento>
<ideContri>
<tpInsc>1</tpInsc>
<nrInsc>01628604</nrInsc>
</ideContri>
<infoContri>
<inclusao>
<idePeriodo>
<iniValid>2018-01</iniValid>
</idePeriodo>
<infoCadastro>
<classTrib>99</classTrib>
<indEscrituracao>1</indEscrituracao>
<indDesoneracao>0</indDesoneracao>
<indAcordoIsenMulta>0</indAcordoIsenMulta>
<indSitPJ>0</indSitPJ>
<contato>
<nmCtt>xxx</nmCtt>
<cpfCtt>xxx</cpfCtt>
<foneFixo>(011)xxx</foneFixo>
<email>xxxx</email>
</contato>
</infoCadastro>
</inclusao>
</infoContri>
</evtInfoContri>
</Reinf>
</evento>
</loteEventos>
</REINF>
</soap:Body>
<Signature xmlns="http://www.w3.org/2000/09/xmldsig#">
<SignedInfo>
<CanonicalizationMethod Algorithm="http://www.w3.org/TR/2001/REC-xml-c14n-20010315" />
<SignatureMethod Algorithm="http://www.w3.org/2001/04/xmldsig-more#rsa-sha256" />
<Reference URI="">
<Transforms>
<Transform Algorithm="http://www.w3.org/2000/09/xmldsig#enveloped-signature" />
<Transform Algorithm="http://www.w3.org/TR/2001/REC-xml-c14n-20010315" />
</Transforms>
<DigestMethod Algorithm="http://www.w3.org/2001/04/xmlenc#sha256" />
<DigestValue>Mv9GEVe2dZ4zstZhfcp0xok3OHolN5kjUNtMDvG1uF4=</DigestValue>
</Reference>
</SignedInfo>
<SignatureValue>DUE7Q+LguONd8cDbPZHqC9ObqG51Um5Oh4v7I5fMw43zIQQeeMbhF2j1GrbKg0poFW134OUB1dFXZyXMEQ/ynrNXkRXT1zZAXgQ0jFtfUA6MIjDCUoE68Q/bNDG7N28A1loXqAILKMmVr8WGg73UA9rDU9wMTTWy9yOfN712yqFA5RJzp53wpVX6BmjC05lfRblslEcSZTvQg2QbgLh+FBRIubznVn+sFuZ5hAB6QggCumSyxVBrpeLPyvlpSPvQM2xHw4IO3bFSq2BqPHSePU3lLyl+8/irlOLUGYSSX77m8hmBOpKzmqBqjX+v8E6QLHLEUpXcN0HT3Y5bGMByrQ==</SignatureValue>
<KeyInfo>
<X509Data>
<X509Certificate>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</X509Certificate>
</X509Data>
</KeyInfo>
</Signature>
</soap:Envelope>
O XML completo de retorno:
<Reinf xmlns="http://www.reinf.esocial.gov.br/schemas/retornoLoteEventos/v1_04_00">
<retornoLoteEventos id="IDCE8ED92E6A9124FCD401AFF46756541E">
<ideTransmissor>
<IdTransmissor>01628604000140</IdTransmissor>
</ideTransmissor>
<status>
<cdStatus>1</cdStatus>
<descRetorno>ERRO</descRetorno>
<dadosRegistroOcorrenciaLote>
<ocorrencias>
<tipo>1</tipo>
<codigo>MS0092</codigo>
<descricao>Versão do lote inválida. Deve ser utilizada a versão 1.04.00.</descricao>
</ocorrencias>
</dadosRegistroOcorrenciaLote>
</status>
</retornoLoteEventos>
<Signature xmlns="http://www.w3.org/2000/09/xmldsig#">
<SignedInfo>
<CanonicalizationMethod Algorithm="http://www.w3.org/TR/2001/REC-xml-c14n-20010315" />
<SignatureMethod Algorithm="http://www.w3.org/2001/04/xmldsig-more#rsa-sha256" />
<Reference URI="#IDCE8ED92E6A9124FCD401AFF46756541E">
<Transforms>
<Transform Algorithm="http://www.w3.org/2000/09/xmldsig#enveloped-signature" />
<Transform Algorithm="http://www.w3.org/TR/2001/REC-xml-c14n-20010315" />
</Transforms>
<DigestMethod Algorithm="http://www.w3.org/2001/04/xmlenc#sha256" />
<DigestValue>gRHi6NxvLDuhFO2Q2jB+Nl7B450LCzRynii+Mo10ZWE=</DigestValue>
</Reference>
</SignedInfo>
<SignatureValue>CmbAd77rYGrwWkFSw+Hx4BqnbV7bgjGfnnw2Y1TkQt1HYJIjoe7JcjBO8v/BFmg0vzHMU4CLr4GuryhMFFP1m6ED/OhAbH1DAqwmPITypfSnLTZno5ILgxtV1mxic88oYY3VebPZnAeINg7OK1u+J3tQMmpEuTx0ew7kY5wIFOW+Fa2bUQXGRqrLP2r5C7Gmt+b/Z0mHxLAaeb/qwdhxqnyDVuCayFo93dI5rREP9WgQhTijwRv8k0uBQ94bOzv12kcF15w00rg3F/1bo6Vy8iSzFc9OtEl0C5tUMC765QnfFChdWdkT9BtQwVnKiOHTGISixa9EHpzjEv24cGvrPA==</SignatureValue>
<KeyInfo>
<X509Data>
<X509Certificate>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</X509Certificate>
</X509Data>
</KeyInfo>
</Signature>
</Reinf>
EDIÇÃO
Tem razão, não havia percebido que a assinatura deve ser por evento e não por lote.
Sobre "A forma correta do seu soap:Envelope deveria ser essa":
fiz a edição conforme indicado, mas continua retornando a mesma mensagem de lote errado:
{<Reinf xmlns="http://www.reinf.esocial.gov.br/schemas/retornoLoteEventos/v1_04_00">
<retornoLoteEventos id="ID54ED726E636E2B6F9FD4FD7077870873">
<ideTransmissor>
<IdTransmissor>01628604000140</IdTransmissor>
</ideTransmissor>
<status>
<cdStatus>1</cdStatus>
<descRetorno>ERRO</descRetorno>
<dadosRegistroOcorrenciaLote>
<ocorrencias>
<tipo>1</tipo>
<codigo>MS0092</codigo>
<descricao>Versão do lote inválida. Deve ser utilizada a versão 1.04.00.</descricao>
</ocorrencias>
</dadosRegistroOcorrenciaLote>
</status>
</retornoLoteEventos>
<Signature xmlns="http://www.w3.org/2000/09/xmldsig#">
<SignedInfo>
<CanonicalizationMethod Algorithm="http://www.w3.org/TR/2001/REC-xml-c14n-20010315" />
<SignatureMethod Algorithm="http://www.w3.org/2001/04/xmldsig-more#rsa-sha256" />
<Reference URI="#ID54ED726E636E2B6F9FD4FD7077870873">
<Transforms>
<Transform Algorithm="http://www.w3.org/2000/09/xmldsig#enveloped-signature" />
<Transform Algorithm="http://www.w3.org/TR/2001/REC-xml-c14n-20010315" />
</Transforms>
<DigestMethod Algorithm="http://www.w3.org/2001/04/xmlenc#sha256" />
<DigestValue>PkXxGAtPdkWYgZ21TLotsVYYE7mPLGnmpmLMXI65gnQ=</DigestValue>
</Reference>
</SignedInfo>
[...]</SignatureValue>
<KeyInfo>
<X509Data>
<X509Certificate> [...]</X509Certificate>
</X509Data>
</KeyInfo>
</Signature>
</Reinf>}
BaseUri: ""
Document: null
FirstAttribute: {xmlns="http://www.reinf.esocial.gov.br/schemas/retornoLoteEventos/v1_04_00"}
FirstNode: {<retornoLoteEventos id="ID54ED726E636E2B6F9FD4FD7077870873" xmlns="http://www.reinf.esocial.gov.br/schemas/retornoLoteEventos/v1_04_00">
<ideTransmissor>
<IdTransmissor>01628604000140</IdTransmissor>
</ideTransmissor>
<status>
<cdStatus>1</cdStatus>
<descRetorno>ERRO</descRetorno>
<dadosRegistroOcorrenciaLote>
<ocorrencias>
<tipo>1</tipo>
<codigo>MS0092</codigo>
<descricao>Versão do lote inválida. Deve ser utilizada a versão 1.04.00.</descricao>
</ocorrencias>
</dadosRegistroOcorrenciaLote>
</status>
</retornoLoteEventos>}
HasAttributes: true
HasElements: true
IsEmpty: false
LastAttribute: {xmlns="http://www.reinf.esocial.gov.br/schemas/retornoLoteEventos/v1_04_00"}
LastNode: {<Signature xmlns="http://www.w3.org/2000/09/xmldsig#">
<SignedInfo>
<CanonicalizationMethod Algorithm="http://www.w3.org/TR/2001/REC-xml-c14n-20010315" />
<SignatureMethod Algorithm="http://www.w3.org/2001/04/xmldsig-more#rsa-sha256" />
<Reference URI="#ID54ED726E636E2B6F9FD4FD7077870873">
<Transforms>
<Transform Algorithm="http://www.w3.org/2000/09/xmldsig#enveloped-signature" />
<Transform Algorithm="http://www.w3.org/TR/2001/REC-xml-c14n-20010315" />
</Transforms>
<DigestMethod Algorithm="http://www.w3.org/2001/04/xmlenc#sha256" />
<DigestValue>PkXxGAtPdkWYgZ21TLotsVYYE7mPLGnmpmLMXI65gnQ=</DigestValue>
</Reference>
</SignedInfo>
<SignatureValue>NaDYeZv3gFDBH41BY5j8hcYXHhpdx+hjFWCIYvLaBxAVWNUUaAZixzV3Bld4Lul3m0yKL6Hq16xmKYiA55tCPbZWWy+YymnCtj3OC7rZPv2lQZZyTK6DiLugIIiUyk7TsTd1rZmT6D55OxwJfpmb1M+rQLK6siYAZmI2pguUX0NvKITIaaxUD6HGZOgO6cc9VGRR7PTxjhkl0j0jcwFPxCEvk8cTbT0PnkkLw1v5zPBC1nvgpz3Kn+fG11qkTwEIIdNonx/O1i7ZdvpLUe7/MXEYNbip/lcrGL/zapBHNTsLgN6NNI7QUUuPIPB4yMhpQcNA9FfzEgzdZVhbDzdLeQ==</SignatureValue>
<KeyInfo>
<X509Data>
<X509Certificate> [...]</X509Certificate>
</X509Data>
</KeyInfo>
</Signature>}
Name: {{http://www.reinf.esocial.gov.br/schemas/retornoLoteEventos/v1_04_00}Reinf}
NextNode: null
NodeType: Element
Parent: null
PreviousNode: null
Value: "016286040001401ERRO1MS0092Versão do lote inválida. Deve ser utilizada a versão 1.04.00.PkXxGAtPdkWYgZ21TLotsVYYE7mPLGnmpmLMXI65gnQ=NaDYeZv3gFDBH41BY5j8hcYXHhpdx+hjFWCIYvLaBxAVWNUUaAZixzV3Bld4Lul3m0yKL6Hq16xmKYiA55tCPbZWWy+YymnCtj3OC7rZPv2lQZZyTK6DiLugIIiUyk7TsTd1rZmT6D55OxwJfpmb1M+rQLK6siYAZmI2pguUX0NvKITIaaxUD6HGZOgO [...]"
Não consigo postar meu XML aqui por limitação de caracteres.
A: O XML por padrão é sensível a maiúsculas e minúsculas (case-sensitive), e você chamou o elemento raiz do lote como <REINF>, mas o esquema XSD diz que o nome deveria ser <Reinf>. Altere isso e veja se resolve.
Referências:
*
*Is XML case-sensitive? - Stack Overflow
*Case sensitivity in XML Schema Definitions - Msdn forums
*The XML FAQ: Which parts of an XML document are case-sensitive?
EDIÇÃO
Eu não havia notado no primeiro momento, mas o problema está na assinatura. Aqui há uma outra resposta que fala sobre isso: Erro de assinatura do XML para o EFD-Reinf.
Mas, basicamente, o seu evento R-1000 deveria estar assim, com a assinatura embutida dentro dele:
<Reinf xmlns="http://www.reinf.esocial.gov.br/schemas/evtInfoContribuinte/v1_04_00">
<evtInfoContri id="20181121110000000000000001">
[...]
</evtInfoContri>
<Signature xmlns="http://www.w3.org/2000/09/xmldsig#">
[...]
</Signature>
</Reinf>
Você está colocando a assinatura depois do soap:Body, o que está errado:
<soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/" xmlns:sped="http://sped.fazenda.gov.br/">
<soap:Header />
<soap:Body>
<Reinf xmlns="http://www.reinf.esocial.gov.br/schemas/envioLoteEventos/v1_04_00">
<loteEventos>
<evento id="ID1">
<Reinf xmlns="http://www.reinf.esocial.gov.br/schemas/evtInfoContribuinte/v1_04_00">
<evtInfoContri id="20181121110000000000000001">
[...]
</evtInfoContri>
</Reinf>
</evento>
</loteEventos>
</REINF>
</soap:Body>
<Signature xmlns="http://www.w3.org/2000/09/xmldsig#">
[...]
</Signature>
</soap:Envelope>
A forma correta do seu soap:Envelope deveria ser essa:
<soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/" xmlns:sped="http://sped.fazenda.gov.br/">
<soap:Header />
<soap:Body>
<Reinf xmlns="http://www.reinf.esocial.gov.br/schemas/envioLoteEventos/v1_04_00">
<loteEventos>
<evento id="ID1">
<Reinf xmlns="http://www.reinf.esocial.gov.br/schemas/evtInfoContribuinte/v1_04_00">
<evtInfoContri id="20181121110000000000000001">
[...]
</evtInfoContri>
<Signature xmlns="http://www.w3.org/2000/09/xmldsig#">
[...]
</Signature>
</Reinf>
</evento>
</loteEventos>
</REINF>
</soap:Body>
</soap:Envelope>
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 8,108
|
require File.expand_path('../../../test_helper', __FILE__)
class RequestTest < ActionDispatch::IntegrationTest
def setup
JSONAPI.configuration.json_key_format = :underscored_key
JSONAPI.configuration.route_format = :underscored_route
Api::V2::BookResource.paginator :offset
$test_user = Person.find(1001)
end
def after_teardown
JSONAPI.configuration.route_format = :underscored_route
end
def test_get
assert_cacheable_jsonapi_get '/posts'
end
def test_large_get
assert_cacheable_jsonapi_get '/api/v2/books?include=book_comments,book_comments.author'
end
def test_post_sessions
session_id = SecureRandom.uuid
post '/sessions', params: {
data: {
id: session_id,
type: "sessions",
attributes: {
survey_id: SecureRandom.uuid,
},
relationships: {
responses: {
data: [
{
type: "responses",
attributes: {
response_type: "single_textbox",
question_id: SecureRandom.uuid,
},
relationships: {
paragraph: {
data: {
type: "responses",
response_type: "paragraph",
attributes: {
text: "This is my single textbox response"
}
}
}
}
},
],
},
},
}
}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_jsonapi_response 201
json_body = JSON.parse(response.body)
session_id = json_body["data"]["id"]
# Get what we just created
get "/sessions/#{session_id}?include=responses"
assert_jsonapi_response 200
json_body = JSON.parse(response.body)
assert(json_body.is_a?(Object));
assert(json_body["included"].is_a?(Array));
assert_equal("single_textbox", json_body["included"][0]["attributes"]["response_type"]["single_textbox"]);
get "/sessions/#{session_id}?include=responses,responses.paragraph"
assert_jsonapi_response 200
json_body = JSON.parse(response.body)
assert_equal("single_textbox", json_body["included"][0]["attributes"]["response_type"]["single_textbox"]);
# Rails 4.2.x branch will not retrieve the responses.paragraph, 5.x branch will - this looks to be a deeper, but unrelated bug
#assert_equal("paragraphs", json_body["included"][1]["type"]);
end
def test_get_inflected_resource
assert_cacheable_jsonapi_get '/api/v8/numeros_telefone'
end
def test_get_nested_to_one
assert_cacheable_jsonapi_get '/posts/1/author'
end
def test_get_nested_to_many
assert_cacheable_jsonapi_get '/posts/1/comments'
end
def test_get_nested_to_many_bad_param
assert_cacheable_jsonapi_get '/posts/1/comments?relationship=books'
end
def test_get_underscored_key
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.json_key_format = :underscored_key
assert_cacheable_jsonapi_get '/iso_currencies'
assert_equal 3, json_response['data'].size
ensure
JSONAPI.configuration = original_config
end
def test_filter_with_value_containing_double_quote
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.json_key_format = :underscored_key
get '/iso_currencies?filter[country_name]=%22'
assert_jsonapi_response 200
ensure
JSONAPI.configuration = original_config
end
def test_get_underscored_key_filtered
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.json_key_format = :underscored_key
assert_cacheable_jsonapi_get '/iso_currencies?filter[country_name]=Canada'
assert_equal 1, json_response['data'].size
assert_equal 'Canada', json_response['data'][0]['attributes']['country_name']
ensure
JSONAPI.configuration = original_config
end
def test_get_camelized_key_filtered
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.json_key_format = :camelized_key
assert_cacheable_jsonapi_get '/iso_currencies?filter[countryName]=Canada'
assert_equal 1, json_response['data'].size
assert_equal 'Canada', json_response['data'][0]['attributes']['countryName']
ensure
JSONAPI.configuration = original_config
end
def test_get_camelized_route_and_key_filtered
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.json_key_format = :camelized_key
assert_cacheable_jsonapi_get '/api/v4/isoCurrencies?filter[countryName]=Canada'
assert_equal 1, json_response['data'].size
assert_equal 'Canada', json_response['data'][0]['attributes']['countryName']
ensure
JSONAPI.configuration = original_config
end
def test_get_camelized_route_and_links
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.json_key_format = :camelized_key
JSONAPI.configuration.route_format = :camelized_route
assert_cacheable_jsonapi_get '/api/v4/expenseEntries/1/relationships/isoCurrency'
assert_hash_equals({'links' => {
'self' => 'http://www.example.com/api/v4/expenseEntries/1/relationships/isoCurrency',
'related' => 'http://www.example.com/api/v4/expenseEntries/1/isoCurrency'
},
'data' => {
'type' => 'isoCurrencies',
'id' => 'USD'
}
}, json_response)
ensure
JSONAPI.configuration = original_config
end
def test_get_multiple_accept_media_types
get '/posts', headers:
{
'Accept' => "application/json, #{JSONAPI::MEDIA_TYPE}, */*"
}
assert_equal 200, status
end
def test_put_single_without_content_type
put '/posts/3', params:
{
'data' => {
'type' => 'posts',
'id' => '3',
'attributes' => {
'title' => 'A great new Post'
},
'relationships' => {
'tags' => {
'data' => [
{'type' => 'tags', 'id' => '503'},
{'type' => 'tags', 'id' => '504'}
]
}
}
}
}.to_json,
headers: {
'CONTENT_TYPE' => nil,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_equal 415, status
end
def test_put_single
put '/posts/3', params:
{
'data' => {
'type' => 'posts',
'id' => '3',
'attributes' => {
'title' => 'A great new Post'
},
'relationships' => {
'tags' => {
'data' => [
{'type' => 'tags', 'id' => '503'},
{'type' => 'tags', 'id' => '504'}
]
}
}
}
}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_jsonapi_response 200
end
def test_post_single_with_wrong_content_type
post '/posts', params:
{
'posts' => {
'attributes' => {
'title' => 'A great new Post'
},
'relationships' => {
'tags' => {
'data' => [
{'type' => 'tags', 'id' => '503'},
{'type' => 'tags', 'id' => '504'}
]
}
}
}
}.to_json,
headers: {
'CONTENT_TYPE' => 'application/json',
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_equal 415, status
end
def test_post_single
post '/posts', params:
{
'data' => {
'type' => 'posts',
'attributes' => {
'title' => 'A great new Post',
'body' => 'JSONAPIResources is the greatest thing since unsliced bread.'
},
'relationships' => {
'author' => {'data' => {'type' => 'people', 'id' => '1003'}}
}
}
}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_jsonapi_response 201
end
def test_post_single_missing_data_contents
post '/posts', params:
{
'data' => {
}
}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_jsonapi_response 400
end
def test_post_single_minimal_valid
post '/comments', params:
{
'data' => {
'type' => 'comments'
}
}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_jsonapi_response 201
assert_nil json_response['data']['attributes']['body']
assert_nil json_response['data']['relationships']['post']['data']
assert_nil json_response['data']['relationships']['author']['data']
end
def test_post_single_minimal_invalid
post '/posts', params:
{
'data' => {
'type' => 'posts'
}
}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_jsonapi_response 422
end
def test_update_relationship_without_content_type
ruby = Section.find_by(name: 'ruby')
patch '/posts/3/relationships/section', params: { 'data' => {'type' => 'sections', 'id' => ruby.id.to_s }}.to_json
assert_equal 415, status
end
def test_patch_update_relationship_to_one
ruby = Section.find_by(name: 'ruby')
patch '/posts/3/relationships/section', params:
{ 'data' => {'type' => 'sections', 'id' => ruby.id.to_s }}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_equal 204, status
end
def test_put_update_relationship_to_one
ruby = Section.find_by(name: 'ruby')
put '/posts/3/relationships/section', params: { 'data' => {'type' => 'sections', 'id' => ruby.id.to_s }}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_equal 204, status
end
def test_patch_update_relationship_to_many_acts_as_set
# Comments are acts_as_set=false so PUT/PATCH should respond with 403
rogue = Comment.find_by(body: 'Rogue Comment Here')
patch '/posts/5/relationships/comments', params: { 'data' => [{'type' => 'comments', 'id' => rogue.id.to_s }]}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_jsonapi_response 403
end
def test_post_update_relationship_to_many
rogue = Comment.find_by(body: 'Rogue Comment Here')
post '/posts/5/relationships/comments', params: { 'data' => [{'type' => 'comments', 'id' => rogue.id.to_s }]}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_equal 204, status
end
def test_put_update_relationship_to_many_acts_as_set
# Comments are acts_as_set=false so PUT/PATCH should respond with 403. Note: JR currently treats PUT and PATCH as equivalent
rogue = Comment.find_by(body: 'Rogue Comment Here')
put '/posts/5/relationships/comments', params: { 'data' => [{'type' => 'comments', 'id' => rogue.id.to_s }]}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_jsonapi_response 403
end
def test_index_content_type
assert_cacheable_jsonapi_get '/posts'
assert_match JSONAPI::MEDIA_TYPE, headers['Content-Type']
end
def test_get_content_type
assert_cacheable_jsonapi_get '/posts/3'
assert_match JSONAPI::MEDIA_TYPE, headers['Content-Type']
end
def test_put_content_type
put '/posts/3', params:
{
'data' => {
'type' => 'posts',
'id' => '3',
'attributes' => {
'title' => 'A great new Post'
},
'relationships' => {
'tags' => {
'data' => [
{'type' => 'tags', 'id' => '503'},
{'type' => 'tags', 'id' => '504'}
]
}
}
}
}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_match JSONAPI::MEDIA_TYPE, headers['Content-Type']
end
def test_put_valid_json
put '/posts/3', params: '{"data": { "type": "posts", "id": "3", "attributes": { "title": "A great new Post" } } }',
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_equal 200, status
end
def test_put_invalid_json
put '/posts/3', params: '{"data": { "type": "posts", "id": "3" "attributes": { "title": "A great new Post" } } }',
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_equal 400, status
assert_equal 'Bad Request', json_response['errors'][0]['title']
assert_match 'unexpected token at', json_response['errors'][0]['detail']
end
def test_put_valid_json_but_array
put '/posts/3', params: '[{"data": { "type": "posts", "id": "3", "attributes": { "title": "A great new Post" } } }]',
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_equal 400, status
assert_equal 'Request must be a hash', json_response['errors'][0]['detail']
end
def test_patch_content_type
patch '/posts/3', params:
{
'data' => {
'type' => 'posts',
'id' => '3',
'attributes' => {
'title' => 'A great new Post'
},
'relationships' => {
'tags' => {
'data' => [
{'type' => 'tags', 'id' => '503'},
{'type' => 'tags', 'id' => '504'}
]
}
}
}
}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_match JSONAPI::MEDIA_TYPE, headers['Content-Type']
end
def test_post_correct_content_type
post '/posts', params:
{
'data' => {
'type' => 'posts',
'attributes' => {
'title' => 'A great new Post'
},
'relationships' => {
'author' => {'data' => {'type' => 'people', 'id' => '3'}}
}
}
}.to_json,
headers: {
"CONTENT_TYPE" => JSONAPI::MEDIA_TYPE
}
assert_match JSONAPI::MEDIA_TYPE, headers['Content-Type']
end
def test_destroy_single
delete '/posts/7', headers: { 'Accept' => JSONAPI::MEDIA_TYPE }
assert_equal 204, status
assert_nil headers['Content-Type']
end
def test_destroy_multiple
delete '/posts/8,9', headers: { 'Accept' => JSONAPI::MEDIA_TYPE }
assert_equal 400, status
end
def test_pagination_none
Api::V2::BookResource.paginator :none
assert_cacheable_jsonapi_get '/api/v2/books'
assert_equal 901, json_response['data'].size
end
def test_pagination_offset_style
Api::V2::BookResource.paginator :offset
assert_cacheable_jsonapi_get '/api/v2/books'
assert_equal JSONAPI.configuration.default_page_size, json_response['data'].size
assert_equal 'Book 0', json_response['data'][0]['attributes']['title']
end
def test_pagination_offset_style_offset
Api::V2::BookResource.paginator :offset
assert_cacheable_jsonapi_get '/api/v2/books?page[offset]=50'
assert_equal JSONAPI.configuration.default_page_size, json_response['data'].size
assert_equal 'Book 50', json_response['data'][0]['attributes']['title']
end
def test_pagination_offset_style_offset_limit
Api::V2::BookResource.paginator :offset
assert_cacheable_jsonapi_get '/api/v2/books?page[offset]=50&page[limit]=20'
assert_equal 20, json_response['data'].size
assert_equal 'Book 50', json_response['data'][0]['attributes']['title']
end
def test_pagination_offset_bad_param
Api::V2::BookResource.paginator :offset
get '/api/v2/books?page[irishsetter]=50&page[limit]=20', headers: {
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_jsonapi_response 400
end
def test_pagination_related_resources_link
Api::V2::BookResource.paginator :offset
assert_cacheable_jsonapi_get '/api/v2/books?page[limit]=2'
assert_equal 2, json_response['data'].size
assert_equal 'http://www.example.com/api/v2/books/1/book_comments',
json_response['data'][1]['relationships']['book_comments']['links']['related']
end
def test_pagination_related_resources_data
Api::V2::BookResource.paginator :offset
Api::V2::BookCommentResource.paginator :offset
assert_cacheable_jsonapi_get '/api/v2/books/1/book_comments?page[limit]=10'
assert_equal 10, json_response['data'].size
assert_equal 'This is comment 18 on book 1.', json_response['data'][9]['attributes']['body']
end
def test_pagination_related_resources_links
Api::V2::BookResource.paginator :offset
Api::V2::BookCommentResource.paginator :offset
assert_cacheable_jsonapi_get '/api/v2/books/1/book_comments?page[limit]=10'
assert_equal 'http://www.example.com/api/v2/books/1/book_comments?page%5Blimit%5D=10&page%5Boffset%5D=0', json_response['links']['first']
assert_equal 'http://www.example.com/api/v2/books/1/book_comments?page%5Blimit%5D=10&page%5Boffset%5D=10', json_response['links']['next']
assert_equal 'http://www.example.com/api/v2/books/1/book_comments?page%5Blimit%5D=10&page%5Boffset%5D=16', json_response['links']['last']
end
def test_pagination_related_resources_links_meta
Api::V2::BookResource.paginator :offset
Api::V2::BookCommentResource.paginator :offset
JSONAPI.configuration.top_level_meta_include_record_count = true
assert_cacheable_jsonapi_get '/api/v2/books/1/book_comments?page[limit]=10'
assert_equal 26, json_response['meta']['record_count']
assert_equal 'http://www.example.com/api/v2/books/1/book_comments?page%5Blimit%5D=10&page%5Boffset%5D=0', json_response['links']['first']
assert_equal 'http://www.example.com/api/v2/books/1/book_comments?page%5Blimit%5D=10&page%5Boffset%5D=10', json_response['links']['next']
assert_equal 'http://www.example.com/api/v2/books/1/book_comments?page%5Blimit%5D=10&page%5Boffset%5D=16', json_response['links']['last']
ensure
JSONAPI.configuration.top_level_meta_include_record_count = false
end
def test_filter_related_resources_relationship_filter
Api::V2::BookCommentResource.paginator :offset
JSONAPI.configuration.top_level_meta_include_record_count = true
assert_cacheable_jsonapi_get '/api/v2/books/1/book_comments?filter[book]=2'
assert_equal 0, json_response['meta']['record_count']
assert_cacheable_jsonapi_get '/api/v2/books/1/book_comments?filter[book]=1&page[limit]=20'
assert_equal 20, json_response['data'].length
assert_equal 26, json_response['meta']['record_count']
ensure
JSONAPI.configuration.top_level_meta_include_record_count = false
end
def test_filter_related_resources
Api::V2::BookCommentResource.paginator :offset
JSONAPI.configuration.top_level_meta_include_record_count = true
assert_cacheable_jsonapi_get '/api/v2/books/1/book_comments?filter[body]=2'
assert_equal 9, json_response['data'].length
assert_equal 9, json_response['meta']['record_count']
ensure
JSONAPI.configuration.top_level_meta_include_record_count = false
end
def test_page_count_meta
Api::V2::BookCommentResource.paginator :paged
JSONAPI.configuration.top_level_meta_include_record_count = true
JSONAPI.configuration.top_level_meta_include_page_count = true
get '/api/v2/books/1/book_comments', headers: {
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_equal 26, json_response['meta']['record_count']
# based on default page size
assert_equal 3, json_response['meta']['page_count']
get '/api/v2/books/1/book_comments?page[size]=5', headers: {
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_equal 26, json_response['meta']['record_count']
assert_equal 6, json_response['meta']['page_count']
ensure
JSONAPI.configuration.top_level_meta_include_record_count = false
JSONAPI.configuration.top_level_meta_include_page_count = false
end
def test_pagination_related_resources_without_related
Api::V2::BookResource.paginator :offset
Api::V2::BookCommentResource.paginator :offset
assert_cacheable_jsonapi_get '/api/v2/books/10/book_comments'
assert_nil json_response['links']['next']
assert_equal 'http://www.example.com/api/v2/books/10/book_comments?page%5Blimit%5D=10&page%5Boffset%5D=0', json_response['links']['first']
assert_equal 'http://www.example.com/api/v2/books/10/book_comments?page%5Blimit%5D=10&page%5Boffset%5D=0', json_response['links']['last']
end
def test_related_resource_alternate_relation_name_record_count
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.default_paginator = :paged
JSONAPI.configuration.top_level_meta_include_record_count = true
assert_cacheable_jsonapi_get '/api/v2/books/1/aliased_comments'
assert_equal 26, json_response['meta']['record_count']
ensure
JSONAPI.configuration = original_config
end
def test_pagination_related_resources_data_includes
Api::V2::BookResource.paginator :offset
Api::V2::BookCommentResource.paginator :offset
assert_cacheable_jsonapi_get '/api/v2/books/1/book_comments?page[limit]=10&include=author,book'
assert_equal "1", json_response['data'].first['relationships']['book']['data']['id']
assert_equal 10, json_response['data'].size
assert_equal 'This is comment 18 on book 1.', json_response['data'][9]['attributes']['body']
end
def test_pagination_empty_results
Api::V2::BookResource.paginator :offset
Api::V2::BookCommentResource.paginator :offset
assert_cacheable_jsonapi_get '/api/v2/books?filter[id]=2000&page[limit]=10'
assert_equal 0, json_response['data'].size
assert_nil json_response['links']['next']
assert_equal 'http://www.example.com/api/v2/books?filter%5Bid%5D=2000&page%5Blimit%5D=10&page%5Boffset%5D=0', json_response['links']['first']
assert_equal 'http://www.example.com/api/v2/books?filter%5Bid%5D=2000&page%5Blimit%5D=10&page%5Boffset%5D=0', json_response['links']['last']
end
# def test_pagination_related_resources_data_includes
# Api::V2::BookResource.paginator :none
# Api::V2::BookCommentResource.paginator :none
# assert_cacheable_jsonapi_get '/api/v2/books?filter[]'
# assert_equal 10, json_response['data'].size
# assert_equal 'This is comment 18 on book 1.', json_response['data'][9]['attributes']['body']
# end
def test_polymorpic_related_resources
assert_cacheable_jsonapi_get '/pictures/1/imageable'
assert_equal 'Enterprise Gizmo', json_response['data']['attributes']['name']
assert_cacheable_jsonapi_get '/pictures/2/imageable'
assert_equal 'Company Brochure', json_response['data']['attributes']['name']
end
def test_flow_self
assert_cacheable_jsonapi_get '/posts/1'
post_1 = json_response['data']
assert_cacheable_jsonapi_get post_1['links']['self']
assert_hash_equals post_1, json_response['data']
end
def test_flow_link_to_one_self_link
assert_cacheable_jsonapi_get '/posts/1'
post_1 = json_response['data']
assert_cacheable_jsonapi_get post_1['relationships']['author']['links']['self']
assert_hash_equals(json_response, {
'links' => {
'self' => 'http://www.example.com/posts/1/relationships/author',
'related' => 'http://www.example.com/posts/1/author'
},
'data' => {'type' => 'people', 'id' => '1001'}
})
end
def test_flow_link_to_many_self_link
assert_cacheable_jsonapi_get '/posts/1'
post_1 = json_response['data']
assert_cacheable_jsonapi_get post_1['relationships']['tags']['links']['self']
assert_hash_equals(json_response,
{
'links' => {
'self' => 'http://www.example.com/posts/1/relationships/tags',
'related' => 'http://www.example.com/posts/1/tags'
},
'data' => [
{'type' => 'tags', 'id' => '501'},
{'type' => 'tags', 'id' => '502'},
{'type' => 'tags', 'id' => '503'}
]
})
end
def test_flow_link_to_many_self_link_put
assert_cacheable_jsonapi_get '/posts/5'
post_5 = json_response['data']
post post_5['relationships']['tags']['links']['self'], params:
{'data' => [{'type' => 'tags', 'id' => '510'}]}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_equal 204, status
assert_cacheable_jsonapi_get post_5['relationships']['tags']['links']['self']
assert_hash_equals(json_response,
{
'links' => {
'self' => 'http://www.example.com/posts/5/relationships/tags',
'related' => 'http://www.example.com/posts/5/tags'
},
'data' => [
{'type' => 'tags', 'id' => '510'}
]
})
end
def test_flow_self_formatted_route_1
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.route_format = :dasherized_route
JSONAPI.configuration.json_key_format = :dasherized_key
assert_cacheable_jsonapi_get '/api/v6/purchase-orders'
po_1 = json_response['data'][0]
assert_equal 'purchase-orders', json_response['data'][0]['type']
assert_cacheable_jsonapi_get po_1['links']['self']
assert_hash_equals po_1, json_response['data']
ensure
JSONAPI.configuration = original_config
end
def test_flow_self_formatted_route_2
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.route_format = :underscored_route
JSONAPI.configuration.json_key_format = :dasherized_key
assert_cacheable_jsonapi_get '/api/v7/purchase_orders'
assert_equal 'purchase-orders', json_response['data'][0]['type']
po_1 = json_response['data'][0]
assert_cacheable_jsonapi_get po_1['links']['self']
assert_hash_equals po_1, json_response['data']
ensure
JSONAPI.configuration = original_config
end
def test_flow_self_formatted_route_3
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.route_format = :underscored_route
JSONAPI.configuration.json_key_format = :underscored_key
assert_cacheable_jsonapi_get '/api/v7/purchase_orders'
assert_equal 'purchase_orders', json_response['data'][0]['type']
po_1 = json_response['data'][0]
assert_cacheable_jsonapi_get po_1['links']['self']
assert_hash_equals po_1, json_response['data']
ensure
JSONAPI.configuration = original_config
end
def test_post_formatted_keys
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.route_format = :dasherized_route
JSONAPI.configuration.json_key_format = :dasherized_key
post '/api/v6/purchase-orders', params:
{
'data' => {
'attributes' => {
'delivery-name' => 'ASDFG Corp'
},
'type' => 'purchase-orders'
}
}.to_json,
headers: {
"CONTENT_TYPE" => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_jsonapi_response 201
ensure
JSONAPI.configuration = original_config
end
def test_post_formatted_keys_different_route_key_1
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.route_format = :dasherized_route
JSONAPI.configuration.json_key_format = :underscored_key
post '/api/v6/purchase-orders', params:
{
'data' => {
'attributes' => {
'delivery_name' => 'ASDFG Corp'
},
'type' => 'purchase_orders'
}
}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_jsonapi_response 201
ensure
JSONAPI.configuration = original_config
end
def test_post_formatted_keys_different_route_key_2
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.route_format = :underscored_route
JSONAPI.configuration.json_key_format = :dasherized_key
post '/api/v7/purchase_orders', params:
{
'data' => {
'attributes' => {
'delivery-name' => 'ASDFG Corp'
},
'type' => 'purchase-orders'
}
}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_jsonapi_response 201
ensure
JSONAPI.configuration = original_config
end
def test_post_formatted_keys_wrong_format
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.route_format = :dasherized_route
JSONAPI.configuration.json_key_format = :dasherized_key
post '/api/v6/purchase-orders', params:
{
'data' => {
'attributes' => {
'delivery_name' => 'ASDFG Corp'
},
'type' => 'purchase-orders'
}
}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_jsonapi_response 400
ensure
JSONAPI.configuration = original_config
end
def test_patch_formatted_dasherized
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.route_format = :dasherized_route
JSONAPI.configuration.json_key_format = :dasherized_key
patch '/api/v6/purchase-orders/1', params:
{
'data' => {
'id' => '1',
'attributes' => {
'delivery-name' => 'ASDFG Corp'
},
'type' => 'purchase-orders'
}
}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_jsonapi_response 200
end
def test_patch_formatted_dasherized_links
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.route_format = :dasherized_route
JSONAPI.configuration.json_key_format = :dasherized_key
patch '/api/v6/line-items/1', params:
{
'data' => {
'id' => '1',
'type' => 'line-items',
'attributes' => {
'item-cost' => '23.57'
},
'relationships' => {
'purchase-order' => {
'data' => {'type' => 'purchase-orders', 'id' => '2'}
}
}
}
}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_jsonapi_response 200
ensure
JSONAPI.configuration = original_config
end
def test_patch_formatted_dasherized_replace_to_many
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.route_format = :dasherized_route
JSONAPI.configuration.json_key_format = :dasherized_key
patch '/api/v6/purchase-orders/2?include=line-items,order-flags', params:
{
'data' => {
'id' => '2',
'type' => 'purchase-orders',
'relationships' => {
'line-items' => {
'data' => [
{'type' => 'line-items', 'id' => '3'},
{'type' => 'line-items', 'id' => '4'}
]
},
'order-flags' => {
'data' => [
{'type' => 'order-flags', 'id' => '1'},
{'type' => 'order-flags', 'id' => '2'}
]
}
}
}
}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_jsonapi_response 200
ensure
JSONAPI.configuration = original_config
end
def test_patch_formatted_dasherized_replace_to_many_computed_relation
$original_test_user = $test_user
$test_user = Person.find(1005)
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.route_format = :dasherized_route
JSONAPI.configuration.json_key_format = :dasherized_key
patch '/api/v6/purchase-orders/2?include=line-items,order-flags', params:
{
'data' => {
'id' => '2',
'type' => 'purchase-orders',
'relationships' => {
'line-items' => {
'data' => [
{'type' => 'line-items', 'id' => '3'},
{'type' => 'line-items', 'id' => '4'}
]
},
'order-flags' => {
'data' => [
{'type' => 'order-flags', 'id' => '1'},
{'type' => 'order-flags', 'id' => '2'}
]
}
}
}
}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_jsonapi_response 200
ensure
JSONAPI.configuration = original_config
$test_user = $original_test_user
end
def test_post_to_many_link
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.route_format = :dasherized_route
JSONAPI.configuration.json_key_format = :dasherized_key
post '/api/v6/purchase-orders/3/relationships/line-items', params:
{
'data' => [
{'type' => 'line-items', 'id' => '3'},
{'type' => 'line-items', 'id' => '4'}
]
}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_equal 204, status
ensure
JSONAPI.configuration = original_config
end
def test_post_computed_relation_to_many
$original_test_user = $test_user
$test_user = Person.find(1005)
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.route_format = :dasherized_route
JSONAPI.configuration.json_key_format = :dasherized_key
post '/api/v6/purchase-orders/4/relationships/line-items', params:
{
'data' => [
{'type' => 'line-items', 'id' => '5'},
{'type' => 'line-items', 'id' => '6'}
]
}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_equal 204, status
ensure
JSONAPI.configuration = original_config
$test_user = $original_test_user
end
def test_patch_to_many_link
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.route_format = :dasherized_route
JSONAPI.configuration.json_key_format = :dasherized_key
patch '/api/v6/purchase-orders/3/relationships/order-flags', params:
{
'data' => [
{'type' => 'order-flags', 'id' => '1'},
{'type' => 'order-flags', 'id' => '2'}
]
}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_equal 204, status
ensure
JSONAPI.configuration = original_config
end
def test_patch_to_many_link_computed_relation
$original_test_user = $test_user
$test_user = Person.find(1005)
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.route_format = :dasherized_route
JSONAPI.configuration.json_key_format = :dasherized_key
patch '/api/v6/purchase-orders/4/relationships/order-flags', params:
{
'data' => [
{'type' => 'order-flags', 'id' => '1'},
{'type' => 'order-flags', 'id' => '2'}
]
}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_equal 204, status
ensure
JSONAPI.configuration = original_config
$test_user = $original_test_user
end
def test_patch_to_one
original_config = JSONAPI.configuration.dup
JSONAPI.configuration.route_format = :dasherized_route
JSONAPI.configuration.json_key_format = :dasherized_key
patch '/api/v6/line-items/5/relationships/purchase-order', params:
{
'data' => {'type' => 'purchase-orders', 'id' => '3'}
}.to_json,
headers: {
'CONTENT_TYPE' => JSONAPI::MEDIA_TYPE,
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_equal 204, status
ensure
JSONAPI.configuration = original_config
end
def test_include_parameter_allowed
assert_cacheable_jsonapi_get '/api/v2/books/1/book_comments?include=author'
end
def test_include_parameter_not_allowed
JSONAPI.configuration.allow_include = false
get '/api/v2/books/1/book_comments?include=author', headers: {
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_jsonapi_response 400
ensure
JSONAPI.configuration.allow_include = true
end
def test_filter_parameter_not_allowed
JSONAPI.configuration.allow_filter = false
get '/api/v2/books?filter[author]=1', headers: {
'Accept' => JSONAPI::MEDIA_TYPE
}
assert_jsonapi_response 400
ensure
JSONAPI.configuration.allow_filter = true
end
def test_sort_parameter_not_allowed
JSONAPI.configuration.allow_sort = false
get '/api/v2/books?sort=title', headers: { 'Accept' => JSONAPI::MEDIA_TYPE }
assert_jsonapi_response 400
ensure
JSONAPI.configuration.allow_sort = true
end
def test_sort_parameter_quoted
get '/api/v2/books?sort=%22title%22', headers: { 'Accept' => JSONAPI::MEDIA_TYPE }
assert_jsonapi_response 200
end
def test_sort_parameter_openquoted
get '/api/v2/books?sort=%22title', headers: { 'Accept' => JSONAPI::MEDIA_TYPE }
assert_jsonapi_response 400
end
def test_sort_primary_attribute
get '/api/v6/authors?sort=name', headers: { 'Accept' => JSONAPI::MEDIA_TYPE }
assert_jsonapi_response 200
assert_equal '1002', json_response['data'][0]['id']
get '/api/v6/authors?sort=-name', headers: { 'Accept' => JSONAPI::MEDIA_TYPE }
assert_jsonapi_response 200
assert_equal '1005', json_response['data'][0]['id']
end
def test_sort_included_attribute
get '/api/v6/authors?sort=author_detail.author_stuff', headers: { 'Accept' => JSONAPI::MEDIA_TYPE }
assert_jsonapi_response 200
assert_equal '1000', json_response['data'][0]['id']
get '/api/v6/authors?sort=-author_detail.author_stuff', headers: { 'Accept' => JSONAPI::MEDIA_TYPE }
assert_jsonapi_response 200
assert_equal '1002', json_response['data'][0]['id']
end
def test_include_parameter_quoted
get '/api/v2/posts?include=%22author%22', headers: { 'Accept' => JSONAPI::MEDIA_TYPE }
assert_jsonapi_response 200
end
def test_include_parameter_openquoted
get '/api/v2/posts?include=%22author', headers: { 'Accept' => JSONAPI::MEDIA_TYPE }
assert_jsonapi_response 400
end
def test_getting_different_resources_when_sti
assert_cacheable_jsonapi_get '/vehicles'
types = json_response['data'].map{|r| r['type']}.sort
assert_array_equals ['boats', 'cars'], types
end
def test_getting_resource_with_correct_type_when_sti
assert_cacheable_jsonapi_get '/vehicles/1'
assert_equal 'cars', json_response['data']['type']
end
def test_get_resource_with_belongs_to_relationship_and_changed_primary_key
worker = Worker.find(1)
access_card = worker.access_card
assert_cacheable_jsonapi_get '/workers/1?include=access_card'
assert_jsonapi_response 200
data = json_response['data']
refute_nil data
assert_equal worker.id.to_s, data['id']
refute_nil data['relationships']
refute_nil data['relationships']['access_card']
refute_nil data['relationships']['access_card']['data']
assert_equal 'access_cards', data['relationships']['access_card']['data']['type']
assert_equal access_card.token, data['relationships']['access_card']['data']['id']
included = json_response['included']
refute_nil included
assert_equal 'access_cards', included.first['type']
assert_equal access_card.token, included.first['id']
end
end
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 9,677
|
{"url":"https:\/\/math.stackexchange.com\/questions\/415384\/domain-of-definition-of-a-rational-automorphism","text":"# Domain of definition of a rational automorphism\n\nI am a bit confused about the following problem from Liu's book on Algebraic geometry. Let $X= \\mathbb{P}^1_\\mathbb{Z}$ and let $f: X_\\mathbb{Q} \\rightarrow X_\\mathbb{Q}$ be an automorphism corresponding to a matrix of $PGL_2 (\\mathbb{Q})$. Determine the domain of definition of the rational map $X \\rightarrow X$ induced by f.\n\nMy question is: How does f induce a rational map? I thought that, by definition , a rational map is defined on an open set. Should I just \"clear the denominators\" ? Second - how do I determine the domain of definition? My intuition says defined everywhere, but I am probably wrong.\n\nIf $X$ and $Y$ are irreducible, then rational maps from $X$ to $Y$ correspond to injections of fields $K(Y) \\hookrightarrow K(X)$. Now $K(X) = K(X_{\\mathbb Q}) = \\mathbb Q(t)$, and $f$ induces an automorphism of $\\mathbb Q(t)$ by the usual linear fractional transformation. This will extend to a (regular, not just rational) automorphism of $\\mathbb P^1_{\\mathbb Q}$ (i.e. $X_{\\mathbb Q}$), but it may not extend regularly over all of $X$, and the problem is to determine whether or not it does.\nAlso, the process of converting a map $K(Y) \\hookrightarrow K(X)$ to a rational map $X \\to Y$ is essentially one of clearing denominators, as you say. But computing the maximal domain of definition of the rational map can be more subtle, since clearing denominators in different ways may lead to extensions over different open pieces of $X$. (E.g. for rational maps between smooth projective varieties, the complement of the domain of definition is always codimension $2$; and although you are not quite in this context, your scheme $X$ is regular of dimension two, and it is fairly analogous; so one thing you'll need to work out is whether in your context the complement of the maximal domain of definition of your rational map has codimension one or two.)\n\u2022 @HeidarSvan: Dear Heidar, Let's take the matrix $(2 0, 0 1),$ so in terms of the usual coordinate on $\\mathbb A^1$, this gives the transformation $x \\mapsto 2 x$. Can you see that this formula might become slightly problematic in characteristic $2$? Regards, \u2013\u00a0Matt E Jun 10 '13 at 18:18","date":"2019-08-23 02:11:59","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9721122980117798, \"perplexity\": 138.2755034474958}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.3, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-35\/segments\/1566027317817.76\/warc\/CC-MAIN-20190823020039-20190823042039-00277.warc.gz\"}"}
| null | null |
Produced by Donald Cummings, Adrian Mastronardi and the
Online Distributed Proofreading Team at http://www.pgdp.net
(This file was produced from images generously made
available by The Internet Archive/American Libraries.)
Transcriber's Note: The author's variable spelling has been preserved,
except in cases where it could be nothing other than a printing error.
[Illustration: LIFE IN THE SOUDAN]
[Illustration: FROM A NEGATIVE BY LOMBARDI, PALL MALL.
Yours very truly
Josiah Williams.]
LIFE IN THE SOUDAN:
ADVENTURES AMONGST THE TRIBES, AND TRAVELS
IN EGYPT, IN 1881 AND 1882.
BY
DR. JOSIAH WILLIAMS, F.R.G.S.
(_Surgeon-Major, Imperial Ottoman Army, 1876-1877_).
ILLUSTRATED.
LONDON:
REMINGTON & CO., PUBLISHERS,
HENRIETTA STREET, COVENT GARDEN.
1884.
[_All Rights Reserved._]
To
SIR SAMUEL WHITE BAKER, F.R.G.S.,
I DEDICATE, WITH HIS PERMISSION,
THIS BOOK,
CONTAINING AN ACCOUNT OF
TRAVELS IN THE SOUDAN
AND EXPLORATION IN THE
BASÉ OR KUNAMA COUNTRY IN 1882.
ILLUSTRATIONS.
THE AUTHOR _Frontispiece._
SOUÂKIN 97
HADENDOWAH ARAB CAMEL-MEN 128
KASSALA MOUNTAIN 160
THE AUTHOR ATTENDING TO ARAB AILMENTS 248
MOUNTAIN PASS NEAR SANHÎT 304
THE CAUSEWAY AT MASSAWAH 312
CONTENTS.
PAGE
CHAPTER I.
Leave England for Paris—Drugs and Clothing Required—A "Sleeby"
German—Turin 5-12
CHAPTER II.
Milan—The Cathedral—Galleria Vittorio Emmanuele—Piazza
d'Armi—Palazzo de Brera—Lake of Como—Bologna—Its Ancient
History—Leaning Towers—The Certosa—Teatro Communale—Brindisi 13-23
CHAPTER III.
P. and O. Steamer _Tanjore_—Arrival of the Mail and
Passengers—Ancient Brindisi—Brindisi to Alexandria—Alexandria
Past and Present—Its Trade 24-37
CHAPTER IV.
The Fertilizing Rivers of Egypt—Leave Alexandria—Incidents
_en route_—Shepheard's Hotel—Ancient and Modern Cairo—The
Donkey-boys—Arab Patients—Dancing Dervishes—The House where
Joseph, Mary, and the Infant Saviour Lived in Old Cairo—The
Boulac Museum—The Petrified Forest—Mokattam Hills—Tombs of the
Caliphs and Citadel—Cairo by Sunset 38-66
CHAPTER V.
A Young American at Shepheard's Hotel—Drive to the Pyramids
of Gizeh—Ascent and Exploration of the Pyramid of Cheops—The
Sphinx 67-80
CHAPTER VI.
Heliopolis—The Shoubra
Road—Bedrashyn—Mitrahenny—Memphis—Sakhara—Apis
Mausoleum—Worship of the Bull Apis—Tomb of King Phta—Meet
the Khedive—Engage Servants for the Soudan 81-91
CHAPTER VII.
The Land of Goshen—Ancient Canals—Suez—Howling
Dervishes—Eclipse of the Moon and Strange Behaviour of
Natives—Leave Suez—Where the Israelites Crossed the Red
Sea—Pass Mount Sinai—Coral Reefs Abundant 92-97
CHAPTER VIII.
Arrival at Souâkin—The Soudan—Bedouin Arab Prisoners in
the Square, Not "on the Square"—Ivory—Engage Camels—Sheik
Moussa—Souâkin Slaves—Tragic End of a Doctor—Hadendowah
Arabs—An Ill-fated Missionary Enterprise 98-110
CHAPTER IX.
The Start Across the Desert—My Camel Serves me a Scurvy
Trick—The Camel, its Habits and Training 111-118
CHAPTER X.
Our First Camp—Torrents of Rain—Jules Bardet—Camel-drivers
Behave Badly—Suleiman in Trouble—Camel-drivers get
Upset—The Desert—Two of Us Lose our Way—Jules Suffers from
Dysentery—Sand-storm—A Pilgrim Dies on the Road; Another in the
Camp—Jules' Illness—Camp Split Up—Lose Our Way—Encamp Several
Days in the Desert—Arab Huts—The Mirage—A Lion 119-143
CHAPTER XI.
Arrive at Kassala—Description of Kassala—We buy Camels
and Horses—The Mudir gives a Dinner—Jules' Death and
Burial—Hyænas—Arab Patients—Mahoom's History—Demetrius
Mosconas on Slavery—Menagerie at Kassala 144-153
CHAPTER XII.
Camels from the Atbara—The Mudir—Gordon Pasha's Character in
the Soudan—Fertility of the Soudan 154-159
CHAPTER XIII.
Leave Kassala—Character of the Country—Meet Beni-Amir Arabs
on the River-bed—The Baobob Tree 160-164
CHAPTER XIV.
Encamp at Heikota—Sheik Ahmed—Herr Schumann and His Zareeba—We
Make a Zareeba—The Mahdi—Excitement in the Village—Horrible
Tragedy—Sheik Ahmed Dines with us—The Magic Lantern—Lions
Visit Us 165-177
CHAPTER XV.
Patients at Heikota—Leave Heikota—Game in the Basé Country—See
our First Lion—A Lion Interviews the Author—typo Tetél, Nellut,
and other Game Killed on the March 178-183
CHAPTER XVI.
We Arrive at the Basé or Kunama Country—The Village of
Sarcella—Murder of Mr. Powell and Party—My Camel and I
Unceremoniously Part Company—The First Basé We See—Encamp at
Koolookoo—Our First Interview with Basé—They make "Aman" with
Us—Their Appearance—Description of Koolookoo and the Basé
People—Their Habits and Customs 184-200
CHAPTER XVII.
We leave Koolookoo, Accompanied by a Number of the Basé—The
Magic Lantern—See Buffalo and Giraffe for the First Time—Two
Buffalos Killed—A Basé Feast—Curious Basé Dance—They Dry their
Meat on Lines in the Sun—A Wounded Buffalo—Hoodoo, Chief Sheik
of the Basé, Visits Us—A Column of Sand—A Leper—The Basé
Squabble over the Meat—We Arrive at Abyssinia 201-214
CHAPTER XVIII.
The Dembelas Attack Us, Mahomet Wounded, Narrow Escape of two
of our Party—Activity in Camp, We Make a Zareeba and Fire
the Country—Hold a Council of War—Our Silent and Dangerous
Ride—Hoodoo's Sagacity—Arrival in Camp of Mahomet, Wounded—We
Retreat—Mahomet's Death and Burial 215-229
CHAPTER XIX.
Messrs. James and Phillipps Start on a Visit to
Rasalulu—Curious Way of Shaving Children's Heads—A Disgusting
Basé—The Camel-drivers become Mutinous—Intended Attack
by Basé—We Fire the Country and Make a Zareeba—Encamp at
Wo-amma—Trouble Again with Camel-men—Lions Disturb Us—Arrival
at Heikota—A Tale of Blood and Slavery 230-243
CHAPTER XX.
Patients Arrive from all Parts—Rough Journeys—Arrive at the
Hamran Settite—Mahomet Sali Deceives us—Crocodiles, Turtle,
and Fish—We Move on to Boorkattan, in Abyssinia—Next Day we
Move off as Abyssinians Approach—We Catch Enormous Quantities
of Fish with the Net—Narrow Escape from a Wounded Buffalo—The
Coorbatch Administered—Scorpions and Snakes—Hamrans Visit
Us—Hamran mode of Hunting and Snaring—Hamran and Basé—The
Hamrans Threaten to Fire on Us—Again Return to the Hamran
Settite—Encamp at Omhagger 244-263
CHAPTER XXI.
A Boa-constrictor Visits Us—The Burton Boat—Moussa's Behaviour
Entails a Thrashing and His Discharge—Great Heat—A Fine
Hippopotamus Killed—Hamran Feast—The White Ants—Another
Hippopotamus Killed—Mahomet Sali Brings Supplies—Native
Music in the Night—Delicate Hints Conveyed to the
Performer—A Remarkably Fine Nellut Shot—Arab and Egyptian
Taxation—Baboons—A Hamran Story—Ali Stung by a Scorpion—On
the March Once More—Rough Journeys 264-278
CHAPTER XXII.
Encamp at Lakatakoora Without the Caravan—Description of
Village—Basé Ladies Visit Me ere I Get Out of Bed—They Receive
Presents and are very Amusing—Enormous Numbers of Doves and
Sand-grouse—Aboosalal to Sogoda—Boa-constrictor Killed—An
Unpleasant Journey, We all Get Separated—Arrive at Heikota
Again 279-284
CHAPTER XXIII.
An Abyssinian Improvisatore and His Little Slave—Prepare for a
March to Massawa—A Strange Basé Breakfast—Patients—Arrive at
Toodloak—Beni-Amirs Encamped on the Gash—Lions and Leopards
are Shot—Our Monkeys in Camp—Baboon Mode of Attacking
Leopards—Crafty Baboons—Lions Abound—Hyæna Method of Attacking
a Lion—Hyæna Interviews Mr. Colvin—Arrival at Amadeb—Departure
from Amadeb—Bareas Attempt an Attack on the Caravan—Beni-Amirs
Watering their Flocks and Herds—We Meet with a young
Elephant—Leopard and Hyæna Shot at Khor-Baraker 285-297
CHAPTER XXIV.
A Lion Near the Camp—The Monks of Chardamba—We Meet Ali Dheen
Pasha, Governor-General of the Soudan—Arrival at Keren, or
Sanhît—The Priests at Keren—Account of Keren—Merissa—Dra, a
Domestic Slave, Made Free—Descent from Sanhît to the Anseba
Valley—The Birds There—Along the River-bed of the Labak—A Big
March—Massawa—Farewell to Camels—Massawa to Souâkin—Take in
Cargo—Farewell to the Soudan—Arrival at Suez 298-314
CHAPTER XXV.
Suez to Cairo—Alexandria—On Board the _Mongolia_—Passengers
on Board—Hibernian Humour—Venice—The Piazza of St. Mark—The
Campanile—The Piazetta—The Zecca, or Mint—The Palace of the
Doges—St. Mark's—The Arsenal 315-330
CHAPTER XXVI.
We Hear of the Murder of Lord Frederick Cavendish and Mr.
Burke—A Grand Serenade on the Grand Canal—My Journey from
Venice to England 331-338
PREFACE.
The Soudan, two years ago, was a name unknown to the million, and I will
venture to say that at that time not one in fifty knew anything about it.
Only those who could afford to obtain Sir Samuel Baker's interesting and
instructive work, "The Nile Sources of Abyssinia," would be acquainted
with the locality and other particulars.
The literature extant on Egypt proper would probably amount to tons, but
that on the Soudan would occupy a very small space indeed on the library
shelf, for the simple reason that so very few have travelled through it.
In November, 1881, I left England to accompany six gentlemen on an
exploring expedition in the Soudan, and, in view of passing events in
Egypt and that locality, I indulged in the hope that an account of my
journey will not be unacceptable to the public. I held the post of
medical officer to the expedition, partly on account of my experience
in the Turkish war, where I was continually brought face to face with
dysentery, ague, and other tropical diseases, which are so easily
recognised without any extraneous assistance, medical or lay, but which
are troublesome to treat, especially when hampered by an ignorant and
fussy interference. Doubtless many faults of omission and commission may
be found in my book; but I trust that those who criticise it will do so
leniently, and remember that it has been written during spare hours,
when the exigencies of practice would allow of my seeking recreation by
the use of my pen. "Oh, that mine enemy would write a book!" was the
heartfelt expression of a vindictive old gentleman, well known for his
great patience. My enemies, I trust, are few; those I have shall be
gratified, though I hope I shall not find any who are utterly callous,
but will use me in a gentlemanly fashion.
I have ventured to describe not only my travels in the Soudan, but the
journey from England and home again, extracted from my journal, which
is most accurate, as I kept it religiously day by day. Much of the
old-world history has been culled from various sources of information.
The Illustrations of Soudan scenery, natives, and objects of interest
are from rude sketches of my own, elaborated by Mr. Fanshawe, a perfect
master in the art. The frontispiece is from a photograph taken by Messrs.
Lombardi and Co., of Pall Mall.
Although I am aware of the fact that Mr. F. L. James has published a
book on the Soudan, I have carefully refrained from reading it, fearing
I might inadvertently use any of his expressions, and also feeling sure
that in _some_ matters we may materially differ in opinion.
Although I have, on some occasions, written for the medical journals, I
am quite aware that there may be many faults of style and finish in this
my first effort at a book; such shortcomings I would ask the reader to
overlook. It is but a plain, unvarnished account of a journey through a
territory hitherto but little known, and as such I trust it may be of
interest to the majority of my readers.
CHAPTER I.
LEAVE ENGLAND FOR PARIS—DRUGS AND CLOTHING REQUIRED—A "SLEEBY"
GERMAN—TURIN.
I was bound for the Soudan, and had arranged to meet my party at Brindisi
on the 21st of November, 1881. I therefore sent on all my heavy baggage
by Peninsular and Oriental steamer to Suez; included in this was a
good-sized medicine chest, well stocked with drugs for the relief or cure
of nearly all the ills that flesh is heir to.
I am an old campaigner, having served as a surgeon-major in the Turkish
Army in 1876 and '77, consequently had a very good idea of what drugs
would be most necessary and useful. Knowing also that we were going to
a very hot part of the globe, I took as few liquids, such as tinctures,
&c., as possible.
Everything that I could have made in the form of pills I got Messrs.
Richardson and Co., of Leicester, to do; their coated pills stood the
journey splendidly, and could always be depended on.
It will not be necessary to enumerate all the contents of the
medicine-chest; but I think it might be useful to those who take a
similar journey if I mention a few things that ought certainly to be
taken, and they are the following: A good stock of quinine, oil of
male-fern, as tape-worm is by no means uncommon; ipecacuanha, for that
formidable complaint, dysentery; castor oil, opium, Dover's powder,
iodoform, chlorodyne, calomel, blue pill, and various other mercurial
preparations, much required for complaints in the Soudan; iodide of
potash, carbonate of soda, powdered alum, sulphate of zinc, sugar of
lead, solution of atropine, solution of ammonia, Epsom salts, a large
bottle of purgative pills, nitrate of silver (lunar caustic), carbolic
acid, lint, a few dozen bandages, and plaster in a tin. Ointments are
useless, as they soon become quite liquid in such a hot climate, and run
all over the medicine chest, making a great mess.
_Clothing._—Of course every gentleman will be provided with the ordinary
European clothing for use in civilized parts, but such things as
nicely-polished boots, collars, neckties, and so forth, may be easily
dispensed with in the Soudan. The most necessary articles are two or
three dozen pairs summer socks, half-a-dozen thin flannel shirts, three
or four silk shirts, three pairs of _brown_ leather lace-up boots, and a
comfortable pair of slippers, three or four suits of thin light clothing,
a strong jean coat and trousers, that will not be easily torn by the
thorns whilst hunting, and a pith helmet.
Soldiers cannot march without easy boots, and travellers cannot travel
with comfort unless they have suitable braces. This may seem a small
matter to talk about, but I have often heard strong language poured forth
at the secession of a trouser-button; and I know from past experience
what a nuisance it is to be obliged to sew on one's trouser-buttons.
A long time is spent in searching for a needle and thread, and a much
longer time, by the unpractised one, in sewing on the button. Now,
fortunately, these annoyances are things of the past, since the invention
of what is known as "the traveller's patent buttonless brace."
It is simplicity itself. Instead of buttons on the trousers, there are
eyelet holes, through which a little bar attached to the brace—instead
of a loop—is slipped, and there is an end for ever of the nuisance of
buttons coming off.
A good supply of soap for washing clothes should be taken, also plenty
for personal use. Pear's Soap, I think, is an excellent one in every
respect. Some of our party took thick woollen pads with them, which they
wore over the spine. I did not, neither do I think them at all necessary.
As I was not due at Brindisi until the 21st November, I decided to
have a ramble through parts of Italy which I had not before visited.
Accordingly, I left England in the early part of the month.
On my way to Paris I made the acquaintance of a German residing in
London. We soon got on conversational terms, and ere long he informed me
that he had not been well lately, and was much concerned about himself,
that one afternoon, feeling rather tired, he lay down on the sofa,
intending to have a nap. He was so unhappy or unwise as to sleep for a
whole week without once awaking. To sink into this blissful state of
oblivion may have its advantages, also its disadvantages. On another
occasion he performed the same feat, but indulged in this lethargic
propensity for a much longer period. If I remember rightly, he observed
this condition during a fortnight. However, I pointed out to him what an
immense advantage this was, as he would not have his mind worried by the
Income Tax, Poor Rate, and other objectionable collectors; also what a
saving in eating and drinking would be effected by this _somnia similima
mortis_ habit of his, and that balmy sleep was kind nature's sweet
restorer. Strange to say, my arguments were ineffective, as he replied
that "Sleeb vas all very vell in its way, but I would rater not sleeb
so much as dat, as I have my business to attend to, for vich I must be
wide-avake."
We were glad to get off the boat that took us from Dover to Calais, as
both of us suffered from that miserable complaint, _mal-de-mer_, to some
extent. We reached la belle Paris in the evening, very glad of a rest.
After spending two days very pleasantly and agreeably in Paris, I took
train at 9 p.m. from the Gare de Lyon for Turin. Fortunately, a French
gentleman and I were the only two occupants of the carriage during the
night. We turned up the arm-rests, each occupied a side of the carriage,
and slept soundly all night. At Maćon we had breakfast, wash and brush
up, then resumed our journey. Passing through grand mountain scenery,
and quite close to the railway, we passed a beautiful lake some miles
in extent, the name of which I forget. When we reached Chambery I lost
my agreeable French companion. In the afternoon we ran through the Mont
Cenis tunnel, the time occupied being just thirty-eight minutes. The
gradient became somewhat steep, and the lovely Alpine scenery glorious
and lonely, now winding through deep gorges, anon running downwards for
miles along the very edge of a fearful precipice.
I reached Turin in the evening succeeding my departure from Paris. The
station is situated in the Piazza Carlo Felice, and is a fine, spacious
building. When my luggage had been duly inspected by Custom House
officials, I was permitted to transport myself and my belongings to an
omnibus from the Hotel Trambetta, whither I was driven just in time for
_table d'hôte_. Immediately after leaving the station the driver was
stopped by an official, who opened the door, asked if I had any complaint
to make, and looked round to see if there were any provisions, as the
octroi duties prevail in Italy. I had no complaint; the door was shut,
and off we went.
As I did not intend to remain long in Turin, I was up the following
morning in good time, determined to see as much of the place as I could
in a short time. The streets are clean and well laid out, the houses
large and handsome generally, and the town comparatively modern, although
it was originally founded by a tribe called the Taurini, was the capital
of Piedmont during the 14th century, and the capital of Italy until 1865.
The population is about 208,000, and the University perhaps the most
important in Italy, there being over 1,500 students.
I should liked to have spent a week in exploring Turin and the
neighbourhood, but had to be content with the short time at my disposal.
I took a walk down the Via Lagrange, and soon reached the Palazzo Madama
(Piazza Castello). This Palace was used for the sittings of the Italian
Senate when Turin was the seat of government (1865). In the early part
of last century the mother of King Amadeus lived in and embellished it.
Opposite this is the Sardinian monument, presented to the city by the
Milanese in 1859, just after the war, on which, _in relievo_, is the
figure of Victor Emmanuel—_Il re galuntuomo_—at the head of his troops.
Just beyond the Palazzo Madama is the Palazzo Reale (Royal Palace). The
exterior is nothing to look at, being plain and heavy, but the interior
is magnificent. From here I extended my walk to the Giardino Reale (Royal
Gardens), then the Cathedral of Turin, Santa Giovanni Battista, which was
erected in the latter part of the 15th century by Pintelli. In the chapel
of St. Sudorio, just behind the high altar, is a small portion of linen
cloth in a glass case. This is a valuable relic, for it is said to be a
portion of the cloth in which the body of the Saviour was embalmed. This
may, or may not, be true; belief in the matter is optional. One really
gets so accustomed in Italy to seeing the bones of deceased saints, a bit
of the true cross, a nail of it, and so on, that the probability is nine
out of ten are sceptical.
CHAPTER II.
MILAN—THE CATHEDRAL—GALLERIA VITTORIO EMMANUELE—PIAZZA
D'ARMI—PALAZZO DE BRERA—LAKE OF COMO BOLOGNA—ITS ANCIENT
HISTORY—LEANING TOWERS—THE CERTOSA—TEATRO COMMUNALE—BRINDISI.
From Turin I went by train to Milan. I ought to have gone direct past
Magenta, but by some mistake I found myself making quite a round-about
journey, _viâ_ Piacenza and Lodi; however, all's well that ends well. I
arrived at the hotel in Milan in time for _table d'hôte_. Now, although I
am writing a book principally on travels and adventures in Egypt and the
Soudan, I dare say my readers will excuse me if I attempt a description
of my travels out and home. All the places I visited were extremely
interesting to me, and I cannot forbear a little gossip and relating
what I know respecting them. Those who have not visited these places
will perhaps be pleased to read my description, and those who have will
be able to compare notes and see if they are correct. I had been told
that the best time to visit Il Duomo—the Cathedral—was at eight or nine
o'clock in the morning, on account of the splendid view obtainable from
the roof; this I did on the morning following my arrival, and was richly
rewarded for my trouble. Il Duomo is certainly a magnificent structure,
inferior in magnitude to St. Peter's at Rome, but in some respects
not an unworthy rival. It is built of white marble, and is one of the
most impressive ecclesiastical edifices in the world. In its present
form it was commenced in 1387, and is not yet entirely completed. Its
form is that of a Latin cross, divided into five naves, terminated by
an octagonal apsis, and supported by fifty-two octagonal pilasters of
uniform size, except four, which, having to bear the cupola, are larger.
Around the exterior are 4,500 niches, of which above 3,000 are already
occupied by statues. In the interior everything is of the most imposing
and gorgeous description. I said everything, but I should except an image
of wax of the Virgin Mary, with the infant Saviour in her arms. The waxen
face and arms looked very dirty, her attire was very commonplace-looking
stuff, and I did not think her rather dirty-looking neck was much
improved by a bit of paltry-looking green ribbon encircling it. This
image would certainly be more suitable at Madame Tussaud's than in this
beautiful cathedral. But I will finish with the exterior. The roof is a
perfect forest of marble pinnacles, nearly all crammed with most valuable
marble statues. The celebrated marble flower-bed contains several
thousand flowers, each distinct and each different in design. I leave
the roof and ascend the tower, from which I obtain a magnificent view of
the Alpine range, Mont Blanc, Monte Rosa, the St. Bernard and Matterhorn
right away to the Superga and Mont Cenis.
In the interior we notice the rich stained-glass windows of the choir,
comprising about 350 subjects of Biblical history, the Gothic decorations
of the sacristy, the candelabra in front of the altar shaped like a
tree, and decorated with jewels, then the Chapel of St. Borromeo, which
is a subterranean chapel of a most gorgeous and costly character, as it
is one mass of jewels. The shrine and walls are silver, all inlaid with
gold and precious stones. If I remember rightly, I paid a franc extra for
my visit here, and had the gratification of seeing the embalmed body of
St. Borromeo, with the valuable rings of office still on his fingers. A
golden crown (presented by the unfortunate Maria Teresa) is suspended
over his head, and a large crucifix of splendid emeralds lies on his
chest—this, I am told, was given by the Empress of Austria.
Of course, in Milan, as in all large towns in Italy, there are any number
of beautiful and remarkable churches. Among the most remarkable edifices
are the church of Sant' Ambrogio, founded by St. Ambrose in 387, the
churches of Sant' Eustargio, San Lorenzo, Santa Maria delle Grazie, with
a cupola and sacristy by Bramante, and the celebrated Last Supper by
Leonardo da Vinci; Santa Maria della Passione, a majestic edifice, with
excellent paintings and a magnificent mausoleum; San Paolo, San Carlo
Borromeo, &c.
Immediately adjoining the Cathedral is a magnificent square, which
was finished on the occasion of the Austrian Emperor's visit to Milan
in 1875. This is called the Piazza del Duomo. From this square I pass
through the Galleria Vittorio Emmanuele, a very fine glass-roofed arcade,
or gallery, connecting the Piazza del Duomo with the Scala Theatre; the
cost of this was about £320,000. It was commenced in 1865 and opened in
1867. The glass canopy is illuminated by 2,000 jets of gas, and when
these and the beautiful and brilliant shops are lighted the effect is
charming. The length of this kind of covered street is 320 yards. La
Scala Theatre was not open for performances when I was there, but by the
judicious disposition of a franc or so I obtained admission just to see
it. It is, I understand, capable of accommodating 3,600 spectators. I
next strolled on to the Piazza d'Armi, which occupies an immense space,
obtained by the demolition of the citadel and its outworks. Part of
it has been converted into an amphitheatre, 800 feet long by 400 feet
broad, used in summer for races and shows, and capable of containing
30,000 spectators. The castle, now a barrack, fronts the Piazza d'Armi
on one side; at the opposite side is the Porta Sempione, with the
fine Arco Sempione, or Arco della Pace. This is a lofty gateway, with
three passages, built of blocks of white marble, adorned with reliefs
and statues, and bearing inscriptions commemorating the emancipation
of Italy. My next visit was to the Palazzo di Brera, or Delle Scienze
Lettere ed Arte, containing the Pinacoteca, or picture-gallery, with a
very valuable collection of paintings and statuary, and containing also
the library of the Academy (170,000 volumes). Besides this library, Milan
possesses the Ambrosian library, the earliest and still one of the most
valuable public libraries in Europe. There is also a valuable museum
of natural history, a conservatory of music, a military college, a
theological seminary, and a veterinary school.
Though Milan is one of the most ancient towns in Lombardy, it has so
often been partially destroyed and rebuilt that few antiquities remain.
It is entered by eleven gates, several of which are magnificent. Its
foundation is attributed to the Insubrian Gauls; but the first distinct
notice of it occurs B.C. 221, when it was subdued by the Romans, under
whom it acquired so much importance that in the division of the empire
attributed to Constantine the Great it ranks as the second city of
Italy. In the middle of the fifth century it was sacked by the Huns,
under Attila, and again in the following century by the Goths; but
greater horrors yet awaited it, for the Goths, who had been driven out
by Belisarius, having regained possession by the aid of the Burgundians,
gave it up to the flames, and put almost all its inhabitants to the
sword. The most important manufactures are tobacco, silks, cottons, lace,
carpets, hats, earthenware, white-lead, jewelry, and articles in gold
and silver. The spinning and throwing of silk employs a large number
of hands, and furnish the staple article of trade. The other principal
articles are corn, rice, cheese, and wines.
In the evening of the second day (whilst engaged in the purchase of
everything Milanese in the way of photographs) I met with a Milan
gentleman, who had lived some years in America, and who could speak
English remarkably well. He was a genial, good-hearted looking kind of
fellow, and we soon got into an animated conversation. I was surprised
to find how well up he was in English politics, and as for the Irish
question, he could hold his own with any Englishman; he was, too, a great
admirer of Lord Beaconsfield. When we had had about an hour's chat I was
about to return to my hotel; he then asked me how long I was going to
remain in Milan. I told him I intended leaving next day for Bologna.
"Have you seen the lake of Como?" said he.
"No," I replied. "I should like to do so very much, but fear I cannot
spare the time, as I have to be at Brindisi on the 21st."
"But you must not leave," said he, "until you have been there; it is only
a run of thirty miles to Como by rail. I live there. Come to-morrow and
visit me, and I will put you in the way of seeing Bologna in half the
time that you would do it in without assistance."
This very kind offer I accepted, and spent next day a very agreeable time
with my new acquaintance, who was most hospitable and friendly. We parted
with mutual protestations of goodwill, and I took train for Bologna,
which is several hours' ride from Milan.
Bologna (anciently Bonovia) is one of the oldest, largest, and richest
cities of Italy. It lies at the foot of the Apennines, between the Rivers
Reno and Savena, 190 miles N.N.W. from Rome. It is five or six miles in
circumference, and is surrounded by an unfortified wall of brick; it has
extensive manufactures of silk goods, velvet, artificial flowers, &c. It
struck me as being a quaint old city. All the houses, or nearly so, are
built out over the shops and pavement, supported by large pillars, and
forming a covered way nearly all over the city which affords shade and
shelter to the foot-passengers.
Bologna was long renowned for its university, founded, according
to tradition, by Theodosias, the younger, in 425, and restored by
Charlemagne, which, in the centuries of barbarism, spread the light
of knowledge all over Europe. It once had 10,000 students, but the
number now averages only 300. The university formerly possessed so much
influence, that even the coins of the city bore its motto—_Bonovia
docet_. During 1400 years every new discovery in science and the
arts found patrons here. The medical school is celebrated for having
introduced the dissection of human bodies, and the scientific journals
prove that the love of investigation is still awake in Bologna. The
chief square in the city, Piazza Maggiore, the forum in the Middle Ages,
is adorned by several venerable buildings. Among them are the Palazzo
Pubblico, which contains some magnificent halls, adorned with statues and
paintings; Palazzo del Padesta, chiefly remarkable as having been the
prison of Eugenis, King of Sardinia, and son of the Emperor Frederick
II. who was captured and kept here by the Bolognese for more than twenty
years, till his death; and the church or Ansilica of St. Petronio, which
was commenced in 1390, and is not yet finished. The palaces and churches
are too numerous to make any remarks on. The leaning towers, Degli
Asmilli and Garisenda, dating from the twelfth century, are among the
most remarkable objects in Bologna. The former is square, and of massive
brick-work, built in three portions, and diminishing in diameter to the
top. Its height is 321 feet, and its inclination from the perpendicular
6ft. 10in. The Garisenda is 161 feet high, and inclines a little more
than 8 feet. Bologna has always been famous for cheap living, and has
been chosen as a residence by many literary men. Gourmands praise it
as the native country of excellent maccaroni, sausages, liquors, and
preserved fruits. The pilgrimage to the Madonna di S. Lucca, whose
church is situated at the foot of the Apennines, half a league distant
from Bologna, and to which an arcade of 640 arches leads, annually
attracts a great number of people from all parts of Italy. Bologna
was founded by the Etruscans under the name of _Felsina_, before the
foundation of Rome. In 189 B.C. it was made a Roman colony, and called
_Bonovia_.
I had been told that the Certosa, or burying ground, was well worth a
visit. It is about 2½ miles outside the city by the Porta St. Isaia,
so I took a cab and was well rewarded for my trouble, for this burying
ground is the most beautiful and remarkable in Italy. Here we can walk
for hours under cover between rows of statues and marble tablets of the
greatest beauty. When I returned to my hotel I found dinner waiting, and
afterwards it struck me that I must seek some more exhilarating mode
of amusement after my visit to the Certosa. I accordingly made my way
to the Teatro Communale, one of the three best theatres in Italy, San
Carlos at Naples and La Scala in Milan taking precedence. The opera was
"Mefistofele," splendidly mounted and well supported by artistes. The
orchestra was large and all that could be desired by the most fastidious
critics, and there are plenty of them in a Bolognese audience. Boxes are
in _every_ tier in the house, and the effect is very pretty.
As I had to start for Brindisi at 3 a.m. on Sunday, November 20th, I had
not much time for sleep, notwithstanding which I got between the sheets
until then, when I was conveyed to the station and finished my nap in the
train.
CHAPTER III.
P. AND O. STEAMER "TANJORE"—ARRIVAL OF THE MAIL AND
PASSENGERS—ANCIENT BRINDISI—BRINDISI TO ALEXANDRIA— ALEXANDRIA
PAST AND PRESENT—ITS TRADE.
I arrived at Brindisi at 10 p.m. and was straightway driven off to the
quay, was soon on board the P. and O. steamship _Tanjore_, commanded by
Captain Briscoe, and not many minutes afterwards in my berth and fast
asleep. My slumber was disturbed at 6 a.m. by the arrival of the Indian
mail and a large number of passengers, who produced a great commotion
over-head quite incompatible with sleep. I therefore turned out, and
was soon on deck watching the busy scene. Some little time after I had
breakfasted I discovered two of the party which I was to accompany,
Messrs. F. L. James and E. L. Phillipps. We were to meet three more at
Cairo, and one at Suez, to complete the party.
No one would care to remain very long in Brindisi, as it is a most
uninteresting place notwithstanding its antiquity. I remember once, in
1877, spending a few hours there, and was then very glad when my train
left for Naples. Brindisi (ancient _Brundusium_) was, if I remember
rightly, the birth-place of Virgil. It is a sea-port and fortified town
45 miles from Taranto. In ancient times it was one of the most important
cities of Calabria. It is said by Strabo to have been governed by its
own kings at the time of the foundation of Tarentum. It was one of the
chief cities of the Sallentines, and the excellence of its port and
commanding situation in the Adriatic were among the chief inducements
of the Romans to attack them. The Romans made it a naval station, and
frequently directed their operations from it. It was the scene of
important operations in the war between Cæsar and Pompey. On the fall of
the Western Empire it declined in importance. In the eleventh century
it fell into the possession of the Normans, and became one of the chief
ports of embarkation for the Crusades. Its importance as a sea-port
was subsequently completely lost, and its harbour blocked. In 1870 the
Peninsular and Oriental Steam Navigation Company put on a weekly line
of steamers between Brindisi and Alexandria for the conveyance of Her
Majesty's eastern mails, and at the same time made it a post of transit
for goods brought from India by these steamers to be forwarded to the
north of Italy by rail. From this cause the imports of Brindisi have
suddenly risen in importance.
About 12 mid-day on the 21st November, we got under way with 110
first-class passengers on board, the weather was fine, much warmer than
in Turin and Milan, and the sea smooth, which I was thankful for; 22nd
the same; 23rd fine and sea smooth until about 4 p.m., when the sea
became rough, and I very uncomfortable, undesirous of dinner and very
desirous of being quietly settled in my berth, which I sought without
loss of time, knowing by a past bitter and sour experience that I should
ere long present a pitiable spectacle. During the night the sea became
so rough that the port-holes of the cabins had to be closed, so that in
addition to feeling excessively sick I was almost suffocated, as the
weather was very warm. On the morning of the 24th, at 10 o'clock, we
landed at the far-famed city of Alexandria.
Even in sunny Italy I had felt the weather, in the neighbourhood of
Turin, Milan, and Bologna, cold and frosty enough in the morning for an
overcoat. At Brindisi it was not so cold, but as we neared the African
coast the sky grew warmer and warmer, and tinged, so to speak, with a
reflection of the Libyan desert, a soft purple hue, rather than the deep
blue of Italy. Only those who have witnessed sunset in Africa can form
any conception of the beautiful tints reflected from the rocks and sands;
there you see the soft purple, lovely crimson, pale gold, rose and violet
colours all shading off into one another in the most charming manner. I
have never seen anywhere such glorious sunsets as in Africa.
Having but a short time to stay in Alexandria, I made good use of it in
exploring the place. Through what strange vicissitudes has this ancient
city passed. Alexandria was founded by Alexander the Great, B.C. 332, on
the site of a village called Rakôtis, or Racoudah. Its founder wished to
make it the centre of commerce between the east and west, and we know how
fully his aspirations have been realized. It stood a little to the south
of the present town, was 15 miles in circumference, and had a population
of 300,000 free inhabitants, and at least an equal number of slaves. So
distinguished was it for its magnificence, that the Romans ranked it next
to their own capital, and when captured by Amru, general of the Caliph
Omar (A.D. 641), it contained 4,000 palaces, 4,000 baths, 400 theatres
or places of amusement, 12,000 shops for the sale of vegetables, and
40,000 tributary Jews. But we are getting on a little too fast. As I
said before, it was founded B.C. 332, by Alexander the Great, who is
said to have traced the plan of the new city himself, and his architect,
Dinarchus or Dinocrates (the builder of the temple of Diana at Ephesus)
directed its execution. The city was regularly built, and traversed by
two principal streets, each 100 feet wide, and one of them four miles
long. Campbell says: "He designed the shape of the whole after that of a
Macedonian cloak, and his soldiers strewed meal to mark the line where
its walls were to rise. These, when finished, enclosed a compass of 80
furlongs filled with comfortable abodes, and interspersed with palaces,
temples, and obelisks of marble porphyry, that fatigued the eye with
admiration. The main streets crossed each other at right angles, from
wall to wall, with beautiful breadth, and to the length (if it may be
credited) of nearly nine miles. At their extremities the gates looked out
on the gilded barges of the Nile, of fleets at sea under full sail, on a
harbour that sheltered navies, and on a lighthouse that was the mariner's
star and the wonder of the world."
One-fourth of the area upon which it was built was covered with temples,
palaces, and public buildings. Conspicuous upon its little isle was the
famous lighthouse of Pharos, the islet being connected with the city by
a mole. Under the Cæsars, Alexandria attained extraordinary prosperity;
large merchant fleets carried on a reciprocal commerce with India and
Ethiopia, and its industrial population were chiefly employed in the
weaving of linen, and the manufacture of glass and papyrus.
The Alexandrians were turbulent, and several times revolted under the
Ptolemies and the Romans. Cæsar was obliged, in B.C. 47, to put down
a terrible insurrection in this city. Under the emperors, Alexandria
suffered a series of massacres, which gradually depopulated it. In 611,
Chosroës, King of Persia, seized it, but his son restored it to the
emperors. In 641, Amru—whom I spoke of just now—took it by storm, after
a siege of 14 months, and a loss of 23,000 men. The Turks captured it in
868 and 1517.
So from time to time Alexandria has been the scene of the greatest
splendour, adorned by marble palaces, temples, and obelisks, also of
great squalor, and covered with mud huts; passing under the sway of
Persian, Greek, Roman, and Turk, and at the time I am writing this
(March, 1884) I think I may safely say under the _sway_ of Great Britain,
although not belonging to this country.
In the early part of this century, under the vigorous, but most
unscrupulous, rule of Mehemet Ali (who was appointed Pasha of Alexandria,
and afterwards of all Egypt), Alexandria became again a thriving and
important place.
It is said that in the character of the population, at least, there
still remains a strong resemblance to the ancient city of the
Ptolemies. Sullen-looking Copts replace the exclusive old Egyptians,
their reputed ancestors. Greeks and Jews, too, swarm as before, both
possibly changed a little for the worse. The mass of Levantines and
(with, of course, honourable exceptions) Franks, who make up the sum of
the population, may, I think, without any exaggeration, be designated
as the off-scourings of their respective countries. The streets swarm
with Turks in many- robes, half-naked, brown-skinned Arabs,
glossy <DW64>s in loose white dresses and vermilion turbans, sordid,
shabby-looking Israelites in greasy black, smart, jaunty, rakish Greeks,
heavy-browed Armenians, unkempt, unmasked Maltese ragamuffins, Albanians
and Europeans of every shade of respectability, from lordly consuls down
to refugee quacks, swindlers, and criminals, who here get whitewashed and
established anew. Here you see a Frank lady in the last Parisian bonnet,
there Egyptian women enveloped to the eyes in shapeless black wrappers,
while dirty Christian monks, sallow Moslem dervishes, sore-eyed beggars,
and naked children covered with flies, present a shifting and everlasting
kaleidoscope of the most undignified phases of Eastern and Western
existence.
The great square, or Grande Place, is the chief place of business and
resort. It is a quarter of a mile long, and 150 feet wide, paved on each
side, with a railed garden in the centre, planted with lime-trees, and
having a fountain at each end. Here are the principal shops and hotels,
the English consulate and church, banks, offices of companies, &c. The
buildings are all in the Italian style, spacious and handsome, or,
rather, were when I visited it. Most of the ancient landmarks are fast
disappearing. The site of Cleopatra's Palace is now occupied by a railway
station for the line to Ramleh, seven miles distant, overlooking the
bay of Abaukir, the scene of Nelson's victory over the French fleet in
1798. Of course, I could not be in Alexandria without paying a visit to
Pompey's pillar, or, more properly, Diocletian's pillar. It is a grand
column, and occupies an eminence 1,800 feet to the south of the present
walls; its total height is 98 feet 9 inches. It is a single block of red
granite on the mounds overlooking the lake Mareotis and the modern city.
An account of the ancient and modern history of Alexandria would fill a
volume of the most stirring interest. I, however, will be content with
giving to my readers a very small portion of a volume on Alexandria, as I
shall have a good deal yet to say on Cairo and neighbourhood, and still
more to say on the Soudan.
It was to Alexandria that science, fostered by the munificence of the
Ptolemies, retired from her ancient seat at Heliopolis. "The sages of the
Museum, who lodged in that part of the palace of the Lagides, might there
be said to live as the priests of the Muses, taking the word in its wide
sense, as the patronesses of knowledge. They had gardens, and alleys,
and galleries where they walked and conversed, a common hall where they
made their repasts, and public rooms where they gave instruction to the
youth who crowded from all parts of the world to hear their lectures."
This museum, a unique establishment in literary history, was founded
by Ptolemy Soter, King of Egypt, who died B.C. 283, and was greatly
enlarged by his son Ptolemy Philadelphus and the succeeding Ptolemies.
In connection with the museum was the Alexandrian Library, the most
famous and the largest collections of books in the world, and the glory
of Alexandria. Demetrius Phalereus, after his banishment from Athens, is
said to have been its first superintendent, when the number of volumes,
or rolls, amounted to 50,000.
If the other Ptolemies were as unscrupulous in obtaining books as
Energetes is said to have been, it is no wonder that the library
increased in magnitude or value. We are told that he refused to sell
corn to the Athenians during a famine unless he received in pledge the
original manuscripts of Aschylus, Sophocles, and Euripides. These were
carefully copied, and the copies returned to the owners, while the
King retained the originals. Various accounts are given of the number
of books contained in the library at its most flourishing period, when
Zenodotus, Callimachus, the poet Eratosthenes, of Cyrene, and Appolinius
Rhodius were its librarians. Seneca states the number at 400,000; Aulus
Gellius makes it 700,000. Some reconcile the discrepancy by making the
statements refer to different periods, while others believe that the
larger figure includes more than one collection. That there were more
than one collection is known. The original, or Alexandrian library _par
excellence_, was situated in the _Brucheion_, a quarter of the city
in which the royal palace stood; and besides this there was a large
collection in the Serapeion, or temple of Jupiter Serapis, but when or by
whom this was founded we do not know. The former was accidentally burned
during the Julius Cæsar's siege of the city, but was replaced by the
library of Pergamus, which was sent by Antony as a present to Cleopatra.
The Serapeion library, which probably included the Pergamean collection,
existed to the time of the Emperor Theodosius the Great. At the general
destruction of the heathen temples, which took place under this emperor,
the splendid temple of Jupiter Serapis was set upon and gutted (A.D. 391)
by a fanatical crowd of Christians at the instigation of the Archbishop
Theophilus, when its literary treasures were destroyed or scattered. The
historian Orosius relates that in the beginning of the fifth century only
the empty shelves were to be seen.
A valuable collection was again accumulated in Alexandria, but was doomed
to suffer the same fate, being burned by the Arabs when they captured
the city under the Caliph Omar in 641. Amru, the captain of the Caliph's
army, would have been willing to spare the library, but the fanatical
Omar disposed of the matter in the famous words:—"If these writings of
the Greeks agree with the Koran, there could be no need of them; if
they disagree, they are pernicious, and ought to be destroyed;" and they
were accordingly used for heating the 4,000 baths in the city. Just
before the time of Mehemet Ali, Alexandria was a miserable place of a few
thousand inhabitants, cut off from the valley of the Nile by the ruin of
the ancient canal. Under his rule it greatly revived in political and
commercial importance, and the re-opening of its canal has restored to
its harbour all the trade of Egypt.
The principal articles of export are cotton, beans, peas, rice, wheat,
barley, gums, flax, hides, lentils, linseed, mother-of-pearl, sesamum,
senna, ostrich feathers, &c.
Those who are not given to pedestrian exercise can easily avail
themselves of a cab or donkey, and they will find the streets, which are
spacious and handsome, very pleasant to traverse, as they are all well
paved in the city; but the dust outside the walls covers the ground from
four to six inches deep, and in combination with the intense glare of the
sun, and the wretched hovels of the natives, produces the ophthalmia so
common, especially among the Arabs. Owing to the want of proper drainage,
what would otherwise be a salubrious site is subject to malarious disease
and the plague.
I have spoken of the Alexandrian library; quite as much may be said of
the Alexandrian school; combined, they may be justly considered the first
academy of arts and sciences.
The grammarians and poets are the most important among the scholars
of Alexandria. These grammarians were philologists and literati, who
explained things as well as words, and may be considered a sort of
encyclopedists. Such were Zenodotus the Ephesian, who established
the first grammar school in Alexandria; Eratosthenes, of Cyrene;
Aristophanes, of Byzantium; Aristarchus, of Samothrace; Crates, of
Mallus; Dionysius the Thracian; Appolonius the sophist; and Zoilus. To
the poets belong Appolonius the Rhodian, Lycophron, Aratus, Nicander,
Emphorion, Callimachus, Theocritus, Philetas, Phanocles, Timon the
Philasian, Scymnus, Dionysius, and seven tragic poets, who were called
_Alexandrian Pleiads_.
The most violent religious controversies disturbed the Alexandrian church
until the orthodox tenets were established in it by Athanasius, in the
controversy with the Arians.
Among the scholars are to be found great mathematicians, as Euclid, the
father of scientific geometry, and whose work, I distinctly recollect,
was a great bore to me in my younger days; Appolonius, of Perga, in
Pamphylia, whose work on conic sections still exists; Nichomachus, the
first scientific arithmetician; astronomers, who employed the Egyptian
hieroglyphics for marking the northern hemisphere, and fixed the images
and names (still in use) of the Constellations, who left astronomical
writings (_e.g._, the _Phœnomena_ of Aratus, a didactic poem; the
_Spherica_ of Menelaus; the anatomical works of Eratosthenes, and
especially the _Magna Syntaxis_ of the geographer Ptolemy), and made
improvements in the theory of the calendar, which were afterwards adopted
into the Julian calendar; natural philosophers, anatomists, as Herophilus
and Erasistratus; physicians and surgeons, as Demosthenes Philalethes,
who wrote the first work on diseases of the eye; Zopyrus and Cratenas,
who improved the art of pharmacy and invented antidotes; instructors
in the art of medicine, to whom Asclepiades, Loranus, and Galen owed
their education; medical theorists and empirics, of the sect founded by
Philinus. All these belonged to the numerous association of scholars
continuing under the Roman dominion and favoured by the Roman emperors,
which rendered Alexandria one of the most renowned and influential seats
of science in antiquity. With this passing glance at Alexandria, we will
journey on to Cairo.
CHAPTER IV.
THE FERTILIZING RIVERS OF EGYPT—LEAVE ALEXANDRIA—INCIDENTS EN
ROUTE—SHEPHEARD'S HOTEL—ANCIENT AND MODERN CAIRO—THE DONKEY
BOYS—ARAB PATIENTS—DANCING DERVISHES—THE HOUSE WHERE JOSEPH,
MARY, AND THE INFANT SAVIOUR LIVED IN OLD CAIRO—THE BAULAC
MUSEUM—THE PETRIFIED FOREST—MOKATTAM HILLS—TOMBS OF THE CALIPHS
AND CITADEL—CAIRO AT SUNSET.
In former times, before the introduction of railways, the traveller
to Cairo had to go by canal, hire a boat, servant; procure a carpet,
mattress, and bedding; lay in a store of provisions, and a variety of
minor articles that would fill a page or two to mention. Now we can go
comfortably by rail in a few hours, the distance being something like 120
miles, I think.
We pass, _en route_, Lake Mareotis and the Mohmoudieh Canal, cultivated
land near Alexandria, then a good deal uncultivated and desert; but as we
approach Cairo, we see large tracts of cultivated land, all accomplished
by irrigation, and I am told that as much as two or three crops in the
year can be obtained off these lands without very great labour. A hot
sun can always be depended on. The agricultural labourer has not to go
through the laborious work of ploughing and manuring as in England.
All he has to do is to scratch the ground, and put in the seed in the
fertilizing alluvium which has been brought down from the rich lands of
Meroe and portions of Abyssinia by the Athara river and its tributaries,
the Salaam, Augrab, and the greater stream, Tacazze or Settite. All these
rivers cut through a large area of deep soil, through which, in the
course of ages, they have excavated valleys of great depth, and in some
places of more than two miles in width. The contents of these enormous
cuttings have been delivered upon the low lands of Egypt at the period of
the inundations. The Athara is the greatest mud-carrier, then the Blue
Nile, which effects a junction with the White Nile at Khartoum.
The White Nile is of lacustrine origin, and conveys no mud, but an excess
of vegetable matter, suspended in the finest particles, and exhibiting
beneath the microscope minute globules of green matter, which have the
appearance of germs. When the two rivers meet at the Khartoum junction,
the water of the Blue Nile, which contains lime, appears to coagulate the
alluminous matter in that of the White Nile, which is then precipitated,
and forms a deposit; after which the true Nile, formed by a combination
of the two rivers, becomes wholesome, and remains comparatively clear,
until it meets the muddy Athara. The Sobat river is a most important
tributary, supposed to have its sources in the southern portion of the
Galla country.
For the foregoing information on these rivers I am indebted to an
article of Sir Samuel Baker's, which I read with great interest in the
_Contemporary Review_; and I daresay many of my readers will thank me for
reproducing it.
After this slight digression, I will continue my journey to Cairo. At
the stations were numbers of women and children with refreshments for
the traveller in this land, where the sun always shines with a burning
heat; women with goolehs of water to sell; children naked, or nearly so,
with sugar-cane, melons, oranges, dates, fresh sugar-cane, figs, &c. Vast
numbers of these poor creatures were afflicted with ophthalmia, their
eyelids covered with flies, which they take no notice of whatever, many
of them blind, or partially so, blind beggars; one and all, whether they
can sell anything or not, continually uttering the cry of "Backsheesh,
backsheesh, howaga," which comes faintly on my ears as the train leaves
the station. As we journey on there is much to be noticed. Now we pass
a camp of Bedouins in the desert; next a large grove of date-palms (the
owner of which has to pay a tax on every tree). Here the domestic buffalo
walks round and round a circle; he is working the sakia or water-wheel,
which winds up the water for irrigation. This is also taxed. Scattered
all over the country are innumerable shadoofs, another mode, and the most
ancient, of obtaining water; there the stately-looking camel strides
along, looking intensely unconcerned. Trotting past him on his little
donkey is an Arab in loose, white, flowing robes, and turbaned head. At
one time we pass squalid, wretched-looking mud-huts; anon Nubians, as
black as coal, working in the fields. We arrived at Cairo in the evening
about seven, and were at once driven off to the well-knewn Shepheard's
Hotel. The _cuisine_ is all that could be desired, and every attention
is paid to insure the comfort of visitors. Mr. Grose, the manager, is a
particularly obliging and attentive gentleman.
Cairo (in Arabic, _Kahira_, which signifies _victorious_) is the capital
city of Egypt. It lies on the east bank of the Nile, in a sandy plain,
and contains old Cairo, Boulac (_the harbour_), and new Cairo, which are,
to a considerable degree, distinct from each other. The city itself,
separate from the gardens and plantations which surround it, is about
10 miles in circuit, has 31 gates, and 240 irregular unpaved streets,
which during the night are, or were, closed at the end of the quarter, to
prevent disturbances. The houses are for the most part built of brick,
with flat roofs, and the interior of many of them is very sumptuous.
The chief square of Cairo, El-Esbekiah, has a magnificent area, the
centre of which is laid out as a garden, and is annually inundated by
the overflowing of the Nile. It is surrounded by the finest palaces.
There is in it a monument to General Kleber. The inhabitants of the
city and suburbs, in 1871 353,851, are Arabs or Mahomedans, Coptish
Christians, Mamelukes, Greeks, Syrians, Armenians, Jews, and natives
of various countries of Europe. The castle, or citadel, situated on a
rock, containing Joseph's Well, 276 feet deep, is the residence of the
Pasha. There are 80 public baths, 400 mosques, two Greek, 12 Coptish,
one Armenian, and one English church, 36 synagogues, and many silk,
camlet, tapestry, gunpowder, leather, linen, and cotton factories.
Among the mosques, which, though many of them are in ruins, form the
most conspicuous edifices of the city, the most remarkable is that of
Sultan Hassan, which is built of blocks of polished marble, obtained from
the outer casing of the pyramids, or pyramid rather, for, if my memory
serves me right, they are from the great pyramid of Cheops at Gizeh.
It has a beautifully ornamented porch, richly corniced walls, and many
tall minarets. Here is also a Mahomedan high school, a printing office
and 25,000 volumes. The largest convent of dervishes is at Cairo. It was
built in 1174. The traffic of Cairo is very great, since it is the centre
of communication between Europe, the Mediterranean Sea, Asia, and the
North of Africa, and is upon the railway from Alexandria to Suez. The
principal bazaars are the Ghoreah and Khan Khalel. Goods are disposed
of there by public auction, and the different bazaars exhibit different
kinds of merchandise. Ibrahim Pasha commenced a public library in 1830,
and in 1842 a European Society, called the Egyptian Literary Association,
was established. Mehemet Ali introduced schools for elementary education,
and the Church of England Missionary Society has two schools.
Cairo was founded by Jauhar, general of the Caliph Moez, in the year of
the Hegira 368, or A.D. 969, on the site of the Egyptian Babylon. Moez
afterwards made it his capital, which distinction it retained until the
overthrow of the Mamelukes by Sultan Selim in 1517. Saladin extended
and fortified it in 1176. It was repeatedly attacked by the Crusaders,
particularly by St. Louis in 1249. It was occupied by the French from
1798 to 1801, when it was recovered by the Turks with the assistance of
the English. A great fire occurred there in February, 1863; advantage was
taken of it to improve the town.
Our military occupation of Egypt (or shall I say that it is simply a
"measure of police?"), and events that are now transpiring there, are a
sufficient excuse (if one were required) for dealing shortly with the
ancient history of Cairo and the neighbourhood.
Soon after our arrival at Shepheard's Hotel, when we had restored
ourselves to our personal comfort, our host provided us with a good
dinner, to which we did ample justice, and as the weather (although the
end of November) was like a summer's evening in England, we enjoyed the
usual after-dinner cigarettes in the balcony, which is a very pleasant
lounge, even in the day time, as it is quite sheltered from the blazing
sun. I soon strolled off to bed with the idea of obtaining a good
night's rest, so that I should awake refreshed and fit for a pilgrimage
to the various shrines of intense interest with which Cairo and its
neighbourhood abounds. I have visited and seen all that was interesting
in Rome, once the mistress of the world—Corinth, once the seat of
learning and the abode of a most polished people; Ephesus; have stood
on the ancient Acropolis of Athens, the plains of Troy, celebrated by
Virgil; explored Misenum, Pateoli, Baiæ, Pompeii, and Herculaneum, all
rich in historical associations; but compared with the remains of ancient
cities near Cairo these places were of yesterday's growth, and were not
even thought of until ages after the glory and high civilization of the
people in the land of the Pharaohs had passed away. When Abraham entered
the Delta from Canaan with his countrymen, moving about in tents and
waggons, the Egyptians were living in cities enjoying all the advantages
of a settled government and established laws; had already cultivated
agriculture, parcelled out their valley into farms, and reverenced a
landmark as a god.
While Abraham knew of no property but herds and movables, they had
invented records and wrote their kings' names and actions on the massive
temples which they raised. They had invented hieroglyphics and improved
them into syllabic writing, and almost into an alphabet. The history
of Greece _begins_ with the Trojan war, but _before_ the time of David
and _before_ the time of the Trojan war, the power and glory of Thebes
had already passed away. About 1,000 years B.C. Shishak the conqueror
of Rehoboam, son of Solomon, governed all Egypt; at his death it was
torn to pieces by civil wars. After a time the kings of Ethiopia reigned
in Thebes, and helped the Israelites to fight against their Assyrian
masters. This unsettled state of things lasted nearly 300 years, during
which, as the Prophet Isaiah foretold, "Egyptians fought against
Egyptians, brother against brother, city against city, and kingdom
against kingdom." At last the city of Sais put an end to this state
of things and under the Sais kings Egypt enjoyed again a high degree
of prosperity. They were more despotic than the kings of Thebes, and
struggled with the Babylonians for the dominion of Judæa.
Probably many of my readers are aware that M. Ferdinand de Lesseps was
not the originator of a canal to the Red Sea, for Pharaoh Necho, one of
the Sais kings, began it from the Nile. His sailors, circumnavigated
Africa; he conquered Jerusalem, and when the Chaldees afterwards drove
back the Egyptian army the remnant of Judah, with the Prophet Jeremiah,
retreated into Egypt to seek a refuge with King Hophra.
523 B.C. the Persians became masters of Egypt, and behaved with great
tyranny. Cambyses plundered the tombs and temples, broke the statues,
and scourged the priests. They ruled for 200 years; then the Greeks,
B.C. 332, the Romans, B.C. 30, and on the division of the Roman Empire,
A.D. 337, Egypt fell to the lot of Constantinople. In A.D. 640, just 670
years after the Roman conquest, Egypt was conquered by the followers
of Mahomet, and now, in this year of grace, A.D. 1884, we are rather
upsetting the late order of things, but whether for good or evil time
will show.
In this age of progress, it may seem strange to say so, but Egyptian
landlords had much the same tastes 3,000 years ago as English landlords
have now. They were much addicted to field-sports. Not only does history
tell us so, but I have seen often in their sculptures and paintings that
this was so. Even on the tomb and chapel of King Phty at Sakkara, which
is said to be over 5,000 years old, I saw scenes of fowling, fishing,
hunting, running down the gazelle, spearing the hippopotamus, of coursing
and netting hares, of shooting wild cattle with arrows, and catching them
with the lasso. They had fish ponds, game preserves, and game laws, they
were fond of horses and dogs, kept good tables, gave morning and evening
parties, amused themselves with games of skill and chance, were proud of
their ancestors, built fine houses and furnished them handsomely, and
paid great attention to horticulture and arboriculture.
This certainly reads like contemporary history; but I will go further.
To use a well-known expression, "would you be surprised to hear" that
the tenants paid the same proportionate rent as the British farmer of
to-day? The average gross produce of a farm here was £8 an acre, average
rent about 32s. an acre—just one-fifth—the exact rent paid by the tenants
of Potiphar, Captain of the Guard, and of Potipherah, Priest of On,
Joseph's father-in-law, and the same was paid to Pharaoh himself by
his tenants. At that time the whole acreage of the country was divided
into rectangular estates. One-third belonged to the king, two-thirds in
equal proportions to the priestly and military castes; and these were
cultivated by another order of men, who, for the use of the land, paid
rent—one-fifth of the gross produce—to the owner.
Altogether I spent nearly a fortnight in Cairo, and feeling a great
interest in the historical associations of this ancient place and the
neighbourhood, I resolved to see and learn as much as I could of them
during my short stay. In the morning, after early breakfast, I amused
myself for a short time by sitting in the shade of the extensive balcony
in front of Shepheard's Hotel, which overlooks the street, and is
contiguous to it. The scene which presented itself to my gaze was truly
Oriental in character. Now I see a few camels stalking silently, slowly,
and sedately on, variously laden—some with baskets of large stones for
building purposes, others with long pieces of timber on each side, others
with skins of water and so on; then an Arab lady on donkey-back, riding
after the manner of men, and covered from head to foot in unsightly black
wrappers, having just a slit in them, through which can be seen a large
pair of lustrous dark eyes, and down the bridge of her nose are some
brass-looking ornaments, resembling as much as anything a row of thimbles
inserted in one another. A Turkish lady's dress and yashmack (covering
worn over the face) is much more becoming, and her nose is not ornamented
by the addition of the thimble arrangement. The Turkish ladies wear (in
Constantinople) quite a thin white muslin yashmack over their faces. This
does not conceal very much of the features, which, as a rule, are very
beautiful. The Egyptian ladies wear a black yashmack, which conceals
all except the eyes. Report says they are ugly; if so, they are quite
right to do so. Next I see a carriage driven along preceded by two sais,
or runners, to clear the way, and it is surprising what a pace they go
at with a long, swinging trot. They are picturesquely and gorgeously
dressed, each bearing a long wand, and wearing a tarboosh (Turkish fez),
the long thick blue tassel of which floats gracefully over the shoulders,
and not at all unlike what some of the ladies in Athens wear, except that
their tassels are black. Then we see blind, or partially blind, beggars,
of whom there are vast numbers, Coptic and Mahomedan women and children,
girls with baskets of flowers and lovely roses, sweet-meat, fly-whisp,
water, and fruit-sellers, conjurers, snake-charmers, one and all
soliciting "backsheesh," dusky, brown-skinned Arabs clad in loose-flowing
robes and white turbans, coal-black Nubians, Jews, Greeks, Armenians,
and Europeans of all shades of colour, religion, and politics. Here, in
fact, in this city of Saladin and of the "Arabian Nights Entertainments"
creations (which once seemed to be so fanciful and visionary) kindle into
life and reality as I look upon everything around me.
The apartments of an Arab house of the well-to-do are decorated with
Arabesque lattices, instead of glass windows. Inside are luxurious divans
heaped with soft cushions, instead of sofas and chairs; and instead
of the rattling of cabs, carts, and tramcars we hear the wild, shrill,
trilling note of the Arabian women indicating some occasion of joy or
sorrow, or hear the equally peculiar long drawn-out note of the muezzin
from some minaret calling the faithful to prayer.
Very near to our hotel, on the opposite side, are always to be found a
number of donkeys ready for hire, and very good little donkeys they are.
I can see the head, legs, and tail of a donkey; the remaining portion of
him is almost concealed by a great padded saddle, to which is attached a
very inconvenient pair of stirrups, into which you _may_ get the tips of
your toes, and sometimes a portion of the foot, but if the foot is not
small, or is so unfortunate as to possess a respectably-sized bunion,
you must be content if you can get the tips of your toes only in the
stirrup; this, again, slips down to the right or left, according as you
put more pressure on one side or the other. There are no girths, but one
long strap placed around the saddle and donkey very insecurely fixes the
former. If my reader has not been accustomed to circus-riding, I assure
him he would experience some difficulty at first in exhibiting his powers
of equitation before the Egyptian public under these circumstances, and
I have seen more than one individual come into ignominious contact with
mother earth; fortunately he has not far to go ere he humbles and tumbles
himself in the dust.
My first experience was this: as soon as I was seated and had rammed
the tip of my boot into the stirrup, the donkey-boy shouts, "Ha—ha."
This warning note the donkey knows full well, and off he goes at a kind
of running trot, which is all right. Soon these ha-ha's increase in
frequency, and ere long I can fancy myself a second Mezeppa. The imp
behind now accompanies his peculiar yell with a sharp prog of a pointed
stick, and the donkey takes a very pointed cognisance of it, for now
"He urges on his wild career." In the wide, open streets this rapid
mode of progression has an exhilarating tendency, but in the narrow
streets of the bazaars unguarded human beings fly to the right of me,
unguarded human beings fly to the left of me, and imprecations, not
loud, but deep, in an unknown tongue, fall on my untutored ear as my
donkey indiscriminately cannons on to the unobservant. A few words about
these donkeys, and donkey-boys so called. Most of the latter are not
boys at all, but full-grown men, notwithstanding which they are always
called donkey-boys. These and their donkeys are quite an institution
in the East. The donkeys own all kinds of popular English names, and
of course (if the owner may be believed) are possessed of every good
quality. Most of the donkey-boys have picked up more or less English,
and in expatiating on the good qualities of their beasts are accustomed
to interlard their speech with the strong language of the West, and
you would be surprised to hear how promptly they will consign a fellow
donkey-boy to an inhospitable and much-warmer region than Cairo, and to
the care of a much blacker individual than themselves. The reader is here
called upon to exercise his or her imagination. I had myself derived
considerable amusement when watching an intending pilgrim securing one of
these donkeys. To be forewarned is to be forearmed; I flattered myself
that by making my selection sure before I got amongst them, my tactics
would be most successful, but as the sequel will show, I was grossly
deceived, having reckoned without my host, or hosts I ought to say.
First intending pilgrim. He descends the steps of Shepheard's Hotel, and
moves towards the donkeys—a fatal movement. Instantly the air is thick
with donkeys and donkey-boys. The latter yell frantically a chorus of
praises concerning the useful quadrupeds, which are most adroitly and
with surprising dexterity brought one after the other under his very
nose, whilst the poor victim is jostled about in the most bewildering
and unpleasant manner. I have been both a spectator of and an actor in
this performance, and I can safely say the spectator derives by far the
greatest amusement.
I resolved to pay a visit to the bazaars and some of the mosques of note.
Having, as I thought, gained some experience by observing the misfortunes
of others, I executed a strategic movement which I fondly imagined would
turn out successful. I had, from a distance selected my donkey; then
cunningly walked up and down the pavement smoking a cigarette, apparently
with no object in view. Suddenly I darted on to the enemy, but alas! I
found myself in an absolute whirlwind of donkeys and their troublesome
two-legged attendants, who yelled into my ears and bumped me about until
I was quite unable to recognise the donkey I had selected. Beauties
were here represented, such as Mrs. Cornwallis West, and Mrs. Langtry;
national names, such as John Bull, and Yankee-doodle; mythical names,
such as Jim Crow and Billy Barlow. One donkey rejoiced in the name of
Dr. Tanner, another in that of Madame Rachel; others, again, had been
honoured with the names of statesmen, such as Prince Bismarck, John
Bright, Sir Stafford Northcote, Lord Randolph Churchill, Mr. Gladstone,
Mr. Parnell, Lord Beaconsfield, and others. "Dr. Tanner, he debbil to
go—he berry good donkey indeed, hakeem," said the owner. However, I
declined him, as he was said to be a FAST one (excuse the joke), and as
this was entirely an Eastern question, I could not help thinking that
Lord Beaconsfield would certainly be the most likely to carry me safely
through. I therefore selected him, and had every reason to be satisfied
with him and his secretary, Lord Rowton _alias_ Ibrahim, the donkey-boy,
whom I employed on several subsequent occasions. He proved a very good
conductor, for he took me through the various bazaars, Tunis, Algiers,
Turkish, Persian, and Arab, &c., pointing out all places of note and
interest _en route_. Ibrahim soon got to know that I was a doctor, and so
indeed did all the attendant Arabs about the hotel. He, like hundreds of
his countrymen, suffered from ophthalmia, and when I was out with him he
said—
"Hakeem, what I do with my eyes? They very bad sometimes."
"Oh!" said I, "you bring me a bottle to-morrow morning, and I will give
you something for them," little thinking of the consequences. The lotion
did his eyes a great deal of good, and two days afterwards a great many
of his friends called, to all of whom I gave lotion. During my stay here,
and some months afterwards on returning from the Soudan, I was, every
morning, employed after breakfast at my medicine chest preparing eye
lotions for my Arab friends, invocations for the blessings of Allah being
my recompense. The poor fellows appeared to be grateful, and I dare say
it was genuine, not like a canting old Irish vagrant woman, who, if you
give a hunk of bread and cheese to, will exclaim—
"Thank yer honour kindly!" and as long as she is in hearing keep
muttering, "Och! sure now, there's a kind jintleman for ye, me darlint.
Sure now he is intirely an illigant jintleman; only for him I would not
have a bite this morning, that's sure for ye. May Heaven guide him and
the blessed Virgin protect him!" Then out of hearing it is, "Och! the
dirty spalpeen! What will I do wid this? May the curse of Cromwell light
on ye for a murthering Sassenach. What will I do honey? and I not had a
sup of gin this blessed day to keep the cowld out of me poor thrimbling
ould body!"
But I am digressing. One day I took a donkey ride to old Cairo, and with
others from the hotel visited the dancing dervishes, and the house said
to have been inhabited by our Saviour. Old Cairo is about two miles
distant from Grand Cairo. It was at old Cairo that the child Jesus,
with Joseph and Mary, lived for a time, having fled from the bloody,
persecuting Herod. The place said to have been His exile home is now a
small Greek church. The steps to the room are very much worn, but great
care is taken of every part of it; silver lamps, hung from the ceiling,
are burning night and day, and no one is allowed to enter without the
presence of a Greek priest. It certainly is not difficult to believe
that, considering the mild Syrian atmosphere, and the absence of rain,
the building may be much more than 1,800 years old.
The dancing dervishes next engaged our attention. When in Constantinople
I visited the dancing dervishes at Pera and the howling dervishes on
the other side of the Bosphorus at Scutari. The dancing dervishes wear
a dress of greyish material, which reaches a little below the knee,
and is confined by a girdle round the waist. When they spin round like
Teetotums this looks like an open umbrella. The head is covered by a
curious-looking, tall, conical felt hat without any brim.
The word itself, Dervish, or Dervise, is of Persian origin, and signifies
poor. It denotes the same amongst Mahomedans as _monk_ with Christians.
The observance of strict forms, fasting and acts of piety, give them a
character of sanctity amongst the people. They live partly together in
monasteries partly alone, and from their number the Imams (priests)
are generally chosen. Throughout Turkey they are freely received, even
at the tables of persons of the highest rank. Among the Hindus they
are called _fakirs_. There are throughout Asia multitudes of these
devotees, monastic and ascetic, not only among the Mahomedans, but
also among the followers of Brahma. There are no less than thirty-two
religious orders now existing in the Turkish Empire, many of whom are
scarcely known beyond its limits; but others, such as the Nakshbendies
and Mevlevies, are common in Persia and India. All these communities are
properly stationary, though some of them send out a portion of their
members to collect alms. The regularly itinerant dervishes in Turkey
are all foreigners or outcasts, who, though expelled from their orders
for misconduct, find their profession too agreeable and profitable to
be abandoned, and therefore set up for themselves, and, under colour of
sanctity, fleece honest people. All these orders, except the Nakshbendies
are considered as living in seclusion from the world; but that order
is composed entirely of persons who, without quitting the world, bind
themselves to a strict observance of certain forms of devotion, and
meet once a week to perform them together. Each order has its peculiar
statutes, exercises, and habits. Most of them impose a novitiate, the
length of which depends upon the spiritual state of the candidate, who
is sometimes kept for a whole year under this kind of discipline. In
the order of the Mevlevies, the novice perfects his spiritual knowledge
in the kitchen of the convent. The numerous orders of dervishes are all
divided into two great classes, the dancing and the howling dervishes.
The former are the Mevlevies, and are held in much higher estimation
than the other class, and are the wealthiest of all the religious bodies
of the Turkish Empire. Their principal monastery is at Konieh, but they
have another at Pera, a suburb of Constantinople, where they may be
seen engaged in their exercises every Wednesday and Thursday. These are
performed in a round chamber, in the centre of which sits their chief
or sheik, the hem of whose garment each dervish reverently kisses on
entering the chamber, after which they go and range themselves round
the chamber with their legs tucked under them. When all the dervishes
have entered and saluted the sheik, they all rise together and go in
procession three times round the room, the sheik at their head. Each time
they do obeisance to the empty seat of the sheik on coming to a certain
part of the room. The procession ended, the sheik again takes his place
in the centre, and all the others begin dancing round him, turning on
themselves at the same time that they move round the room. The arms are
extended, the palm of the right turned upward and the palm of the left
downward, to indicate that what they receive from heaven with the right
they give away to the poor with the left, while sounds of music are
heard from a neighbouring gallery. The movement at first is slow, but
as the dervishes become excited they become more animated, and revolve
so quickly that they look like tops spinning round; at last they sink
exhausted on the floor. After a while they renew their exertions, and
repeat it several times. The whole is concluded by a sermon.
The howling dervishes do not confine themselves in their exercises to
the dancing just described. They accompany them with loud vociferations
of the name of Allah, and violent contortions of the body such as are
seen in persons seized with epileptic fits. And even these extravagances
are not so bad as those which were formerly practised, when the
dervishes, after working themselves into a frenzy, used to cut and
torture themselves in various ways with apparent delight. The sheiks
of all orders have the credit of possessing miraculous powers. The
interpretation of dreams, the cure of diseases, and the removal of
barrenness, are the gifts for which the dervishes are most in repute. Had
I to live in such a hot climate as Cairo, I should feel thankful that our
religion does not necessitate such violent bodily exertion as that which
these dervishes indulge in. The road to old Cairo was very, very dusty,
and the weather excessively hot, as it always is in the day time. We left
the dancing dervishes after remaining about half-an-hour, and rode back
to our hotel in the afternoon too late for any further explorations that
day. On the following day I spent some hours in a very enjoyable and also
instructive manner, namely in inspecting the priceless articles in the
Baulac Museum. This museum, I suppose, contains some of the most ancient
things in this world, and I regret very much that I could not devote a
week to inspecting the contents of it instead of a few hours. I should
have seen the treasures contained here, and known very little concerning
them (as there was no catalogue), had I not been so fortunate as to get
into conversation with Brusch Bey, the curator, a most intelligent and
obliging gentleman, whose heart is enthusiastically in his work. He was
kind enough to spend about two or three hours with me and enlighten me
on very many things which would have been a sealed book to me but for
him. There lay before us one grand discovery of 32 kings and queens,
who had ruled Egypt in the dim distant ages long ago. The gilding on the
inner coffins was as perfect and untarnished as it was the week they
were executed, although thousands of years have rolled by since the
handy craftsman was engaged on them. They were covered with information
that none but an Egyptologist could decipher. In this museum was pointed
out to me a picture said to be the most ancient in the world, it was a
painted picture of Egyptian geese, as well done, I should imagine, as
any ordinary painter of the present day could do it. There were bronzes
and polished marble statuary as perfect in appearance as when they left
the workmen's hands, and, as far as I could judge, as well finished as
they would be by workmen of the present day, although 2,000 or 3,000
years old. An ingenious and strong little cabinet engaged my attention
some time; the doors of hard wood were well carved and the joints as
exquisitely dove-tailed in as any man of the present day could make
them. In a glass case I saw basket-work, a chair, rope, twine, seals,
rings, javelins, slings, food and seeds as they were found in an ancient
tomb, the mason's mallet cut out of a solid piece of wood, precisely the
same shape and size as those in use here at the present time, jewellery
well-finished and solid-looking, and many other things too numerous to
mention. On carefully examining this valuable and interesting collection,
some of which were 3,000, 4,000, or 5,000 years old, I could not help
thinking that they served well to illustrate the highly civilized
condition of the people at so remote a period.
To give details of all the interesting things in this museum would occupy
too much time to the exclusion of other matter, but there are two things
that call for notice on account of their very great antiquity. One is a
wooden statue, which has been carved out of a solid block of very hard
wood, and is that of a man about 5ft. 7in. in height. As one stands in
front of that wooden statue gazing for a short time, he almost appears
to be endowed with a soul and the power of speech, so excellent is the
execution of the figure, and so expressive the face; no one can doubt
for a moment that he was the creation of a high civilization. It was
found in a tomb at Sakhara and belongs to one of the early dynasties
of the old primæval monarchy, and is absolutely untarnished by the
thousands of years it has been reposing in that tomb; there is actually
no sign of decay. The antiquity of that statue astonishes me, and I
dare say it will my readers. Brusch Bey told me that it was supposed
to be 5,400 years old, and that probably it was older than that. The
other statue, that of Chephren, the builder of the second Pyramid, with
his name inscribed upon it, is in Diorite, one of the hardest kind of
stones, carefully executed and beautifully polished. These Egyptians were
evidently people of considerable forethought, and when they wanted their
names and deeds to live long after them engraved on tablets of stone,
they selected the most durable they could, and it is more than probable
that had they contemplated building such houses of Parliament as we have
built in London, they would have selected a hard, not a soft stone, that
continually requires patching up. Well, the features of Chephren's statue
are uninjured, and Brusch Bey and I gazed on them just as they were seen
by Chephren and his court 5,000 years ago. It was discovered by Mariette
Bey, at the bottom of a well, which supplied the water used for sacred
purposes in the sepulchral temple attached to Chephren's Pyramid. It was
no doubt originally erected in the temple, and was probably thrown into
the well by the barbarous Hyksos or iconoclastic Persians.
During the late military operations, or "police measures," grave
apprehensions for the safety of the Baulac Museum arose, but fortunately
it escaped the violence of the mob. The greater part of one day was
occupied by a visit with my familiar Ibrahim to the mosques of note,
the citadel, tombs of the Caliphs and Mamelukes. Another day I got a
companion from the hotel to accompany me to the petrified forest, some
miles out in the desert. It covers an area of about 15 miles. All this
space is pretty thickly strewed over with what appears to be trunks and
branches of trees. I took hold of what appeared exactly like the wooden
branch of a tree, and so it had once been, but for ages it had lain here,
a solid piece of very hard stone. The place is an absolutely desolate one
in the desert, with not a sign of vegetation in sight. Whether these had
been washed here during the flood or had once grown in the neighbourhood
or not, or how they came there, I never could ascertain, although I have
sought for information on the subject in all directions. No one seems to
be able to tell me anything about the origin of this petrified forest,
and I have not hitherto found a book containing any allusion to it. We
returned to Cairo by the Mokhottam hills behind the citadel somewhat late
in the afternoon, consequently had to urge on our donkeys so that we
should see Cairo by sunset. We were here just in time to do so, as there
is scarcely any twilight in the East; the transition from day to night
does not occupy very many minutes. The picturesque panorama that opened
out to our view well repaid us for our trouble. There before and beneath
us lay Cairo with its innumerable mosques and minarets, the Nile with the
peculiar Nile boats called dahabeahs floating peacefully on its surface.
Here and there the stately camel strides silently on, veiled women and
turbaned Arabs in loose flowing robes, groves of palm trees, while nearer
to us we see the half-ruined tombs of the Caliphs and Mamelukes, the
citadel and the beautiful mosque of Mehemet Ali full of carved columns of
alabaster. To the late burning heat which we encountered in the desert
succeeds a soft, balmy, dry air, and the beautiful and varied hues of
the setting sun is reflected from the glittering mosques and minarets,
rocks and sands, presenting a picture which will not soon fade from my
memory, and which requires the poetry, eloquence, and pen of a Byron to
adequately describe. In striking contrast to the beautiful scene we had
just enjoyed was the wretched-looking houses of the Arabs, the squalor,
dirt and miserable pathways on the hill-side which we encountered
immediately afterwards as we pursued our homeward journey.
CHAPTER V.
A YOUNG AMERICAN AT SHEPHEARD'S HOTEL—DRIVE TO THE PYRAMIDS
OF GIZEH—ASCENT AND EXPLORATION OF THE PYRAMID OF CHEOPS—THE
SPHINX.
We arrived at our hotel rather tired, and felt it quite a relief to
stretch our legs out straight after having them cramped up so long whilst
on our donkeys. Having partaken of a good dinner, I adjourned to the
balcony with a cigarette, sank into an easy lounge, and communed with
my own thoughts. I had not been here long before I discovered sitting
near me an individual, apparently about 23 years of age, whose nether
extremities rested on the back of a chair, his feet being parallel with
his chin. He was dressed in a somewhat _outré_ manner, the lower limbs
being encased in check prolongations; the body in a brown coat, something
like a sack in shape; the throat was surrounded by a loose, turn-down
collar, and loose neckerchief, whilst the summit of this curious
specimen of humanity was crowned by a huge felt hat, with an enormous
brim. The clouds of smoke which he emitted from his mouth rivalled a
young volcano; he was smoking a cigar, and did not forget to expectorate
in a most profuse and dangerous manner, so much so that, feeling in
somewhat dangerous proximity to the fire of his artillery, I got up with
the intention of escaping any little salivary accidents; but my silent
companion had his eye on me, and thus suddenly addressed me in the
decidedly nasal accent and twang peculiar to the inhabitants of America—
"Stranger, I guess this Cairo is a tarnation rummy place?"
Seeing no reason to dispute this by no means rash assertion, I readily
conceded the point; and, by way of carrying on the conversation, ventured
to remark that—
"It certainly is a very curious and interesting old place, and the
inhabitants no less so."
_He_: "That's so, sirree; they _are_ queer beggars, and so _are_ their
wimen."
This also was an indisputable fact, and I acknowledged that they were a
strange race, strongly wedded to old customs, and as strongly opposed to
innovations.
_He_: "Stranger, yew don't roost here, I guess?"
_I_: "No; I am just travelling for a few months, and shall leave Cairo in
two or three days' time."
_He_: "In what line may you be travelling, stranger?"
Now, of course I knew what he meant, but thought his remarks were so
original, not to say impertinent, that I must not omit this opportunity
of extracting some amusement, and provide material for my diary. I
therefore replied—
"Oh! I came by the P. and O. line to Alexandria, by rail here, and now my
lines have fallen in pleasant places."
"Guess yew don't quite fathom me. What's yer business, and where are you
going tew?" said he.
I then gave him the names of a number of places in Egypt and the Soudan,
enumerating them as rapidly as I could, so that I am quite sure my nasal
friend was very little the wiser for the information.
He enshrouded himself in a huge cloud of smoke, vigorously expectorated
once more, and regarding me fixedly for a moment, exclaimed—
"By Jupiter! stranger, that's a large order. Opening up a trade or
colon_ize_, I guess."
I suppose, because I told him I was travelling with six other gentlemen,
he thought we were going to start a colony somewhere, and then annex
all the adjacent country, which, by the way, would certainly be a very
good thing for the Egyptians and the Soudanese, and very probably for
ourselves also. However, I gave him to understand that we were simply
travelling for pleasure, exploration, and sport. Notwithstanding this, my
Yankee acquaintance was determined to turn me inside out if he could; he,
therefore, was so complimentary as to say—
"Well, now, I guess you are a gentleman?"
To this I answered—
"Thanks; I trust your surmise is a correct one;" and I might have said,
but I did not, "Sorry I cannot return the compliment."
I have often heard of the pertinacity of an American reporter, but it
appears to me that the bump of inquisitiveness is not by any means
confined to them, but pervades the whole community. There was no
shilly-shallying, no delicate, nicely-worded hints and adroitly-put
questions; but my interrogator was determined to find out all about me if
he could, and so he asked me how long I had been in Cairo, how old I was,
if I was married or single, how many children I had, if I lived on my
money, and lots of the most impertinent questions, and finally finished
up by saying, "Guess you are a Britisher?"
Having, as he thought, pumped me pretty considerably, he was good enough
to take me into his confidence, and tell me all about himself, and his
belongings, and "_hew_ his father had left him a pile," adding, "Guess I
spend some, and move about a bit." I could not help saying—
"I think you are wise to pursue that course; travel will improve you a
good deal, and, like the marble statuary in the Baulac Museum, it will
put on a little polish."
He eulogised the States and the inhabitants thereof, and was apparently
under the impression that America was the only place worth speaking of,
winding up with the quite unnecessary announcement—
"I'm 'Merican."
"Oh, yes," I replied; "I knew at once you were an American."
"Yes; is that so?" said he. "Hew did ye know that, stranger?"
"Well," said I, "by your accent, the estimate you form of your country,
and, pardon me for saying so, but no one but an American would have asked
me such questions as you have, or manifested such a desire to find out
all about me and my affairs."
He did not appear to be at all annoyed at this remark, but merely said—
"By thunder! stranger, you are a queer <DW53>. Will you come and liquor?"
I declined with thanks, and left young America to ponder over the
inscrutable ways and manners of the "darned Britisher." He was evidently
the offspring of a parent who, perchance, had "struck ile," and had never
before forsaken his ancestral home in search of travel and adventure;
and, if such was the case, we must excuse the young man. As soon as I
left him I sought my bedroom, chronicled the above conversation in my
diary, and retired to bed, where I slept soundly.
The following day I and three others formed a party for a visit to the
far-famed pyramids of Gizeh. We chartered a carriage, taking our lunch
with us; and from the time we left Shepheard's Hotel until we returned
that hateful word, "Backsheesh," resounded in our ears; indeed, I should
say that there is no word in the Egyptian language so frequently on the
tip of an Arab tongue as that. I should suppose that the pyramids of
Gizeh are about ten miles from Cairo. There is a pretty good road, which
was constructed by the former Khedive, Ismail, specially to accommodate
the Prince of Wales when he visited the place some years ago. During
our drive we could almost have imagined that a line of sentries had
been posted all along the road specially to utter that horrid word,
"Backsheesh," so continuously were our tympanums offended with it.
Arrived at the base of the Great Pyramid, we are immediately surrounded
by a considerable number of Arabs, who are all anxious to assist us
in the ascent, of course, for a small consideration; but one of our
party having been there once before, knew how to set about matters in a
business-like way, so he demanded at once the presence of the Pyramid
Sheik, who very soon came. We told him we did not want all this crowd
of Arabs, but two each would be sufficient. Accordingly he allotted us
these, but as he suggested that a third would be desirable to push us up
from behind, we had him. Those who have ascended the Great Pyramid are
not likely to forget the dusky demons who accompany them.
I commence the ascent with my body-guard, who appear now to look upon me
as a piece of brittle china, and are most anxious to prevent me using
my limbs in my own way—they will not let me take a step without their
assistance. Directly I had started I found my body-guard considerably
augmented, and notwithstanding repeated warnings that I did not want
them, and that they would not get any backsheesh, they stuck to me
all the way up and back. For the time being I belong solely to these
energetic, incessantly-chattering Arabs, whose most strenuous efforts
were now put forth to damaging my ball and socket joints.
I have to ascend 203 steps—the lower steps are about four feet high, and
few of them less than three feet anywhere. The two Arabs in front get on
to the step I have to land on; each seizes an arm, one gets behind, and
the hoisting process begins. The latter gives the cue, and with a loud
"Ha-hu," up I go from one step to the other. This game goes on with great
rapidity, until I had got about half-way up, where I think it advisable
to rest awhile. So down I sat, but soon found that instead of three Arabs
I was at once surrounded by about a dozen, all talking most vehemently
to me at the same time. It was in vain to protest—all had curiosities,
scarabei, little images, and ancient coins, some of them curious, no
doubt dating back to the time of Adam. For a small consideration they
were all anxious to place these in my possession, and all were shouting
into my ear, "Autica efendi, autica." At last, for peace sake, I bought
a small image of a defunct Pharaoh from one, and from another three or
four copper coins, all, of course, the only genuine. In vain I protested
against having any more. I had no peace until I had bought something from
each one; in fact, I had no quiet until I turned all my pockets inside
out, showing conclusively that I had spent every piaster with them.
After resting awhile, we continued the ascent, and after a succession of
ha-hus, tugs, and hoistings, I at last found myself on the summit of the
Great Pyramid, and well rewarded for the trouble I had taken.
Here, in this bright, clear atmosphere, I saw stretching out for miles
on the west the Libyan Desert, and reaching out before and around us in
vast extent the classic and historic hills, rivers, and plains of Ham and
Mizraim, Heliopolis, Memphis, Mount Mokattam, Sakhara, the beautiful city
of Cairo, with its numberless mosques and slender minarets, skirted by
the outstretched Nile, bearing on its placid bosom hundreds of dahabeahs,
and on its banks tall waving palm-trees. Nearer is the village of Gizeh,
and closer still the remaining pyramids of Gizeh, the granite temple,
and the sphinx, the whole forming a picture that cannot be effaced in a
life-time.
It is said with some truth, "Time tries all," but I have also heard it
said that "the pyramids try time," and, upon my word, it almost seems
so, when we think of their great antiquity. Here they have stood for
thousands of years in majestic grandeur, looking down on many Pharaohs
and many dynasties, and witnessing the rise, greatness, and decline of a
once mighty nation. Abraham, Isaac, Jacob, Joseph, and Moses have gazed
on these huge piles of masonry, which raised their lofty heads long, long
before Abraham in a day of famine sought bread at the hands of Pharaoh.
When I had spent some time in gazing again and again on this beautiful
scene, and thoroughly succeeded in obtaining a mental photograph of it, I
commenced the descent, and found I could get down much more comfortably
without assistance than with; but this the pertinacious Arabs would not
hear of, they said, on account of the danger to me. But my own private
opinion was that they wanted to earn good backsheesh by persistent
attention. I resigned myself to my fate, and at last reached the foot
of the pyramid by a series of jumps and bumps very trying to my spinal
column, and which joggled my internal economy most unpleasantly.
After a short rest, we explored the interior, a rather difficult
achievement in some parts. We had brought a good substantial luncheon
with us from the hotel, which we thankfully disposed of at a house, or
palace, near by. This was specially built, I believe, either for the
Prince of Wales or the Empress Eugenie, I really forget which. After
lunch we visited the sphinx, two or three tombs, and the other two
pyramids, settled up with the sheik, and drove off to our hotel; and
not until I reached the steps of the hotel did I hear the last of that
hateful word "Backsheesh." When I retired to rest I dreamt of a pocketful
of large copper coins and scarabei, an armful of defunct Pharaohs, an
army of lithe, sinewy, swarthy, impecunious Arabs, amongst whom I had
scattered a ship-load of piasters, and "still they were not happy."
Before I have done entirely with the pyramids, I think I ought to say
something about them, as those at Gizeh are the most remarkable. This
group consists of nine, and comprises three of the most remarkable
monuments in existence—those of Cheops, Cephren, and that of Mycerinus,
the last-named much smaller than the other two. Herodotus, who was born
about 500 B.C., tells us that in building the great pyramid of Cheops
it took 100,000 men working incessantly for 30 years to complete it;
10 years of this 30 was spent in making a causeway 3,000 feet long, to
facilitate the transportation of the stone from the Turah quarries.
Herodotus describes the method of building by steps, and raising the
stones from layer to layer by machines, and finally of facing the
external portion from the top down. Its present height is 460 feet, the
original height was 480 feet.
The extent of solid masonry has been estimated at 82,111,000 cubic feet.
It at present covers 12 acres. The only entrance is on the north face,
49 feet above the base, though the masonry has been so much broken away
that the _débris_ reaches nearly up to it. A passage, 3 feet 11 inches
high and 3 feet 5½ inches wide, conducts from the entrance down a <DW72>
at an angle of 26° 41´, a distance of 320 feet 10 inches to the original
sepulchral chamber, commonly known as the subterraneous apartment; it
is carried, reduced in dimensions, beyond this a distance of 52 feet
9 inches into the rock, though for what purpose remains a matter of
conjecture. The sepulchral chamber is 46 feet long by 27 feet wide, and
11½ feet high. From the entrance passage another branches off and leads
to several other passages and chambers. One of the latter, known as the
_Queen's Chamber_, is situated about the middle of the pyramid, 67 feet
above the base; it has a groined roof, and measures 17 feet broad by
18 feet 9 inches long and 20 feet 3 inches high. The other, called the
_King's Chamber_, is reached by an offshoot from the Queen's Passage, 150
feet long. Its dimensions are 34 feet 3 inches long by 17 feet 1 inch
wide, and 19 feet 1 inch high. The chamber is lined with red granite
highly polished, single stones reaching from the floor to the ceiling,
and the ceiling itself is formed of nine large slabs of polished granite
extending from wall to wall. The only contents of the apartment is a
sarcophagus of red granite, which, judging by its dimensions, must have
been introduced when the building was proceeding. It is supposed to have
contained a wooden coffin with the mummy of the king, and that these long
since disappeared when the pyramids were first opened and plundered. We
do not see these pyramids as they originally were. The outer casing of
polished stone has been removed and utilized in constructing the mosque
of Sultan Hassan. These pyramids were built between 5,000 and 6,000 years
ago. Great was the antiquity of Thebes before European history begins to
dawn. It was declining before the foundations of Rome were laid, but the
building of the great pyramids of Gizeh preceded the earliest history
of Thebes by 1,000 years. Whilst speaking of Thebes, I'll just mention
that there are to be seen to-day the tomb of the great Sethos, Joseph's
Pharaoh, of his greater son, Rameses II., and of Menophres or Meneptha,
in whose reign the Exodus took place. In the tomb of Sethos,
sculptures cover 320 feet of the excavation. There is to be seen the
draughtsman's handiwork in red colour, showing the designs that were to
be executed by the sculptor, and the corrections in black ink of the
superintendent of such works, and although these sketches were made 3,000
years ago, they are still quite clear and fresh-looking. On the east side
of the pyramid, half buried in sand, is the wonderful colossal Sphinx,
his head 25 feet high and back 100 feet long, all one stone.
CHAPTER VI.
HELIOPOLIS—THE SHOUBRA
ROAD—BEDROSKYN—MITRAHENNY—MEMPHIS—SAKHARA—APIS
MAUSOLEUM—WORSHIP OF THE BULL APIS—TOMB OF KING PHTA—MEET THE
KHEDIVE—ENGAGE SERVANTS FOR THE SOUDAN.
My next visit was to Heliopolis on donkey-back. I was told that it would
be a nice ride, but nothing to see except an obelisk when I got there.
Notwithstanding this, I felt very desirous of visiting this ancient seat
of learning, where Moses had lived and "become learned in all the wisdom
of the Egyptians." Accordingly Ibrahim and I started off. Leaving the
citadel and tombs of the Caliphs on my right, I had a pleasant ride of
about two hours or so from Cairo through avenues of acacias and tamarisk
trees, a large plain covered with a luxuriant growth of sugar-cane,
citrons, lemons, oranges, ricinus, cactuses, olive trees and palms.
Before reaching the mounds of Heliopolis is a well of fine water on
the border of a grove of citrons and palms, and in the midst of these
is a venerable old sycamore enclosed by palisades and regarded with
veneration by the Copts, as the place where Joseph, Mary and the infant
Saviour rested on their flight into Egypt. Although a very aged tree,
it cannot be, of course, as old as the legend affirms. It is, however,
a very pretty spot, sheltered from the busy hum of life, embowered in
citron thickets, which resound with the music of birds, and with tall,
waving palm trees, on the trembling branches of which large vultures
rock to and fro. I approach the site of Heliopolis on a dead level, and
find that it stood formerly on an artificial elevation, overlooking
lakes which were fed by canals communicating with the Nile. With what
history does this place teem! Here, or in the vicinity, Jeremiah wrote
his Lamentations. Thales, Solon, Pythagoras and Plato studied here. From
the learned priests of Heliopolis, Plato—who studied here for several
years—is believed to have derived the doctrine of the immortality of the
soul and of a future state of rewards and punishments. This neighbourhood
was probably the scene of the Exodus of the Israelites, and here was
the most celebrated university in the world for philosophy and science.
It was here that Potipherah, the priest or Prince of On, resided. Here
Joseph married his daughter Asenath, who became the mother of Ephraim
and Manasseh. Now what do I see? This once famous city of the sun, the
Heliopolis of Herodotus and Strabo, the On of Joseph, the Bethshemesh of
Jeremiah, the university of the world at that time, with its collection
of colleges and temples, avenues of sphinxes and extensive dwellings of
the learned priests, dazzling palaces, obelisks and splendid edifices
has been almost blotted out, and as I stood there absorbed in thought,
and feebly endeavouring to picture to myself this place as it once
stood, teeming with life, wealth and power, those beautiful words of
Shakespeare, our immortal bard, came floating through my mind as very
descriptive of what I now saw—
The cloud-capt towers,
The gorgeous palaces,
The solemn temples,
The great globe itself,
Yea, all which it inherit,
Shall dissolve,
And like the baseless fabric of a vision,
Leave not a wreck behind.
All was now desolation, if I except the massive foundations of the Temple
of the Sun, which are still visible in a few places. The one solitary
object that serves to mark this once celebrated city is an obelisk of
solid granite, 62 feet high, the last monument of a temple that once
vied in magnificence with those of Karnak or Baalbeck, and which has
been pointing to the sky from the time of the old monarchy for more than
4,000 years. It bears the name of Osirtesen I. (Joseph's contemporary),
the first great name in Theban history, builder of the older and smaller
part of the great temple of Karnak and King of Upper and Lower Egypt,
and probably where I then stood looking at, but unable to decipher the
hieroglyphics on this obelisk, Joseph and Moses (who had both been
admitted to the priest cast) had stood before me. _Sic transit gloria
mundi._
I had now seen all there was to see, and was pleased that I had made this
visit, so I mounted my donkey and got back to Cairo. It happened to be
Friday, the Mahomedan Sunday. On this day all the rank and fashion can be
seen between four and six driving up and down the Shoubra Road. This is
lined by a splendid avenue of trees, which meet over-head, thus forming
a delightful shade. It was now about 4 p.m.; I performed a hasty toilet
and set off for a carriage drive down this road. I found it thronged with
visitors and a goodly sprinkling of officers, amongst whom I saw the now
famous Arabi Pacha. Mounted sentries also were posted at intervals each
side of the road as the Khedive usually takes a drive there every Friday
about 4 or 5 p.m. I had not been there long ere he came sweeping down
with his escort.
Next day I devoted to exploring the ancient (probably the most ancient
city in the world), Memphis, the Noph of the Bible, and its necropolis,
Sakhara. According to Herodotus its foundation was ascribed to Menes, the
first King of Egypt. If this was so it would be about 6,000 years old,
and it is said that the art of building was known centuries before his
time.
It is quite a good day's work to perform this journey in the blazing sun.
I get an early breakfast and leave at 7.30 on my donkey, accompanied by
Ibrahim on another donkey, in possession of my luncheon. The distance
to the railway station is about two miles. Here I procure tickets for
ourselves and the two donkeys, proceed to Bedrashyn, a distance of about
ten miles, then remount and pass through the village of Mitrahenny, then
a very fine palm-grove, on to the site of ancient Memphis, once a large,
rich, and splendid city, remarkable for its temples and palaces. As late
as 524 B.C., at the time of the conquest of Cambyses it was the chief
commercial centre of the country, and was connected by canals with the
Lakes Mœris and Mareotis. Some distance from the village of Mitrahenny I
saw near the pathway a colossal statue of Rameses the Great in excellent
preservation. It is composed of a single block of red granite, polished.
It was originally 50 feet in length, but has been mutilated, and now
does not measure more than 48 feet. It lies on its side in a pit by the
wayside, which, during the inundation of the Nile, is filled with water.
On its subsidence the alluvial deposit is scraped off sufficiently to
show the statue to travellers. Vast mounds of broken pottery and statuary
are to be seen about here and Sakhara, probably burying the ancient city.
Sakhara is about two miles or so from Memphis, and the greater part of
the ride lies through sandy desert. It lies, in fact, on the edge of the
Lybian Desert. It is remarkable for its ancient monuments, among which
are 30 pyramids. The great step pyramid is said to be even older than the
pyramids of Gizeh. Besides these 30 there are the ruins of a great many
others, and numberless grottoes, sarcophagi, the Ibis catacombs, and Apis
Mausoleum, which was discovered by Mariette Bey. He observed the head of
a sphinx protruding from the sand, and remembering that Strabo described
the Serapeum of Memphis as approached by an avenue of sphinxes, he at
once commenced his explorations in search of the temple in which Apis was
worshipped when alive and the tomb in which it was buried when dead. The
sand-drift, after immense exertions, was cleared away, and the avenue was
laid bare from a superincumbent mass, which was in some places 70 feet
deep. Conceive, if you can, the splendour of this imposing approach; no
less than 141 sphinxes were discovered _in situ_, besides the pedestals
of others. The temple to which they led has disappeared, but the tomb
remains.
I go down hill, nearly up to my knees in sand, with my guide. A great
door is unlocked and thrown open, we then light our candles and explore.
We proceed a considerable distance through a passage or tunnel, and
then find ourselves in a large vault or tunnel some 200 or 300 yards in
length. Chambers lead out of it on either side as large as an ordinary
sitting-room, and about 12 feet high, in each of which is a ponderous
granite sarcophagus, polished. Placed on the sarcophagus like a lid was a
granite slab of great size and weight, the whole weighing about 20 tons.
Near the subterranean cemetery of the bulls are the groves or pits of the
sacred Ibis also formerly worshipped. These are enclosed in earthenware
vases; the bones and broken urns now lie scattered all around. These huge
blocks of granite were actually transported from the quarries near Syene
to Memphis, a distance of nearly 600 miles! I carefully examined one
sarcophagus containing the embalmed dead deity. It was carved all over
with sacred hieroglyphics, sharp and clear in their outlines, and the
polish on the marble bright as it was 3,000 years ago. I saw between 30
and 40 of these sarcophagi here.
The worship of the bull Apis was celebrated with great pomp and
splendour, and he was regarded as the representative of Osiris.
His interment would cost as much as that of any king or conqueror. It
was necessary that he should be black with a triangle of white on the
forehead, a white spot in the form of a crescent on the right side,
and a sort of knot like a beetle under his tongue. When a bull of this
description was found he was fed four months, in a building facing the
east. At the new moon he was led to a splendid ship with great solemnity
and conveyed to Heliopolis, where he was fed 40 days more by priests and
women, who performed before him various indecent ceremonies. After this
no one was suffered to approach him. From Heliopolis the priests carried
him to Memphis, where he had a temple, two chapels to dwell in, and a
large court for exercise. He had a prophetic power which he imparted to
the children about him. The omen was good or bad according as he went
into one stable or the other. His birthday was celebrated every year
when the Nile began to rise; the festival continued seven days. A golden
patera was thrown into the Nile, and it was said that the crocodile was
tame as long as the feast continued. He was only suffered to live 25
years, and at his death he was embalmed and buried in these sarcophagi
amidst universal mourning till the priest had found a successor.
When I emerged once more from this mausoleum and struggled up through
the sand I paid a visit to the tomb of King Phty or Phta, said to be
5,400 years old. His sarcophagus is similar to those I had just visited,
and is contained in a nice lofty room, the walls of which, as are the
walls of the chapel outside, plentifully and excellently sculptured,
and quite fresh in appearance, though so ancient. I do not remember
all I saw represented on the walls and tombs, but amongst other things
there were lions, giraffes, ostriches, sacred Ibis, owls, crocodiles,
elephants, buffaloes, a boat floating on the water with a man in it,
and in the water fish of different kinds, Egyptians fishing, harpooning
the hippopotamus, agricultural pursuits, ploughing and sowing, treading
out the corn just as they do now, the butcher sharpening his knife, the
butcher killing the animal whilst another holds him down, hunting,
battle scenes, &c., &c. Some figures on the wall had been painted red;
the paint is still good and not at all frayed. In another excavation,
after leaving this tomb, I saw a mummy; but I must not expend too much
time over this place, although I feel quite disposed to keep on talking
of it. We cannot leave the plain of Memphis without recurring to the most
memorable event in all its eventful history. It was probably here that
Moses and Aaron stood before Pharaoh and demanded that he should let the
people go. This was the spot where "Pharaoh rose up in the night, he and
all his servants, and all the Egyptians; and there was a great cry in
Egypt, for there was not a house in which there was not one dead."
Ruminating on the mutability of human affairs, I mounted my donkey, had
a long ride through beautiful palm groves, and finally emerged from
the village of Gizeh on to the main road from the pyramids and over
a handsome bridge across the Nile to my hotel. When half-way across
the Nile, I observed the Khedive and his escort coming along, so I
got off my donkey to watch him pass. I took off my hat to him, and he
acknowledged my salutation with a gracious bow. As I returned homewards,
in imagination I saw these glorious cities of old Egypt peopled. I tried
to picture to myself—feebly, I dare say—the splendour and wealth of
those people, the magnificence of the designs carried out, the result of
which was that neither before nor since has the sun shone on anything
like such superb, massive, and imposing temples, palaces, and tombs in
the world. Thebes, with its hundred gates, was perhaps the most splendid
city in the world for many centuries. Then there were Luxor, Karnak,
Philæ, Elephantine, Baalbeck, Dendera, Aba-Simbal, Abydos, Esneh,
Edfau, Silsilis, and other places, all decorated with palaces, temples,
pyramids, tombs, and sphinxes, &c., on the same magnificent scale; but
all have shared the same fate, and their stupendous ruins are all that
remain to strike the stranger with awe and wonder.
About two days after our arrival in Cairo, our party was augmented by the
arrival of Mr. W. D. James, Mr. A. James, and Mr. Percy Aylmer, Mahoom,
a black boy; who had been rescued from the Soudan some years beforehand;
Jules, George, and Anselmia, the three latter European servants. Here we
engaged Suleiman as a sort of general manager for the caravan; he had
travelled through the Soudan with Sir Samuel Baker; Ali, a very good
cook, and Cheriffe, who made a very good butler, and had been accustomed
to travel as a kind of steward on the Nile boats.
CHAPTER VII.
THE LAND OF GOSHEN—ANCIENT CANALS—SUEZ—HOWLING
DERVISHES—ECLIPSE OF THE MOON AND STRANGE BEHAVIOUR OF
NATIVES—LEAVE SUEZ—WHERE THE ISRAELITES CROSSED THE RED
SEA—PASS MOUNT SINAI—CORAL REEFS ABUNDANT.
Our next move was on to Suez by rail, a day's journey through another
very interesting portion of Egypt, the land of Goshen, the home of the
Israelites for 430 years. A good deal of country near the line of railway
is now under good cultivation, supplied by the Sweet Water Canal. The
earliest attempt that we are acquainted with to construct a canal was by
Rameses the Great. It was between 50 and 60 miles in length, and left
the Nile at Bubastis, reaching into the neighbourhood of Lake Timsah.
Upon it Rameses built his two treasure cities, Pithom and Raamses near
Ismailia, mentioned in the 1st chapter of Exodus, and there is little
doubt that the Israelites, who were then in bondage, laboured at these
cities, and the canal 3,000 years ago. It is probable also that the
canal dated far back beyond this time, for the Egyptians had been great
in canal making 1,000 years or more before then. One of the greatest
marks of Rameses was the covering the whole of Egypt with a net-work of
waterways in connection with the river. They served a double purpose—they
greatly extended the supply of water and the area of cultivation, and
were invaluable for defensive purposes. Many centuries after this Pharaoh
Necho took this canal in hand 500 or 600 years B.C. He undertook to adapt
it for navigation and prolong it to the head of the Arabian Gulf. He is
the only Egyptian monarch whose name appears connected with maritime
enterprise, and he was so zealous as to perfect the formation of a ship
canal connecting the Nile with the Red Sea. He carried the great work as
far as the Bitter Lakes, and then abandoned it, warned by an oracle to
desist, after expending the lives of 120,000 fellahs. Herodotus actually
_saw_ the docks, which as a part of the plan, he had constructed on the
Red Sea. One conqueror succeeded another, and the works got neglected
and the canal choked up. The Romans again carried on extensive repairs
and alterations, but on the downfall of the Roman Empire anarchy and
confusion prevailed, and all public works were allowed to fall into
dilapidation. The canals were choked up, and remained unnavigable till
the Arab conquest of Egypt. Under the vigorous administration of Amrou
they were re-opened, and corn and other provisions were conveyed along
them for the use of Mecca, Medina, and other Arabian towns. A very great
deal could be said about their ancient canals, but I have only time to
glance _en passant_ at a little of the ancient history of the places I
passed by. In the evening we arrived at Suez, 76 miles east of Cairo.
There is very little to interest or amuse at Suez, but here we were
obliged to remain for nearly a week by reason of stoppages in the canal,
which are frequent. The day after our arrival we took donkey rides down
the Mole, which is 850 yards long, to see after our provisions, tents,
&c., which Mr. James and his friends had got together for our campaigning
in the Soudan. We found them, and there sure enough was a stambouk (a
native boat something like a fishing smack) not only full but piled
up with everything that we could possibly require, and the collecting
of which must have necessitated a great deal of forethought. Two days
after our arrival, Mr. J. B. Colvin, of Monkham's Hall, Waltham Abbey,
arrived by steamer from Australia, to join us, thus completing the party.
During our stay here there happened to be an eclipse of the moon. This
appeared to have a very disturbing influence on the native element, as I
should think that every tom-tom in Suez was called into requisition and
incessantly beaten all over the town during the eclipse to drive away
the evil spirits. If it did not succeed I have no hesitation in saying
that all the good spirits (ourselves) would very soon have vanished if
we could. We had ample time to explore the town both by day and night,
and amuse ourselves as well as we could by donkey rides down the Mole,
boating, fishing and bathing, but whilst bathing we were careful not
to go far from shore for a header or remain in long, as sharks are so
plentiful in the Red Sea. One evening, two or three of us were wandering
about at night and heard strange noises issuing from a small building.
We were sufficiently inquisitive to go up a narrow passage to ascertain
the cause. There we found about a dozen very dirty howling dervishes in
the odour of sanctity (a decidedly strong odour we thought) performing
their senseless and absurd mode of worship with great energy. They were
in a dirty room, having a damp, uneven, earthen floor, the dimensions of
which were about 7 feet high, 7 feet wide, and perhaps 10 feet long.
Very little light or air could find its way in. The weather was very hot,
and the sudoriferous glands of these unsavoury gentry were in an abnormal
state of activity. Need I say that we remained here a very short time?
We were all thoroughly tired of Suez, and anxious to get on to Souâkin,
but unfortunately, amongst all the steamers blocked in the Suez Canal, we
could not hear of a single one bound for Souâkin. The _Agra_, a British
India steamer, was bound for Jeddah, on the opposite coast, so Mr. James
telegraphed to London, asking the Company to let us be taken to Souâkin.
They acceded to the request. Accordingly, on the 8th December, we got on
board, unloaded the stambouk, and started off for Souâkin, _the_ port
of Nubia, and indeed of Central Africa, since made historical by our
slaughter of thousands of Arabs in that neighbourhood. The places of
interest pointed out to us on the Red Sea coast were Moses' Well, Mount
Sinai, and the spot where the Israelites crossed. Here the arm of the sea
is 12 miles wide, and just here Pi-hahiroth before Baal-zephon is the one
and only opening in the mountains. Here one million and a half of the
Israelites—men, women, and children—passed through in the night, whilst
the army of Egypt pursued them. After a most agreeable but very warm
voyage (90° F. in the shade) of 3½ days we reached Souâkin. During our
last day at sea Captain Smith was very careful in his navigation, as the
Red Sea, particularly in that last day's voyage, abounds in coral-reefs.
[Illustration: LANDING PLACE AT SOUÂKIN.]
CHAPTER VIII.
ARRIVAL AT SOUÂKIN—THE SOUDAN—BEDOUIN ARAB PRISONERS IN
THE SQUARE, NOT "ON THE SQUARE"—IVORY—ENGAGE CAMELS—SHEIK
MOUSSA—SOUÂKIN—SLAVES—TRAGIC END OF A DOCTOR—HADENDOWAH
ARABS—AN ILL-FATED MISSIONARY ENTERPRISE.
We were now just about to land in the Soudan, and as that word is, as I
am writing, in everyone's mouth, it would be as well to say something
about it before I go any further.
The Soudan, or Beled-es-Sudan, Land of the Blacks, has since the Middle
Ages been the common name of the vast extent of country in Central
Africa, which stretches southward from the Desert of Sahara to the
Equator. The name was originally applied by the Arabs, but with great
latitude of signification, different authors giving it to the different
parts of the territory with which the varying routes across the desert
made them acquainted. Later geographers divide it into High and Low
Soudan. Many include Senegambia in it. High Soudan stretches from the
sources of the Niger, Senegal and Gambia, to the Upper Nile, or, at all
events, to the south of Lake Chad, and embraces the mountains of Kong
and of Upper Senegambia, the kingdoms of Ashantee, Dahomey, Mandingo,
Houssah, and Feelah. All this country is richly watered and wooded,
distinguished by a luxuriant tropical vegetation and by deposits of gold.
Low Soudan stretches on the north of High Soudan, eastward to Kordofan,
and northward to the desert. This district is partly level, partly
undulating, and partly broken by chains of lofty hills rising within
its own limits. Its situation between the desert on the north and the
mountains which border it on the south, with a climate destructive to
foreigners, and a lawless and predatory population, make it one of the
most inaccessible regions in the world. In the south, where it is watered
by the Niger, Lake Tchad, and their tributaries, it assumes a fertile
and cultivated appearance. The inhabitants contain numerous nations of
different races, chiefly of the <DW64>, Fulde, or Fellatah stems, together
with many Arab colonists.
This is what Sir Samuel Baker says about the Soudan in the _Contemporary
Review_: "Before the White Nile annexation the Soudan was accepted as
a vague and unsatisfactory definition as representing everything south
of the first cataract at Assouan, without any actual limitation; but
the extension of Egyptian territory to the Equator has increased the
value of the term, and the word Soudan now embraces the whole of that
vast region which comprises the Deserts of Libya, the ancient Merve,
Dongola, Kordofan, Darfur, Senaar, and the entire Nile Basin, bordered
on the east by Abyssinia, and elsewhere by doubtful frontiers. The Red
Sea alone confines the Egyptian limit to an unquestionable line. Wherever
the rainfall is regular the country is immensely fertile; therefore the
Soudan may be divided into two portions—the great deserts which are
beyond the rainy zone, and consequently arid, and the southern provinces
within that zone, which are capable of great agricultural development.
Including the levels of the mighty Nile, a distance is traversed of about
3,300 miles from the Victoria N'yanza to the Mediterranean; the whole of
this region throughout its passage is now included in the name 'Soudan.'"
We had on board Captain Gascoigne and Dr. Melidew, of the Royal Horse
Guards. They were also bent on a shooting expedition in the Soudan,
but did not accompany us farther than Souâkin. There were several other
passengers on board bound for India.
We landed at Souâkin on the quay, in a large open square. One side is
occupied by what is absurdly called the palace, a large building in which
the Governor transacts his official duties, the opposite side by the
custom-house, the other by a guard-house, whilst the opposite side was
not occupied by any building, but was open to and contiguous to the Red
Sea; it was, in fact, the quay.
Here I saw nine tons of elephants' tusks ready for shipment. The average
weight of each pair of tusks would be somewhere about 36lbs. I computed
that about 560 elephants would have been slaughtered to make up nine
tons of ivory; and if elephants are killed at that rate, people may well
exclaim about the scarcity of ivory. What next attracted my attention was
about 60 Bedouin Arabs in heavy chains, wandering about in this large
open square. These poor fellows had to pay their gaolers 100 dollars
a month. The Maria Theresa dollar which is in use in the Soudan, and
preferred to any other coin, is worth 4s. of our money. They had to
find their own food, or rather their tribe did so. I was told that at
one time they were a strong tribe, and had come over from Arabia. They
had at one time 8,000 camels, but they had dwindled down to 2,000, as
whenever they failed to pay the taxes some of their camels were seized.
I cannot speak with any certainty of their offence, but somehow or other
they had incurred the anger of the then Governor of the Soudan, Ali Riza
Pacha, about a year beforehand. He clapped them into irons, and there
they seemed likely to remain, unless some more kindly-disposed Governor
superseded him. This fortunately happened not long before our return to
Souâkin in the following April, when Ali Dheen Pacha was appointed, who
soon liberated them.
The inhabitants of Souâkin are principally Arabs, a few Greek and Italian
merchants, and two Englishmen. The Government usually have a garrison
of about 300 Nubian troops stationed in an undefended barrack on the
mainland, about a mile from the town.
Blind to their own interest, the Egyptian Government obstructs traffic by
the heavy duties which it levies. Cattle and sheep, which can be obtained
from the tribes in the neighbourhood, are sent by hundreds annually to
Suez by sea. Were it not for the heavy duties imposed, I should say that
a large trade ought to be done with Suez, which is but three and a half
days from Souâkin. There is a telegraph line to Kassala. They have large
numbers of camels for sale or hire, but no horses, mules, or donkeys. The
water is collected during the wet season in a large reservoir about a
mile from the town; there are also two or three wells at the same place.
We soon introduced ourselves to Mr. Brewster, an Englishman, and head
of the custom-house; and he in turn sent for Achmet Effendi, the Civil
Governor of Souâkin, to whom he introduced us. Of course, there followed
the inevitable salaaming, coffee and cigarettes, so customary in the
East. Our business was very soon explained; we wanted about 80 camels
provided without delay to transport ourselves and our baggage across
the desert to Kassala. The camel sheik, Moussa, was sent for, and soon
appeared—a really picturesque, handsome-featured man, almost black,
possessed of gleaming, regular teeth, wearing a snow-white turban and
loose white robe, precisely like the ancient Roman toga. _En passant_, I
cannot help thinking that the slang word "togs" is derived from the word
toga.
The Sheik Moussa promised to provide us with the camels within three
days; and, strange to say, he did so, a singular instance of a man
keeping his word to one in the East. I know that my experience amongst
the officials in Turkey was very different—there everything was put off
until to-morrow. A day would be fixed for me to call at the Seraskierat,
or War Office, and when I went I was usually met with the reply, "Yarrin
sabbah, effendi" (to-morrow, sir), or "Ywash, ywash" (by and bye), not
once or twice, but I daresay five or six times. Another inconvenient
phrase which is always on their lips if one wants any money from them,
and which is spoken trippingly on the tongue, is "Para yok" (no paras),
in English, "I haven't a farthing."
It soon became known that there was a "Hakeem Ingelese," as they called
me, in our party, and I very soon had many patients, amongst whom was a
child of one of the Bedouin Arabs.
In the afternoon I improved my acquaintance with Mr. Brewster, who had
officially resided here four years, and, of course, knew most of the
people and the customs of the place. There are a great many good and
curiously-built houses with flat roofs, built of blocks of white coral,
and a great many tent-like structures constructed with reeds, stalks of
palm leaves, and matting, which is very cheap and abundant, made by the
natives out of palm leaves. Mr. Brewster was good enough to escort me
over Souâkin, and give me all the information he could about the place
and people. As we strolled on he pointed out the home of a slave-dealer,
who then had several slaves—children and young girls. These could easily
be transferred as ivory, dhurra, or something of the kind, as old Achmet
Effendi connived at slave-dealing, and would shut his eyes to the
transaction provided his palm was crossed with a couple of dollars per
head. The little children realize from 30 to 40 dollars a head, and young
girls 70, 80, or 100 dollars.
"Why," said I, "in England it is supposed that the slave trade has been
abolished in Egypt long ago. When in Cairo I saw the slave-market, but
was told no slaves have been sold there for the past three or four year."
"Ah," said he, "you will find, when you get further into Africa, that it
is still carried on, and more openly than it is here. When they have been
captured they are driven across the desert just like cattle to some quiet
place on the Red Sea coast, where there is a stambouk waiting; there
shipped and taken across to Jeddah in a day or so, and sold by public
auction." The only other Englishman resident at Souâkin was Mr. Bewlay;
he had at once lived in Jeddah for a time, and he assured me that he had
often seen slaves sold there. _Apropos_ of my profession, Mr. Brewster
related a very interesting, and, to me, a very instructive anecdote,
which served to enlighten me considerably as to the peculiar line of
thought which sometimes permeates the native brain, and to the still more
peculiar line of action which it leads to. He told me that about three
years or so before our arrival a German doctor, who had settled there,
whilst attending a native, had occasion to perform some trivial operation
which was not attended with the success which he desired or anticipated,
as unfortunately for the native, and subsequently for the doctor, the
former was so inconsiderate as to expire a day or two afterwards. The
doctor could truly say after this, "A doctor's lot is not a happy one,"
inasmuch as the friends of the defunct Arab paid him a visit, and in a
marked but highly objectionable manner, showed what they thought of the
doctor's services in a way that did not commend itself to me, and which,
for want of a better illustration, we will call "a new way of paying
old debts." The worthy leech was requested, in so pressing a manner
that refusal was out of the question, to accompany these friends of the
deceased, and _nolens volens_, they escorted him to a large open space
just outside the town, where dhurra and other things were sold, and there
they remunerated him, not in dhurra, not in sheep, not in goats, not
even in money, but in a most cutting manner, for they fell upon him with
their knives and literally chopped him to pieces. Reader, "would you be
surprised to hear," that on learning this I was extremely careful not to
perform any rash operations, and that my ministrations to the lame, the
halt, the sick, and the blind, should be successful. At all events, it is
a source of great gratification to me that they were not so unsuccessful
as to necessitate the sudden and unlooked-for departure of any of my
patients to their happy hunting-grounds.
The Hadendowah Arabs are the most numerous tribe in the neighbourhood
of Souâkin, and are, for the most part, good-looking men; they are very
dark, approaching to blackness, have good, well-formed features, large
dark eyes, arched black eyebrows, and face, on which as a rule there is
little or no hair, and nearly every Arab, here and elsewhere, that I met
with, is possessed of the most beautifully white, regular, and sound
teeth possible. There is little doubt but that this is due to the simple
manner in which they live; their chief food is dhurra (sorghum vulgare).
This contains 11½ per cent. of gluten, our wheat only ten per cent.
This is the wheat of Egypt, and is the food of camels, horses, and men.
Camels, however, get very little of it, as a rule, unless on a forced
march, or are owned by a man who can afford it. It grows to the height of
nine or ten feet, and is very prolific. I never counted the seeds in a
head of this sorghum, but Sir Samuel Baker did, and he says that in one
single head he found 4,840 grains. The Arabs, speaking generally, are not
big-boned men, but are lithe, active, and sinewy. Their hair is bushy,
frizzly, long, and black, which they wear very curiously; they often take
as much trouble with it as any West-end dandy would do. A parting is made
around the crown from one temple to the other; the hair on the top is
combed up and kept short—perhaps an inch long—the rest is combed down,
and stands out in a bush all round the head to a distance of three or
four inches; a thin piece of stick, like a skewer slightly bent towards
the sharp point, is stuck through the hair at the top, and is often used
to stir up the population, which is no doubt very numerous. I have often
seen their hair white with fat, which they plaster on most abundantly
when they can get it, and as few wear any covering over their shoulders
when they are exposed to the heat of the blazing sun, this drips down on
to them. They wear a bundle of charms secured just above the elbow, a
tope, or loin-cloth round the waist, which reaches down to their knees,
and very many a ring in one nostril. Nearly all of them carry a shield
and a long spear weighted at one end. The Hadendowhas are much given to
lying and laziness.
During the time that we remained here we were fully occupied in
preparing for our journey across the desert from Souâkin to Kassala,
a distance of about 280 miles; we cut up old boxes, made new ones,
and sorted out what provisions, &c., we should require. I arranged my
medicine-chest and surgical instruments so that I could get at what I
might want easily. We got a little shooting, sand-grouse, flamingoes,
pelicans, and herons; wandered about the town and frightened all the
children in the place, who thought we were slave-dealers come to steal
them. The principal slave supply is obtained from the White Nile and
Darfour; Khartoum, I believe, is the principal slave mart.
At nights we stretched ourselves out on the divan that ran round the room
in the palace, and slept head to feet all round. This room adjoined and
looked out on the square in which the Bedouin prisoners were confined;
frequently in the early morning they woke us up with their clanking
chains, or by indulging in their peculiar mode of devotion. The day
before we started on our journey, Mr. Brewster said—
"Well, Doctor, I hope you will all return alive and well, and not be so
unfortunate as a party that Dr. Felkin accompanied a year or two ago."
"I am sure I quite indulge in the hope of returning to England in a sound
state," I replied. "But tell me about the misfortunes of the party you
speak of."
"That is done in a very few words," said he. "Six missionaries went from
Souâkin and six from Zanzibar, meeting eventually in the wilds of Africa,
sent out by the English Church Mission Society, to reclaim lost sheep.
They were not happy in the selection of a suitable spot for evangelising,
as only three of them and Dr. Felkin returned to Souâkin, looking
considerably the worse for wear; the others had succumbed to fever,
dysentery, and spears. Indeed, I am not quite sure that some of them were
not eaten."
CHAPTER IX.
THE START ACROSS THE DESERT—MY CAMEL SERVES ME A SCURVY
TRICK—THE CAMEL, ITS HABITS AND TRAINING.
Three days after our arrival at Souâkin there were some very heavy
showers of rain. Mr. Brewster informed me that it was eighteen months
since it last rained there.
On the fourth day after our arrival about 80 hired camels were brought
into the large open square to be laden with the tents and baggage of
every description. I wish I could adequately describe the scene that
ensued—the camels groan and bellow without any provocation, as if
they were the most ill-used animals in existence; the Arabs shout and
wrangle with each other as they adjust the loads on the haweias (a kind
of pack-saddle), clutch one another by the hair of the head, after the
manner of women when quarrelling, and shake the offending head about most
vigorously. Our head-man, Suleiman, walks round and distributes his
favours very impartially—a tug of the hair for one, a box on the ears
for another, and a flick of the coorbatch (a whip made of hippopotamus
hide) for another. This scene lasted for about three hours, and when at
last they did start, they formed a very long hamlah, or caravan. The
head of one camel is tied to the tail of the one in front, a long piece
of rope intervening to allow for the long stride of the camel. We posted
our letters—the last for some time to come—for England, to say that we
were just starting on our Arab life across the Nubian desert. The caravan
having started, each of us sees to his riding camel being got ready. We
are some time in starting, getting our makloufas (camel saddles) properly
and securely adjusted, and our little belongings, such as rifles,
revolvers, saddle-bags, travelling satchels, &c., fixed on them. Each one
has a zanzimeer hung on to a strap by the side of the camel. The word
zanzimeer requires explanation; it is a large leathern bottle, capable
of holding three or four quarts of water. As, in our journey across
the desert, we should perhaps be sometimes two or three days before we
came to any well, we had to provide a water-camel, whose business was
to carry two large barrels full of water for domestic purposes. Each of
these had a padlock on them, so that the Arabs could not get at them
just whenever they felt inclined—a very necessary precaution, as they
are so very careless, would take the spigot out of the barrel, quench
their thirst, and as likely as not insecurely replace the plug, and let
the water waste, which would be a very serious calamity. The mode of
mounting and sitting on a camel is peculiar; my legs don't hang down each
side of him in stirrups, but hang down in front of the saddle each side
of his neck or crossed over the neck. No stirrups are used. The camel,
of course, is on the ground, with his legs tucked under him; I approach
his side and give a sudden vault or spring on to the makloufa. This must
be done with great dexterity and quickness, unless the attendant has one
foot placed on his fore-leg, as the camel gets up _instantly_ as soon as
I leave the ground, so of course, unless I am quick and dexterous, the
result is disasterous; in other words, the camel gets on to his legs,
and I go off mine on to my back. I watched the process of mounting very
carefully, as it was my first experience of camel riding. I attempted
and succeeded in doing the same as my pattern, and when my camel got up
(which he did pretty quickly, and not without considerable danger and
inconvenience to me), I felt that I occupied a very high and somewhat
precarious position. However, I soon got accustomed to the peculiar
motion of a camel. A hygeen, dromedary, or riding camel, can go on a
shuffling kind of trot (which is infinitely preferable to a fast trot or
walk) at the rate of about five miles an hour, and I am sure that anyone
who rides 25 or 28 miles a day, under the burning rays of an African sun,
will think he has done quite enough, although on some occasions we have
made forced marches and travelled 30 or 33 miles in one day. There were
no hygeens at Souâkin; we therefore rode our caravan camels. A hamlah,
or caravan camel, is capable of carrying considerably over 3 cwt. for
very long distances, travels at the rate of 2½ miles per hour, and will
go steadily on for 12, 14, or 16 hours without stopping to eat or drink.
He only requires water every fourth day, and can go without (on a pinch)
5 or 6 days, but when he does drink it is as well to let out his girths
a few inches, or he will burst them. The twigs and leaves of the mimosa
and kittar bushes, the scanty herbage of the desert, is all he requires,
except whilst making forced marches, when he requires a certain amount
of dhurra, because he has no time for grazing. This useful animal may
well be called the ship of the desert, for if it were not for him, the
enormous extent of burning sand which separates the fertile portion of
the Soudan from Lower Egypt would be like an ocean devoid of vessels,
and the deserts would be a barrier absolutely impassable by man. During
the season when fresh pasture is abundant camels can go for weeks without
water, provided they are not loaded or required to make extraordinary
exertions; the juices of the plants which form their food are then
sufficient to quench their thirst. The flesh of the young animal is one
of the greatest luxuries; of the skins tents are made; the various sorts
of hair or wool shed by the camel are wrought into different fabrics;
and its dried dung constitutes excellent fuel, the only kind, indeed,
to be obtained throughout vast extents of country. In order to qualify
camels for great exertions and the endurance of fatigue, the Arabs begin
to educate them at an early age. They are first taught to bear burdens
by having their limbs secured under their belly, and then a weight
proportioned to their strength is put on; this is not changed for a
heavier load till the animal is thought to have gained sufficient power
to sustain it. Food and drink are not allowed at will, but given in small
quantity, at long intervals. They are then gradually accustomed to long
journeys and an accelerated pace until their qualities of fleetness and
strength are fully brought into action. They are taught to kneel, for
the purpose of receiving or removing their load. When too heavily laden
they refuse to rise, and by loud cries complain of the injustice. Those
which are used for speed alone are capable of travelling from 60 to 90
miles a day: Instead of employing blows or ill-treatment to increase
their speed, the camel-drivers sing cheerful songs, and thus urge the
animals to their best efforts. When a caravan of camels arrives at a
resting or halting-place, they kneel, and the cords sustaining the loads
being untied, the bales slip down on each side. They generally sleep on
their bellies: In an abundant pasture they generally browse as much in
an hour as serves them for ruminating all night, and for their support
during the next day. But it is uncommon to find such pasturage, and they
are contented with the coarsest fare, and even prefer it to more delicate
plants. Breeding and milk-giving camels are exempted from service, and
fed as well as possible, the value of their milk being greater than
that of their labour. The milk is very thick, abundant, and rich, but
of rather a strong taste. Mingled with water it forms a very nutritive
article of diet. The young camel usually sucks for twelve months, but
such as are intended for speed are allowed to suck and exempted from
restraint for two or three years. The camel attains the full exercise of
its functions within four or five years, and the duration of its life is
from forty to fifty years. The hump or humps on the back of a camel are
mere accumulations of cellular substance and fat, covered by skin and a
longer hair than that on the general surface. During long journeys, in
which the animals suffer severely from want of food, and become greatly
emaciated, these protuberances become gradually absorbed, and no trace
of them left, except that the skin is loose and flabby where they were
situated. In preparing for a journey, it is necessary to guard the humps
from pressure or friction by appropriate saddles, as the slightest
ulceration of these parts is followed by the worst consequences: insects
deposit their larvæ in the sores, and sometimes extensive and destructive
mortification ensues. I have often seen crows pecking away at sores on
a camel's side, and was surprised to see how little notice it takes of
them. After all, I must say of the camel, that he not only groans and
roars when he is too heavily laden, but at all times without the least
occasion, and although it may appear mild, docile, and patient, it is
frequently perverse and stupid. The males especially are at certain
times dangerous. It is sure-footed, too, as I have often experienced in
travelling over mountains so precipitous that no animal but a camel could
have carried such heavy loads as I have seen it do without accident. All
breeds of camels could not do so, but those belonging to the Hadendowah
Arabs, between the Red Sea and Taka, are very sure-footed. The camels
most highly thought of in the Soudan are the Bishareen; they are very
strong and enduring, but not so large as many others. There is quite
as much difference in the breeds of camels as of horses, and as much
difference in riding a hygeen and baggage camel as there would be in
riding a nice springy cob and a cart horse. Amongst the Arabs a good
"hygeen," or riding dromedary, is worth from 50 to 150 dollars; the
average value of a baggage camel is about 15 dollars, but I believe our
average ran up to 30 or 35 dollars.
CHAPTER X.
OUR FIRST CAMP—TORRENTS OF RAIN—JULES BARDET—CAMEL-DRIVERS
BEHAVE BADLY—SULEIMAN IN TROUBLE—CAMEL-DRIVERS GET
UPSET—THE DESERT—TWO OF US LOSE OUR WAY—JULES SUFFERS FROM
DYSENTERY—SAND-STORM—A PILGRIM DIES ON THE ROAD, ANOTHER IN THE
CAMP—JULES' ILLNESS—CAMP SPLIT UP—LOSE OUR WAY—ENCAMP SEVERAL
DAYS IN THE DESERT—ARAB HUTS—THE MIRAGE—A LION.
After this digression and short dissertation on the camel, I will return
to the subject of our journey. We now formed a tolerably numerous
company, ourselves seven, three European servants, Suleiman, Mahoom,
Cheriff, and Ali, the cook, with an assistant, four or five native
servants, and nearly thirty camel-drivers. George, one of our servants,
and I had some trouble in getting our makloufas properly adjusted on our
camels; consequently, we were behind the others in starting. I also
made a call on Mr. Bewlay, who pressed me to remain to luncheon. As I
knew that this was to be a short march of about three hours, I did so.
I then bade adieu to Mr. Bewlay (one of the nicest and most gentlemanly
fellows to be met with), and commenced my journey, thinking I should
soon overtake my comrades, but in this I was greatly mistaken. I had
reached the middle of the town, amongst the bazaars, when the eccentric
conduct of my camel was quite alarming, exciting grave apprehensions
respecting the safety of my limbs, I being quite a novice in the art of
camel-riding. Down he flopped without the least preliminary warning,
whilst I held on to the makloufa as if I had been in a hurricane. I
plied my coorbatch on his tough hide; the only effect it produced was
to make him open his mouth (to such a width that it could easily have
accommodated a human head) and groan away with most stentorian voice. At
last an Arab succeeded in getting him on his legs, and away he went at
such a jolting pace that I experienced the greatest difficulty in keeping
my seat. Down he flopped again in the same unceremonious manner as before
just in front of a projecting part of the Police Station.
"Well," I mentally ejaculated, "this is, indeed, too much. I will not be
placed in such jeopardy as this any longer."
I lost no time in dismounting; Sheik Moussa was sent for, and at once
promised to find me a tractable beast. George remained with me. We had
no sooner unburdened the camel, and got under the projecting roof of the
Police Station, than down came the rain in torrents; then I felt thankful
that my camel had proved so awkward and disobedient. Two hours and a half
elapsed ere a respectable camel was brought. By that time the rain had
ceased, and George and I resumed our journey in comfort.
When we arrived at camp at 6 p.m., we found the tents pitched and
everyone changing their clothes, except Jules; they had all been
drenched to the skin. This was a favourable opportunity for me to
deliver a lecture on sanitary precautions. I therefore did so, warning
all Europeans to remember that we were not now in England, but in the
tropics, where the days were excessively hot and the nights not only
cool, but often very cold at this time of the year; always to change wet
clothing as soon as we got to camp; never to expose themselves to the
burning rays of a tropical sun without helmets; and last, but not by any
means least, to be extremely careful as to the quality of water they
drank, and always to see that the zanzimeers were well washed out before
they were replenished. Well, I know that in England, whilst practising
my profession, I have met with extremely clever people who not only know
their own business, but that of everyone else, and are most ready with
their unasked-for advice. They are quite encyclopedias of knowledge, or,
at least, they would have one think so. They apparently listen, with
folded arms and the head a little bit on one side, in the most attentive
manner, literally drinking in all the doctor is telling them when he
forbids this and orders that, and yet will use their own judgment or
sense—presuming, of course, that they have any—and the moment his back is
turned they exclaim—
"Pooh! what an old fidget that doctor is. I know that when poor Mrs.
Smith was ill her doctor didn't do ought like that, but let her have a
glass of stout for dinner, and ordered her a glass of hot whisky and
water at bed-time, poor thing, and that was what kep her up."
"When the doctor very impressively says, "Now, Mrs. Thompson, your friend
is very ill—I wish you to be careful to give her so-and-so and avoid
so-and-so," Mrs. Thompson says, "Yes, doctor—I quite understand;" and
Mrs. T., being a very garrulous, and also a very knowing personage, will
begin a long rigmarole about her first husband's case some 20 years
before, and how beautifully she nursed him through an illness of "seven
week," as she calls it, and brought him round, she, of course, not having
had her clothes off for four weeks, nor a wink of sleep for ten nights,
till she was a perfect "shada," but still able to articulate, poor thing.
Unless the poor doctor now bolts off, she will then confidentially
commence a history of three or four other cases in which she was, of
course, eminently successful. These very clever people, so wise in their
own conceit, are really very dangerous people, and I always look after
them well. Of course, Mrs. Thompson may think the medicine "strong enough
for a horse," as she expresses herself, and will administer it if _she_
thinks it suits the case, and exercise her very discriminating faculties
in the way of diet, and matters of that kind; but at the end of a week
Mrs. Thompson—who has, of course, seen many similar cases—expresses to
her neighbours and confidants (who look upon her utterances as oracular)
her dissatisfaction with that ere doctor, and is determined on his next
visit to favour him with what she is pleased to call a bit of her mind.
She does as promised—
"Well now, doctor, what do you think _is_ the matter with poor Mrs.
Smith? She don't seem to get on at all. I remember when poor Mrs.
Rodgers, my second husband's first wife's cousin, was laid up with—"
But, reader, you may imagine the rest; I can very well. I have used the
preceding imaginary conversation "to point a moral and adorn my tale."
In our camp I had a very headstrong Mr. "Cleverity," if I may say so,
to deal with. Jules, before we started, was working away, sorting the
baggage, &c., in his shirt sleeves after passing through the rain,
getting thirsty, and drinking bad claret and beer, such as he could
obtain in the place. Indeed, his absorbing powers were remarkable—he
resembled a huge dry sponge, which, when dipped into a basinful of water,
absorbs it _all_. I ascertained, from one who knew him well, that this
absorbing tendency was not altogether induced by the heat of the climate,
but that it was his normal condition which he always suffered from in
England, where he lived a life of comparative ease and indulgence. I only
knew Jules absorb water when he could not get anything stronger. I had
warned him at Souâkin not to get wet, as the evenings were so cold, and
now, on arriving at camp, here he was again wet to the skin, helping to
pitch tents and put things ship-shape; but, with a thirst unquenchable,
he was continually drinking water which was the colour of pea-soup, but
not _quite_ so thick.
"Now, Jules," said I, "remember what I told you at Souâkin. You are going
the right way to get dysentery."
He replied—
"Oh, I am all right, doctor. I am not an old woman, or a piece of
barley-sugar. I shall take no harm."
The sequel will show how disastrous was his disregard of my repeated
warnings, and very much grieved I was for two reasons: one was the
loss of a really good-hearted fellow, who had proved a faithful and
affectionate servant to his master, who thought very much of him, for
many years; the other was, that although I used every effort to save him,
and many a time was unable to sleep on account of the anxiety the case
caused me, so much so that I frequently visited his tent in the night,
yet all was of no avail. Added to this, I was excessively and incessantly
annoyed by the fussy interference of two amateur doctors in camp, who,
as educated men, ought to have known better than to worry me seven or
eight times a day with useless suggestions of a shadowy character as to
the treatment of a complaint of which they knew absolutely nothing. They
were great examples of an old adage, "A _little_ knowledge is a dangerous
thing."
In the evening of our first day's camping out, just after dinner (we
dined at 7 p.m.), down came the rain again, causing us all to scamper
off to our respective tents; spades were out, and trenches dug round,
and there we remained until morning. At 6 a.m. we were up, and saw no
more rain for several months; indeed, not until I reached Venice in the
following May.
It was about 10 a.m. next day ere our caravan started. The sun blazed
out with a scorching heat, causing us to feel as if we were in a Turkish
bath from the evaporation which took place, and our solid leather
portmanteaus, which were thoroughly saturated the day before, to curl
up like match-boxes. Before we started on our second day's march across
the desert our camel men were told they were to go on until 6 p.m., and
Suleiman was commissioned to see this order carried out. We often went on
in front of them in the morning, on the look out for a shot at a desert
gazelle; but it was singularly noticeable that about 1.30 p.m. we were
all somewhere in the neighbourhood of Cheriff, our butler, who was in
the charge of the canteen on a camel. Unless we made a forced march,
we usually breakfasted about 7 a.m., luncheon at 1.30, dined at 7, and
retired to rest at 9 or 9.30 p.m. After luncheon we frequently lay down
on our rugs, smoking cigarettes and reading some book, long after the
caravan had passed us. This day we did so, but judge of our astonishment
when, at four o'clock, we came upon some of our camels browsing; others
had not been unburdened, and nearly all the camel-drivers were in a
circle, with uplifted spears.
We soon ascertained the cause of this; there was poor Suleiman, our
head boss, the centre of attraction for these Hadendowab Arabs, with
their uplifted spears, who were angrily jabbering away. To the question,
"What's the meaning of this, Suleiman? they were to have gone on until
six o'clock," he replied, "Yes, I know, gentlemen, that I tell them they
no stop till I say, and I catch hold of one mans to stop him take the
load off the camel, and now they say they spear me if I don't leave them
alone."
[Illustration: HADENDOWAH ARAB CAMEL-MEN.]
He pointed out the ringleaders of what looked like mutiny against
authority, and as soon as he had done so, in true old English fashion,
a few well-directed blows put about five Arabs in the prone position;
all pulled out revolvers, and made them pile their spears, which were
at once secured, tied in a bundle, and given in charge to the English
servants. They were then made to re-load all the camels, and, at great
inconvenience to ourselves, we re-start at 6.30, and march until nearly
ten, just to let them see that they could not do as they liked, and that
we were masters and not they. This assertion of authority had a most
beneficial effect on the native mind. It was past eleven that night ere
we dined, and I retired to rest at half-past twelve, with a feeling of
general bruising and dislocated vertebræ easily accounted for, as I was
unaccustomed to the peculiar motion of a camel, which has a knack of
shaking up one's liver in a most effectual manner. Referring to my diary,
I find that on our third day's march, Dec. 17th, the temperature was
82° F. in the shade at 1.30 p.m. I generally took the temperature when
we halted for luncheon, which would usually be about one or half-past.
We could do with the dry heat very well as we were mounted, but now, in
consequence of the late heavy rains, we felt it very relaxing, and just
like a Russian vapour-bath. The Red Sea was still visible to the east
of us; to the west, a large tract of desert, backed up by impassable
rocky mountains. We now saw desert gazelles for the first time, and one
of the party brought one down, thus providing dinner for the evening.
We marched from 9 a.m. until 6 p.m., came to water then, and pitched
our tents near to it. We generally had very good water, but here it had
a brackish taste; still, with the aid of four bottles of champagne, we
managed to slake our thirst tolerably well. So far the mimosa and kittar
bushes were abundant, particularly during the first two days, but on the
fourth day we saw very few indeed, and marched through absolute desert,
saw nothing but the burning sands, and huge rocks of volcanic origin. We
filled our barrels, zanzimeers and girbas with water before we started,
and again marched from ten till six; temp. 81° in the shade. A girba is
the skin of a gazelle dressed. It is dressed in the following way by the
Arabs:—They get the chopped red bark of the mimosa tree, and put in the
skin with water, it is allowed to remain there for three or four days
and then it is converted into leather. This day we encamped at a place
called Settareb. On the fifth day we again made the usual march, and shot
two gazelles. We started off at nine—and left the caravan to follow.
About 11 a.m. one of our servants caught us up with the information that
some of the camels had been lost. Messrs. A. and W. James returned to
see about them, and found that it was a dodge of the camel-drivers, who
thought they would try to sneak back to Souâkin. The camels were easily
found; the two camel-drivers were tied together, marched into camp, duly
admonished and punished. At 5 p.m. we come to water, turn out all the
brackish water, fill our barrels, &c., and march until 6 p.m.; temp. 82°
in shade. Dine at 7.30, bed at 10 p.m., but before going to bed we had
all the camel-drivers up, some of whom appeared inclined to be mutinous.
We gave them a sound lecturing, and let them distinctly understand that
we would not stand this kind of thing any more, and that the next offence
would be punished with the coorbatch. Our camping ground is called Wadi
Osier. The next day, our sixth in the desert, Mr. F. L. James and I had a
somewhat unpleasant experience. After luncheon, as usual, we all rested
awhile, allowing the caravan to go on. Mr. F. L. J. and I, who were
absorbed with our books, remained long after our comrades had proceeded
on their journey. When at last we did start, we were surprised to find
how late it was getting. Knowing that there is little or no twilight in
these parts, we hurried on, hoping to catch the caravan ere darkness
overtook us, but could not do so. Darkness comes on—a most profound
darkness, too—and we lose the track; we dismount and light matches to see
if we can find it again. We don't, however, succeed in doing so. Nothing
now remains but to remount our camels and trust to them and Providence.
On we go, at the rate of four miles an hour. The silence of the tomb and
the darkness of Erebus surround us; not a glimmer of light could be seen
in any direction, not the sound of a wild animal, of a bird, or even
the rustling of a leaf, or the sigh of the softest zephyr. When we had
gone on thus for about an hour, neither seeing a light, nor hearing a
sound, we began to get uneasy, not knowing if we were going in the right
direction, but knowing full well that it might prove to be a serious
matter if we strayed off into the limitless waste of the desert. Every
now and again I fired a shot from my revolver, but I might as well have
used a pop-gun. Now the stars begin to make their appearance; by them we
see that we are, as we think, pursuing some track. We now dismount, and
finding that revolvers are useless, Mr. James gets his rifle and lets
off one barrel. We wait, and anxiously look for a corresponding flash;
hear we could not, as by this time a slight breeze had sprung up, and was
blowing from us towards our caravan. Another barrel is now fired, but no
reply. We were now rapidly coming to the conclusion that we should have
to tie our camels to a mimosa bush, and sleep out without food, and what
was still worse, without water, as both our zanzimeers were nearly empty.
Still we perseveringly jogged on, and after a time discharged another
barrel. In a few minutes' time we see a slight flash, which appears to
be so far off that we cannot make out whether it is in the heavens or on
the earth. Not a sound reaches our ears. Our cartridges are also nearly
exhausted, and we have all but made up our minds to sleep out, but try
the rifle once more. This time both barrels are discharged one after
the other. We look out anxiously; not a sound reaches us, but we see a
corresponding double flash a long way off, and are convinced that this
comes from our camp. We see certain stars over the spot, and for these
we steer. When we had jogged on for another hour and a-half, we see a
glimmering light like the flicker of a lantern (it really was a huge
bonfire)—another half-hour, and we can plainly see lanterns moving about,
and to our great relief and that of our friends, we gain the camp at 9
p.m., thoroughly hungry, thirsty, and tired. The temperature this day was
86° in the shade; my ears and nose were quite scorched, and smarting from
the heat of the sun.
On arriving in camp I found Jules very ill indeed. On the fourth day he
came to me in the evening complaining of a bilious attack. I gave him
something for it. In some respects he was better by the evening of the
fifth, and on the morning of the 6th bilious vomiting had ceased, but
in the evening, judging from the symptoms, I was afraid that formidable
complaint, dysentery, was setting in. However, I kept this to myself,
at present, not wishing to alarm the camp, and hoping that treatment
might prove beneficial. We dined at 10.30, and retired to bed at 12 p.m.
When I say we retired to bed I literally mean that, not to a shake-down
sort of thing with a rug over me and a portmanteau for a pillow, but a
comfortable bed with a comfortable pillow, in a comfortable tent, and
cocoa-matting on the ground. There really was an air of comfort about
all our surroundings. We had comfortable shut-up and open-out chairs, a
comfortable folding-up table, each a nice portable india-rubber bath,
and, whenever we encamped by water we each had a bath before breakfast
and another before dinner. As for eatables and drinkables, the most
fastidious would not turn up their noses at those. We had sufficient
champagne and claret to last us during the whole campaign; a freezing
machine, so that we could have these iced in the hottest weather. We had
gozogenes and any amount of seltzer-water, and when we fell short of that
we used Eno's fruit-salt in the gozogenes. We had Peek Frean and Co's.
biscuits, Cross and Blackwell's excellent Chutnee pickles and pickalili,
tomato sauce, asparagus, green peas, plum puddings, French jams, minced
collops, kidneys, tinned soups from Fortnum and Mason's, Piccadilly, and
everything else one could think of to insure comfort in eating, drinking,
and sleeping. Some ascetics would say we were of the earth, earthy, but
I maintain that if you mean to keep a _mens sana in corpore sano_ one
may just as well—and a great deal better—build up the waste tissue from
time to time and travel with every comfort, if one can afford it, as
do the reverse. I have tried both; whilst campaigning in Turkey, when
I have been on the march with the army, I have indulged in the luxury
of a small onion and a limited piece of somewhat indifferent bread for
breakfast, washed down with a drop of water, the same again for luncheon,
and the same again for dinner, "and still I was not happy," for I could
comfortably have disposed of breakfast, luncheon, and dinner at one
sitting without the slightest inconvenience. The ground was my bed, the
canopy of heaven was my tent, the twinkling stars my lantern, and a stone
or water-jug with a coat rolled round it was my pillow. I shall endeavour
to avoid the slightest exaggeration in this book, and will go so far as
to say that the former mode of travelling is by far the most comfortable,
and, in my humble opinion, the most conducive to health.
Again I find myself branching off, and cannot give any guarantee but what
I may still do so. All I ask is that my readers will overlook this little
failing of mine.
7th day.—We found water here, and of course replenished everything
with this valuable fluid. The mimosas were scanty and very stunted
here. During the night and all this day a great wind has been blowing,
producing a most blinding sand-storm, fortunately at our backs, or we
should not have been able to proceed. No one can form any idea of the
intense discomfort of a sand-storm, unless he has been in one. I may
close up my tent and be roasted inside. I may lock up my portmanteau,
which fits pretty closely, and have it in my tent, the lock covered with
leather, yet when I go to bed I find the sheets brown with sand, the most
secret recesses of my portmanteau and the lock filled with sand, and my
writing-case also, which is inside. I open my mouth to speak, and I can
masticate sand, if so disposed. I eat—all my food is full of sand. I
drink, not water, but water and sand. In fact, sand is everywhere; eyes,
nose, mouth, ears, hair, brains, and everything else has a mixture of
sand about it. I chance to leave a book, a pair of boots, or anything
else outside my tent, they soon become invisible, and are covered inches
deep in sand. Here we found great difficulty in pitching our tents, as
there was nothing but sand to drive the pegs into, and then we came to
rocks. Three or four days ago a lame woman and a man joined our caravan,
and two days ago two men, all bound for Kassala, all pilgrims from Mecca.
They were allowed to accompany us, and we fed them. To-day we miss the
woman, and on inquiry find that she was knocked up _en route_ yesterday,
and so her companion left her to die, and probably when we discovered
this she had been picked clean by jackals and vultures. Such is the value
put upon human life out here.
8th day.—The sand-storm still rages with unabated violence. We decide not
to go on, but encamp here to-day. We are, however, obliged to move our
tents to a place that is a little more sheltered, as at present it is
absolutely miserable. Jules still very ill. Temperature 86° in the shade.
In the day time the fierce heat of the sun rendered the interior of the
tents like ovens. Outside the sand reflected the heat. Although producing
great personal discomfort, our sufferings were nothing to what poor Jules
endured, who is now unmistakably suffering from dysentery badly. Under
any circumstances this is a grave complaint to have, but under present
circumstances, doubly so; that which he requires is impossible to obtain,
namely, absolute rest and a suitable diet. The poor fellow complains
to-day of incessant thirst, and everything he gets to eat or drink is
impregnated with sand, which it is impossible to avoid.
About 12 meridie, Mr. Phillipps, who was passing across the camp, saw
the two pilgrims whom we had allowed to join the caravan, two brothers.
One was supporting the head of the other in the blazing sun. The poor
fellow's eyes, nose, ears, and hair, &c. were full of sand. He said his
brother was ill. I was at once called to him, and found him _in articulo
mortis_. Very little could be done for him, and in twenty minutes' time
he died. His brother borrowed a spade, dug a shallow grave near the camp
and buried him, putting a mound of little white stones on the grave.
In my journey across the desert I frequently came across these graves,
sometimes two or three together, sometimes 20, 50, or 100. Occasionally
skeletons of camels were met with. In the present instance, the poor
fellow who died looked very emaciated and weak, probably exhausted by
constant marching and a deficient supply of food. But he had accomplished
the pilgrimage to Mecca, and I suppose he died a happy man.
9th day.—Poor Jules is so ill to-day that I cannot consent to have him
removed. The camp is accordingly split up, Mr. Phillipps and I, with a
few camels and attendants, remaining behind. The sand-storm is abating,
but the heat is very great and trying to Jules. A gazelle was shot
to-day. I cannot say that gazelle is a particularly toothsome morsel
under our circumstances. We are obliged to cook it on the same day that
it has been killed. The flesh of a desert gazelle is hard, and has very
little flavour. Our comrades left us about 10 a.m., and directly they had
gone down came the vultures for pickings.
10th day.—Jules still very ill, but in some respects a trifle better.
We decide on advancing to-day, if possible, and encamp a little longer
when we get to water. Accordingly we strike our tents and help the camel
men to load, send them on, then see to our own. We do not get off until
4.30 p.m. Half-an-hour afterwards we come to a dry river course, on each
side of which are dhoum palms and other trees. We saw a couple of jackals
sneaking off here, but did not get a shot at them. We trusted to one of
our Arabs to show us the way. When we had gone on for about an hour, he
suddenly stopped in the middle of a great sandy plain, said he was not
sure of the way, and as it was getting dark, thought we had better stop
until daylight. On hearing this Mr. Phillipps retraced his steps, and
was absent about two hours. I now became anxious about him, and every
now and then fired off my revolver. Fortunately I happened to have a box
of matches with me, and kindled a fire, then Mahoom and I tore up all
the stuff that would ignite. Half-an-hour afterwards Mr. Phillipps found
us, but he had been unsuccessful in his search for the road. However, we
kept up the fire, hoping some of our camel men would see the signals of
distress, which fortunately they did after a time; at last one of them
found us. In the meantime Jules was lying on the ground exhausted, with a
rug thrown over him. Our man led us to where the other camels were. Now
we had another bother: one of the camels had thrown his load off; the old
fellow who was in charge was lying on the ground, said he had got a pain
in his stomach, and we must stop there, as he could not possibly go on.
We roused him up, gave him a good shaking, and made him come on. But he
soon stopped again, and laid down to sleep, most coolly saying he could
not go any further. The fact is that just before we started he had eaten
a large quantity of raw meat, had, in fact, thoroughly gorged himself.
However, there we left him, and went on another two or three miles.
Halted at 10 p.m. and kindled a fire, had a cup of cocoa, a bit of bread,
rolled ourselves up in rugs, and lay on the ground. Jules suffered much
from this, as the nights were so cold.
11th day.—Up early, feeling stiff, cold, and hungry. Marched until 10
a.m. (four hours), intending to rest during the excessive heat of the
day, as my poor invalid was almost too weak to set up. About 3 p.m.
Mr. F. L. James appears on the scene, and tells us that the camp is
only about four miles off, at a place called Waudy. We get there about
6.30 p.m., and find the camp pitched near a well surrounded by dhoum
palms. Temperature to-day, 88° in the shade. This being Christmas Day,
we had some excellent plum puddings, made by Crosse and Blackwell, iced
champagne, and other luxuries for dinner.
12th day.—Jules was very ill indeed to-day, thoroughly prostrated by
his complaint, which had increased in intensity—it was quite out of the
question for him to attempt to move. We held a council, and decided
that as there were a few huts and goats, and a well, that it would be
advisable to let Jules rest here awhile, for now we could get a little
milk for him twice a day. Accordingly on the 13th day the camp was split
up. Messrs. A. and W. James, Colvin, and Aylmer went on to Kassala,
whilst Jules, Messrs. Phillipps, F. L. James, and I remained behind. Here
we rested for five days, and what with treatment, diet, and rest Jules
improved daily.
On the 16th day we rigged up an augarip (a kind of litter), with an
awning of matting and palm leaves to keep off the sun, and on the 17th
day this was slung across a camel. Jules got into it, and off we started
at 7.30 a.m., marching until 7.30 p.m. Much too long a journey for Jules,
who was again thoroughly knocked up and exhausted. I suggested now that
such marches were too long, and that our best plan was for me to start
off early with Jules, say 6 a.m., and march until 10, then rest until 4
and go on until 7 or 8 p.m. This was agreed to.
18th day. January 1st, 1883.—I visited Jules at 6 a.m.; found him no
worse. We started at 8, halt at 12, rest until 2, and go on until we
catch up the caravan, at 8.30 p.m. Jules complained bitterly of these
long journeys, which were so exhausting to an invalid. Medicine was now
out of the question, as the rolling motion of the camel made him very
sick. At the mid-day halt we found some empty huts in the desert. These
we explored, and found rather interesting. In several of them I found a
hole in the floor, the use of which is rather singular. The good wife of
the house uses this. She gets certain fragrant barks and frankincense,
burns them in the hole, then stands over them, having her dress drawn
round her, to fumigate herself and make herself acceptable to her
husband. In England, of course, this is not at all necessary. We passed
through a fine palm-grove to-day by a khor, and shot three gazelles.
19th day.—March again about 12 hours. Jules worse. Again I pointed out
the bad effect of these long marches on the invalid.
20th day.—Ten hours' march to-day; halt near a deep well and a large
palm-grove. Here I shot a fine golden-crested eagle. Jules frightfully
done up, and rapidly going the wrong way.
21st day.—On the march at 9 a.m. We marched the greater part of this day
across an awful desert, where no living thing except ourselves could be
seen. No shelter was attainable for the mid-day lunch. Temperature 92° in
what shade we could manufacture. During several hours of the day I saw
that optical illusion which so often mocks the thirsty traveller, called
the mirage—mirage, called by the Arabs, Bahr esh Sheitan, "The Devil's
Sea." By a strange refraction of the atmosphere, plains of arid sands
seem to be rippling lakes of water as far as the eye can reach, lapping
the base of stupendous mountains of rocks, and bathing the roots of the
stunted mimosa bushes. This day marched nearly 14 hours. Jules takes
scarcely anything, is rapidly sinking, and again complains of these long
marches.
22nd day.—Another 12 hours' march. See mirage again for hours. Encamp at
Fillick. Here there is a military station and a telegraph office.
23rd day.—Mirage again. Shot two gazelles, four bustards, and five
guinea-fowl. Appear to be getting into a better country. Jules much
weaker, pulse scarcely perceptible. Ten hours' march to-day.
24th day.—Eleven hours' march to-day, and, I am thankful to say, the last
day's march across the desert. Temperature 93° in the shade. Since 11
a.m. we have travelled through much better country, and after our late
experience it was quite refreshing to see a luxuriant vegetation once
more, such as dhoum palms, colocynth, tamarisks, nebbucks, heglecks—not
stunted mimosa bushes now, but different kinds of mimosa trees and
various trees and shrubs. The place I am speaking of was quite like a
gentleman's park. Here also were ariels, gazelles, bustards, parroquets,
eagles, vultures, and jackals. About seven, and pitch-dark, we, for the
first time, heard the roar of a lion not far off. Our sensations were of
a creepy character, and would, perhaps, have been more so had we known
what we did when we got to Kassala—that he had lately dined, at separate
times, on four human beings.
CHAPTER XI.
ARRIVED AT KASSALA—DESCRIPTION OF KASSALA—WE BUY CAMELS
AND HORSES—THE MUDIR GIVES A DINNER—JULES' DEATH AND
BURIAL—HYÆNAS—ARAB PATIENTS—MAHOOM'S HISTORY—DEMETRIUS MOSCONAS
ON SLAVERY—MENAGERIE AT KASSALA.
We arrived at Kassala at 8 p.m., and found the camp pitched about a
quarter of a mile, or less, from it, close to a garden full of fine trees
of various kinds; in fact, between this garden and a very wide river bed.
This river is here called the Gash, but nearer to Abyssinia it is called
the Mareb. Jules' exhaustion and pulseless condition was most alarming. I
succeeded in obtaining some eggs and milk; these I mixed with brandy, and
had this mixture administered to him every half-hour.
Whilst we were dining, at 9 p.m., we heard the peculiar cry of hyænas,
which literally swarm round the camp at night and make an awful row. I
daresay there would be 150 or 200 come round every night after dark, and
when we retired to our tents, and the lights were put out, they would
not only come close to, but actually into the camp; I can assure the
reader that I am not, as it is called, drawing the long bow when I say
that I have seen one poke his head in at my tent door more than once. We
remained here many days, and sometimes the hyænas would be so troublesome
and noisy that it was a by no means uncommon thing for one of us to
get up in the night, go to the edge of the camp in our night-shirt and
discharge the contents of one or two barrels into the noisy crowd. This
had a quieting effect, and we often found one or two dead hyænas in the
morning, the rest having scampered off.
We found Kassala a very warm place, for in the middle of January the
thermometer registered 90° in the shade. It is situated about 1,900 feet
above the level of the sea, is surrounded by a wall made of mud bricks
baked in the sun, and plastered over with mud and the refuse of cows.
The wall is loop-holed for musketry, and surrounded by a deep fosse. The
exports of the Soudan are ivory, hides, gum arabic, senna, bees' wax, and
honey—the latter obtained chiefly from the Abyssinian border.
The Kassala Mountain, which is just outside Kassala, is an enormous,
almost perpendicular, mass of granite, several thousand feet in height,
rising straight out of the plain, and can be seen for many miles in all
directions. The population was in 1882, something like 25,000, without
reckoning the garrison, which consisted of about 1,000 Nubian troops.
There are large numbers of cows, goats, sheep, and camels in the
neighbourhood, and a great deal of camel-breeding is carried on here.
When I left England, the thought occurred to me that it would be a good
idea to take out some knives, razors, beads, and so forth, both as
presents and for barter. I, therefore, provided myself with some common
knives, about a dozen of a better class, and half-dozen hunting-knives
from Mr. T. B. Hague, of Sheffield; a couple of dozen of Mappin and
Webb's, and Heiffer's shilling razors, which were much prized by the
Arabs. These I found very useful at Kassala, as I bartered some of them
for a dozen beautiful long ostrich feathers, and a handful of shorter
ones. The natives were well pleased, and so was I.
January 8th.—We were now comfortably encamped, but, alas! too late for
Jules, who was fearfully emaciated and prostrate. I visited his tent
twice in the night, at one and four o'clock, but could do very little
more for him, and I fear he will soon go. All the camels that we hired
at Souâkin will have to return there with their drivers. One reason our
friends preceded us to Kassala was to let it be known that we wanted
to buy or hire camels. The result was not exactly what we anticipated,
for they kept them back awhile. When at last they were brought for
inspection, the most extravagant prices were demanded, whilst many
of them were absolute screws. We also required a few little horses
for hunting purposes; the dealers were as knowing as horse-dealers in
England, and that is saying a great deal.
January 9th.—Jules is evidently sinking fast. I visited his tent five
times in the night, but could do little for him beyond giving him drink
twice. Mr. W. James and Mr. Aylmer had taken lessons in photography
before leaving England, and were each provided with a good apparatus.
With these they took many interesting views in different parts of the
country. This day the Mudir (Governor) of Kassala sent his two little
boys and ponies to be photographed. Whilst we were at breakfast to-day
an Arab brought two playful little leopards, which he had stolen from
their nest. I could have bought them for a couple of dollars each, and
probably should have done so had I been on my way home. The Mudir paid
us a visit at noon, inviting us all to dine with him; but Mr. Phillipps
and I could not go on account of Jules's illness, which now will be of
short duration. I understood afterwards from those who did go that the
dinner consisted of 15 or 18 courses. About 5 p.m. I visited Jules, and
found him asleep, but evidently sinking. Mahoom, who was my servant
during the whole campaign—and a very good boy he was, too—attended well
to him, and frequently sat up at nights with him. At 10 p.m. poor Jules
breathed his last. The particulars of his illness, together with other
curious and interesting medical notes, can be found in an article of mine
in the _British Medical Journal_, September 23rd and 30th, 1882. As soon
as he was dead I washed and laid him out. M. Demetrius Mosconas, a Greek
living in Kassala, was good enough to at once see about some kind of
coffin, covered with black cloth, shaped thus—[Illustration] This was to
be sent early in the morning.
January 10th.—At 10 a.m. the coffin was brought. Jules was put in it. On
the corpse were laid sprays of green shrubs all round. At 11.30 he was
carried to his last resting-place by natives, all of us following; the
Union Jack being placed on the coffin. The weather was fearfully hot,
and the roads very dusty. He was buried in a garden where three other
Christians had been buried. Mr. F. L. James read the burial service, and
we remained by the grave until it was filled up, which was very quickly
done in the following manner: The earth had been thrown up each side of
the grave, eight Arabs stood each side with their backs to the grave, and
as soon as the word was given they, with their hands, pushed the earth
between their legs, filling the grave within about ten minutes. A cross
was afterwards made of ebony by Mr. Phillipps, and placed at the head of
the grave. This concluded our last duty to poor Jules.
When we returned to camp, four horses and four camels were bought; after
lunch Messrs. Colvin and A. James, with Suleiman and a few servants,
started off for the Atbara in quest of camels, as these people were
holding theirs back in the hopes of making a good thing by so doing.
Korasi, on the Atbara, is considered one of the cheapest and best places
to buy camels.
January 11th.—Soon after my arrival at Kassala it became known that there
was a "Hakeem Ingelese," as they called me, and I very soon found that
my patients daily increased in number. Every morning after breakfast
there was I, wearing my pith helmet, in the broiling sun, with Mahoom
as interpreter, for two hours or more attending to a large number of
Arabs—men, women, and children—who squatted round my tent on their
haunches in a semicircle. I frequently saw 60, 70, or 80 patients a day.
I did not charge them anything; probably, had I done so I should have
materially thinned out the applicants for medical and surgical relief. It
is a strange thing, but human nature is (in some respects) pretty much
the same in Central Africa as it is in England.
Ali Mahoom's history, poor boy, was not, in early life, a very bright
one, as he was stolen by the slave-dealers. He said—
"I remember very well, sir, when dey took me. My mother was out when a
lot of mens come down and took all de little childrens dey see. Dey took
me with dem to Khartoum, and dere my mother found me. Dey let her stop
with me for a long time. She begged dem to let her have her little boy
back, but dey say, 'No, unless you steal two little children; den you
shall have him back.' Den I was sold to somebody else; after dat Gordon
Pasha find me, and he take me and give me to Mr. Felkin, and he has
been good to me ever since." He gave me the name of the country he came
from, but I forget it, adding, "It is the next country but one to the
Niam-Niams."
I had heard they were cannibals, so I said to Mahoom—
"What do they eat, Mahoom?"
"Dey eat de flesh of beobles," he replied.
"But," I said, "why don't they eat antelopes and other animals?"
"Dey say de flesh of beobles is much nicer," he replied.
He, of course, was disgusted with them. I found also that if a relation
dies they bury that relation in close relation to them, that is, at the
entrance to the hut; and if a baby, they send it to the relatives they
have most respect for, and these relatives, to show their respect for
their friends, eat the baby—cooked, I presume, but that I am not sure of.
About noon M. Demetrius Mosconas (who spoke English fairly well, and
is, in a certain sense, a brother) asked me to go and see his son, who
was very ill. I did so, and found him suffering from acute rheumatism.
Mosconas, it seems, was engaged by the Government in sinking wells in
some outlying districts, and his son, who partly superintended this work,
must have contracted this disease by sleeping out, as, although the days
were very hot, the nights were often excessively cold, and I have often
known the thermometer sink from 90° at 1.30 p.m. to 45° or lower by 11
p.m.
I had a talk with Mosconas about slavery, and learnt very conclusively
that it exists pretty openly in this part of the Soudan. He informed
me that the woman (his servant) who had just brought us two small cups
of coffee was a slave, one of a dozen owned by an Arab woman, and she
realized a living by letting them out on hire. He paid three dollars
per month for the hire of this slave. He also told me that Georgie Bey,
an Arab Army doctor, who left Kassala for Khartoum the day before our
arrival, sold two of his slaves before he went. This Georgie Bey was
since killed with Hicks Pasha's army near Souâkin.
Another bit of information which Mosconas gave me was that some Arabs and
Greeks breed from slave-women, who are kept for profit just as cattle are
in England. The children born under these circumstances are sold by their
fathers.
Having had a long chat with Mosconas, I next paid a visit to Herr
Schumann's collection of wild animals. I found a large building and yard
occupied by them. Schumann himself was away some distance from Kassala,
amongst the Beni-Amir tribe, where he had a zareeba. He collected animals
there, and sent them on to his manager at Kassala. Just before the rainy
season commenced, about May, he would go on himself, with quite an army
of attendants, across the Nubian Desert to Souâkin with his collection;
then take them to London, Liverpool, Hamburg, Vienna, and America, where
they would be sold. Here he had four giraffes, four gazelles, three fine
antelopes, as tame as lambs, a nice, amiable little baby elephant, five
young lions, chained to rickety old posts, which rattled up, as they
darted at a passer-by, in a very alarming manner, nine infant leopards,
two or three young hyænas, eight ostriches, some wild hogs, baboons,
tiger-cats, and other animals.
CHAPTER XII.
CAMELS FROM THE ATBARA—THE MUDIR—GORDON PASHA'S CHARACTER IN
THE SOUDAN—FERTILITY OF THE SOUDAN.
January 15th.—Messrs. Colvin and A. James, with their attendants,
returned this afternoon from the Atbara, having purchased thirty-four
camels. This rather astonished the natives here, who had no idea that
we could do without theirs, and quite thought we should be obliged
eventually to buy their camels, and of course give them a good price.
Whilst our friends were gone to the Atbara, we bought sixteen camels,
eight ponies, a number of sheep, and some milk-giving goats, so that we
could have milk with our porridge every morning for breakfast. It was
decided to-day that we must hire a few more camels for our march from
here to the Beni-Amirs, and leave Kassala on the 17th.
January 16th.—Mosconas dined with us to-night; he says that a year ago
King John's nephew, of Abyssinia, whilst fighting with some of the Arab
tribes, was killed, and that a Shukeryiah Arab took out his heart and ate
it on the spot. He said that these Hadendowahs, Beni-Amirs, Shukeriyahs,
Hamrans, and others, are always fighting, either with the Abyssinians or
amongst themselves, and that in consequence of the frequent raids made by
the Egyptian governors on the Arabs for taxes, the latter frequently hide
their money in the ground, putting oil on it to prevent discolouration.
He also informed me that the present Mudir of Kassala was a very sharp
fellow, and extremely successful in squeezing money out of the poor
Arabs, but that Gordon Pasha, when he was Governor of the Soudan, was
very kind and lenient, frequently remitting taxes when he thought they
were unduly pressed.
Now my own opinion of Gordon Pasha is that he was a just, honest, and
honourable man, whose sole aim was, not to extort all he could from these
poor Arabs, and so make himself popular at head-quarters, but to do that
which was simply right between man and man, just to those who employed
him, and to those whom he ruled. I believe I should not be far wrong
in saying that all Egyptian governors who preceded and succeeded him
were _extortionists_. What a charm his name had in the Soudan, amongst
different tribes in different parts! I have noticed that when Gordon's
name was mentioned the Arab's countenance would become radiant with
pleasure, as if calling up recollections of a good friend in bygone days,
and with a significant, "Ah! Gordon Pasha," they would begin to expatiate
on his good qualities in such a way that I could not help thinking that
he had more influence amongst them than any man living. The fact is, that
these poor down-trodden Arabs had, unknowingly, adopted one of our own
texts, "Prove all things, hold fast that which is good." Gordon Pasha
was kind to them, and when he passed his word he would keep it, whether
it clashed with the interests of the Khedive or not. He was the only man
as a governor whom they trusted as children would their father, and who
never forfeited their confidence. All his actions were summed up in the
words justice, truth, and duty.
A wit's a feather,
A chief's a rod,
An honest man's the noblest work of God.
As I am writing this (March 30th, 1884), I cannot help thinking that, had
Gordon Pasha been sent out to the Soudan nine months ago, before the wave
of rebellion had increased to raging billows, sweeping over the land, we
should have been spared the painful events which have taken place there
quite lately, and by this time the Mahdi would have been little heard
of. Then Gordon Pasha would have come on them like the noon-day sun, and
all the tribes would have flocked round him as a deliverer, whilst the
Mahdi would have been powerless, and relegated to the obscurity from
which he had sprung. What a country Egypt and the Soudan might become
under British rule! In Lower Egypt we should form a net-work of canals in
communication with the Nile, as in the days of Pharaoh; then thousands
upon thousands of acres, which to-day look sterile deserts, would be
made to yield enormous quantities of sugar-cane, cotton, flax, dhurra,
&c., and at least two crops in the year could be obtained. Politics I
have nothing to do with. I am simply giving an expression to an opinion
which I formed when in the Soudan and Egypt. This opinion I expressed
in January, 1882, when in the Soudan, and subsequently very many times
since in England. It is a source of great satisfaction to me that so
eminent an authority as Sir Samuel Baker, F.R.G.S., entertains the same
ideas. As things now are, extensive cultivation in the Soudan would be
almost useless, the only means of exporting their products being by
the very slow method of camels to the nearest sea-port; taxes would
increase, and the only people who would derive any benefit would be the
Egyptians, not the Arabs. If we ruled there we would appoint an English
governor, who would let the Arabs live on their own industry, if they had
any, and, if not, bring the fallaheen there, and take only a reasonable
share of taxation. We should put down a railway from Souâkin to Kassala,
from there to Massawah and elsewhere. We should sink wells, and erect
engines which would pump up sufficient water to irrigate thousands of
acres of the most fertile land. These lands require no manuring, nothing
but clearing, scratching the rich alluvial deposit, and putting in the
seeds. The crops when gathered could then be transported by rail to
Souâkin, which would soon become a flourishing port, could, in fact,
be transferred direct from central Africa to any English port within
three weeks or so. Senna, coffee, tobacco, cotton, sugar, flax, dhurra,
wheat, oats, oranges, lemons, and I don't know what, could be grown
there without the least trouble, besides which, hides, honey, beeswax,
and other things would be abundant. I was so struck, soon after we left
Kassala, with the immensely fertile appearance of the soil, and the vast
extent of country capable of cultivation, and of producing two crops a
year at least, that I mentioned this to Mr. Colvin, as we rode side by
side on our camels, afterwards making a note of the same in my diary.
Every night of our stay at Kassala the monotonous beating of the tom-tom
would commence about 8 p.m. and continue incessantly till about twelve,
accompanied by the most extraordinary shrill trilling note of a female
every now and then. It seems there had been a death in the family,
and on all occasions of great joy and sorrow this unpleasant musical
entertainment is at once provided. Mosconas tells me that, after a
death, the mourners go on in this kind of way every night for about a
month, and a very irritating performance it is to our untutored English
auditory nerves. We found the white ants very troublesome here. They are
very destructive little insects, and, I believe, will destroy everything
but iron and stone. If a strong, solid, leather portmanteau be left on
the ground for two or three nights, they will destroy it, but if two
stones, sufficient to raise it an inch or two from the ground, be placed
underneath, they will not touch it, and if we encamp on sand we are safe
from them. They destroyed all the matting laid down in our tents during
the few days that we were here, although it was taken up and beaten every
day. Scorpions as well as white ants were frequently found underneath the
matting. This is very cheap, indeed; we therefore secured a fresh lot.
CHAPTER XIII.
LEAVE KASSALA—CHARACTER OF THE COUNTRY—MEET BENI-AMIR ARABS ON
THE RIVER BED—THE BAOBOB TREE.
All being ready, we started off from Kassala at 3 p.m. on the 17th
January, intending to reach Heikota within three days. We encamped at
8 p.m., and next day were off at 9 a.m., marching until 6 p.m.—temp.
90° in shade. Fine trees became more numerous; the country looked much
greener, and water was usually obtainable every day simply by digging
a few feet in the sandy river-bed. This day we passed through a large
field of dhurra. To guard this from elephants, birds, and buffaloes
several platforms were erected in different parts of the field about
10 or 12 feet high, and on these boys and men spent the whole day
unsheltered from the scorching rays of the sun—which was like a ball of
fire—cracking whips and uttering hideous noises. We came across distinct
and recent tracks of a lion, two full-grown and one young elephant. I
also saw a few monkeys, a musk cat, several young tiger-cats and hundreds
of guinea-fowl. We shot on the march one eagle, two ariels, and two
gazelles; the latter are not so hard and dry as those in the desert.
[Illustration: THE KASSALA MOUNTAIN.]
On the 19th we made the usual march, encamping at Ashberra. The general
character of the country has now become much more varied and interesting.
This day we travelled through tall grass, about 10 feet high, for a
long time—perhaps an hour. When we got out of this we found ourselves
encountering the prickly thorns of the mimosa and kittar bushes. We had
now done with caravan routes entirely, and my camel-riding capabilities
were fully tested as I go up and down hill; now over rocky hills, down
steep banks and across dry river-courses, then through a forest of dhoum
palms, dodging as I go the great projecting strong branches, which appear
strong enough almost to decapitate or sweep me off my camel. Then, by
way of variety, we pass through a mile or two of the horrid cruel thorns
of the mimosa and kittar trees, which every now and then bury themselves
in my flesh, and tear my clothes and helmet as I duck my head to avoid
having my face lacerated. The camel, of course, walks on in the most
unconcerned manner, just as if he was on open ground, taking no notice
whatever of these obstructions, but brushing past them as if they were
twigs or straws. Everything, even the smallest of the mimosa tree, is
armed with long, strong, very sharp thorns. Each thorn is as sharp as a
needle and about an inch long; indeed, the native women use them as a
cobbler does his awl, and I have often seen a woman using the thorn to
pierce a girba and shreds of the palm leaf to sew it up with. The thorns
of the kittar bushes are quite semicircular in shape, very near to each
other, not long but very strong, and each successive thorn crooks in a
different direction to its predecessor—one crooks up and the other down.
When they catch hold of anyone they stick to him as close as a brother.
If my clothes get entangled, I must stay and pick myself out, or if I
elect to go on without doing so, I must submit to having my clothes, and
perchance my flesh, effectually torn across. We saw _en route_ a great
many baboons, vultures of course always, eagles, thousands of doves,
guinea-fowl, and recent tracks of elephants, lions, and leopards. Mr.
Phillipps and I, whilst stalking some gazelles in a large palm-grove,
lost the caravan for hours, and just as we emerged from it on to the
wide river-bed of the Mareb we came upon a large number of the Beni-Amir
tribe in that semi-nude condition, which is so fashionable amongst them,
watering their goats and cattle. We dismounted and joined the sable
throng, made them understand that we should like some goat's milk, which
they gave us, after which we showed them our pocket and hunting knives,
revolvers, and watches. We were at once surrounded by an admiring throng
of our new acquaintances, who seemed greatly pleased with what they saw,
but when I applied my watch to the ear of one his surprise and delight
was immense, for he had never in his life seen a watch before. The
ticking tickled him greatly, so much so that he pushed all his friends
forward one after the other to participate in his joy. However, as the
long bushy hair of these fellows was streaming with fat, I observed
caution.
January 20th.—We were off at 9.30, and had not far to go ere we reached
Heikota, where the Beni-Amir tribe then lived. This was the shortest
march we ever made, for we arrived there at 11 a.m., amongst the most
luxuriant vegetation, encamping just by a huge baobob tree (_Adamsonia
digitata_), 51 feet in girth. Large as this may seem, it is not by any
means as large as they grow—they are frequently 60 to 85 feet in girth.
The trunk is not above 12 feet high ere the branches are put forth. The
flowers are in proportion to the size of the tree, and followed by a
fruit about 10 inches long. This looks like a greenish pod or capsule,
having a bloom on it such as we see on a plum; on breaking the capsule
we find a large number of granular-like substances very much resembling
pieces of white starch packed closely together, which have an agreeable
sub-acid flavour. When this white substance, which is very thin, is
dissolved, and it does so readily in the mouth, we came to a dark,
brownish little stone, very much like a tamarind stone in appearance,
but smaller. When dry, the pulp, by which the seeds are surrounded, is
powdered and brought to Europe from the Levant, under the name of _terra
sigillata lemnia_—the seeds are called _goui_.
CHAPTER XIV.
ENCAMP AT HEIKOTA—SHEIK AHMED—HERR SCHUMANN AND HIS ZAREEBA—WE
MAKE A ZAREEBA—THE MAHDI—EXCITEMENT IN THE VILLAGE—HORRIBLE
TRAGEDY—SHEIK AHMED DINES WITH US—THE MAGIC LANTERN—LIONS VISIT
US.
Ere we could pitch our tents we had to cut down a number of young palm
trees, and clear away a quantity of tall grass, &c. Whilst doing so Sheik
Ahmed, of the powerful Beni-Amir tribe, who paid us a visit, gave us
a good deal more of manual labour by advising us to make a zareeba (a
fence of prickly trees), assigning as a reason, and a very substantial
one, the fact that lions came down every night, and often made such a
noise as to disturb his slumbers, but that we had nothing to fear from
his tribe. He said further that boa-constrictors and scorpions were very
common, leopards also. We found traces of the latter whilst clearing.
Sheik Ahmed, or Achmet, is said to be one of the most powerful Sheiks
in the Soudan; he certainly was far and away the best sample of a Sheik
that I have seen anywhere. His head was kept shaved; on it he wore a
tight-fitting white skull-cap. He was almost black, and of a determined
aspect, but his features were good, and his teeth white, sound and
regular; his eyes were keen, black and glittering as a hawk's; he was
dressed in spotless white and scrupulously clean; quick in action,
thought, speech, and appearance. One might almost say really that he was
an educated man, for he could both write and read, and certainly looked a
remarkable, shrewd, and intelligent man. During the piping times of peace
he could be a merry fellow of infinite jest; and he and I cracked many
a joke together by the aid of an interpreter. He could also be a fierce
warrior when necessary, and bore marks, some deep ones, too, of many a
skirmish he had been engaged in. "Beware of entrance to a quarrel, but
being in at it, bear thyself, that thine opponent may beware of thee."
The latter half of this proverb, I think, would be quite applicable to
our friend, Sheik Ahmed, as all his wounds were in front; the first
part, I fear, he would be rather regardless of, probably, indeed, more
of the temper of an Irishman handling a shillelagh at Donnybrook fair,
exclaiming—"Will ye jist thread on the tail of me coat, now?" And I
assure you, reader, that had you known Sheik Ahmed you would hesitate
ere you trod on his caudal appendage, if you discovered it. I have heard
that he is one of the most powerful vassals under the Khedive, and that
should occasion require he could put in the field about 10,000 horsemen
and tribesmen.
These people, over whom the Sheik seemed to possess great control and
authority, are quite pastoral in their pursuits, and own such enormous
flocks and herds as would astonish any ordinary mortal. These are every
night driven in from grazing to a large zareeba on the river-bed. I was
irresistibly reminded here of the patriarchs of old. Here was this Sheikh
with four wives and I don't know how many children, the leader or petty
sovereign of a large and powerful tribe, over whom he possessed absolute
power, and, as I said before, owning these flocks and herds. On the
river-bed of the Mareb the tribe, or a part of it rather, lived, their
dwellings simply consisting of a few stakes driven into the sand, over
and around which is a covering of tall grass and matting made from the
palm leaves. They live in this neighbourhood as long as there is anything
for their flocks and herds to eat; when there is not, like locusts,
they move off a few miles to pastures new. Then, when the wet season
commences, they clear off to the mountains or desert, else the tetse fly
would destroy the animals. They seem a contented lot, and may truly say,
as a deceased M.P. once said, "My riches consist, not in the vastness of
my possessions, but in the fewness of my wants."
They live simply, on milk, honey, and dhurra principally, and to that
fact may be attributed the beautifully white sound teeth they possess.
I think I ought to say that the Sheik was good enough to ask me to stay
with the tribe for three or four years, and as an inducement was good
enough to say that if I would he would give me four wives, thus placing
me on a par with himself. However, I neither embraced this tempting offer
nor the sable females, and here I think the utterances of the deceased
M.P. would be peculiarly applicable. When the Sheik, who was very
friendly, left us, he accepted our invitation to dinner at 7 p.m. Quite
close to our camp was another zareeba; in this dwelt Herr Schumann and
his wild animals. He had, when we were there, three young elephants; two
females, and one rogue (the latter, being a rather fierce little fellow,
was chained by the leg), a few young lions, wild cats, leopards, and 15
young ostriches about the size of Dorking fowls.
_Apropos_ of the Mahdi, I find the following in to-day's paper:
"April 2nd, '84—An Austrian dealer in wild animals, writing from Kassala
to friends in Vienna, gives some information about the Mahdi, whom he
knows personally, and with whom he has frequently transacted business,
the Mahdi himself having for years past dealt in wild beasts for the
different European Zoological Gardens. He is described by the writer
as a very cunning impostor, and as an instance, it is related that a
short time ago he suddenly appeared with a number of warts on his right
cheek, these having been artificially produced by the aid of a German
called Schandorper, formerly a clown, and afterwards a hairdresser, now
in the service of the Mahdi. The reason was that the legends about the
expected Mahdi speak of him as having such marks. Like the beasts he
formerly dealt in, the Mahdi sleeps in the daytime and transacts business
during the night." I have no doubt whatever that our old acquaintance,
who was the only animal collector I ever met with, is the author of the
preceding, and I think a very credible man.
In the evening, just before dinner, we heard near our camp a great number
of women and children, accompanied by the inevitable beating of the
tom-toms, and that wild, peculiar trilling note of a woman to which I
have before alluded. Being desirous to find out as much as I could of the
habits and customs of these people, I got Mr. Colvin to accompany me to
ascertain the cause. We went and found a great number of women shouting
and chanting, whilst a number of their braves were executing a war dance
with spear and shield, others in the meantime sharpening their spears.
Of course we were at a loss to account for the extreme activity and
evident war-like preliminaries, and returned to camp not feeling certain
whether these sharpened spears would not on the morrow make unpleasant
incisions in our intercastal spaces—at least Mr. Colvin, who was a very
facetious and witty fellow, humorously suggested this. On returning
to camp we passed Herr Schumann's zareeba, and told him what we had
witnessed, asking him the meaning of it all. He narrated the following
tale of blood: The day before our arrival at Heikota, when we were in
the immediate neighbourhood, probably about the time we were encamping,
a number of the Basé people from the village of Sarcella had come upon
the children of the Beni-Amirs driving the flocks and herds in for the
night. They then perpetrated a deed which makes one shudder to think of,
for they were not satisfied with simply slaughtering these unoffending
children, but doing so in a most horrible manner; in short, they ripped
them open with their knives, and drove off about 2,000 head of cattle. In
consequence of this the Sheik had given the word to his men to prepare
for action, and they were now doing so, intending to make an attack on
the village of Sarcella in the morning.
In the evening we had a champagne dinner; the Sheik studiously avoided
the champagne, and had the shocking bad taste to prefer raspberry vinegar
and water. Herr Schumann also joined us, but, like a Christian, partook
of champagne.
We pretended not to know anything of this slaughtering business, so asked
the Sheik what was the reason of all the commotion amongst the tribe. He
related the same story, adding that he should get his men together in
the morning and attack these Basé (that means kill all they could lay
their hands on) and get his cattle back again. How ably he carried out
his destructive intentions I will tell the reader later on. The customs
are somewhat peculiar in this part of the world. Supposing I and my
party, who are not Beni-Amirs, enter the Basé country from the Beni-Amir
tribe, and they should be at enmity with them at that time, they would
regard us as enemies. Knowing this, we told the Sheik we were very sorry
this had occurred just now, as we intended to explore that country,
and his fighting might make it a very difficult, if not impossible
matter to do so. However, with the true instincts of a gentleman sheik,
he accommodated himself to all parties, very readily acquiesced in our
views, and was good enough to postpone his bloodthirsty intentions for a
few days.
After dinner we chatted round the camp fire for awhile, smoking the
"calumet of peace," to use a Cooperian phrase, and retired to our
different tents to rest 9.30 p.m. The Sheik, ere he left us, accepted
an invitation to breakfast next day at 7 a.m. If we are late birds in
England, we are early ones in the Soudan.
January 21st.—True to his appointment the Sheik breakfasted with us this
morning. He was not only punctual, but he literally _did_ breakfast;
there was no finiking and fiddling about with his food, for he disposed
of it in a most straightforward manner. Imagine an opening in the
pavement for the reception of coals, and you have a pretty good idea of
the rapid disappearance of food down the œsophagus of our friend. We
commenced with porridge and milk; a dish evidently highly appreciated by
the Sheik; then we had minced collops, kippered herrings, gazelle, stewed
kidneys, wild honey, French jam and coffee, to all of which Sheik Ahmed
did ample justice. After breakfast many warriors drop into camp, and,
in their fashion, squat round in a circle on their haunches. One of the
spears was covered with leather as a sign that they were at peace with
us. A great and long pow-wow ensued as to our future journey; we should
want to buy or hire camels for going through the Basé country, as those
we had hired at Kassala would have to return from Heikota.
Last night we were rather disturbed by the noise of lions, and this
morning, within about a dozen yards of my tent, I found their footprints,
fortunately outside the zareeba. I dare say I spent about an hour or
so at my tent this morning attending to a large number of natives, and
afterwards visited others in their own tents on the river-bed. In some
instances I was obliged to crawl in on my hands and knees. When I had
finished my morning's work I took up my shot gun and strolled off in
quest of some beautifully plumaged birds which were abundant here and
brought home an eagle, paroquet, laughing-bird, falcon, and shreik, which
I skinned after luncheon.
We again invited Sheik Ahmed and Herr Schumann to join us at the festive
board at 7 p.m.; we also told the former to let his people know that at
about 8.30 p.m. there would be what they call a _fantasia_. Just before
7 p.m. the Sheik arrived and behaved himself in quite a gentlemanly
manner. He was dressed in spotless white, and was so particular as to
borrow a pen-knife from me to clean his nails with (a great instance of
the civilizing effect of Englishmen). Although we drank iced champagne
and claret, he stuck to raspberry vinegar and water, which he consumed
with great relish. He was rather clumsy with a knife and fork; indeed,
almost the only breach of manners that he perpetrated was to finish up
the repast (just before coffee was brought) by plunging the teaspoon
into the preserve, scooping out as much as it would conveniently hold,
conveying it to his mouth and replacing the spoon in the preserve; this
mode of eating has its inconveniences.
Another peculiarity of his was a singular habit that one requires to get
thoroughly accustomed to to really appreciate; he generally indulged in
it largely at meal times when conversing, and having his face directed
to the object of attack. I scarcely know how to describe it, and perhaps
ought not to do so in polite society, but that I wish to tell my readers
exactly what kind of a man this was. It was a method (not unfamiliar
even to English ears) of producing a peculiar vibration or concussion of
the atmosphere by a noise proceeding from the mouth; some polite people
would call it an eructation, but that is not sufficiently explanatory. It
is familiarly and vulgarly known as "belching," and so frequently did it
occur at meal times that it became known amongst ourselves as "the genial
belch of the Sheik." I suggested that probably it was a complimentary
proceeding on his part, but I must say if it was so we could readily
have forgiven this _too_ frequent formality. After dinner a great many
of his people assembled (no women, and very few children) to witness the
mysteries and wonders of the magic lantern, or _fantasia_. Would that I
had the pencil of an artist to delineate the picture which the _Graphic_
or any other illustrated paper would have been glad to have reproduced.
Here we were encamped in equatorial Africa; we had five tents pitched
amongst waving dhoum palms, tamarisk, and tamarind, nebbuck, baobob,
hegleek, ebony, and other trees, and the usual luxuriant growth of tall
grass and young palms. About three hundred of these dusky-skinned, almost
black, agile-looking fellows, wearing simply the tope or loin-cloth,
the foremost squatting on their haunches, the rest standing behind, the
Europeans in white clothing, and the picturesquely-dressed Sheik in his
white turban and robes. It was a weird, wild scene, when viewed by the
flickering light of the lanterns as they moved about the camp, but
when the moon shone out, shedding a soft, bright light on the scene, it
certainly was a most charming and interesting picture. Amidst it all
could be seen three hundred glittering spear-heads, making the picture
complete. How easily, had they been so disposed, could these wild sons of
the Soudan have made an end of us, but I am happy to say this ceremony
was not included in the evening's programme. We placed a wet sheet across
the entrance to one of the bell-tents, and as the Queen (whom they called
the Sultana) the Prince and Princess of Wales, the elephant, lion,
rhinoceros, hippopotamus, giraffe, ostrich, crocodile (snapping his jaws
together), and other animals with which they were familiar, appeared on
the canvas, the delight of these grown-up children was manifested by loud
expressions of approval. When the Sheik, his retinue and people took
their departure, we further astonished them by letting off rockets and
illuminating their way with red and blue fire. If I went out there again
I should certainly take out a galvanic battery, which I am sure would
astonish and amuse immensely. We here engaged fresh camel-men, huntsmen,
horse-boys, and servants, at rather high wages, on account of the
rumoured ferocious character of the Basé, the Sheik taking a pretty good
share of the wages himself. All camels were bought, not hired; when we
wanted to hire we were cheerfully assured by the owners that we should
very likely all be killed by the Basé or Kunama people, and they would
lose their camels. The Sheik was presented with a capital bell-tent,
a rifle, and a good musical-box, which played six airs, others, with
razors, butchers' knives in sheaths, topes, beads, knives, scissors,
small portable looking-glasses, &c., all of which were productive of
great wonder and joy. Sheik Ahmed, in return, sent us a present of ten
sheep and ten milk-giving goats, so that now we had sixteen goats, which
furnished us with plenty of milk every morning to our porridge. As we
intended resuming our journey on the morrow, we were all busy writing
letters to England, which Herr Schumann engaged to forward to Kassala.
CHAPTER XV.
PATIENTS AT HEIKOTA—LEAVE HEIKOTA—GAME IN THE BASÉ COUNTRY—SEE
OUR FIRST LION—A LION INTERVIEWS THE AUTHOR—TETÉL, NELLUT, AND
OTHER GAME KILLED ON THE MARCH.
On the 22nd January we were up in good time, as there was a good deal
to be seen to ere we continued our march. We intended to return to
Heikota after exploring the Basé country, which we thought would occupy
about four or five weeks. It would not, therefore, be necessary to
take all our baggage with us; accordingly, a considerable quantity was
left behind in Herr Schumann's zareeba until our return—assuming that
we should do so. I was, as usual, busily occupied after breakfast in
attending to my patients, who not only came from close by, but from long
distances on camels. It had got noised abroad from Kassala that there
was a "Hakeem Ingelese" travelling with these gentlemen, and whenever
we encamped anywhere for a day or two many patients came to visit me.
They appeared inordinately fond of my pills, and would swallow them
with as much avidity as boys in our country swallow lollipops. To judge
from what was expected of me, they must have thought that I was endowed
with almost supernatural powers. One boy was brought to me whose hip
had been dislocated a year or so before; another person who had been
positively blind from ophthalmia two years, hoped I could let in a stream
of welcome light: Alas! poor fellow, I could not make the blind see, or
the lame walk, under such circumstances. However, I was often able to
effect cures in some and relief in other cases, and when we returned to
Heikota many grateful patients came to thank me; one would give me some
dhurra, another a skin of milk, an Arab knife, a spear, a sheep, and so
on. Gratitude even is pleasing to a doctor, although sometimes a scarce
commodity. We did not succeed in making a start until 4 p.m.; halted at
six. The Sheik, who came part way with us, on returning to his tribe,
said he would join us in the morning, and see us well on the way ere
he interviewed the Basé at Sarcella, whom he had an account to settle
with. During our various conversations with him he informed us that we
should find abundance of shooting of every kind in the country—elephants,
lions, leopards, porcupines, wild cats, hyænas, buffaloes, jackals,
giraffes, ostriches, rhinoceros, antelopes of different kinds, gazelles,
oterops, ariels, maarifs, mehedehét, tetél, nellut, dick-dick, baboons
and monkeys; all kinds of birds; falcons, Egyptian hawks, rollo-birds,
paroquets, eagles, vultures, doves, quail, partridges, sand-grouse,
guinea-fowl, and I don't know what besides—all of which was quite true;
there was really enough of shooting of every description to satisfy the
most ardent sportsman. He also advised us, when we got into the Basé
country, not to have our guns, rifles, and revolvers in cases, but ready
at a moment's notice, night and day, and this advice we strictly followed
during the whole of our journey.
On the 23rd we marched nine hours, encamping at a place called Toodlook.
Our sleep was rather disturbed in the night by the noise of lions and
hyænas, which came very near the camp. We marched to-day through varied
scenery and pretty country—now along the Mareb, then for two hours across
country, through jungle, again coming on to the Mareb, across it, and
over a plain studded with trees and shrubs, finally encamping by the side
of the Mareb. Whilst our tents were being pitched, Messrs. A. and W.
James and I reconnoitred, soon coming near to a place where there was
some water. Suddenly we discovered, about two hundred yards from us, a
fine lion lying down on a little elevated land, no doubt on the look-out
for some unsuspecting antelope coming to drink. Mr. A. James ran back
to camp for his rifle, crept up, without arousing the suspicion of the
noble beast, and fired, but not being near enough, missed him. The lion
simply got up and calmly turned off into the jungle, where it was deemed
unadvisable to follow him. On our way back to camp we saw one place where
there had evidently been a desperate struggle between a lion and his
prey; the former evidently had the best of it, as we saw a long trail, he
having dragged his supper into some long grass and young palms.
On the 25th we were up and off in good time, leaving Suleiman and the
English servants to follow in charge of the caravan. Last night a
rather curious adventure occurred to me, which might have had a curious
termination. When we arrived at a camping-ground I usually selected the
spot for my tent, quite regardless of where the others were going to be
pitched. On this occasion I had done so, and ordered it to be pitched
under some trees close to young palms and tall grass, some distance from
the others. Suleiman remonstrated with me for doing so, saying that
the Basé or lions might come down in the night. However I would have
it so. Every day whilst we dined a large camp-fire was lighted, as the
nights were very chilly, although the heat was so great in the daytime.
Around this we smoked and chatted over politics, English friends and the
events of the day, and plans for the future, skinned birds or animals,
wrote letters, or posted up diaries. At half-past nine or ten o'clock we
gradually melted away one by one to bed. On this night I was the last,
having stayed to have an extra pipe. At last I lighted my lantern, was
walking off to, and had nearly reached, my tent, when I was startled by
a low growl issuing from a thick growth of young palms, about a dozen
yards from my tent; there was no mistaking the nature of the growl, and
I rapidly executed a retrograde movement, poked my head into the nearest
tent, calling out to the semi-sleeping occupants thereof, "I say, did you
hear that salutation just as I was going to my tent?" Answer by Mr. F. L.
James and Mr. Phillipps, "No; what was it, doctor? We were just going to
sleep." "Why, it is a lion close to my tent, and there is no mistaking
it." They laughed immensely, and seemed to think it a good joke, but
jumped up and came with me towards my tent, I think slightly incredulous.
Their incredulity was, at all events, quickly dispelled, as the lion, by
another louder expression of opinion, gave us distinctly to understand
that he was not only in unpleasant proximity to, but had his eye on us.
Again an _extremely_ rapid retrograde movement by the trio ensued, and
a joking remark from Lort Phillipps, "Doctor, you will be dragged off
to-night, as sure as fate," and a consoling remark from Mr. F. James that
the lion was perhaps hungry. We seized some burning brands from the fire,
and piled on a large number of dried palm leaves in front of my tent. I
then retired to rest in peace, and when I arose in the morning my friends
were, I hope, pleased to find I was not in pieces. We heard both lions
and panthers in the night pretty near to us, but so long as they did not
visit the camp we did not care. In the morning at breakfast the Sheik
was highly amused by an account of my night's experience, and extremely
jocular over it. This day we killed two tetél on the march, and caught
fifty-seven sand-grouse in a net, but only kept sufficient for dinner and
luncheon. One of our courses at dinner was an omelette of ostrich eggs.
CHAPTER XVI.
WE ARRIVE AT THE BASÉ OR KUNAMA COUNTRY—THE VILLAGE OF
SARCELLA—MURDER OF MR. POWELL AND PARTY—MY CAMEL AND I
UNCEREMONIOUSLY PART COMPANY—THE FIRST BASÉ WE SEE—ENCAMP AT
KOOLOOKOO—OUR FIRST INTERVIEW WITH BASÉ—THEY MAKE "AMAN" WITH
US—THEIR APPEARANCE—DESCRIPTION OF KOOLOOKOO AND THE BASÉ
PEOPLE—THEIR HABITS AND CUSTOMS.
From January 26th to 28th nothing of importance took place. A day seldom
passed without nellut, tetél, and gazelle falling to someone's rifle. We
were all busy during leisure hours in writing letters for England, as we
should not be able to do so or receive any again until our return from
the Basé country to Herr Schumann's zareeba, where Mr. James had arranged
all letters and papers from Kassala were to be brought. We often saw
baboons gambolling about, also tracks of elephants, lions, and panthers.
The Shiek says we are now in the neighbourhood of giraffes, ostriches,
and buffalos. This day he leaves us to fight or come to terms with the
Basé, and graciously condescends to act as our postman as far as Heikota,
promising to see that our letters are forwarded from there to Kassala.
He tells us that his men are at Sarcella, where they have about 1,000 of
the offending Basé shut up in a cave; they were now waiting for him ere
they took any further action. What transpired we did not learn until we
returned to Heikota. That I will describe when we return there.
January 29th.—We are now in Basé territory. We have for days past done
with caravan routes or paths, and travel over rocky mountains, large
plains, jungle, river-beds, and through a forest of tamarind, tamarisk,
palm, baobob, nebbuck, hegleek, and mimosa trees. On the branches of the
latter we frequently saw lumps of gum arabic, as large as walnuts, which
had exuded through the bark.
At 4 p.m. we saw a Basé village on fire, and rightly surmised, as we
found out afterwards, that the Beni-Amirs had been the authors of this.
Just after breakfast, before the camels were brought, we shot eight
partridges and ten quail, which were handed over to the cook. Some were
prepared for dinner, and some for luncheon next day. We also shot on
the march a buck tetél; the prime bit was, of course, reserved for our
dinner, and was more like roast beef than the flesh of any other animal
I tasted; the rest was given to our attendants. We encamped by the side
of the river-bed, where we found water on digging to the depth of 7 or
8 feet. Ere we could encamp we had to set to with axes and clear away
a number of young palms and mimosa bushes, make a zareeba, and before
retiring for the night look to our rifles and revolvers and see that we
had plenty of cartridges ready in case of emergency, as we flattered
ourselves that we were like the Bristol, Sheffield, or any other boys
in England who slept with one eye open. At all events we had heard
sufficient of this country to know that it would be unwise for us to
be caught napping, especially as we noticed some of the natives spying
about soon after we had pitched our tents. About mid-day we sighted the
village of Sarcella, the inhabitants of which Sheik Ahmed had gone to
interview, and whom Suleiman designated as "a very bad peoples." There
is very little doubt that Suleiman was right, if all we heard about
them was true, for in 1869 or '70—I am not sure which—Mr. Powell, wife,
little boy, and all the Europeans were spitted on their long spears.
They now lie buried in Bassaleg churchyard, near Newport, having been
brought home by his brother, Mr. Powell, M.P., who fearfully avenged his
brother's death. Whilst I was in the Soudan I saw by a newspaper which we
got, that this gentleman lost his life in a balloon.
January 30th.—Made a short march to-day, namely, from 10.30 a.m. until
4.30 p.m., encamping at a place called Wo-amma, playfully christened
and ever afterwards known as "Whoa Emma." The country was to-day very
mountainous and difficult for the caravan, to say nothing of ourselves.
I distinctly recollect that on this very day, whilst travelling along
a plane, one of the horse-boys came trotting quickly along, causing
my camel not only to shy, but to bolt when I was quite unprepared for
any such _contretemps_. A spectator would, doubtless, have been much
amused. I was not. For the space of about 20 yards I bounded like an
india-rubber ball on the makloufa; then came suddenly to grief from
my lofty elevation, the distance from the camel's hump to his feet
being considerable. I fell with a regular bang on to my hip, which
felt very painful for some days afterwards, and had the mortification
to see my belongings gradually parting company with the camel—my rug,
then my satchel, a basket, zanzimeer, &c. The camel was caught after
some trouble, whilst I and they were gradually picked up, I with rage
in my heart, for this camel, being a bolter, had served me several
scurvy tricks before; for instance, if we came to any little declivity,
the beast would persist in making a trot of it, greatly exciting my
apprehensions. Again, when we came to a narrow pathway I would duck my
head where there were overhanging boughs of prickly shrubs; he, thinking
I was going to thrash him, would at once bolt, and when he had rushed
through I should find my head and hands were like a pincushion. I could
then knock my helmet out of the trees and at my leisure pick out the
horrid thorns with which my head and hands abounded. Once I was nearly
swept off the wretched beast as he bolted in this way. A strong chain of
cactus was across our way, catching me in the middle. I saw the danger
in time, and clutched hold of the makloufa with all my might, on I might
have been found suspended amongst the trees. It broke, fortunately, and
I escaped, but I never shall forget how angry that camel frequently made
me, what self-restraint I was obliged to exercise, for if I chastised
him he would bellow and bolt again, to my great danger and annoyance.
I had in England extracted many human teeth in my time, but this day
I extracted an elephant's tooth, and brought it home as a curiosity.
However, I think I ought to say the elephant was not alive, and that on
the march we passed the skeletons of two others, which, I have no doubt,
furnished some excellent repasts for the natives. These were not all we
passed calling for notice, for some of our men came upon two of the Basé
people; the first sample of these curiosities we had seen. One was in
a baobob tree gathering the pods and throwing them down to the other,
who was collecting them in a basket. These so-called ferocious savages
appeared terribly alarmed when our men came upon them. The one on _terra
firma_, with the true instinct that "self-preservation is the first law
of nature," bolted like a shot, but our men captured him. The other
was afraid to come down until one of the English servants discharged
the contents of one barrel of his rifle, and let him know by the aid
of Beyrumfi, our interpreter, that the contents of the other would be
lodged in his frail tenement of clay unless he was more sociable. This
persuasive kind of argument appeared very effective, for down he came,
and I am sure both he and his companion, who, doubtless, were accustomed
to be hunted like wild beasts, were agreeably surprised when they each
received a pocket-knife and a bit of bread and meat. Remembering the
injunctions of Sheik Ahmed, and with Powell on the brain, we took
precautions against surprises—set to work and made a zareeba round the
camp, lighted camp fires, and looked well to rifles and revolvers. The
latter we kept under our pillows; the former at the heads of our beds,
ready at a moment's notice. Sentries were posted, and occasionally
relieved, whilst one of us (whoever chanced to awake) went round to see
that they were doing their duty. The man who was not, felt unhappy next
morning, as he received an intimation from the coorbatch before breakfast
that there had been a certain dereliction of duty on his part.
January 31st.—We marched eight hours to-day, encamping at Fodie on the
dry river-bed, close to some wells. This was a very fatiguing march
for the caravan, on account of our luggage, which was much obstructed
by trees. We travelled through quite a forest of these. Where there
were no trees the grass had been burnt for miles round. Many quail and
sand-grouse were shot to-day.
February 1st.—After a short march of four hours only, we encamped on
the broad, sandy river-bed of the Mareb, very near to the village of
Koolookoo, which we could see high up amongst the rocks of a mountain
on the opposite side. Each side of the Mareb was plentifully lined
with overhanging trees of all kinds, and amongst the twigs of some
could be seen many hundreds of the beautifully constructed nests of the
weaver-bird. Not very far from the Mareb, at the base of the mountain, in
which these Basé at Koolookoo lived, were the remains of a mud house in
which Mr. Powell had once lived. This was, I believe, as far, or nearly
so, as he had penetrated.
The Kunama, or Basé country, is quite a _terra incognita_, and, as far as
we could ascertain, we are the first and only Europeans who have explored
that country at all. This being so, I shall expend a good deal of time in
saying all I can about this country and people. We shall have to thank
Messrs. W. D. James and Percy Aylmer for a map of that country, and also
for some photographs of the people and scenery, which will be found in
Mr. F. L. James's book. This book I have not yet read, and shall not do
so until my own is in the publisher's hands, for fear I may unwittingly
adopt any of his theories or expressions, but rather prefer to be
perfectly independent of it, and give _my own_ ideas and description
in my own way, be they good, bad, or indifferent. No two men agree on
any subject, and it is _very_ probable that Mr. F. L. James and I may
materially differ on _many_. So that I shall not be termed a copyist, I
shall neither reproduce the map or photographs, but trust to sketches
taken to the best of my poor ability, but which, I hope, will convey
a pretty good idea of the kind of place and people that we sojourned
amongst for a while. Very well, then, after this exordium—probably the
longest I shall make—I will continue my narrative, as the dog would say
of his caudal appendage.
This was the first Basé village we had come to, and ere we could go any
further it was necessary that we should interview the Shiek of this
village, and explain the object of our visit. We made an ostentatious
display of our rifles and guns, twenty-four in number, and placed them
against the bank ready for immediate use if necessary, whilst each of us
sported a six-chambered revolver in our waist-belts. When we had—as we
thought—taken sufficient precautions against surprises or treachery, we
were curious to see these much-dreaded savages, whom report said were
capable of any sanguinary deed (could, in fact, murder with a smiling
face), and although their neighbours lived on their borders, they
appeared to know little more about them than we did ourselves. Whilst
we lunched within easy reach of our rifles, we sent forth one Beyrumfi,
"our guide, philosopher, and friend" (and the only man who knew anything
of the language) to the village. When we had finished our luncheon, we
got our field-glasses, and on the very summit of the rocky mountain
we saw all the women and children, and a few of the men, looking down
on us. Half an hour afterwards, winding round by a circuitous pathway,
on sloping ground, and occasionally hidden by trees, we could now see
Beyrumfi, accompanied by seven or eight of the Basé, each carrying his
spear and shield. When they appeared on the edge of the river-bed in
single file, headed by Beyrumfi, the Sheik's son (a fine, strapping,
well-made fellow, who took his father's place during his absence) dropped
his shield, and, without stopping, drove his spear quivering into the
sand; his example was followed by all the others. They all marched
briskly across the river-bed, whilst we, in our English fashion, stood
up and shook hands all round, which, under such circumstances, was much
more agreeable than kissing all round. Sheik junior, if I may call him
so, was about 5ft. 10in. in height, as straight as a dart, and not by
any means over-dressed, for he wore nothing but a bit of soft leather,
very much the shape and size of a man's bathing drawers. He got the
twig of a tree and broke it with us as a sign of friendship. All then
squatted round on their haunches, with their knees under their chins
(their customary mode of resting themselves), and Beyrumfi explained the
object of our visit. This was satisfactory. The Sheik then borrowed a
two-edged sword from Beyrumfi, placed it on the ground with the point
directed towards us, put his _naked_ foot on it, and delivered a short
harangue, the purport of which was that we were in his country now, and
as long as we remained neither he nor his people would harm, but do all
they could to assist us, and that we were now his brothers. However, he
could really only speak for his village. This is what is called making
"Aman"—that is, swearing peace and friendship, and that we will trust
one another; but we didn't. On hospitable thoughts intent, we ordered a
large bowl of cooked meat; our new acquaintances soon squatted round, and
judging from the rapid disappearance of the food, I should imagine that
a larger bowl would have done very well. We gave each of these fellows
small presents, amongst other things an empty claret bottle each, which
was much prized, but to the Sheik's son we gave a few extra things, such
as a tope or loin-cloth, a razor, a knife in sheath, needles, pins and
thread, a velvet necklet, and a waistcoat striped yellow and black. He
at once invested himself with the order of the tope and yellow and black
waistcoat, to the great admiration of his friends, who continually made
a clucking noise with the mouth, just as we do to urge on a horse; from
their point of view it meant how wonderful, how nice, and what a swell
you are. The claret- lead-capping of bottles, which had been
thrown on the ground, they gathered up, using them to decorate their hair
with, or as an addition to their necklaces. Our rifles and guns were
still leaning against the bank, just to show how well armed we were. Now,
finding the natives were so friendly, and that they had left their spears
on the other side of the river-bed, we ordered our rifles to be taken
into our tents; still, however, retaining our revolvers. Of course a long
pow-wow ensued. Whilst this was going on the women and children were not
idle in the village, for they stood out on various places of vantage,
looking down on their braves. We lent the Basé field-glasses to look at
them, and it was most amusing to hear their expressions of surprise,
with any amount of the clucking accompaniment, as they saw how near the
glass brought their friends to view. After a while they returned to their
village, upon which several of their friends, finding not only that we
appeared reasonable beings, but that we had given several presents,
paid us a visit, no doubt hoping that we would serve them in the same
way. Of course the wonderful Ingelese exhibited to all these visitors
their rifles and revolvers, accompanied by an elaborate explanation of
their killing powers. Beyrumfi explained all this amidst a shower of
cluckings. We had been told by Sheik Ahmed that the Basé were no better
than beasts, that they lived in holes in the ground and in caves; we
resolved to see for ourselves, and so told the Sheik that we would pay
him a visit on the morrow, which we did. They don't absolutely live in
holes in the ground like rabbits, but where the rocks lean against one
another, or project out, forming an awning, they utilize these accidents
to convert such a place into a dwelling; they also have many well-made
huts. In these particulars they differ from wild beasts, but I think in
most other particulars they very much resemble them. As for their being
the ferocious savages represented to us, I must say that they appeared
more afraid of us than we were of them. I formed an idea that they had a
cowed, hunted look, and well they may have, as the Egyptians squeeze all
they can out of them on one side, and the Abyssinians on the other, and
the reason they live in such places amongst rocks difficult of access is
that if attacked, they can roll these rocks down on their assailants. The
attire of both men and women is extremely simple and scanty. The women
wear a short skirt reaching from the waist to the knees, most of them a
large ring in one nostril. Many of them are not bad looking; their black
hair is not profuse, but inclined to be frizzly; this is plaited down,
whilst bits of metal, brass rings or beads, are frequently interlaced.
All have lovely teeth. In stature they are rather short and when young
possess rather graceful, well-formed figures. Either beads, metal, or
some other ornament surrounds the neck, the arm, just above the elbow,
the wrists and ankles. Very many, both men and women (the Arabs as well),
have the scars of burns about the size of a shilling. I do not know
whether it is so in all cases, but in very many, if they are in pain in
any part of the body, they apply a hot iron button (technically known
as the actual cautery). A very common custom is to decorate the chest,
abdomen, and back (sometimes one of these, sometimes all of them) with
a series of little cuts, into which a dye called kohl is rubbed in.
Kohl is also, much used by the Basé to stain their eyelids all round,
which produces a bluish-black stain. Whilst speaking of this dye, I may
say that it is supposed this was the very thing which Jezebel used to
improve her personal appearance. The difference between the Basé men
and women in the matter of dress and ornaments is that the men, instead
of a short tope or skirt, wear a bit of thin leather round their loins
(like a rather scant pair of bathing drawers), and a scratcher in their
hair. I saw some moderately big boys attired in the most inexpensive suit
conceivable; namely, an anklet and bracelet of metal, and a bit of a
porcupine's quill in the left nostril.
Speaking generally, the men are well-formed, agile-looking fellows.
These Basé people are quite hemmed in in their small country, on the
one hand by the Abyssinians and on the other sides by different tribes
of Arabs, with whom they appear to have little or no communication or
dealings, and if they venture out of their own country they are hunted
down by the Arabs just like wild animals. The Arabs of the Soudan are
darker than the Abyssinians, but the Basé are much darker than the Arabs
and speak a different language. The Basé are quite a different race to
their neighbours, and more nearly approach the <DW64> type. They are
blacker than the Arabs, but not the coal-black of the <DW64>; their hair
is shorter, more crisp and woolly, than the Arabs, but not the absolute
wool of the <DW64>. The Arabs have good regular features, lips and noses
like our own; the Basé are the contrary, and more resemble the <DW64>
in this respect and their high cheek-bones, but they are not nearly so
pronounced as the <DW64>. Their foreheads, as a rule, are rather narrow
and receding. I was obliged perforce to depend on Sheik Ahmed, and
more particularly on Beyrumfi, for all the information I could glean
respecting these people. They say they have no religion. Sheik Ahmed,
speaking very contemptuously of them, says "that they have a rain-maker
who promises rain, when it is pretty sure to come; but if he makes
several promises and the rain does not come, he goes"—to that bourne from
whence no traveller shall return. In the little matter of marriage, their
laws and ceremonies are extremely simple, for they marry their sisters,
their daughters, their cousins, and their aunts, possibly their mothers
and grandmothers. Courtship is brief and primitive. A Basé man fancies
a Basé girl (presumably not his own daughter); he tells the nearest
male relatives so, father or brother—good; he then presents him with a
few yards of calico or some skins, the same also to his bride, and she
becomes his.
Now with regard to their diet. I cannot help thinking that this admits
of considerable improvement. As they are not possessed of large flocks
and herds like their neighbours, the Beni-Amirs, they have not much milk
or meat, neither have they so much dhurra as an article of diet. They
obtain meat occasionally when they can ensnare an animal; the _kind_ of
meat is rather a secondary consideration for they will eat the meat of
lion, panther, elephant monkeys, lizards, or giraffes with as much gusto
as that of antelope or buffalo. They are not so particular, either, as
they ought to be, for they consume all except the skin and bones. They
also eat the roots of young palm trees, the outer covering of the dhoum
palm nut, nebbuck, and hegleek nuts, the fruit of the baobob, wild honey,
and a certain, or rather an uncertain, quantity of milk and dhurra. They
do not indulge in baked baby, and I am quite sure that their carnal
longings are never satiated with cold or roast missionary, as there are
no missionaries there, but it has occurred to me that this place is
virgin soil for missionary enterprise, as there does not appear to be any
religion that requires eradicating from their minds.
In the evening of the 2nd February a dirty-looking old fellow (a sheik
from Aidaro), paid us a visit, bringing with him a gourd of wild honey
as a peace offering, made "aman" with us, and of course received his
presents.
I was much struck when visiting the village with their beautifully made
baskets; so closely woven are they as to enable them to carry milk or
water in them without a drop oozing through.
CHAPTER XVII.
WE LEAVE KOOLOOKOO, ACCOMPANIED BY A NUMBER OF THE BASÉ—THE
MAGIC LANTERN—SEE BUFFALO AND GIRAFFE FOR THE FIRST TIME—TWO
BUFFALOS KILLED—A BASÉ FEAST—CURIOUS BASÉ DANCE—THEY DRY THEIR
MEAT ON LINES IN THE SUN—A WOUNDED BUFFALO—HOODOO, CHIEF SHEIK
OF THE BASÉ, VISITS US—A COLUMN OF SAND—A LEPER—THE BASÉ
SQUABBLE OVER THE MEAT—WE ARRIVE AT ABYSSINIA.
On the 3rd February we made a further advance, starting at 11 a.m., and
encamping at 4 p.m., again on the river-bed, at Aibara. This day we
marched for the space of five hours through a forest; the heat was very
great, and the ground over which we travelled was full of large, deep
cracks, often two or three inches wide, caused by the contraction of the
earth, which had been subjected to a continuous baking by the hot sun
since the rainy season. Oftentimes could be seen the great footprint of
an elephant, now quite a moulded one, having been there since the rainy
season. On leaving Koolookoo we were accompanied by about 80 or 100 of
the inhabitants, having nothing with them but spear and shield. We knew
what this meant—that we should have to provide them with food—a rather
large undertaking considering that our own party, including camel-men,
horse-boys, and servants, numbered about 40 or 50. Accordingly a delicate
hint was conveyed to our new body-guard, that our own people would first
of all have to be provided with food; then if there was plenty of meat to
spare they would be quite welcome to it. To this arrangement they amiably
acceded. On _terra firma_ we could have made a good stand with our rifles
and revolvers in case of attack, but had these Basé thought proper,
at a preconcerted signal, to make an onslaught on our long straggling
caravan, I am afraid we should have fared very badly, notwithstanding
our being well armed. However, I think their principal reason for coming
with us was to have a continual feast of meat, an article of diet they
were capable of stowing away as capaciously as a lion would do, and with
as little ceremony. In the evening three sheiks paid us a visit, each
going through the ceremony of "aman." After dinner the magic lantern was
exhibited, and this excited their astonishment even more than it did
that of the Beni-Amirs. I do not intend to go into a description much of
hunting-scenes, as they would occupy too much space, and I do not think
that the frequent repetition of such scenes would be interesting to the
generality of my readers; besides which I have no doubt Mr. F. L. James
has done this in his book. Suffice it to say that as there was abundance
of game of every description, scarcely a day passed without plenty being
brought into camp.
February 4th.—Off at 10.30 a.m., halt at 5 p.m.; pitch tents at
Maissasser, on the bank of the Mareb, and quite close to jungle. About 12
o'clock, as our camels were slowly winding along the bed of the Mareb,
a grand bull buffalo, an enormous beast, dashed right across in front
of us, raising quite a cloud of dust. This was the first buffalo we
had seen; at half-past 4 p.m. we saw three more, and just afterwards a
giraffe. There was a good deal of chuckling now at the prospect of sport
in store, and we resolved to encamp here for the next two or three days.
To-day we saw miles of country on fire. The country looks much greener in
this neighbourhood; trees and jungle abundant, and water much nearer to
the surface.
February 5th.—Last night, about 11 o'clock, just as I had gone to
sleep, I was considerably startled by several rifle shots, one after
the other. In an instant I was out of bed, rifle in hand, rushed out
of my tent in my slippers and night shirt, not knowing what to think;
the first idea naturally was that we were being attacked. Messrs. F. L.
James and Phillipps, who slept in another tent, were also out, clad in
the same airy costume as myself, and, like me, each with a rifle in his
hand. All this was the work of a minute—we had scarcely time to say,
"Whatever is the meaning of it all?" when close behind my tent, amongst
the thick stems of the tall grass, there was a sound as of a rushing
mighty wind. This was enough; the whole affair was explained at once;
we knew directly that this was nothing less than a herd of buffalos,
and I am very thankful that they just avoided my tent, which could as
easily have been upset by them as a box of matches. It seems that just
after we had gone to bed, the others, Messrs. Colvin, P. Aylmer, W. D.
and A. James, "from information received," took their rifles (it was
a bright moonlight night) and stole out cautiously to the edge of the
jungle. There they saw a herd of buffalos drinking, and into it they
discharged their rifles with pretty good effect, for about 11 o'clock
this morning one of the herd was found dead in the jungle pretty near
to camp. A camel was sent to the spot to bring home the quarters for
food, and the head, of course, as a trophy. A great number of the Basé
were pretty quickly on the spot; then there ensued such a scene as I had
never before witnessed, and which almost baffles description. I will,
however, endeavour to describe it, as some of my readers might like to be
furnished with particulars. Invalids, persons of delicate organization
and others, might, however, like to omit this little account of a Basé
feast, which I assure them will not have an appetising effect. I may here
say that there is not the least occasion for me to draw on my imagination
and indulge in what some people facetiously call "crackers," which I have
not and shall not do, as there is no necessity for doing so, there being
abundance of material of a strictly veracious character which I cull from
my diary, written carefully down at the time. Incredible as some accounts
may appear, I must ask my readers to accept these facts without the usual
formula _cum grano salis_. Very well, then, I will write down, and you,
reader, can read, mark, and inwardly digest (if you please) _without_ the
usual proverbial pinch of salt, a description of a scene that I was an
eye-witness of, and if I should somewhat interfere with your enjoyment,
when called from labour to refreshment, don't blame me, but blame the
Basé. All I can say is, that this is not what incredulous people call "a
traveller's tale," but a "true story." Do not say, "It strikes me that he
doth protest too much."
I recollect to have seen somewhere or other a pamphlet entitled "The
Stomach and Its Trials." That useful organ in the human body of Basé does
not appear to be subject to usual inconveniences, but accommodates itself
to circumstances, not unlike an india-rubber bag. The only trials I saw
them suffer was trying how much they could stow away without causing a
rupture of that viscera.
Well, to continue. As soon as the animal was opened they fell upon the
intestines like hungry wolves. Oh! such a scramble for tit-bits. There
were our dusky friends very soon ankle-deep in the viscera, and about
20 pairs of hands clutching at them. Two would perhaps get hold of the
same piece, and pull away like a couple of dogs, until a knife produced
a solution of continuity. Another group could be seen hacking away at
pieces of the liver, and cramming the warm, quivering morsels into their
mouths. One could be seen stuffing a lump of fat into his mouth with
one hand, the other at the same time would be industriously employed in
rubbing this adipose tissue into his hair. Another appeared to have a
predilection for kidneys; and so this disgusting feast went on, until
the whole interior of the animal was consumed, without such absurd
preliminaries as cooking.
One would naturally suppose that I should be busy at my medicine-chest
next day, but not one of them even so much as troubled me for a pill
afterwards. They might truly say, "We are fearfully and wonderfully
made;" and after this exhibition of digestive powers, I should be
obliged to coincide with them. When they had gorged themselves like
boa-constrictors, I should not exaggerate if I said that they presented
a most filthy and disgusting spectacle. Their proportions were quite
aldermanic, and their mouths, faces, and arms up to the elbows were
smeared with fat and gore. Had this buffalo lived a month or two longer
she would have become a mother.
We do not consider very young veal wholesome, but whether the Basé
thought this _very_ young buffalo would be a delicacy they must not touch
I know not; at all events it was brought into camp. We gave it the Basé,
who appeared quite pleased. In less than ten minutes afterwards we saw
three of them engaged in tearing it limb from limb, and eating it without
going through the formality of cooking. The quarters of buffalo, senior,
were divided between our men and Basé; the hide was cut up into sections
and given to the sheiks and others to make into shields.
Messrs. Colvin and Aylmer shot to-day two mehedehét. This is a very
beautiful antelope, possessing a very rough coat, a fine pair of horns,
slightly curved and annulated; is about 13 hands high, and in colour much
resembles the red deer. Messrs. F. L. and W. James stalked an ostrich for
two hours, but did not succeed in bringing him down.
We were encamped on a little plateau by the side of the Mareb, close to a
great jungle. On the opposite side of this wide river-bed were very many
trees of different kinds, and on both sides rocky mountains. Just by our
camp, on the sandy river-bed, the Basé were encamped. Notwithstanding
their mid-day feast on the uncooked internals of the buffalo, they were
ready and willing for another set to in the evening—this time of cooked
meat. Whether this second gorge had a stimulating and intoxicating effect
on them I don't know, but just as we were off to our various tents for
the night, at 9.30 p.m., we heard strange noises issuing from the Basé
camp. Messrs. A. James, Colvin, and I were curious to ascertain the
cause, so down we went amongst them, and this is what we saw, and what I
have some difficulty in describing:—
All the Basé were engaged in a peculiar dance, four or six abreast, and
so close to each other as to be almost touching; those behind always
treading in the footsteps of those in front, whilst each one held his
spear aloft at arm's-length. What they were saying I don't exactly
know, but it was a dance of joy celebrating their feast of meat. A few
words as a solo would be sung, then all would join in chorus. This
went on for about half an hour; then they broke up and went through a
wild war-dance—now flying forwards and darting out their spears at an
imaginary enemy whilst protecting themselves with their shields, then
nimbly retreating, crouching and springing like wild cats. It was a novel
and singular spectacle to see nearly a hundred of these black savages,
with their glittering spears, agile as monkeys, leaping in and out
between about 30 flickering fires on the river-bed. Like the Pharisee of
old, I could not help (mentally) exclaiming, "Thank God I am not as these
men are."
I then retired to rest, and slept peacefully and soundly until the
following morning.
February 6th.—Soon after breakfast we saw, stretched from tree to tree
near the camp, what appeared like clothes-lines with stockings suspended
on them to dry. The Basé had made ropes out of the palm-leaves, and on
these the meat which they could not then dispose of hung in strips and
festoons to dry. When dried, they would stuff them into gazelle skins,
or bags of some kind, for future use. To-day about 40 of them returned to
Koolookoo, well-charged with meat, both in their own skins and that of
gazelles; the rest remained with us.
This morning another buffalo, which was wounded on the night of the 4th,
was found, but not dead. He, however, received his _coup de grace_ from
Mr. Colvin's rifle, but not until he had made a furious charge, though
so badly wounded, fortunately without any ill-results. A wounded buffalo
is about as dangerous and fierce an animal as can be met with, and will
charge most savagely if he is only within five minutes of death, provided
his limbs will support him. We had plenty of meat brought into camp, for
in addition to this buffalo, two nellut, a mora, and two buffalos were
shot. Temperature at 1.30, solar thermometer 150°, wet bulb 66°, dry bulb
in shade 90°.
This has been rather an exciting day, as Mr. W. James saw and stalked
three giraffes, but was not successful in getting near enough for a shot.
Sali, the tracker, saw three ostriches and a rhinoceros, the latter
pretty close to him, and I two full-grown elephants a distance off, but
none of them were bagged. No doubt had these elephants been followed up
for a day or so they could have been got at, but they were not. Messrs.
A. James and Colvin followed them up next day for many miles, but not
far enough.
February 7th.—Messrs. Colvin and A. James, who went in quest of the
two elephants, returned about 4 p.m., without having seen them. In the
evening Hoodoo, chief sheik of the Basé, paid us a visit, bringing with
him a pot of wild honey as a present. He went through the formality of
making "aman" with us, after which he squatted on his haunches in the
usual native fashion.
During a long pow-wow which ensued, I was busy making mental notes of
Hoodoo, not by any means complimentary to that august personage. He was a
dirty-looking old fellow, as scantily dressed as his colleagues, nearly
black, with an ill-favoured, sinister cast of countenance, and not by
any means a man whom I should place unbounded confidence in. He received
several presents, amongst others a bernouse and a rather gorgeous-looking
abia (a cloak-looking kind of thing), with gold braid ornamentation
around the neck. He seemed mightily pleased with these. He then joined
his comrades' camp, and we went to our dinner.
This was rather a nice camping-ground, but quite unsheltered from the
sun by trees. However, we provided a shelter by cutting down some young
trees, fixing them in the ground and making a covering of matting, tall
grass, and palm leaves, which were obtainable close by. So great was the
heat now that the ink dried in my pen ere I could write three lines on
a page of foolscap. However, I was fortunately provided with Mappin and
Webb's stylographic pen, which is really invaluable in such hot climates.
Always about 12 or 1 o'clock a slight breeze would spring up, producing
occasionally a very curious phenomenon. A very high column of dust
(perhaps half a mile in height) would come whirling and waltzing along
right through the middle of the camp, and so long as it was not hidden
by trees or mountains I could see it spinning on and on, looking in the
distance like a column of smoke. A good deal of sport was obtained in
this neighbourhood, chiefly buffalos and antelopes of different kinds.
February 8th.—We struck our tents and were in marching order by 11 a.m.
After an easy journey through some pretty good country, where vegetation
was abundant, we encamped at 3 p.m. on a nice bit of land by the Mareb
amongst many tall trees and shrubs, which afforded a good shade. Here
we purposed remaining for a week at least, as big game of all kinds was
plentiful, and here for the first time we found rhinoceros' tracks. This
place is called Maiambasar, and is situated on the border of Abyssinia.
Water is near the surface here. During the journey I have noticed that
as we have got nearer to Abyssinia we have found the water nearer the
surface. We heard panthers, hyænas, and lions last night near camp, but
lions are not so plentiful as they were a few days ago.
February 9th.—Abdullah, a black boy, who looks after my camel, has been
walking very lame during the last few days, having considerable swelling
at the knee. I find he has a large abscess, produced by a guinea-worm. He
comes from Algeden, which is about five days from here between Kassala
and Souheet. He says guinea-worm is very common there and on the White
Nile.
Strange to say, all returned to camp in the evening without having
obtained game of any kind, although out all day. Mr. Aylmer, whilst in
search of game, suddenly came upon a rather curious scene. There, on
the mountain side, scarcely sheltered from the burning rays of the sun,
was an old man suffering from leprosy, miles away, apparently, from any
human being or habitation. Food and water had been placed near him, to
which he could help himself. Mr. Aylmer informed me that the surrounding
atmosphere was charged with the stench arising from the decomposed food
which was scattered around the place. I should say that most probably
that old man furnished a meal for one of the wild beasts ere long.
February 10th.—Two buffalos and two nellut killed this day. One, a bull
buffalo, was an enormous beast, and probably the hero of many a fight,
for one of his horns had been knocked short off, one eye knocked out,
whilst his forehead was covered with scars—evidently a disreputable,
cantankerous old buffalo. His carcase was given to the Basé, who were
well pleased with the donation. The Koolookoo Basé who came with us
through the country would, I daresay, have divided the meat amicably
between them, as their Sheik was with them, but their number had been
materially increased by other Basé.
In the evening, during the division of the spoil, just outside our camp,
a great difference of opinion prevailed as to _meum_ and _tuum_. Knives,
clubs, spears, and staves were freely brandished amid a chorus of yells
and shouts, ending in a scramble for the joints and pieces of meat.
Some of them secured a reasonable share, and trotted with it; others,
again, not so fortunate, would intercept a fugitive caressing, perhaps,
the thigh of the deceased buffalo. Then a desperate struggle would
ensue between them, five or six pulling away in different directions.
Fortunately all was settled without bloodshed, peace reigned in the camp,
and we all retired to our respective tents at a respectable hour.
CHAPTER XVIII.
THE DEMBELAS ATTACK US, MAHOMET WOUNDED,—NARROW ESCAPE OF TWO
OF OUR PARTY—ACTIVITY IN CAMP, WE MAKE A ZAREEBA AND FIRE
THE COUNTRY—HOLD A COUNCIL OF WAR—OUR SILENT AND DANGEROUS
RIDE—HOODOO'S SAGACITY—ARRIVAL IN CAMP OF MAHOMET, WOUNDED—WE
RETREAT—MAHOMET'S DEATH AND BURIAL.
February 11th, 1882, was the most memorable day of the whole campaign.
Thinking it was not safe to leave the camp without protectors, Messrs. W.
D. and A. James and I remained in camp, whilst Messrs. F. L. James and
Colvin went out in one direction and Phillipps and Aylmer in another,
in search of big game. Each party went out mounted on ponies, which had
been bought for the purpose at Kassala and Heikota. Each party took an
agreegeer, or huntsman, a horse-boy, camel-boy, and camel with them, the
latter for the purpose of carrying home the game. They started soon
after 8 a.m., Messrs. Phillipps and Aylmer went in the direction of some
mountains on the Abyssinian border, whilst Messrs. James and Colvin took
the opposite side of the Mareb. We amused ourselves in camp in reading,
writing letters, or posting up diaries. Keeping my diary carefully and
correctly posted up day by day was a duty which I most religiously
attended to before ever I retired to rest, however fatigued I may have
been by the day's march. Incidents and impressions written down at the
time are more likely to be correct than if left to memory. From my diary
I quote the following particulars:—
"About 1.30 p.m., just as we were about to sit down to luncheon, Messrs.
Aylmer and Phillipps came into camp looking considerably chop-fallen and
exhausted, and having only one horse between them. Of course we did not
expect either party home until 5 or 6 p.m., so I said to Aylmer—
"'Hallo! how is it you are back so soon, and looking so precious serious?'
"'I can tell you, doctor, this is no laughing matter,' said he, 'for we
have been attacked by the Abyssinians or Dembelas, and very likely they
will soon be down on our camp.'
"This certainly did look a serious business, especially as we had no
zareeba round the camp; so I said—
"'Well, the best thing we can do is to have our luncheon at once; then we
shall be more fit for work.'"
The wisdom of this suggestion was apparent, and at once acted upon.
Whilst it was being brought I strapped on my revolver, brought out my
diary and entered the above conversation. Mahomet Sali and others were
at once sent out as scouts in search of Messrs. James and Colvin, with a
promise of five dollars each to those who brought them into camp. We then
sat down to luncheon, and the following account was given of this affair,
and was duly entered in my diary immediately afterwards, as we did not
know when we might be attacked, and I was desirous of leaving my diary
posted up complete to latest date.
Aylmer's story:—"We had got about eight or nine miles from camp on the
sandy river-bed, quite in a hollow, precipitous rocks and trees on each
side of us, when suddenly about 30 strangers, who turned out to be
Dembelas, appeared. We thought they were Abyssinians, because they were
so much lighter in colour than Arabs, and, of course, quite different in
every respect from the Basé. Some of them seized our hands and commenced
kissing them profusely, exclaiming 'Aman, aman,' at the same time
beckoning us to lay down our rifles. Now, although we thought they were
friendly, we did not think it wise to be so confiding as this, until
Mahomet, the agreegeer (who, we supposed, knew more of the customs of
these people than we did), lay down his, beckoning us to do the same,
saying it would be better to do so. One fellow, wearing a felt hat, was
more demonstrative even than the others. Well, we followed Mahomet's
example; no sooner had we done so (we had four rifles with us and about
50 cartridges) than they were immediately seized, and a struggle ensued
for their possession. The man wearing the felt hat seized a valuable
elephant rifle, vaulted on to the back of Mr. Phillipps's horse and
galloped off. Attached to the saddle of that horse was his revolver, a
number of cartridges, and a field-glass. The horse-boy vanished like
smoke, whilst his horse was taken possession of by another of the enemy.
They attempted to spear the camel-boy, missed him and speared the camel.
Phillipps received a blow from the butt-end of a rifle which would have
prostrated him had not his helmet protected his head. He, however, turned
round, closed with his assailant, and succeeded in wrenching the rifle
from him. I pulled the trigger of my revolver, but on account of sand,
which obstructed it, I could not discharge it.
"In the hubbub which ensued Mahomet could nowhere be seen, and about 8 or
10 Basé (who had accompanied us, with the intention of having a feast and
cutting up the animal into quarters) vanished at once. We were now alone,
and by this time there were about 100 yelling demons brandishing their
spears, whooping and leaping about. Under such circumstances we thought
discretion the better part of valour, so we fled with one horse between
us, and have made the best of our way to camp, riding and walking by
turns."
This really was very alarming news, and we quite expected that we should
soon be fully occupied in defending ourselves from an attack.
We now resolved to fire the country all round and construct a strong
zareeba. Before I fell to with the axe I once more sought out my diary
and chronicled the above. All then fell to with a will, cutting down all
the prickly trees in the neighbourhood, dragging them round the camp
and so forming a very strong zareeba. This was no joke when the solar
thermometer registered about 150°, and the heat was 100° in the shade.
Whilst we were thus employed the horse-boy made his appearance, streaming
with blood, his flesh being torn by the cruel thorns as he rushed
blindly on. We now set fire to all the tall grass and bushes in the
neighbourhood. The terrific speed with which it spread was surprising,
miles of country were soon in flames. The crackling of the grass and
trees resembled more than anything else the most fierce hailstorm I ever
saw.
Now the camel-boy turned up and this was his account of the affair:—Just
as he leaped off the camel he saw Mahomet on the ground, whilst one
of the Dembelas darted his spear at him several times; then left him
writhing. He went to his assistance, and helped him along some distance,
whilst poor Mahomet supported his intestines (which had gushed out) in
his tope. At last the poor fellow sank down under a tree, saying—
"I can go no further; I shall soon die. Save yourself; make haste to camp
and tell the gentlemen to make a strong zareeba at once as they will
certainly be attacked, probably to-night."
By this time Messrs. F. James and Colvin, who had been found by the
scouts, arrived. We at once held a council of war, and determined not
only to go in search of Mahomet, but to attack the Dembelas if we could
find them, leaving two of our company in camp to command the men in case
of attack. All the neighbouring Basé, we found, had bolted—Elongi, the
Sheik from Koolookoo, and his men, alone remained, and promised to stand
by us and fight for us if necessary.
Our armoury consisted of about 22 shot-guns and rifles, and about a dozen
six-chambered revolvers. All the camel-men and native servants were
armed with spear and shield. Having provided ourselves each with rifle,
revolver, and cartridges, our two European servants; Suleiman, Ali, and
Cheriff also; we called up our native servants. To the most trusty of
these, including Beyrumfi, our guide, Mahomet Sali, Sali, the tracker,
and Ali Bacheet, the head camel-man, we entrusted the remainder of our
firearms, but unfortunately most of them had to be instructed in the use
of them.
I provided myself with a few bandages, lint, and my pocket case of
surgical instruments, and feeling that we were embarking on a very
perilous enterprise, left instructions respecting my immortal diary,
which would convey full information up to the time of our departure from
camp.
Just as we were getting into the saddle who should turn up but the Chief
Sheik, Hoodoo, who left our last camping place the morning after his
arrival. He _appeared_ surprised at all this commotion. Whether he was
really so or not I do not know, but I am sure, from the very first, I
did not feel that I should repose in him the trustful confidence of an
innocent child.
Our little army consisted now of Messrs. F. L. and A. James, Phillipps,
Aylmer, myself, six of our men with firearms, the Koolookoo Sheik, and 15
of his men with spear and shield. We started off, on what _might_ prove
to be our last journey, about 4 p.m. And I think we shall all remember
that silent ride of eight miles up the sandy river-bed of the Mareb to
Abyssinia, shaded often by trees which thickly adorned the banks. We well
knew, as we approached Abyssinia, that each bush may conceal an enemy,
who might at any moment spring out on us unawares, and knowing this,
each one clutched his rifle with a firm grip, ready for instant use, and
determined to sell his life dearly if the worst came to the worst. I felt
that our present position was something like that of Fitz-James, when he
held the interesting conversation with Rhoderick Dhu which Sir Walter
Scott so graphically describes—
Instant, through copse and heath, arose
Bonnets, and spears, and bended bows;
On right, on left, above, below,
Sprang up at once the lurking foe;
From shingles grey their lances start,
The bracken-bush sends forth the dart,
The rushes and the willow-wand
Are bristling into axe and brand,
And every tuft of broom gives life
To plaided warrior armed for strife.
That whistle garrisoned the glen
At once with full five hundred men,
As if the yawning hill to heaven
A subterranean host had given.
No one spoke above a whisper as we stole silently and quickly on, until
at last we arrived at the scene of the scuffle.
It was a wild and strange retreat,
As e'er was trod by outlaw's feet.
The dell, upon the mountain's crest,
Yawned like a gash on warrior's breast.
Here, and in the neighbourhood, we searched about for Mahomet or his
assailants. We spent two hours thus unsuccessfully, until darkness warned
us it was time to return to camp.
The shades of eve come quickly down,
The woods are wrapped in deeper brown,
The owl awakens from her dell,
The fox is heard upon the fell;
Enough remains of glimmering light
To guide the wanderer's steps aright,
Yet not enough from far to show
His figure to the watchful foe.
As we were returning, our attention was attracted to a large baobob tree
full of vultures. Sali and Mahomet Sali thought we might find Mahomet's
bones, picked clean by these foul birds, near the tree. We, therefore,
searched that neighbourhood, but found him not.
Darkness had come on ere we had retraced our way any distance, and we
returned as silently as we had advanced, keeping well on the alert until
we neared the camp.
In dread, and danger, all alone,
Famished and chilled, through ways unknown,
Tangled and steep, we journeyed on;
Till, as a rock's huge point we turned,
Our camp fire close before us burned.
It was about 10 p.m. by the time we got to camp. Dinner was soon on the
table. This we at once discussed, also our plans for the next day. The
line of action determined on was this—sentries were to be posted about
the camp, and a few outside to guard against surprise. These, again,
would be looked after by Suleiman. We should load up in the morning, and
return to our former camp, first of all having another hunt for Mahomet.
We could not divest our minds of the idea that we ought to attack these
Dembelas if we could find them, and thought that perhaps our dirty old
friend Hoodoo would assist us with some of the Basé. Accordingly, whilst
seated round the camp fire after dinner, he was sounded on the matter,
and promised £100 if he would lend a 100 of his people next morning.
Hoodoo mentally said, "Hoo dont." As a cautious look stole over his black
face he raised his eyes from the camp fire for a moment, stealing a
furtive glance at us; then, as he slowly shook his head, replied—
"I and my people will be here after you are gone, and if I was to do so
the Abyssinians would come down on us, burn our villages, kill our men,
and take our women and children as slaves."
I assure you, reader, that when that old man shook his head he did not
shake all the sense out of it. There was a good deal of logic in his
remarks, and it is highly probable that had the old man accepted the
offer made, he and many of his braves would soon have been translated (to
use an orthodox phrase), and the merry dollars would have danced off into
possession of the Dembelas; therefore the old man "deserved well of his
country." Indeed, his diplomatic action would almost entitle him to the
appellation of a grand old man, though he did not look it.
February 12th.—The stillness of the night fortunately was not broken by
the clash of arms, and I awoke, refreshed by a sound sleep, at 6 a.m.
Whilst we were at breakfast Mr. F. James proposed that we again ask Sheik
Hoodoo to let us have 200 of his men to assist us in making a raid on
the enemy, and if he would not, that we should send the caravan back to
the last camping place, and take about 20 of our own, as we were not at
all satisfied to leave poor Mahomet (who might still be living) to his
fate. Hoodoo was accordingly approached with the same result as before,
he adding—
"I am at peace with them, and I cannot make your quarrel mine."
The old man was about right. Six camels had yesterday been sent to
Amadeb, a garrison town, for dhurra. We therefore had to divide their
loads with the remaining camels. When most of the camels had their loads
on, we told our men that we should want some of them to go with us. This
(with the exception of about half-a-dozen) they flatly refused to do,
saying that they were engaged as camel-drivers not to fight the Dembelas.
They were evidently bent on retreating from the Abyssinian frontier
whether we were or not. Suleiman, who knew these people pretty well, now
stepped forward—
"What good you gentlemen go fight Dembelas? You only six or seven, the
Dembelas hundreds. You do no good. Mahomet, he dead now, and the vultures
eat him. If you go, these men go off with their camels. How, then, we get
out of the country? The Basé and Abyssinians then turn round and kill us
all."
This was good reasoning. Abdullah now brought my camel ready for
starting, so—pending the settlement of "to be or not to be"—I spread
my rug on the ground and lay down to read a book, with my rifle by
my side. It was now about 11 o'clock. I had not been here above ten
minutes when I saw everyone rushing across the camp, rifle in hand,
shouting—"Hakeem," and "Doctor, doctor—quick!" I was up in an instant,
rifle in hand, and darted across in the same direction as the others,
naturally thinking—"The time has come at last. We are in for it now with
the Dembelas."
It was not an attack at all, but a most pitiful sight. There was poor
Mahomet, who had managed to crawl into camp, then sank exhausted on the
ground. The poor fellow turned his large soft-looking eyes piteously on
to me. He was supporting with his hands and tope as much viscera—covered
with sand dried on it, and quite adherent—as would fill a good-sized
washing-basin. My rifle was at once dropped for my dressing-case; water
was obtained, the opening (made by a spear) slightly enlarged, the
viscera washed, replaced in the cavity of the abdomen, and the opening
secured by a suture or two. He was in a state of collapse, and death
pretty certain. He had also a wound in the fleshy part of the arm, and
two others in the muscles of the back, just by the spine. Whilst I
attended to these, some extract of beef and brandy was obtained, a fire
soon kindled, and a good supply of brandy and beef-tea, administered,
which soon revived him. An angarep was then rigged up, and twelve of
the Basé (six at a time) were told off at a dollar each to carry him on
to the next camp. I followed close by, giving him brandy and beef-tea
every half-hour; but all was of no avail, for he died at 9 o'clock next
morning. He was a Mahomedan. A tope as a shroud, with some needles and
thread, were given to the friends. They would not, however, use the
needle and thread, preferring the shreds of the palm leaf. A grave
was dug near the camp, and there the poor fellow was buried; Suleiman
remarking, "These Basé, they soon have Mahomet up; they not leave that
good tope there long." And I have no doubt he was right in his prediction.
Mahomet's account of the scrimmage, and his escape to camp, was this:
He stated that when the Abyssinians, or Dembelas, said "aman" it meant
"Put down your arms and take your lives; we are stronger than you, so
you must give up all you have." He noticed that they were tremendously
out-numbered, so thought it was the best thing to do. When he saw Mr.
Phillipps trying to recover his rifle, another Abyssinian was about to
strike him with the butt-end of a rifle; he rushed to his assistance,
and it was then he received the fatal spear thrust in his abdomen,
causing him to fall down on the sand, where he received two or three
more, as detailed before. When everyone ran away, he tried to struggle
on, succeeding ere night came on in getting pretty near to camp by
alternately walking and resting awhile, until at last he sat under a
tree quite exhausted. He says that he both saw and heard us when we were
returning after our search for him, and that he cried out, but could
not make us hear. The greater part of the way to camp next morning he
accomplished by crawling on his hands and knees.
Who can imagine the sufferings of the poor fellow—out all night, sitting
with his back to a tree? He said he felt the cold very much; and well
he might, as, although the thermometer registered 101°F. in the shade
at 1 p.m., it dropped to 37° by 11 p.m. He passed the night in constant
expectation of a wild beast coming to tear him in pieces. We were very
glad that we had him with us during his last few hours, where he received
every attention that we could bestow.
February 13th.—Mahomet was buried, as I said before. Little was done
to-day; one nellut and three gazelles were shot.
CHAPTER XIX.
MESSRS. JAMES AND PHILLIPPS START ON A VISIT TO
RASALULU—CURIOUS WAY OF SHAVING CHILDREN'S HEADS—A DISGUSTING
BASÉ—THE CAMEL-DRIVERS BECOME MUTINOUS—INTENDED ATTACK
BY BASÉ—WE FIRE THE COUNTRY AND MAKE A ZAREEBA—ENCAMP AT
WO-AMMA—TROUBLE AGAIN WITH CAMEL-MEN—LIONS DISTURB US—ARRIVAL
AT HEIKOTA—A TALE OF BLOOD AND SLAVERY.
February 14th.—Made a rather short march, and encamped at Aibara, on some
table-land by the Mareb. Ere doing so we had to clear away a quantity of
mimosa bushes and young palms; then construct a zareeba. Mr. Phillipps,
at the request of one of the Basé, shot a monkey to-day. This was skinned
and eaten by them in the evening, and was, no doubt, looked upon as a
delicate morsel, probably as much so as grouse or partridge is with us.
February 15th.—This morning, at 9 a.m., Messrs. Phillipps and F. James
went off to Amadeb, to complain to Rasalulu, a deputy of King John of
Abyssinia, about our late attack, and endeavour to get their rifles back.
Whether they ever succeeded in doing so I don't know; but I should think
probably not.
To-day we lost another camel; this makes the sixth we have lost in the
Basé country. A camel is a particularly stupid kind of animal, and does
not seem to know what is good for him, or rather, what is bad for him,
for he will frequently eat a very poisonous green-looking shrub, called
"heikabeet." This appears to produce considerable pain, and, as far as I
could make out, inflammation of the intestines. I brought some of it home
with the intention of having it analysed, but somehow or other it has got
lost.
February 16th.—The Basé women and children, when we first came here, were
rather shy, and ran away from us as if we were monsters of iniquity; now
they appear to be getting quite tame, and are continually hanging about
the camp. The heads of the children are curiously trimmed, according to
fancy, just as they are at Kassala. All kinds of fantastical devices are
arranged, with the aid of a razor, just as a gardener operates on a box
bush in England. I have seen a child's head shaved completely, with the
exception of a tuft of hair just over the right temple; another will have
a tuft on each side, whilst a third will have those and one on the crown
in addition; another will have several other little islands, and another
a tuft running from the forehead to the back of the head, just for all
the world like a clown in a circus, and so on.
Ali Bacheet to-day injured his foot with an axe. I bathed it, and whilst
getting a bandage one of the Basé diligently employed himself in sucking
it, then rinsed his mouth two or three times with the bloody water which
had washed his foot. This I thought was a somewhat nasty proceeding, but
I did not waste my breath in expostulating with these men of primitive
habits.
Five tetél were shot to-day. In the evening our men with the dhurra from
Amadeb returned.
February 17th.—Last night our camel-drivers, with their singing, and
hyænas howling and laughing, much disturbed our slumbers. This morning
the Basé here were very uneasy in their minds, being under the impression
that we had sent to Amadeb for Turkish soldiers. However, I think we made
them believe—what really was the case—that Messrs. James and Phillipps
had gone to lay a complaint about the Dembelas.
Just after dinner, whilst we were sitting round the camp fire smoking
the pipe of peace, the camel-men whom we had hired at Kassala came in a
body to us, saying they wanted to return to Kassala, stating as a reason
that they were afraid of the Basé and Abyssinians, they being so few
in number. We gave them distinctly to understand that we were neither
afraid of them nor the Basé; for the latter we had plenty of bullets if
they interfered with us, and for our camel-drivers who did so we had the
coorbatch, and so we dismissed them to chew the cud of reflection.
Two tetél shot to-day by Messrs. Colvin and A. James, and several
beautiful birds by me. We are passing a very peaceful and calm existence
at present, little to do except to amuse ourselves as fancy dictates.
Some go out on horseback in search of antelopes or buffalos; I generally
content myself just here with taking out a shot-gun after breakfast,
prowling round in quest of some of the beautiful plumaged birds which are
so numerous, and in the afternoon write up my diary and prepare letters
for post. After that read one of the many interesting books which we have
until 6 p.m., when we all have our evening bath, just before dinner,
which was always ready at 7 p.m. After dinner we sit round the camp fire
and chat over the social pipe, when some go to bed, and I skin and
prepare my birds to bring home.
We had a capital library with us, and were never short of most
interesting works, such as Macaulay's Essays, Sir Samuel Baker's "Nile
Tributaries," Trollope's, Dicken's, Thackeray's, Disraeli's, and other
works.
February 18th.—A young baboon and a small monkey were captured yesterday;
this day they are quite tame, allowing us to stroke them without
exhibiting any signs of fear. Unfortunately the young baboon had been
injured in the thigh by a spear which severed the muscles, causing the
wound to gape very much. The flies annoyed him so much that I determined
to put him under chloroform, and bring the edges together by means of
two or three silver sutures. I therefore put him on the table, where he
lay as quietly and sensibly as any human being, looking up at me with
his nice brown eyes in a very human-being like kind of way. He almost
seemed to say, "I know it is for my good, doctor; don't hurt me more
than you can help, and be quick about it." He took the chloroform very
well, and when complete anæsthesia had been produced I relinquished the
post of chloroformist to an assistant, with suitable instructions. He,
however, was so intent in watching the operation that sufficient air
was not admitted with the anæsthetic, the result being that just as I
had finished putting in the last suture our poor little friend looked to
all appearance dead. I at once set up artificial respiration, but to no
purpose—the vital spark had fled.
Two Basé sheiks from Kokassie visited our camp to-day. They had a short
pow-wow both on their arrival and departure. They kissed our hands
profusely—overdid it, we thought; we were apt to look with suspicion on
an excessive manifestation of friendship.
February 19th.—Just after breakfast I picked up my gun, intending to
take a stroll in the neighbourhood, when Elongi, the Koolookoo Sheik,
taking hold of my arm, led me to Beyrumfi, to whom he communicated some
important information, which he in turn communicated to Suleiman in
Arabic, and the latter to me thus—
"You not go out this morning, doctor. The Sheik, he say, 300 or 400 bad
Basé have come about the mountains by us, and they come bym-bye to kill
us all."
I regret to say that Suleiman's indignation caused him to indulge in
profane language, and he expressed a strong wish to know "What the d—l
dese black rascals meant. We find them plenty meat; we give plenty
presents to them; we kind to them always, and now dey want to kill us
all." Then, turning abruptly to Beyrumfi and a cluster of Basé, he opened
a box full of rifle cartridges, and very angrily said, "Tell dese black
d—s, and dey can tell de other Basé, that we will give them some of dese
bullets, and that we kill one, two hundred of dem in five minutes."
Beyrumfi translated this pleasing intelligence to his hearers, who, in
due time, I dare say, passed it on. Elongi and his men swore they would
stick to us, and I believe they would; but for all that we did not allow
any Basé to sleep within our zareeba. We had become rather lax in the
matter of zareebas lately, and had not constructed one here; but I need
hardly say that on hearing this all in camp were soon set in motion, I
remarking what a fine field this would be for Mr. Gladstone to indulge in
his tree-felling propensities. He would have found some ebony trees well
worthy of his grand old arm.
We had a great deal of very fatiguing work for hours, not only in cutting
down and dragging in a sufficient number of trees to form our zareeba,
but also in felling young palm trees just round the camp. When all this
had been completed the country was set on fire. This quickly spread for
miles. In the midst of it all Messrs. James and Phillipps returned
from Amadeb much surprised at the activity in camp. We soon gave them
all the news, and I cannot say that we were altogether surprised at the
information we received in the morning, as we had observed a good many
camp fires in the night—all over the hills—where no camp fires should be.
February 20th.—Last night we went to bed, leaving sentries posted round
the camp, and well prepared to give a good account of ourselves should
the Basé have conceived the idea of attacking us. Perhaps Suleiman's
timely admonition and explanation respecting the penetrating power of
our bullets deterred them; at all events we were not attacked, which was
satisfactory both to us and the Basé. Had they done so, I computed that
with our 22 rifles and guns, and about a dozen revolvers, protected by
our strong zareeba, we could have polished off about 100 of these poor
savages every five minutes, which would have been no satisfaction to
them or us. Looking at the matter again in another light, had they come
in sufficient numbers, or laid siege to the camp, we should inevitably
have gone to the bad, which would have been a decided inconvenience to
us, to say the least of it. Our comrades informed us that when they
arrived at Amadeb they heard that our late disaster had been telegraphed
to Kassala, Cairo, and, of course, to England. I then felt glad I had
sent a true version of the affair to England, knowing full well that
wild reports, of a most unreliable character, were more likely to get
abroad than true ones. From my youth up I have remembered the story of
the three black crows; also that David once made a very pungent remark,
"I have said in my heart all men are liars," and Carlyle, "There are so
many millions of people in the world mostly fools." However, respecting
the latter remark, I should say that—speaking from experience—they are
frequently not such fools as they look. The former remark was rather a
sweeping one, not _quite_ adapted to the present day.
To-day we moved on to Onogooloo, about two hours beyond Koolookoo. On
passing the latter place Elongi and many of the Basé remained behind, but
his father, a quiet, peaceable-looking old fellow, came on with us. This
was a short march of about seven hours only.
February 22nd.—This day, after a march of about six hours, we arrived at
our old camping placed, called by the festive name of Wo-amma, familiarly
known as Whoa Emma. There we found that, within the past 12 hours,
quite a drove of elephants had been past, and, of course, we were so
unfortunate as to miss them. The Basé are thinning off, but Elongi has
rejoined us to-day. To-day my rifle barrel was so hot at 5 p.m. that no
one could grasp it.
February 23rd.—Breakfast at 7 a.m. On the march at 10, and encamp at
Gebel-Moussa at 5.40 p.m. _En route_ we observed a large tract of country
on fire, and suddenly came upon a herd of buffalos, which raised a
tremendous cloud of dust. Of course we gave chase for a short distance,
and of course did not get near them, for they can go at a tremendous pace.
February 24th.—Life is more enjoyable, if we have some difficulties
to overcome occasionally, and succeed in doing so; and if we do not,
perhaps (speaking as a philosopher) it is better than having a quiet run
of prosperity. To-day, like the past few days, has been warm, 95° in
the shade. Our journey was short, namely, from 10 a.m. till 1.15 p.m.,
encamping at Abion. _En route_ we came across many elephant tracks, a
lion and lioness, and after that a lion, lioness, and three cubs, but did
not succeed in bagging any of them, but three tetél, a nellut, gazelle,
and two bustards were shot. The latter are remarkably fleshy, and very
good eating. Seldom a day passed without tetél, nellut, gazelles, maarif,
mehedehét or dick-dick being shot. The latter is a beautiful little
antelope of the smallest kind. I shot many very small, beautifully
plumaged sun-birds to-day—less than half the size of wrens—but only
managed to bring two or three of them home, as the shot, small as it was,
blew them all to pieces; they ought really to be shot with sand.
It became known in camp that we purpose to-morrow cutting across country
for the river Settite, or Tacazze, amongst the Hamran or sword-hunting
Arabs, _viâ_ Sarcella. In consequence of this we were told (just
before dinner) that after that meal we should receive a deputation of
camel-drivers and horse-boys to enter a protest against this plan.
Accordingly, just after dinner they came in a body, saying that nothing
would induce them to pass the village of Sarcella, as the Basé there were
bad people, and they had just heard that they had sworn to spear every
man, woman, or child of the Beni-Amirs that they encountered, on account
of the raid which Sheik Ahmed had made on them the other day, just after
he left us on our march into their territory. This was the first news
we had of his performances there. They said that after making "aman"
with the Basé, he speared three or four hundred of the men and took all
the women and children as slaves. We reproved them for their cowardice,
saying that they were not old women or children, they had their spears
and shields, whilst we had rifles and revolvers, and were strong enough
to make a two days' march through their territory, instead of one day.
Our arguments were fruitless; they were quite willing to go with us from
Hiekota to the Settite, but they would not, on any account, pass by
Sarcella. We, therefore, made a virtue of necessity, and gave up the idea.
February 25th.—To-day we encamped at Toodloak, having made a journey of
seven and a half miles. I captured a chameleon on the road. Panthers
rather disturbed us last night, and at 4 a.m. a hyæna close to my tent
exercised his risible faculties so much that I, not seeing exactly where
the laugh came in, got up and saluted him with a shower of stones. About
5 a.m. lions were heard; some of us got up and went in quest of them,
came within about 40 yards of one, but he turned off into the jungle when
he caught sight of us. However, during our stay in the Basé country 18
buffalos and about 60 antelopes, besides other game, were shot by members
of the party. We could easily have secured elephants had we remained long
enough and followed them up, and many more buffalos and antelopes had we
remained longer in the country, and, of course, giraffes and ostriches.
The only rhinoceros, or tracks of one, we did not find until we reached
Abyssinia. I have not enlarged much on hunting scenes, fearing that my
book would become bulky, and that the _generality_ of my readers would
scarcely care to read a repetition of such scenes.
February 26th.—Heat getting great, 94° in shade to-day. Another 7½ hours
brought us to Heikota. There we found quite a heap of letters, papers,
and periodicals from England.
The contents of the letters were, of course, greedily devoured by us all,
and as for the newspapers and periodicals, they furnished enough of news
for days. Although many of them were fully two months old, the contents
were new to us.
About an hour after our arrival Sheik Ahmed appeared and received us
literally with open arms, at the same time kissing us on either cheek.
This I could have put up with under different circumstances, but I must
say this mode of salutation is not acceptable to me. We found from Herr
Schumann that the wildest rumours respecting us had reached them—five had
been killed by the Abyssinians, two taken prisoners and put in irons, all
our men killed, whilst our camels and everything else had been annexed.
The Sheik says that had he known of the attack in time he would willingly
have put 1,400 men in the field at once to assist us. He gave us an
account of his revenge on the Basé at Sarcella after he left us, but
there were some unpleasant little details which he prudently omitted,
thinking probably that they would shock our English susceptibilities. The
particulars Herr Schumann furnished us with.
His tale was this—When the Sheik left us to join his men at Sarcella they
had about 500 of the Basé in a cave; the Sheik arrived there quietly,
beseiged them for about 10 days, of course cutting them off from water
and food.
During this time they ate their goats and sheep raw, quenching their
thirst with the blood of these animals. Finally the only course left open
to the beseiged was to place themselves at the mercy of their merciless
conquerors; so, driven by hunger, thirst, and the smell of their dead,
they crawled out, weakened by want, in threes and fours. All the men to
the number of about 300, were speared on the spot, whilst about 200 women
and children were taken into captivity and sold as slaves, realizing
30, 40, 50, 60, and 70 dollars each. About 30 remained unsold on our
arrival; these I saw next day. All the cattle, sheep, goats, dhurra, and
everything else the Beni-Amirs could lay their hands on were seized. Now
we could understand why the idea of passing through or by Sarcella was so
repugnant to our men. I have many patients to attend to, who literally
appear to hunger and thirst after my pills and medicine.
CHAPTER XX.
PATIENTS ARRIVE FROM ALL PARTS—ROUGH JOURNEYS—ARRIVE AT THE
HAMRAN SETTITE—MAHOMET SALI DECEIVES US—CROCODILES, TURTLE,
AND FISH—WE MOVE ON TO BOORKATTAN, IN ABYSSINIA—NEXT DAY WE
MOVE OFF, AS ABYSSINIANS APPROACH—WE CATCH ENORMOUS QUANTITIES
OF FISH WITH THE NET—NARROW ESCAPE FROM A WOUNDED BUFFALO—THE
COORBATCH ADMINISTERED—SCORPIONS AND SNAKES—HAMRANS VISIT
US—HAMRAN MODE OF HUNTING AND SNARING—HAMRAN AND BASÉ—THE
HAMRANS THREATEN TO FIRE ON US—AGAIN RETURN TO THE HAMRAN
SETTITE—ENCAMP AT OMHAGGER.
February 27th.—I was well employed at my medicine chest again this
morning. Amongst some of my patients was a man who had followed us about
for weeks from Kassala, but had always arrived too late to come up with
us. Many others whom I had attended before we entered the Basé country
also visited me, expressing their thanks for what I had done for them,
and presenting me with a spear, a shield, an Arab knife, a gourd of wild
honey, a sheep, and other things; indeed, I met with more gratitude
amongst those poor Arabs than I have in much more favoured climes where
people are well educated, and where the sentiment often is very scarce,
as well as the money.
February 28th.—This morning, about seven o'clock, a great number of
women and children came close to camp making a great noise with the
accompaniment of the tom-toms and that peculiar trilling note to which
I have before alluded. It seems that this was a complimentary serenade,
and that they were rejoicing at our deliverance from the hands of the
Philistines.
Yesterday was occupied a good deal in making arrangements with Sheik
Ahmed for a march on to the Basé Settite. Mahomet Sali, who knows the
country well, will be our principal guide there. We have not seen
a flowing river since we left Kassala; we hope soon to do so, and
are told that we shall find any number of nellut, gazelles, tetél,
buffalos, giraffes, hippopotamus, lions, leopards, crocodiles, and
plenty of fishing, besides monkeys, baboons, golden-crested and toke or
fish eagles, paroquets; rollo-birds, and grouse, doves, guinea-fowl,
partridges, king-fishers, &c., surely a sufficient assortment of sport
to satisfy the most ardent sportsman.
A start was not effected until two. Sheik Ahmed, with some minor chiefs
and a number of his people, accompanied us a part of the way, which was
an uninteresting monotonous journey of about 10 miles over a dry, dusty
plain, the only vegetation being a great number of mimosa bushes, not
trees. The only game observed on the way was a few gazelles. Encamped at
Falookoo, in Basé territory, at 5.30 p.m.
February 29th.—Marched from 10 to 4; encamping on a wretched plain, where
the fine dust was about an inch thick, pitched our tents near to a deep
well at Sogoda. Several Basé came to salute us. They do not seem _quite_
so wild as those we have lately been amongst, and most of them wear a
tope. This was not by any means an enjoyable journey, as the roads were
bad and mountainous, and covered with intensely prickly trees, through
which my camel rushed me, and which lacerated my poor face, legs, arms,
clothes, and helmet in a dreadful manner. Needles and thread were in
great request after dinner to repair the damage done to clothes.
March 1st.—This has been a long, tedious march from 10 a.m. to 11 p.m.
All the discomforts and thorn-scratchings of yesterday intensified six
fold; frequently men had to go on in front and cut down trees to enable
the caravan to proceed. At 5 p.m. we arrived at a river-bed, dug a well,
filled our barrels with water and resumed our journey. It was past 12 at
midnight ere we dined, and 2 a.m. before we retired to our much-needed
rest, which we had very little of, as we were up at 6 a.m.
March 2nd.—Feeling stiff, sore, and tired. A bath would have been most
refreshing, but this morning we were obliged to deny ourselves the luxury
because all the water obtainable was in our barrels. Although our clothes
and flesh have not been so lacerated to-day, the march has been tedious
and very monotonous. For nine successive hours our route lay through
an immense forest of young mimosa trees; these and a quantity of dry,
withered grass was all that we saw during the time, except a few wild
hogs, one of which was chased and speared by a native. By 5 p.m. all our
water was gone, and the thirst of every one was excessive, the heat being
so great, 94° F. in the shade. After travelling 13 hours, we encamped at
9.30 p.m. by a broad, noble-looking river, the Tacazze or Settite which
lay like a lake in front of our camp, either side being fringed with
shrubs and trees of all kinds, amongst the branches of which brilliant
plumaged birds unsuspiciously roosted, little thinking that I should be
looking after them on the morrow.
This river was to us a most refreshing sight after travelling hundreds
of miles over burning deserts and khors or dry river-courses, never
seeing water except by digging for it. It was 12 o'clock this night ere
we got our dinner. All are very angry with Mahomet Sali, our guide,
who professed, and no doubt _does_ know, the whole of this country and
neighbourhood well, for he has brought us, not to the Basé, but the
Hamran Settite, where there is not very much game, as the Hamrans, or
sword-hunters, have destroyed it. This fellow told us on the way that he
would take us to the Basé Settite, within two days or less of Abyssinia,
where there would be plenty of game of all kinds, and here he brings us
about three days out of the way. An unpleasant interview and discourse
will ensue with Mahomet Sali on the morrow. Mr. F. L. James thinks,
rightly or wrongly I don't know, that Sheik Ahmed has instructed Mahomet
Sali and Beyrumfi not to take us to the Basé Settite, fearing we might
get into trouble, either with them, or, what was more likely, with the
Abyssinians. This place is called Geebau.
[Illustration: THE AUTHOR ATTENDING TO ARAB AILMENTS.]
March 3rd.—Mahomet Sali was summoned before the council just after
breakfast, and his delinquencies forcibly pointed out. What he said for
himself I do not know, as I picked up a shot-gun and rifle, taking one of
the boys with me to carry the rifle and anything I shot. This river is
full of crocodiles, turtle, and fish. I was not long out ere a crocodile
received a bullet from my rifle. I also shot a sacred ibis, a crocodile
bird, and a beautiful golden-crested eagle.
In the afternoon the big net was sent on a camel some considerable
distance down the river, where it was rather deep, accompanied by most of
our men and ourselves. The river was dragged, and about 100lbs. of fine
fish were secured, some weighing 10 or 12 lbs. The addition of fish to
our dinner was much appreciated.
March 4th.—The nights are now getting decidedly warm, so much so that I
sleep now with simply a sheet covering me and both ends of my tent open.
To-day it is 94° in the shade. Last night Sali saw a wild beast pass
quite close to the camp just where he was sleeping. Presently he heard
him washing himself, and indulge in a little vocal display. There was no
mistaking his note—it was a lion. Sali at once called Mr. Phillipps, but
could not get any intelligible reply from him, as he was so excessively
sleepy, and knew nothing about it until next morning. Others, however,
were much vexed with Sali for not letting them know.
Mahomet Sali was sent off this morning with five camels and two men to
the Hamrans to procure dhurra. We at the same time, 10 a.m., strike our
tents and start for the Basé Settite, to the great disgust of our men,
who manifest a decided disinclination to visit that locality. We lunched
at Khor-Maiateb on a nice piece of table-land overlooking a beautiful
sheet of water, and shaded by tamarisk, tamarind, and other trees. This
place simply swarmed with crocodiles. I saw a great many Marabou-storks
and two Egyptian geese; one of the latter I managed to bag, and part of
the skeleton of a hippopotamus—the carcase of which, I doubt not, had
provided a rare feast for his slayers.
After lunch we pushed on, and very soon travelled over some vile country.
First of all over very stony road, then down a very steep declivity, over
rocks and big stones, next up a mountain side of the same character—no
road, no pathway even; then along a mountainous pathway, through an
awfully sterile country, covered with nothing but leafless trees,
withered grass, and precipitous rocks, finally encamping at Boorkattan,
above the most gigantic rocks of basalt, of great extent, these again
overlooking the river. We can see, probably a day's march from here, an
immense tall mountain in Abyssinia, on the summit of which is said to be
a fortress. We found the large footprints of the hippopotamus in the
sand by the river, and quite expect to have him in the morning.
We find before night that we are in Abyssinia, so that it is quite
evident Mahomet Sali has not adhered strictly to the truth, as here we
are positively in Abyssinia in one day's march. I indulged three times
after our arrival in a bath in the river. I dare not dive into any of
the pools, fearing that a crocodile might consider me a delicate morsel,
but picked out a kind of cradle on the edge, where I could lie down
comfortably.
March 5th.—We hear that there is an Abyssinian village about seven miles
from here, and that our men are determined they will not proceed any
further than this camp. They also think that our camp, pitched as it is
on such elevated ground, can be plainly seen by Abyssinians, who they
quite expect will make an attack on us to-day. Should they do so, we
should come off badly, as there are no means of forming a zareeba. It is
quite apparent that they did not feel easy in their minds last night, as
a considerable number near my tent were chattering away half the night,
instead of going to sleep or allowing me to do so.
After breakfast Beyrumfi was sent for. A great pow-wow ensued, he
definitely stating that we can go no further without getting right among
Abyssinians, that the country is so rocky, wild, and mountainous that
hunting is impossible, and the camels cannot travel there. Accordingly
orders are given to load up the camels and return to Khor-Maiateb, where
we lunched yesterday.
I had just fixed my bag, rifle, &c., on my camel, when Mr. Aylmer came
running to me, rifle in hand, saying, "Doctor, get your rifle and
revolver ready. Some of our men say that they have seen a large body of
Abyssinians coming down on us, some on horseback and some on foot; but
at present they are a pretty good distance off." Our caravan was nearly
ready to start. Of course we all armed ourselves pretty quickly; then
saw some of our camel-drivers (one old fellow particularly) working
themselves up into a frenzy of excitement, leaping about like lunatics,
at the same time brandishing their spears in a most threatening manner,
indicative of what would be the fate of the enemy should they appear. As
we looked upon this performance as so much waste of time, we scruffed
these fellows, boxed their ears, and told them to make haste and
load up their camels, which they did with a will. As a rule they are
generally a couple of hours loading, but now they were wonderfully quick,
accomplishing the work in half the usual time. We got off in safety, and
arrived at Khor-Maiateb in the afternoon. Temperature, 95° in the shade.
March 6th.—Crocodiles are rather too common here to be pleasant, and
interfere with the luxury of the morning and evening bath. To avoid any
unpleasant _contretemps_, I generally collected together several big
stones by the side of a large pool, threw them in one after the other to
frighten the crocodiles away, then threw myself in. This device proved
eminently successful, enabling me to enjoy a plunge and a short swim. I
need scarcely say I did not fool about long in the water, fearing they
might return to see what white object was swimming about.
To-day we used the large net, and landed 210lbs. of different kinds of
fish. Keeping sufficient for our dinner, the rest was divided between our
men, who ate what they wanted, throwing the remainder into the bushes or
anywhere round the camp, causing an insufferable stench next day, which
we did not get rid of until the fish had been all gathered up and thrown
into the river.
March 7th.—The little canvas boat is in great request, and enables us to
go a good way up and down the river. The net was used to some purpose
to-day, for we landed 360lbs. of fish and one turtle. At 12 a.m. about
20 Basé came to us with information that elephants are not far off, as
they saw and heard them; also that on the 5th, near our last camp, whilst
they were looking for wild honey, the Abyssinians swooped down on them,
killed several, including the Sheik's son, and stole three women and a
few children. No doubt these were the very fellows who were coming down
on us. When we discreetly and gracefully retired, they found us gone,
and so seized the Basé. After lunch Messrs. A. and W. James and Colvin
mounted their horses and went in search of the elephants. Temperature,
96° in the shade.
March 8th.—About 11 a.m. two or three Basé (who accompanied Messrs. James
and Colvin yesterday) returned, saying that the latter had shot two
buffalos, one of which was killed, the other wounded only, and that they
had seen plenty of elephant tracks but no elephants. At 4 p.m. they all
returned, having tracked and secured the wounded buffalo, and an ariel. A
crocodile, fish-eagle, and an enormous horned-owl fell to my gun to-day.
Temperature, 98° in shade.
March 9th.—Temperature, 100° F. dry bulb, wet bulb 71°, solar thermometer
156°. So far this is the hottest day I have ever experienced. Whilst
bathing to-day I put my towel near the water's edge to stand on, as the
stones were like hot coals to the feet. We have cleared many of our
followers and men out of camp to-day. The Basé, Mahomet Sali, Beyrumfi,
and all the Beni-Amirs have been discharged for misleading us. Messrs.
F. James and Phillipps have gone in quest of game, and Messrs. Colvin,
A. and W. James have returned to the same place as before in search of
elephants.
March 10th.—Last night seven or eight rifle reports from the other camp
reached us. At 1 p.m. they were accounted for by Messrs. Colvin and
company, who arrived in camp. They had shot at and wounded two buffalos
(one a bull). Two or three lions had attacked one of the wounded, leaving
very distinct marks of the struggle; still, the buffalo had managed to go
on. They tracked him for some distance, but the heat became so great by
mid-day—101° in the shade—that they had to desist. Another, wounded in
the night, they followed up nearly to the jungle, when suddenly he darted
on to them, charging most furiously in Mr. A. James' direction. He,
however, saluted him with two eight-bore bullets in the chest, which had
the effect of turning him from his purpose, and causing him to change his
plan, for he turned and then charged Mr. Colvin, who, very fortunately
for him, happened to give him a bullet in the fore leg at very close
quarters, as the buffalo fell right against him with some violence, and
sent him reeling on the ground. I should think this was about as close to
an enraged wounded buffalo as Mr. Colvin or any other man in his senses
could desire.
We could very frequently get a shot at a crocodile when in the water,
but seldom on land; they seem much too wary to be caught there. I have
often seen them basking in the sun on the bank of the river, crept
cautiously up, and whether they have seen me, smelt me, or I have trodden
on a twig I know not, but before I could get near enough they have all
disappeared in the water. They come up to the surface often. We see a
dark spot in the middle of a quiet-looking pool, and take a pot-shot, but
seldom get the reptile until next day, when we find him floating, but
so mutilated that he is not worth securing. To-day, however, Mr. Aylmer
shot one in the water near the camp, and was fortunate enough to secure
him by the aid of a native, who dived into the pool with a rope, which
he slipped over his upper jaw. I fancy crocodiles prefer white skins to
black, for these black fellows plunge into the water and swim about where
we would not dare to go.
Before the crocodile episode—in fact, just after breakfast—our court
of justice sat. This consisted of ourselves, who were the judges, the
jurors, and the counsel; and I venture to say that strict justice was
dealt out with an even hand. The culprit was a fine, strapping, rather
good-looking fellow of about six feet, a camel-driver. He had been
troublesome on two or three previous occasions, but last night he passed
the bounds of discretion. His brother roused him up in the night to take
his turn at sentry-duty; in return for this he warned his brother that
he would make him suffer in the morning—which he certainly did, as he
got him under some trees and there chastised him severely with a stick.
When we heard of this, the culprit, prosecutor, and witnesses had to
appear before the tribunal. The charge was proved, and the culprit was
ordered 20 lashes of the coorbatch, to be administered by Suleiman, four
camel-men to hold him down. He at once dispensed with the assistance of
the camel-men, and without making any bother at all, laid down on the
sand, face downwards, whilst Suleiman went in search of the coorbatch.
The castigation was duly administered, the fellow taking it without
flinching an atom. When finished he got up, brushed the dust from his
tope, and walked off in his usual manner. He seemed not to bear the least
malice, for some time afterwards he was as busy as anyone helping to land
the crocodile.
March 11th.—Two bull buffalos, a tetél, and nellut were shot to-day.
Scorpions are too plentiful here; we are continually finding them in our
tents, but so far none of us have received any of their dreadful stings.
They belong to the class _Arachnida_. A scorpion has what looks like a
claw in his long tail, through which the poison, which lies in a bag
at the bottom of it, is projected. This tail, preparatory to taking the
offensive, lies curled up on his back, not unlike a squirrel. He can at
will bring this down with considerable force, but only in a straight
line—he cannot twist it to strike.
Whilst strolling up the river-bed with my gun in the afternoon I came
upon Mr. W. D. James, who had just met with a rather curious, and
not altogether agreeable, adventure. He had brought his photographic
apparatus with him, and planted it within a convenient distance from a
pool, intending to photograph gazelles when they came to drink. He was
successful in obtaining a good picture of two—one drinking, the other
looking straight at the camera. Whilst waiting patiently for them,
seated on some rocks under a large baobob tree, he heard a hissing noise
behind him. On turning his head, he saw a snake waving about in an erect
position, with tongue out, looking as if he was about to strike. Mr. W.
D. James did not sit on the stone any longer, but seized a stick, and was
lucky enough to kill it ere he was able to bite.
About a week ago we set some mustard and cress; to-day we had a good
quantity for luncheon, and found it a very agreeable addition.
March 12th.—A few Hamrans called to-day, and are very anxious to
persuade us to go towards Abyssinia, saying they are friendly with the
Abyssinians, and can show us hippos. The offer is not accepted. I find
these fellows do not by any means confine their hunting tendencies to
simply the use of the sword, as I have often found very ingeniously
constructed snares plentifully placed in runs leading to the river.
Doubtless when the animals are thus ensnared they are despatched with the
sword or spear.
I will try and describe the kind of snare: They get a strong branch of a
tree that will bend, not break, into a circle; this they firmly secure.
They have a number of strips of wood, broad at the base, and gradually
getting narrow, converging until they meet in the middle of the circle
bent downwards on one side; these again are firmly secured to the circle.
A hole, perhaps a foot deep, and half a foot or a foot in diameter, is
dug in the ground where the run is. On the top of this hole is placed
the snare, covered with earth, attached by a strong rope to a great log,
or the trunk of a tree. The unsuspecting animal comes to drink, puts
his foot on this, and it slips in. He cannot pull his leg out, for the
harder he pulls the more firmly is he secured, as the sharp spokes stick
into his flesh. It is, in fact, just like a wheel: the tire is the outer
circle, and the axle represents the hole through which his leg goes.
The reason the Hamrans are called sword-hunters is this—I am quite sure
that neither I nor any of our party can speak from experience, as we
never saw the feat performed, but Sir Samuel Baker has: Whilst hunting
the elephant, or giraffe, a Hamran on horseback gallops in front of an
enraged elephant armed with a sword, whilst one behind, similarly armed,
gallops after him. The elephant may elect to turn round and chase the
one behind—in any case, he is between two evils, for eventually the one
behind, whilst the horse is going at full gallop, will, when he is near
enough, jump off and with great dexterity hamstring the elephant with his
long two-edged sword; then, of course, he can easily be despatched. His
tusks are cut off and sold, and his carcase provides a good feast.
Two Basé who had remained in our camp slipped off to the mountains on
seeing the Hamrans in our camp, returning again after their departure. It
is quite evident they do not regard them as friends.
March 13th.—This morning several Hamrans, with the late Sheik's son,
interviewed us, and seem very desirous of acting as pioneers in this
part of the country. We declined their services. This seems displeasing
information to them, and they also express anger at our having two Basé
in camp. On leaving us they went towards the Basé country. Suleiman
explained this by saying, "The two Basé in our camp, they go soon as they
see these Hamrans. Now he go after the Basé; they kill his father long
time ago. Now he kill all the Basé he find if he strong enough and have
plenty of mens with him." This was really the case, and the Hamrans were
now hunting the Basé just as we would wild animals. However, they had a
good start, and probably made the best of their way to their country.
Our camels were now being loaded, as we had resolved to move our camp.
Whilst we were preparing, a Hamran came, saying he was sent to tell us
that we had given the Heikota sheik a number of presents, but they had
nothing to do with him. We were now in the Hamran's country, or soon
would be, and they were not willing for us to kill their game without
permission from their Sheik, adding that if we advanced they would fire
on us. Our reply was, "Tell your friends that our camels are loaded, and
so are we. We are coming your way in less than half-an-hour, and strongly
advise them to save their powder, as that is a game that two can play
at. Threats don't alarm us in the least. If they are ready to commence
hostilities so are we." I suppose they thought better of it. Shortly
afterwards we started without hindrance until we got just beyond our old
camping ground, by the Settite, where our tents were then pitched. There
were two hippopotami in the river just here, one of which we saw. The
river was dragged, but they slipped under the net.
March 14th.—Marched from 8.30 a.m. until 2.30 p.m., encamping on a high,
flat table-land overlooking a beautiful sheet of water plentifully
bordered on the bank by trees and bushes in which could be found any
number of beautiful birds and doves. At the back of our camp was a
large wood on perfectly level ground, which gave shelter to myriads of
guinea fowls, doves, and other birds, also vast numbers of baboons.
The occupants of the water were crocodiles, turtle, and very numerous
different kinds of fish. The shore, a little way from camp, was
frequented by Marabou storks, flamingoes, ibis, cranes, storks, Egyptian
geese, herons, crocodile-birds, &c., &c.
This was the most enjoyable camping ground we had yet come to. It was
also the hottest place we had hitherto found, for the temperature at 2
p.m. to-day was 105° Fah. in the shade. During such hot weather a bath
was of course a most delicious thing to indulge in, but I must say I did
so with some trepidation, as the pool in front of us was frequented
by some good-sized crocodiles whom it was as well not to trifle with.
I therefore contented myself, as a rule, with lying down in the water
on the edge where it was shallow. When feeling inclined for a plunge
and swim I invariably adopted the preliminary caution of hurling in
several big stones; on these occasions I was sufficiently discreet not to
remain long in the water, having conceived a very wholesome objection to
furnishing any of these scaly monsters with such a repast as a Williams.
The water was quite tepid, of course from the great heat. This place is
called Omhagger, not far from the village of Ombrager.
CHAPTER XXI.
A BOA-CONSTRICTOR VISITS US—THE BURTON BOAT—MOUSSA's BEHAVIOUR
ENTAILS A THRASHING AND HIS DISCHARGE—GREAT HEAT—A FINE
HIPPOPOTAMUS KILLED—HAMRAN FEAST—THE WHITE ANTS—ANOTHER
HIPPOPOTAMUS KILLED—MAHOMET SALI BRINGS SUPPLIES—NATIVE
MUSIC IN THE NIGHT—DELICATE HINTS CONVEYED TO THE
PERFORMER—A REMARKABLY FINE NELLUT SHOT—ARAB AND EGYPTIAN
TAXATION—BABOONS—A HAMRAN STORY—ALI BITTEN BY A SCORPION—ON THE
MARCH ONCE MORE—ROUGH JOURNEYS.
In the evening, whilst George and Anselmia, our two European servants,
were dining by their tent, George called out, "A snake, a snake." A
little terrier, named "Tartar" (which Mr. W. D. James had brought from
England) began barking furiously, whilst we sallied forth with anything
we could lay our hands on—Mr. Phillipps and I each with a spear, Mr.
Colvin with an Abyssinian sword—darted off, just in time to see a great
boa-constrictor gliding through the grass and into some thorny bushes
where we could not pursue him.
George said, "I heard something hissing," and said to Anselmia, "What the
devil is that?" looked round and saw an enormous snake about a yard off
in the tree behind me, hissing away, with head up. I was off in quick
sticks. Last night a lion came so close to my tent, and made such a noise
that he woke us all up, and produced quite a stampede amongst the horses
and camels. Some of the natives sleeping just outside my tent threw
firebrands at him. Unfortunately the moonlight was wanting at the time or
we might, perhaps, have bagged him.
Two tetél, four ariel, and several birds were shot to-day.
March 15th.—Hyænas were rather noisy last night, but I have never known
them so troublesome anywhere as at Kassala. The heat to-day was 106° in
the shade—so far the hottest day I ever experienced. Of course the Burton
boat is in frequent use now. To-day, whilst quietly punting about near
the bushes with my gun laid across the seat, I observed some beautiful
and strange birds. I quietly seized my gun, and found the barrel so
excessively hot that I positively could not hold it until I wrapped my
pocket-handkerchief round it. I succeeded in bagging two fine spotted
giant king-fishers. This morning Moussa, a mischievous young rascal,
whom we had brought with us from Kassala, was severely thrashed with
the coorbatch, then sent away with our head camel-man (who was going
for dhurra) to the Sheik at Ombrager, to be forwarded on to Kassala. It
seems that he had quarrelled with Idrees, a native servant from Keren;
then, whilst struggling together, he whipped out a pair of scissors and
with it snipped out several bits of flesh from his arms and chest. This
was not his first offence, for on the 13th he received 20 lashes of the
coorbatch. Then he laid himself down at once, face downwards, and took it
without flinching; to-day he got it severely, and yelled most vigorously.
His offence on the former occasion was this: Whilst Sali was running to
camp, rifle in hand, this impudent young scamp struggled with him for
the possession of it (this was just after dark). In the struggle the
rifle went off, and might have lodged a bullet in Sali or anyone in camp.
Unfortunately the coorbatch is the only remedy for these natives—the only
way of keeping up discipline. If treated with kindness and forbearance
they think we are getting lax and easy, will at once take advantage of
it, skulk about and do nothing, but the coorbatch at once brings them to
reason.
Mr. A. James found a man's skull to-day, also a gigantic tortoise shell.
Mr. Phillipps angled and caught an enormous gamout, weighing 31 lbs. Two
ariel, a nellut, and calf buffalo were shot.
March 16th.—Temperature 105°, wet bulb 71°, solor 160°. Last night hyænas
and wild cats exercised a disturbing influence on our slumbers. Soon
after breakfast a Hamran Sheik, with attendants, called, presenting us
with a good quantity of milk and a sheep. During the day a crocodile was
shot, also a very fine hippopotamus. The latter was observed poking his
head above water in a large pool a little way from camp, little thinking
of the danger awaiting him. He had no sooner done so than crash went a
hardened bullet into his skull. Down he went, but could not stay long,
for he must come up to breathe. On his reappearance he received another
leaden messenger in his skull. On his coming to the surface a third
time he spouted up a quantity of blood and water, and received one more
bullet, after which he disappeared mortally wounded. The next time he
came to the surface a floating corpse. A few hours afterwards ropes were
obtained and fastened round him by some Arabs, who dived, and he was
dragged to shore amid universal rejoicing, they knowing right well that
a feast was in store.
We find the white ants very troublesome here. Should anyone be careless
enough to leave his satchel or portmanteau on the bare ground he would
regret it in the morning. Anyone who has visited Central Africa will
come away with very distinct recollections of the white ants. They are
really wonderful little creatures, and the structures they erect are
often on a colossal scale. Quite near to our dining place here is one
of their buildings. As a rule they are built of a conical or sugar-loaf
form, but I have seen them of the form of turrets. They are worked up
from the soil of the country by the ants, and are of the consistency of
stone, and so strong that a buffalo or leopard has been known to take
up its position on the top for the purpose of observation. During one
portion of our journey, on emerging from a wood, I saw what I at first
took to be a village on a large plain; the habitations resembled huts,
in some cases 15 feet high, and proportionately large at the base. They
were only a colony of white ants, and I dare say their village consisted
of 200 of these ant hills. The white ants (_Termes bellicosus_) are not
true ants; that is, they do not belong to the order _Hymenoptera_, which
embraces the industrious bee and the crafty ichneumon, but belong to
the order _Neuroptera_, which embraces the brilliant, though voracious
dragon-fly, the ephemeral may-fly, and the wily ant-lion. They are
called ants because they are similar to them in their habits and in the
constitution of their colonies. Their antennæ are larger than the head,
their mandibles are well-developed, and the inferior pair of wings is
generally as large as the superior.
There are four classes found in the colony of the white ant—the king
and queen, who live together in a central chamber near the ground,
after having lost their wings; the workers, who build and nurse their
young; the soldiers, who never build or nurse, whose duty consists in
defending the nest when attacked. Neither the workers nor the soldiers
have wings. The largest worker is supposed to be a fifth of an inch long.
The soldiers, which have an enormous head and formidable mandibles, are
at least twice as long, and are said to weigh as much as thirty workers,
attaining the length of nearly four-fifths of an inch, while the female,
when she has become a queen, and about to form an extensive colony,
attains the length of six inches, and lays eggs at the rate of sixty
a minute, or more than eighty thousand a day. The white ants are most
destructive to houses, furniture, clothes, and books; they will, in
fact, destroy anything but stone and metal. Anything that is reducible
to powder will, where they have located themselves, fall to certain
destruction. They work unseen. I have often noticed twigs, leaves,
branches of trees, and so on, destroyed by them. They plaster them over
with mud, and underneath this cover they work. Wooden pillars and beams
are continually made perfect shells by their operations, and the safety
of houses is frequently affected, though externally they would appear
strong and good. The library at the Faurah Bay Church Missionary College,
Sierra Leone, was in a great measure destroyed by their instrumentality.
In 1879 the Bishop of Sierra Leone appealed for funds in order to repair
the churches, which, he said, "are ant-eaten." Now, although the white
ants are so annoying that hardly anything is proof against their attacks,
they are a great blessing in tropical climes, their office being, in the
economy of nature in these hot countries, to hasten the decomposition
of the woody and decaying parts of vegetation, which, without their
intervention, would render these regions uninhabitable by breeding a
pestilence. The remains of the white ant in the Coal Measures is an
evidence that it was, to a certain extent, through their destructive
agency that the tropical vegetable matter was accumulated which went
to form our coal. The white ant, by hastening the decomposition of
vegetable substances, has ever proved a friend to man; the true ants also
have proved themselves a boon to the inhabitants of tropical climes by
destroying what are popularly classed as vermin.
March 17th.—I am happy to say that the temperature has dropped to 97°,
and that is quite as hot as one cares about. Whilst breakfasting,
about 6 a.m., a native came, saying "Assint effendi," at the same time
jerking his thumb towards the river. This meant a hippopotamus. He was
accordingly sought for, found and killed before 8 a.m., not far from
camp. To see the Arabs then cutting up the carcase, wallowing in gore,
and stuffing lumps of meat or fat into their mouths, and rubbing the
latter on their heads was a most disgusting sight, almost as bad as the
Basé. Festoons of meat soon ornamented every tree in the neighbourhood of
the camp. The natives themselves looked like a number of dips melting in
the sun. Every head to-day is dripping with fat, which melts and drops
about all over the shoulders. Feeling a little Mark Twainish, I cannot
help remarking that this plastering of the head with fat is, no doubt,
a very ancient custom, in which Aaron, his friends, and contemporaries
were accustomed to indulge to a great extent. Do we not read? "And they
annointed his head with oil, which ran down even unto his beard." This
was probably the fat obtained from hippopotamus, buffaloes, &c., and
clearly indicates that these old gentlemen were, more or less, affected
with sporting proclivities; but it is more than probable that they were
not at that time in possession of eight-bore rifles and hardened bullets.
During the greater part of this day we have observed a peculiar hazy
appearance on the other side of the river—extending for miles—resembling
mist in appearance. It was really a very fine dust. In the evening this
was followed by a good deal of wind, which much increased towards night.
March 18th.—Last night and this morning was cooler than any for weeks
past. Temperature at 1.30 p.m. 92° in the shade. Mahomet Sali put in an
appearance to-day with a camel-load of flour and bread from Kassala.
He also brought the skin of a very fine boa-constrictor, which one of
our party was not long in annexing. It really was intended for George,
but the annexer gave the owner two dollars for it, and so it became his
pretty quickly. The bread we have is like the Cairo bread we brought with
us. It is baked hard in squares, resembling dog-biscuits. When wanted at
meal-times they are dipped in water and then put over the fire for a few
minutes, when they become spongy and eatable.
Two or three little matters conspired to ruffle me rather last night.
When retiring for the night my first business was to kill two scorpions
and some gigantic spiders which I found in my tent. Having accomplished
these murders to my entire satisfaction, I undressed, lastly taking off
my socks. I had no sooner put my feet on the matting than the soles of my
feet felt as if they were pin-cushions, receiving a thousand prickling
sensations—sometimes not unlike those produced by a galvanic battery.
I found on inspection of the matting swarms of large black ants. This
was not all. I got into bed, but was not allowed to sleep without the
accompaniment of a stringed instrument, somewhat resembling a banjo,
which a wretched Arab had constructed out of an empty preserved meat tin.
He had stretched a piece of skin tightly across this, attached a bridge,
three strings, a finger-board, and all the rest of it, then strummed and
fingered away close to my tent, producing the most monotonous sounds
for an hour or two after I was in bed, evidently as much in love with
his instrument as someone else would be with his violoncello. I had not
done anything deserving such torture as this wretch thought proper to
inflict on me, so would not stand it any longer. I therefore got up and
delicately conveyed a hint (in the shape of a boot which I hurled at him)
that this mode of serenading in the middle of the night was not only
unappreciated by me, but decidedly objectionable. The stringed instrument
ceased for a while until I was just going to sleep, when this demon in
human form again started. I immediately threw out three more rather
forcible hints. They were another boot and two empty claret bottles; and
I rather think the last two hints appealed forcibly to his feelings, as
the light guitar was then laid aside, and I was allowed to sink into a
calm sleep.
During the day Mr. W. D. James shot a magnificent buck nellut, which had
the finest head we have yet seen. His horns, taking the direct length,
measured 39½ inches; taking the curves, 53 inches. An enormous tortoise
was brought to camp this morning alive (abugeddir, as they called him).
Two stout men can stand on his back, and he walks away with them as if
they were two straws.
March 19th.—Temperature, 92° in shade. The hazy appearance noticed during
the last two days has passed away; the wind also has subsided. I think we
may expect a hot day again.
This morning a son of the Hamran Sheik came to camp, demanding a tax
of eight dollars on our guide. The latter is to receive 25 dollars per
month, out of which the Heikota Sheik wants eight dollars. To begin at
the fountain-head, the Egyptian Government have a head-tax—every young
person on reaching a certain age is taxed. The owner of every date
palm-tree has to pay a tax, the same with the owner of a "sageer," or
sakia (a water-wheel); in fact, I believe everybody and everything is
taxed. The Government look to the Governor-General of the Soudan for
a good round sum; he, in turn, looks to the Mudir, or governor of a
district. He squeezes the necessary out of the sheiks of the various
tribes, and they in turn (to use a metaphor, suck the orange dry) screw
out of the poor Arabs of the tribe what they require. If the sheik fails
to produce the sum required of him by the Mudir, the latter swoops down
on his camels, flocks, and herds and sells a sufficient number of them to
produce the required sum; but if the sheik has no camels, &c., he himself
is seized and put in durance vile until the tribe find the necessary
number of dollars.
Some worms there are who feed on men;
Others there are who feed on them.
These lesser worms have worms to bite 'em;
Thus worm eats worm _ad infinitum_.
How can Egypt ever prosper under such a system? What inducement have
these poor Arabs to accumulate anything more than is sufficient for their
daily wants? None. When we engaged servants at Heikota, at, say, 12
dollars per month, the first thing the sheik did was to take two dollars
from each man, and very probably as much, or more, at the termination of
their services. The beasts, the ants, the reptiles, and birds prey on one
another; crocodiles on big fish, and big fish on little ones. There is no
Salvation Army there, and if there were (I don't want to be ironical, but
_Byron_-ical) I do not think there exists a more preying community.
Two nellut and a maarif shot to-day.
March 20th.—This morning, just after breakfast, I took up my gun and went
about 100 yards from camp, with the intention of shooting a baboon. But
my heart smote me—they looked so awfully human—and I desisted; but I sat
down and derived much amusement from watching enormous baboons and little
monkeys gambolling by the water's edge.
On returning to camp I found Suleiman conversing with a man wearing a
belt full of cartridges (a rather uncommon spectacle). The conversation
lasted some time. I was told afterwards that when he learned from
Suleiman that we had been again by or in Abyssinian territory, he
exclaimed, "You have to thank the good God that you came away when you
did. Had the Abyssinians seen your tents and all those boxes they would
certainly have come down on you and killed everyone of you for the sake
of the tents alone, to say nothing of the boxes." He also informed him
that about a month ago they killed a party of Hamrans, who went up there
hunting from this neighbourhood, just for the sake of a gun or two, and
whatever else they could lay their hands on.
A buck nellut, two wild boars, killed; two buffalos wounded. Temperature,
97° F. in shade.
March 21st.—Temperature, 100°. Nothing of interest to-day. A tetél, two
nellut, and mehedehét shot; maariff wounded, but not secured. To-morrow
we turn our faces to the Red Sea coast, and expect to reach Massawa in
about three weeks time.
March 22nd.—Last night, just as I was about to retire for the night, I
was sent for to Ali, the cook, who had just been stung on the thigh by a
scorpion. He was evidently suffering great pain. I gave him a strong dose
of ammonia and some brandy, at the same time advising him to poultice the
wound after I had cauterised it. Suleiman having more faith in a more
heroic mode of treatment, obtained a razor, and with it made a series of
little gashes, remarking at the same time, "There, now Ali better after
that; the bad blood come from him now." Of course I did not interfere,
but allowed them both to have their own way, to their mutual satisfaction.
As we were now about to take our farewell of the river, I indulged in
a swim at 6 a.m., not forgetting at the time to hurl in several large
stones preparatory to my dive, as a warning to all crocodiles to vanish
for a time.
At 10 a.m. the hamlah was on the move. Our journey this day was pursued
under very unpleasant conditions, as we travelled frequently through
large bushes of mimosa and kittars, tearing our helmets and clothes to
pieces, and inflicting not a few scratches on our bodies. At 7 p.m. we
encamped at Khor-Maiatah, about five miles from our old camp. The day's
sport was a crocodile and buffalo.
CHAPTER XXII.
ENCAMP AT LAKATAKOORA WITHOUT THE CARAVAN—DESCRIPTION OF
VILLAGE—BASÉ LADIES VISIT ME ERE I GET OUT OF BED—THEY RECEIVE
PRESENTS AND ARE VERY AMUSING—ENORMOUS NUMBERS OF DOVES AND
SAND-GROUSE—ABOOSALAL TO SOGODA—BOA-CONSTRICTOR KILLED—AN
UNPLEASANT JOURNEY, WE ALL GET SEPARATED—ARRIVE AT HEIKOTA
AGAIN.
March 23rd.—Leaving Khor-Maiatah at 8.45 a.m., we had an exceptionally
unpleasant day of it. We had seen the last of that fine river, the
Tacazze; now if water was wanted it could only be obtained by digging a
few feet in the sandy river-beds. We travelled over mountains, plains,
valleys, river-beds, and nearly all day through a forest of those horrid
mimosas, finally arriving at Lakatakoora, in the Basé country, at 7 p.m.,
_without_ the caravan. At about 11 p.m. Cheriff, with the canteen, Ali
the cook, and a few only of the camels arrived.
As we had not tasted food since about 1 p.m., Cheriff's canteen was soon
surrounded by us, and the contents of it cleared out in a very short
time. Our dinner (which was a scanty one this time) did not appear until
12.30 a.m. Whilst this was being prepared by Ali we fired off rifles,
burnt blue lights, and lighted a beacon fire for Suleiman with the hamlah
to see where we were; but all to no purpose—they had lost their way.
The moon retired for the night, and so many trees had to cut be down to
enable them to come on, that at last they gave up the idea of attempting
to find us, so slept out.
We also had rather a hard time of it. There was no choice of a
camping-ground—there was but one. This was a large open space devoid of
vegetation, but covered with a thick layer of impalpable dust, about an
inch or so in thickness, infested with white ants. Fortunately our tents
and bedding arrived. I did not wait for my tent to be pitched, but placed
my bedding on the canvas covering of my tent on the ground, and there I
managed to get through the night in a rather unsatisfactory manner. The
other members of the party elected to remain up until their tents were
pitched.
March 24th.—I passed a somewhat uncomfortable night amongst the white
ants, lulled to sleep by the music of hyænas, some of whom seem to have
been intensely amused at our situation, if one may judge from the bursts
of merriment issuing from _their_ camp, as they were evidently excited
by uncontrollable fits of laughter, making the woods in the immediate
neighbourhood resound by the exercise of their risible faculties.
I awoke about 6.30 a.m., but did not arise until 7. Quite near to our
camp I observed on a precipitous mountain side enormous basalt rocks,
some _single_ rocks as big as a good-sized house. This was the village
of Lakatakoora; and amongst these rocks, concealed from view, lived some
Basé. I had not been awake long ere many of the Basé ladies, covered
with beads, their eyelids and lips stained with kohl, rings in their
noses and ears, and a strip of cloth around their waists, came and shook
hands with me, murmuring "Mida" (good-day). Both men and women came in
such increasing numbers that I decided to get up, and had to perform my
toilet in their presence. They watched the whole performance with evident
interest. Such a contrivance as a tooth-brush and tooth-powder elicited
expressions of wonder and admiration, but a hair-brush and comb pleased
them still more. Water was scarce. I therefore had to wash _à la_ Turk.
Mahoom stood by me with a salmon tin full of water, pouring out little
driblets of water into my hands, finally douching my head with the
remaining drop, about two ounces.
Our guide had been sent off at 6 a.m. in search of the lost portion of
the caravan. He found it, and piloted it into our camp at 8.30 a.m.
The Basé were very friendly and obliging in the way of water, for they
brought us this invaluable liquid in very beautifully worked baskets,
so closely woven that not a drop escaped, slung on the shoulder, like
a pair of scales. They also brought us several gourds of wild honey,
which we bought. Many beads and small looking-glasses were given to the
ladies, who appeared highly delighted and amused when they saw their own
faces reflected from a looking-glass for the first time in their lives.
They crowded round the fortunate recipient of one of these reflectors,
peeping over one another's shoulders, giggling and laughing at their own
reflection in a most amusing manner.
All being packed up by 10 a.m., we moved off a short distance, halting
at Aboosalal at 12.30 p.m. on the sandy bed of a khor, surrounded on
either side by lofty, precipitous rocks, along which scampered hundreds
of enormous baboons and monkeys. Both men and camels seemed completely
knocked up. A little way from camp was a little pool of water in this
khor; and in the neighbourhood, without exaggeration, were thousands
upon thousands of doves and sand-grouse. I took my gun down there in the
evening when they came to drink, and stationed myself behind a huge rock.
In less than half-an-hour I bagged about 15 brace of sand-grouse and five
brace of doves.
The day before we arrived here the natives had killed an elephant. They
and the vultures had picked his bones pretty clean, as nothing but his
skeleton remained when I saw it. His tusks, of course, had been taken
away. Later on they were offered to me by a Sheik's son; but as they were
damaged, small, and the price excessive, I was not a purchaser.
March 25.—Made a long march from Aboosalal to Sogoda, where water could
be found. We started soon after 8 a.m., and long before we reached our
destination darkness came on. This was a most unpleasant journey through
prickly trees, again tearing our patched-up clothes and helmets. In one
place we all got separated, each one selecting the way he thought best.
I lost my way in a forest of kittar and mimosa bushes. I was obliged to
dismount my camel, and presently got in such a fix that I could scarcely
move either backwards or forwards. Noises became indistinct, and finally,
I could not hear a sound. Others were, apparently, in the same position
as myself, for shots and revolvers resounded on all sides. Eventually we
reached the camping ground in detachments at 8, 9, and 10 o'clock, dining
at 11 p.m., bed at 12.30. In the evening, before dark, we came across the
trail of a boa-constrictor, followed it up, and successfully despatched
the reptile, which measured 12 feet in length. The next day we pitched
our tents at Fahncoub, on very good ground, surrounded by fine trees.
Sogoda was a horrid place, the ground being covered with fine dust.
March 26th.—Starting from Fahncoub at 8.30 a.m., we once more reached
Heikota at 11.30, having travelled a distance of 10 miles only. We found
Herr Schumann, the animal collector, had gone, calling at Kassala, and
taking all the animals with him to Souâkin. I take the opportunity
of sending a letter on to Kassala by Alki, who has been a very good
trustworthy fellow, but who is now leaving us, as he has a bad whitlow.
CHAPTER XXIII.
AN ABYSSINIAN IMPROVISATORE AND HIS LITTLE SLAVE—PREPARE FOR A
MARCH TO MASSAWA—A STRANGE BASÉ BREAKFAST—PATIENTS—ARRIVE AT
TOODLOAK—BENI-AMIRS ENCAMPED ON THE GASH—LIONS AND LEOPARDS
ARE SHOT—OUR MONKEYS IN CAMP—BABOON MODE OF ATTACKING
LEOPARDS—CRAFTY BABOONS—LIONS ABOUND—HYÆNA METHOD OF ATTACKING
A LION—HYÆNA INTERVIEWS MR. COLVIN—ARRIVAL AT AMADEB—DEPARTURE
FROM AMADEB—BAREAS ATTEMPT AN ATTACK ON THE CARAVAN—BENI-AMIRS
WATERING THEIR FLOCKS AND HERDS—WE MEET WITH A YOUNG
ELEPHANT—LEOPARD AND HYÆNA SHOT AT KHOR-BARAKER.
March 27th.—A great part of the morning was occupied with sorting out the
heads of antelopes, buffalos, &c. I was particularly engaged in attending
to about 50 patients in the morning, and perhaps 20 or so in the evening.
Many of them had come from long distances to see the "Hakeem."
In the afternoon a little Abyssinian boy, about 12 years of age (an
improvisatore), made his appearance, accompanied by his slave (a small
black boy), perhaps seven or eight years of age, carrying a rude native
instrument with three or four strings on it, something like a banjo. He
favoured us with a song, in which the word "Ingelese, Ingelese" occurred
very frequently. The interpretation of the whole song I do not know, but
it referred principally to the prowess of the English in the Abyssinian
war. On being asked if he would sell this instrument, he replied—
"No, I cannot do that; it is my father and my mother."
The young rascal had saved up all his money and bought the little black
boy as his slave. He was delighted on seeing a scrap-book and a small
musical box. These Mr. James presented him with, also a dollar, which the
youngster handed over to his slave with quite the air of a superior.
March 28th.—To-morrow we hope to be on the march once more. To-day we
are much engaged in securing fresh camels and camel-men for the march to
Massawa. I was employed for fully two hours in the morning, and again in
the evening, in relieving the several necessities and tribulations of my
Arab friends.
March 29th.—Physic again greedily sought for by Hamrans, Beni-Amirs,
Shukeriyahs, and Hadendowahs. I was engaged in dispensing from breakfast
time until near luncheon time, and again all the evening until dark. If
we do not move away from Heikota the medicine chest will soon be empty,
for these people seem to positively enjoy mixtures and pills. In the
evening, after dinner, all our late camel-men and horse-boys were paid
off, each one receiving five dollars backsheesh, and a good knife and
a razor, which are much prized. I have already spoken of the wonderful
digestive powers of the Basé people, but they exceeded my wildest
expectations, for on the 26th a circumstance occurred whilst we were on
the march that convinced me they could digest anything from a boot to a
pair of trousers, even with a man inside of them. Had anyone recounted
the anecdote to me I daresay I might have been somewhat incredulous, and
I cannot blame anyone, who does not know me, for being the same. However,
they can please themselves as to whether they believe what I am going to
state or not. I can vouch for the fact as I saw it myself.
In front of my camel, whilst on the march, were two Basé. One of them
found that a portion of his sandal (made of buffalo hide) had worn away
and came off; he was a careful Basé and wasted nothing. Feeling rather
hungry, the owner of the sandal pounded the bit of seceding sandal
between two pieces of stone, and having done so began to masticate it.
The other Basé man, who had secured a bit of the toothsome morsel,
evidently thought this a superfluous proceeding on the part of his
comrade, so incontinently proceeded to masticate it without this
preliminary precaution. Their teeth certainly are good, even beautiful,
and would be envied by any drawing-room belle, and I am sure their
digestive powers would also be envied by any individual living—no need of
Richardson's Peptacolos here. I would recommend the party who wrote about
"The Stomach and Its Trials" to peruse the above instructive anecdote.
March 30th.—This morning, after breakfast, my tent was surrounded again
with the patients, who had arrived from all parts, far and near, on
camels, on donkeys, and on foot. Amongst them was one old man who, after
the approved native fashion, squatted on his haunches, holding out a
little squeaking chicken, about a week or two old, at arm's length,
calling out "Hakeem howaga" (doctor, sir). This he continued at intervals
for about a quarter of an hour, until I had got rid of many patients.
What he intended this diminutive chicken for, unless as a present, I
was at a loss to understand; but when, at last, his turn came, I soon
discovered through Mahoom why the old man was so importunate. Yesterday
he came complaining of his eyes. I then told him he must procure a small
bottle for some lotion. "And now," said Mahoom, laughing till he could
hardly stand, "he bring you dis little chicken." What he thought I was
going to do with the chicken I don't know, unless he expected me to smash
it up, and with it make a lotion for his eyes. On my previous visit I
gouged out some diseased bone from a boy's foot, straightened the leg
of another boy who had suffered from a contracted knee, and so on. My
fame spread; hence the influx of patients suffering from all kinds of
diseases, curable and incurable. At 11.30 a.m. we left Heikota, reaching
Toodloak at 7 p.m., where we encamped on very dusty ground.
March 31st.—Here we found a very large village of Beni-Amirs encamped
on the Gash, who may be likened to locusts, as they remained at Heikota
until the animals had eaten up everything there; then they moved away to
pastures new. Toodloak is quite a lion neighbourhood. An immense zareeba
had been constructed on the river-bed in a large circle. Within this was
an inner circle formed by the huts of the tribe, and into that inner
circle many hundreds of camels, sheep, cattle, and goats are every night
driven in for protection from the wild beasts; yet, notwithstanding this,
a lion will sometimes leap the barrier and bound off with, a sheep or a
goat. Temp. 100° F. in the shade. The nights are now much warmer than
they were three weeks ago.
April 1st.—Last night two lions, a lioness, and two leopards were shot,
and all secured except one lion. We saw no less than seven or eight last
night. The lion was a very fine fellow, measuring nine feet two inches
from nose to tip of his tail. Sali cut a bit of the lion's liver, to make
him brave, as he said—a quite unnecessary proceeding, for he once was
plucky enough to enter the jungle in search of a wounded lion, and that
is what few would care to do.
April 2nd.—Lions are plentiful here. Last night they made a great noise
round the camp, probably attracted thither by the sheep and goats. The
Arabs had to mount guard over these, walking round in couples with spears
and shield, and shouting to keep them away. Last night the party went out
in couples at considerable distances apart. Each couple was ensconced
in a small zareeba, in front of which was a goat or sheep, tethered to
a stake driven in the ground, to tempt a lion. Opposite Mr. Colvin's
zareeba, whilst he was patiently waiting inside for a lion to make his
appearance, he had the mortification to see one leap on to his sheep,
seize him, and bound off again with the sheep, pegs and rope ere he could
get a fair shot at him; he discharged his rifle, but missed his lion. The
heat is so great to-day that it is quite uncomfortable to walk across the
river-bed in slippers—104° F. in the shade. To-morrow we shall probably
start afresh, and pitch our moving tents a day's march nearer home.
April 3rd.—All astir this morning preparing for the march to Amadeb,
where we expect to find letters and newspapers from England. We have some
tame monkeys and baboons in camp, which are very amusing and playful.
When we are at breakfast some of them come and watch us, thankfully
receiving any crumbs of comfort we choose to throw them. Although their
movements are quite unfettered, they make no attempt to leave us and
join their comrades who have not been civilized. Baboons, which are very
large and strong, will sometimes form themselves into a limited liability
company for the purpose of attacking a leopard, relieving one another at
intervals; then, at last, when they think a favourable opportunity has
arrived, all suddenly swoop down on him at once. They will often lay in
wait for the goats and sheep going to water, then go and suck them whilst
they are drinking; after which refreshing performance these robbers,
not liking to return to their mountain fastnesses empty-handed, will
take the liberty of borrowing a kid from the flock. Last night, about 10
o'clock, we heard lions near camp. One came pretty close to the horses,
causing quite a stampede amongst them; at the same time all the camels
got up, showing signs of great uneasiness. Lions are very numerous here,
hyænas few. When hyænas predominate they will often attack a lion, and
in a very systematic way, too. A number will get in front of him, and
act as designing persons in England sometimes do. They throw dust in
his eyes, then the whole pack will fall upon and make an end of him. At
such times both the lion and hyænas make a great uproar. We heard them
one night, and Sali, who is an excellent tracker, told us that hyænas
were then attacking a lion. We hear that Sheik Ahmed, with a number of
the Beni-Amirs, have made arrangements with the Basé for a shooting
expedition in their country, intending to go as far as Maiambasar, then
cross from there to the Settite. They have a curious way of making
themselves agreeable out there. To make themselves acceptable, they will
take wives from amongst the Basé; then when they leave the country they
leave them behind, at the same time presenting them with a piece of
cloth, some dhurra, and, perhaps, a few dollars.
April 4th.—Marched about nine hours to-day, encamping in a very
mountainous region. Whilst the tents were being pitched, Mr. Colvin
picked up a book, went about a hundred yards from camp, and sat down
under a tree to read. He had not been there long ere he heard something
breathing near him; looked round, and saw a hyæna within a _very_ easy
distance. He naturally got up and made tracks for camp, as he had no
rifle with him. His unwelcome interviewer followed him pretty closely
until he approached the camp. Soon after this, very near to camp, one of
them made for Dra, who drew his sword and shouted. George saw him, and at
once gave him the contents of his rifle.
April 5th.—Marched from 8 a.m., reaching Amadeb, a garrison town, at 8.30
p.m. We travelled a very mountainous region to-day—in one part, for the
distance of a mile or two, over large boulders of marble on the roadway.
On arriving at Amadeb, all were soon busy in reading their letters from
England. The Bey here treated us with great consideration and kindness,
providing us with tea, sugar, milk, biscuits, and cigarettes whilst our
tents were being pitched and we were reading our letters. I had almost
forgotten to mention one very exciting incident during our journey
to-day, and it was this—All our party, except Mr. Colvin and myself,
had gone on considerably in advance of the caravan (we were travelling
along a dry river-bed, consequently on very open ground), when Suleiman
came up in great haste, saying, "Doctor, have you got your rifle ready?
The caravan is going to be attacked in the rear;" and off he went to
inform the others. Colvin and I soon got our rifles and revolvers in
readiness, and trotted back to the rear, where we found our men in a
great state of excitement, leaping about, brandishing their spears, and
insanely yelling, "Whoop! whoop!" which, of course, was no use, but an
inconsiderate way of spending time and ventilating their feelings. Facing
them were, perhaps, 80 or 100 Bareas (akin to the Basé), indulging in
the same lunatic kind of performance. Our reinforcements arrived, in the
shape of George, Anselmia, and the rest of the party. The Bareas, seeing
our strength, then vanished like chaff before the wind. This was destined
to be our last scare. From the information we could gather, it seems
that the Bareas thought our numerous boxes were filled with dollars, and
(according to Dra and Girgas) began pointing their spears towards them
as if on hostile intentions bent. They retaliated by pointing rifles
at them, which had the effect of inciting these fellows into a kind of
war-like dance, accompanied by a significant brandishing of spears; and
there is very little doubt that blood would have been spilt had we not
promptly appeared on the scene.
April 6th.—Amadeb appears to be a tolerably large place. The houses are
made of mud bricks baked in the sun, and are thatched. The whole town
is surrounded by a mud wall. As we did not start until 3 p.m., I had
many patients to attend. At 5 p.m. we encamped on pretty good ground,
surrounded by trees, in which were thousands of doves, about 16 brace of
which Mr. F. James and I soon knocked over, thus providing luncheon for
the next day or two. We found water here.
April 7th.—We marched from 8 a.m. until 8 p.m., a distance of about 30
miles. At noon we passed a place where some thousands of camel, cattle,
sheep, and goats were being watered. Of course, at this time of the
year, all the river-courses are dry, and the water has to be drawn up
from deep wells. The Beni-Amirs, and other neighbouring tribes, are a
pastoral people, and owners of enormous herds, on which they wholly
subsist. They produce nothing, and seem to require nothing, except a few
yards of cotton. Their mode of watering the herds is a very laborious
one. On this sandy river-bed I observed about 20 very large but shallow
mud-basins, and close to each basin a well. At the top of each well is an
Arab, provided with a large antelope skin (dressed), fastened around the
sides, and attached to a long rope. This skin he drops into the well, and
hauls up, hand-over-hand, full of water, which he tips into the mud-basin
until it is full. With such insufficient means, of course, a long time is
occupied in filling the basins.
In the afternoon George, who had gone on ahead, came back much excited,
saying there was a young elephant in a wood which could be easily shot,
adding that if he had had an elephant rifle with him he should have been
tempted to let fly. It was fortunate that he had not, for presently we
met the owner of this captured elephant, accompanied by a number of
natives. They had halted for a short time, and had secured the elephant
to a tree by a rope round the neck and leg. He had been captured at
Forfar. His tusks were just protruding. Just as I approached him a native
ran up, saying, "Batal, batal howaga" (bad, bad, sir). The man was right,
for this infant had already wounded five men severely. I, therefore, kept
a respectful distance. The country through which we travelled to-day
was very mountainous, and here and there big plains covered with bushes;
still nothing like so bad as the road before we got to Amadeb.
April 8th.—Marched from 8 a.m. until 1 p.m., encamping at Gargee,
situated on Khor-Baraker. We stay here to-night, hoping to secure another
lion. Just after dinner we heard jackals barking very much like dogs.
CHAPTER XXIV.
A LION NEAR THE CAMP—THE MONKS OF CHARDAMBA—WE MEET ALI DHEEN
PASHA, GOVERNOR-GENERAL OF THE SOUDAN—ARRIVAL AT KEREN, OR
SANHÎT—THE PRIESTS AT KEREN—ACCOUNT OF KEREN—MERISSA—DRA, A
DOMESTIC SLAVE, MADE FREE—DESCENT FROM SANHÎT TO THE ANSEBA
VALLEY—THE BIRDS THERE—ALONG THE RIVER-BED OF THE LABAK—A BIG
MARCH—MASSAWA—FAREWELL TO CAMELS—MASSAWA TO SOUÂKIN—TAKE IN
CARGO—FAREWELL TO THE SOUDAN—ARRIVAL AT SUEZ.
April 9th.—About 9 o'clock we were off again, halting at 5.30 p.m. at
Adatur, on the River Bogoo. A fine leopard and hyæna were shot last
night. About 9 p.m., whilst I was writing my diary, alone in camp, the
others having gone out in search of lions, a native came up to me,
whispering, "Asset, asset, howaga" (a lion, a lion, sir). It was a lovely
moonlight night, so light that I could see to read or write distinctly.
I looked on to the river-bed in the direction indicated by the Arab, and
there, 40 yards off, I plainly saw a noble-looking lion. I was not long
in obtaining a rifle from my tent, and following him up the river-bed;
but to no purpose, as he was too far off to risk a shot at him. Finally
he turned off into a large palm grove on the opposite side, where, of
course, I did not follow him. I am told that about 14 years ago all this
neighbourhood belonged to Abyssinia, but now belongs to Egypt. From
this spot, not very far off, we noticed a most precipitous mountain,
called Chardamba, in Abyssinia; it is almost inaccessible, and can only
be ascended by taking off one's shoes and stockings, on account of the
smoothness of the stones, at least, so Mr. Phillipps says. He and some of
his party had the curiosity to climb it last year, and nearly got killed
by stones being rolled down on them, for on the very summit of this
mountain dwell some monks who have quite eschewed the pomps and vanities
of this world. Many of them are withered, shrivelled old fellows who have
lived there without once coming down for the space of 30 or 40 years,
and, to all appearance, are one or two hundred years old. They possess a
deep well, a chapel, each one a separate apartment or den, and they grow
a little dhurra.
The next day we marched from 7.30 a.m. until 5.30 p.m., encamping
at Ashdera. On the road we met, about 1 p.m., Ali Dheen Pasha, the
Governor-General of the Soudan (who succeeded Ali Riza Pasha, a most
unpopular governor). He was coming from Massawa, and, I believe, going
on to Kassala. We had a very long pow-wow with him. He informed us that
our Abyssinian affair had created quite a sensation in Cairo, that the
Minister of the Interior had telegraphed to him asking if we were safe,
and that the Mudir of Kassala had been dismissed for not sending soldiers
with us, which, by the way, was not his fault, for we would not have
them. The palace at Massawa was built by Gordon Pasha when Governor of
the Soudan. This Ali Dheen kindly placed at our disposal, until a boat
could take us to Suez.
On the 10th April a march of about 10 miles only brought us to Keren,
or Sanhît. Soon after our arrival Réschid Pasha sent an officer, whom
Ali Dheen Pasha had sent on, to inquire if we had arrived safe, and
were comfortable. From here we sent telegrams to Massawa respecting
boats to Suez. Although our journey was a short distance, it took us
a considerable time to accomplish it. I should think, for about the
distance of a mile, we traversed a most precipitous ascent in a zig-zag
fashion; this was not far from Keren. On reaching the summit we found
ourselves in quite a different kind of country. We soon paid a visit to
the priests at Keren, who have an excellent garden. They kindly sent
us some fine cabbages leeks, carrots, potatoes, and lettuce. As we had
not seen anything of this kind for months they were, of course, a great
luxury for dinner in the evening. In the afternoon the army doctor (who
could speak French tolerably well) called on me, and as I had almost
finished my campaign I gave him the greater part of the contents of my
medicine chest. In the afternoon the tom-toms were set going, accompanied
by the peculiar trilling note of the women, to express their joy at our
arrival. They had heard that we had all been killed by the Abyssinians,
our men also, many of whom came from here—Ali Bacheet, Dra, Girgas,
and Mahomet Zanzimeer. The latter was thus called because he generally
attended to our zanzimeers when we arrived in or started from camp. Pere
Picard and another priest dined with us in the evening. After dinner
about 60 children, well clothed, nice and clean, from their schools, came
to witness the mysteries of the magic lantern; not only they, but Réschid
Pasha, with a number of officers, came also, and were much pleased. When
they took their departure rockets, together with red and blue fires, were
let off.
Keren, or Sanhît, is the capital and only town of Bogos. It is 4,469 feet
above the level of the sea, on the edge of the highlands of Abyssinia,
and on one side of a table-land, about four miles long by two broad,
surrounded by hills. Being situated in such a high position, it is
healthy at all times of the year.
This elevated plateau has for its southern boundary the Abyssinian
mountains, which come down to within about two miles of the town, which
is built just outside the gates of the fort, and consists of two short,
broad streets of very poor Greek stores, and some clusters of Abyssinian
houses and Arab huts.
About a mile distant across the plateau is the French monastery. With
the exception of the Pasha's residence, the monastery and its adjacent
buildings are the only respectable-looking habitations there. They
have an Amharic printing press for publishing Bibles in the native
tongue, schools for educating Abyssinian boys for the priesthood, also
for the education of girls. When I was there, about four priests and a
few sisters fed, clothed, lodged, and educated upwards of 80 boys and
50 girls. They had a good schoolroom, chapel, dormitories, pharmacy,
and harmonium, which latter one of the sisters play. The children, who
looked very clean and orderly, sang very nicely a hymn in French for our
especial benefit. These priests print, bind books, have carpenter's and
blacksmith's shops, a dairy, vegetable garden, and appear to be a very
useful community, exercising, I should say, a great deal of self-denial
by residing in such a place as they do. They cannot certainly have a very
festive time of it, for the Egyptian authorities regard them as spies,
while, on the other hand, the Abyssinians, who distrust them, will not
allow them into their country. I must speak of these priests as I found
them. None of us were Roman Catholics, but they behaved with very great
kindness to us; and we must award them a great degree of credit for the
civilizing influence they exercised amongst these waifs and strays. The
thought that occurred to me was that if Roman Catholic priests could
be doing all this good, why not English missionaries also? There is
generally a garrison of about 1,000 soldiers at Keren. There is scarcely
any trade, as merchants will not risk their lives and goods, knowing the
hostility of the Abyssinians, who are anxious for the recovery of Keren
and the lost province of Bogos, added to which the Egyptian policy is
to isolate Abyssinia, thereby preventing the importation of arms and
ammunition for King John's army. The Beni-Amirs and other neighbouring
tribes, who are all owners of large herds, on which they wholly subsist,
are the Egyptian subjects. Fuel is scarce, as it has to be brought from
the Anseba valley, four miles distant. Water also has to be brought
from a distance, but there is a well inside the fort which supplies the
garrison. Dhurra is dear, as the continual dread of Abyssinian raids
prevents people from cultivating, also from building decent houses. There
are no camels, and very few horses and mules.
[Illustration: MOUNTAIN PASS NEAR KEREN OR SANHÊT]
I once tasted at Kassala a kind of beer made from dhurra, called merissa,
and thought it exceptionally nasty. It is probably the same as that
mentioned by Herodotus, and is in common use in the Soudan and Upper
Egypt. I was again induced by a native, who thought he would be attentive
to me for some little professional attendance I had given to taste this
vile liquid. Thinking, perhaps, it might not be so nauseous as my last
dose, I drank some, and acted like a stoic and philosopher, for I did not
exhibit any outward and visible sign of the rebellious feelings taking
place within. But "no matter;" I mentally resolved that in whatever part
of the Soudan I might at some future time find myself, I would never
again degrade my esophagus by allowing any merissa to glide down it,
for I would just as soon have taken one of my own vile black draughts,
and infinitely have preferred a bottle of Bass's pale ale. I have no
hesitation in saying that if merissa—a very euphonious name, so much
like Nerissa—was the usual beverage in England, the Blue Ribbon Army
would find in me a very ardent supporter, as it is more unpalatable,
thick stuff than schliva (a Bosnian drink, made from the juice of prunes,
possessing a strong asafœtida flavour). Merissa is made from dhurra,
which is allowed to germinate in the sun, then reduced to flour by
hand-mills. This, again, is converted into dough, boiled, and left to
ferment. I hope that if I ever again pollute my lips with merissa—not
Nerissa—I may be served in the same way, that is, that I may germinate in
the sun, be ground into powder, made into a pulp, boiled, and allowed to
ferment.
To-day a circumstance occurred in camp, illustrative of a curious custom
existing in the Soudan. Dra, who has been with us as a native servant
ever since we left Kassala, is, we find, a slave. Some altercation took
place this morning between Dra and another native, when the latter called
him a slave. This excited Dra's anger to such an extent that he talked
about knifeing his accuser. However, Messrs. James and Co. enquired
into the matter, and found that he was what they call a slave—what we
should call a domestic slave or bondsman—and that it came about in this
way—Dra's father had, some time ago, "pinched" a cow, _i.e._, purloined
it. He was ordered to pay back two within a month; failed to do so. The
fine was then doubled every month, until he owed about a hundred cows.
As there was just about as much probability of his being able to pay his
fines as there would be of my paying the National Debt, he and all his
children and children's children became bondsmen and bondswomen of the
creditor. Should Dra, who was single, desire to marry, he would find it
difficult, under such circumstances, to obtain a wife; but should he
succeed in doing so, and children be born to him, all the males would
be slaves, and the females _compelled_ to become public women. This
little matter was talked over in the evening at the dinner table, when it
was ascertained that his freedom could be secured by the payment of 35
dollars, which my colleagues gave to the priests to effect his freedom
with. They say that papers will have to be made out and signed by the
Pasha at Keren, and that on payment of the 35 dollars he will be free. A
man then goes round the town with a trumpet, and proclaims the fact in
all directions. This, I am told, is quite typical of the system in vogue
here.
On the 11th of April, at 11 a.m., we were once more on the road to
Massawa, encamping at 5 p.m. at Gubana. Ere we left Keren a donation in
aid of the schools was given by the party, and the priests very kindly
sent us a sack of potatoes, carrots, and parsnips. As we had been without
fresh vegetables so long, these were really quite a treat. We descended
from Sanhît into the Anseba valley. About a mile from the town, in the
lowlands, were the gardens of the priests, full of vegetables, growing in
the most luxuriant manner. The trees, too, flourished, and were covered
with beautiful green foliage. Here, on reaching the river-bed, we found a
little water actually on the surface.
April 12th.—To-day we travelled 10 hours through quite a different kind
of country to what we have lately been accustomed. A great variety of
lovely plumaged birds are abundant in the Anseba valley. Even the doves
here are tinged with bright yellow, green and blue. I managed to secure
some specimens of them, the brilliant plumaged rollo-birds, paroquets,
and a fine eagle. Soon after starting our route lay up the side of a
frightfully precipitous mountain. The ascent was made by a most tortuous
pathway, where in very many places the camels walked on the brink of a
fearful abyss. Having at last reached the summit of this mountain in
safety, our descent was made by an equally horrible pathway, after which
we travelled for the greater part of our journey along a dry river-bed,
surrounded on each side by steep, precipitous mountains covered with
large cactus trees in full bloom, and gaunt, leafless, but enormous
baobob trees, and, of course, the never-to-be-forgotten prickly mimosas,
but I saw much less gum arabic exuding from them there than I did in the
Basé country. Large tamarisk trees, also, were numerous. The branches of
all kinds of trees lay scattered about in the greatest profusion, the
natives having chopped them off for the goats, sheep, and cows to feed on.
On the 13th April, another 10 hours' march brought us to Kalamet, where
we pitched our tents for the night, having travelled through very
wild and picturesque scenery, the greater part of the distance being
along the river-bed of the Labak. Water was very near to, and in some
places on, the surface. For a distance of about two miles we journeyed
along a very deep pass; on either side of us extremely rugged, lofty,
precipitous crags, clothed with cactus trees and other bushes, through
which we occasionally discerned jackals and wild goats peeping at us with
astonishment. Scarcely any birds of any description were seen this day.
The pass, I believe, is called El Ain; a representation of it appeared
some months ago in the _Graphic_. Our camping and marching, like Bedouin
Arabs, is now nearly at an end, for in another day or two we expect to
reach Massawa, once more set our eyes on the Red Sea, and a steamer, don
decent apparel, and return to civilized life.
Our next day, April 14th, was occupied in marching to near Kamfar, a
distance of about 27 miles. This was an easy road for the camels, as
the greater part of the distance was along level ground, covered with
mimosas, more like the desert near Souâkin. We came across a few jackals,
ariels, gazelles, falcons, and extremely beautiful diminutive sun-birds,
so small that they ought to be shot with sand.
April 16th.—At 7.30 a.m. we once more, and for the last time, mounted our
camels, and made a good march of about 32 miles, arriving at Massawa at
8.30 p.m. Some of our Arabs had never before had an opportunity of seeing
the Red Sea; now that they did, their delight seemed great. We should not
have made such a long march to-day but that a messenger had been sent
from Massawa to meet us with the information that an Italian steamer
(Rubattino line) would be leaving on the morrow, and that if we missed
that we might, perhaps, have to wait two or three weeks ere we should get
another. However, we did not intend to wait, for we should have chartered
a stambouk from there to Aden, there catching a P. and O. boat for Suez.
Massawa, the principal sea-port of Abyssinia, is on a small barren
coral island in the Red Sea, about a mile long, by 400 yards broad, at
the northern extremity of the Bay of Arkeeko. It is connected with the
mainland by a causeway a mile in length, built across the shallow water.
There are barracks and huts for about 1,000 soldiers. It is badly off
for water. This has to be carried two miles to the town. The town is
built partly of stone and coral, but most of the houses are constructed
of poles and bent grass, and surrounded by a reed fence. The most
considerable buildings are the mosques, the houses of the traders, and a
few warehouses, which are built of coral, and the bazaar.
The surrounding country is little cultivated, and industry paralysed
by fear of raids by the Abyssinians. But little trade is done with the
Soudan; formerly a brisk trade was done with slaves, and that seems
likely to be the case again. Most of the imports come from Abyssinia.
They are grain, gold, cotton, manufactures, glass-wares, spices,
arms, cutlery, hides, butter, wines and spirits. Principal exports,
rhinoceros-horns, gold, ivory, honey and wax.
As usual, vexatious export and import duties are imposed by Egypt. No
guns or powder are allowed to be imported. There are very few camels, and
horses and mules are scarce and dear. The population has been estimated
at about 4,000. The island is dependent on Egypt, and is ruled by a
governor appointed by the Khedive. The chief inhabitants of Massawa are
a few Greeks and Italians, an Italian and French Consul, one Englishman,
and a small colony of Banians, through whose hands almost all the trade
passes.
On arriving at Massawa I took my leave of camel riding, and cannot say I
was sorry, having ridden, probably about 2,000 miles on caravan camels,
which roared at me when I mounted them, roared when I dismounted, roared
if I passed them, and roared if I looked at them. My knickerbockers were
just about done for, and I should not like to say how many hours I had
spent in repairing them, on two occasions, with a girba cut up. The Bey
soon put in an appearance; had a bit of a pow-wow; gave us coffee and
cigarettes, then showed us to the palace, where we had our camp-beds once
more unfolded.
April 17th.—So much had to be done to-day—paying servants, selling
camels, and tents, &c., that we found it impossible to get away. Word to
this effect was sent to the Italian steamer, so the Captain has postponed
his departure until to-morrow. Abdullah, a <DW64>, who had always seen
to getting my camel ready, helping to pitch and take down my tent, and
otherwise proved a faithful henchman, presented me with his spear,
and kissed my hand. I rewarded him with a small pin-fire revolver, 50
cartridges, a good knife, a razor, two dollars, and my knickerbockers.
His delight was unbounded, possibly at receiving the last-named
unmentionables. I am told that at an island near here a considerable
pearl fishery is carried on, and that at another island close by large
quantities of fine melons are grown.
[Illustration: THE CAUSEWAY AT MASSAWAH.]
At 8.15 a.m. the next morning we took our leave of the Soudan. The Bey
had received orders to pay us every attention. He accordingly sent a
gunboat, manned by sailors of the navy, with an officer, and we were at
once rowed off to our steamer, bound for Suez. In the evening of the
next day, at 5.30 p.m., we anchored at Souâkin. I and two others went
ashore and spent the evening with Mr. Bulay (who was always most kind
and obliging), returning to the ship at 10.30 p.m. All sorts of dreadful
reports about the attack of the Dembelas on us had reached him, which,
fortunately, were not true. Other startling news he gave us, such as that
the Queen had been again shot at; that Arabi Pasha was getting a power
in the State, and that the Khedive was likely to be dethroned, and Arabi
substituted.
About an hour before we dropped anchor at Souâkin some of our boxes
arrived from Kassala. We are likely to remain here two or three days
taking in cargo. The day before we left Souâkin a French Consul
and two American gentlemen (a doctor and missionary, who were at
Shepheard's Hotel in November last at the same time that we were), came
on board. They had been to the White Nile. We have quite a menagerie
on board—parrots, paroquets, tiger-cats, jackals, monkeys, baboons,
extraordinary looking geese, two enormous tortoises, and other animals.
We took on board a good many cattle for Suez, and left the Soudan for
good at 9 a.m. next day, April 22nd, arriving, after a very pleasant, but
warm voyage, at Suez about 3 a.m. At 8 we went ashore and breakfasted at
the Suez Hotel. Of course Mr. Clarke, the excellent and obliging manager
of this hotel, related the usual tale of our horrible massacre, &c.
CHAPTER XXV.
SUEZ TO CAIRO—ALEXANDRIA—ON BOARD THE "MONGOLIA"—PASSENGERS
ON BOARD—HIBERNIAN HUMOUR—VENICE—THE PIAZZA OF ST. MARK—THE
CAMPANILE—THE PIAZETTA—THE ZECCA, OR MINT—THE PALACE OF THE
DOGES—ST. MARK'S—THE ARSENAL.
April 27th.—Heads and skins had to be sorted out, turpentined, packed and
sent by sea from Suez, together with Mahoom and Girgas, the latter an
Abyssinian whom Mr. Phillipps is taking home with him as a servant. On
the 28th we left Suez for Cairo, arriving there at about 5 p.m., where I
found several letters awaiting me—some of rather old dates. Of course the
wildest reports of our massacre had reached Cairo, and been the topic of
the day at the time. Our stay in Cairo was of short duration this time,
as we found the Peninsular and Oriental Company's steamboat _Mongolia_
would be leaving Alexandria on the 4th May. Messrs. Colvin and Aylmer
went on to India, but the rest of us started for England.
Leaving Alexandria on the 4th, with a goodly number of passengers, about
120, we had a pleasant voyage to Venice, passing on the 5th the Morea,
Navarino, and Caudia, on the 6th Xante and Cephalonia, and on the 7th
arrived at Brindisi, viewing Montenegro and Corfu in the distance. There
we got rid of the mails, and fully half the passengers, and at 6 p.m.
on the 8th steamed up the grand canal, and soon arrived at Venice, the
Queen of the Adriatic, the home of poetry and song. How pleasant it was
to find myself, after all this Arab life, comfortably housed at the Hotel
d'Italie, amongst civilized people, I will leave the reader to judge.
There were a great many notables on board, amongst them several ladies
connected with officials in Cairo. We knew that matters were a little
unsettled in Egypt at this time, and so drew our own conclusions. These
ladies were being sent out of the way, and within 3 or 4 weeks after I
had seen the last of the great square in Alexandria it was in ruins.
There were on board big men and little men, both in stature and in their
own estimation. There were fat men and thin men, agreeable and chatty
men, disagreeable and morose men, humble and meek men, busy and sleepy
men, easy-going looking men, one or two of the "Ah! I see, thanks, I'll
not twouble you" kind of fellows, Colonels, Lieut.-Colonels, and other
officers, Governors and Judges returning home on leave of absence, and
genial, good-hearted, jolly sort of fellows. I acted here, as I always
do at home, avoided the starchy "Ah! I see—not-twouble-you kind of
fellows," full of their own importance, whose brains are concentrated
in their nicely-polished boots, &c., and fraternised with the sociable,
sensible, good-hearted kind. Amongst them was one of my own profession,
brimful of Hibernian humour and mirth. He was a brigade-surgeon in the
68th in India, where he had been for 25 years, and was now on leave of
absence. Dr. Kilkelly and I conceived a mutual regard for one another.
He and I, with a Judge from Cawnpore, a Colonel and Lieut.-Colonel,
generally got together on the deck, enjoying ourselves very comfortably
until we parted. I cannot remember all the jokes and witticisms of our
friend, Dr. Kilkelly, but I do remember one circumstance that amused us
all immensely, and caused great laughter, as much in the way of saying
it as the thing that was said. We had been having a great talk about
the Soudan. When I happened to say "Two of our party are going on from
Cairo to India, and will not be in England until this time next year,"
the doctor exclaimed, "Sure, ye don't say they are going on there now? I
could not have thought a man in his senses would be going to India now.
Do ye know what it is like this time of the year?"
"Hot, I suppose," said I, whilst the others smoked their pipes and looked
amused, evidently expecting some "rale Irish joke."
"Well, then, I'll tell ye," said our humorous friend, with a merry
twinkle in his eye, and a really comic aspect; "d'ye know, docthor, when
I have been in India this time of the year I have often made the natives
dig a grave for me to lie in, half fill it with grass and pour buckets
of cold wather on me to keep me from melting. I'll tell ye another
thing—cholera is so bad at this time of the year, that, by the Viceroy's
orders, coffins of all sizes are kept ready at the railway stations, and
when the ticket-collector goes round, saying, 'Yere tickets, plase,' he
finds a poor divil in the corner who does not respond; looks at him,
finds him dead, pulls him out, finds a coffin the right size, puts him
in, and by St. Pathrick he's buried before the sun sets. Now what d'ye
think of that? That's what India's like this time of the year."
Of course we all roared with laughter at the voluble and comical way in
which this was said, and I mentally made a note that I should not start
for India in May for my first visit.
Amongst our passengers were two sons of Sir Salar Jung, the Prime
Minister of the Nizam of Hyderabad, on a visit to England. The elder
one, though young, was a very Colossus, and an extremely intelligent,
agreeable fellow, who spoke English fairly well, and was very chatty.
He invited me, if I visited India, to visit him, and promised I should
have some tiger-hunting. Whether I shall ever do so, or he would remember
his promise, I don't know—probably not. Dr. Kilkelly and I put up at the
same hotel (the Hôtel d'Italie), and spent a few days very pleasantly. I
cannot say I should like to live in a place where, if I enter my front
door, I must step out of a gondola, or if I want to visit a friend I must
cross the street in a gondola; but it is a charming place to pay a visit
to for a few days, especially for a person with a romantic and poetic
turn of mind, and although romance has, to a great extent been knocked
out of me, I still have sufficient of the poetic temperament to have been
highly pleased with my visit to Venice, short though it was.
Pursuing the course I have hitherto adopted, I will not leave Venice
without a brief sketch of it and my visits to various places of great
interest, although, perhaps, repeating an oft-told tale. The man who
ventures on a description of a visit to Venice ought to be thoroughly
imbued with romance and poetry ere he can do justice to his subject.
Under such circumstances I cannot hope to rival many another; but, as
the Yankees say, "I'll do my level best." On the evening of my arrival,
I met, by appointment, one of the officers of the _Mongolia_, whom I
accompanied to St. Mark's Place. The side of the Piazza facing St. Mark
is a line of modern building erected by the French, somewhat in the style
of the Palais Royal at Paris, but yet having some sort of keeping with
the edifices on the south side. They are termed the Procuratie Nuove,
and form the south side, the Procuratie Vecchie the north side. The end
is composed of a French façade uniting the two. Near the east end of the
Procuratie Nuove, just by the point where it makes an angle with the
Piazetta, stands the Campanile of St. Mark. It is, in fact, the belfry
of the Cathedral, although it stands some considerable distance from it.
The separation of the belfry from the church is very common in Italy, and
there are a few instances of it in our own country. On the summit of the
Campanile is a large open belfry to which you ascend in the inside by
means of a series of inclined planes. The sides of the belfry are formed
by sixteen arches, four facing each quarter of the heavens. A gallery
with a parapet runs round the outside. I was told that the First Napoleon
ascended the inclined plane by means of a donkey. I, however, had to
walk it, and was well recompensed for my trouble by the magnificent view
obtained from the summit. Southward lies the noble Adriatic, with the
Pyrenees to the right; northward the Tyrolese Alps; immediately spreading
round this singular post of observation lies the city of Venice,
map-like, with its canals and neighbouring isles; and just under the eye,
to the east, is St. Mark's Church, considerably below, with its fine
domes, its four bronze horses, its numerous pinnacles, and in front of it
its three tall, red standards.
It is impossible to describe the effect produced on the mind, on a
summer's evening, as the sun is going down in his glory over the mainland
beyond the lagoons, lighting them up with his parting rays, while the
murmurs of the crowd assembled in St. Mark's Place ascend like the hum of
bees around the hive door, and the graceful gondolas are seen noiselessly
gliding along the canals. Traversing the Piazza, we find ourselves in
the Piazetta running down from the east end of the great one by St.
Mark's Church to the water-side, where the eye ranges over the lagoons
and isles. The next side of this open space contains a continuation
of the walk under arcades, which surround St. Mark's Place. The upper
part exhibits a specimen of the Italian style, designed by Sansovino.
The whole belonged to the royal palace, or Palace of the Doges, which
extends along the south and west sides of the Piazza. Turning round the
west corner of the Piazetta, on the Mole, with the canal in front, we
see another of Sansovino's works, called the Zecca, or Mint, from which
the gold coin of the Republic derived the name of Zecchino. In front of
the open space and landing steps of the Piazetta are two lofty columns,
which appear so prominently in the pictures of that part of Venice.
They are of granite, and came from Constantinople—trophies of Venetian
victories in the Turkish wars. The right hand column, looking towards the
sea, is surmounted by a figure of St. Mark, standing on a crocodile. The
left hand is surmounted by the lion of St. Mark. The west front of the
ducal palace forms the east side of the Piazetta; the south front runs
along the whole, and looks out upon the sea. They are its most ancient
portions. The front, overlooking the Piazetta, is composed of two rows of
arcades, one above the other; the lower a colonnade, the upper a gallery,
surmounted by a very large and lofty surface of wall of a reddish
marble, pierced by fine large windows. One gentleman says of it, "The
ducal palace is even more ugly than anything I have previously mentioned."
Mr. Ruskin, on the other hand, says that, "Though in many respects
imperfect, it is a piece of rich and fantastic colour, as lovely a dream
as ever filled the imagination, and the proportioning of the columns and
walls of the lofty story is so lovely and so varied that it would need
pages of description before it could be fully understood."
Having done the Campanile, and strolled round the Piazza and Piazetta, we
took our seats in the Piazza. On the west and south sides, as well as the
north, the lower part of the buildings under the arcades is appropriated
to shops or cafés. The latter are particularly celebrated. Towards sunset
the area of St. Mark's Place is overspread with tables and chairs, where
ladies and gentlemen are seated at their ease, as if in a drawing-room,
taking refreshments. A space in the middle is left for promenaders, and
when the military band is playing, which it does two or three times a
week, the concourse is immense, and the scene very lively and charming,
enabling one to realise the saying of Bonaparte, "The Place of St. Mark
is a saloon of which the sky is worthy to serve as a ceiling."
Having enjoyed the sweet strains discoursed by the military band to our
heart's content, we took our departure, my companion to his ship, and I
to my hotel.
The following day was occupied in various ways by Dr. Kilkelly and
myself. In the first place we, of course, paid a visit to the Palace
of the Doges. If those walls could speak, how many tales of horror and
cruelty they could unfold! Our visit to some portions of the Palace
enabled us to vividly imagine some of them. Of course, in many of the
trials here, whatever may be thought of the sentence inflicted, guilt,
and that of a heavy kind, was proved against the accused. The place was
not always a slaughter-house for innocence, a butchery for men guilty of
light offence. Grave crimes against the State were here disclosed, and
the memory especially dwells on that night in the April of 1855, when
Marino Faliero, a traitor to the Government of which he was the head,
was arraigned before his old companions in office, and when the sword
of justice, covered with crape, was placed on the throne which he had
been wont to fill. A very minute inspection of the Doge's Palace was
not practicable, for two reasons; one was want of time, the other the
impatience of my friend. Whilst in the Council Chamber of the Senate,
and for a minute or so looking at the largest painting I ever saw in
my life, "The Day of Judgment," my hasty friend seized me by the arm,
exclaiming, "Come along, do;" and soon afterwards, when I was deeply
engaged in the futile endeavour _apparently_ of dislocating my neck by
looking at the painted ceilings, and getting up the requisite enthusiasm
for the marvellous productions of some of the masterpieces of Titian,
Paulo Veronese, and Tintoretto, I was told to "Come along, docthor. Sure
ye'll have a crick in yer neck, and not be able to eat yer dinner at
all, at all, if ye stand looking at the ceiling in that kind of way;"
and so I allowed my volatile friend to rush me through the Palace of
the Doges, coming away with a hazy recollection of thousands of books,
wondrous paintings, the Council Chamber of the Senate, before whom an
undefended prisoner had formerly appeared; the Council of Ten, where he
generally got deeper in the mire; the Council of Three, whose decrees
were like the laws of the Medes and Persians; and, finally, the dungeon,
or condemned cell, just by one end of the Bridge of Sighs, where he was
strangled within, I think, three days of his condemnation. I was also
shown a dungeon, but not so low down as the condemned cell, where no
ray of light could be admitted, but where the poor wretch had a stone
slab, such as we have in our cellars, to lie upon, and let into the wall
was an opening through which the Grand Inquisitor could calmly gaze on
the torturing process produced by the rack and thumb-screw, and other
ingenious but painful arrangements constructed for blood-letting. Some
of the blood of deceased victims was shown to me on the walls, possibly
like the blood of Rizzio on the floor in Holyrood Palace, renewed once
a year. Of course there were many objects of great interest in the
Doge's Palace that I should have, no doubt, made many notes of had I
been by myself, but mental notes were all I was permitted to take. Many
people give a free rein to their fancy, and argue much on the origin
of species. This is a free country, and I may form my own idea of the
General Post-Office. Suppose I were to say that it originated from the
Doge's Palace? but fortunately for us, with a more agreeable class of
men as letter-carriers. I remember to have seen a lion, griffin, or
"goblin damned," at the head of one staircase, with open mouth, whose
sole object was to receive accusations, true and false, against citizens
of the State, and woe betide him if he came before the Council of Three,
from whom there was no appeal. Here our accusers have to prove us guilty,
there the accused had to prove himself innocent; and I doubt not that,
in those dark ages of cruelty, such a mode had its inconveniences,
necessitating a considerable amount of trouble for nothing on the part of
the accused.
We passed on from the Doge's Palace to St. Mark's. This church is very
ancient; it was begun in the year 829, and after a fire rebuilt in the
year 976. It was ornamented with mosaics and marble in 1071. Its form is
of Eastern origin, and it is said its architects, who were ordered by the
Republic to spare no expense, and to erect an edifice superior in size
and splendour to anything else, took Santa Sophia, in Constantinople, for
their model, and seem to have imitated its form, its domes, and its bad
taste. But if riches can compensate the absence of beauty, the Church of
St. Mark possesses a sufficient share to supply the deficiency, as it is
ornamented with the spoils of Constantinople, and displays a profusion of
the finest marbles, of alabaster, onyx, emerald, and of all the splendid
jewellery of the East. The celebrated bronze horses stand on the portico
facing the Piazza. These horses are supposed to be the work of Lysippus;
they ornamented successively different triumphal arches at Rome, were
transported by Constantine to his new city, and conveyed thence by
Venetians, when they took and plundered it in 1206. They were erected on
marble pedestals over the portico of St. Mark, where they stood nearly
six hundred years, a trophy of the power of the Republic, until they
were removed to Paris by Napoleon in the year 1797, and placed on stone
pedestals behind the Palace of the Tuileries, where they remained some
time, until they were again restored to Venice. In St. Mark's I was shown
two pillars of alabaster, two of jasper, and two of verde antique, said
to have been brought from King Solomon's temple, also two magnificent
doors, inlaid with figures of gold and silver, and a very large crucifix
of gold and silver, brought from Santa Sophia. I was also shown the
tomb of St. Mark the Evangelist. How true all this is I cannot say, but
perhaps many of my readers would like to know why St. Mark should be so
much thought of in Venice, so much so as to become the patron saint, and
have his name given to the most celebrated and splendid of its churches.
Over a thousand years ago—to be precise, in the year eight hundred and
twenty-nine—two Venetian merchants, named Bano and Rustico, then at
Alexandria, contrived, either by bribery or by stratagem, to purloin
the body of St. Mark, at that time in the possession of Mussulmen, and
to convey it to Venice. On its arrival it was transported to the Ducal
Palace, and deposited, by the then Doge, in his own chapel. St. Mark was
shortly after declared the patron and protector of the Republic; and the
lion, which, in the mystic vision of Ezekiel, is supposed to represent
this evangelist, was emblazoned on its standards and elevated on its
towers. The Church of St. Mark was erected immediately after this event,
and the saint has ever since retained his honours. But the reader will
learn with surprise that notwithstanding these honours the body of the
Evangelist was, in a very short space of time, either lost or _privately
sold_ by a tribune of the name of Carozo, who had usurped the dukedom,
and to support himself against the legitimate Doge, is supposed to have
plundered the treasury and to have alienated some of the most valuable
articles. Since that period the existence of the body of St. Mark has
never been publicly ascertained, though the Venetians firmly maintain
that it is still in their possession, and, as I said before, positively
show the tomb which, they say, covers him.
Our next visit was to the arsenal. This occupies an entire island, and is
fortified, not only by its ramparts, but by the surrounding sea; it is
spacious, commodious, and magnificent.
Before the gate stand two vast pillars, one on each side, and two immense
lions of marble, which formerly adorned the Piræus of Athens. They are
attended by two others of smaller size, all, as the inscription informs
us, "Triumphali manu Piræo direpta" ("Torn from the Piræus by the hand of
Victory"). The staircase in the principal building is of white marble,
down which the French (who invaded Venice) rolled cannon balls, an act of
wanton mischief quite inexcusable, at the same time they dismantled the
Bucentaur, the famous State galley of the Republic—a very Vandal-like act.
Venice, when in the zenith of its fame, might justly be said to bear
a striking resemblance to Rome. The same spirit of liberty, the
same patriot passion, the same firmness, and the same wisdom that
characterized the ancient Romans seemed to pervade every member of the
rising State, and at that time it might truly be said of Venice—
Italia's empress! queen of land and sea!
Rival of Rome, and Roman majesty!
Thy citizens are kings; to thee we owe
Freedom, the choicest gift of Heaven below.
By thee barbaric gloom was chased away,
And dawn'd on all our lands a brighter day.
But _tempora mutantur_.
CHAPTER XXVI.
WE HEAR OF THE MURDER OF LORD FREDERICK CAVENDISH AND MR.
BURKE—A GRAND SERENADE ON THE GRAND CANAL—MY JOURNEY FROM
VENICE TO ENGLAND.
Every evening during our stay in Venice, just about the time we finished
dining in the evening, a gondola full of serenaders would take up their
position just beneath our open window, and sing some of their charming
Italian ballads in a very pleasing style, undisturbed by the rattling
of cabs and omnibuses. Indeed, it seemed very strange, as we wandered
about this town of waterways, spanned by about 360 bridges, never to see
a vehicle or living thing except human beings. Of course, the quietude
that reigned all round was very favourable to the serenaders. One very
favourite song, both with the serenaders and visitors, was called "Santa
Lucia." This, I think, we had every night, for, if they left this out
of their programme, someone at the hotel would be certain to ask for it.
Venice, as all the world knows, is noted for its glass manufactories. To
one of these, owned by Dr. Salviati, my friend and I wended our steps.
We were much interested on going over the place, and much burdened on
coming out of it, as each of us emerged with an armful of purchases. It
was fortunate for me that my stay was of short duration, as each time I
returned to my hotel after a stroll, I generally did so with an armful of
purchases of some description or other.
Dr. Kilkelly, whose residence was in Dublin, had arranged with me that
we should return to England by a different route to that by which I had
come by. We intended to travel through the Brenner Pass to Munich, then
to Bingen on the Rhine, down that river to Cologne, and on to England,
but as we were to have a grand serenade on the Grand Canal at night,
we postponed the journey until the morrow. Soon after this, whilst
walking along, a large poster of the _Standard_ newspaper attracted
our attention, announcing "The Murder of Lord Frederick Cavendish." A
paper was soon procured, and there we read the account of his murder
and that of Mr. Burke's, dastardly deeds that all respectable Irishmen
blush to think of, and which excited the indignation of my friend to
boiling-point. This at once altered his plans entirely. Said he, "I
must start home to-morrow. As head of the family I must get to Dublin at
once; perhaps the place will be under martial law." I then relinquished
the idea of taking our route, not caring to go by myself, and resolved to
spend another day or so in Venice, returning _viâ_ Turin.
In the evening the doctor, Mrs. and Miss P., and myself made our way to
the water-side, with the intention of engaging a gondola to witness this
grand aquatic _fête_. How we should have got on with the gondolier I
don't know, as he could only speak his own language; but, fortunately,
I knew just sufficient Italian to pull us out of the difficulty. In
whatever country I remain a few days, my first business is to know the
value of the coins and be able to count in the language of the country.
This I have always found extremely useful, particularly in Turkey,
where the coinage is very confusing to a stranger. A piaster is worth
about 2d. There they have silver, copper, metallic, and paper piasters;
and unless one knows all about the rate of exchanging a Medjidie, the
trusting individual may possibly and probably be the victim of misplaced
confidence.
Having secured our gondola, we pulled up opposite Danielli's Hotel, a
little way above the Doge's Palace. Here we found a large floating
stage, occupied by those who were to take part in the serenade. It was
profusely and very prettily decorated with festoons of flowers and
evergreens, among which were interlaced a vast number of variegated
lamps (the centre piece forming quite a tree of these little lamps).
Under this stood the conductor, whilst around him was the orchestra
and singers. This great stage started from opposite Danielli's Hotel,
drawn by two boats. Following it were hundreds of gondolas jostling one
another on the Grand Canal, each one trying to get as near as possible
to the stage. Those on shore took as much interest in the proceedings
as those on the canals, for at every stoppage—and these occurred very
frequently—a performance took place. Opposite each stoppage there was a
grand pyrotechnic display by those on shore. Our first halt was opposite
the Palace of the Doges and the Piazetta of St. Mark, where numberless
Roman candles, Bengal lights, rockets, &c., were let off, brilliantly
illuminating the far-famed old Palace, Piazza, Campanile, St. Mark's,
and all the surroundings, and so on past the hotels (once gorgeous
marble palaces); the Church of Santa Maria della Salute, on the opposite
shore; the Palazzo Foscari, the Academy of Arts, and other palaces,
winding up with a grand scene at the Ponte Rialto. The time occupied
was about three hours, from 9 until 12 p.m. No amount of word-painting
can convey to the reader an adequate description of the scene, which
was most enjoyable throughout. It was a beautiful summer's night—no
moon, not a cloud; the blue sky studded with bright twinkling stars,
the stage adorned with flowers, evergreens, and hundreds of variegated
lamps; no sound but the splash of the gondoliers' oar and jostling of
the gondolas; a stop, then sweet strains of music arising from stringed
and wind instruments and two or three dozen well-trained male and female
voices; and every now and then the banks on either side lighting up, by
the illuminations, the grand old churches and fine old marble palaces of
the old Venetian nobility, to each of which is attached a history. It
was a scene which I shall always remember, but which I feel quite unable
to describe as I should wish. Our hotel was soon reached when all was
over, and I went to bed, lulled to sleep by the sweet Italian music of
gondoliers which came floating on the midnight air as they returned home
after this grand serenade.
On the following day my friend, the doctor, started for England. Soon
after his departure Mr. P. and I hired a gondola, and paid a visit to
the Academy of Arts, some of the principal churches, palaces, and
various places of interest, and saw some grand old sculpture, the tombs
of Canova and Titian, and paintings by Rubens, Titian, Tintoretto,
Paulo Veronese and others. I had resolved on the morrow to forsake this
dreamland under the clear blue sky of Italy, and once more rouse myself
to the stern realities of life. Accordingly, I found myself in the train
next day at 9 a.m., with a ticket for Paris. Passing Verona, Padua, and
other interesting places, I arrived in Paris at 5.30 p.m. on the evening
of the next day. A day's rest there, and I was on my way to dear old
England, which I reached in due time. A trip abroad is mentally and
bodily beneficial, but after wanderings in various countries, I have come
to the conclusion that the most comfortable place to _permanently_ reside
in, provided one is not absolutely devoid of the "almighty dollar"—as the
Americans would say—is "perfidious Albion."
I have travelled through the waving forests of Austria, miles of charming
vineclad <DW72>s in Hungary, acres of maize, rice, and tobacco fields near
Salonica, the beautiful cypress groves of Scutari, near Constantinople,
roamed over the wild mountains of Bosnia and Montenegro, through classic
Greece and Italy, and traversed the burning sands of Africa; but,
go where I will, nowhere is the general appearance of the country so
beautiful as in old England, where we find the little cottage of the
rustic so prettily embowered amidst fruit trees, shrubs and flowers,
whilst all around are undulating green fields, rippling brooks, and
winding rivers. Nowhere else is there anything to compare to our pretty
country lanes and variegated hedgerows, covered with sweet-smelling
hawthorn, the wild rose, honeysuckle, and the red berries of the ash,
whilst the banks are adorned with foxgloves and beautiful ferns, or
white with primroses, cowslips, and a thousand other wild flowers which
surround fields of waving golden ears of corn and the well-wooded estates
of the landed gentry, that in turn give shelter to the fox, who will
afford sport in the winter, and to the hares, rabbits, partridges, and
pheasants, who will assist in satiating our gastronomic propensities.
It is an Englishman's privilege to grumble, and whilst living here we
often find a great deal to grumble about, in politics particularly; but I
don't think there are many who, having travelled abroad continuously for
six, twelve, or eighteen months, will not say with me, on returning home
once more, "England, with all thy faults, I love thee still."
"A plain unvarnished tale I have unfolded," and as such, at this
particular time, I trust it will meet with the approbation of the
_majority_ of my readers.
Many faults, I am sure, may be picked out, as I have not only written,
but revised the book myself, instead of employing (as some do) a skilled
and experienced reader. Even had I done so I should still be able to say—
"Whoever thinks a faultless piece to see,
Thinks what ne'er was, nor is, nor e'er shall be."
FINIS.
ELEVENTH EDITION. NOW READY.
THE STORY OF CHINESE GORDON.
BY A. EGMONT HAKE.
_Demy 8vo., handsomely bound, with Portraits and Maps, 15s._
SOME PRESS OPINIONS.
The Times.
"The story of Chinese Gordon's Life, full as it has been of
adventure and stirring incident, cannot fail to appeal to a
wide circle of readers.… The record of Gordon's career would
be impressive under any circumstances; but it will be allowed
to his biographer that he has turned to good use the copious
materials placed at his disposal, and that this volume is
worthy the fame of its subject."
Saturday Review.
"The contents of this book mainly relate to the two most
prominent questions of the day—the position of China as a
fighting Power, and the condition of the Soudan; and it appears
therefore at a most opportune moment.… The present volume will
prove a valuable guide to politicians at the present crisis,
as well as a welcome source of information to those who desire
to learn more than can be gathered from newspaper reports of
the condition of the Chinese and Army, and of the forces of the
Madhi."
Spectator.
"A fresh and connected account of his marvellous campaigns
against the Taipings and against Zebehr's black brigade of
Slavedealers in the Soudan is heartily welcome.… His story is
not only rich in humility, abnegation, contempt for merely
objective human pleasures; it reveals to us a singularly happy
man."
Graphic.
"A volume which we should like to see in the hands of every
sub. and of every boy with military aspirations."
Standard.
"A strong and vivid biography of Chinese Gordon."
Daily News.
"Few careers of our own or any other time will compare in
picturesqueness with 'The Story of Chinese Gordon,' by Mr.
Egmont Hake."
REMINGTON & CO., HENRIETTA ST., COVENT GARDEN, LONDON.
End of the Project Gutenberg EBook of Life in the Soudan, by Josiah Williams
***
|
{
"redpajama_set_name": "RedPajamaBook"
}
| 965
|
Q: Characters not displaying correctly in different browsers I used certain characters in website such as • — " " ' ' º ©.
I found that when testing to see what my website looked like under different browsers (BrowserLab)
the afore-mentioned characters are replaced with �.
I then changed the charset in the webpage header from:
<meta http-equiv="Content-Type" content="text/html; charset=windows-1252">
to
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
Suddenly all the pages have the above mentioned characters replaced with a ?.
Even more puzzling is this is not always consistent across and even within the same page, as some sections display the character • and © correctly.
In particular, I need to replace the character • with one that will display across browsers, can anyone help me with the answer? Thanks.
A: You should save your HTML source as UTF8.
Alternatively, you can use HTML entities instead.
A: The source code needs to be saved in the same encoding as you're instructing the browser to parse it in. If you're saving your files in UTF-8, instruct the browser to parse it as UTF-8 by setting an appropriate HTTP header or HTML meta tag (headers preferable, your web server may be setting one without you knowing). Use a decent editor that clearly tells you what encoding you're saving the file as. If it doesn't display correctly, there's a discrepancy between what you're telling your browser the file is encoded in and what it's really encoded in.
A: Check to see if Apache is setup to send the charset. Look for the directive "AddDefaultCharset" and set it to Off in .htaccess or your config file.
Most/all browsers will take what is sent in the HTTP headers over what is in the document.
A: If you're using Notepad++, I suggest You to use Edit Plus editor to copy the text (which has the special characters) and paste it in your file. This should work.
A: Yes I had this problem too in notepad++ copy and pasting wasn't working with some symbols
I think SLaks is right
HTML entities for copyright symbol ©
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 9,785
|
Pseudeboda africana is a species of moth of the family Tortricidae. It is found in South Africa.
References
Endemic moths of South Africa
Moths described in 1964
Tortricini
Taxa named by Józef Razowski
Moths of Africa
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 2,903
|
Ely Minnesota Education Attainment Charts
This section of charts contains education data for Ely Minnesota based mainly on the latest year 2020 American Community Survey census data but also the survey from Common Core Data available for Public Elementary through Secondary Schools. In Figure 1, the percent of all people aged 25 years or older, who have either graduated from high school or completed the Graduate Equivalency Degree (GED) or some equivalent certification/credential. Ely depicts it has a High School Grad or higher of 95% which is the third most percent high school graduates or better of all other places in the greater Ely region. The city with the highest percent high school graduates or better in the area is Tower which shows a high school graduates of 97% ( only slightly larger).
In Figure 2, the percentage of people aged 25 years or older who have graduated from college/university with at least a bachelor's degree is provided. Note that the bachelor's degree is also called a four-year degree because it normally takes four years of full-time study to finish the course curriculum required to obtain the degree. This chart portrays the proportion of the population in this region who are college graduates with at least a bachelor degree or higher. In many ways, this analysis alongside the prior chart are very quick measures of the level of education in any particular area. Ely depicts it has a Bachelors Degree or higher of 33% which is at the top of all other places in the greater Ely region.
Figure 4 shows the percentage of people 25 years of age or older who either have no schooling at all or dropped out of school before being able to complete high school. Additionally, these are people who also do not have a Graduate Equivalency Degree (GED) or some other high school level type credential/student achievement. This analysis, along with the last two charts, provides another high level review and comparative data on the Ely Minnesota educational level. Ely depicts it has a School Dropout Rate of 5% which is less than most other places in the metro area. The city with the highest percent who dropped out of school in the area is Winton which shows a percent who dropped out of school of 17% (approximately 3.8 times bigger).
Figure 5 provides a more detailed look at the educational attainment for Ely Minnesota. This chart provides the proportion of people aged 25 years of age or older and what was their level of educational attainment. The chart provides 5 broad categories including: No Education/No School, Some High School, High School or equivalent, Some college or Associates Degree, and Bachelors Degree or higher. Ely has the percentage of percent of people with no schooling less than most other places in the metro area at 4% of the total. Second, it has the smallest proportion of percent of people with less than a high school education at 20% of the total. Third, it has the largest proportion of percent of people with some college or an associates degree at 33% of the total and is ranked #1.
The next chart shows a break down of people who have received a bachelor's degree or higher advanced degree generally in a campus learning environment. In particular, Figure 6, provides the proportional breakdown of all the people who have received a postsecondary education along with what the level of advanced degree that was obtained. Note that these categories do not include any type professional development type activities such as those related to maintaining professional credentials in workshop lessons. Note Professional Degree includes medical, dental, lawyers, etc. Ely has one of the largest proportions of percent of people with an associate degree at 48% of the total and is ranked #2. The only larger city being Tower with 49%. Second, it has in the mid range of other places in the area in terms of percent of people with a masters degree at 1% of the total. Third, it has the largest proportion of percent of people with a professional school degree at 2% of the total and is ranked #1.
The chart in Figure 7 shows the broad area of academic concentration or the discipline for people who have received a bachelors degree. This high level classification is essentially the field of study for which a degree was obtained. Ely has one of the largest proportions of percent of people with a degree in a science or engineering related at 29.7% of the total and is ranked #2. The only larger city being Tower with 31.7%. Second, it has one of the largest proportions of percent of people with an arts, humanities, or other degree at 21.6% of the total and is ranked #2. The only larger city being Hoyt Lakes with 26.4%.
The next chart (Figure 8) provides a more detailed deep dive on the category of major degree obtained for people aged 25 years or older who earned a bachelor's degree or higher. Specifically, this frequency distribution details out what the major field of study was the degree obtained. Ely has one of the largest proportions of percent of people with a degree in computers, mathematics and statistics at 12.3% of the total and is ranked #3. Only #2 Hoyt Lakes (15.1%), and #1 Soudan CDP (25.0%) are larger. Second, it has one of the largest proportions of percent of people with a degree in science and engineering related fields at 21.6% of the total and is ranked #2. The only larger city being Hoyt Lakes with 26.4%. Third, it has the largest proportion of percent of people with a degree in education at 6.9% of the total and is ranked #1. Also, it has the largest proportion of percent of people with a degree in liberal arts and history at 5.0% of the total and is ranked #1.
Figure 9 provides comparative data between the places in the greater Ely Minnesota region for broad educational attainment. This analysis uses provides five broad education attainment categories including: No Education/No School, Some High School, High School or equivalent, Some college or Associates Degree, and Bachelors Degree or higher. Ely has the smallest proportion of percent of people with less than a high school degree at 20.2% of the total. Second, it has the largest proportion of percent of people with some college or an associates degree at 32.9% of the total and is ranked #1.
A more detailed frequency distribution of educational attainment is provided in Figure 10. In particular this illustration breaks out the highest levels of university educational opportunities beyond the four year college degree. Included in the breakout are the relative proportion of masters degrees, PhD/Doctorate/Doctorial holders, and professional degrees such as medicine, dentistry, lawyers, etc. Ely has the percentage of percent of people with no schooling less than most other places in the greater region at 3.7% of the total. Second, it has the smallest proportion of percent of people with less than high school at 20.2% of the total. Third, it has the smallest proportion of percent of people high school graduates (or ged) at 22.6% of the total. Also, it has the largest proportion of percent of people with an associate degree at 25.2% of the total and is ranked #1. In addition, it has the largest proportion of percent of people with a professional school degree (e.g. law or medicine) at 0.9% of the total and is ranked #1. The next exhibit (Figure 11) provides detailed cross tabulation analysis that provides education success data broken out or cross tabulated by age group. Please note that the columns add to 100% and you must use the pagination buttons at the bottom of the table to see all the rows. Figure 12 is a cross tabulation analysis that shows large educational success categories and is broken out or cross tabulated by racial group. Please note that the columns add to 100% and you must use the pagination buttons at the bottom of the table to see all the rows.The final cross tabulation analysis is provided in Figure 13 and shows education success broken out by gender. Please note that the columns add to 100% and you must use the pagination buttons at the bottom of the table to see all the rows.
Ely Minnesota School Enrollment Charts
The next section of chart resources look at school enrollment by a variety of educational institutions and are categorized into a number of other groupings. Figure 14 provides the overall school enrollment by broad range of school age/level groupings. Ely has the smallest proportion of percent of children in kindergarten at 14.4% of the total. Second, it has the smallest proportion of percent of children in grade 1 to 4 at 12.0% of the total. Third, it has the second smallest as measured by percent of children in kindergarten of all the other places in the local area as measured by percent of children in grades 5 to 8 at 17.9% of the total. Also, it has the largest proportion of percent of children in grades 9 to 12 at 52.6% of the total and is ranked #1.
Figure 15 provides a simple high level comparison of the proportion of students that are enrolled in public schools versus students enrolled in private schools in the Ely Minnesota region. Ely depicts percent enrolled in public schools markedly bigger as the percent enrolled in private schools.
The next comparison provided in Figure 16 shows the proportion of students enrolled in private schools versus students enrolled public schools for students enrolled in Kindergarten through 8th grade (grade 8) or (K through 8). Ely shows percent of children enrolled in public k-8 grades very much bigger as the percent of children enrolled in private k-8 grades.
The next chart (Figure 18) provides a similar analysis to the last chart but provides the data for public post secondary education versus private higher education colleges. Ely shows percent enrolled in a public college considerably bigger as the percent enrolled in a private college.
The next chart in this series of resources shown in Figure 20 looks at the total number of students enrolled in any educational institution for each place in the greater Ely region. (Total enrollment in this case includes all students from preschool all the way through students enrolled in graduate school.) Ely shows it has a Total Enrolled of 683 which is ranked #1 of all places in the greater Ely region. Comparing total population enrolled in school to the United States average of 81,084,866, Ely is only about 0.0% the size. Also, measured against the state of Minnesota, total population enrolled in school of 1,395,300, Ely is only about 0.0% the size.
The final chart in this series is Figure 22 and provides analysis data for the school year shown. The chart shows the proportion of kids/children that are enrolled in a pre-school public district versus those kids/children who are enrolled in a private pre-school oriented tuition institution (private salary/superintendent/etc.) Ely Minnesota has the largest proportion of public preschool enrollment at 28% of the total and is ranked #1.
Ely Minnesota Area Schools Charts
Figure 23 lists all the schools in the area along with the school district, county location and other program information/credential such as if they are a public charter school or private charter school or magnet school. Some of the Area Schools are: Memorial Secondary, Washington Elementary, Babbitt Elementary, Northeast Range Secondary, and Tower-Soudan Elementary. The next illustration in Figure 24 shows the total child school enrollment for all grades (through 12th grade) at the school shown using NCES data (Common Core of Data, Public Elementary-Secondary School Universe Survey.) Looking at Enrollment for Area Schools we find that Virginia Secondary ranks the largest with a value of 713 enrolled students. The next largest values are for: Parkview Elementary (602), Mesabi East Elementary (522), Mesabi East Secondary (407), and Roosevelt Elementary (380). The difference between the highest value (Virginia Secondary) and the next highest (Parkview Elementary) is that the enrolled students is about 18.4% larger.
Figure 25 show the ratio of the number of students to the number of teachers in the classroom. A good student to teacher ratio should be low because it indicates that there are less students for any one teacher to educate in a class and generally a better learning environment, better success and optimal teaching excellence. Teachers includes all educational staff such as special education teachers and any other educator. Note that distance learning (online learning/remote learning) is not included in these values. Looking at Student to Teacher Ratio for Area Schools we find that Northland Learning Center 020 ranks the largest with a value of 5.2 student to teacher ratio. The next largest values are for: East Range Academy Of Tech-Science (10.0), Tower-Soudan Elementary (10.3), Northland Learning Center 010 (13.3), and Northeast Range Secondary (13.6). The difference between the lowest value (Northland Learning Center 020) and the next lowest (East Range Academy Of Tech-Science) is that the student to teacher ratio is about twice as large.
The next chart, Figure 26, shows the racial mix of students at each location in this district of the state of Minnesota department of education.
Ely, Minnesota Education Data
Education Attainment
School Enrollment
Figure 1: Ely, MN At least High School Education
Figure 2: Ely, MN Bachelors Degree or Better Education
Figure 3: Advertisement
Figure 4: Ely, MN School Dropout Rate
Figure 5: Ely, MN Education Attainment Breakdown
Figure 6: Higher Education Attainment (100%=All People with Bachelor or better)
Figure 7: Ely, MN Bachelors Degrees Field of Study
Figure 8: Ely, MN Bachelors Degree Obtained
Figure 9: Ely, MN Education Attainment by Level Comparison (Age 25+)
Figure 10: Ely, MN Education Attainment Detailed Comparison (Age 25+)
Figure 11: Ely, MN Detailed Education Attainment Breakout by Age Group (Age 18+)
Figure 12: Ely, MN Detailed Education Attainment Breakout by Race (Age 25+)
Figure 13: Ely, MN Detailed Male and Female breakdown of Educational Attainment
Figure 14: Ely, MN School Enrollment by Aggregate Categories
Figure 15: Ely, MN Overall Public vs. Private School Enrollment
Figure 16: Ely, MN Public vs. Private K-8 School Enrollment
Figure 17: Ely, MN Public vs. Private High School Enrollment
Figure 18: Ely, MN Public vs. Private College Enrollment
Figure 19: Ely, MN Public vs. Private Graduate or Professional School Enrollment
Figure 20: Ely, MN Total Enrolled in Schools
Figure 21: Advertisement
Figure 22: Ely, MN Public vs. Private Preschool
Figure 23: List of Schools in the Ely, MN Area (2013)
Figure 24: Ely, MN School Enrollment (2013)
Figure 25: Student to Teacher Ratios (2013) - Low Scores Are Better
Figure 26: Ely, MN School Racial Mix (2013)
Cities marked with an asterisk ("*") should resemble a city or town but do not have their own government (i.e. Mayor, City Council, etc.) These places should be recognizable by the local community but their boundaries have no legal status. Technically these include both Census Designated Places (CDP) and Census County Divisions (CCD) which are defined by the Census Bureau along with local authorities. (For more information, see: Census Designated Place or "CDP") and Census County Division "CCD".)
For comparison purposes, the US national average and the state average value are provided. Additionally, the "Combined Statistical Area" or CSA is shown that is closest to the city, county, or zip code shown. A CSA is a large grouping of adjacent metropolitan areas that identified by the Census Bureau based on social and economic ties. (See: Combined Statistical Area)
Data sources - Mouse over icon in upper right corner of each chart for information.
Near Ely, MN
Select a City-PlaceVirginiaElyHoyt LakesGilbertAuroraBabbittBiwabikTowerSoudan CDPMcKinleyWinton
Select a Zipcode5579255731557055575055790557065573255708557825560755725
Select a CountySt Louis CountyItasca CountyDouglas CountyCarlton CountyPine CountyAitkin CountyAshland CountyBayfield CountyKoochiching CountyLake CountyCook County
See All Minnesota:
Minnesota Cities A-Z
Minnesota Zipcodes, all
Minnesota Counties, A-Z
Education Metrics Ranked:
Minnesota Rankings
All State Rankings
Top 100 US Cities Ranked
Search All US:
Index of All States
Towncharts.com © | Privacy Policy
Towncharts Think Tank
Top 25 Cities Ranked for Every State
Free Market Research with Towncharts
Developing a Customer Profile
Retail Site Selection using Towncharts.com
Occupation Earnings - Data over 500 Job Titles
Top cities in the US ranked by Demographics
Top cities in the US ranked by Education
Top cities in the US ranked by Economy
About Towncharts
Citation Instructions
Email Towncharts
Agreement & Disclaimer
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 4,011
|
Updated family home adjacent to quiet neighborhood park has largest backyard on the block. 2-story living room has vaulted ceiling & new wood laminate flooring that comes w/ lifetime warranty. New flooring extends throughout family hub bathed in cheery morning light. Stainless steel appliances include new stove & fridge. Slab granite on counters & island. Convenient eating space & pantry. Kitchen opens to family room w/ gas fireplace. New paint & new carpet throughout upstairs. Master suite has 5-piece bath & walk-in closet. Secondary bedrooms are serviced by full bath. Sunny loft works great as a home office, exercise or hobby room. Main floor powder room has granite counter. Finished basement has new carpet in the family room plus a 4th bathroom w/ new paint & tile floor. Deep backyard has huge Trex deck (new in 2015), utility shed, dog run & a water feature. New roof in 2015. New water heater in 2012. New furnace in 2010. Walk/bike to elementary & middle school, minutes to highway.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 3,540
|
\section{Introduction}
\label{secIntro} Observations of the Cosmic Microwave Background
(CMB) provide some of the most compelling support for the currently
favored $\Lambda$CDM, or \emph{concordance}, cosmological model. The
concordance framework predicts that the CMB should posses
temperature fluctuations which are both statistically isotropic
(i.e. stationary over the celestial sphere) and Gaussian
\citep{Guth1982,Starobinskij1982,Bardeen1983}. Measurements by the
Wilkinson Microwave Anisotropy Probe (WMAP)
\citep{Bennett2003,Hinshaw2009} have undergone extensive statistical
analysis, much of which has confirmed the concordance model but with
some indications of departures that may be significant; see for
example \cite{Yadav2008}. More specifically, there is some evidence
for hemispherical power asymmetry
\citep{Eriksen2004a,Park2004,Eriksen2007,Hoftuft2009,Hansen2009} and
also a Cold Spot has been identified \citep{Vielva2004,Cruz2005}. In
other words there is some evidence of an anisotropic universe, i.e.
one in which the background cosmology may not be described by the
standard Friedman-Robertson-Walker (FRW) metric. Of course the
background cosmology for a non-isotropic universe may still be
described by the FRW metric, but this would require a non-standard
topology which we do not consider in this analysis.
The Bianchi classification provides a complete characterization of
all the known homogeneous but anisotropic exact solutions to General
Relativity. The classification was first proposed by Bianchi and
later applied to General Relativity \citep{Ellis1969}. Initial
studies used the lack of large-scale asymmetry in the CMB
temperature to put strong constraints on the possible Bianchi models
\citep{Barrow1985, Bunn1996, Kogut1997}. However, simulations of the
CMB from Bianchi universes not only show a preferred direction, but
models with negative spatial curvature (such as the types V and
VII$_h$) can produce localized features \citep{Barrow1985}. So more
recently attention has shifted to reproducing a Cold Spot such as
that claimed to exist in the WMAP data. Initially, Type VII$_h$ was
the favored model to best reproduce the anomaly \citep{Jaffe2005,
Jaffe2006a,Jaffe2006b}, and this has subsequently been investigated
quite thoroughly \citep{McEwen1,McEwen2,Pontz1,Pontz2,Sung2010},
although more recent work has also looked at the Bianchi Type V
which also produces localized features \citep{Sung2009}.
The most interesting range of anisotropic structures is produced in
Bianchi Types VII$_h$, VII$_0$ and V. These different Bianchi types
have the effect of focusing and/or twisting the initial quadrupole
over time (see Figure \ref{figBT}). In this paper we study the
behavior of these Bianchi models so as to identify characteristics
of the radiation fields they produce and develop methods that can be
used to identify more general forms of anisotropy. Understanding the
characteristics identified in these particular cases will hopefully
help us find better and more systematic ways of constraining the
level of global symmetry present in the real Universe. Note we
consider just characteristics observable in the CMB temperature; we
shall return to a study of the polarization radiation component in
later work.
We consider two statistical measures of anisotropy in some detail in
this paper. Neither of these is entirely new and both have
previously been applied to observed CMB maps. However, the general
philosophy behind previous applications of these methods has been
simply to look for departures from the (composite) null hypothesis
of statistical isotropy and Gaussianity (or more recently they have
been developed to look at universes with multiply-connected
topologies \citep{Bielewicz2009}). In other words, they have been
used to construct hypothesis tests with the concordance cosmology
but their performance has not hitherto been evaluated on models with
built-in anisotropy.
For example, if the concordance model is correct, the {\em phases}
of the spherical harmonic coefficients of the CMB should be
independently random and uniformly distributed. Recent studies have
suggested some deviation from this
\citep{Coles2004,sc2005,dc2005,Chiang2007,ccno7} but it is not clear
whether they indicate global anisotropy or departures from
Gaussianity, let alone whether these are of cosmic or instrumental
origin. Here we examine the use of phase correlations in quantifying
the temperature patterns generated in models with known levels of
global inhomogeneity.
Multipole vectors were first introduced over a century ago
\citep{Maxwell1891}. There have since been attempts to understand
the multipole vectors in order to explain the CMB anomalies reported
at large angular scales \citep{Katz2004, Schwarz2004, Copi2004,
Land2005a, Land2005b, Land2005c, Land2005d, Land2005e, Copi2006,
Copi2007} since is not clear how to quantify and verify such
properties from the CMB anomalies in spherical harmonics. They have
been used in a number of studies to show anomalies, such as
alignments of multiples \citep{Abramo2006} in a similar plane to the
axis of evil \citep{Land2005d,Land2007}. Our aim here is to examine
the behavior of the multipole vectors in cases where the form of
anisotropy is known {\em priori} in order to assess their potential
to act as more general descriptors.
Two points are worth making before we continue. First, any
realistic cosmology (whether of FRW or Bianchi type) will possess
random fluctuations on top of a smooth background. If these
fluctuations are stationary Gaussian then they will add correlated
``noise'' to any signal arising from the background model and will
thus hamper the performance of any statistical analysis method,
especially at smaller angular scales. This Gaussian ``noise'' (which is
equivalent to stationary Gaussian fluctuations, and not to be
confused with instrumental noise) is completely characterized by
second-order statistical quantities (i.e. the power spectrum in
harmonic space or the autocorrelation function in pixel space). The
statistical descriptors we explore are {\em independent} of the
power-spectrum, so adding Gaussian noise will not produce any
systematic response in them.
We also restrict ourselves to looking at just the large-scale features
because the patterns in the temperature maps resulting from the Bianchi models
is over large scales. Therefore, by looking at large scales only, there
is more chance of detecting the anisotropy. However, it goes without saying we are
not claiming that these Bianchi models are in themselves complete
alternatives to the concordance cosmology. Rather we think of them
as representing possible perturbation modes of the FRW background.
The layout of this article is as follows. In Section \ref{secPixel}
we look at pixel distributions of the CMB maps to show how the
statistical anisotropy present in these models produces a form of
non-Gaussianity in the pixel distribution over the celestial sphere.
We then introduce phase correlations in Section \ref{secPhase} to
provide characterization of the anisotropy displayed by the models.
In Section \ref{secMultipole} we look at the behavior of the
multipole vectors as characteristics of the anisotropy of the same
maps. Finally, Section \ref{secConclu} summarizes the conclusions.
\begin{figure}
\begin{centering}
\includegraphics[width=58mm]{MapVz000.ps}
\includegraphics[width=58mm]{MapVII0z000.ps}
\includegraphics[width=58mm]{MapVIIhz000.ps}.
\caption{Simulated maps of the the CMB temperature, at redshift z = 0, using Bianchi type cosmologies. From left to right the Bianchi types are: V, VII$_0$ and VII$_h$. The colour scale is marked in milliKelvin. All the maps started as a quadruple at z = 500. The Bianchi V map shows a focused feature, the Bianchi VII$_0$ map has a twisted feature and the Bianchi VII$_h$ map has both focusing and twisting in the resulting temperature pattern.}
\label{figBT}
\end{centering}
\end{figure}
\section{Statistical Descriptors of non-Gaussianity}
\label{secPixel} \cite{Sung2009} discussed how localized features in
the CMB temperature pattern, perhaps similar to the Cold Spot
observed in the WMAP data \citep{Vielva2004,Cruz2005}, can be
generated in models with negative spatial curvature, i.e. Bianchi
types V and VII$_{h}$. In the standard cosmological framework the
temperature fluctuations are described by a Gaussian random process
over the sky, so a feature like the cold spot corresponds to an
extreme event in the tail of the distribution of fluctuations. In a
Bianchi model, however, it is not stochastic at all but produced
{\em coherently} as a result of the geometry of the space-time.
Clearly we need a systematic way to characterize the relationship
between rare events like this and their origin through either
non-Gaussianity or global anisotropy. Analysis of the temperature
patterns using standard descriptors in non-standard scenarios will
produce signals different from what one would see in the presence of
stationary Gaussian noise. To illustrate this issue we study the
pixel distribution function of temperature values. This is an
obvious way to test for non-Gaussianity in a random field of
temperature values, but a coherent fluctuation field also possesses
a one-point distribution that yields some diagnostic information. In
this sense, all the Bianchi models are inherently non-Gaussian but
their non-Gaussianity is simply a manifestation of the presence of
anisotropy.
We calculated the pixel distribution function, which is simply a
frequency count of the pixel (or temperature) values, for each of
the Bianchi maps and used it to plot the histogram seen in Figure
\ref{PDF0}. A perfectly homogeneous and isotropic map, such as that
predicted by the concordance model, would have a constant value over
the whole map - remember we are considering maps without
fluctuations - and therefore a histogram of the pixel distribution
histogram for this map would give a delta function at the mean. Our
results show some deviation from this prediction. In Figure
\ref{PDF0} we see the plots for the Bianchi V and VII$_h$ types have
strongly peaked features at the mean, but still with some non zero
variance; the type VII$_0$ model has a nearly uniform distribution
across the whole temperature range.
Although not demonstrated in this diagram, another point to note is
that at early times the histograms of the pixel distribution
functions of the three different Bianchi types are almost identical;
the different values of the temperature pixels are roughly even over
the range. As redshift\footnote{Note, redshift is defined here as
proportional to the inverse geometric mean of three scale factors.}
decreases, the temperature patterns for Bianchi V and VII$_h$
gradually start to focus \citep[see][]{Sung2009}, and their
histograms of the pixel distribution functions become successively
more peaked. For the Bianchi type VII$_0$ the temperature pattern
just twists, reorganizing the pattern on the sky while the histogram
stays roughly the same. These observations help to explain the
features in the histograms. As the temperature patterns become more
focused, more of the rest of the map becomes uniform and so the
histogram is tending towards a delta function.
In summary, what we would hope to discover from the pixel
distribution histogram is that it gives us some clues about the
homogeneity and isotropy of the maps i.e. the degree of
concentration around the mean might tells us about the homogeneity
of the parameters or the asymmetry of the distribution might give
information about the anisotropy. But as it stands, the information
from the pixel distribution function is not that clear. All we can
say is that the histograms differ from a delta function, so the maps
are not perfectly homogeneous and isotropic, and that the histograms
are clearly non-Gaussian in shape. However the shape of this
one-point pixel distribution does not furnish us with a complete
description of the pattern because it does not take into angular
correlations between the pixels.
\begin{figure}
\begin{centering}
\includegraphics[scale=0.4]{PDHtypes.ps}
\caption{The pixel distribution function of the normalized temperature for the Bianchi CMB maps. The y axis indicates the normalized pixel number which is the whole range of temperature covered by the plot, but normalized so they can be plotted on the same axes. The colour cyan represents type V, magenta for VII$_h$, and green for VII$_0$. The plot shows that the Bianchi V and VII$_h$ types have strongly peaked features at the mean, whereas the type VII$_0$ has a nearly uniform distribution across the temperature range. }
\label{PDF0}
\end{centering}
\end{figure}
\section{Phase correlations of Bianchi CMB Maps}
\label{secPhase}
We now move on from pixel distributions to consider the spherical harmonics of the temperature maps, or more
specifically the phases of the spherical harmonic coefficients.
\subsection{Spherical Harmonics}
\label{secSH} The temperature of the CMB, $T_{\theta,\phi}$, is defined on a
sphere where $\theta \in [0,\pi]$ and $\phi\in [0,2\pi]$ are the
polar and azimuthal angles. Therefore one way of describing the
temperature anisotropies, $\Delta T_{\theta,\phi}$, is to extract the
corresponding spherical harmonic coefficients ($a_{\lm}$):
\begin{eqnarray}
\Delta T_{\theta,\phi} = \frac{T_{\theta,\phi} - \bar T}{\bar T} = \sum_{\ell=0}^{\infty} \sum_{m=-\ell}^{\ell} |a_{\lm}| e^{i\Phi_{\lm}} Y_{\lm}(\theta,\phi),
\label{eqnALM}
\end{eqnarray}
where $ |a_{\lm}|$ and $\Phi_{\lm}$ are the amplitudes and phases of the
spherical harmonic coefficients, and $Y_{\lm}$ are the spherical
harmonics which are defined here as:
\begin{eqnarray}
Y_{\lm} = (-1)^{m} \sqrt{\frac{2{\ell}+1}{4 \pi} \frac{({\ell}-m)!}{({\ell}+m)!}} P_{{\l m}} (\cos \theta) e^{im\phi} ,
\label{eqnYLM}
\end{eqnarray}
where $P_{{\l m}}$ is the associated Legendre Polynomial. Note that
this definition of spherical harmonics includes a phase factor of
$(-1)^{m}$, also known as the Condon-Shortley phase.
In the standard cosmological model, the temperature fluctuation
field is produced by stochastic fluctuations which are Gaussian and
statistically stationary over the celestial sphere. In this case the
phases $\Phi_{\lm}$ of each spherical harmonic mode $a_{\lm}$ are independent
and uniformly random on the interval $[0,2\pi]$ \citep{Coles2004}.
If instead the temperature pattern on the sky is produced by a
Bianchi geometry then the $a_{\lm}$ are no longer stochastically
generated but can be directly calculated from parameters of the
model. Analytical forms for the temperature pattern can be used to
obtain the spherical harmonic phases \citep{McEwen1,McEwen2}, but it
is clumsy to transform these between different coordinate systems
\citep{Coles2004}. In the following we therefore obtain
distributions of $\Phi_{\lm}$ from Bianchi maps generated using
the method described by \cite{Sung2010}.
\subsection{Visualising phase correlations}
\label{secVPC} To visualize the information held in the phases,
$\Phi_{\lm}$, of the spherical harmonic coefficients, $a_{\lm}$, we plotted
them over all ${\ell}$ and $m$. Rather than using a 3D plot, colour has
been used to represent the $\Phi_{\lm}$ following \cite{cc2000}. The
colours equate to the angle on a colour wheel: red ($\Phi_{\lm} = 0$),
green ($\Phi_{\lm} =\pi/2$), cyan ($\Phi_{\lm} =\pi$), and purple
($\Phi_{\lm}=3\pi/2$). To understand these plots, first consider
what we would expect to see in the case of an isotropic and
homogeneous universe as predicted by the concordance model. This
would be a uniform map (as we are not at this point considering
fluctuations) but in spherical harmonics this only has power in one
mode (${\ell}=m=0$), so there is no phase for the other modes. Better to
consider a map with Gaussian fluctuations as later in the section we
will move on to add noise to the Bianchi maps. Figure
\ref{figRandom} shows the phases ($\Phi_{\lm}$) for a homogeneous and
isotropic map with Gaussian fluctuations. The phases are random over
the space i.e. there are no visible patterns in the distribution of
colours in the plot. Note that for all the maps, $\Phi_{\lm}$ = 0 or $\pi$
for $m$ = 0 because the $a_{\lm}$ coefficients are defined so that
$a_{\lm}=a_{l,-m}$. Other than this, the distribution of $\Phi_{\lm}$ is
random. Also, note that for all $|m| > {\ell}$, $\Phi_{\lm}$ = 0.
\begin{figure}
\begin{centering}
\includegraphics[width=60mm]{Random_CPP.ps}
\caption{Example of the spherical harmonic phases ($\Phi_{\lm}$) we would expect for the concordance model where ${\ell}$, $m$ $\in$ [0, 20]. The distribution of $\Phi_{\lm}$ is random, apart from for $m=0$, where the phases can only be 0 or $\pi$ by definition. The colours represent the different values of the $\Phi_{\lm}$ (red ($\Phi_{\lm} = 0$), green ($\Phi_{\lm} =\pi/2$), cyan ($\Phi_{\lm} =\pi$), and purple ($\Phi_{\lm}=3\pi/2$)). }
\label{figRandom}
\end{centering}
\end{figure}
We extracted the $\Phi_{\lm}$ from each of the Bianchi maps using H{\sc ealpix }
\citep{heal} and plotted them in the same way as we have described;
the results are shown in Figure \ref{figPhase}. The plots show
that the $\Phi_{\lm}$ are not random but have patterns, i.e. the harmonic
modes manifest some form of phase correlation. For all the Bianchi
types, $\Phi_{\lm}$ = 0 for all odd $m$. For the VII$_0$ and V types, all
the $\Phi_{\lm}$ are orthogonal i.e. they are either 0, $\pi/2$,
$\pi$, or $3\pi/2$. Both the VII$_0$ and VII$_h$ types show
sequences of increasing/decreasing phases, which are particularly
prominent for $m=2$.
\begin{figure}
\begin{centering}
\includegraphics[width=41mm]{Vz000_CPP.ps}
\includegraphics[width=41mm]{VII0z000_CPP.ps}
\includegraphics[width=41mm]{VIIhz000_CPP.ps}
\caption{Phases of the spherical harmonic coefficients ($\Phi_{\lm}$) for Bianchi types V (left), VII$_0$ (middle) and VII$_h$ (right) where ${\ell}$, $m$ $\in$ [0, 20] and z = 0. Note that ${\ell}$ is plotted against the x axis, increasing from left to right, and $m$ is plotted against the y axis, increasing from bottom to top. The distributions are not random (as in Figure \ref{figRandom}) but exhibit some distinctive features. All the $\Phi_{\lm}$ for the VII$_0$ and V types are orthogonal, and there are sequences of colours in the type VII$_h$ (see $m=2$).}
\label{figPhase}
\includegraphics[width=41mm]{Vz000_CPPD.ps}
\includegraphics[width=41mm]{VII0z000_CPPD.ps}
\includegraphics[width=41mm]{VIIhz000_CPPD.ps}
\caption{$\Delta\Phi_{\lm}$ for Bianchi types V (left), VII$_0$ (middle) and VII$_h$ (right) where ${\ell}$, $m$ $\in$ [0, 20] and z = 0. Note that ${\ell}$ is plotted against the x axis, increasing from left to right, and $m$ is plotted against the y axis, increasing from bottom to top. Like the phases (Figure \ref{figPhase}) the distributions are not random but exhibit some distinctive features. All the $\Delta\Phi_{\lm}$ for the V type are either 0 or $\pi$. The $\Delta\Phi_{\lm}$ for the VII$_0$ type are again orthogonal but in a more correlated way. Similarly, the sequences of colours in the type VII$_h$ are now even more prominent (see $m=2$).}
\label{figPhaseDif}
\end{centering}
\end{figure}
While some patterns are apparent in these plots, an even better way
to visualize the phase correlations is to look at the phase
differences which are defined here as:
\begin{eqnarray}
\Delta \Phi_{\lm} = \Phi_{\lm} -\Phi_{{\ell},m-1}
\end{eqnarray}
The phase differences are shown in Figure \ref{figPhaseDif} and the
correlations are much more apparent compared to the plots of $\Phi_{\lm}$.
All the $\Delta\Phi_{\lm}$ for the V type are lined up, i.e. either 0 or
$\pi$. The $\Delta\Phi_{\lm}$ for the VII$_0$ type are again orthogonal,
but whereas in the phases the distribution of 0, $\pi/2$,
$\pi$, and $3\pi/2$ seemed some what random, in the phase
differences we see similar values `clump' together. Similarly, the
sequences of colours in the type VII$_h$ (see $m=2$ for example) are
now even more prominent.
So, strong correlations are observed in the phases and phases
differences of the simulated Bianchi CMB maps. But we have only
looked at large angular scales where there are only a small number
of independent data points. Even without noise, it is important to
ask the question whether these correlations are likely to be
statistically significant. One way to quantify this is to use a
\textit{Kolmogorov-Smirnov test}. This is a non-parametric
statistical test which measures the maximum distance of a given
distribution from a reference probability distribution. In this case
we want to show deviation from a random set of $\Delta \Phi_{\lm}$, i.e. a
uniform distribution, which is predicted by the concordance model.
To calculate the Kolmogorov-Smirnov test statistic a set of phase
differences $\Delta \Phi_{\lm}$ are separated into bins of equal size
between 0 and $2\pi$. The number of $\Delta \Phi_{\lm}$ which fall into
each bin are counted and a cumulative distribution derived. If the
distribution is uniform, as in the case of the reference probability
distribution, then the number of $\Delta \Phi_{\lm}$ in each of the bins
should increase roughly linearly across the bins. The difference
between both the sample and uniform cumulative distributions is
found for each bin and the biggest difference is the
Kolmogorov-Smirnov statistic \textbfit{D}.
To deduce the significance of \textbfit{D}, a set of ten thousand
tests have been run to generate sets of random angles of equal size
to the sample sets. \textbfit{D} was found for each of these sets
and this data was used to find the significance of \textbfit{D} for
the sample distributions from the Bianchi maps.
The Kolmogorov-Smirnov statistic \textbfit{D}, and the derived
probability of that statistic P(\textbfit{D}), for all the Bianchi
maps are detailed in Table \ref{tableKS}.
\begin{table}
\begin{centering}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
Map & z & \textbfit{D} & P(\textbfit{D}) \% \\ \hline
VII$_h$ & 500 & 0.11 & 94.5 \\
VII$_h$ & 200 & 0.09 & 77.1 \\
VII$_h$ & 60 & 0.14 & 99.2 \\
VII$_h$ & 10 & 0.24 & $>$99.9 \\
VII$_h$ & 3 & 0.27 & $>$99.9 \\
VII$_h$ & 1 & 0.38 & $>$99.9 \\
VII$_h$ & 0 & 0.27 & $>$99.9 \\
VII$_0$ & 0 & 0.28 & $>$99.9 \\
V & 0 & 0.73 & $>$99.9 \\
\hline
\end{tabular}
\caption{Results from the Kolmogorov-Smirnov test comparing the distribution of phase differences in the Bianchi CMB maps with a random distribution of phases as predicted by the concordance model. \textbfit{D} is the Kolmogorov-Smirnov statistic found by comparing the phase differences $(\Delta\Phi$). $P(\textbfit{D})$ is the Monte Carlo estimate of the probability of getting the value of \textbfit{D}, or less, found for the Bianchi models, from a random selection of phase differences. These are computed by forming an empirical distribution of \textbfit{D} from sets of random simulations and counting what fraction of the ensemble gives the results obtained for the Bianchi maps. For example, in the case of the $P(\textbfit{D})$ for the VII$_h$ map (z = 500) we find that, out of 10000 simulations, 9450 have a value of \textbfit{D} {\em less than} 0.11. Given the probable sampling accuracy of around one percent, we have rounded the results.}
\label{tableKS}
\end{centering}
\end{table}
This table shows that there is indeed a significant deviation from
a uniform distribution for the phase differences for all Bianchi
types. Of the 10000 random sets of data, none showed a value for
\textbfit{D} as high as seen for the Bianchi cases.
The Bianchi VII$_h$ type was also considered at different redshifts
to see how the correlations changed with time. Table \ref{tableKS}
shows that in general value of \textbfit{D} gets more significant
over time i.e. the correlations in the phase differences of the
Bianchi maps become stronger over time.
\subsection{Rotating maps and adding noise}
\label{secRN} In Section \ref{secVPC} we applied the
Kolmogorov-Smirnov test to a ``clean'' map that is perfectly aligned
with the vertical axis. This section addresses how noise and
rotation affect the identification of correlations in the phases of
the spherical harmonics of CMB maps from Bianchi models.
First we consider rotation. Phases of spherical harmonic coefficients are not rotation-invariant. Rotating the coordinate system used to represent a CMB map in $\phi$ (which is equivalent to rotation around the z axis) would increment each of the spherical harmonic phases by $\phi$, so the phase differences would remain the same. Therefore rotation in $\phi$ would have no effect on the value of the Kolmogorov-Smirnov statistic \textbfit{D}. Rotation in $\theta$ is more complicated to express so we used an empirical approach to quantity the effect on \textbfit{D}. The Bianchi CMB maps were rotated by a small angle, $\theta = \pi/8$, and then the spherical harmonic coefficients were derived and used to calculate \textbfit{D}. The results in Table \ref{tableKSRot} show that the values of \textbfit{D} for each of the maps are even higher than in maps that hadn't been rotated, indicating the presence of even stronger correlations. This suggests that, at least for small rotations off the axis, the correlations are just as significant, if not more so.
\begin{table}
\begin{centering}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
Map & z & \textbfit{D} & P(\textbfit{D}) \% \\ \hline
VII$_h$ & 0 & 0.46 & $>$99.9 \\
VII$_0$ & 0 & 0.38 & $>$99.9 \\
V & 0 & 0.77 & $>$99.9 \\
\hline
\end{tabular}
\caption{Results from the Kolmogorov-Smirnov test comparing the distribution of phase differences in the Bianchi CMB maps, rotated by $\theta = \pi/8$, with a random distribution of phase differences as predicted by the concordance model. \textbfit{D} is the Kolmogorov-Smirnov statistic found when considering the phase differences. $P(\textbfit{D})$ is the Monte Carlo estimate of the probability of getting the value of \textbfit{D}, or less, found for the Bianchi models, from a random selection of phase differences. These are computed by forming an empirical distribution of \textbfit{D} from sets of random simulations and counting what fraction of the ensemble gives the results obtained for the Bianchi maps. For example, in the case of the $P(\textbfit{D})$ for the VII$_h$ map we find that, out of 10000 simulations, over 9999 have a value of \textbfit{D} {\em less than} 0.46.}
\label{tableKSRot}
\end{centering}
\end{table}
As an aside, the colour plots of the phase differences for Bianchi
maps rotated by a number of different $\theta$ in the range 0 to
2$\pi$ were generated. These plots have been condensed together into
movies\footnote{The movies can be found at
http://www.astro.cardiff.ac.uk/research/theoreticalcosmology/?page=research}
which show that the correlations in the VII$_0$ and V maps are
visible across all $\theta$ and for the VII$_h$ map are visible within
about $\pi/3$ of the preferred axis. .
\begin{figure}
\begin{centering}
\includegraphics[width=41mm]{VrotWhiteNoise_CPP.ps}
\includegraphics[width=41mm]{VrotWMAPNoise_CPP.ps}
\includegraphics[width=41mm]{VrotflucNoise_CPP.ps}
\caption{$\Delta\Phi_{\lm}$ for ${\ell}$, $m \in$ [0, 20] (Bianchi type V map at z = 0 with white (left), WMAP (middle), and $\Lambda$CDM fluctuations (right) noise maps , $\theta=\pi/8$). Note that ${\ell}$ is plotted against the x axis, increasing from left to right, and $m$ is plotted against the y axis, increasing from bottom to top. Correlations can still be observed as lines of similar colours.}
\label{figRotNoisePhaseDif}
\end{centering}
\end{figure}
\begin{table}
\begin{centering}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
Map & z & \textbfit{D$_{white}$} & P(\textbfit{D})\% & \textbfit{D$_{wmap}$} & P(\textbfit{D})\% & \textbfit{D$_{\Lambda CDM}$} & P(\textbfit{D})\% \\ \hline
VII$_h$ & 0 & 0.15 & 99.2 & 0.16 & 99.8 & 0.17 & 99.8 \\
VII$_0$ & 0 & 0.09 & 77.1 & 0.08 & 66.1 & 0.07 & 52.0 \\
V & 0 & 0.19 & $>$99.9 & 0.18 & 99.9 & 0.14 & 98.2 \\
\hline
\end{tabular}
\caption{Results from the Kolmogorov-Smirnov test comparing the distribution of phase differences in the Bianchi CMB maps rotated by $\theta = \pi/8$ with white, WMAP, and $\Lambda$CDM noise maps (z = 0) . \textbfit{D} is the Kolmogorov-Smirnov statistic found when considering the phase differences. $P(\textbfit{D})$ is the Monte Carlo estimate of the probability of getting the value of \textbfit{D}, or less, found for the Bianchi models, from a random selection of phase differences. These are computed by forming an empirical distribution of \textbfit{D} from sets of random simulations and counting what fraction of the ensemble gives the results obtained for the Bianchi maps. For example, in the case of the $P(\textbfit{D})$ for the VII$_h$ map with white noise we find that, out of 10000 simulations, 9920 have a value of \textbfit{D} {\em less than} 0.15.}
\label{tableKSRotNoise}
\end{centering}
\end{table}
Now to investigate the effect of noise, we considered three different types of noise. Firstly we tried the simplest form by just adding white noise to the Bianchi map. A map of random Gaussian noise (white noise) was generated. Using Healpix the spherical mode resolution was reduced to $\ell$ $\le$ 20. Then the ``noise'' map was modified to have zero mean and variance half that of the Bianchi map. The second ``noise'' map was derived from a product available on the WMAP L\textsc{ambda}\footnote{http://lambda.gsfc.nasa.gov/} website which provides the effective number of observations per pixel. A map of random Gaussian noise was again generated. The variance was modified per pixel so that it was inversely proportional to the square of the number of observations in that pixel. Using Healpix the spherical mode resolution was reduced to $\ell$ $\le$ 20. Then the noise map was modified to have zero mean and variance half that of the Bianchi map. The final ``noise'' map used a simulation of $\Lambda$CDM fluctuations of the CMB (as performed by \cite{Eriksen2005}). Again the noise map was modified to reduce the spherical mode resolution to $\ell$ $\le$ 20 and have variance half that of the Bianchi map.
Each of these ``noise'' maps was added to each of the rotated Bianchi maps. We see from the example in Figure \ref{figRotNoisePhaseDif} that the spherical harmonic coefficients derived still have visible correlations in the phases for the Bianchi V map. The results of the Kolmogorov-Smirnov
test (see Table \ref{tableKSRotNoise}) show that the correlations are still detectable and significant for the Bianchi V and VII$_h$ maps but not so well for the VII$_0$ maps. So the method is better for detecting focused features that twisted features.
We see that the effect of adding fluctuations here is not dissimilar to adding just Gaussian noise. The
concordance model predicts fluctuations which are stationary and
Gaussian, as discussed in the Introduction (Section
\ref{secIntro}). Although these fluctuations are correlated on the sky, they have random phases so are incoherent with respect to what our statistic measures.
The ``noise'', or fluctuation, maps are added to the Bianchi maps so that the ratio of the variances is of order unity. However, any ratio is possible; this specific choice is just for illustrative purposes to demonstrate the proposed methods. Nevertheless, if a random-phase (Gaussian) signal is superimposed on the Bianchi template, the phase coherence of the resulting map will still be degraded. If the Gaussian component is too large, the overall map will be indistinguishable from one with purely random phases.
In the examples we have shown, the Gaussian ``noise'' or fluctuation maps are added to the Bianchi maps in such a way that the ratio of the overall variance is of order unity. Our method still functions well with this level of ``contamination'', but if the noise variance is much higher than that of the Bianchi maps the method begins to struggle.
So summarizing, the phases of the spherical harmonic coefficients are a very effective way of identifying focusing
features in CMB maps, as long as the noise is not excessive, and can be used to give quantifiable
significances. However, like the pixel distributions, the variation
from the expectation of the concordance model only gives us an
indication of non-Gaussianity. It is not clear in what form the
non-Gaussianity occurs, such as an anisotropy or inhomogeneity.
Hence the next section looks at multipole vectors which are built
from spherical harmonic coefficients but can be used to give results
in pixel (as opposed to harmonic) space, which is more meaningful
from the point of view of diagnosing the presence of a preferred
direction.
\subsection{Application to WMAP 5 Year Data}
\label{secWMAP} For pedagogical interest, the methods described in
Section \ref{secVPC} are applied here to the WMAP 5 year Internal
Linear Combination (ILC) map. The results of the Kolmogorov-Smirnov
test on the ILC map in the galactic coordinate system show very low
significance correlations in $\Delta\Phi$ (see Table
\ref{tableKSRotNoise2}).
However Section \ref{secRN} showed that to
see the phase correlations, the Bianchi maps needed to be rotated
close to the preferred axis. There have been studies that have found
a preferred direction in the WMAP data, highlighted by the alignment
of at least the quadrupole (${\ell}$ = 2) and octopole (${\ell}$ = 3). This
preferred axis is known as the Axis of Evil. Therefore the methods
from Section \ref{secVPC} are applied to the ILC map rotated so that
the vertical axis aligns with the Axis of Evil. These $\Delta\Phi$
plotted in Figure \ref{figPhaseDifWMAP} do not show any visual
correlations. For comparison the figure also includes a plot of the
same ILC data but with the phases replaced with random angles (i.e.
so as to not affect the magnitude of the amplitudes of the
$a_{l,m}$). Whilst the Kolmogorov-Smirnov test (Table
\ref{tableKSRotNoise2}) finds higher significance results than when
the map was in Galactic coordinates, the results are still at a low
significance.
The results show there is no significant detection, so if we do live in an anisotropic universe then it must be obscured with considerable ``noise'' (fluctuations). However the fact that the significance of the results do increase when the map is aligned with the Axis of Evil is intriguing.
\begin{table}
\begin{centering}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
Map & Axis & \textbfit{D} & P(\textbfit{D}) \% \\ \hline
ILC & Galactic & 0.06 & 66.20 \\
ILC & Evil & 0.07 & 86.05 \\
\hline
\end{tabular}
\caption{Results from the Kolmogorov-Smirnov test comparing the distribution of phase differences in the WMAP ILC map, rotated to align with either the galactic axis or axis of evil, with a random distribution of phase differences as predicted by the concordance model. \textbfit{D} is the Kolmogorov-Smirnov statistic found when considering the phase differences. $P(\textbfit{D})$ is the Monte Carlo estimate of the probability of getting the value of \textbfit{D}, or less, found for the Bianchi models, from a random selection of phase differences. These are computed by forming an empirical distribution of \textbfit{D} from sets of random simulations and counting what fraction of the ensemble gives the results obtained for the Bianchi maps. For example, in the case of the P(\textbfit{D}) for the ILC map in the galactic plane we find that, out of 10000 simulations, 6620 have a value of \textbfit{D} {\em less than} 0.06.}
\label{tableKSRotNoise2}
\end{centering}
\end{table}
\begin{figure}
\begin{centering}
\includegraphics[width=41mm]{WMAProt_CPPD.ps}
\includegraphics[width=41mm]{Random_CPPD.ps}
\caption{$\Delta\Phi$ for ${\ell}$, $m$ $\in$ [0, 20] for the ILC map with the Axis of Evil aligned with the preferred axis (left) and for the same map but with the phases replaced with random phases (right). Note that ${\ell}$ is plotted against the x axis, increasing from left to right, and $m$ is plotted against the y axis, increasing from bottom to top. No correlations are visible in either plot.}
\label{figPhaseDifWMAP}
\end{centering}
\end{figure}
\section{Multipole Vectors from Bianchi Universe}
\label{secMultipole} As shown in the previous section, the
properties of spherical harmonic coefficients provide us with a
generally effective way of identifying anisotropy through
correlations in CMB maps. However the geometric interpretation of
the mode correlations seen in harmonic space is by no means easy to
interpret geometrically. In an effort to use the spherical harmonics
to provide more meaningful explanation of non-Gaussianities found,
we now consider an alternative approach, based on multipole vectors
These can be constructed from spherical harmonics, using the $a_{\lm}$
coefficients derived from CMB maps, but they give results in real
(i.e. pixel) space. For a summary of the main terminology for the
multipole vectors, using the polynomial interpretation approach
which was introduced by \cite{Katz2004}, please see Appendix
\ref{appMV}.
\subsection{Results for Bianchi maps}
\begin{figure}
\begin{centering}
\includegraphics[scale=0.15,angle=90]{RmvsVz0Di.ps}
\includegraphics[scale=0.15,angle=90]{RmvsVz0Quad.ps}
\includegraphics[scale=0.15,angle=90]{RmvsVz0Oct.ps}
\caption{The multipole vectors from Bianchi V map: dipole (left), quadrupole (middle) and octopole (right).}
\label{MvsVz0}
\end{centering}
\end{figure}
Figure \ref{MvsVz0} shows the multipole vectors from the Bianchi V map which serves as a good example to show how strongly the multipoles are correlated. The dipole (left) lies exactly at the top of the sphere, which is at the centre of the image. The two quadrupole vectors (middle) are located on the same spots on which two of octopole vectors (right) are placed. The remaining octopole vector is in the centre, i.e. the same place as dipole. Now we plot
the dipole, quadrupole and octopole on the same 2-sphere for all the Bianchi maps, at different redshifts, to see how exactly they overlap (see Figure \ref{MvsMaps}). The background colours indicate if any of the multipoles over-lap; yellow for no overlapped multipoles, green for overlapped dipole and octopole, light yellow for overlapped quadrupole and octopole, and light blue if all the multipole vectors are overlapped. The z-axis is into the page and
the x-y plane is the large marked circle.
\begin{figure}
\begin{centering}
\includegraphics[scale=0.15,angle=90]{MvsVz0.ps}
\includegraphics[scale=0.15,angle=90]{MvsVIIhz0.ps}
\includegraphics[scale=0.15,angle=90]{MvsVII0z0.ps}\\
\includegraphics[scale=0.15,angle=90]{MvsVz2.ps}
\includegraphics[scale=0.15,angle=90]{MvsVIIhz2.ps}
\includegraphics[scale=0.15,angle=90]{MvsVII0z2.ps}\\
\includegraphics[scale=0.15,angle=90]{MvsVz500.ps}
\includegraphics[scale=0.15,angle=90]{MvsVIIhz500.ps}
\includegraphics[scale=0.15,angle=90]{MvsVII0z500.ps}\\
\caption{The multipole vectors from the Bianchi V (left), VII$_h$ (middle) and VII$_0$ (right) maps from early stage (bottom panel) to late time (top panel). The vectors are represented by dots: dark blue for the dipole, light red for the quadrupole and brown for the octopole. Background colours also indicate if any of the multipoles over-lap; yellow for no overlapped multipoles (right column and bottom of middle column), green for overlapped dipole and octopole, light yellow (bottom of left column) for quadrupole and octopole and light blue (top of left column) if all the multipole vectors are overlapped.}\label{MvsMaps}
\end{centering}
\end{figure}
First of all, in all types of the Bianchi models, we see the
quadrupole and octopole vectors lie on the same plane, except for
one of the octopole vectors which is always located in the centre of
the image. For the Bianchi V and VII$_h$ types, the dipole vectors
lie very near the centre in the early stage but not exactly on it.
However, as time goes on, the dipole vector is overlapped by one of
the octopole vectors in the centre. The dipole vector of Bianchi
VII$_0$ type is different from Bianchi VII$_h$. In the Bianchi
VII$_0$ type, there is no particular correlation between the dipole
and other multipoles since the dipole is not coupled with the other
multipoles (quadrupole and octopole). For the Bianchi V (left), at
the beginning of the stage, the dipole vector is `almost' on the
z-axis and one of octopole vectors is exactly on the z-axis. However
the two later-time Bianchi V cases on the top left of Figure
\ref{MvsMaps} show the dipole and octopole are on the same spot, in
the middle of the image, which is exactly on the z-axis. Meanwhile,
one of components of quadrupole and octopole vectors are on the same
spot on the x-y plan. This means that the later Bianchi V models
have more extreme correlation between the multipoles.
\subsection{Results for WMAP Data}
Again for pedagogical interest, the multipole vectors are applied to the WMAP 5 year Internal Linear Combination (ILC) map in Figure \ref{figmvWMAP}. At first glance, the results in the top row look like they might be clustered in a similar way to those in the Bianchi models (see Figure \ref{MvsMaps}). However when we look at the results in the second row, where the map is orientated in the usual galactic coordinates, we see the multipoles line up along the x-y plane. This suggests an alternative explanation that the clustering near the x-y plane could be a feature of the residual galactic contamination which is known to remain in the full sky ILC maps.
\begin{figure}
\begin{centering}
\includegraphics[width=41mm]{Rvec2mapl=1rot.ps}
\includegraphics[width=41mm]{Rvec2mapl=2rot.ps}
\includegraphics[width=41mm]{Rvec2mapl=3rot.ps}
\\\includegraphics[width=41mm]{Rvec2mapl=1.ps}
\includegraphics[width=41mm]{Rvec2mapl=2.ps}
\includegraphics[width=41mm]{Rvec2mapl=3.ps}
\caption{The multipole vectors from the WMAP 5 year ILC map: dipole (left), quadrupole (middle) and octopole (right). The top row shows results where the z-axis is into the page and the x-y plane is the large marked circle. The bottom row shows results where the z-axis is the vertical line and the x-y plane is the horizontal line across the centre of the map.}
\label{figmvWMAP}
\end{centering}
\end{figure}
\section{Discussion and Conclusions}
\label{secConclu} The aim of this article was to explore some
simple ways of characterizing the large-scale temperature patterns
in CMB maps generated in anisotropic Bianchi type V, VII$_h$ and
VII$_0$ universes. The ultimate purpose of investigating this
behavior is to find ways of quantifying the global properties of the
pattern produced in order to isolate the effect of anisotropy from
that of non-Gaussianity. We repeat that when we talk about
non-Gaussianity here is not related to a stochastic field; there are
no fluctuations in the Bianchi maps.
We first discussed perhaps the simplest and perhaps the most obvious
possible descriptive statistics, the histogram of the pixel values,
primarily with the aim of demonstrating how non-Gaussianity of a
sort can arise from asymmetry. We evaluated the pixel distribution
functions for each of the maps and compared them to results expected
in a universe consistent with the concordance model. The type
VII$_0$ maps show the strongest deviation from the null hypothesis;
but types V and VII$_h$ behaved in a similar fashion to each other,
and closer to that of the null hypothesis. The reason these two gave
lesser indications of the presence of anomalies was because the
focussing effect produces a pattern that covers only a smaller part
of the celestial sphere, which tends to get lost when averaged over
the whole sky. This method is therefore useful to characterize
coherent signals extended over a large region, such as a spiral
pattern, but not if they are concentrated.
Phase analysis is a relatively new technique, and has consequently
not been used to quantify many alternative situations to the
concordance model. The phases of the spherical harmonic coefficients
provide a generic way of looking at correlations in harmonic space
that could arise from non-stationarity or non-Gaussianity. While
this is a potential strength of the approach - while phase
correlations will not just be useful for identifying anisotropies
specific to the Bianchi models, but in theory any isotropy
introduced to the CMB - it could also prove a weakness, in that more
general methods may lack the power to discriminate very specific
models.
The phase correlations identified in our Bianchi maps using this
technique were much stronger than we at first expected; given the
generic nature of the metric it was not expected to yield good
results. In addition to this, the strong correlations were found to
be robust to both rotation and moderate noise. Significant correlations in
both twisted and focusing features were also identified. However
using the same methods on the WMAP 5 year data shows little evidence
of non-Gaussianity. Given that the diagnostics are identified in
harmonic space, it is difficult to say whether any of the anomalies
identified this way are down to isotropy or homogeneity.
The analysis of multipole vectors is also a relatively new
technique. It has been used to identify non-Gaussianities in the
WMAP data, and has been particularly successful in identifying
anisotropies (i.e. asymmetries and/or preferred directions). The
multipole vectors are calculated from spherical harmonic
coefficients which, as we have already shown, themselves provide a
very effective way of identifying correlations in Bianchi (and
presumably other anisotropic) patterns. The multipole vectors must
include at least some of the information needed to describe these
mode correlations. The advantage of multipole vectors over the
spherical harmonics themselves, however, is that they give results
in real (i.e. pixel) space which is much more informative to the
user. The results when applied to the Bianchi maps show very strong
correlations between the directions of the multipole vectors for low
${\ell}$, often with them entirely overlapping, and hence showing
preferred directions. Since these vectors would not be aligned in
the case of a stationary stochastic field over the sky, these
results demonstrate that they are sensitive to departures from the
standard cosmological model.
It remains the case that the standard cosmological model is a good
fit to a huge range of observational data. Nevertheless, it is
important that tools are developed that are sufficiently sensitive
to hunt efficiently for possible anomalies in the next generation of
observations. There are many ways that the CMB temperature pattern
could be anomalous other than through the presence of Bianchi
perturbation modes such as those we have studied here. Just as there
are many ways a distribution can be non-Gaussian, so are there also
many ways a fluctuation field can be non-stationary. Testing for
departures from the standard model will require not one but a
battery of statistical techniques each sensitive to particular
aspects of the distribution.
This has been a very preliminary analysis, aimed at establishing
whether the diagnostics described in this paper are {\em in
principle} capable of uncovering evidence of underlying anomalies in
CMB data. Of course these patterns represent somewhat extreme
departures from the standard framework so it is no real surprise
that they register strongly in the descriptors used. However, in all
cases our analysis has involved only a relatively small number of
quantities, so the fact that we see quantifiable effects emerging is
very encouraging.
\section*{Acknowledgments}
We acknowledge the use of the Legacy Archive for Microwave
Background Data Analysis (L\textsc{ambda}) and many of the
calculations in this paper made use of the H{\sc ealpix } package
\citep{heal}. Jo Short receives funding from an STFC studentship.
Rockhee Sung acknowledges an Overseas Scholarship from the Korean
government.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 7,008
|
Q: Getting any AudioPlayer to stream or play music Bit at a loss here. I am on Xcode 7 and Swift 2 and I am trying to find and example of streaming audio that actually works.
Essentially I want to be able to stream, but I also tried this: Swift Radio Streaming AVPlayer
Which uses AVAudioPlayer. The github code works, but copying the code into my project with the same audio file crashes the app : Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: '-[SwiftBasicMVC_Git.AudioStreamProxy copyWithZone:]: unrecognized selector sent to instance 0x14fdea100'
I also tried some ACPlayer examples and AVQueuePlayer, without success of playing any music.
A: I use singleton for my basic radio application because need to reach and handle it on every page. It's so simple to create, let's start with declare singleton and variables. I will explain everything step by step;
import AVKit
import AVFoundation
private var radioContext: UInt8 = 1
final class RadioModel: NSObject {
dynamic var radioItem:AVPlayerItem? = nil
dynamic var radioPlayer:AVPlayer? = nil
static let sharedModel = RadioModel()
private final let dataModel = BasicRadioApp.DataManager
private override init() {
super.init()
do {
_ = try AVAudioSession.sharedInstance().setCategory(AVAudioSessionCategoryPlayback)
_ = try AVAudioSession.sharedInstance().setActive(true)
UIApplication.sharedApplication().beginReceivingRemoteControlEvents()
} catch {
print("Error")
}
}
If you want to use AirPlay mode for podcast or radio, you have to control your sound object with KVO (Key Value Observer) and it's not supporting for Swift. So, our AVPlayerItem and AVPlayer objects are defined by dynamic which means that according to documentation;
Apply this modifier to any member of a class that can be represented by Objective-C. When you mark a member declaration with the dynamic modifier, access to that member is always dynamically dispatched using the Objective-C runtime. Access to that member is never inlined or devirtualized by the compiler.
Because declarations marked with the dynamic modifier are dispatched using the Objective-C runtime, they're implicitly marked with the objc attribute.
I'm getting data from server in my other Singleton object BasicRadioApp.
I need to control my audio session interacts with others (etc. Apple Watch) so need to set your AVAudioSessionCategory. Also i want to remote my audio on lockscreen and Glance in Apple Watch music app.
In AVAudioSession class, this function declared like that;
/* set session category */
public func setCategory(category: String) throws
As you see, they throw exception so you need to set them in error handling like do/try concept for >Swift 2.0.
Let's try to play sound;
final func playRadio() {
do {
try removeObserverFromRadioItem()
} catch {
print("Can not remove observer from AVPlayerItem object!")
}
radioItem = AVPlayerItem(URL: NSURL(string: dataModel.radioStreamUrl)!)
radioPlayer = AVPlayer(playerItem: radioItem!)
radioItem!.addObserver(self, forKeyPath: "status", options: [.New, .Initial, .Old], context: &radioContext)
if MPNowPlayingInfoCenter.defaultCenter().nowPlayingInfo == nil {
let albumArt = MPMediaItemArtwork(image: UIImage(named: "AlbumArt")!)
let songInfo: Dictionary = [MPMediaItemPropertyTitle: "Now listening Hard Rock Bla Bla FM",
MPMediaItemPropertyArtist: "My Radyo",
MPMediaItemPropertyAlbumTitle: "105.5 FM",
MPMediaItemPropertyArtwork: albumArt]
MPNowPlayingInfoCenter.defaultCenter().nowPlayingInfo = songInfo
}
}
Before declaring my AVPlayerItem, i need to remove KVO on it if it exist. You'll see what removeObserverFromRadioItem() is doing next. I need to register my AVPlayerItem to KVO object and it'll handle to play or not. As i said, i need to remote my audio on lockscreen so i'm using MPNowPlayingInfoCenter for that.
final func stopRadio() {
MPNowPlayingInfoCenter.defaultCenter().nowPlayingInfo = nil
radioPlayer?.pause()
}
final func removeObserverFromRadioItem() throws {
radioItem?.removeObserver(self, forKeyPath: "status", context: &radioContext)
}
The most important part is here;
override func observeValueForKeyPath(keyPath: String?, ofObject object: AnyObject?, change: [String : AnyObject]?, context: UnsafeMutablePointer<Void>) {
if object as? AVPlayerItem == radioPlayer?.currentItem && keyPath! == "status" {
if self.radioPlayer?.currentItem!.status == .Failed {
print("STREAM FAILED")
} else if self.radioPlayer?.currentItem!.status == .ReadyToPlay {
self.radioPlayer!.play()
radioDelegate?.RadioDidStartPlay()
}
return
}
super.observeValueForKeyPath(keyPath, ofObject: object, change: change, context: context)
}
How do you start your radio from any class?
class RadioController: UIViewController {
let sharedRadioPlayer = RadioModel.sharedModel
..
override func viewDidLoad() {
super.viewDidLoad()
// check if radio is NOT playing
if sharedRadioPlayer.radioPlayer?.rate < 1 {
sharedRadioPlayer.playRadio()
}
}
Last step is remote my app from lockscreen. Stop, play etc.
For remote that, i'm calling my method in RadioModel from AppDelegate. In AppDelegate;
override func remoteControlReceivedWithEvent(event: UIEvent?) {
RadioModel.sharedModel.handleRemoteControlEvent(event!)
}
and in RadioModel;
func handleRemoteControlEvent(responseEvent: UIEvent) {
if responseEvent.type == UIEventType.RemoteControl {
switch(responseEvent.subtype) {
case UIEventSubtype.RemoteControlPlay:
playRadio()
break
case UIEventSubtype.RemoteControlPause:
stopRadio()
break
case UIEventSubtype.RemoteControlTogglePlayPause:
if self.radioPlayer!.rate > 0 {
stopRadio()
} else {
playRadio()
}
break
default:
print("Remote Error")
}
}
}
Don't forget that in your app TARGETS > Capabilities > switch Background Modes on and select Audio, AirPlay and Picture in Picture. Then check your info.plist and search "Requires background modes". If it's not exist, create new one as Array and set item that "App plays audio or streams audio/video using AirPlay". That's it now you can play airPlay radio/stream.
A: DONT forget to add : App Transport Security Settings
As of XCode 7.x AND iOS 9.x
Needed in the pList settings : Add Arbitrary as YES
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 4,248
|
La Guerra ruso-polaca de 1792 fue un conflicto bélico que enfrentó a la República de las Dos Naciones con la Confederación de Targowica y el Imperio ruso que se oponía la Constitución del 3 de mayo de 1791.
Guerra
El 18 de mayo de 1792, sin declaración de guerra previa, un ejército ruso con más de 97.000 hombres cruzó la frontera de Polonia. Estanislao II Poniatowski, rey de la República de las Dos Naciones, envío a un ejército de 70.000 hombres para hacerles frente. Las fuerzas de la República de las Dos Naciones, leales al rey y al Gran Sejm, defendieron la Constitución del 3 de mayo frente a los opositores de la Confederación de Targowica y el Imperio ruso.
En los primeros días de la guerra, Prusia rompió su alianza con Polonia y el comandante prusiano originario del Gran Ducado de Lituania, Luis de Württemberg traicionó la causa polaco-lituana. El ejército polaco dirigido por el sobrino del rey, el príncipe Józef Antoni Poniatowski, se enfrentó a una fuerza enemiga muy superior y logró la victoria en Zieleńce el 18 de junio. Luego, las tropas polacas se retiraron al río Bug Occidental, donde lucharon bajo las órdenes de Tadeusz Kościuszko en la batalla de Dubienka (:en:Battle of Dubienka), que acabó en empate. No obstante, a las tropas rusas se les concedió libre paso por los territorios austriacos.
La derrota final de los polacos se debió a la traición de su propio rey Estanislao II Poniatowski, quien se unió a la confederación de Targowica. Esta acción precipitó la segunda partición de Polonia el 21 de enero de 1793. Tras la Sejm de Grodno, la población polaca se redujo a un tercio de lo que tenía cuando empezó la partición en 1772. El territorio quedó ocupado por tropas extranjeras y su independencia se redujo en gran medida.
El último intento de Polonia por recuperar su independencia se produjo el 4 de abril de 1794 con la insurrección de Kościuszko que tras un año de batallas dio como resultada una victoria rusa que acabó definitivamente con el Estado de Polonia.
Referencias
Rusopolaca de 1792
Rusopolaca de 1792
Historia de Polonia (1569-1795)
Años 1790 en Polonia
Años 1790 en Lituania
Conflictos en 1792
Guerras del Imperio ruso (1721-1917)
Años 1790 en Rusia
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 7,888
|
Q: Enforcing range in an array in C Quick, probably super basic question. If I declare an array of 10 doubles and prompt a user to input how many double they want in the index (between 1 and 10, obviously), how would I enforce that range? I assume with an if/else statement, but is there a more optimal way to enforce that range?
Again, probably really simple. I'm new so not yet familiar C++ or JavaScript.
A: Get that no. of elements from user till that no. is not within your range.
Say n is number that you want in range 1 to 10.
Solution:
int n = 0;
while(n<1 || n>10) {
printf("Enter Correct value on n i.e. within range 1 to 10:\t");
scanf("%d", &n);
}
A: Another solution maybe if you wrap the value within the required range using % (modulo) operator.
int n;
int arr[10];
printf("Enter a number : ");
scanf("%d",&n);
printf("%d",arr[n%10]);
The expression n%10 will always result in a value between 0 to 9.
Edit (for a better validation):
#include<stdio.h>
main()
{
int x, n;
int arr[]={0,1,2,3,4,5,6,7,8,9,10,11,12};
printf("Enter a number : ");
if( scanf("%d",&n)!=1 )
{
printf("Not a valid integer!!");
return;
}
n=(n<0)?-n:n;
printf("%d",arr[n%10]);
}
A: First you should read the documentation because this is a very basic example.
Second avoid requesting code here. You should always try to found solution then post your code and your error here to correct it.
You can try this:
#include<stdio.h>
int main(){
int n = 0;
while(!(n > 1 && n < 10)){
printf("Enter an integer between 1 and 10: \n");
scanf("%d", &n);
}
}
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 3,059
|
Q: Can't change a file's group owner in Cygwin I'm trying to do:
chown :Users filename
And I get
chown: invalid grup: «:Users»
What am I doing wrong? Is there something I don't understand?
A: You are trying to keep the same user and change the group, saying chown :group instead of chown user:group.
The syntax is wrong, so you can either use chown indicating the current user or directly use chgrp instead:
chgrp Users filename
A: Try:
chgrp Users /path/to/filename
PS: keep in mind that "Users" is different of "users".
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 2,515
|
{"url":"http:\/\/math.stackexchange.com\/questions\/402042\/properties-of-quotient-categories","text":"Properties of quotient categories.\n\nLet $\\mathcal{A}$ be an abelian category and $\\mathcal{C}$ a localizing subcategory in the sense of Gabriel. (A Serre subcategory or \"thick\" subcategory, such that the quotient functor $T\\colon \\mathcal{A}\\rightarrow\\mathcal{A}\/\\mathcal{C}$ admits a right adjoint, the \"section functor\".) Then we can form the quotient category $\\mathcal{A}\/\\mathcal{C}$.\n\nWhich properties inherits $\\mathcal{A}\/\\mathcal{C}$ from $\\mathcal{A}$? To be more precise:\n\n1. If $\\mathcal{A}$ has enough injectives (resp. projectives), does $\\mathcal{A}\/\\mathcal{C}$ too? If not, under which conditions?\n2. If $A\\in \\mathcal{A}$ is injective (resp. projective), is it $T(A)$, too? If not, under which conditions?\n3. If $A\\in \\mathcal{A}$ is a cogenerator, is it $T(A)$ too? If not, under which conditions?\n4. If $\\mathcal{A}$ is complete, is it $\\mathcal{A}\/\\mathcal{C}$ too? If not, under which conditions?\n\nI know that:\n\n1. If $\\mathcal{A}$ is cocomplete then so is $\\mathcal{A}\/\\mathcal{C}$. ($T$ is a left adjoint)\n2. If $\\{U_i\\}$ is a set of generators then so is $\\{T(U_i)\\}$.\n3. If $\\mathcal{A}$ is AB5 then so is $\\mathcal{A}\/\\mathcal{C}$. ($T$ commutes with limits and one can prove that taking directed limits is exact.)\n-\nThis question continues on mathoverflow.net\/questions\/132334\/\u2026 . \u2013\u00a0archipelago Jun 2 '13 at 11:31","date":"2016-04-30 23:17:37","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9798609614372253, \"perplexity\": 410.28860042190973}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2016-18\/segments\/1461860113010.58\/warc\/CC-MAIN-20160428161513-00019-ip-10-239-7-51.ec2.internal.warc.gz\"}"}
| null | null |
Patronyme
Auguste-Henry Berthoud (1829-1887), peintre suisse ;
Ferdinand Berthoud (1725-1807), horloger neuchâtelois ;
Fritz Berthoud (1812-1890), homme politique suisse ;
Henri Berthoud (1877-1948), homme politique suisse ;
Gabrielle Berthoud (1907-1987), historienne suisse ;
Jean Berthoud (1961-), banquier suisse ;
Jean-Édouard Berthoud (1846-1916), homme politique suisse ;
Louis Berthoud (1754-1813), neveu de Ferdinand, horloger français ;
Paul François Berthoud (1870-1939), peintre, sculpteur et graveur français ;
Samuel-Henri Berthoud (1804-1891), écrivain et journaliste français.
Toponyme
Berthoud, commune suisse du canton de Berne ;
Berthoud, municipalité américaine du Colorado.
Voir également
Berthoud, constructeur français de pulvérisateurs pour l'agriculture et la viticulture ;
Le berthoud est un plat de fromage fondu de la région chablaisienne dans les Alpes franco-suisses.
Référence
Homonymie de patronyme
Patronyme suisse
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 3,392
|
Also known as "Drunken Noodles" or "Drunkard's Noodles" - most likely because of the "late night snack" status of this dish in Thailand This is probably my favorite Thai dish (and no, it's not because I'm a drunkard, but it helps...). It's wonderful with seafood, but in my opinion, better as a vegetarian dish, if done right.
Soak noodles for 30-40 minutes in hot water until tender but not completely soft. Drain and toss with 1 TB oil. Fry garlic, shallots, chiles, and cilantro root in remaining oil for 30 seconds. Add onion and pepper and cook 5 minutes. Mix sauce ingredients and stir until sugar dissolves. Add noodles, sauce, and basil to wok, and stir until basil is wilted and sauce is absorbed by noodles. Remove from wok and garnish with cilantro, tomato wedge, and cucumber slices.
Like this vegetarian Pad Kee Mao recipe? Add to "My Recipe Box"
Did you enjoy this recipe for Pad Kee Mao? Would you change it somehow? Let us know your thoughts!
I've been driving 60 miles to buy 6 dollars worth of Drunkard noodles, they're that good. This recipe is spot on but I'd lower the thai chiles to 1, they really are killers. My local Asian market had all the herbs and sauces and was extremely low-priced.
I have just accidentally come across your page, I really like what I\'m reading. I\'m definitely going to try some of your recipes when I have some time.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 9,334
|
{"url":"https:\/\/www.greenemath.com\/Algebra1\/48\/RationalEquationsTest1.html","text":"Sections:\n\n# Rational Equations Test #1\n\nAbout:\nAdditional Resources:\n\nIn this section, we learn about Rational Equations. Essentially we are learning how to solve equations with rational expressions involved. Recall that to solve an equation with fractions, we can clear the fractions by multiplying both sides by the LCD. In the case of an equation with rational expressions, we can use the same trick. We would begin by finding the LCD for the rational expressions in our equation. We then multiply both sides of our equation by the LCD to clear all denominators. Once this is done, we can solve our equation as we normally would. We must pay close attention to our solution. If we check the proposed solution and it results in a denominator of zero, we must reject this solution. Recall that we can never divide by zero.\n+ Show More +","date":"2020-01-26 23:29:49","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8010491132736206, \"perplexity\": 197.49306276915974}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": false}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-05\/segments\/1579251694071.63\/warc\/CC-MAIN-20200126230255-20200127020255-00307.warc.gz\"}"}
| null | null |
The U.S. Army Signal Corps advertised for bids for a two-seat observation aircraft and in 1908 and 1909, the Wrights flew the Wright 1908 Military Flyer at official Army trials at Fort Myer, Virginia. On September 17, 1908, Orville Wright and Lieutenant Thomas Etholen Selfridge made a flight as a part of these trials. While in the air, a blade on the propeller broke and the aircraft fell from 75 feet up and landed on Wright and Selfridge. Both men sustained severe injuries. Wright broke a leg and several ribs, and Selfridge was taken off the field unconscious. Selfridge died shortly after the accident from his injuries and became powered flight's first fatality. The Army was to purchase the Wright (Co) Type A Military (Signal Corps No. 1) for $30,000 in 1909.
Collection Item Long Description:
Wright (Co) Type A Military (Signal Corps No. 1)
Clime, W. S
Wright, Orville 1871-1948
Knab, Frederick 1865-1918
Taylor, Charles Edward 1868-1956
Lahm, Frank Purdy 1877-1963
Selfridge, Thomas Etholen
.5 cubic feet (1 folder)
Fort Myer (Va.)
This collection consists of 45 black and white photographs, of varying sizes and including some duplicates, taken by photographer W. S. Clime at the 1908-1909 official army trials at Fort Myer, Virginia. Shown in the images are the Wright 1908 Military Flyer in flight and on the ground; the crash that injured Orville Wright and killed Thomas Selfridge; and Thomas S. Baldwin with an Army dirigible. Other persons shown in the photographs include mechanic Charles W. Taylor; Major George O. Squier; Lieutenant Frank P. Lahm; Lieutenant Benjamin D. Foulois; Augustus Post; and Mrs. M. Longworth. The collection also contains one unrelated photograph of A. Roy Knabenshue making a flight over Washington, D.C.
Collection name, Accession XXXX-0904, National Air and Space Museum, Smithsonian Institution
National Air and Space Museum, Archives Division, MRC 322, Washington, DC, 20560
Collection descriptions
XXXX-0904
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 8,791
|
5 Days for the Homeless
March 2, 2017 —
Five University of Manitoba students will live outside for five chilly nights as part of the national 5 Days for the Homeless campaign, which raises funds and awareness for homeless youth.
The participants – or Sleepers – will spend their nights in a shelter on the east side of University Centre on the U of M's Fort Garry campus from March 13 to 17. The Sleepers are:
Raelene Roy
Dara Hallock
George Semchuk
Abel Nyamori
Santiago Godoy
During the campaign, the students must attend all their classes and meet their academic duties and deadlines. They will not be able to shower and carry any cash or identification, and can only eat food that has been donated to them.
This marks the tenth consecutive year that students from the U of M's Asper School of Business have run this campaign. The U of M's 5 Days campaign is raising funds for Resource Assistance for Youth, Inc. (RaY), a local charity that helps homeless and marginalized young people achieve independence.
Several events have been planned to help the campaign reach this year's goal of $30,000:
Fivehole for Five Days Street Hockey Tournament
Saturday, March 4, 2017 – 10:00 am
ACE & AC Parking Lots at the University of Manitoba's Fort Garry Campus
5 Days Trivia Night
Saturday, March 11, 2017 – 8:00 pm
The Hub Social Club, University Centre at University of Manitoba's Fort Garry Campus
Monday, March 13, 2017 – 12:00 pm (noon)
Asper School of Business, 181 Freedman Crescent
Monday, March 13, 2017 to Friday, March 17, 2017
5 Acts for the Homeless
Friday, March 17, 2017 – 8:00 pm
The Riverside Inn, 531 St Mary's Road
Closing Ceremonies
Saturday, March 18, 2017 – 12:00 pm (noon)
Media and guests are invited to attend the Opening Ceremonies on Monday, March 13 at 12:00 (noon) in Drake Centre (181 Freedman Crescent) to meet the Sleepers and listen to distinguished guests share their experiences working with homeless youth in Winnipeg.
Media and guests are also invited to meet with the 'Sleepers' and follow their story throughout the week of March 13 to 17 at their shelter on the east side of University Centre.
5 Days for the Homeless was founded as a local initiative by University of Alberta School of Business students. The initiative expanded to a national scale in 2008. To date, the campaign has raised over $1,893,525 for charitable organizations across the country and has raised awareness about youth homelessness from coast to coast.
Resource Assistance for Youth, Inc. (RaY) is a non-profit street-level agency working with street-entrenched and homeless youth up to the age of 29. RaY is non-judgmental and non-partisan, employing a harm reduction approach to all interactions with youth in need.
Johnathan Reynolds
Head Marketing Coordinator
johnnyreynolds26 [at] gmail [dot] com
Nicholas Anandranistakis
External Marketing Coordinator
nicholas [dot] stakis [at] gmail [dot] com
Follow the campaign at:
twitter.com/5D4H_WPG
instagram.com/5D4H_WPG
facebook.com/5dayswinnipeg
http://5days.ca/schools/university-of-manitoba/
Asper Staff
Asper School of Business, Asper Students, community engagement, Students
An Entrepreneur's Journey: Ogo Okwumabua
Community is at the heart of Okwumabua's vision for Zueike.
Asper Alumni, Asper School of Business, Bisons, Stu Clark Centre for Entrepreneurship
Lei Lu reappointed as Bryce Douglas Chair in Corporate Finance
Dr. Lei Lu has been reappointed as the Bryce Douglas Chair in Corporate Finance.
Asper Research, Asper School of Business, donor relations
Family enterprise case competition a milestone win for Asper
The I. H. Asper School of Business is celebrating an exciting milestone with our first international graduate case competition victory
Alumni, Asper Case Competitions, Asper Experiential Education, Asper MBA, Asper School of Business, experiential learning, graduate students, Stu Clark Graduate School
Negotiation and Consensus Building February 7, 2023
Speaking as a Leader February 8, 2023
Alumni and community
Stu Clark Graduate School
Email: asper_info@umanitoba.ca
181 Freedman Crescent
University of Manitoba, Winnipeg, MB
R3T 5V4 Canada
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 7,642
|
Q: The em dash, and correct usage in the sentence that follows As I walked through the channel — I knew it was over.
Is this a correct usage of the em dash?
A: No. You can use a dash to connect two clauses that aren't complete sentences, but in this case where the second clause flows naturally from the first one, you would be better served using a comma. Dashes are usually used to indicate a change in tone or direction (often to a facetious or informal one).
After three weeks on set, the cast was fed up with his direction — or, rather, lack of direction.
If you're using a dash instead of a comma because you feel that a comma wouldn't give the second clause enough weight, you'll probably need to change your wording. Something like this could work:
As I walked through the channel, I realized I had to give up. It was over.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 2,935
|
package com.sqisland.android.advanced_textview;
import android.app.Activity;
import android.content.Context;
import android.graphics.LinearGradient;
import android.graphics.Matrix;
import android.graphics.Paint;
import android.graphics.Shader;
import android.os.Bundle;
import android.text.SpannableString;
import android.text.TextPaint;
import android.text.format.DateUtils;
import android.text.style.CharacterStyle;
import android.text.style.UpdateAppearance;
import android.view.animation.LinearInterpolator;
import android.widget.TextView;
import com.nineoldandroids.animation.FloatEvaluator;
import com.nineoldandroids.animation.ObjectAnimator;
import com.nineoldandroids.animation.ValueAnimator;
import com.nineoldandroids.util.Property;
public class AnimatedRainbowSpanActivity extends Activity {
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_animated_rainbow_span);
final TextView textView = (TextView) findViewById(R.id.text);
String text = textView.getText().toString();
AnimatedColorSpan span = new AnimatedColorSpan(this);
final SpannableString spannableString = new SpannableString(text);
String substring = getString(R.string.animated_rainbow_span).toLowerCase();
int start = text.toLowerCase().indexOf(substring);
int end = start + substring.length();
spannableString.setSpan(span, start, end, 0);
ObjectAnimator objectAnimator = ObjectAnimator.ofFloat(
span, ANIMATED_COLOR_SPAN_FLOAT_PROPERTY, 0, 100);
objectAnimator.setEvaluator(new FloatEvaluator());
objectAnimator.addUpdateListener(new ValueAnimator.AnimatorUpdateListener() {
@Override
public void onAnimationUpdate(ValueAnimator animation) {
textView.setText(spannableString);
}
});
objectAnimator.setInterpolator(new LinearInterpolator());
objectAnimator.setDuration(DateUtils.MINUTE_IN_MILLIS * 3);
objectAnimator.setRepeatCount(ValueAnimator.INFINITE);
objectAnimator.start();
}
private static final Property<AnimatedColorSpan, Float> ANIMATED_COLOR_SPAN_FLOAT_PROPERTY
= new Property<AnimatedColorSpan, Float>(Float.class, "ANIMATED_COLOR_SPAN_FLOAT_PROPERTY") {
@Override
public void set(AnimatedColorSpan span, Float value) {
span.setTranslateXPercentage(value);
}
@Override
public Float get(AnimatedColorSpan span) {
return span.getTranslateXPercentage();
}
};
private static class AnimatedColorSpan extends CharacterStyle implements UpdateAppearance {
private final int[] colors;
private Shader shader = null;
private Matrix matrix = new Matrix();
private float translateXPercentage = 0;
public AnimatedColorSpan(Context context) {
colors = context.getResources().getIntArray(R.array.rainbow);
}
public void setTranslateXPercentage(float percentage) {
translateXPercentage = percentage;
}
public float getTranslateXPercentage() {
return translateXPercentage;
}
@Override
public void updateDrawState(TextPaint paint) {
paint.setStyle(Paint.Style.FILL);
float width = paint.getTextSize() * colors.length;
if (shader == null) {
shader = new LinearGradient(0, 0, 0, width, colors, null,
Shader.TileMode.MIRROR);
}
matrix.reset();
matrix.setRotate(90);
matrix.postTranslate(width * translateXPercentage, 0);
shader.setLocalMatrix(matrix);
paint.setShader(shader);
}
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 3,282
|
Ust-Dscheguta () ist eine Stadt in der nordkaukasischen Republik Karatschai-Tscherkessien in Russland mit Einwohnern (Stand ).
Geografie
Die Stadt liegt am Nordrand des Großen Kaukasus etwa 15 km südlich der Republikhauptstadt Tscherkessk am rechten Ufer des Kuban bei der Mündung des Flüsschens Dscheguta. Der Kuban ist oberhalb der Stadt zur kleinen Ust-Dschegutinsker Talsperre angestaut, welche den hier abzweigenden, der Bewässerung dienenden Großen Stawropoler Kanal speist.
Ust-Dscheguta ist Verwaltungszentrum des gleichnamigen Rajons.
Die Stadt ist über eine 67 Kilometer langen Zweigstrecke mit Tscherkessk sowie der Hauptstrecke der Nordkaukasus-Eisenbahn (Station Selentschuk) verbunden. Die Station der Stadt heißt Dscheguta.
Geschichte
Der Ort entstand 1861 als Staniza Ust-Dschegutinskaja von Kosaken, die vom Unterlauf des Kuban hierher umsiedelten, und erhielt 1975 das Stadtrecht. Die Bezeichnung ist vom russischen Wort ustje für Mündung sowie den karatschaischen Wörtern dschoge (Linde) und tei (Fluss) abgeleitet.
Bevölkerungsentwicklung
Anmerkung: Volkszählungsdaten
Wirtschaft
Ust-Dscheguta ist Zentrum der Baumaterialienwirtschaft: hier sind Werke für Stahlbetonfertigteile, Zement, Kalk und Silikatziegel angesiedelt.
Söhne und Töchter der Stadt
Dima Bilan (* 1981), Popsänger
Weblinks
Webseite der Stadtverwaltung (russisch)
Ust-Dscheguta auf mojgorod.ru (russisch)
Einzelnachweise
Ort in Karatschai-Tscherkessien
Stadtrechtsverleihung 1975
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 9,035
|
Peter Arne Jöback (Estocolmo, 4 de junio de 1971), es un cantante y actor teatral de origen sueco.
Biografía
Peter tiene un hermano Mikael Jöback y tres medios hermanos Stefan, Håkan y Göran.
Desde el 25 de junio de 2010 está casado con Oscar Jöback, la pareja tiene dos hijas adoptivas Vega Lee e Iggy-Lou.
Carrera
Peter hizo su debut profesional en la producción sueca de "Sonrisas Y Lágrimas" a la edad de diez años, interpretando el papel de Kurt Von Trapp. Otras de sus primeras apariciones, incluyen trabajos en "West Side Story" (como Tony), "El Libro De La Selva" (como Gray), "Here And Now" (como Ben), "Cavalry Maiden" (como Astahof), "La Belle Epoque" (como Morgan) y "Mio My Mio" como Lollo).
Más tarde, conseguiría hacerse popular en la producción realizada en Estocolmo de "Fame", como Nick Piazza. Fue nominado al Swedish Golden Mask Award al Mejor Actor en un Musicla por este papel. Más tarde, participaría en el doblaje de la película estadounidense "Aladino", trabajo que volvería a repetir en la serie de dibujos animados del mismo nombre.
Como mucho otros artistas de su país, intentó participar en Eurovisión, presentándose al Melodifestivalen 1990 con el tema "En Sensation". Desafortunadamente, sólo consiguió alcanzar la novena posición.
Durante el comienzo de los años '90, comenzó su carrera discográfica, con la publicación de dos sencillos dentro de la discográfica DoReMe. Ambos trabajos estaban compuestos en inglés y, con uno de ellos, "This Time" se hizo con la victoria en el Baltic Song Contest de 1992. Poco después, llegaría a grabar diez temas en inglés para un álbum que jamás se llegó a publicar.
Sin embargo, firmó un contrato con la compañía Sony Records, realizando un dúo con Towe Jaarnek, "More Than A Game", que se convirtió en la canción oficial de los campeonatos de fútbol celebrados en Suecia en 1992.
Con la colaboración de su hermano Mikael, publicó un álbum con el mismo título ("More Than A Game") en 1993. Desafortunadamente, este disco no tuvo una gran repercusión en el mercado sueco.
Junto a Lisa Nilsson, Lizette Pålsson y Andreas Lundstedt formó parte del grupo Stage Four, participando en diferentes programas de televisión y acudiendo al 125 aniversario del rotativo sueco Dagens Nyheter.
Pero, sin lugar a dudas, su momento de gloria llegó en 1995 al ser seleccionado para el papel de Robert en el musical "Kristina Från Duvemåla", creado por Björn Ulvaeus y Benny Andersson. El sencillo "Guldet Blev Till Sand", extraído de dicho musical e interpretado por Peter, entró en la lista de éxitos de Suecia] y permaneció en el puesto número 1 durante 36 semanas.
Su talento había sobrepasado las fronteras suecas, lo que le supuso viajar a Londres en 1997 para interpretar el papel de Chris en "Miss Saigon" durante ocho meses. A su regreso en septiembre del mismo año, comenzó a trabajar en un nuevo álbum, así como en una nueva gira. El primer sencillo de este trabajo discográfico, "En Sång Om Oss" ("Una Canción Sobre Nosotros") fue publicado en el mes de noviembre, al que le siguió el álbum "Personliga Val" ("Decisión Personal") en diciembre. Consiguió ser disco de platino con más de 100.000 copias vendidas, y alcanzó el puesto número dos en la lista de éxitos.
En 1998 recreó de nuevo el papel de Robert en la première en Estocolmo del musical con el que había alcanzado la fama. Dicho papel, lo compaginó con su rol de Jesús en los conciertos basados en el musical "Jesucristo Superstar" celebrados en la capital y en Gotemburgo. Más tarde, crearía su propio show en Börsen, Estocolmo, un espectáculo que duró 3 meses y en el que pudo demostrar sus dotes como bailarín y cantante.
El éxito continúo en 1999, empezando con tres conciertos en Finlandia. Igualmente, hizo su primer debut en el cine con la película "Där Regnbågen Slutar" ("Allí Donde El Arcoiris Termina"). Un año más tarde, doblaría al personaje principal de la película "Stuart Little", y en junio se trasladó al West End de nuevo para interpretar el papel de Michael en el musical "Las Brujas de Eastwick".
Tras su colaboración en dicho musical,, publicó en septiembre de 2000 su álbum en inglés "Only When I Breathe", que alcanzó nuevamente el número uno en su país. En marzo de 2002, su segundo álbum en inglés, "I Feel Good And I'm Worth It" vio la luz, llegando al número 2 en la lista de los más vendidos.
En ese mismo año, participó en tres conciertos en homenaje al grupo ABBA, junto a otros artistas del país, así como con la legendaria Dionne Warwick. Con motivo de la Navidad, publicó su álbum "Jag Kommer Igen Till Jul" ("Vengo De Nuevo A Casa Por Navidad"), alcanzando lo más alto de la lista de ventas durante cinco semanas, y convirtiéndose en el séptimo trabajo discográfico más vendido en 2002.
Después de varios conciertos durante 2003, y pequeñas colaboraciones en diferentes musicales, publica en 2004 dos álbumes diferentes. El primero "Det här är platsen" ("Éste Es El Lugar") en abril y, como con sus últimos trabajos, consiguió éxito de crítica y público. El segundo "Storybook", publicado más tarde, estaba totalmente compuesto por canciones en inglés.
Tras la publicación de estos nuevos trabajos, realizó dos conciertos para promorcionar sus trabajos en Gotemburgo y en Estocolmo. Durante el mes de diciembre, participó como artista invitado en varios conciertos de la artista noruega Sissel. En 2010 volvió a presentarse al Melodifestivalen, con el tema "Hollow".
Filmografía
Series de televisión
Películas
Escritor
Compositor
Véase también
Melodifestivalen 1990
Referencias
Enlaces externos
Peter Jöback página oficial (en sueco)
Página No Oficial (en inglés)
Página oficial del musical Kristina från Duvemåla (en sueco e inglés)
Letra de su canción En Sensation (en sueco)
Cantantes masculinos de Suecia
Nacidos en Estocolmo
Personas LGBT de Suecia
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 6,647
|
The Hanyu Shuiping Kaoshi (HSK; ), translated as the Chinese Proficiency Test, is the standardized test of Standard Chinese (a type of Mandarin Chinese) language proficiency of Mainland China for non-native speakers such as foreign students and overseas Chinese. The test is administered by Hanban, an agency of the Ministry of Education of the People's Republic of China.
The test cannot be taken in Taiwan, where only Taiwan's TOCFL exam can be taken. In turn, the TOCFL exam is not available in Mainland China.
Background
Development of the HSK test began in 1984 at Beijing Language and Culture University, and was officially made a national standardized test in 1992. By 2005, over 120 countries had participated as regular host sites and the tests had been taken around 100 million times (domestic ethnic minority candidates included). In 2011, Beijing International Chinese College became the first HSK testing center to conduct the HSK test online.
The HSK test is analogous to the English TOEFL, and an HSK certificate is valid without any limitation in China. The test aims to be a certificate of language proficiency for higher educational and professional purposes. It is not uncommon to refer to a standard or level of proficiency by the HSK level number, or score. For example, a job description might ask for foreign applicants with "HSK5 or better."
The HSK is administered solely in Mandarin and in simplified Chinese characters; however, if the exam is paper-based, the test-taker can choose to write the writing assignments in simplified or in traditional characters, at their discretion. The test can be either paper-based or Internet-based, depending on what the specific test center offers. With an Internet-based test, the writing part with characters (from HSK 3 on) is subjectively and slightly easier, as one types the pinyin and selects the right character from the list, while with a paper-based test, one must remember the characters, their strokes and their order, and write them out.
Test takers with outstanding results can win a scholarship for short-term language study in China.
Structure
From 2021
In 2020, an internal Chinese academic paper discussed that the Chinese Proficiency Standards would undergo a change: a hybrid paradigm of "Three Stages and Nine Levels" characterized by all-in-one integration. The Ministry of Education of the People's Republic of China announced further details regarding HSK 3.0 on April 1, 2021. Among the information shared was a wordlist for each individual stage, and a date for the introduction of the new test—July 1, 2021. As of April 2022 no HSK 3.0 tests have been rolled out and only a single preparation book for the basic levels has been published, additionally the HSK 2.0 test format is still in use. An update on the official HSK website suggests, that the April 2021 date only signified the start date for the conception of the new HSK 3.0. New practice materials for the intermediate and advanced levels are expected to be released in the third or fourth quarter of 2022. One of the main problems with the current HSK test is that it does not follow the Common European Framework of Reference for Languages (CETR). For example, HSK 6 was nowhere close to Near Native C2 in English, and so on.
The new test is meant to follow the latest research in the field of language studies and testing. The basic levels (roughly corresponding to CEFR A1/A2) should test from 500 to 2245 words, the intermediate levels (roughly corresponding to CEFR B1/B2) from 3245 to 5456 words, the advanced levels (roughly corresponding to CEFR C1/C2) from about 6000 to 11000 words.
The intermediate levels should test the ability to understand slightly modified authentic materials or authentic materials written/spoken in a clear manner and intended for educated mother tongue speakers: materials concerning everyday topics, simple articles, simple commentaries and critiques found in newspapers and magazines.
The advanced levels should test the ability to understand longer, more complicated and abstract materials: extracts from authentic sources such as textbooks for Chinese university students, Chinese magazine and essays, extracts from modern Chinese literature, interviews and extracts from Chinese media. Moreover, Classical Chinese expressions and grammatical structures used in modern formal Chinese should be included too.
Between 2010–2021
The previous format was introduced in 2010, with a philosophy of testing "comprehensive language and communication ability". Most notable are the inclusion of written segments at all levels (not just in the Advanced levels as in the pre-2010 test), a reform of the ranking system, and the use of new question structures. Complete vocabulary lists, previous tests, and simulated tests are available as preparation materials. A minor update of the vocabulary lists was made in 2012.
The HSK consists of a written test and an oral test, which are taken separately. This oral test is also known as the HSKK or .
Written test
The Listening, Reading and Writing tests each have a maximum score of 100. HSK 1 and 2 therefore have a maximum score of 200 with 120 points required to pass. HSK 3 and 4 have a maximum of 300 points with 180 points required to pass. There is no minimum number of points required for each of the sections as long as the sum is over 120 or 180 points respectively.
HSK 5 and 6 also have a maximum of 300 points and originally required a score of 180 points to pass. However, since a decision made in February 2013, there has been no official passing score for either HSK 5 or 6.
Hanban provides examples of the exam for the different levels together with a list of words that need to be known for each level. These examples are also available (together with the audio for the Listening Test) on the websites of the Confucius Institute at QUT and HSK Academy.
Online test
The written version is now available in two forms, a computer and a paper based test. Both tests are still held at test centers, the differences between the two are as follows:
Not every test center has the facilities for conducting computer-based tests
Computer-based tests allow you to input characters using the keyboard
Results of computer-based tests are published two weeks after the exam, paper-based test results take one month
Oral test
The HSKK test is a separate test. However, the three HSKK levels correspond with the six HSK levels of the written test.
Comparison with CEFR levels
In 2010, Hanban asserted that the HSK's six levels corresponded to the six levels of the Common European Framework of Reference for Languages (CEFR). However, the German and French associations of Chinese language teachers reject this equivalency, arguing that HSK Level 6 is only equivalent to CEFR Level B2 or C1.
Before 2010
Ranking
Formerly, there were 11 possible ranks (1-11) and 3 test formats (Basic, Elementary/Intermediate, and Advanced). A rank of between 3 and 8 was needed to enroll in a Chinese university, depending on the subject being studied. A score of 9 or higher was a common business standard.
A student taking the Basic test (HSK) could attain a rank of 1 through 3 (1-3), or fail to meet requirements and thus not receive a rank. The Elementary/Intermediate test (HSK) covered ranks 3-8 (3-8), with ranks below 3 not considered. Likewise, the Advanced test (HSK) covered ranks 9-11 (9-11), with scores below 9 not considered.
Content
The previous format for both Basic and Elementary/Intermediate HSK included four sections: listening comprehension, grammar structures, reading comprehension, and written expressions. Aside from the written expressions portion (which requires writing of Chinese characters), these two tests were completely multiple-choice. The Advanced HSK however, added an additional two portions: spoken and written.
Test dates and locations
The HSK is held at designated test centers in China and abroad. A list of test centers can be found at the HSK website. Test dates are published annually and written tests are more frequently held than spoken ones, generally around once a month, depending on the test center. Test registration is usually open until 30 days prior to the actual test date for the paper-based test or around 10 days prior the actual test date for the computer-based test. Results are generally available around 30 days after completion (but no definite date is given for results).
The test cannot be taken in Taiwan (The Republic of China). In Taiwan, only the TOCFL exam can be taken. Conversely, the TOCFL can not be taken in Mainland China, Macau and Hong Kong.
See also
Test of Chinese as a Foreign Language – the Chinese language test used in Taiwan
ZHC – a Chinese language written test intended for native speakers in China
Putonghua Proficiency Test – a Chinese language oral test intended for native speakers in China
CTCSOL - Certificate for Teachers of Chinese to Speakers of Other Languages
List of language proficiency tests
References
External links
Official HSK Website
News on publishing the word list for the 2021 version of HSK (March 31, 2021)
2021 HSK word list (plain text file that can be imported into Pleco)
Comprehensive guide with complete list of characters, words and sounds for all 6 HSK levels
Official HSK Website at Hanban
List of Chinese characters needed to be known to pass HSK
List of HSK words by level on Wiktionary
Exam Centre: Yale-China Chinese Language Centre, The Chinese University of Hong Kong
Passing the HSK Exam
HSK Anti mix-up tool
The Ultimate HSK Guide including all 5000 vocabularies with Pinyin and translation
Standard Chinese
Chinese language tests
Language tests
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 3,045
|
\section{Introduction}
Coupled dynamical systems have been extensively studied over the last
few decades as they provide us a framework for modelling many complex
systems \cite{bio_osc,ecosystem_own}. The focus of investigations have mostly been
on phenomena emerging under variation of the local dynamics of the
units and the interactions among them. Most studies have assumed the
interactions among the nodes to be invariant over time. However in
recent times there have been efforts to incorporate a time-varying
links, namely changing connections between the units in a dynamical
network
\cite{links1,links2,links3,links4,links5,links6,links7,links8,links9,links10}. Such
time varying interactions model the evolution of connections over
time, and are commonly found in physical, biological, social and
engineered systems
\cite{timevarying1,timevarying2,timevarying3,timevarying4,timevarying5}.
Studies so far have considered the variation in links as a function of
time, while the form of coupling remains the same. Here we will
explore a new direction in time-varying interactions: we will study
the {\em effect of switched coupling forms} on the emergent
behaviour.
We consider two coupling functions, one diffusive and the other
conjugate coupling between the two oscillators. Coupling via conjugate
variables is natural in a variety of experimental situations where
sub-systems are coupled by feeding the output of one into the
other. An example from the recent literature is provided by the
experiments of Kim and Roy on coupled semiconductor laser systems
\cite{laser1}, where the photon intensity fluctuation from one laser
is used to modulate the injection current of the other, and vice
versa. Hybrid coupling \cite{hybrid1,hybrid2,hybrid3} also has
relevance in ecological models, where migration and cross-predation
(analogous to conjugate coupling) \cite{conj_coup} occurs between two
population patches, namely over some time migration or diffusive
coupling may be dominant, while at other times cross-predation between
the two patches is prevalent.
The primary goal of this study is to demonstrate the non-trivial
dynamical states arising out of the temporal interplay between two
coupling forms. The test-bed of our inquiry will be a generic system
of coupled oscillators, which we describe in detail below.
\section{Coupled oscillators}
A general form of coupled dynamical oscillators is given by:
\begin{eqnarray}
\dot{X}_i = F(X_i) + K G_i(X_j,X_j',X_i)
\end{eqnarray}
where $X_i$ denotes the set of $m$ dynamical variables of the $i$th oscillator. The matrix $K$ of dimension $m \times m$ contains information on the coupling topology. $G_i$ is the coupling function that represents the nature of the interaction and the variables involved in the interaction term, with the superscript primes ($'$) on $X$ denoting conjugate or ``dissimilar'' variables \cite{mixed}.
Now, complex systems often undergo Hopf bifurcations and sufficiently close to
such a bifurcation point, the variables which have slower time-scales
can be eliminated. This leaves us with a couple of simple first order
ordinary differential equations, popularly known as the Stuart-Landau
system \cite{kuramoto}. In this work we will consider two diffusively
coupled oscillators of this form, namely two coupled Stuart-Landau
systems.
Specifically, when the coupling between the oscillators is through {\em
similar} variables, the dynamical equations are given by:
\begin{eqnarray}
\label{eqn_similar}
\dot{x_1}&=& f^x(x_1,y_1) + \varepsilon(x_2-x_1) \nonumber \\
\dot{y_1}&=& f^y(x_1,y_1) \\
\dot{x_2}&=& f^x(x_2,y_2) + \varepsilon(x_1-x_2) \nonumber \\
\dot{y_2}&=& f^y(x_2,y_2) \nonumber
\end{eqnarray}
and when the oscillators are coupled to each other through
{\em dissimilar} variables, the dynamical equations are given by:
\begin{eqnarray}
\label{eqn_dissimilar}
\dot{x_1}&=& f^x(x_1,y_1) + \varepsilon(y_2-x_1) \nonumber \\
\dot{y_1}&=& f^y(x_1,y_1) \\
\dot{x_2}&=& f^x(x_2,y_2) + \varepsilon(y_1-x_2) \nonumber \\
\dot{y_2}&=& f^y(x_2,y_2) \nonumber
\end{eqnarray}
with
\begin{eqnarray}
\label{eqn_stuart}
f^x&=& x ( 1- (x^2 + y^2 ) ) -5y \\
f^y&=& y ( 1- (x^2 + y^2 ) ) +5x \nonumber
\end{eqnarray}
Coupled Stuart-Landau oscillators have been a good model for the study of oscillation death \cite{dissimilar1,dissimilar2} and amplitude death \cite{ad1,ad2,ad3,ad4}. When the coupling is through similar variables (cf. Eqn.~\ref{eqn_similar}), the oscillators get synchronized at a very low coupling strengths, and remain oscillatory and synchronized up to very high coupling strength. However when coupling is through dissimilar variables (cf. Eqn.~\ref{eqn_dissimilar}), the system shows oscillatory behaviour for small coupling strengths, and increasing coupling strengths give rise to amplitude death and subsequently oscillation death \cite{dissimilar3}.
\section{Time-varying Coupling Form}
Now we consider a scenario where the form of the coupling between the
oscillators is time dependent, namely, the form of coupling {\em
switches between the similar and dissimilar variables}. The
switching of the coupling form may be periodic or probabilistic.
\subsection{Periodic Switching of Coupling Forms}
Here the oscillators change their coupling form periodically. If the
time period of the switch is $T$, we consider the system to be coupled
via similar variables for fraction $f_{sim}$ of the cycle, followed by
coupling to dissimilar variables for the remaining part. So when
$f_{sim}=0$, the oscillators are always coupled to dissimilar
variables and when $f_{sim}=1$ the oscillators are coupled through
similar variable for all time. For $0 < f_{sim} < 1$, the oscillators
experience coupling through similar variables for time $f_{sim} T$,
followed by coupling through dissimilar variables for time $T (1 -
f_{sim})$, in each cycle of period $T$.
Fig.~\ref{fig:bif_periodic_fixed_T} shows the bifurcation diagram for
a system of two coupled oscillators with periodically changing
coupling form, with respect to coupling strength, for different
$f_{sim}$. At $\varepsilon=0$, i.e. for uncoupled Stuart-Landau
oscillators, one naturally obtains period $1$ oscillations. Increasing
the coupling strength results in suppression of oscillations.
Interestingly though, the window of coupling strength over which
oscillations are suppressed depends non-monotonically on $f_{sim}$. At
first, as $f_{sim}$ increases the fixed point window increases
(cf. Fig.~\ref{fig:bif_periodic_fixed_T}a for $f_{sim}=0.2$ vis-a-vis
Fig.~\ref{fig:bif_periodic_fixed_T}c at $f_{sim}=0.6$). However, when
$f_{sim}$ gets even larger this window vanishes entirely
(cf. Fig.~\ref{fig:bif_periodic_fixed_T}d), namely the oscillations
are no longer suppressed anywhere.
\begin{figure}[H]
\centering
\includegraphics[width=1.0\linewidth]{bif_periodic_x1_t_switch_0.02}
\caption{Bifurcation diagram displaying variable $x$
of one of the oscillators, with respect to coupling
strength $\varepsilon$. Here the coupling switches
between similar and dissimilar variables
periodically, with similar-variable coupling (for
time $f_{sim} T$) followed by dissimilar-variable
coupling (for time $(1-f_{sim})T$), for (a)
$f_{sim}$=$0.2$, (b) $f_{sim}$=$0.4$, (c)
$f_{sim}$=$0.6$ and (d) $f_{sim}$=$0.8$, and time
period of switching $T=0.02$. The two colours
represent symmetric solutions around the unstable fixed
point, arising from different initial conditions.}
\label{fig:bif_periodic_fixed_T}
\end{figure}
Further, notice from Fig.~\ref{fig:bif_periodic_fixed_T}a-b that low-amplitude oscillations are restored at higher coupling strengths again, for intermediate $f_{sim}$. Fig.~\ref{fig:osc_amplitude_periodic} shows the effect of the frequency of switching coupling forms upon the amplitude of the revived oscillations. Increase in frequency of switching leads to the reduction of oscillation amplitude, and the results approach those arising from effective mean-field like dynamical equations (cf. Eqn. \ref{eqn_effective}) which will be presented in Section V (namely, Fig.~\ref{fig:bif_periodic_fixed_T}(b) approaches Fig.~\ref{fig:bif_effective_fixed_fsim}(b) obtained from an approximate effective description of the system).
\begin{figure}[H]
\centering
\includegraphics[width=0.85\linewidth]{periodic_osc_width_f_sim_0.4_cp_9.0}
\caption{Dependence of $A = x_{max}-x_{min}$ on the frequency of switching (where $x_{max}$/$x_{min}$ are the maximum/minimum values of variable $x$ of an oscillator). Here $f_{sim}=0.4$, $\varepsilon=9.0$, and the coupling switches between similar and dissimilar variables periodically, with similar-variable coupling for time $f_{sim} T$, followed by dissimilar-variable coupling for time $(1-f_{sim})T$, where $1/T$ is the frequency of switching. }
\label{fig:osc_amplitude_periodic}
\end{figure}
\subsection{Probabilistic Switching of Coupling Forms}
Here the oscillators change their coupling form at intervals of time
$T$, with the coupling form chosen probabilistically. We consider the
probability for the oscillators to be coupled via similar variables to
be $p_{sim}$, and the probability of coupling mediated via dissimilar
variables to be $(1-p_{sim})$. For $0 < p_{sim} < 1$, at the time of
switching, the similar-variable coupling form is chosen with
probability $p_{sim}$ and the dissimilar-variable form is chosen with
probability $(1-p_{sim})$. So larger $p_{sim}$ favours coupling
through similar variables and smaller $p_{sim}$ favours
dissimilar-variable coupling, with the oscillators always experiencing
dissimilar-variable coupling for the limiting case of $p_{sim}=0$ and
similar-variable coupling for $p_{sim}=1$. Here the probability
$p_{sim}$ plays a role equivalent to $f_{sim}$ in the case of periodic
switching of coupling forms.
Fig.~\ref{fig:bif_probabilistic_fixed_T} shows the bifurcation diagram,
with respect to coupling strength, for different $p_{sim}$. Again it
is evident that the window of coupling strength over which
oscillations are suppressed, depends non-monotonically on $p_{sim}$,
as it did under variation of $f_{sim}$ for the case of periodically
switched coupling forms
(cf. Fig.~\ref{fig:fp_width_t_switch_0.02}). First, as $p_{sim}$
increases from zero, the fixed point window increases, as seen from
Fig.~\ref{fig:bif_probabilistic_fixed_T}a for $p_{sim}=0.2$ vis-a-vis
Fig.~\ref{fig:bif_probabilistic_fixed_T}c for $p_{sim}=0.6$. However,
when $p_{sim}$ gets even larger this window vanishes entirely, as
evident from Fig.~\ref{fig:bif_probabilistic_fixed_T}d, and the
oscillations are no longer suppressed anywhere. This suggests that
when the probability of coupling through similar variables and
dissimilar variables is similar (i.e. $p_{sim} \sim 0.5$) oscillations
are suppressed to the greatest degree.
\begin{figure}[H]
\centering
\includegraphics[width=1.0\linewidth]{bif_probabilistic_x1_t_switch_0.02}
\caption{Bifurcation diagram displaying
variable $x$ of one of the oscillators, with
respect to coupling strength
$\varepsilon$. Here the coupling form
probabilistically switches at time intervals
of $T$ between similar and dissimilar
variables, with the probability of
similar-variable coupling being $p_{sim}$,
and the probability of dissimilar-variable
coupling being $(1-p_{sim})$, for (a)
$p_{sim}=0.2$, (b) $p_{sim}=0.4$, (c)
$p_{sim}=0.6$ and (d) $p_{sim}=0.8$ (with
$T=0.02$). The two colours represent
emergent dynamics from different initial
conditions}
\label{fig:bif_probabilistic_fixed_T}
\end{figure}
Fig.~\ref{fig:fp_width_t_switch_0.02} shows the width
of fixed point window, with respect to $f_{sim}$ for
the case of periodic switching of coupling forms, and
$p_{sim}$ for the case of probabilistic switching of
coupling forms. As shown, the width of the fixed point
window is non-monotonic, and has a maximum at $f_{sim}
(p_{sim}) \sim0.58$. Namely, the largest window of
fixed point dynamics arises where there is balance in
the probability of occurrence of the coupling forms.
\begin{figure}[H]
\centering
\includegraphics[width=0.85\linewidth]{fp_width_t_switch_0.02}
\caption{Width of the fixed point window as a function of $f_{sim}$ (blue) for the case of periodic switching of coupling forms, and $p_{sim}$ (red) for the case of probabilistic switching of coupling forms, with switching period $T=0.02$.}
\label{fig:fp_width_t_switch_0.02}
\end{figure}
\section{Global stability measure}
The commonly employed linear stability analysis, based on the
linearization in the neighbourhood of fixed points, provides only local
information about the stability at the fixed point. It cannot
accurately indicate the stability for large perturbations, nor the
basin of attraction of the dynamics, especially in the presence of
other attractors in phase space.
Here we calculate the Basin Stability of the dynamical states
\cite{basin}. This is a more robust and global estimate of stability,
and effectively incorporates non-local and non-linear effects on the
stability of fixed points. Specifically, Basin Stability is calculated
as follows: we choose a large number of random initial conditions,
spread uniformly over a volume of phase space, and find what fraction
of these are attracted to stable fixed points.
Figs~\ref{fig:bs_periodic_fixed_t_switch} shows the Basin Stability of
the fixed point state, in the parameter space of $f_{sim}$ and
coupling strengths, for different time periods of switching
$T$. Clearly one obtains oscillation suppression in windows of
coupling strength and $f_{sim}$, and oscillation revival again beyond
the window. The window of coupling strengths that gives rise to fixed
points is very sensitive to the frequency of switching, at low
frequencies. After a high enough switching frequency (i.e. low enough
$T$), the fixed point region remains unchanged, as evident through the
fact that Fig.~\ref{fig:bs_periodic_fixed_t_switch}a and
Fig.~\ref{fig:bs_periodic_fixed_t_switch}b are identical.
Now, the dependence of the fixed point window on $f_{sim}$ is actually
quite counter-intuitive, as already indicated in the bifurcation
diagrams. For rapidly switched coupling forms, at large coupling
strengths, the oscillation suppression occurs at an {\em intermediate}
value of $f_{sim}$. Namely, as the dominance of similar-variable
coupling $f_{sim}$ increases the oscillations are first suppressed and
then after a point the oscillations are revived again, with the window
of fixed points shifting towards higher $f_{sim}$, as coupling
strength increases. This is counter-intuitive, as similar-variable
coupling is known to only allow oscillations, while
dissimilar-variable coupling can yield some windows of oscillation
suppression. Also interestingly for low $f_{sim}$, as we increase the
coupling strength, first we encounter oscillation suppression and then
on further increase of coupling strength the oscillations are
restored. So there exists an intermediate window of coupling strength
that yields fixed point dynamics.
Fig~\ref{fig:bs_periodic_fixed_f_sim} shows the Basin Stability of the
fixed point state, in the parameter space of the frequency of
switching and coupling strengths, for different $f_{sim}$. It is clear
again that after a critical switching frequency the dynamics does not
depend on the rate at which the coupling form is
changed. Significantly, it is also evident that {\em fast changes in
coupling form, namely lower time period for change, yields large
fixed point regions in parameter space.}
However, at low switching frequencies the emergent dynamics is
sensitive to how rapidly the coupling form varies.
\begin{figure}[H]
\centering
\includegraphics[width=1.0\linewidth]{bs_periodic_t_switch_fixed}
\caption{Basin Stability of the fixed point state of coupled oscillators, in the parameter space of coupling strength $\varepsilon$ and $f_{sim}$. Here the coupling periodically switches between similar and dissimilar variables, with similar-variable coupling (for time $f_{sim} T$), followed by dissimilar-variable coupling (for time $(1-f_{sim})T$), for (a) $T=0.01$, (b) $T=0.10$, (c) $T=1.00$ and (d) $T=2.00$. The region in yellow represents fixed point dynamics (i.e. oscillation suppression), and the region in black represents oscillatory dynamics.}
\label{fig:bs_periodic_fixed_t_switch}
\end{figure}
\begin{figure}[H]
\centering
\includegraphics[width=1.0\linewidth]{bs_periodic_f_sim_fixed}
\caption{Basin Stability of the fixed point
state of coupled oscillators, in the
parameter space of coupling strength
$\varepsilon$ and frequency of
switching. Here the coupling periodically
switches between similar and dissimilar
variables, with similar-variable coupling
(for time $f_{sim} T$), followed by
dissimilar-variable coupling (for time
$(1-f_{sim})T$), for (a) $f_{sim}=0.3$, (b)
$f_{sim}=0.4$, (c) $f_{sim}=0.5$ and (d) $f_{sim}=0.7$. The
region in yellow represents fixed point
dynamics (i.e. oscillation suppression), and
the region in black represents oscillatory
dynamics.}
\label{fig:bs_periodic_fixed_f_sim}
\end{figure}
Fig~\ref{fig:bs_probabilistic_fixed_t_switch} shows
the Basin Stability of the fixed point state for the
case of probabilistically varying coupling form, in
the parameter space of $p_{sim}$ and coupling
strengths. {\em Interestingly again, as we increase
the coupling strength, oscillations first get
suppressed and then restored.} Also notice the
marked similarity of
Fig.~\ref{fig:bs_probabilistic_fixed_t_switch}a and
Fig.~\ref{fig:bs_periodic_fixed_t_switch}a. Namely
frequent periodic switching of coupling forms yields
the same result as the frequent probabilistic
switching of coupling forms.
\begin{figure}[H]
\centering
\includegraphics[width=1.0\linewidth]{bs_probabilistic_t_switch_fixed}
\caption{Basin Stability of the fixed point state of coupled oscillators, in the parameter space of coupling strength $\varepsilon$ and $p_{sim}$. Here the coupling probabilistically switches between similar and dissimilar variables, with the probability of similar-variable coupling being $p_{sim}$, and the probability of dissimilar-variable coupling being $(1-p_{sim})$ with the period of switching (a) $T=0.01$, (b) $T=0.02$, (c) $T=0.10$ and (d) $T=1.00$. The region in yellow represents fixed point dynamics (i.e. oscillation suppression), and the region in black represents oscillatory dynamics. Notice the marked similarity of panel (a) with Fig.~\ref{fig:bs_periodic_fixed_t_switch}a.}
\label{fig:bs_probabilistic_fixed_t_switch}
\end{figure}
Fig.~\ref{fig:bs_probabilistic_fixed_p} shows the
Basin Stability of the fixed point state, in the
parameter space of frequency of switching and coupling
strengths. Clearly the effects of the frequency of
switching are pronounced over a larger range of
switching frequency for probabilistic switching, as
compared to periodic switching. But significantly
again, it is evident that {\em fast changes in
coupling form, namely lower time period for change,
yields large fixed point regions in parameter
space.} Lastly, it is also clear that as the
frequency of switching increases, the fixed point
region moves towards higher values of $p_{sim}$ where
similar-variable coupling dominates. This is again
surprising, as similar-variable coupling is known to
support only oscillations, while dissimilar-variable
coupling has more propensity towards oscillation
suppression.
\begin{figure}[H]
\centering
\includegraphics[width=1.0\linewidth]{bs_probabilistic_p_sim_fixed}
\caption{Basin Stability results for coupled oscillators, in the parameter space of coupling strength $\varepsilon$ and frequency of switching, when coupling switches between similar and dissimilar variables probabilistically, with a probability of similar coupling $p_{sim}$, and probability of dissimilar coupling $(1-p_{sim})$, for (a) $p_{sim}=0.2$, (b) $p_{sim}=0.3$, (c) $p_{sim}=0.5$ and (d) $p_{sim}=0.7$. The region in yellow represents fixed point dynamics (i.e. oscillation suppression), and the region in black represents oscillatory dynamics.}
\label{fig:bs_probabilistic_fixed_p}
\end{figure}
\section{Effective model for time dependent coupling}
Now we attempt to rationalize our results through an effective
phenomenological model for the dynamics. The idea is to mimic the time-dependent
coupling by a coupling form where the similar and dissimilar coupling
forms are appropriately weighted by $f_{sim}$. This is given by:
\begin{eqnarray}
\label{eqn_effective}
\dot{x_1}&=& f^x(x_1,y_1) + \varepsilon [ f_{sim} (x_2-x_1) + (1-f_{sim})(y_2-x_1) ] \nonumber \\
\dot{y_1}&=& f^y(x_1,y_1) \\
\dot{x_2}&=& f^x(x_2,y_2) + \varepsilon [ f_{sim} (x_1-x_2) + (1-f_{sim})(y_1-x_2) ] \nonumber \\
\dot{y_2}&=& f^y(x_2,y_2) \nonumber
\end{eqnarray}
This effective picture is expected to hold true when the frequency of switching is very high (namely $T$ is very small). Completely equivalent results can be obtained with $p_{sim}$ in place of $f_{sim}$ in the equations above.
\begin{figure}[H]
\centering
\includegraphics[width=1.0\linewidth]{bif_effective_x1}
\caption{Bifurcation diagram displaying variable $x$ of one of the oscillators of the effective coupled system given by Eqn.~\ref{eqn_effective}, with respect to coupling strength $\varepsilon$, at different $f_{sim}$ values: (a) $f_{sim}$=$0.2$, (b) $f_{sim}$=$0.4$, (c) $f_{sim}$=$0.6$ and (d) $f_{sim}$=$0.8$. The two colours represent symmetric solutions around the unstable fixed point, arising from different initial conditions.}
\label{fig:bif_effective_fixed_fsim}
\end{figure}
Fig.~\ref{fig:bif_effective_fixed_fsim} displays the
bifurcation diagram of the effective coupled system given by
Eqn.~\ref{eqn_effective}, and provides insight into the
suppression and revival of oscillations in the coupled
oscillator system with time-varying coupling forms. In
particular, notice the marked similarity of the fixed point
region in Fig.~\ref{fig:bif_effective_fixed_fsim} with the
fixed point regions evident in
Figs.~\ref{fig:bif_periodic_fixed_T} and
\ref{fig:bif_probabilistic_fixed_T}.
\begin{figure}[H]
\centering
\includegraphics[width=0.85\linewidth]{ev_max_effective}
\caption{Results from Linear Stability analysis of the effective model of the coupled oscillators given by Eqn.~\ref{eqn_effective}, in the parameter space of coupling strength $\varepsilon$ and $f_{sim}$. The color code represents the value of the maximum eigen value of the Jacobian corresponding to Eqn.~\ref{eqn_effective}, $\lambda_{max}$. The region in pink, where $\lambda_{max} < 0$, represents stable fixed points. The region in green, where $\lambda_{max} > 0$, represents unstable fixed points. Notice the marked similarity with Figs.~\ref{fig:bs_periodic_fixed_t_switch}a and \ref{fig:bs_probabilistic_fixed_t_switch}a.}
\label{fig:linear_stability_effective}
\end{figure}
Increasing the coupling strength, makes the fixed point at
zero in Eqn.~\ref{eqn_effective} unstable, and leads to
creation of new fixed
points. Fig.~\ref{fig:linear_stability_effective} shows
results from the linear stability analysis of
Eqn.~\ref{eqn_effective}, in the neighbourhood of the fixed
point at zero. The region in pink represents the stable fixed
point, where the maximum eigen value of the Jacobian
corresponding to Eqn.~\ref{eqn_effective}, $\lambda_{max}$, is
negative. The region in green represents unstable fixed
points, as $\lambda_{max}>0$ there. Notice the marked
similarity of these results with
Fig.~\ref{fig:bs_periodic_fixed_t_switch}a (or equivalently
Fig.~\ref{fig:bs_probabilistic_fixed_t_switch}a), namely the
fixed point region is completely well-described by the
analysis of Eqn.~\ref{eqn_effective} when the frequency of
switching is high ($\sim100$ Hz). Significantly then, the
results from the global estimates of the basin of stability of
the fixed point, for rapid periodic and probabilistic
switching of coupling form, are {\em recovered accurately
through the linear stability analysis of a set of effective
dynamical equations.}
\section{Conclusions}
While the variation in links, namely the connectivity matrix, as a
function of time has been investigated in recent times, the dynamical
consequences of time-varying coupling forms is still not
understood. In this work we have explored this new direction in
time-varying interactions, namely we have studied the effect of
switched coupling forms on the emergent behaviour. The test-bed of our
enquiry is a generic system of coupled Stuart-Landau oscillators,
where the form of the coupling between the oscillators switches
between the similar and dissimilar (or conjugate) variables. We
consider two types of switching, one where the coupling function
changes periodically and one where it changes probabilistically. When
the oscillators change their coupling form periodically, they are
coupled via similar variables for fraction $f_{sim}$ of the cycle,
followed by coupling to dissimilar variables for the remaining
part. In the case of probabilistic switching, the probability for the
oscillators to be coupled via similar variables is $p_{sim}$, and the
probability of coupling mediated via dissimilar variables is
$(1-p_{sim})$.
We find that time-varying coupling forms suppress oscillations in a
window of coupling strengths, with the window increasing with the
frequency of switching. That is, {\em more rapid changes in coupling
form leads to large windows of oscillation suppression}, with the
window of amplitude death saturating after a high enough switching
frequency. Interestingly, for low $f_{sim}$ ($p_{sim}$), the {\em
oscillations are revived} again beyond this window. That is, too low
or too high coupling strengths yield oscillations, while coupling
strengths in-between suppress oscillations.
Also interestingly, the width of the coupling strength window
supporting oscillation suppression is non-monotonic with respect to
$f_{sim} (p_{sim})$, and has a maximum at $f_{sim} (p_{sim}) \sim
0.58$. Namely, the largest window of fixed point dynamics arises
where there is balance in the probability of occurrence of the
coupling forms.
Focusing on the dependence of the window of oscillation suppression,
at fixed coupling strengths and varying predominance of coupling
forms, we observe the following: for rapidly switched coupling forms,
at large coupling strengths, the oscillation suppression occurs at an
{\em intermediate} value of $f_{sim}$ ($p_{sim}$). Namely, as the
dominance of similar-variable coupling increases the oscillations are
first suppressed, and then after a point the oscillations are revived
again. The fixed point window shifts towards a higher probability of
similar-variable coupling, as coupling strength increases. This is
counter-intuitive, as purely similar-variable coupling yields
oscillatory behaviour, while dissimilar-variable coupling supports
oscillation suppression.
Lastly, we have suggested an effective dynamics that successfully
yields the observed behaviour for rapidly switched coupling forms,
including an accurate estimate of the fixed point window through
stability analysis. Thus our results will potentially enhance the
broad understanding of coupled systems with time-varying connections.
\bigskip
\bigskip
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 4,459
|
{"url":"https:\/\/asa2.silverchair.com\/anesthesiology\/article\/85\/5\/961\/35875\/Is-Voluntary-Reporting-of-Critical-Events?searchresult=1","text":"Key words: Complications. Monitoring. Quality assurance.\n\nIn this issue of Anesthesiology, another study adds to evidence that quality assurance (QA) programs based on voluntary reporting identify only a small fraction of critical events. [1]We learn that anesthesiologists, as has been shown for other clinicians, do not report most events meeting criteria that they themselves had defined as relevant to QA. This shouldn't be surprising. Why should we expect clinicians to report their own adverse outcomes if reporting might jeopardize their career? The prospect of public embarrassment or retribution could be a sufficient deterrent to satisfactory self-reporting. Perhaps equally dissuading is that the benefit to patient care of anesthesia QA systems has not been established rigorously, especially for QA indicators such as deviations in vital signs. This study also indicates the potential for using the Automated Anesthesia Record for Quality Assurance: Is this tool sufficiently practical and useful to be worth the cost for this application?\n\nSanborn et al. [1]provide data that question the effectiveness of voluntary reporting to measure accurately the true incidence of undesirable anesthesia clinical events. Using a commercial Automated Anesthesia Record (AAR, essentially equivalent to an Automated Information Management System), they documented deviations from predefined limits of vital signs. These deviations had been agreed on by their department for its QA program. Eliminating artifacts and nonclinically-significant deviations, they found 434 intraoperative events within 5,454 anesthesia records. Relatively few of these events (4.1%) were reported in a parallel, electronic system that clinicians used for entry of QA data. They further demonstrated that these vital signs' deviations were associated with (although not necessarily the cause of) intraoperative and postoperative mortality.\n\nSeveral other studies support the argument that QA programs based on voluntary reporting detect relatively few of all events. Most recently, Cullen et al. [2]reported that only 6% of hospital adverse drug events were reported through the hospital incident reporting system, and in other studies, researchers reported similar findings. [3,4]There is some evidence to the contrary (e.g., an innovative, confidential program for voluntary reporting by house officers appears to have been efficacious. [5]As many events were reported as were identified via a chart review and at a lower cost). But, there are good reasons to believe that voluntary reporting does not reliably capture the true nature or extent of adverse events. Sanborn et al. offer some possible reasons for the differences between the electronic record and voluntary reporting. Yet, without a measure from the participants (e.g., via a survey or qualitative study), we are left to speculate why the clinicians didn't report more of the events. A general explanation is that better incentives are needed to encourage anesthesiologists to report their outcomes, mistakes, and other system problems. At least two steps toward that are necessary: 1) demonstrate the value of reporting, and 2) change the culture that attributes error to negligence.*\n\nTracking substantive deviations in vital signs should, in principle, be useful to indicate something about the process of anesthesia care. But, the value in documenting these events, whether reported voluntarily or by an AAR, has yet to be demonstrated. Will decreasing the rate of hypotension, hypertension, tachycardia and bradycardia, hypoxemia and hypothermia, presuming that it can be done, make any difference to patient outcome? There is some anecdotal evidence that comprehensive use of an AAR is associated with reductions in occurrence of such indicators and even of indicators that are not documented automatically.** But, there still is no well controlled study to suggest that this will result in reductions in the rate of serious adverse outcomes. Perhaps it is better care if rates of events such as hypotension or hypoxemia of lasting duration are reduced from some baseline; common sense suggests that flying further above the trees provides a larger margin for error. Or, perhaps these events are just normal perturbations. If nothing else, reporting the rates among centers that use AAR systems will give comparative data for benchmarking between hospitals and clinicians. Sanborn et al. and others with access to an AAR have the opportunity to lead in this effort. They must examine not only event frequencies, but also measure what it takes to improve performance indicators, with or without the assistance of expensive technology. There actually is some evidence that self-reporting can provide useful information for anesthesia quality improvement if the criteria for reporting are set by the individual clinician for the individual case. [6]To be effective, such a program requires investigation of system issues and a greater expenditure of resources to learn which events are important and to devise remedial strategies.\n\nAlthough it was not one of the objectives of the study, it may have been helpful had the investigators examined the most serious outcomes more closely, to identify patterns or system problems. The information in the AAR may have provided new insights. There probably is a payoff to examining such sentinel events [2]-the Australians think so. Their legal system protects from disclosure; they collect data and analyze it via study committees. They have shown improvements in outcome during the 30 years this has been done. [7]As Gaba [8]advocated for anesthesia and Leape [9]more recently for medicine and health care in general, it is the identification and correction of system problems that is more likely to yield benefits. But, for that to happen, a culture change must be made.\n\nThe culture of health care is not forgiving of errors by clinicians, especially physicians. As Leape wrote recently,\n\n\". . .In everyday hospital practice, the message is .. clear: mistakes are unacceptable. Physicians are expected to function without error, an expectation that physicians translate into the need to be infallible. . . .This kind of thinking lies behind a common reaction by physicians: How can there be an error without negligence?\" [9]\n\nNeither the public nor physicians themselves are tolerant of medical error; indeed, physicians generally do not feel comfortable talking about their errors, nor can colleagues generally be expected to offer support. [10,11]\n\nFor meaningful change toward more openness in reporting errors and system problems, the culture must change. Deviations and errors need to be accepted as the inevitable result of imperfect systems and the normal limitations of men and women. A culture of safety understands that reports of problems, errors, and accidents are the material from which learning and improvement are born. Fortunately, there is a growing body of literature in which this important topic is being aired.*** [9,11]But, more open dialogue about error is needed for quality assurance to have its intended impact.\n\nWithout such a substantial change in the culture of health care, voluntary reporting of events will continue to underperform. Perhaps what is needed is more protection from unwarranted public disclosure, as is afforded to pilots through the Aviation Safety Reporting System or for anesthesiologists in Australia who report through the Australian Incident Monitoring System. [12]Can clinicians in the United States be protected from retribution for reporting problems for which they may have some responsibility? Despite efforts to offer such protection, there has not yet been effective action to motivate increased levels of reporting and discussion of adverse events. In New Jersey, there is a ray of hope. A state law that requires reporting of adverse events also calls for protecting the events from public disclosure. Although there has not yet been a court challenge to test the law, a recent request for disclosure was rebuffed.**** We are probably a long way from a national system to report medical error and system problems similar to the Aviation Safety Reporting System. In the meantime, development of local reporting systems, with appropriate legal protection, should be an attainable and useful goal. Those systems must include provisions for analysis and feedback of the information and for instituting system changes that generate real improvements in the system for patient care.\n\nThe many potential benefits of the AAR have been described. [13,14]An AAR system was a necessary instrument for the current study, and its use uncovered deviations associated with mortality. Should departments now rush to purchase this costly technology? Manual recordkeeping itself is suspect for documenting events. Substantial discrepancies have been demonstrated between automatic and manual reporting of intraoperative blood pressure. [15]Yet, the greater accuracy, neatness, and ability to track noncritical events are unlikely to be sufficient motivation to justify the current cost of approximately $6 per procedure.***** Because a benefit for outcome or quality cannot be shown clearly, implementing an AAR will continue to have a tough battle when competing in the new corporate environment of health care. Quality assurance alone will not be the driving force for the AAR except for the more forward-thinking administrators. Its acceptance is more likely to be market-driven. As a hospital recognizes that it needs information about utilization of the operating room resources, as many already have, automated data-collection will be the only way to get that reliably and consistently. Anesthesia departments will be more driven to modernize their production facility when the hospital information system really works and can connect to it. Many are already doing so; it is likely that all will in the not-too-distant future. Quality assurance may well be a beneficiary. Can the AAR be useful and cost effective as a tool for QA? Sanborn et al. provide evidence of the potential for that. What needs to be shown now is how the data can be used to direct and stimulate effective and long-lasting changes. But, that will not be so easy. In the larger arena of outcomes research, despite efforts to do so, it has proven difficult to design effective tools to consistently change behavior. As has been said about motivating behavior change for compliance of national guidelines for care, \". . .it is unlikely that progress will be made . . . ..without input from the fields of behavioral science and organizational behavior.\" [16] When it comes to getting human error out in the open and making patient safety a priority, anesthesiology is the leading medical specialty. Despite the deficiencies in reporting systems, anesthesiologists can feel pride in all they have done to assume responsibility for iatrogenic outcomes. The development of the AAR and the publication of studies such as Sanborn et al. in Anesthesiology are further demonstrations of a historical commitment of this specialty to identify and talk about its contribution to adverse outcomes. More of the same is needed to lower barriers that continue to dissuade clinicians from openly revealing the imperfections that are the data for learning how to improve. Is it time to abandon self-reporting of incidents or events as a QA tool? Probably not. Even if only a few percent of actual deviations are reported, some important problems are likely to surface. [2]These reports will come from those who recognize the importance of fixing system problems and who are willing to identify incidents in which they have been involved or even helped to cause. Some departments seem to have successfully implemented QA programs using voluntary reporting. [6]But, as Sanborn et al. suggest, the data probably do not represent a thorough picture of the extent of the problem. In addition, effective monitoring and improvement in quality must be based on a process that includes thorough investigation of sentinel events, feedback of data to individuals, and analysis of system problems. Most of all, a change is needed in the culture of all of health care. Rather than assigning blame, we must encourage open discussion of problems and protect clinicians from unwarranted public scrutiny. Reward people for doing the right thing: being honest about mistakes and so working to prevent them from happening again. Jeffrey B. Cooper, Ph.D., Department of Anesthesia and Critical Care, Massachusetts General Hospital, 55 Fruit St., Boston, Massachusetts 02114. * Heylan R: The relevant cause in anesthetic death. Critique on concrete singular causal opinions in research on avoidable anesthetic mortality and morbidity highlighted by legal and philosophical causation theories. R. Heylan, St. Jansziekenhuis, Schiepse Bos 2, B-3600 Genk Belgium, 1995 (PhD dissertation). ** Edsall DW, Deshane PD: Getting the most out of an information system: Annual meeting of the New England Society of Anesthesiologists, Chatham, Massachussetts, September 27-28, 1996. *** Heylan, op. cit. **** Barbara Andrews, NJ State Department of Health, and Ervin Moss, M.D.: (personal communicaton). ***** Basis of estimate:$20,000\/rm; 7 yr useful life; 5% annual service and software maintenance; 10% annual administrative management cost; total of approximately $6,000\/yr\/rm; for 1,000 procedures\/rm, [nearly equal]$6\/procedure.\n\n1.\nSanborn KV, Castro J, Kuroda M, Thys DM: Detection of intraoperative incidents by electronic scanning of computerized anesthesia records: Comparison with voluntary reporting. Anesthesiology 1996; 85:977-87.\n2.\nCullen DJ, Bates DW, Small SD, Cooper JB, Nemeskal AR, Leape LL: The incident reporting system does not detect adverse drug events: A problem for quality improvement. Jt Comm J Qual Improv 1995; 21:541-8.\n3.\nAllan EL, Barker KN: Fundamentals of medication error research. Am J Hosp Pharm 1990; 47:555-71.\n4.\nShannon RC, DeMuth JE: Comparison of medication error detection methods in the long term care facility. Consulting Pharmacy 1987; 2:148-151.\n5.\nO'Neil AC, Peterson LA, Cook FE, Bates DW, Lee TH, Brennan TA: Physician reporting compared with medical-record review to identify adverse medical events. Ann Intern Med 1993; 119:370-6.\n6.\nPosner KL, Kendall-Gallagher D, Wright IH, Glosten B, Gild WM, Cheney FW: Linking process and outcome of care in a continuous quality improvement program for anesthesia services. Am J Med Qual 1994; 9:129-57.\n7.\nWarden JC, Borton CL, Horan BF: Mortality associated with anaesthesia in New South Wales, 1984-90. Med J Aust 1994; 161:585-93.\n8.\nGaba DM: Human error in dynamic medical environments, Human Error in Medicine. Edited by Bogner MS. Hillsdale, NJ, Lawrence Erlbaum Associates, 1994, pp 197-224.\n9.\nLeape LL: Error in medicine. JAMA 1994; 272:1851-7.\n10.\nNewman MC: The emotional impact of mistakes on family physicians. Arch Fam Med 1996; 5:71-5.\n11.\nEly JW: Physicians' Mistakes: Will your colleagues offer support? Arch Fam Med 1996; 5:76-7.\n12.\nWebb RK, Currie, CA Morgan, Williamson JA, Mackay P, Russell WJ, Runciman WB: The Australian incident monitoring study: An analysis of 2000 incident reports. Anaesth Intensive Care 1993; 21:520-8.\n13.\nEdsall DW: Quality assessment with a computerized Anesthesia Information Management System (AIMS). Q Rev Bull 1991; 17:182-193.\n14.\nEichhorn J, Edsall DW: Computerization of anesthesia information management. J Clin Monit 1991; 7:71-82.\n15.\nCook RI, McDonald JS, Nunziata E: Differences between handwritten and automatic blood pressure records. Anesthesiology 1989; 71:385-90.\n16.\nGuadagnoli E, McNeil BJ: Outcomes research: Hope for the future or the latest rage? Inquiry 1994; 31:14-24.","date":"2021-12-05 12:04:46","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.3162657618522644, \"perplexity\": 6138.012007509467}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-49\/segments\/1637964363157.32\/warc\/CC-MAIN-20211205100135-20211205130135-00105.warc.gz\"}"}
| null | null |
{"url":"https:\/\/www.opgenorth.net\/post\/2013-03-09-sublime-text-2-and-arduino\/","text":"# Sublime Text 2 and Arduino\n\nIf you\u2019re looking to get into Arduino, and you\u2019re a programmer, the first thing that will jump out at you is the Arduino IDE. It\u2019s best described as \u201cspartan\u201d (to say the least). As I\u2019m used to full featured IDE\u2019s I started looking for a replacement to the default Arduino IDE.\n\nThere are extensions to use Visual Studio, but that means me starting up a VM to run Windows which I don\u2019t really want to do for Arduino development. There is a another IDE which looks promising called Maria Mole - but it\u2019s Windows only so not really a contender for me. I need something for OS X. I looked at setting up Eclipse as my default IDE, but ran into some issues with that. Nothing to major, but as I don\u2019t like Eclipse in the first place I wasn\u2019t to motivated to sort things out, so I abandoned Eclipse as an IDE choice.\n\nThe next thing I tried was Sublime Text. There is an Ardunio plugin called Stino that turns Sublime into a not bad IDE. In terms of writing your programs, Stino can pretty much do everything the Arduino IDE can do: compile programs, upload them to your Arduino board, import libraries, etc.\n\n2. In Sublime Text, go to Package Control, and type Install Package.\n3. Search using the keyword Arduino, you should see a couple of packages - one contains Stino and the other are a couple of snippets.\n4. Next open a .ino file in Sublime Text. When you do this the Arduino menu should appear in the Sublime toolbard.\n5. From the Arduino menu, specify the location of the Ardunio directory, i.e. \/Applications\/Arduino.app.","date":"2021-10-16 00:11:30","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.22810350358486176, \"perplexity\": 2922.6328453499523}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-43\/segments\/1634323583087.95\/warc\/CC-MAIN-20211015222918-20211016012918-00471.warc.gz\"}"}
| null | null |
#
**MURRAY R. SPIEGEL** received the MS degree in Physics and the PhD in Mathematics from Cornell University. He had positions at Harvard University, Columbia University, Oak Ridge, and Rensselaer Polytechnic Institute, and had served as mathematical consultant at several large companies. His last position was Professor and Chairman of Mathematics at the Rensselaer Polytechnic Institute, Hartford Graduate Center. He was interested in most branches of mathematics, especially those which involved applications to physics and engineering problems. He was the author of numerous journal articles and 14 books on various topics in mathematics.
**DR. ROBERT E. MOYER** has been teaching mathematics and mathematics education at Southwest Minnesota State University in Marshall, Minnesota, since 2002. Before coming to SMSU, he taught at Fort Valley State University in Fort Valley, Georgia, from 1985 to 2000, serving as head of the Department of Mathematics and Physics from 1992–1994. Prior to teaching at the university level, Dr. Moyer spent seven years as the mathematics consultant for a five-county Regional Educational Service Agency in central Georgia and twelve years as a high school mathematics teacher in Illinois. He has developed and taught numerous inservice courses for mathematics teachers. He received his Doctor of Philosophy in Mathematics Education from the University of Illinois (Urbana-Champaign) in 1974. He received his Master of Science in 1967 and his Bachelor of Science in 1964, both in Mathematics Education from Southern Illinois University (Carbondale).
Copyright © 2014 by McGraw-Hill Education. All rights reserved. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written permission of the publisher.
ISBN: 978-0-07-182585-6
MHID: 0-07-182585-1
The material in this eBook also appears in the print version of this title: ISBN: 978-0-07-182181-0, MHID: 0-07-182181-3.
eBook conversion by codeMantra
Version 1.0
All trademarks are trademarks of their respective owners. Rather than put a trademark symbol after every occurrence of a trademarked name, we use names in an editorial fashion only, and to the benefit of the trademark owner, with no intention of infringement of the trademark. Where such designations appear in this book, they have been printed with initial caps.
McGraw-Hill Education eBooks are available at special quantity discounts to use as premiums and sales promotions, or for use in corporate training programs. To contact a representative please visit the Contact Us page at www.mhprofessional.com.
Trademarks: McGraw-Hill Education, the McGraw-Hill Education logo, Schaum's, and related trade dress are trademarks or registered trademarks of McGraw-Hill Education and/or its affiliates in the United States and other countries, and may not be used without written permission. All other trademarks are the property of their respective owners. McGraw-Hill Education is not associated with any product or vendor mentioned in this book.
**TERMS OF USE**
This is a copyrighted work and McGraw-Hill Education and its licensors reserve all rights in and to the work. Use of this work is subject to these terms. Except as permitted under the Copyright Act of 1976 and the right to store and retrieve one copy of the work, you may not decompile, disassemble, reverse engineer, reproduce, modify, create derivative works based upon, transmit, distribute, disseminate, sell, publish or sublicense the work or any part of it without McGraw-Hill Education's prior consent. You may use the work for your own noncommercial and personal use; any other use of the work is strictly prohibited. Your right to use the work may be terminated if you fail to comply with these terms.
THE WORK IS PROVIDED "AS IS." McGRAW-HILL EDUCATION AND ITS LICENSORS MAKE NO GUARANTEES OR WARRANTIES AS TO THE ACCURACY, ADEQUACY OR COMPLETENESS OF OR RESULTS TO BE OBTAINED FROM USING THE WORK, INCLUDING ANY INFORMATION THAT CAN BE ACCESSED THROUGH THE WORK VIA HYPERLINK OR OTHERWISE, AND EXPRESSLY DISCLAIM ANY WARRANTY, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO IMPLIED WARRANTIES OF MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE. McGraw-Hill Education and its licensors do not warrant or guarantee that the functions contained in the work will meet your requirements or that its operation will be uninterrupted or error free. Neither McGraw-Hill Education nor its licensors shall be liable to you or anyone else for any inaccuracy, error or omission, regardless of cause, in the work or for any damages resulting there from. McGraw-Hill Education has no responsibility for the content of any information accessed through the work. Under no circumstances shall McGraw-Hill Education and/or its licensors be liable for any indirect, incidental, special, punitive, consequential or similar damages that result from the use of or inability to use the work, even if any of them has been advised of the possibility of such damages. This limitation of liability shall apply to any claim or cause whatsoever whether such claim or cause arises in contract, tort or otherwise.
## PREFACE
In the third edition, the comprehensiveness of the second edition is maintained so that all of the topics commonly taught in college algebra are contained in a single source. Recognizing that the use of logarithm tables and determinants is declining, the material on these two areas was reduced, with the two chapters on determinants in the second edition collapsed into a single chapter in the third edition. The material on solving problems using logarithms manually was retained for those who want to learn how to do these problems prior to using a calculator to solve them. Also, the proofs of the properties of determinants were retained to underscore the foundation of the properties used in evaluating determinants.
The book is complete in itself and can be used equally well by those who are in a class studying college algebra for the first time as well as those who wish to review the fundamental principles and procedures of college algebra on their own. Students who are studying advanced algebra in high school will be able to use the book as a source of additional examples, explanations, and problems. The thorough treatment of the topics of algebra allows an instructor to use the book as the textbook for a course, as a resource for material on a specific topic, or as a source for additional problems.
Each chapter contains a summary of the necessary definitions and theorems followed by a set of solved problems. These solved problems include the proofs of theorems and the derivations of formulas. The chapters end with a set of supplementary problems and their answers.
The choice of whether to use a calculator or not is left to the student. A calculator is not required, but it can be used in conjunction with the book. There are no directions on how to use a graphing calculator to do the problems, but there are several instances of the general procedures to be used and the student needs to consult the manual for the calculator being used to see how to implement the procedures on that particular calculator.
DR. ROBERT E. MOYER
_Associate Professor of Mathematics
Southwest Minnesota State University_
## CONTENTS
**CHAPTER 1 Fundamental Operations with Numbers**
1.1 Four Operations
1.2 System of Real Numbers
1.3 Graphical Representation of Real Numbers
1.4 Properties of Addition and Multiplication of Real Numbers
1.5 Rules of Signs
1.6 Exponents and Powers
1.7 Operations with Fractions
**CHAPTER 2 Fundamental Operations with Algebraic Expressions**
2.1 Algebraic Expressions
2.2 Terms
2.3 Degree
2.4 Grouping
2.5 Computation with Algebraic Expressions
**CHAPTER 3 Properties of Numbers**
3.1 Sets of Numbers
3.2 Properties
3.3 Additional Properties
**CHAPTER 4 Special Products**
4.1 Special Products
4.2 Products Yielding Answers of the Form _a_ n ± _b_ n
**CHAPTER 5 Factoring**
5.1 Factoring
5.2 Factorization Procedures
5.3 Greatest Common Factor
5.4 Least Common Multiple
**CHAPTER 6 Fractions**
6.1 Rational Algebraic Fractions
6.2 Operations with Algebraic Fractions
6.3 Complex Fractions
**CHAPTER 7 Exponents**
7.1 Positive Integral Exponent
7.2 Negative Integral Exponent
7.3 Roots
7.4 Rational Exponents
7.5 General Laws of Exponents
7.6 Scientific Notation
**CHAPTER 8 Radicals**
8.1 Radical Expressions
8.2 Laws for Radicals
8.3 Simplifying Radicals
8.4 Operations with Radicals
8.5 Rationalizing Binomial Denominators
**CHAPTER 9 Simple Operations with Complex Numbers**
9.1 Complex Numbers
9.2 Graphical Representation of Complex Numbers
9.3 Algebraic Operations with Complex Numbers
**CHAPTER 10 Equations in General**
10.1 Equations
10.2 Operations Used in Transforming Equations
10.3 Equivalent Equations
10.4 Formulas
10.5 Polynomial Equations
**CHAPTER 11 Ratio, Proportion, and Variation**
11.1 Ratio
11.2 Proportion
11.3 Variation
11.4 Unit Price
11.5 Best Buy
**CHAPTER 12 Functions and Graphs**
12.1 Variables
12.2 Relations
12.3 Functions
12.4 Function Notation
12.5 Rectangular Coordinate System
12.6 Function of Two Variables
12.7 Symmetry
12.8 Shifts
12.9 Scaling
12.10 Using a Graphing Calculator
**CHAPTER 13 Linear Equations in One Variable**
13.1 Linear Equations
13.2 Literal Equations
13.3 Word Problems
**CHAPTER 14 Equations of Lines**
14.1 Slope of a Line
14.2 Parallel and Perpendicular Lines
14.3 Slope–Intercept form of Equation of a Line
14.4 Slope–Point Form of Equation of a Line
14.5 Two-point Form of Equation of a Line
14.6 Intercept Form of Equation of a Line
**CHAPTER 15 Simultaneous Linear Equations**
15.1 Systems of Two Linear Equations
15.2 Systems of Three Linear Equations
**CHAPTER 16 Quadratic Equations in One Variable**
16.1 Quadratic Equations
16.2 Methods of Solving Quadratic Equations
16.3 Sum and Product of the Roots
16.4 Nature of the Roots
16.5 Radical Equations
16.6 Quadratic-type Equations
**CHAPTER 17 Conic Sections**
17.1 General Quadratic Equations
17.2 Conic Sections
17.3 Circles
17.4 Parabolas
17.5 Ellipses
17.6 Hyperbolas
17.7 Graphing Conic Sections with a Calculator
**CHAPTER 18 Systems of Equations Involving Quadratics**
18.1 Graphical Solution
18.2 Algebraic Solution
**CHAPTER 19 Inequalities**
19.1 Definitions
19.2 Principles of Inequalities
19.3 Absolute Value Inequalities
19.4 Higher Degree Inequalities
19.5 Linear Inequalities in Two Variables
19.6 Systems of Linear Inequalities
19.7 Linear Programming
**CHAPTER 20 Polynomial Functions**
20.1 Polynomial Equations
20.2 Zeros of Polynomial Equations
20.3 Solving Polynomial Equations
20.4 Approximating Real Zeros
**CHAPTER 21 Rational Functions**
21.1 Rational Functions
21.2 Vertical Asymptotes
21.3 Horizontal Asymptotes
21.4 Graphing Rational Functions
21.5 Graphing Rational Functions using a Graphing Calculator
**CHAPTER 22 Sequences and Series**
22.1 Sequences
22.2 Arithmetic Sequences
22.3 Geometric Sequences
22.4 Infinite Geometric Series
22.5 Harmonic Sequences
22.6 Means
**CHAPTER 23 Logarithms**
23.1 Definition of a Logarithm
23.2 Laws of Logarithms
23.3 Common Logarithms
23.4 Using a Common Logarithm Table
23.5 Natural Logarithms
23.6 Using a Natural Logarithm Table
23.7 Finding Logarithms Using a Calculator
**CHAPTER 24 Applications of Logarithms and Exponents**
24.1 Introduction
24.2 Simple Interest
24.3 Compound Interest
24.4 Applications of Logarithms
24.5 Applications of Exponents
**CHAPTER 25 Permutations and Combinations**
25.1 Fundamental Counting Principle
25.2 Permutations
25.3 Combinations
25.4 Using a Calculator
**CHAPTER 26 The Binomial Theorem**
26.1 Combinatorial Notation
26.2 Expansion of ( _a_ \+ _x_ ) _n_
**CHAPTER 27 Probability**
27.1 Simple Probability
27.2 Compound Probability
27.3 Mathematical Expectation
27.4 Binomial Probability
27.5 Conditional Probability
**CHAPTER 28 Determinants**
28.1 Determinants of Second Order
28.2 Cramer's Rule
28.3 Determinants of Third Order
28.4 Determinants of Order _n_
28.5 Properties of Determinants
28.6 Minors
28.7 Value of a Determinant of Order _n_
28.8 Cramer's Rule for Determinants of Order _n_
28.9 Homogeneous Linear Equations
**CHAPTER 29 Matrices**
29.1 Definition of a Matrix
29.2 Operations With Matrices
29.3 Elementary Row Operations
29.4 Inverse of a Matrix
29.5 Matrix Equations
29.6 Matrix Solution of a System of Equations
**CHAPTER 30 Mathematical Induction**
30.1 Principle of Mathematical Induction
30.2 Proof by Mathematical Induction
**CHAPTER 31 Partial Fractions**
31.1 Rational Fractions
31.2 Proper Fractions
31.3 Partial Fractions
31.4 Identically Equal Polynomials
31.5 Fundamental Theorem
31.6 Finding the Partial Fraction Decomposition
**APPENDIX A Table of Common Logarithms**
**APPENDIX B Table of Natural Logarithms**
**INDEX**
## CHAPTER 1
Fundamental Operations with Numbers
### **1.1 FOUR OPERATIONS**
Four operations are fundamental in algebra, as in arithmetic. These are addition, subtraction, multiplication, and division.
When two numbers _a_ and _b_ are added, their sum is indicated by _a_ \+ _b_. Thus 3 + 2 = 5.
When a number _b_ is subtracted from a number _a_ , the difference is indicated by _a – b_. Thus 6 – 2 = 4.
Subtraction may be defined in terms of addition. That is, we may define _a_ – _b_ to represent that number _x_ such that _x_ added to _b_ yields _a_ , or _x_ \+ _b_ = _a_. For example, 8 – 3 is that number _x_ which when added to 3 yields 8, i.e., _x_ \+ 3 = 8; thus 8 – 3 = 5.
The product of two numbers _a_ and _b_ is a number _c_ such that _a_ × _b_ = _c_. The operation of multiplication may be indicated by a cross, a dot or parentheses. Thus 5 × 3 = 5 · 3 = 5(3) = (5)(3) = 15, where the factors are 5 and 3 and the product is 15. When letters are used, as in algebra, the notation _p_ × _q_ is usually avoided since × may be confused with a letter representing a number.
When a number _a_ is divided by a number _b_ , the quotient obtained is written
where _a_ is called the dividend and _b_ the divisor. The expression _a/b_ is also called a fraction, having numerator _a_ and denominator _b_.
Division by zero is not defined. See Problems 1.1( _b_ ) and ( _e_ ).
Division may be defined in terms of multiplication. That is, we may consider _a/b_ as that number _x_ which upon multiplication by _b_ yields _a_ , or _bx_ = _a_. For example, 6/3 is that number _x_ such that 3 multiplied by _x_ yields 6, or 3 _x_ – 6; thus 6/3 = 2.
### **1.2 SYSTEM OF REAL NUMBERS**
The system of real numbers as we know it today is a result of gradual progress, as the following indicates.
(1) _Natural numbers_ 1, 2, 3, 4,... (three dots mean "and so on") used in counting are also known as the positive integers. If two such numbers are added or multiplied, the result is always a natural number.
(2) _Positive rational numbers_ or positive fractions are the quotients of two positive integers, such as 2/3, 8/5, 121/17. The positive rational numbers include the set of natural numbers. Thus the rational number 3/1 is the natural number 3.
(3) _Positive irrational numbers_ are numbers which are not rational, such as , _π_.
(4) _Zero_ , written 0, arose in order to enlarge the number system so as to permit such operations as 6 – 6 or 10 – 10. Zero has the property that any number multiplied by zero is zero. Zero divided by any number ≠ 0 (i.e., not equal to zero) is zero.
(5) _Negative_ integers, negative rational numbers and negative irrational numbers such as –3, –2/3, and – , arose in order to enlarge the number system so as to permit such operations as 2 – 8, _π_ – 3π or 2 – 2 .
When no sign is placed before a number, a plus sign is understood. Thus 5 is +5, is + . Zero is considered a rational number without sign.
The real number system consists of the collection of positive and negative rational and irrational numbers and zero.
_Note_. The word "real" is used in contradiction to still other numbers involving , which will be taken up later and which are known as _imaginary_ , although they are very useful in mathematics and the sciences. Unless otherwise specified we shall deal with real numbers.
### **1.3 GRAPHICAL REPRESENTATION OF REAL NUMBERS**
It is often useful to represent real numbers by points on a line. To do this, we choose a point on the line to represent the real number zero and call this point the origin. The positive integers +1, +2, +3,... are then associated with points on the line at distances 1, 2, 3,... units respectively to the _right_ of the origin (see Fig. 1-1), while the negative integers –1, –2, –3,... are associated with points on the line at distances 1, 2, 3,... units respectively to the _left_ of the origin.
**Fig. 1-1**
The rational number 1/2 is represented on this scale by a point _P_ halfway between 0 and +1. The negative number – 3/2 or –1 is represented by a point _R_ 1 units to the left of the origin.
It can be proved that corresponding to each real number there is one and only one point on the line; and conversely, to every point on the line there corresponds one and only one real number.
The position of real numbers on a line establishes an order to the real number system. If a point _A_ lies to the right of another point _B_ on the line we say that the number corresponding to _A_ is _greater_ or _larger_ than the number corresponding to _B_ , or that the number corresponding to _B_ is _less_ or _smaller_ than the number corresponding to _A_. The symbols for "greater than" and "less than" are > and < respectively. These symbols are called "inequality signs."
Thus since 5 is to the right of 3, 5 is greater than 3 or 5 > 3; we may also say 3 is less than 5 and write 3 < 5. Similarly, since –6 is to the left of –4, –6 is smaller than –4, i.e., –6 < –4; we may also write –4 > –6.
By the absolute value or numerical value of a number is meant the distance of the number from the origin on a number line. Absolute value is indicated by two vertical lines surrounding the number. Thus |–6| = 6, |+4| = 4, |–3/4| = 3/4.
### **1.4 PROPERTIES OF ADDITION AND MULTIPLICATION OF REAL NUMBERS**
(1) _Commutative property for addition_ The order of addition of two numbers does not affect the result.
Thus
_a_ \+ _b_ = _b_ \+ _a_ , 5 + 3 = 3 + 5 = 8.
(2) _Associative property for addition_ The terms of a sum may be grouped in any manner without affecting the result.
_a_ \+ _b_ \+ _c_ = _a_ \+ ( _b_ \+ _c_ ) = ( _a_ \+ _b_ ) + _c_ , 3 + 4 + 1 = 3 + (4 + 1) = (3 + 4) + 1 = 8
(3) _Commutative property for multiplication_ The order of the factors of a product does not affect the result.
_a_ · _b_ = _b_ · _a_ , 2 · 5 = 5 · 2 = 10
(4) _Associative property for multiplication_ The factors of a product may be grouped in any manner without affecting the result.
_abc_ = _a_ ( _bc_ ) = ( _ab_ ) _c_ , 3 · 4 · 6 = 3(4 · 6) = (3 · 4)6 = 72
(5) _Distributive property for multiplication over addition_ The product of a number _a_ by the sum of two numbers ( _b_ \+ c) is equal to the sum of the products _ab_ and _ac_.
_a_ ( _b_ \+ _c_ ) = _ab_ \+ _ac_ , 4(3 + 2) = 4 · 3 + 4 · 2 = 20
Extensions of these laws may be made. Thus we may add the numbers _a, b, c, d, e_ by grouping in any order, as ( _a_ \+ _b_ ) + _c_ \+ ( _d_ \+ _e_ ), _a_ \+ ( _b_ \+ _c_ ) + ( _d_ \+ _e_ ), etc. Similarly, in multiplication we may write ( _ab_ ) _c_ ( _de_ ) or _a_ ( _bc_ )( _de_ ), the result being independent of order or grouping.
### **1.5 RULES OF SIGNS**
(1) To add two numbers with like signs, add their absolute values and prefix the common sign. Thus 3 + 4 = 7, (–3) + (–4) = –7.
(2) To add two numbers with unlike signs, find the difference between their absolute values and prefix the sign of the number with greater absolute value.
**EXAMPLES 1.1**. 17 + (–8) = 9, (–6) + 4 = –2, (–18) + 15 = –3
(3) To subtract one number _b_ from another number _a_ , change the operation to addition and replace _b_ by its opposite, – _b_.
**EXAMPLES 1.2**. 12 – (7) = 12 + (–7) = 5, (–9) – (4) = –9 + (–4) = –13, 2 – (–8) = 2 + 8 = 10
(4) To multiply (or divide) two numbers having like signs, mutiply (or divide) their absolute values and prefix a plus sign (or no sign).
**EXAMPLES 1.3**.
(5) To multiply (or divide) two numbers having unlike signs, multiply (or divide) their absolute values and prefix a minus sign.
**EXAMPLES 1.4**.
### **1.6 EXPONENTS AND POWERS**
When a number _a_ is multiplied by itself _n_ times, the product _a_ · _a_ · _a_ ··· _a_ ( _n_ times) is indicated by the symbol _a n_ which is referred to as "the _n_ th power of _a_ " or " _a_ to the _n_ th power" or " _a_ to the _n_ th."
**EXAMPLES 1.5**. 2 · 2 · 2 · 2 · 2 = 25 = 32, (–5)3 = (–5)(–5)(–5) = –125
2 · _x_ · _x_ · _x_ = 2 _x_ 3, _a_ · _a_ · _a_ · _b_ · _b_ = _a_ 3 _b_ 2, ( _a_ – _b_ )( _a_ – _b_ )( _a_ – _b_ ) = ( _a_ – _b_ )3
**In** _a n_, the number _a_ is called the _base_ and the positive integer _n_ is the _exponent_.
If _p_ and _q_ are positive integers, then the following are laws of exponents.
(1) _a p_ · _a q_ = _a p_+ _q_
(2)
(3) ( _a p)q_ = _a pq_
(4)
### **1.7 OPERATIONS WITH FRACTIONS**
Operations with fractions may be performed according to the following rules.
(1) The value of a fraction remains the same if its numerator and denominator are both multiplied or divided by the same number provided the number is not zero.
**EXAMPLES 1.6**.
(2) Changing the sign of the numerator or denominator of a fraction changes the sign of the fraction.
**EXAMPLE 1.7**.
(3) Adding two fractions with a common denominator yields a fraction whose numerator is the sum of the numerators of the given fractions and whose denominator is the common denominator.
**EXAMPLE 1.8**.
(4) The sum or difference of two fractions having different denominators may be found by writing the fractions with a common denominator.
**EXAMPLE 1.9**.
(5) The product of two fractions is a fraction whose numerator is the product of the numerators of the given fractions and whose denominator is the product of the denominators of the fractions.
**EXAMPLES 1.10**.
(6) The reciprocal of a fraction is a fraction whose numerator is the denominator of the given fraction and whose denominator is the numerator of the given fraction. Thus the reciprocal of 3 (i.e., 3/1) is 1/3. Similarly the reciprocals of 5/8 and –4/3 are 8/5 and 3/–4 or –3/4, respectively.
(7) To divide two fractions, multiply the first by the reciprocal of the second.
**EXAMPLES 1.11**.
This result may be established as follows:
### **Solved Problems**
**1.1** Write the sum _S_ , difference _D_ , product _P_ , and quotient _Q_ of each of the following pairs of numbers:
( _a_ ) 48, 12; ( _b_ ) 8, 0; ( _c_ ) 0, 12; ( _d_ ) 10, 20; ( _e_ ) 0, 0.
**SOLUTION**
( _a_ ) _S_ = 48 + 12 = 60, _D_ = 48 – 12 = 36, _P_ = 48(12) = 576,
( _b_ ) _S_ = 8 + 0 = 8, _D_ = 8 – 0 = 8, _P_ = 8(0) = 0, _Q_ = 8 ÷ 0 or 8/0
But by definition 8/0 is that number _x_ (if it exists) such that _x_ (0) = 8. Clearly there is no such number, since any number multiplied by 0 must yield 0.
( _c_ ) _S_ = 0 + 12 = 12, _D_ = 0 – 12 = –12, _P_ = 0(12) = 0,
( _d_ ) _S_ = 10 + 20 = 30, _D_ = 10 – 20 = –10, _P_ = 10(20) = 200,
( _e_ ) _S_ = 0 + 0 = 0, _D_ = 0 – 0 = 0, _P_ = 0(0) = 0, _Q_ = 0 ÷ 0 or 0/0 is by definition that number _x_ (if it exists) such that _x_ (0) = 0. Since this is true for _all_ numbers _x_ there is no one number which 0/0 represents.
From ( _b_ ) and ( _e_ ) it is seen that division by zero is an undefined operation.
**1.2** Perform each of the indicated operations.
( _a_ ) 42 + 23, 23 + 42
( _b_ ) 27 + (48 + 12), (27 + 48) + 12
( _c_ ) 125 – (38 + 27)
( _d_ ) 6 · 8, 8 · 6
( _e_ ) 4(7 · 6), (4 · 7)6
( _f_ ) 35 · 28
( _g_ ) 756 ÷ 21
( _h_ )
( _i_ ) 72 ÷ 24 + 64 ÷ 16
( _j_ ) 4 ÷ 2 + 6 ÷ 3 – 2 ÷ 2 + 3 · 4
( _k_ ) 128 ÷ (2 · 4), (128 ÷ 2) · 4
**SOLUTION**
( _a_ ) 42 + 23 = 65, 23 + 42 = 65. Thus 42 + 23 = 23 + 42.
This illustrates the commutative law for addition.
( _b_ ) 27 + (48 + 12) = 27 + 60 = 87, (27 + 48) + 12 = 75 + 12 = 87. Thus 27 + (48 + 12) = (27 + 48)+ 12.
This illustrates the associative law for addition.
( _c_ ) 125 – (38 + 27) = 125 – 65 = 60
( _d_ ) 6 · 8 = 48, 8 · 6 = 48. Thus 6 · 8 = 8 · 6, illustrating the commutative law for multiplication.
( _e_ ) 4(7 · 6) = 4(42) = 168, (4 · 7)6 = (28)6 = 168. Thus 4(7 · 6) = (4 · 7)6.
This illustrates the associative law for multiplication.
( _f_ ) (35)(28) = 35(20 + 8) = 35(20) + 35(8) = 700 + 280 = 980 by the distributive law for multiplication.
( _g_ )
_h_ )
( _i_ ) Computations in arithmetic, by convention, obey the following rule: Operations of multiplication and division precede operations of addition and subtraction.
Thus 72 ÷ 24 + 64 ÷ 16 = 3 + 4 = 7.
( _j_ ) The rule of ( _i_ ) is applied here. Thus 4 ÷ 2 + 6 ÷ 3 – 272 + 3 · 4 = 2 + 2 – 1 + 12 = 15.
( _k_ ) 128 ÷ (2 · 4) = 128 ÷ 8 = 16, (128 ÷ 2) · 4 = 64 · 4 = 256
Hence if one wrote 128 ÷ 2 · 4 without parentheses we would do the operations of multiplication and division in the order they occur from left to right, so 128 ÷ 2 · 4 = 64 · 4 = 256.
**1.3** Classify each of the following numbers according to the categories: real number, positive integer, negative integer, rational number, irrational number, none of the foregoing.
**SOLUTION**
If the number belongs to one or more categories this is indicated by a check mark.
**1.4** Represent (approximately) by a point on a graphical scale each of the real numbers in Problem 1.3.
Note: 3 _π_ is approximately 3(3.14) = 9.42, so that the corresponding point is between +9 and +10 as indicated. is between 2 and 3, its value to three decimal places being 2.236.
**1.5** Place an appropriate inequality symbol (< or >) between each pair of real numbers.
( _a_ ) 2, 5
( _b_ ) 0, 2
( _c_ ) 3, –1
( _d_ ) –4, +2
( _e_ ) –4, –3
( _f_ ) _π_ , 3
( _g_ )
( _h_ ) – , –1
( _i_ ) –3/5, –1/2
**SOLUTION**
( _a_ ) 2 < 5 (or 5 > 2), i.e., 2 is _less than_ 5 (or 5 is _greater than 2_ )
( _b_ ) 0 < 2 (or 2 > 0)
( _c_ ) 3 > –1 (or –1 <3)
( _d_ ) –4 < +2 (or +2 > –4)
( _e_ ) –4 < –3 (or –3 > –4)
( _f_ ) π > 3 (or 3 < π)
( _g_ )
( _h_ ) – < –1 (–1 > – )
( _i_ ) –3/5 < –1/2 since –.6 < –.5
**1.6** Arrange each of the following groups of real numbers in ascending order of magnitude.
( _a_ ) –3, 22/7, , –3.2, 0
( _b_ )
**SOLUTION**
( _a_ ) –3.2 < –3 < 0 < < 22/7
( _b_ )
**1.7** Write the absolute value of each of the following real numbers.
**SOLUTION**
We may write the absolute values of these numbers as
which in turn may be written 1, 3, 2/5, , 3.14, 2.83, 3/8, _π_ , 5/7 respectively.
**1.8** The following illustrate addition and subtraction of real numbers.
( _a_ ) (–3) + (–8) = –11
( _b_ ) (–2) + 3 = 1
( _c_ ) (–6) + 3 = –3
( _d_ ) –2 + 5 = 3
( _e_ ) –15 + 8 = –7
( _f_ ) (–32) + 48 + (–10) = 6
( _g_ ) 50 – 23 – 27 = 0
( _h_ ) –3 – (–4) = –3 + 4 = 1
( _i_ ) –(–14) + (–2) = 14 – 2 = 12
**1.9** Write the sum S, difference D, product _P_ , and quotient Q of each of the following pairs of real numbers:
( _a_ ) –2, 2;
( _b_ ) –3, 6;
( _c_ ) 0, –5;
( _d_ ) –5, 0
**SOLUTION**
( _a_ ) _S_ = –2 + 2 = 0, _D_ = (–2) – 2 = –4, _P_ = (–2)(2) = –4, _Q_ = –2/2 = –1
( _b_ ) _S_ = (–3) + 6 = 3, _D_ = (-3) – 6 = –9, _P_ = (–3)(6) = –18, _Q_ = –3/6 = –1/2
( _c_ ) _S_ = 0 + (–5) = –5, _D_ = 0 – (–5) = 5, _P_ = (0)(–5) = 0, _Q_ = 0/–5 = 0
( _d_ ) _S_ = (–5) + 0 = –5, _D_ = (–5) – 0 = –5, _P_ = (–5)(0) = 0, _Q_ = –5/0 (an undefined operation, so it is not a number).
**1.10** Perform the indicated operations.
( _a_ ) (5)(–3)(–2) = (5)(–3) = (–15)(–2) = 30
= (5)[(–3)(–2)] = (5)(6) = 30
The arrangement of the factors of a product does not affect the result.
( _b_ ) 8(–3)(10) = –240
( _c_ )
( _d_ )
**1.11** Evaluate the following.
( _a_ ) 23 = 2 · 2 · 2 = 8
( _b_ ) 5(3)2 = 5 · 3 · 3 = 45
( _c_ ) 24 · 26 = 24+6 = 210 = 1024
( _d_ ) 25 · 52 = (32)(25) = 800
( _e_ )
( _f_ )
( _g_ ) (23)2 = 23·2 = 26 = 64
( _h_ )
( _i_ )
( _j_ )
**1.12** Write each of the following fractions as an equivalent fraction having the indicated denominator.
( _a_ ) 1/3; 6
( _b_ ) 3/4; 20
( _c_ ) 5/8; 48
( _d_ ) –3/7; 63
( _e_ ) –12/5; 75
**SOLUTION**
( _a_ ) To obtain the denominator 6, multiply numerator and denominator of the fraction 1/3 by 2.
Then
( _b_ )
( _c_ )
( _d_ )
( _e_ )
**1.13** Find the sum _S_ , difference _D_ , product _P_ , and quotient Q of each of the following pairs of rational numbers: ( _a_ ) 1/3, 1/6; ( _b_ ) 2/5, 3/4; ( _c_ ) –4/15, –11/24.
**SOLUTION**
( _a_ ) 1/3 may be written as the equivalent fraction 2/6.
( _b_ ) 2/5 and 3/4 may be expressed with denominator 20: 2/5 = 8/20, 3/4 = 15/20.
( _c_ ) –4/15 and –11/24 have a least common denominator 120: –4/15 = –32/120, –11/24 = –55/120.
**1.14** Evaluate the following expressions, given _x_ = 2, _y_ = –3, _z_ = 5, _a_ = 1/2, _b_ = –2/3.
( _a_ ) 2 _x_ \+ _y_ = 2(2) + (–3) = 4 – 3 = 1
( _b_ ) 3 _x_ – 2 _y_ – 4 _z_ = 3(2) – 2(–3) – 4(5) = 6 + 6 – 20 = –8
( _c_ ) 4 _x_ 2 _y_ = 4(2)2(–3) = 4 · 4 · (–3) = –48
( _d_ )
( _e_ )
### **Supplementary Problems**
**1.15** Write the sum _S_ , difference _D_ , product _P_ , and quotient _Q_ of each of the following pairs of numbers:
( _a_ ) 54, 18;
( _b_ ) 4, 0;
( _c_ ) 0, 4;
( _d_ ) 12, 24;
( _e_ ) 50, 75.
**1.16** Perform each of the indicated operations.
( _a_ ) 38 + 57, 57 + 38
( _b_ ) 15 + (33 + 8), (15 + 33) + 8
( _c_ ) (23 + 64) – (41 + 12)
( _d_ ) 12 · 8, 8 · 12
( _e_ ) 6(4 · 8), (6 · 4)8
( _f_ ) 42 · 68
( _g_ ) 1296 ÷ 36
( _h_ )
( _i_ ) 45 ÷ 15 + 84 ÷ 12
( _j_ ) 10 ÷ 5 – 4 ÷ 2 + 15 ÷ 3 + 2 · 5
( _k_ ) 112 ÷ (4 · 7), (112 ÷ 4) · 7
( _l_ )
**1.17** Place an appropriate inequality symbol (< or >) between each of the following pairs of real numbers.
( _a_ ) 4, 3
( _b_ ) –2, 0
( _c_ ) –1, 2
( _d_ ) 3, –2
( _e_ ) –8, –7
( _f_ ) 1,
( _g_ )
( _h_ ) –1/3, –2/5
**1.18** Arrange each of the following groups of real numbers in ascending order of magnitude.
( _a_ )
( _b_ )
**1.19** Write the absolute value of each of the following real numbers:
**1.20** Evaluate.
( _a_ ) 6 + 5
( _b_ ) (–4) + (–6)
( _c_ ) (–4) + 3
( _d_ ) 6 + (–4)
( _e_ ) –8 + 4
( _f_ ) –4 + 8
( _g_ ) (–18) + (–3) + 22
( _h_ ) 40 – 12 + 4
( _i_ ) –12–(–8)
( _j_ ) –(–16) – (–12) + (–5) – 15
**1.21** Write the sum _S_ , difference _D_ , product _P_ , and quotient _Q_ of each of the following pairs of real numbers:
( _a_ ) 12, 4;
( _b_ ) –6, –3;
( _c_ ) –8, 4;
( _d_ ) 0, –4;
( _e_ ) 3, –2.
**1.22** Perform the indicated operations.
( _a_ ) (–3)(2)(–6)
( _b_ ) (6)(–8)(–2)
( _c_ ) 4(–1)(5) + (–3)(2)(–4)
( _d_ )
( _e_ ) (–8) ÷ (–4) + (–3)(2)
( _f_ )
**1.23** Evaluate.
( _a_ ) 33
( _b_ ) 3(4)2
( _c_ ) 24 · 23
( _d_ ) 42 · 32
( _e_ )
( _f_ )
( _g_ )
( _h_ ) (32)3
( _i_ )
( _j_ )
( _k_ )
( _l_ )
**1.24** Write each of the following fractions as an equivalent fraction having the indicated denominator.
( _a_ ) 2/5; 15
( _b_ ) –4/7; 28
( _c_ ) 5/16; 64
( _d_ ) –10/3; 42
( _e_ ) 11/12; 132
( _f_ ) 17/18; 90
**1.25** Find the sum _S_ , difference _D_ , product _P_ , and quotient _Q_ of each of the following pairs of rational numbers:
( _a_ ) 1/4, 3/8;
( _b_ ) 1/3, 2/5;
( _c_ ) –4, 2/3;
( _d_ ) –2/3, –3/2.
**1.26** Evaluate the following expressions, given _x_ = –2, _y_ = 4, _z_ = 1/3, _a_ = –1, _b_ = 1/2.
( _a_ ) 3 _x_ – 2 _y_ \+ 6 _z_
( _b_ ) 2 _xy_ \+ 6 _az_
( _c_ ) 4 _b_ 2 _x_ 3
( _d_ )
( _e_ )
( _f_ )
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**1.15** ( _a_ ) _S_ = 72, _D_ = 36, _P_ = 972, _Q_ = 3
( _b_ ) _S_ = 4, _D_ = 4, _P_ = 0, _Q_ undefined
( _c_ ) _S_ = 4, _D_ = –4, _P_ = 0, _Q_ = 0
( _d_ ) _S_ = 36, _D_ = –12, _P_ = 288, _Q_ = 1/2
( _e_ ) _S_ = 125, _D_ = –25, _P_ = 3750, _Q_ = 2/3
**1.16** ( _a_ ) 95, 95
( _b_ ) 56, 56
( _c_ ) 34
( _d_ ) 96, 96
( _e_ ) 192, 192
( _f_ ) 2856
( _g_ ) 36
( _h_ ) 30
( _i_ ) 10
( _j_ ) 15
( _k_ ) 4, 196
( _l_ ) 3
**1.17** ( _a_ ) 3 < 4 or 4 > 3
( _b_ ) –2 < 0 or 0 > –2
( _c_ ) –1 < 2 or 2 > –1
( _d_ ) –2 < 3 or 3 > –2
( _e_ ) –8 < –7 or –7 > –8
( _f_ )
( _g_ )
( _h_ ) –2/5 < –1/3 or –1/3 > –2/5
**1.18** ( _a_ )
( _b_ )
**1.19**
**1.20** ( _a_ ) 11
( _b_ ) –10
( _c_ ) –1
( _d_ ) 2
( _e_ ) –4
( _f_ ) 4
( _g_ ) 1
( _h_ ) 32
( _i_ ) –4
( _j_ ) 8
**1.21** ( _a_ ) _S_ = 16, _D_ = 8, _P_ = 48, _Q_ = 3
( _b_ ) _S_ = –9, _D_ = –3, _P_ = 18, _Q_ = 2
( _c_ ) _S_ = –4, _D_ = –12, _P_ = –32, _Q_ = –2
( _d_ ) _S_ = –4, _D_ = 4, _P_ = 0, _Q_ = 0
( _e_ ) _S_ = 1, _D_ = 5, _P_ = –6, _Q_ = –3/2
**1.22** ( _a_ ) 36
( _b_ ) 96
( _c_ ) 4
( _d_ ) 20
( _e_ ) –4
( _f_ ) 1
**1.23**
( _a_ ) 27
( _b_ ) 48
( _c_ ) 128
( _d_ ) 144
( _e_ ) 54 = 625
( _f_ ) 3
( _g_ ) 1/49
( _h_ ) 36 = 729
( _i_ ) 1/2
( _j_ ) –4/3
( _k_ ) 5
( _l_ ) –201
**1.24** ( _a_ ) 6/15
( _b_ ) –16/28
( _c_ ) 20/64
( _d_ ) –140/42
( _e_ ) 121/132
( _f_ ) 85/90
**1.25** ( _a_ ) _S_ = 5/8, _D_ = –1/8, _P_ = 3/32, _Q_ = 2/3
( _b_ ) _S_ = 11/15, _D_ = –1/15, _P_ = 2/15, _Q_ = 5/6
( _c_ ) _S_ = –10/3, _D_ = –14/3, _P_ = –8/3, _Q_ = –6
( _d_ ) _S_ = –13/6, _D_ = 5/6, _P_ = 1, _Q_ = 4/9
**1.26** ( _a_ ) –12
( _b_ ) –18
( _c_ ) –8
( _d_ ) 14
( _e_ ) 16/5
( _f_ ) 48
## **CHAPTER 2
Fundamental Operations with Algebraic Expressions**
### **2.1 ALGEBRAIC EXPRESSIONS**
An algebraic expression is a combination of ordinary numbers and letters which represent numbers.
Thus
are algebraic expressions.
A term consists of products and quotients of ordinary numbers and letters which represent numbers. Thus 6 _x_ 2 _y_ 3, 5 _x_ /3 _y_ 4, –3 _x_ 7 are terms.
However, 6 _x_ 2 \+ 7 _xy_ is an algebraic expression consisting of two terms.
A monomial is an algebraic expression consisting of only one term. Thus 7 _x_ 3y4, 3 _xyz_ 2, 4 _x_ 2/ _y_ are monomials.
Because of this definition, monomials are sometimes simply called terms.
A binomial is an algebraic expression consisting of two terms. Thus 2 _x_ \+ 4 _y_ , 3 _x_ 4 – 4 _xyz_ 3 are binomials.
A trinomial is an algebraic expression consisting of three terms. Thus 3 _x_ 2 – 5 _x_ 2, 2 _x_ \+ 6 _y_ – 3 _z_ , _x_ 3 – 3 _xy/z_ – 2 _x_ 3 _z_ 7 are trinomials.
A multinomial is an algebraic expression consisting of more than one term. Thus 7 _x_ \+ 6 _y_ , 3 _x_ 3 \+ 6 _x_ 2 _y_ 7 _xy_ \+ 6, 7 _x_ \+ 5 _x_ 2/ _y_ – 3 _x_ 3/16 are multinomials.
### **2.2 TERMS**
One factor of a term is said to be the coefficient of the rest of the term. Thus in the term 5 _x_ 3 _y_ 2, 5 _x_ 3 is the coefficient of _y_ 2, 5 _y_ 2 is the coefficient of _x_ 3, and 5 is the coefficient of _x_ 3 _y_ 2.
If a term consists of the product of an ordinary number and one or more letters, we call the number the numerical coefficient (or simply the coefficient) of the term. Thus in –5 _x_ 3 _y_ 2, –5 is the numerical coefficient or simply the coefficient.
Like terms, or similar terms, are terms which differ only in numerical coefficients, For example, 7 _xy_ and –2 _xy_ are like terms; 3 _x_ 2 _y_ 4 and – _x_ 2 _y_ 4 are like terms; however, –2 _a_ 2 _b_ 3 and –3 _a_ 2 _b_ 7 are unlike terms.
Two or more like terms in an algebraic expression may be combined into one term. Thus 7 _x_ 2 _y_ – 4 _x_ 2 _y_ \+ 2 _x_ 2 _y_ may be combined and written 5 _x_ 2 _y_.
A term is integral and rational in certain literals (letters which represent numbers) if the term consists of
( _a_ ) positive integer powers of the variables multiplied by a factor not containing any variable, or
( _b_ ) no variables at all.
For example, the terms 6 _x_ 2 _y_ 3, –5 _y_ 4, 7, –4 _x_ , and are integral and rational in the variables present. However, is not rational in _x_ , 4/ _x_ is not integral in _x_.
A polynomial is a monomial or multinomial in which every term is integral and rational.
For example, 3 _x_ 2 _y_ 3 – 5 _x_ 4 _y_ \+ 2, 2 _x_ 4 – 7 _x_ 3 \+ 3 _x_ 2 – 5 _x_ \+ 2, 4 _xy_ \+ _z_ , and 3 _x_ 2 are polynomials. However, 3 _x_ 2 – 4/ _x_ and are not polynomials.
### **2.3 DEGREE**
The degree of a monomial is the sum of all the exponents in the variables in the term. Thus the degree of 4 _x_ 3 _y_ 2 _z_ is 3 + 2 + 1 = 6. The degree of a constant, such as 6, 0, – , or _π_ , is zero.
The degree of a polynomial is the same as that of the term having highest degree and non-zero coefficient. Thus 7 _x_ 3 _y_ 2 – 4 _xz_ 5 \+ 2 _x_ 3 _y_ has terms of degree 5, 6, and 4 respectively; hence the degree of the polynomial is 6.
### **2.4 GROUPING**
A symbol of grouping such as parentheses ( ), brackets [ ], or braces { } is often used to show that the terms contained in them are considered as a single quantity.
For example, the sum of two algebraic expressions 5 _x_ 2 – 3 _x_ \+ _y_ and 2 _x_ – 3 _y_ may be written (5 _x_ – 3 _x_ \+ _y_ ) + (2 _x_ – 3 _y_ ). The difference of these may be written (5 _x_ 2 – 3 _x_ \+ _y_ ) – (2 _x_ – 3 _y_ ), and their product (5 _x_ 2 – 3 _x_ \+ _y_ )(2 _x_ – 3 _y_ ).
Removal of symbols of grouping is governed by the following laws.
(1) If a + sign precedes a symbol of grouping, this symbol of grouping may be removed without affecting the terms contained.
Thus
(2) If a – sign precedes a symbol of grouping, this symbol of grouping may be removed if each sign of the terms contained is changed.
Thus
(3) If more than one symbol of grouping is present, the inner ones are to be removed first.
Thus
### **2.5 COMPUTATION WITH ALGEBRAIC EXPRESSIONS**
Addition of algebraic expressions is achieved by combining like terms. In order to accomplish this addition, the expressions may be arranged in rows with like terms in the same column; these columns are then added.
**EXAMPLE 2.1**. Add 7 _x_ \+ 3 _y_ 3 – 4 _xy_ , 3 _x_ – 2 _y_ 3 \+ 7 _xy_ , and 2 _xy_ – 5 _x_ – 6 _y_ 3.
Subtraction of two algebraic expressions is achieved by changing the sign of every term in the expression which is being subtracted (sometimes called the subtrahend) and adding this result to the other expression (called the minuend).
**EXAMPLE 2.2.** Subtract 2 _x_ 2 – 3 _xy_ \+ 5 _y_ 2 from 10 _x_ 2 – 2 _xy_ – 3 _y_ 2.
We may also write (10 _x_ 2 – 2 _xy_ – 3 _y_ 2) – (2 _x_ 2 – 3 _xy_ \+ 5 _y_ 2) = 10 _x_ 2 – 2 _xy_ – 3 _y_ 2 – 2 _x_ 2 \+ 3 _xy_ – 5 _y_ 2 = 8 _x_ 2 \+ _xy_ – 8 _y_ 2.
Multiplication of algebraic expressions is achieved by multiplying the terms in the factors of the expressions.
(1) To multiply two or more monomials: Use the laws of exponents, the rules of signs, and the commutative and associative properties of multiplication.
**EXAMPLE 2.3.** Multiply –3 _x_ 2 _y_ 3 _z_ , 2 _x_ 4 _y_ , and –4 _xy_ 4 _z_ 2.
Write
(–3 _x_ 2 _y_ 3 _z_ )(2 _x_ 4 _y_ )(–4 _xy_ 2 _z_ 2).
Arranging according to the commutative and associative laws,
Combine using rules of signs and laws of exponents to obtain
24 _x_ 7 _y_ 8 _z_ 3.
Step ( _1_ ) may be done mentally when experience is acquired.
(2) To multiply a polynomial by a monomial: Multiply each term of the polynomial by the monomial and combine results.
**EXAMPLE 2.4.** Multiply 3 _xy_ – 4 _x_ 3 \+ 2 _xy_ 2 by 5 _x_ 2 _y_ 4.
Write
(3) To multiply a polynomial by a polynomial: Multiply each of the terms of one polynomial by each of the terms of the other polynomial and combine results.
It is very often useful to arrange the polynomials according to ascending (or descending) powers of one of the letters involved.
**EXAMPLE 2.5.** Multiply –3 _x_ \+ 9 + _x_ 2 by 3 – _x_.
Arranging in descending powers of _x_ ,
Division of algebraic expressions is achieved by using the division laws of exponents.
(1) To divide a monomial by a monomial: Find the quotient of the numerical coefficients, find the quotients of the variables, and multiply these quotients.
**EXAMPLE 2.6.** Divide 24 _x_ 4 _y_ 2 _z_ 3 by –3 _x_ 3 _y_ 4 _z_.
Write
(2) To divide a polynomial by a polynomial:
( _a_ ) Arrange the terms of both polynomials in descending (or ascending) powers of one of the variables common to both polynomials.
( _b_ ) Divide the first term in the dividend by the first term in the divisor. This gives the first term of the quotient.
( _c_ ) Multiply the first term of the quotient by the divisor and subtract from the dividend, thus obtaining a new dividend.
( _d_ ) Use the dividend obtained in ( _c_ ) to repeat steps ( _b_ ) and ( _c_ ) until a remainder is obtained which is either of degree lower than the degree of the divisor or zero.
( _e_ ) The result is written:
**EXAMPLE 2.7.** Divide _x_ 2 \+ 2 _x_ 4 – 3 _x_ 3 \+ _x_ – 2 by _x_ 2 – 3 _x_ \+ 2.
Write the polynomials in descending powers of _x_ and arrange the work as follows.
Hence
### **Solved Problems**
**2.1** Evaluate each of the following algebraic expressions, given that _x_ = 2, _y_ = –1, _z_ = 3, _a_ = 0, _b_ = 4, _c_ = 1/3.
( _a_ ) 2 _x_ 2 – 3 _yz_ = 2(2)2 – 3(–1)(3) = 8 + 9 = 17
( _b_ ) 2 _z_ 4 – 3 _z_ 3 \+ 4 _z_ 2 – 2 _z_ \+ 3 = 2(3)4 – 3(3)3 \+ 4(3)2 – 2(3) + 3 = 162 – 81 + 36 – 6 + 3 = 114
( _c_ ) 4 _a_ 2 – 3 _ab_ \+ 6 _c_ = 4(0)2 – 3(0)(4) + 6(1/3) = 0 – 0 + 2 = 2
( _d_ )
( _e_ )
( _f_ )
**2.2** Classify each of the following algebraic expressions according to the categories: term or monomial, binomial, trinomial, multinomial, polynomial.
( _a_ ) _x_ 3 \+ 3 _y_ 2 _z_
( _b_ ) 2 _x_ 2 – 5 _x_ \+ 3
( _c_ ) 4 _x_ 2 _y/z_
( _d_ ) _y_ \+ 3
( _e_ )
( _f_ ) 5 _x_ 3 \+ 4/ _y_
( _g_ )
( _h_ )
( _i_ ) _a_ 3 \+ _b_ 3 \+ _c_ 3 – 3 _abc_
**SOLUTION**
If the expression belongs to one or more categories, this is indicated by a check mark.
**2.3** Find the degree of each of the following polynomials.
( _a_ ) 2 _x_ 3 _y_ \+ 4 _xyz_ 4. The degree of 2 _x_ 3 _y_ is 4 and that of 4 _xyz_ 4 is 6; hence the polynomial is of degree 6.
( _b_ ) _x_ 2 \+ 3 _x_ 3 – 4. The degree of _x_ 2 is 2, of 3 _x_ 3 is 3, and of –4 is 0; hence the degree of the polynomial is 3.
( _c_ ) _y_ 3 – 3 _y_ 2 \+ 4 _y_ – 2 is of degree 3.
( _d_ ) _xz_ 3 \+ 3 _x_ 2 _z_ 2 – 4 _x_ 3 _z_ \+ _x_ 4. Each term is of degree 4; hence the polynomial is of degree 4.
( _e_ ) _x_ 2 – 105 is of degree 2. (The degree of the constant 105 is zero.)
**2.4** Remove the symbols of grouping in each of the following and simplify the resulting expressions by combining like terms.
( _a_ ) 3 _x_ 2 \+ ( _y_ 2 – 4 _z_ ) – (2 _x_ – 3 _y_ \+ 4 _z_ ) = 3 _x_ 2 \+ _y_ 2 – 4 _z_ – 2 _x_ \+ 3 _y_ – 4 _z_ = 3 _x_ 2 \+ _y_ 2 – 2 _x_ \+ 3 _y_ – 8 _z_
( _b_ ) 2(4 _xy_ \+ 3 _z_ ) + 3( _x_ – 2 _xy_ ) – 4( _z_ – 2 _xy_ ) = 8 _xy_ \+ 6 _z_ \+ 3 _x_ – 6 _xy_ – 4 _z_ \+ 8 _xy_ = 10 _xy_ \+ 3 _x_ \+ 2 _z_
( _c_ ) _x_ – 3 – 2{2 – 3( _x_ – _y_ )} = _x_ – 3 – 2{2 – 3 _x_ \+ 3y} = _x_ – 3 – 4 + 6 _x_ – 6 _y_ = 7 _x_ –6 _y_ –7
( _d_ )
**2.5** Add the algebraic expressions in each of the following groups.
( _a_ ) _x_ 2 \+ _y_ 2 – _z_ 2 \+ 2 _xy_ – 2 _yz_ , _y_ 2 \+ _z_ 2 – _x_ 2 \+ 2 _yz_ – 2 _zx_ , _z_ 2 \+ _x_ 2 – _y_ 2 \+ 2 _zx_ – 2 _xy_ , 1 – _x_ 2 – _y_ 2 – _z_ 2
**SOLUTION**
( _b_ )
**SOLUTION**
**2.6** Subtract the second of each of the following expressions from the first.
( _a_ ) _a_ – _b_ \+ _c_ – _d_ , _c_ – _a_ \+ _d_ – _b_.
**SOLUTION**
Otherwise: ( _a_ – _b_ \+ _c_ – _d_ ) – ( _c_ – _a_ \+ _d_ – _b_ ) = _a_ – _b_ \+ _c_ – _d_ – _c_ \+ _a_ – _d_ \+ _b_ = 2 _a_ – 2 _d_
( _b_ ) 4 _x_ 2 _y_ – 3 _ab_ \+ 2 _a_ 2 – _xy_ , 4 _xy_ \+ _ab_ 2 – 3 _a_ 2 \+ 2 _ab_.
**SOLUTION**
**2.7** In each of the following find the indicated product of the algebraic expressions.
( _a_ ) (–2 _ab_ 3)(4 _a_ 2 _b_ 5)
( _b_ ) (–3 _x_ 2 _y_ )(4 _xy_ 2)(–2 _x_ 3 _y_ 4)
( _c_ ) (3 _ab_ 2)(2 _ab_ \+ _b_ 2)
( _d_ ) ( _x_ 2 – 3 _xy_ \+ _y_ 2)(4 _xy_ 2)
( _e_ ) ( _x_ 2 – 3 _x_ \+ 9)( _x_ \+ 3)
( _f_ ) ( _x_ 4 \+ _x_ 3 _y_ \+ _x_ 2 _y_ 2 \+ _xy_ 3 \+ _y_ 4)( _x_ – _y_ )
( _g_ ) ( _x_ 2 – _xy_ \+ _y_ 2)( _x_ 2 \+ _xy_ \+ _y_ 2)
( _h_ ) (2 _x_ \+ _y_ – _z_ )(3 _x_ – _z_ \+ _y_ )
**SOLUTION**
( _a_ ) (–2 _ab_ 3)(4 _a_ 2 _b_ 5) = {(–2)(4)}{( _a_ )( _a_ 2)}{ ( _b_ 3)( _b_ 5)} = –8 _a_ 3 _b_ 8
( _b_ ) (–3 _x_ 2 _y_ )(4 _xy_ 2)(–2 _x_ 3 _y_ 4) = {(–3)(4)(–2)}{( _x_ 2)( _x_ )( _x_ 3)}{( _y_ )( _y_ 2)( _y_ 4)} = 24 _x_ 6 _y_ 7
( _c_ ) (3 _ab_ 2)(2 _ab_ \+ _b_ 2) = (3 _ab_ 2)(2 _ab_ ) + (3 _ab_ 2)( _b_ 2) = 6 _a_ 2 _b_ 3 \+ 3 _ab_ 4
( _d_ ) ( _x_ 2 – 3 _xy_ \+ _y_ 2)(4 _xy_ 2) = ( _x_ 2)(4 _xy_ 2) + (–3 _xy_ )(4 _xy_ 2) + ( _y_ 2)(4 _xy_ 2) = 4 _x_ 3 _y_ 2 – 12 _x_ 2 _y_ 3 \+ 4 _xy_ 4
( _e_ )
( _f_ )
( _g_ )
( _h_ )
**2.8** Perform the indicated divisions.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
Thus
( _g_ )
Arrange in descending powers of _x_.
( _h_ )
Arrange in descending powers of a letter, say _x_.
**2.9** Check the work in Problems 2.7( _h_ ) and 2.8( _g_ ) by using the value _x_ = 1, _y_ = –1, _z_ = 2.
**SOLUTION**
From Problem 2.7( _h_ ), (2 _x_ \+ _y_ – _z_ )(3 _x_ – _z_ \+ _y_ ) = 6 _x_ 2 \+ 5 _xy_ – 5 _xz_ – 2 _yz_ \+ _z_ 2 \+ _y_ 2.
Substitute _x_ = 1, _y_ = –1, _z_ = 2 and obtain
[2(1) + (–1) – 2][3(1) – (2) – 1] = 6(1)2 \+ 5(1)(–1) – 5(1)(2) – 2 (–1)(2) + (2)2 [–1]2
or
[–1][0] = 6 – 5 – 10 + 4 + 4 + 1, i.e. 0 = 0.
From Problem 2.8( _g_ ),
Put _x_ = 1 and obtain
Although a check by substitution of numbers for variables is not conclusive, it can be used to indicate possible errors.
### **Supplementary Problems**
**2.10** Evaluate each algebraic expression, given that _x_ = –1, _y_ = 3, _z_ = 2, _a_ = 1/2, _b_ = –2/3.
( _a_ ) 4 _x_ 3 _y_ 2 – 3 _xz_ 2
( _b_ ) ( _x_ – _y_ ) ( _y_ – _z_ )( _z_ – _x_ )
( _c_ ) 9 _ab_ 2 \+ 6 _ab_ – 4 _a_ 2
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
**2.11** Determine the degree of each of the following polynomials.
( _a_ ) 3 _x_ 4 – 2 _x_ 3 \+ _x_ 2 – 5
( _b_ ) 4 _xy_ 4 – 3 _x_ 3 _y_ 3
( _c_ ) _x_ 5 \+ _y_ 5 \+ _z_ 5 – 5 _xyz_
( _d_ )
( _e_ ) –103
( _f_ ) _y_ 2 – 3 _y_ 5 – _y_ \+ 2 _y_ 3 – 4
**2.12** Remove the symbols of grouping and simplify the resulting expressions by combining like terms.
( _a_ ) ( _x_ \+ 3 _y_ – _z_ ) – (2 _y_ – _x_ \+ 3 _z_ ) + (4 _z_ – 3 _x_ \+ _2y_ )
( _b_ ) 3( _x_ 2 – 2 _yz_ \+ _y_ 2) – 4( _x_ 2 – _y_ 2 3 _yz_ ) + _x_ 2 \+ _y_ 2
( _c_ ) 3 _x_ \+ 4 _y_ \+ 3{ _x_ – 2( _y_ – _x_ ) – _y_ }
( _d_ ) 3 – {2 _x_ – [1 – ( _x_ \+ _y_ )]+ [ _x_ – 2 _y_ ]}
**2.13** Add the algebraic expressions in each of the following groups.
( _a_ ) 2 _x_ 2 \+ _y_ 2 – _x_ \+ _y_ , 3 _y_ 2 \+ _x_ – _x_ 2, _x_ – 2 _y_ \+ _x_ 2 – 4 _y_ 2
( _b_ ) _a_ 2 – _ab_ \+ 2 _bc_ \+ 3 _c_ 2, 2 _ab_ \+ _b_ 2 – 3 _bc_ – 4 _c_ 2, _ab_ – 4 _bc_ \+ c2 – _a_ 2, _a_ 2 \+ 2 _c_ 2 \+ 5 _bc_ – 2 _ab_
( _c_ ) 2 _a_ 2 _bc_ – 2 _acb_ 2 \+ 5 _c_ 2 _ab_ , 4 _b_ 2 _ac_ \+ 4 _bca_ 2 – 7 _ac_ 2 _b_ , 4 _abc_ 2 – 3 _a_ 2 _bc_ – 3 _ab_ 2 _c_ , _b_ 2 _ac_ – _abc_ 2 – 3 _a_ 2 _bc_
**2.14** Subtract the second of each of the following expressions from the first.
( _a_ ) 3 _xy_ – 2 _yz_ \+ 4 _zx_ , 3 _zx_ \+ _yz_ – 2 _xy_
( _b_ ) 4 _x_ 2 \+ 3 _y_ 2 – 6 _x_ \+ 4 _y_ – 2, 2 _x_ – _y_ 2 \+ 3 _x_ 2 – 4 _y_ \+ 3
( _c_ ) _r_ 3 – 3 _r_ 2 _s_ \+ 4 _rs_ 2 – _s_ 3, 2 _s_ 3 \+ 3 _s_ 2 _r_ – 2 _sr_ 2 – 3 _r_ 3
**2.15** Subtract _xy_ – 3 _yz_ \+ 4 _xz_ from twice the sum of the following expressions: 3 _xy_ – 4 _yz_ \+ 2 _xz_ and 3 _yz_ – 4 _zx_ – 2 _xy_.
**2.16** Obtain the product of the algebraic expressions in each of the following groups.
( _a_ ) 4 _x_ 2 _y_ 5, –3 _x_ 3 _y_ 2
( _b_ ) 3 _abc_ 2, –2 _a_ 3 _b_ 2 _c_ 4, 6 _a_ 2 _b_ 2
( _c_ ) –4 _x_ 2 _y_ , 3 _xy_ 2 – 4 _xy_
( _d_ ) _r_ 2 _s_ \+ 3 _rs_ 3 – 4 _rs_ \+ _s_ 3, 2 _r_ 2 _s_ 4
( _e_ ) _y_ – 4, _y_ \+ 3
( _f_ ) _y_ 2 – 4 _y_ \+ 16, _y_ \+ 4
( _g_ ) _x_ 3 \+ _x_ 2 _y_ \+ _xy_ 2 \+ _y_ 3, _x_ – _y_
( _h_ ) _x_ 2 \+ 4 _x_ \+ 8, _x_ 2 – 4 _x_ \+ 8
( _i_ ) 3 _r_ – _s_ – _t_ 2, 2 _s_ \+ _r_ \+ 3 _t_ 2
( _j_ ) 3 – _x_ – _y_ , 2 _x_ \+ _y_ \+ 1, _x_ – _y_
**2.17** Perform the indicated divisions.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
**2.18** Perform the indicated divisions.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
**2.19** Perform the indicated operations and check by using the values _x_ = 1, _y_ = 2.
( _a_ ) ( _x_ 4 \+ _x_ 2 _y_ 2 \+ _y_ 4)( _y_ 4 – _x_ 2 _y_ 2 \+ _x_ 4)
( _b_ )
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**2.10**
( _a_ ) –24
( _b_ ) –12
( _c_ ) –1
( _d_ ) 90
( _e_ ) 11/5
( _f_ ) -8
( _g_ ) –1/6
( _h_ ) –24/5
**2.11**
( _a_ ) 4
( _b_ ) 6
( _c_ ) 5
( _d_ ) 3
( _e_ ) 0
( _f_ ) 5
**2.12**
( _a_ ) 3 _y_ – _x_
( _b_ ) 8 _y_ 2 \+ 6 _yz_
( _c_ ) 12 _x_ – 5 _y_
( _d_ ) _y_ – 4 _x_ \+ 4
**2.13**
( _a_ ) 2 _x_ 2 \+ _x_ – _y_
( _b_ ) _a_ 2 \+ _b_ 2 \+ 2 _c_ 2
( _c_ ) _abc_ 2
**2.14**
( _a_ ) 5 _xy_ – 3 _yz_ \+ _zx_
( _b_ ) _x_ 2 \+ 4 _y_ 2 – 8 _x_ \+ 8 _y_ – 5
( _c_ ) 4 _r_ 3 – _r_ 2 _s_ \+ _rs_ 2 – 3 _s_ 3
**2.15** _xy_ \+ _yz_ – 8 _xz_
**2.16**
( _a_ ) –12 _x_ 5 _y_ 7
( _b_ ) –36 _a_ 6 _b_ 5 _c_ 6
( _c_ ) –12 _x_ 3 _y_ 3 \+ 16 _x_ 3 _y_ 2
( _d_ ) 2 _r_ 4 _s_ 5 \+ 6 _r_ 3 _s_ 7 – 8 _r_ 3 _s_ 5 \+ 2 _r_ 2 _s_ 7
( _e_ ) _y_ 2 – _y_ – 12
( _f_ ) _y_ 3 \+ 64
( _g_ ) _x_ 4 – _y_ 4
( _h_ ) _x_ 4 \+ 64
( _i_ ) 3 _r_ 2 \+ 5 _rs_ \+ 8 _rt_ 2 – 2 _s_ 2 – 5 _st_ 2 – 3 _t_ 4
( _j_ ) _y_ 3 – 2 _y_ 2 – 3 _y_ \+ 3 _x_ \+ 5 _x_ 2 – 3 _xy_ – 2 _x_ 3 – _x_ 2 _y_ \+ 2 _xy_ 2
**2.17**
( _a_ )
( _b_ )
( _c_ )
( _d_ )
**2.18**
( _a_ ) 9 _s_ 2 \+ 12 _s_ \+ 16
( _b_ )
( _c_ )
( _d_ ) _x_ 2 \+ _xy_
**2.19**
( _a_ ) _x_ 8 \+ _x_ 4 _y_ 4 \+ _y_ 8. Check: 21(13) = 273.
( _b_ ) _x_ 2 \+ _y_ 2. Check: 35=7 = 5.
## **CHAPTER 3
Properties of Numbers**
### **3.1 SETS OF NUMBERS**
The set of counting (or natural) numbers is the set of numbers: 1, 2, 3, 4, 5,....
The set of whole numbers is the set of counting numbers and zero: 0, 1, 2, 3, 4,....
The set of integers is the set of counting numbers, zero, and the opposites of the counting numbers:..., –5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5,....
The set of real numbers is the set of all numbers that correspond to the points on a number line. The real numbers can be separated into two distinct subsets: the rational numbers and the irrational numbers.
The set of rational numbers is the set of real numbers that can be written in the form _a/b_ , where _a_ and _b_ are integers and b is not zero. The rational numbers can be thought of as the set of integers and the common fractions. The numbers and 15/1 are examples of rational numbers.
The set of irrational numbers is the set of real numbers that are not rational numbers. The numbers and the mathematical constants _π_ and e are examples of irrational numbers.
### **3.2 PROPERTIES**
A set has closure under an operation if the result of performing the operation with two elements of the set is also an element of the set. The set X is closed under the operation * if for all elements _a_ and _b_ in set X, the result _a*b_ is in set X.
A set has an identity under an operation if there is an element in the set that when combined with each element in the set leaves that element unchanged. The set X has an identity under the operation * if there is an element _j_ in set X such that _j*a_ = _a*j_ = _a_ for all elements _a_ in set X.
A set has inverses under an operation if for each element of the set there is an element of the set such that when these two elements are combined using the operation, the result is the identity for the set under the operation. If a set does not have an identity under an operation, it cannot have the inverse property for the operation. If X is a set that has identity _j_ under operation *, then it has inverses if for each element _a_ in set X there is an element _a_ ′ in set X such that _a*a′ = j_ and _a′*a = j_.
Sets under an operation may also have the associative property and the commutative property, as described in Section 1.4. If there are two operations on the set then the set could have the distributive property, also described in Section 1.4.
**EXAMPLE 3.1.** Which properties are true for the counting numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers under the operation of addition?
### **3.3 ADDITIONAL PROPERTIES**
There are some properties that sets of numbers have that do not depend on an operation to be true. Three such properties are order, density, and completeness.
A set of numbers has order if given two distinct elements in the set one element is greater than the other.
A set of numbers has density if between any two elements of the set there is another element of the set.
A set of numbers has completeness if the points using its elements as coordinates completely fill a line or plane.
**EXAMPLE 3.2.** Which properties are true for the counting numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers?
### **Solved Problems**
**3.1** Which of the properties closure, identity, and inverse does the set of even integers have under addition?
**SOLUTION**
Since even integers are of the form _2n_ where n is an integer, we let 2 _m_ and _2k_ be any two even integers. The sum of two even integers is 2 _m + 2k = 2(m + k_ ). From Example 3.1 we know that _m + k_ is an integer, since m and k are integers. Thus, 2 _(m + k)_ is 2 times an integer and is even, so _2m + 2k_ is even. Therefore, the even integers are closed under addition.
Zero is an even integer since 2(0) = 0. 2 _m +_ 0 = 2 _m +_ 2(0) = 2( _m +_ 0) = 2 _m_. Thus, 0 is the identity for the even integers under addition.
For the even integer 2 _m_ , the inverse is –2 _m_. Since _m_ is an integer, – _m_ is an integer. Thus, –2 _m_ = 2(– _m_ ) is an even integer. Also, 2 _m_ \+ (–2 _m_ ) = 2( _m_ \+ (– _m_ )) = 2(0) = 0. Therefore, each even integer has an inverse.
**3.2** Which of the properties closure, identity, inverse, associativity, and commutativity are true under multiplication for the sets counting numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers?
**SOLUTION**
**3.3** Which of the properties closure, identity, and inverse does the set of odd integers have under multiplication?
**SOLUTION**
Since the odd integers are of the form 2 _n_ \+ 1 where _n_ is an integer, we let 2 _m_ \+ 1 and _2k_ \+ 1 be any two odd integers. The product of two odd integers is represented by (2 _m_ \+ 1)(2 _k_ \+ 1) = 4 _mk_ \+ 2 _m_ \+ 2 _k_ \+ 1 = 2(2 _mk_ \+ _m_ \+ _k_ ) + 1. Since the integers are closed under both addition and multiplication, (2 _mk_ \+ _m_ \+ _k_ ) is an integer and the product (2 _m_ \+ 1)(2 _k_ \+ 1) is equal to 2 times an integer plus 1. Thus, the product is an odd integer. Therefore, the odd integers are closed under multiplication.
One is an odd integer, since 2(0) + 1 = 0 + 1 = 1. Also (2 _m_ \+ 1)(1) = (2 _m_ )(1) + (1)(1) = 2 _m_ \+ 1. Thus, 1 is the identity for the odd integers under multiplication.
Seven is an odd integer, since 2(3) + 1 = 7. Also, 7(1/7) = 1, but 1/7 is not an odd integer. Thus, 7 does not have an inverse under multiplication. Since there is at least one odd integer that does not have an inverse under multiplication, the set of odd integers under multiplication does not have the inverse property.
**3.4** Does the set of even integers have the order, density, and completeness properties?
**SOLUTION**
Given two distinct even integers 2 _m_ and 2 _k_ where _m_ and _k_ are integers, we know that either _m_ > _k_ or _k_ > _m_. If _m_ > _k_ , then 2 _m_ > 2 _k_ , but if _k_ > _m_ , then 2 _k_ > 2 _m_. Thus, the set of even integers has the order property, since for two distinct even integers 2 _m_ and 2 _k_ either 2 _m_ > 2 _k_ or 2 _k_ > 2 _m_.
The numbers 2 _m_ and 2 _m_ +2 are even integers. There is no even integer between 2 _m_ and 2 _m_ \+ 2, since 2 _m_ \+ 2 = 2( _m_ \+ 1) and there is no integer between _m_ and _m_ \+ 1. Thus, the even integers do not have the density property.
Between the two even integers 8 and 10 is the odd integer 9. Thus, the even integers do not represent the coordinates for all points on a number line. Therefore the even integers do not have the completeness property.
**3.5** Let K = {– 1, 1}. ( _a_ ) Is K closed under multiplication? ( _b_ ) Does K have an identity under multiplication? ( _c_ ) Does K have inverses under multiplication?
**SOLUTION**
( _a_ ) (1)(1) = 1, (–1)(–1) = 1, (1)(–1) = – 1, and (–1)(1) = – 1. For all possible products of two elements in K, the result is in K. Thus, K is closed under multiplication.
( _b_ ) 1 is in K, (1)(1) = 1, and (1)(–1) = – 1. Thus 1 is the identity for K under multiplication.
( _c_ ) Since (1)(1) = 1 and (–1)(–1) = 1, each element of K is its own inverse.
### **Supplementary Problems**
**3.6** Which of the properties closure, identity, and inverse does the set of even integers have under multiplication?
**3.7** Which of the properties closure, identity, and inverse does the set of odd integers have under addition?
**3.8** Does the set of odd integers have the order, density, and completeness properties?
**3.9** Which of the properties closure, identity, inverse, associativity, and commutativity are true under subtraction for the sets of counting numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers?
**3.10** Which of the properties closure, identity, inverse, associativity, and commutativity are true under non-zero division for the set of counting numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers?
**3.11** Which of the properties closure, identity, inverse, associativity, and commutativity are true for the set of zero, {0}, under ( _a_ ) addition, ( _b_ ) subtraction, and ( _c_ ) multiplication?
**3.12** Which of the properties closure, identity, inverse, associativity, and commutativity are true for the set of one, {1}, under ( _a_ ) addition, ( _b_ ) subtraction, ( _c_ ) multiplication, and ( _d_ ) division?
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**3.6** Closure: yes; identity: no; inverse: no.
**3.7** Closure: no; identity: no; inverse: no.
**3.8** Order: yes; density: no; completeness: no.
**3.9**
**3.10**
**3.11**
**3.12**
## **CHAPTER 4
Special Products**
### **4.1 SPECIAL PRODUCTS**
The following are some of the products which occur frequently in mathematics, and the student should become familiar with them as soon as possible. Proofs of these results may be obtained by performing the multiplications.
I. Product of a monomial and a binomial
_a_ ( _c_ \+ _d_ ) = _ac_ \+ _ad_
II. Product of the sum and the difference of two terms
( _a_ \+ _b_ )( _a_ – _b_ ) = _a_ 2 – _b_ 2
III. Square of a binomial
( _a_ \+ _b_ )2 = _a_ 2 \+ 2 _ab_ \+ _b_ 2
( _a_ – _b_ )2 = _a_ 2 – 2 _ab_ \+ _b_ 2
IV. Product of two binomials
( _x_ \+ _a_ )( _x_ \+ _b_ ) = _x_ 2 \+ ( _a_ \+ _b_ ) _x_ \+ _ab_
( _ax_ \+ _b_ )( _cx_ \+ _d_ ) = _acx_ 2 \+ ( _ad_ \+ _bc_ ) _x_ \+ _bd_
( _a_ \+ _b_ )( _c + d_ ) = _ac_ \+ _bc_ \+ _ad_ \+ _bd_
V. Cube of a binomial
( _a_ \+ _b_ )3 = _a_ 3 \+ 3 _a_ 2 _b_ \+ 3 _ab_ 2 \+ _b_ 3
( _a_ – _b_ )3 = _a_ 3 – 3 _a_ 2 _b_ \+ 3 _ab_ 2 – _b_ 3
VI. Square of a trinomial
( _a_ \+ _b_ \+ c)2 = _a_ 2 \+ _b_ 2 \+ c2 \+ 2 _ab_ \+ 2 _ac_ \+ 2 _bc_
### **4.2 PRODUCTS YIELDING ANSWERS OF THE FORM _a n_ ± _b n_**
It may be verified by multiplication that
( _a_ – _b_ )( _a_ 2 \+ _ab_ \+ _b_ 2) = _a_ 3 – _b_ 3
( _a – b_ )( _a_ 3 \+ _a_ 2 _b_ \+ _ab_ 2 \+ _b_ 3) = _a_ 4 – _b_ 4
( _a – b_ )( _a_ 4 \+ _a_ 3 _b_ \+ _a_ 2 _b_ 2 \+ _ab_ 3 \+ _b_ 4) = _a_ 5 – _b_ 5
( _a – b_ )( _a_ 5 \+ _a_ 4 _b_ \+ _a_ 3 _b_ 2 \+ _a_ 2 _b_ 3 \+ _ab_ 4 \+ _b_ 5) = _a_ 6 – _b_ 6
etc., the rule being clear. These may be summarized by
VII. ( _a_ – _b_ )( _a_ _n_ –1 \+ _a_ _n_ –2 _b_ \+ _a_ _n_ –3 _b_ 2 \+ ··· + _ab_ _n_ –2 \+ _b_ _n_ –1) = _a n_ – _b n_
where _n_ is _any positive integer_ (1, 2, 3, 4,...).
Similarly, it may be verified that
( _a_ \+ _b_ )( _a_ 2 – _ab_ \+ _b_ 2) = _a_ 3 \+ _b_ 3
( _a_ \+ _b_ )( _a_ 4 – _a_ 3 _b_ \+ _a_ 2 _b_ 2 – _ab_ 3 \+ _b_ 4) = _a_ 5 \+ _b_ 5
( _a_ \+ _b_ )( _a_ 6 – _a_ 5 _b_ \+ _a_ 4 _b_ 2 – _a_ 3 _b_ 3 \+ _a_ 2 _b_ 4 – _ab_ 5 \+ _b_ 6) = _a_ 7 \+ _b_ 7
etc., the rule being clear. These may be summarized by
VIII. ( _a + b_ )( _a_ _n_ –1 – _a_ _n_ –2 _b_ \+ _a_ _n_ –3 _b_ 2–··· _ab_ n–2 \+ _b_ _n_ –1) = _a_ n \+ _b_ _n_
where _n_ is _any positive odd integer_ (1, 3, 5, 7,...).
### **Solved Problems**
Find each of the following products.
**4.1**
( _a_ ) 3 _x_ (2 _x_ \+ 3 _y_ ) = (3 _x_ )(2 _x_ ) + (3 _x_ )(3 _y_ ) = 6 _x_ 2 \+ 9 _xy_ , using I with _a_ = 3 _x_ , _c_ = 2 _x_ , _d_ = _3y_.
( _b_ ) _x_ 2 _y_ (3 _x_ 3 – 2 _y_ \+ 4) = ( _x_ 2 _y_ )(3 _x_ 3) + ( _x_ 2 _y_ )(–2 _y_ ) + ( _x_ 2 _y_ )(4) = 3 _x_ 5 _y_ – 2 _x_ 2 _y_ 2 \+ 4 _x_ 2y
( _c_ ) (3 _x_ 3 _y_ 2 \+ 2 _xy_ – 5)( _x_ 2 _y_ 3) = (3 _x_ 3 _y_ 2)( _x_ 2 _y_ 3) + (2 _xy_ )( _x_ 2 _y_ 3) + (–5)( _x_ 2 _y_ 3)
= 3 _x_ 5 _y_ 5 \+ 2 _x_ 3 _y_ 4 – 5 _x_ 2 _y_ 3
( _d_ ) (2 _x_ \+ 3 _y_ )(2 _x_ – 3 _y_ ) = (2 _x_ )2 – (3 _y_ )2 = 4 _x_ 2 – 9 _y_ 2, using II with _a_ = 2 _x_ , _b_ = 3 _y_.
( _e_ ) (1 – 5 _x_ 3)(1 + 5 _x_ 3) = (1)2 – (5 _x_ 3)2 = 1 – 25 _x_ 6
( _f_ ) (5 _x_ \+ _x_ 3 _y_ 2)(5 _x_ – _x_ 3 _y_ 2) = (5 _x_ )2 – (x3y2)2 = 25 _x_ 2 – _x_ 6 _y_ 4
( _g_ ) (3 _x_ \+ 5 _y_ )2 = (3 _x_ )2 \+ 2(3 _x_ )(5 _y_ ) + (5 _y_ )2 = 9 _x_ 2 \+ 30 _xy_ \+ 25 _y_ 2, using III with _a_ = 3 _x_ , _b_ = 5 _y_.
( _h_ ) ( _x_ \+ 2)2 = _x_ 2 \+ 2( _x_ )(2) + 22 = _x_ 2 \+ 4 _x_ \+ 4
( _i_ ) (7 _x_ 2 – 2 _xy_ )2 = (7 _x_ 2)2 – 2(7 _x_ 2)(2 _xy_ ) + (2 _xy_ )2
=49 _x_ 4 – 28 _x_ 3 _y_ \+ 4 _x_ 2 _y_ 2, using III with _a_ = 7 _x_ 2, _b_ = 2 _xy_.
( _j_ ) ( _ax_ – 2 _by_ )2 = ( _ax_ )2 – 2( _ax_ )(2 _by_ ) + (2 _by_ )2 = _a_ 2 _x_ 2 – 4 _axby_ \+ 4 _b_ 2 _y_ 2
( _k_ ) ( _x_ 4 \+ 6)2 = ( _x_ 4)2 \+ 2( _x_ 4)(6) + (6)2 = _x_ 8 \+ 12 _x_ 4 \+ 36
( _l_ ) (3 _y_ 2 – 2)2 = (3 _y_ 2)2 – 2(3 _y_ 2)(2) + (2)2 = 9 _y_ 4 – 12 _y_ 2 \+ 4
( _m_ ) ( _x_ \+ 3)( _x_ \+ 5) = _x_ 2 \+ (3 + 5) _x_ \+ (3)(5) = _x_ 2 \+ 8 _x_ \+ 15, using IV with _a_ = 3, _b_ = 5.
( _n_ ) ( _x_ – 2)( _x_ \+ 8) = _x_ 2 \+ (–2 + 8) _x_ \+ (–92)(8) = _x_ 2 \+ 6 _x_ – 16
( _o_ ) ( _x_ \+ 2)( _x_ – 8) = _x_ 2 \+ (2 – 8) _x_ \+ (2)(–8) = _x_ 2 – 6 _x_ – 16
( _p_ ) ( _t_ 2 \+ 10)( _t_ 2 – 12) = ( _t_ 2)2 \+ (10 – 12) _t_ 2 \+ (10)(–12) = _t_ 4 – 2 _t_ 2 – 120
( _q_ ) (3 _x_ \+ 4)(2 _x_ – 3) = (3)(2) _x_ 2 \+ [(3)(–3) + (4)(2)] _x_ \+ (4)(–3)
= 6 _x_ 2 – _x_ – 12, using IV with _a_ = 3, _b_ = 4, _c_ = 2, _d_ = –3.
( _r_ ) (2 _x_ \+ 5)(4 _x_ – 1) = (2)(4) _x_ 2 \+ [(2)(–1) + (5)(4)] _x_ \+ (5)(–1) = 8 _x_ 2 \+ 18 _x_ – 5
( _s_ ) (3 _x_ \+ _y_ )(4 _x_ – 2 _y_ ) = (3 _x_ )(4 _x_ ) + ( _y_ )(4 _x_ ) + (3 _x_ )(–2 _y_ ) + ( _y_ )(–2 _y_ )
= 12 _x_ 2 – 2 _xy_ – 2 _y_ 2, using V with _a_ = 3 _x_ , _b_ = _y_ , _c_ = 4 _x_ , _d_ = –2 _y_.
( _t_ ) (3 _t_ 2 _s_ – 2)(4 _t_ – 3 _s_ ) = (3 _t_ 2 _s_ )(4 _t_ ) + (–2)(4 _t_ ) + (3 _t_ 2s)(–3 _s_ ) + (–2)(–3 _s_ )
= 12 _t_ 3 _s_ – 8 _t_ – 9 _t_ 2 _s_ 2 \+ 6 _s_
( _u_ ) (3 _xy_ \+ 1)(2 _x_ 2 – 3 _y_ ) = (3 _xy_ )(2 _x_ 2) + (3 _xy_ )(–3 _y_ ) + (1)(2 _x_ 2) + (1)–3 _y_ )
= 6 _x_ 3 _y_ – 9 _xy_ 2 \+ 2 _x_ 2 – 3 _y_
( _v_ ) ( _x_ \+ _y_ \+ 3)( _x_ \+ _y_ – 3) = ( _x_ \+ _y_ )2 – 32 = _x_ 2 \+ 2 _xy_ \+ _y_ 2 – 9
( _w_ ) (2 _x_ – _y_ – 1)(2 _x_ – _y_ \+ 1) = (2 _x_ – _y_ )2 – (1)2 = 4 _x_ 2 – 4 _xy_ \+ _y_ 2 – 1
( _x_ ) ( _x_ 2 \+ 2 _xy_ \+ _y_ 2)( _x_ 2 – 2 _xy_ \+ _y_ 2) = ( _x_ 2 \+ _y_ 2 \+ 2 _xy_ )( _x_ 2 \+ _y_ 2 – 2 _xy_ )
= ( _x_ 2 \+ _y_ 2)2 – (2 _xy_ )2 = _x_ 4 \+ 2 _x_ 2 _y_ 2 \+ _y_ 4 – 4 _x_ 2 _y_ 2 = _x_ 4 – 2 _x_ 2 _y_ 2 \+ _y_ 4
( _y_ ) ( _x_ 3 \+ 2 + _xy_ )( _x_ 3 – 2 + _xy_ ) = ( _x_ 3 \+ _xy_ \+ 2)( _x_ 3 \+ _xy_ – 2)
= ( _x_ 3 \+ _xy_ )2 – 22 = _x_ 6 \+ 2( _x_ 3)( _xy_ ) + ( _xy_ )2 – 4 = _x_ 6 \+ 2 _x_ 4 _y_ \+ _x_ 2 _y_ 2 – 4
**4.2** (a) ( _x_ \+ 2 _y_ )3 = _x_ 3 \+ 3( _x_ )2(2 _y_ ) + 3( _x_ )(2 _y_ )2 \+ (2 _y_ )3
= _x_ 3 \+ 6 _x_ 2 _y_ \+ 12 _xy_ 2 \+ 8 _y_ 3, using V with _a_ = _x_ , _b_ = 2 _y_.
( _b_ ) (3 _x_ \+ 2)3 = (3 _x_ )3 \+ 3(3 _x_ )2(2) + 3(3 _x_ )(2)2 \+ (2)3 = 27 _x_ 3 \+ 54 _x_ 2 \+ 36 _x_ \+ 8
( _c_ ) (2 _y_ – 5)3 = (2 _y_ )3 – 3(2 _y_ )2(5) + 3(2 _y_ )(5)2 – (5)3
= 8 _y_ 3 – 60 _y_ 2 \+ 150 _y_ – 125, using V with _a_ = 2 _y_ , _b_ = 5.
( _d_ ) ( _xy_ – 2)3 = ( _xy_ )3 – 3( _xy_ )2(2) + 3( _xy_ )(2)2 – (2)3 = _x_ 3 _y_ 3 – 6 _x_ 2 _y_ 2 \+ 12 _xy_ – 8
( _e_ ) ( _x_ 2 _y_ – _y_ 2)3 = ( _x_ 2 _y_ )3 – 3( _x_ 2 _y_ )2(y2) + 3( _x_ 2 _y_ )( _y_ 2)2 – ( _y_ 2)3 = _x_ 6 _y_ 3 – 3 _x_ 4 _y_ 4 \+ 3 _x_ 2 _y_ 5 – _y_ 6
( _f_ ) ( _x_ – 1)( _x_ 2 \+ _x_ \+ 1) = _x_ 3 – 1, using VII with _a_ = _x, b_ = 1.
If the form is not recognized, multiply as follows.
( _x_ – 1)( _x_ 2 \+ _x_ \+ 1) = _x_ ( _x_ 2 \+ _x_ \+ 1) – 1( _x_ 2 \+ _x_ \+ 1) = _x_ 3 \+ _x_ 2 \+ _x_ – _x_ 2 – _x_ – 1 = _x_ 3 – 1
( _g_ ) ( _x_ – 2 _y_ )( _x_ 2 \+ 2 _xy_ \+ 4 _y_ 2) = _x_ 3 – (2 _y_ )3 = _x_ 3 – 8 _y_ 3, using VII with _a_ = _x_ , _b_ = 2 _y_.
( _h_ ) ( _xy_ \+ 2)( _x_ 2 _y_ 2 – 2 _xy_ \+ 4) = ( _xy_ )3 \+ (2)3 = _x_ 3 _y_ 3 \+ 8, using VIII with _a_ = _xy_ , _b_ = 2.
( _i_ ) (2 _x_ \+ 1)(4 _x_ 2 – 2 _x_ \+ 1) = (2 _x_ )3 \+ 1 = 8 _x_ 3 \+ 1
( _j_ ) (2 _x_ \+ 3 _y_ \+ _z_ )2 = (2 _x_ 2) + (3 _y_ )2 \+ ( _z_ )2 \+ 2(2 _x_ )(3 _y_ ) + 2(2 _x_ )( _z_ ) + 2(3 _y_ )( _z_ )
= 4 _x_ 2 \+ 9 _y_ 2 \+ _z_ 2 \+ 12 _xy_ \+ 4 _xz_ \+ 6 _yz_ , using VI with _a_ = 2 _x_ , _b_ = 3 _y_ , _c_ = _z_.
( _k_ ) (u3 – _v_ 2 \+ 2 _w_ )2 = ( _u_ 3)2 \+ (– _v_ 2)2 \+ (2 _w_ )2 \+ 2( _u_ 3)(– _v_ 2) + 2( _u_ 3)(2 _w_ ) + 2(- _v_ 2)(2 _w_ )
= _u_ 6 \+ _v_ 4 \+ 4 _w_ 2 – 2 _u_ 3 _v_ 2 \+ 4 _u_ 3 _w_ – 4 _v_ 2 _w_
**4.3** ( _a_ ) ( _x_ – 1)( _x_ 5 \+ _x_ 4 \+ _x_ 3 \+ _x_ 2 \+ _x_ \+ 1) = _x_ 6 – 1, using VII with _a_ = _x, b_ = 1, _n_ = 6.
( _b_ ) ( _x_ – 2 _y_ )( _x_ 4 \+ 2 _x_ 3 _y_ \+ 4 _x_ 2 _y_ 2 \+ 8 _xy_ 3 \+ 16 _y_ 4) = _x_ 5 – (2 _y_ )5
= _x_ 5 – 32 _y_ 5, using VII with _a_ = _x, b_ = 2 _y_.
( _c_ ) (3 _y_ \+ _x_ )(81 _y_ 4 – 27 _y_ 3 _x_ \+ 9 _y_ 2x2 _–_ 3 _yx_ 3 \+ _x_ 4) = (3 _y_ )5 \+ _x_ 5
= 243 _y_ 5 \+ _x_ 5, using VIII with _a_ = 3 _y_ , _b_ = _x_.
**4.4** ( _a_ ) ( _x_ \+ _y_ \+ _z_ )( _x_ \+ _y_ – _z_ )( _x_ – _y_ \+ _z_ )( _x_ – _y_ – _z_ ). The first two factors may be written as
( _x_ \+ _y_ \+ _z_ )( _x_ \+ _y_ – _z_ ) = ( _x_ \+ _y_ )2 – _z_ 2 = _x_ 2 \+ 2 _xy_ \+ _y_ 2 – _z_ 2,
and the second two factors as
( _x_ – _y_ \+ _z_ )( _x_ – _y_ – _z_ ) = ( _x_ – _y_ )2 – _z_ 2 = _x_ 2 – 2 _xy_ \+ _y_ 2 – _z_ 2.
The result may be written
( _x_ 2 \+ _y_ 2 – _z_ 2 \+ 2 _xy_ )( _x_ 2 \+ _y_ 2 – _z_ 2 – 2 _xy_ ) = ( _x_ 2 \+ _y_ 2 – _z_ 2)2 – (2 _xy_ )2
= ( _x_ 2)2 \+ ( _y_ 2)2 \+ (– _z_ 2)2 \+ 2( _x_ 2)( _y_ 2) + 2( _x_ 2)(– _z_ 2) + 2( _y_ 2)(– _z_ 2) _–_ 4 _x_ 2y2
= _x_ 4 \+ _y_ 4 \+ _z_ 4 \+ 2 _x_ 2y2 – 2 _x_ 2 _z_ 2 – 2 _y_ 2 _z_ 2
( _b_ ) ( _x_ \+ _y_ \+ _z_ \+ 1)2 = [( _x + y_ ) + ( _z + 1_ )]2 = ( _x + y_ )2 \+ 2( _x + y_ )( _z_ \+ 1) + ( _z_ \+ 1)2
= _x_ 2 \+ 2 _xy_ \+ _y_ 2 \+ 2 _xz_ \+ 2 _x_ \+ 2 _yz_ \+ 2 _y_ \+ _z_ 2 \+ 2 _z_ \+ 1
( _c_ ) ( _u_ – _v_ )3( _u_ \+ _v_ )3 = [( _u_ – _v_ )( _u_ \+ _v_ )]3 = ( _u_ 2 – _v_ 2)3
= ( _u_ 2)3 – 3( _u_ 2)2 _v_ 2 \+ 3( _u_ 2)( _v_ 2)2 – ( _v_ 2)3 = _u_ 6 – 3 _u_ 4 _v_ 2 \+ 3 _u_ 2 _v_ 4 – _v_ 6
( _d_ ) ( _x_ 2 – _x_ \+ 1)2( _x_ 2 \+ _x_ \+ 1)2 = [( _x_ 2 – _x_ \+ 1)( _x_ 2 \+ _x_ \+ 1)]2= [( _x_ 2 \+ 1 – _x_ )( _x_ 2 \+ 1 + _x_ )]2
= [( _x_ 2 \+ 1)2 – _x_ 2]2 = [ _x_ 4 \+ 2 _x_ 2 \+ 1 _– x_ 2]2 = ( _x_ 4 \+ _x_ 2 \+ 1)2
= ( _x_ 4)2 \+ ( _x_ 2)2 \+ 12 \+ 2( _x_ 4)( _x_ 2) + 2( _x_ 4)(1) + 2( _x_ 2)(1)
= _x_ 8 \+ _x_ 4 \+ 1 + 2 _x_ 6 \+ 2 _x_ 4 \+ 2 _x_ 2 = _x_ 8 \+ 2 _x_ 6 \+ 3 _x_ 4 \+ 2 _x_ 2 \+ 1
( _e_ ) ( _e_ y \+ 1)( _e_ y _–_ 1)( _e_ 2 _y_ \+ 1)( _e_ 4 _y_ \+ 1)( _e_ 8 _y_ \+ 1) = ( _e_ 2 _y_ _–_ 1)( _e_ 2 _y_ \+ 1)( _e_ 4 _y_ \+ 1)( _e_ 8 _y_ \+ 1)
= ( _e_ 4 _y_ _–_ 1)( _e_ 4 _y_ \+ 1)( _e_ 8 _y_ \+ 1) = ( _e_ 8 _y_ – 1)( _e_ 8 _y_ \+ 1) = _e_ 16 _y_ _–_ 1
### **Supplementary Problems**
Find each of the following products.
**4.5** (a) 2 _xy_ (3 _x_ 2 _y_ – 4 _y_ 3) = 6 _x_ 3 _y_ 2 _–_ 8 _xy_ 4
( _b_ ) 3 _x_ 2 _y_ 3(2 _xy – x_ – 2 _y_ ) = 6 _x_ 3y4 _–_ 3 _x_ 3y3 _–_ 6 _x_ 2y4
( _c_ ) (2 _st_ 3 – 4 _rs_ 2 \+ 3 _s_ 3 _t_ )(5 _rst_ 2) = 10 _rs_ 2 _t_ 5 – 20 _r_ 2 _s_ 3 _t_ 2 \+ 15 _rs_ 4 _t_ 3
( _d_ ) (3 _a_ \+ 5 _b_ )(3 _a_ – 5 _b_ ) = 9 _a_ 2 – 25 _b_ 2
( _e_ ) (5 _xy_ \+ 4)(5 _xy_ – 4) = 25 _x_ 2 _y_ 2 – 16
( _f_ ) (2 – 5 _y_ 2)(2 + 5 _y_ 2) = 4 _–_ 25 _y_ 4
( _g_ ) (3 _a_ \+ 5 _a_ 2 _b_ )(3 _a_ – 5 _a_ 2 _b_ ) = 9 _a_ 2 – 25 _a_ 4 _b_ 2
( _h_ ) ( _x_ \+ 6)2 = _x_ 2 \+ 12 _x_ \+ 36
( _i_ ) ( _y_ \+ 3 _x_ )2 = _y_ 2 \+ 6 _xy_ \+ 9 _x_ 2
( _j_ ) ( _z_ – 4)2 = _z_ 2 – 8 _z_ \+ 16
( _k_ ) (3 _–_ 2 _x_ 2)2 = 9 – 12 _x_ 2 \+ 4 _x_ 4
( _l_ ) ( _x_ 2y _–_ 2 _z_ )2 = _x_ 4 _y_ 2 – 4 _x_ 2 _yz_ \+ 4 _z_ 2
( _m_ ) ( _x_ \+ 2)( _x_ \+ 4) = _x_ 2 \+ 6 _x_ \+ 8
( _n_ ) ( _x_ – 4)( _x_ \+ 7) = _x_ 2 \+ 3 _x_ – 28
( _o_ ) ( _y_ \+ 3)( _y_ – 5) = _y_ 2 – 2 _y_ – 15
( _p_ ) ( _xy_ \+ 6)( _xy_ – _4_ ) = _x_ 2 _y_ 2 \+ 2 _xy_ – 24
( _q_ ) (2 _x_ – 3)(4 _x_ \+ 1) = 8 _x_ 2 – 10 _x_ – 3
( _r_ ) (4 + 3 _r_ )(2 – _r_ ) = 8 + 2 _r_ – 3 _r_ 2
( _s_ ) (5 _x_ \+ 3 _y_ )(2 _x_ – 3 _y_ ) = 10 _x_ 2 – 9 _xy_ – 9 _y_ 2
( _t_ ) (2 _t_ 2 \+ _s_ )(3 _t_ 2 \+ 4 _s_ ) = 6 _t_ 4 \+ 11 _t_ 2 _s_ \+ 4 _s_ 2
( _u_ ) ( _x_ 2 \+ 4 _y_ )(2 _x_ 2 _y_ – _y_ 2) = 2 _x_ 4 _y_ \+ 7 _x_ 2 _y_ 2 – 4 _y_ 3
( _v_ ) _x_ (2 _x_ – 3)(3 _x_ \+ 4) = 6 _x_ 3 – _x_ 2 – 12 _x_
( _w_ ) ( _r_ \+ _s_ – 1)( _r_ \+ _s_ \+ 1) = _r_ 2 \+ 2 _rs_ \+ _s_ 2 – 1
( _x_ ) ( _x_ – 2 _y_ \+ _z_ )( _x_ – 2 _y_ – _z_ ) = _x_ 2 – 4 _xy_ \+ 4 _y_ 2 – _z_ 2
( _y_ )( _x_ 2 \+ 2 _x_ \+ 4)( _x_ 2 – 2 _x_ \+ 4) = _x_ 4 \+ 4 _x_ 2 \+ 16
**4.6** ( _a_ ) (2 _x_ \+ 1)3 = 8 _x_ 3 \+ 12 _x_ 2 \+ 6 _x_ \+ 1
( _b_ ) (3 _x_ \+ 2 _y_ )3 = 27 _x_ 3 \+ 54 _x_ 2 _y_ \+ 36 _xy_ 2 \+ 8 _y_ 3
( _c_ ) ( _r_ – 2 _s_ )3 = _r_ 3 – 6 _r_ 2 _s_ \+ 12 _rs_ 2 – 8 _s_ 3
( _d_ ) ( _x_ 2 – 1)3 = _x_ 6 – 3 _x_ 4 \+ 3 _x_ 2 – 1
( _e_ ) ( _ab_ 2 – 2 _b_ )3 = _a_ 3 _b_ 6 – 6 _a_ 2 _b_ 5 \+ 12 _ab_ 4 – 8 _b_ 3
( _f_ ) ( _t_ – 2)( _t_ 2 \+ 2 _t_ \+ 4) = _t_ 3 – 8
( _g_ ) ( _z_ – _x_ )( _x_ 2 \+ _xz_ \+ _z_ 2) = _z_ 3 – _x_ 3
( _h_ ) ( _x_ \+ 3 _y_ )( _x_ 2 – 3 _xy_ \+ 9 _y_ 2) = _x_ 3 \+ 27 _y_ 3
**4.7** ( _a_ ) ( _x_ – 2 _y_ \+ _z_ )2 = _x_ 2 – 4 _xy_ \+ 4 _y_ 2 \+ 2 _zx_ – 4 _zy_ \+ z2
( _b_ ) ( _s_ – 1)( _s_ 3 \+ _s_ 2 \+ _s_ \+ 1) = _s_ 4 – 1
( _c_ ) (1 + _t_ 2)(1 – _t_ 2 \+ _t_ 4 – _t_ 6) = 1 – _t_ 8
( _d_ ) (3 _x_ \+ 2 _y_ )2(3 _x_ – 2 _y_ )2 = 81 _x_ 4 – 72 _x_ 2 _y_ 2 \+ 16 _y_ 4
( _e_ ) ( _x_ 2 \+ 2 _x_ \+ 1)2( _x_ 2 – 2 _x_ \+ 1)2 = _x_ 8 – 4 _x_ 6 \+ 6 _x_ 4 – 4 _x_ 2 \+ 1
( _f_ ) ( _y_ – 1)3( _y_ \+ 1)3 = _y_ 6 – 3 _y_ 4 \+ 3 _y_ 2 – 1
( _g_ ) ( _u_ \+ 2)( _u_ – 2)( _u_ 2 \+ 4)( _u_ 4 \+ 16) = _u_ 8 – 256
## **CHAPTER 5**
Factoring
### **5.1 FACTORING**
The factors of a given algebraic expression consist of two or more algebraic expressions which when multiplied together produce the given expression.
**EXAMPLES 5.1.** Factor each algebraic expression.
( _a_ ) _x_ 2 – 7 _x_ \+ 6 = ( _x_ – 1)( _x_ – 6)
( _b_ ) _x_ 2 \+ 8 _x_ = _x_ ( _x_ \+ 8)
( _c_ ) 6 _x_ 2 – 7 _x_ – 5 = (3 _x_ – 5)(2 _x_ \+ 1)
( _d_ ) _x_ 2 \+ 2 _xy_ – 8 _y_ 2 = ( _x_ \+ 4 _y_ )( _x_ – 2 _y_ )
The factorization process is generally restricted to finding factors of polynomials with integer coefficients in each of its terms. In such cases it is required that the factors also be polynomials with integer coefficients. Unless otherwise stated we shall adhere to this limitation.
Thus we shall not consider ( _x_ – 1) as being factorable into because these factors are not polynomials. Similarly, we shall not consider ( _x_ 2 – 3 _y_ 2) as being factorable into because these factors are not polynomials with integer coefficients. Also, even though 3 _x_ \+ 2 _y_ could be written we shall not consider this to be a factored form because is not a polynomial with integer coefficients.
A given polynomial with integer coefficients is said to be prime if it cannot itself be factored in accordance with the above restrictions. Thus _x_ 2 – 7 _x_ \+ 6 = ( _x_ – 1)( _x_ – 6) has been expressed as a product of the prime factors _x_ – 1 and _x_ – 6.
A polynomial is said to be factored completely when it is expressed as a product of prime factors.
_Note 1_. In factoring we shall allow trivial changes in sign. Thus _x_ 2 – 7 _x_ \+ 6 can be factored either as ( _x_ – 1)( _x_ – 6) or (1 – _x_ )(6 – _x_ ). It can be shown that factorization into prime factors, apart from the trivial changes in sign and arrangement of factors, is possible in one and only one way. This is often referred to as the Unique Factorization Theorem.
_Note 2_. Sometimes the following definition of prime is used. A polynomial is said to be prime if it has no factors other than plus or minus itself and ±1. This is in analogy with the definition of a prime number or integer such as 2, 3, 5, 7, 11,... and may be seen to be equivalent to the previous definition.
_Note 3_. Occasionally we may factor polynomials with rational coefficients, e.g., _x_ 2 – 9/4 = ( _x_ \+ 3/2)( _x_ – 3/2). In such cases the factors should be polynomials with rational coefficients.
_Note 4_. There are times when we want to factor an expression over a specific set of numbers, e.g., _x_ 2 – 2 = over the set of real numbers, but it is prime over the set of rational numbers. Unless the set of numbers to use for the coefficients of the factors is specified it is assumed to be the set of integers.
### **5.2 FACTORIZATION PROCEDURES**
In factoring, formulas I-VIII of Chapter 4 are very useful. Just as when read from left to right they helped to obtain _products_ , so when read from right to left they help to find _factors_.
The following procedures in factoring are very useful.
A. Common monomial factor. Type: _ac_ \+ _ad_ = _a_ ( _c_ \+ _d_ )
**EXAMPLES 5.2.** ( _a_ ) 6 _x_ 2 _y_ – 2 _x_ 3 = 2 _x_ 2(3 _y_ – _x_ )
( _b_ ) 2 _x_ 3 _y_ – _xy_ 2 \+ 3 _x_ 2 _y_ = _xy_ (2 _x_ 2 – _y_ \+ 3 _x_ )
B. Difference of two squares. Type: _a_ 2 – _b_ 2 = ( _a + b_ )( _a — b)_
**EXAMPLES 5.3.** ( _a_ ) _x_ 2 _–_ 25 = _x_ 2 – 52 = ( _x_ \+ 5)( _x_ – 5) where _a_ = _x_ , _b_ = 5
( _b_ ) 4 _x_ 2 – 9 _y_ 2 = (2 _x_ )2 – (3 _y_ )2 = (2 _x_ \+ 3 _y_ )(2 _x_ – 3 _y_ ) where _a_ = 2 _x_ , _b_ = 3 _y_
C. Perfect square trinomials. Types:
It follows that a trinomial is a perfect square if two terms are perfect squares and the third term is numerically twice the product of the square roots of the other two terms.
**EXAMPLES 5.4.** ( _a_ ) _x_ 2 \+ 6 _x_ \+ 9 = ( _x_ \+ 3)2
( _b_ ) 9 _x_ 2 – 12 _xy_ \+ 4 _y_ 2 = (3 _x_ – 2 _y_ )2
D. Other trinomials. Types:
**EXAMPLES 5.5.** ( _a_ ) _x_ 2 – 5 _x_ \+ 4 = ( _x_ – 4)( _x_ – 1) where _a_ = – 4, _b_ = – 1 so that their sum ( _a_ \+ _b_ ) = –5 and their product _ab_ = 4.
( _b_ ) _x_ 2 \+ _xy_ – 12 _y_ 2 = ( _x_ – 3 _y_ )( _x_ \+ 4 _y_ ) where _a_ = – 3 _y_ , _b_ = 4 _y_
( _c_ ) 3 _x_ 2 – 5 _x_ – 2 = ( _x_ – 2)(3 _x_ \+ 1). Here _ac_ = 3, _bd_ = — 2, _ad_ \+ _bc_ = — 5; and we find by trial that _a_ = 1, _c_ = 3, _b_ = — 2, _d_ = 1 satisfies _ad + bc_ = –5.
( _d_ ) 6 _x_ 2 \+ _x_ – 12 = (3 _x_ – 4)(2 _x_ \+ 3)
( _e_ ) 8 – 14 _x_ \+ 5 _x_ 2 = (4 – 5 _x_ )(2 – _x_ )
E. Sum, difference of two cubes. Types:
**EXAMPLES 5.6.**
F. Grouping of terms. Type: _ac_ \+ _bc_ \+ _ad_ \+ _bd_ = _c_ ( _a + b_ ) + _d_ ( _a_ \+ _b_ ) = ( _a_ \+ _b_ )( _c_ \+ _d_ )
**EXAMPLE 5.7.** 2 _ax_ – 4 _bx_ \+ _ay_ – 2 _by_ = 2 _x_ ( _a_ – 2 _b_ ) + _y_ ( _a_ – 2 _b_ ) = ( _a_ – 2 _b_ )(2 _x_ \+ _y_ )
G. Factors of _a n_ ± _b n_. Here we use formulas VII and VIII of Chapter 4.
**EXAMPLES 5.8.** ( _a_ ) 32 _x_ 5 \+ 1 = (2 _x_ )5 \+ 15 = (2 _x_ \+ 1)[(2 _x_ )4 – (2 _x_ )3 \+ (2 _x_ )2 – 2 _x_ \+ 1] = (2 _x_ \+ 1)(16 _x_ 4 – 8 _x_ 3 \+ 4 _x_ 2 – 2 _x_ \+ 1)
( _b_ ) _x_ 7 – 1 = ( _x_ – 1)( _x_ 6 \+ _x_ 5 \+ _x_ 4 \+ _x_ 3 \+ _x_ 2 \+ x + 1)
H. Addition and subtraction of suitable terms.
**EXAMPLE 5.9.** Factor _x_ 4 \+ 4.
Adding and subtracting 4 _x_ 2 (twice the product of the square roots of _x_ 4 and 4), we have
_x_ 4 \+ 4 = ( _x_ 4 \+ 4 _x_ 2 \+ 4) – 4 _x_ 2 = ( _x_ 2 \+ 2)2 – (2 _x_ )2
= ( _x_ 2 \+ 2 + 2 _x_ )(x2 \+ 2 – 2 _x_ ) = ( _x_ 2 \+ 2 _x_ \+ 2)( _x_ 2 – 2 _x_ \+ 2)
I. Miscellaneous combinations of previous methods.
**EXAMPLES 5.10.**
### **5.3 GREATEST COMMON FACTOR**
The greatest common factor (GCF) of two or more given polynomials is the polynomial of highest degree and largest numerical coefficients (apart from trivial changes in sign) which is a factor of all the given polynomials.
The following method is suggested for finding the GCF of several polynomials. ( _a_ ) Write each polynomial as a product of prime factors. ( _b_ ) The GCF is the product obtained by taking each factor to the lowest power to which it occurs in any of the polynomials.
**EXAMPLE 5.11.** The GCF of 2332( _x_ – _y_ )3( _x_ \+ 2 _y_ )2, 2233( _x_ – _y_ )2( _x_ \+ 2 _y_ )3, 32( _x_ – _y_ )2( _x_ \+ 2 _y_ ) is 32( _x_ – _y_ )2( _x_ \+ 2 _y_ ).
Two or more polynomials are _relatively prime_ if their GCF is 1.
### **5.4 LEAST COMMON MULTIPLE**
The least common multiple (LCM) of two or more given polynomials is the polynomial of lowest degree and smallest numerical coefficients (apart from trivial changes in sign) for which each of the given polynomials will be a factor.
The following procedure is suggested for determining the LCM of several polynomials. ( _a_ ) Write each polynomial as a product of prime factors. ( _b_ ) The LCM is the product obtained by taking each factor to the highest power to which it occurs.
**EXAMPLE 5.12.** The LCM of 2332( _x_ – _y_ )3( _x_ \+ 2 _y_ )2, 2233( _x_ – _y_ )2( _x_ \+ 2 _y_ )3, 32( _x_ – _y_ )2( _x_ \+ 2 _y_ ) is 2333( _x_ – _y_ )3( _x_ \+ 2 _y_ )3.
### **Solved Problems**
**Common Monomial Factor**
Type: _ac_ \+ _ad_ = _a_ ( _c_ \+ _d_ )
**5.1** ( _a_ ) 2 _x_ 2 – 3 _xy_ = _x_ (2 _x_ – 3 _y_ )
( _b_ ) 4 _x_ \+ 8 _y_ \+ 12 _z_ = 4( _x_ \+ 2 _y_ \+ 3 _z_ )
( _c_ ) 3 _x_ 2 \+ 6 _x_ 3 \+ 12 _x_ 4 = 3 _x_ 2(1 + 2 _x_ \+ 4 _x_ 2)
( _d_ ) 9 _s_ 3 _t_ \+ 15 _s_ 2 _t_ 3 – 3 _s_ 2 _t_ 2 = 3 _s_ 2 _t_ (3 _s_ \+ 5 _t_ 2 – _t_ )
( _e_ ) 10 _a_ 2 _b_ 3 _c_ 4 – 15 _a_ 3 _b_ 2 _c_ 4 \+ 30 _a_ 4 _b_ 3 _c_ 2 = 5 _a_ 2 _b_ 2 _c_ 2(2 _bc_ 2 – 3 _ac_ 2 \+ 6 _a_ 2 _b_ )
( _f_ ) 4 _a_ _n_ +1 – 8 _a_ 2 _n_ = 4 _a_ _n_ +1(1 – 2 _a_ _n_ —1)
**Difference of Two Squares**
Type: _a_ 2 – _b_ 2 = ( _a + b_ )( _a – b_ )
**5.2** ( _a_ ) _x_ 2 – 9 = _x_ 2 – 32 = ( _x_ \+ 3)( _x_ – 3)
( _b_ ) 25 _x_ 2 – 4 _y_ 2 = (5 _x_ )2 – (2 _y_ )2 = (5 _x_ \+ 2 _y_ )(5 _x_ – 2 _y_ )
( _c_ ) 9 _x_ 2 _y_ 2 – 16 _a_ 2 = (3 _xy_ )2 – (4 _a_ )2 = (3 _xy_ \+ 4 _a_ )(3 _xy_ – 4 _a_ )
( _d_ ) 1 – _m_ 2 _n_ 4 = 12 – ( _mn_ 2)2 = (1 + _mn_ 2)(1 – _mn_ 2)
( _e_ ) 3 _x_ 2 – 12 = 3( _x_ 2 – 4) = 3( _x_ \+ 2)( _x_ – 2)
( _f_ ) _x_ 2 _y_ 2 – 36 _y_ 4 = _y_ 2[ _x_ 2 – (6 _y_ )2] = _y_ 2( _x_ \+ 6 _y_ )( _x_ — 6 _y_ )
( _g_ ) _x_ 4 – _y_ 4 = ( _x_ 2)2 – ( _y_ 2)2 = ( _x_ 2 \+ _y_ 2)( _x_ 2 – _y_ 2) = ( _x_ 2 \+ _y_ 2)( _x_ \+ _y_ )( _x_ – _y_ )
( _h_ ) 1 – _x_ 8 = (1 + _x_ 4)(1 – _x_ 4) = (1 + _x_ 4)(1 + _x_ 2)(1 – _x_ 2) = (1 + _x_ 4)(1 + _x_ 2)(1 + _x_ )(1 – _x_ )
( _i_ ) 32 _a_ 4 _b_ – 162b5 = 2 _b_ (16 _a_ 4 – 81 _b_ 4) = 2 _b_ (4 _a_ 2 \+ 9 _b_ 2)(4 _a_ 2 – 9 _b_ 2) = 2 _b_ (4 _a_ 2 \+ 9 _b_ 2)(2 _a_ \+ 3 _b_ )(2 _a_ – 3 _b)_
( _j_ ) _x_ 3 _y_ – _y_ 3 _x_ = _xy_ ( _x_ 2 – _y_ 2) = _xy_ ( _x_ \+ _y_ )( _x_ – _y_ )
( _k_ ) ( _x_ \+ 1)2 – 36 _y_ 2 = [( _x_ \+ 1) + (6 _y_ )][( _x_ \+ 1) – (6 _y_ )] = (x + 6 _y_ \+ 1)( _x_ – 6 _y_ \+ 1)
( _l_ ) (5 _x_ \+ 2 _y_ )2 – (3 _x_ – 7 _y_ )2 = [(5 _x_ \+ 2 _y_ ) + (3 _x_ – 7 _y_ )][(5 _x_ \+ 2 _y_ ) – (3 _x_ — 7 _y_ )] = (8 _x_ – 5 _y_ )(2 _x_ \+ 9 _y_ )
**Perfect Square Trinomials**
Types: _a_ 2 \+ 2 _ab_ \+ _b_ 2 = ( _a_ \+ _b_ )2
_a_ 2 – 2 _ab_ \+ _b_ 2 = ( _a_ – _b_ )2
**5.3** ( _a_ ) _x_ 2 \+ 8 _x_ \+ 16 = x2 \+ 2( _x_ )(4) + 42 = ( _x_ \+ 4)2
( _b_ ) 1 + 4 _y_ \+ 4 _y_ 2 = (1 + 2 _y_ )2
( _c_ ) _t_ 2 – 4 _t_ \+ 4 = _t_ 2 – 2( _t_ )(2) + 22 = ( _t_ – 2)2
( _d_ ) _x_ 2 – 16 _xy_ \+ 64 _y_ 2 = (x – 8 _y_ )2
( _e_ ) 25 _x_ 2 \+ 60 _xy_ \+ 36 _y_ 2 = (5 _x_ \+ 6 _y_ )2
( _f_ ) 16 _m_ 2 – 40mn + 25 _n_ 2 = (4 _m_ – 5 _n_ )2
( _g_ ) 9 _x_ 4 – 24 _x_ 2 _y_ \+ 16 _y_ 2 = (3 _x_ 2 – 4 _y_ )2
( _h_ ) 2 _x_ 3 _y_ 3 \+ 16 _x_ 2 _y_ 4 \+ 32 _xy_ 5 = 2 _xy_ 3( _x_ 2 \+ 8 _xy_ \+ 16 _y_ 2) = 2 _xy_ 3( _x_ \+ 4 _y_ )2
( _i_ ) 16 _a_ 4 – 72 _a_ 2 _b_ 2 \+ 81 _b_ 4 = (4 _a_ 2 – 9 _b_ 2)2 = [(2 _a_ \+ 3 _b_ )(2 _a_ – 3 _b_ )]2 = (2 _a_ \+ 3 _b_ )2(2 _a_ – 3 _b_ )2
( _j_ ) ( _x_ \+ 2 _y_ )2 \+ 10( _x_ \+ 2 _y_ ) + 25 = ( _x_ \+ 2 _y_ \+ 5)2
( _k_ ) _a_ 2 _x_ 2 – 2 _abxy_ \+ _b_ 2 _y_ 2 = ( _ax_ – _by_ )2
( _l_ ) 4 _m_ 6 _n_ 6 \+ 32 _m_ 4 _n_ 4 \+ 64 _m_ 2 _n_ 2 = 4 _m_ 2 _n_ 2( _m_ 4 _n_ 4 \+ 8 _m_ 2 _n_ 2 \+ 16) = 4 _m_ 2 _n_ 2( _m_ 2 _n_ 2 \+ 4)2
**Other Trinomials**
Types: _x_ 2 \+ ( _a_ \+ _b_ ) _x_ \+ _ab_ = ( _x_ \+ _a_ )( _x_ \+ _b_ )
_acx_ 2 \+ ( _ad_ \+ _bc_ ) _x_ \+ _bd_ = ( _ax_ \+ _b_ )( _cx_ \+ _d_ )
**5.4** ( _a_ ) _x_ 2 \+ 6 _x_ \+ 8 = ( _x_ \+ 4)( _x_ \+ 2)
( _b_ ) _x_ 2 – 6 _x_ \+ 8 = ( _x_ – 4)( _x_ – 2)
( _c_ ) _x_ 2 \+ 2 _x_ – 8 = ( _x_ \+ 4)( _x_ – 2)
( _d_ ) _x_ 2 – 2 _x_ – 8 = ( _x_ – 4)( _x_ \+ 2)
( _e_ ) _x_ 2 – 7 _xy_ \+ 12 _y_ 2 = ( _x_ – 3 _y_ )( _x_ – 4 _y_ )
( _f_ ) _x_ 2 \+ _xy_ – 12 _y_ 2 = ( _x_ \+ 4 _y_ )( _x_ – 3 _y_ )
( _g_ ) 16 – 10 _x_ \+ _x_ 2 = (8 – _x_ )(2 – _x_ )
( _h_ ) 20 – _x_ – _x_ 2 = (5 + _x_ )(4 – _x_ )
( _i_ ) 3 _x_ 3 – 3 _x_ 2 – 18 _x_ = 3 _x_ ( _x_ 2 – _x_ – 6) = 3 _x_ ( _x_ – 3)( _x_ \+ 2)
( _j_ ) _y_ 4 \+ 7 _y_ 2 \+ 12 = ( _y_ 2 \+ 4)( _y_ 2 \+ 3)
( _k_ ) _m_ 4 \+ _m_ 2 – 2 = ( _m_ 2 \+ 2)( _m_ 2 – 1) = ( _m_ 2 \+ 2)( _m_ \+ 1)( _m_ – 1)
( _l_ ) ( _x_ \+ 1)2 \+ 3( _x_ \+ 1) + 2 = [( _x_ \+ 1) + 2][( _x_ \+ 1) + 1] = ( _x_ \+ 3)( _x_ \+ 2)
( _m_ ) _s_ 2 _t_ 2 – 2 _st_ 3 – 63 _t_ 4 = _t_ 2(s2 – 2 _st_ – 63 _t_ 2) = _t_ 2( _s_ – 9 _t_ )( _s_ \+ 7 _t_ )
( _n_ ) _z_ 4 – 10 _z_ 2 \+ 9 = ( _z_ 2 – 1)( _z_ 2 – 9) = ( _z_ \+ 1)( _z_ – 1)( _z_ \+ 3)( _z_ – 3)
( _o_ ) 2 _x_ 6 _y_ –6 _x_ 4 _y_ 3–8 _x_ 2 _y_ 5=2 _x_ 2 _y_ ( _x_ 4–3 _x_ 2 _y_ 2–4 _y_ 4)=2 _x_ 2 _y_ ( _x_ 2+ _y_ 2)( _x_ 2–4 _y_ 2)=2 _x 2y_( _x_ 2+ _y_ 2)( _x_ +2 _y_ )( _x_ –2 _y_ )
( _p_ ) _x_ 2 – 2 _xy_ \+ _y_ 2 \+ 10( _x_ – _y_ ) + 9 = ( _x_ – _y_ )2 \+ 10( _x_ – _y_ ) + 9
= [( _x_ – _y_ ) + 1][ _x – y_ ) + 9] = ( _x_ – _y_ \+ 1)( _x_ – _y_ \+ 9)
( _q_ ) 4 _x_ 8 _y_ 10 – 40 _x_ 5 _y_ 7 \+ 84 _x_ 2 _y_ 4 = 4 _x_ 2 _y_ 4( _x_ 6 _y_ 6 – 10 _x_ 3 _y_ 3 \+ 21) = 4 _x_ 2 _y_ 4( _x_ 3 _y_ 3 – 7)( _x_ 3 _y_ 3 – 3)
( _r_ ) _x_ 2a – _x_ a – 30 = ( _x_ a – 6)( _x_ a \+ 5)
( _s_ ) _x_ _m_ +2 _n_ \+ 7 _x_ _m_ + _n_ \+ 10 _x m_ = _x m_( _x_ 2 _n_ \+ 7 _x n_ \+ 10) = _x m_( _x n_ \+ 2)( _x n_ \+ 5)
( _t_ ) _a_ 2( _y_ –1) – 5 _a_ _y_ –1 \+ 6 = ( _a_ _y_ –1 – 3)( _a_ _y_ –1 – 2)
**5.5** ( _a_ ) 3 _x_ 2 \+ 10 _x_ \+ 3 = (3 _x_ \+ 1)( _x_ \+ 3)
( _b_ ) 2 _x_ 2 – 7 _x_ \+ 3 = (2 _x_ – 1)( _x_ – 3)
( _c_ ) 2 _y_ 2 – _y_ – 6 = (2 _y_ \+ 3)( _y_ – 2)
( _d_ ) 10s2 \+ 11 _s_ – 6 = (5 _s_ – 2)(2 _s_ \+ 3)
( _e_ ) 6 _x_ 2 – _xy_ – 12 _y_ 2 = (3 _x_ \+ 4 _y_ )(2 _x_ – 3 _y_ )
( _f_ ) 10 – _x_ – 3 _x_ 2 = (5 – 3 _x_ )(2 + _x_ )
( _g_ ) 4 _z_ 4 – 9 _z_ 2 \+ 2 = ( _z_ 2 – 2)(4 _z_ 2 – 1) = ( _z_ 2 – 2)(2 _z_ \+ 1)(2 _z_ – 1)
( _h_ ) 16 _x_ 3 _y_ \+ 28 _x_ 2 _y_ 2 – 30 _xy_ 3 = 2 _xy_ (8 _x_ 2 \+ 14 _xy_ – 15 _y_ 2) = 2 _xy_ (4 _x_ – 3 _y_ )(2 _x_ \+ 5 _y_ )
( _i_ ) 12( _x_ \+ _y_ )2 \+ 8( _x_ \+ _y_ ) – 15 = [6( _x_ \+ _y_ ) – 5][2( _x_ \+ _y_ ) + 3] = (6 _x_ \+ 6 _y_ – 5)(2 _x_ \+ 2 _y_ \+ 3)
( _j_ ) 6 _b_ 2 _n_ +1 \+ 5 _b_ _n_ +1 – 6 _b_ = _b_ (6 _b_ 2 _n_ \+ 5b _n_ – 6) = _b_ (2 _b_ _n_ \+ 3)(3 _b_ _n_ – 2)
( _k_ ) 18 _x_ 4 _p_ + _m_ – 66 _x_ 2 _p_ + _m_ _y_ 2 – 24 _x_ m _y_ 4 = 6 _x_ m(3 _x_ 4 _p_ – 11 _x_ 2 _p_ _y_ 2 – 4 _y_ 4)
= 6 _x_ m(3 _x_ 2 _p_ \+ _y_ 2)( _x_ 2 _p_ – 4 _y_ 2) = 6 _x_ _m_ (3 _x_ 2 _p_ \+ _y_ 2)( _x_ _p_ \+ 2 _y_ )( _x_ _p_ – 2 _y_ )
( _l_ ) 64 _x_ 12 _y_ 3 – 68 _x_ 8 _y_ 7 \+ 4 _x_ 4 _y_ 11 = 4 _x_ 4 _y_ 3(l6 _x_ 8 – 17 _x_ 4 _y_ 4 \+ _y_ 8) = 4 _x_ 4 _y_ 3(l6 _x_ 4 – _y_ 4)( _x_ 4 – _y_ 4)
= 4 _x_ 4 _y_ 3(4 _x_ 2 \+ _y_ 2)(4 _x_ 2 – _y_ 2)( _x_ 2 \+ _y_ 2)( _x_ 2 – _y_ 2)
= 4 _x_ 4y3(4 _x_ 2 \+ _y_ 2)(2 _x_ \+ _y_ )(2 _x_ – _y_ )( _x_ 2 \+ _y_ 2)( _x_ \+ _y_ )( _x_ – _y_ )
**Sum of Difference of Two Cubes**
Types: _a_ 3 \+ _b_ 3 = ( _a_ \+ _b_ )( _a_ 2 – _ab_ \+ _b_ 2)
_a_ 3 – _b_ 3 = ( _a_ – _b_ )( _a_ 2 \+ _ab_ \+ _b_ 2)
**5.6** ( _a_ ) _x_ 3 \+ 8 = _x_ 3 \+ 23 = ( _x_ \+ 2)( _x_ 2 – 2 _x_ \+ 22) = ( _x_ \+ 2)( _x_ 2 – 2 _x_ \+ 4)
( _b_ ) _a_ 3 – 27 = _a_ 3 – 33 = ( _a_ – 3) =( _a_ 2 \+ 3 _a_ \+ 32) = (a – 3)( _a_ 2 \+ 3 _a_ +9)
( _c_ ) _a_ 6 \+ _b_ 6 = ( _a_ 2)3 \+ ( _b_ 2)3 = ( _a_ 2 \+ _b_ 2)[( _a_ 2)2 – _a_ 2 _b_ 2 \+ ( _b_ 2)2] = ( _a_ 2 \+ _b_ 2)( _a_ 4 – _a_ 2 _b_ 2 \+ _b_ 4)
( _d_ ) _a_ 6 – _b_ 6 = ( _a_ 3 \+ _b_ 3)( _a_ 3 – _b_ 3) = ( _a_ \+ _b_ )( _a_ 2 – _ab_ \+ _b_ 2)( _a_ – _b_ )( _a_ 2 \+ _ab_ + _b_ 2)
( _e_ ) _a_ 9 \+ _b_ 9 = ( _a_ 3)3 \+ ( _b_ 3)3 = ( _a_ 3 \+ _b_ 3)[( _a_ 3)2 – _a_ 3 _b_ 3 \+ ( _b_ 3)2] = ( _a + b_ )(a2 – _ab_ \+ _b_ 2)( _a_ 6 – _a_ 3 _b_ 3 \+ _b_ 6)
( _f_ ) _a_ 12 \+ _b_ 12 = ( _a_ 4)3 \+ ( _b_ 4)3 = ( _a_ 4 \+ _b_ 4)( _a_ 8 – _a_ 4 _b_ 4 \+ _b_ 8)
( _g_ ) 64 _x_ 3 \+ 125 _y_ 3 = (4 _x_ )3 \+ (5 _y_ )3 = (4 _x_ \+ 5 _y_ )[(4 _x_ )2 – (4 _x_ )(5 _y_ ) + (5 _y_ )2]
=(4 _x_ \+ 5 _y_ )(16 _x_ 2 – 20 _xy_ \+ 25 _y_ 2)
( _h_ ) ( _x_ + _y_ )3 – _z_ 3 = ( _x_ + _y_ – _z_ )[( _x_ + _y_ )2 \+ ( _x_ + _y_ ) _z + z_ 2]= ( _x_ + _y_ – _z_ )( _x_ 2 \+ 2 _xy_ \+ _y_ 2 \+ _xz_ \+ _yz_ \+ _z_ 2)
( _i_ )( _x – 2) 3 \+ 8 _y_ 3 = (x -_ 2)3 \+ _(2 _y_ )3 = (x – 2 + 2 _y_ )[(x -_ 2)2 – _(x – 2)(2 _y_ ) + (2 _y_ )2]_
= _(x – 2 + 2 _y_ )(x2 -_ 4 _x_ \+ 4 – 2 _xy_ \+ 4 _y_ \+ 4 _y_ 2)
( _j_ ) _x_ 6 – 7 _x_ 3 – 8 = ( _x_ 3 – 8)( _x_ 3 \+ 1) = ( _x_ 3 – 23)( _x_ 3 \+ 1) = ( _x_ – 2)( _x_ 2 \+ 2 _x_ \+ 4)( _x_ \+ 1)( _x_ 2 – _x_ \+ 1)
( _k_ ) x8y – 64 _x_ 2y7 = _x_ 2y( _x_ 6 – 64 _y_ 6) = _x_ 2y( _x_ 3 \+ 8 _y_ 3)( _x_ 3 – 8 _y_ 3) = _x_ 2 _y_ [ _x_ 3 \+ (2 _y_ )3][ _x_ 3 – (2 _y_ )3]
= _x 2 _y_ ( _x_ \+ 2 _y_ )( _x_ 2 – 2 _xy_ \+ 4 _y_ 2)( _x_ – 2 _y_ )( _x_ 2 \+ 2 _xy_ \+ 4 _y_ 2)_
( _l_ ) 54 _x_ 6 _y_ 2 – 38 _x_ 3 _y_ 2 – 16 _y_ 2 = 2 _y_ 2(27 _x_ 6 – 19 _x_ 3 – 8) = 2 _y_ 2(27 _x_ 3 \+ 8)( _x_ 3 – 1) = 2 _y_ 2(3 _x_ )3 \+ 23
= 2 _y_ 2(3 _x_ \+ 2)(9 _x_ 2 – 6 _x_ \+ 4)( _x_ – 1)( _x_ 2 \+ _x_ \+ 1)
**Grouping of Terms**
Type: _ac_ \+ _bc_ \+ _ad_ \+ _bd_ = _c_ ( _a_ \+ _b_ ) + _d_ ( _a_ \+ _b_ ) = ( _a_ \+ _b_ )( _c_ \+ _d_ )
**5.7** ( _a_ ) _bx_ – _ab_ \+ _x_ 2 – _ax_ = _b_ ( _x_ – _a_ ) + _x_ ( _x_ – _a_ ) = ( _x_ – _a_ )( _b_ \+ _x_ ) = ( _x_ – _a_ )( _x_ \+ _b_ )
( _b_ ) 3 _ax_ – _ay_ – 3 _bx_ \+ _by_ = _a_ (3 _x_ – _y_ ) – _b_ (3 _x_ – _y_ ) = (3 _x_ – _y_ )( _a_ – _b_ )
( _c_ ) 6 _x_ 2 – 4 _ax_ – 9 _bx_ \+ 6 _ab_ = 2 _x_ (3 _x_ – 2 _a_ ) – 3 _b_ (3 _x_ – 2 _a_ ) = (3 _x_ – 2 _a_ )(2 _x_ – 3 _b_ )
( _d_ ) _ax_ \+ _ay_ \+ _x_ \+ _y_ = _a_ ( _x_ \+ _y_ ) + ( _x_ \+ _y_ ) = ( _x_ \+ _y_ )( _a_ \+ 1)
( _e_ ) _x_ 2 – 4 _y_ 2 \+ _x_ \+ 2 _y_ = ( _x_ \+ 2 _y_ )( _x_ – 2 _y_ ) + ( _x_ \+ 2 _y_ ) = ( _x_ \+ 2 _y_ )( _x_ – 2 _y_ \+ 1)
( _f_ ) _x_ 3 \+ _x_ 2 _y_ \+ _xy_ 2 \+ _y_ 3 = _x_ 2( _x_ \+ _y_ ) + _y_ 2( _x_ \+ _y_ ) = ( _x_ \+ _y_ )( _x_ 2 \+ _y_ 2)
( _g_ ) _x_ 7 \+ 27 _x_ 4 – _x_ 3 – 27 = _x_ 4( _x_ 3 \+ 27) – ( _x_ 3 \+ 27) = ( _x_ 3 \+ 27)( _x_ 4 – 1)
= ( _x_ 3 \+ 33)( _x_ 2 \+ 1)( _x_ 2 – 1) = ( _x_ \+ 3)( _x_ 2 – 3 _x_ \+ 9)( _x_ 2 \+ 1)( _x_ \+ 1)( _x_ – 1)
( _h_ ) _x_ 3 _y_ 3 – _y_ 3 \+ 8 _x_ 3 – 8 = _y_ 3( _x_ 3 – 1) + 8( _x_ 3 – 1) = ( _x_ 3 – 1)( _y_ 3 \+ 8)
= ( _x_ – 1)( _x_ 2 \+ _x_ \+ 1)( _y_ \+ 2)( _y_ 2 – 2 _y_ \+ 4)
( _i_ ) _a_ 6 \+ _b_ 6 – _a_ 2 _b_ 4 – _a_ 4b2 = _a_ 6 – _a_ 2b4 \+ _b_ 6 – _a_ 4b2 = _a_ 2( _a_ 4 – _b_ 4) – _b_ 2( _a_ 4 – _b_ 4) = ( _a_ 4 – _b_ 4)(a2 – _b_ 2)
= ( _a_ 2 \+ _b_ 2)( _a_ 2 – _b_ 2)( _a_ \+ _b_ )( _a_ – _b_ ) = ( _a_ 2 \+ _b_ 2)( _a_ \+ _b_ )( _a_ – _b_ )( _a_ \+ _b_ )( _a_ – _b_ )
= ( _a_ 2 \+ _b_ 2)( _a_ \+ _b_ )2( _a_ – _b_ )2
( _j_ ) _a_ 3 \+ 3 _a_ 2 – 5 _ab_ \+ 2 _b_ 2 – _b_ 3 = ( _a_ 3 – _b_ 3) + (3 _a_ 2 – _5 _ab_ \+ 2 _b_ 2)_
= ( _a_ – _b_ )( _a_ 2 \+ _ab_ \+ _b_ 2) + ( _a_ – _b_ )(3 _a_ – 2 _b_ )
= ( _a_ – _b_ )( _a_ 2 \+ _ab_ \+ _b_ 2 \+ 3 _a_ – 2 _b_ )
**Factors of _a_ n ± _b_ n**
**5.8** _a n_ \+ _b n_ has _a_ \+ _b_ as a factor if and only if _n_ is a positive odd integer. Then
_a n_ \+ _b n_ = ( _a_ \+ _b_ )( _a_ _n_ –1 – _a_ _n_ –2 _b_ \+ _a_ _n_ –3 _b_ 2 – ··· – _ab_ _n–_ 2 \+ _b_ _n_ –1).
( _a_ ) _a_ 3 \+ _b_ 3 = ( _a_ \+ _b_ )( _a_ 2 – _ab_ \+ _b_ 2)
( _b_ ) 64 + _y_ 3 = 43 \+ _y_ 3 = (4 + _y_ )(42 – 4 _y_ \+ _y_ 2) = (4 + _y_ )(16 – 4 _y_ \+ _y_ 2)
( _c_ ) _x_ 3 \+ 8 _y_ 6 = _x_ 3 \+ (2 _y_ 2)3 = ( _x_ \+ 2 _y_ 2)[ _x_ 2 – _x_ (2 _y_ 2) + (2 _y_ 2)2] = ( _x_ \+ 2 _y_ 2)( _x_ 2 – 2 _xy_ 2 \+ 4 _y_ 4)
( _d_ ) _a_ 5 \+ _b_ 5 = ( _a_ \+ _b_ )( _a_ 4 – _a_ 3 _b_ \+ _a_ 2 _b_ 2 – _ab_ 3 \+ _b_ 4)
( _e_ ) 1 + _x_ 5 _y_ 5 = 15 \+ ( _xy_ )5 = (1 + _xy_ )(1 – _xy_ \+ _x_ 2 _y_ 2 – _x_ 3y3 \+ _x_ 4 _y_ 4)
( _f_ ) _z_ 5 \+ 32 = _z_ 5 \+ 25 = ( _z_ \+ 2)( _z_ 4 – 2 _z_ 3 \+ 22 _z_ 2 – 23 _z_ \+ 24) = ( _z_ \+ 2)( _z_ 4 – 2 _z_ 3 \+ 4 _z_ 2 – 8 _z_ \+ 16)
( _g_ ) _a_ 10 \+ _x_ 10 = ( _a_ 2)5 \+ ( _x_ 2)5 = ( _a_ 2 \+ _x_ 2)[( _a_ 2)4 – ( _a_ 2)3x2 \+ ( _a_ 2)2( _x_ 2)2 – ( _a_ 2)( _x_ 2)3 \+ ( _x_ 2)4]
= ( _a_ 2 \+ _x_ 2)( _a_ 8 – a6 _x_ 2 \+ _a_ 4 _x_ 4 – a2 _x_ 6 \+ _x_ 8)
( _h_ ) _u_ 7 \+ _v_ 7 = ( _u_ \+ _v_ )( _u_ 6 – _u_ 5 _v_ \+ _u_ 4 _v_ 2 – _u_ 3 _v_ 3 \+ _u_ 2 _v_ 4 – _uv_ 5 \+ _v_ 6)
( _i_ ) _x_ 9 \+ 1 = ( _x_ 3)3 \+ 13 = ( _x_ 3 \+ 1)( _x_ 6 – _x_ 3 \+ 1) = ( _x_ \+ 1)( _x_ 2 – _x_ \+ 1)( _x_ 6 – _x_ 3 \+ 1)
**5.9** _a n_ – _b n_ has a – b as a factor if _n_ is any positive integer. Then
_a n_ – _b n_ = ( _a_ – _b_ )( _a_ _n–_ 1 + _a_ _n–_ 2 _b_ \+ _a_ _n–_ 3 _b_ 2 \+ ··· + _ab_ _n–_ 2 \+ _b_ _n–_ 1).
If n is an even positive integer, _a n_ – _b n_ also has _a + b_ as factor.
( _a_ ) _a_ 2 – _b_ 2 = ( _a_ – _b_ )( _a_ \+ _b_ )
( _b_ ) _a_ 3 – _b_ 3 = ( _a_ – _b_ )( _a_ 2 \+ _ab_ \+ _b_ 2)
( _c_ ) 27 _x_ 3 – _y_ 3 = (3 _x_ )3 – _y_ 3 = (3 _x_ – _y)[(3 _x_ )2_ \+ (3 _x_ ) _y_ \+ _y_ 2] = (3 _x_ – _y_ )(9 _x_ 2 \+ 3 _xy_ \+ _y_ 2)
( _d_ ) 1 – _x_ 3 = (1 – _x_ )(12 \+ 1 _x_ \+ _x_ 2) = (1 – _x_ )(1 + _x_ \+ _x_ 2)
( _e_ ) _a_ 5 – 32 = _a_ 5 – 25 = ( _a_ – 2)( _a_ 4 \+ _a_ 3 · 2 + _a_ 2 · 22 \+ _a_ · 23 \+ 24) = ( _a_ – 2)( _a_ 4 \+ 2 _a_ 3 \+ 4 _a_ 2 \+ 8 _a_ \+ 16)
( _f_ ) _y_ 7 – _z_ 7 = ( _y_ – _z_ )( _y_ 6 \+ _y_ 5 _z_ \+ _y_ 4 _z_ 2 \+ _y_ 3 _z_ 3 \+ _y_ 2 _z_ 4 \+ _yz_ 5 \+ _z_ 6)
( _g_ ) _x_ 6 – _a_ 6 = ( _x_ 3 \+ _a_ 3)( _x_ 3 – _a_ 3) = ( _x_ \+ _a_ )( _x_ 2 – _ax_ \+ _a_ 2)( _x_ – _a_ )( _x_ 2 \+ _ax_ \+ _a_ 2)
( _h_ ) _u_ 8 – _v_ 8 = ( _u_ 4 \+ _v_ 4)( _u_ 4 – _v_ 4) = ( _u_ 4 \+ _v_ 4)( _u_ 2 \+ _v_ 2)( _u_ 2 – _v_ 2) = ( _u_ 4 \+ _v_ 4)( _u_ 2 \+ _v_ 2)( _u_ \+ _v_ )( _u_ – _v_ )
( _i_ ) _x_ 9 – 1 = ( _x_ 3)3 – 1 = ( _x_ 3 – 1)( _x_ 6 \+ _x_ 3 \+ 1) = ( _x_ – 1)( _x_ 2 \+ _x_ \+ 1)( _x_ 6 \+ _x_ 3 \+ 1)
( _j_ ) _x_ 10 – _y_ 10 = ( _x_ 5 \+ _y_ 5)( _x_ 5 – _y_ 5) = ( _x_ \+ _y_ )( _x_ 4 – _x_ 3 _y_ \+ _x_ 2 _y_ 2 – _xy_ 3 \+ _y_ 4)( _x_ – _y_ )( _x_ 4 \+ _x_ 3 _y_ \+ _x_ 2 _y_ 2 \+ _xy_ 3 \+ _y_ 4)
**Addition and Subtraction of Suitable Terms**
**5.10** ( _a_ ) _a_ 4 \+ _a_ 2 _b_ 2 \+ _b_ 4 (adding and subtracting _a_ 2 _b_ 2)
= ( _a_ 4 \+ 2 _a_ 2 _b_ 2 \+ _b_ 4) – _a_ 2 _b_ 2 ( _a_ 2 \+ _b_ 2)2 – ( _ab_ )2
= ( _a_ 2 \+ _b_ 2 \+ _ab_ )( _a_ 2 \+ _b_ 2 – _ab_ )
( _b_ ) 36 _x_ 4 \+ 15 _x_ 2 \+ 4 (adding and subtracting 9 _x_ 2)
= (36 _x_ 4 \+ 24 _x_ 2 \+ 4) – 9 _x_ 2 = (6 _x_ 2 \+ 2)2 – (3 _x_ )2
= [(6 _x_ 2 \+ 2) + 3 _x_ ][(6 _x_ 2 \+ 2) – 3 _x_ ] = (6 _x_ 2 \+ 3 _x_ \+ 2)(6 _x_ 2 – 3 _x_ \+ 2)
( _c_ ) 64 _x_ 4 \+ _y_ 4 (adding and subtracting 16 _x_ 2 _y_ 2)
= (64 _x_ 4 \+ 16 _x_ 2y2 \+ y4) – 16 _x_ 2 _y_ 2
= (8 _x_ 2 \+ y2)2 – (4 _xy_ )2= (8 _x_ 2 \+ y2 \+ 4 _xy_ )(8 _x_ 2 \+ y2 – 4 _xy_ )
( _d_ ) _u_ 8 – 14 _u_ 4 \+ 25 (adding and subtracting 4 _u_ 4)
= ( _u_ 8 – 10 _u_ 4 \+ 25) – 4 _u_ 4 = ( _u_ 4 – 5)2 – (2 _u_ 2)2
= ( _u_ 4 – 5 + 2 _u_ 2)( _u_ 4 – 5 – 2 _u_ 2) = ( _u_ 4 \+ 2 _u_ 2 – 5)( _u_ 4 – 2 _u_ 2 – 5)
**Miscellaneous Problems**
**5.11** ( _a_ ) _x_ 2 – 4 _z_ 2 \+ 9 _y_ 2 – 6 _xy_ = ( _x_ 2 – 6 _xy_ \+ 9 _y_ 2) – 4 _z_ 2
= ( _x_ – 3 _y_ )2 – (2 _z_ )2 = ( _x_ – 3 _y_ \+ 2 _z_ )( _x_ – 3 _y_ – 2 _z_ )
( _b_ ) 16 _a_ 2 \+ 10 _bc_ – 25 _c_ 2 – _b_ 2 = 16 _a_ 2 – ( _b_ 2 – 10 _bc_ \+ 25 _c_ 2)
= (4 _a_ )2 – ( _b_ – 5 _c_ )2 = (4 _a_ \+ _b_ – 5 _c_ )(4 _a_ – _b_ \+ 5 _c_ )
( _c_ ) _x_ 2 \+ 7 _x_ \+ _y_ 2 – 7 _y_ – 2 _xy_ – 8 = ( _x_ 2 – 2 _xy_ \+ _y_ 2) + 7( _x_ – _y_ ) – 8
= ( _x_ – _y_ )2 \+ 7( _x_ – _y_ ) – 8 = ( _x_ – _y_ \+ 8)( _x_ – _y_ – 1)
( _d_ ) _a_ 2 – 8 _ab_ – 2 _ac_ \+ 16 _b_ 2 \+ 8 _bc_ – 15 _c_ 2 = ( _a_ 2 – 8 _ab_ \+ 16 _b_ 2) – (2 _ac_ – 8 _bc_ ) – 15 _c_ 2
= (a – 4b)2 – 2c(a – 4b) – 15c2 = (a – 4b – 5c)(a – 4b + 3c)
( _e_ ) _m_ 4 – _n_ 4 \+ _m_ 3 – _mn_ 3 – _n_ 3 \+ _m_ 3 _n_ = ( _m_ 4 – _mn_ 3) + ( _m_ 3 _n_ – _n_ 4) + ( _m_ 3 – _n_ 3)
= _m_ ( _m_ 3 – _n_ 3) + _n_ ( _m_ 3 – _n_ 3) + ( _m_ 3 – _n_ 3)
= ( _m_ 3 – _n_ 3)( _m_ \+ _n_ \+ 1) = ( _m_ – _n_ )( _m_ 2 \+ _mn_ \+ _n_ 2)( _m_ \+ _n_ \+ 1)
**Greatest Common Factor and Least Common Multiple**
**5.12** ( _a_ ) 9 _x_ 4 _y_ 2 = 32 _x_ 4 _y_ 2, 12 _x_ 3 _y_ 3 = 22 · 3 _x_ 3 _y_ 3
GCF = 3 _x_ 3 _y_ 2, LCM = 22 · 32 _x_ 4 _y_ 3 = 36 _x_ 4 _y_ 3
( _b_ ) 48 _r_ 3 _t_ 4 = 24 · 3 _r_ 3 _t_ 4, 54 _r_ 2 _t_ 6 = 2 · 33 _r_ 2 _t_ 6, 60 _r_ 4 _t_ 2 = 22 · 3 · 5 _r_ 4 _t_ 2
GCF = 2 · 3 _r_ 2 _t_ 2 = 6 _r_ 2 _t_ 2, LCM = 24 · 33 · 5 _r_ 4 _t_ 6 = 2160 _r_ 4 _t_ 6
( _c_ ) 6 _x_ – 6 _y_ = 2 · 3( _x_ – _y_ ), 4 _x_ 2 – 4 _y_ 2 = 22( _x_ 2 – _y_ 2) = 22( _x_ \+ _y_ )( _x_ – _y_ )
GCF = 2( _x_ – _y_ ), LCM = 22 · 3( _x_ \+ _y_ )( _x_ – _y_ )
( _d_ ) _y_ 4 – 16 = ( _y_ 2 \+ 4)( _y_ \+ 2)( _y_ – 2), _y_ 2 – 4 = ( _y_ \+ 2)( _y_ – 2), _y_ 2 – 3 _y_ \+ 2 = ( _y_ – 1)( _y_ – 2)
GCF = _y_ – 2, LCM = ( _y_ 2 \+ 4)( _y_ \+ 2)( _y_ – 2)( _y_ – 1)
( _e_ ) 3 · 52( _x_ \+ 3 _y_ )2(2 _x_ – y)4, 23 · 32 · 5( _x_ \+ 3 _y_ )3(2 _x_ – _y_ )2, 22 · 3 · 5( _x_ \+ 3 _y_ )4(2 _x_ – _y_ )5
GCF = 3 · 5( _x_ \+ 3 _y_ )2(2 _x_ – _y_ )2, LCM = 23 · 32 · 52( _x_ \+ 3 _y_ )4(2 _x_ – _y_ )5
### **Supplementary Problems**
Factor each expression.
**5.13** ( _a_ ) 3 _x_ 2 _y_ 4\+ 6 _x_ 3 _y_ 3
( _b_ ) 12s2 _t_ 2 – 6s5 _t_ 4 – 4 _s_ 4 _t_
( _c_ ) 2 _x_ 2yz – 4 _xyz_ 2 \+ 8 _xy_ 2 _z_ 3
( _d_ ) 4 _y_ 2 –100
( _e_ ) 1 – _a_ 4
( _f_ ) 64 _x_ – _x_ 3
( _g_ ) 8 _x_ 4 – 128
( _h_ ) 18 _x_ 3 _y_ – 8 _xy_ 3
( _i_ ) (2 _x_ \+ _y_ )2 – (3 _y_ – _z_ )2
( _j_ ) 4( _x_ \+ 3 _y_ )2 – 9(2 _x_ – _y_ )2
( _k_ ) _x_ 2 \+ 4 _x_ \+ 4
( _l_ ) 4 – 12 _y_ \+ 9 _y_ 2
( _m_ ) _x_ 2y2 – 8 _xy_ \+ 16
( _n_ ) 4 _x_ 3 _y_ \+ 12 _x_ 2 _y_ 2 \+ 9 _xy_ 3
( _o_ ) 3 _a_ 4 \+ 6 _a_ 2 _b_ 2 \+ 3 _b_ 4
( _p_ ) ( _m_ 2 – _n_ 2)2 \+ 8( _m_ 2 – _n_ 2 \+ 16
( _q_ ) _x_ 2 \+ 7 _x_ \+ 12
( _r_ ) _y_ 2 – 4 _y_ – 5
( _s_ ) _x_ 2 – 8 _xy_ \+ 15 _y_ 2
( _t_ ) 2 _z_ 3 \+ 10 _z_ 2 – 28 _z_
( _u_ ) 15 + 2 _x_ – x2
**5.14** ( _a_ ) _m_ 4 – 4 _m_ 2 – 21
( _b_ ) _a_ 4 – 20 _a_ 2 \+ 64
( _c_ ) 4 _s_ 4 _t_ – 4s3 _t_ 2 – 24 _s_ 2 _t_ 3
( _d_ ) _x_ 2 _m_ +4 \+ 5 _x_ _m_ +4 – 50 _x_ 4
( _e_ ) 2 _x_ 2 \+ 3 _x_ \+ 1
( _f_ ) 3 _y_ 2 – 11 _y_ \+ 6
( _g_ ) 5 _m_ 3 – 3 _m_ 2 – 2 _m_
( _h_ ) 6 _x_ 2 \+ 5 _xy_ – 6 _y_ 2
( _i_ ) 36 _z_ 6 – 13 _z_ 4 \+ _z_ 2
( _j_ ) 12( _x_ – _y_ )2 \+ 7( _x_ – _y_ ) – 12
( _k_ ) 4 _x_ 2 _n_ +2 – 4 _x_ _n_ +2 – 3 _x_ 2
**5.15** ( _a_ ) _y_ 3 \+ 27
( _b_ ) _x_ 3 – 1
( _c_ ) _x_ 3 _y_ 3 \+ 8
( _d_ ) 8 _z_ 4 – 27 _z_ 7
( _e_ ) 8 _x_ 4y – 64 _xy_ 4
( _f_ ) _m_ 9 – _n_ 9
( _g_ ) _y_ 6 +1
( _h_ ) ( _x_ – 2)3 \+ ( _y_ \+ 1)3
( _i_ ) 8 _x_ 6 \+ 7 _x_ 3 – 1
**5.16** ( _a_ ) _xy_ \+ 3 _y_ – 2 _x_ – 6
( _b_ ) 2 _pr_ – _ps_ \+ 6 _qr_ – 3 _qs_
( _c_ ) _ax_ 2 \+ _bx_ – _ax_ – _b_
( _d_ ) _x_ 3 – _xy_ 2 – _x_ 2 _y_ \+ _y_ 3
( _e_ ) _z_ 7 – 2 _z_ 6 \+ _z_ 4 – 2 _z_ 3
( _f_ ) _m_ 3 – _mn_ 2 \+ m2 _n_ – _n_ 3 \+ _m_ 2 – _n_ 2
**5.17** ( _a_ ) z5 \+ 1
( _b_ ) _x_ 5 \+ 32 _y_ 5
( _c_ ) 32 – _u_ 5
( _d_ ) _m_ 10 – 1
( _e_ ) 1 – _z_ 7
**5.18** ( _a_ ) _z_ 4 \+ 64
( _b_ ) 4 _x_ 4 \+ 3 _x_ 2 _y_ 2 \+ _y_ 4
( _c_ ) _x_ 8 – 12 _x_ 4 \+ 16
( _d_ ) _m_ 2 – 4 _p_ 2 \+ 4 _mn_ \+ 4 _n_ 2
( _e_ ) 6 _ab_ \+ 4 – _a_ 2 – 9 _b_ 2
( _f_ ) 9 _x_ 2 – _x_ 2y2 \+ 4 _y_ 2 \+ 12 _xy_
( _g_ ) x2 \+ y2 – 4z2 \+ 2 _xy_ \+ 3 _xz_ \+ 3 _yz_
**5.19** Find the GCF and LCM of each group of polynomials.
( _a_ ) 16 _y_ 2 _z_ 4, 24 _y_ 3 _z_ 2
( _b_ ) 9 _r_ 3 _s_ 2 _t_ 5, 12 _r_ 2 _s_ 4 _t_ 3, 21 _r_ 5 _s_ 2
( _c_ ) x2 – 3 _xy_ \+ 2 _y_ 2, 4 _x_ 2 – 16 _xy_ \+ 16 _y_ 2
( _d_ ) 6 _y_ 3 \+ 12 _y_ 2z, 6 _y_ 2 – 24z2, 4 _y_ 2 – 4 _yz_ – 24z2
( _e_ ) x5 – x, x5 – x2, x5 – x3
#### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**5.13** ( _a_ ) 3 _x_ 2y3(y + 2 _x_ )
( _b_ ) 2 _s_ 2 _t_ (6 _t_ – 3 _s_ 3 _t_ 3 \+ 2 _s_ 2)
( _c_ ) 2 _xyz_ ( _x_ – 2 _z_ \+ 4 _yz_ 2)
( _d_ ) 4( _y_ \+ 5)( _y_ – 5)
( _e_ ) (1+ _a_ 2)(1 + _a_ )(1 – _a_ )
( _f_ ) _x_ (8 + _x_ )(8 – _x_ )
( _g_ ) 8( _x_ 2 \+ 4)( _x_ \+ 2)( _x_ – 2)
( _h_ ) 2 _xy_ (3 _x_ \+ 2 _y_ )(3 _x_ – 2 _y_ )
( _i_ ) (2 _x_ \+ 4 _y_ – z)(2 _x_ – 2 _y_ \+ z)
( _j_ ) (8 _x_ \+ 3 _y_ )(9 _y_ – 4 _x_ )
( _k_ ) ( _x_ \+ 2)2
( _l_ ) (2 – 3 _y_ )2
( _m_ ) (xy – 4)2
( _n_ ) _xy_ (2 _x_ \+ 3 _y_ )2
( _o_ ) 3( _a_ 2 \+ _b_ 2)2
( _p_ ) ( _m_ 2 – _n_ 2 \+ 4)2
( _q_ ) ( _x_ \+ 3)( _x_ \+ 4)
( _r_ ) ( _y_ – 5)( _y_ \+ 1)
( _s_ ) ( _x_ – 3 _y_ )( _x_ \+ 5 _y_ )
( _t_ ) 2 _z_ ( _z_ \+ 7)( _z_ – 2)
( _u_ ) (5 – _x_ )(3 + _x_ )
**5.14** ( _a_ ) ( _m_ 2 – 7)( _m_ 2 \+ 3)
( _b_ ) ( _a_ \+ 2)( _a_ – 2)( _a_ \+ 4)( _a_ – 4)
( _c_ ) 4 _s_ 2 _t_ ( _s_ – 3 _t_ )( _s_ \+ 2 _t_ )
( _d_ ) _x_ 4( _x_ m – 5)( _x_ _m_ \+ 10)
( _e_ ) (2 _x_ \+ 1)(x + 1)
( _f_ ) (3 _y_ – 2)( _y_ – 3)
( _g_ ) _m_ (5 _m_ \+ 2)( _m_ – 1)
( _h_ ) (2 _x_ \+ 3 _y_ )(3 _x_ – 2 _y_ )
( _i_ ) _z_ 2(2 _z_ \+ 1)(2 _z_ – 1)(3 _z_ \+ 1)(3 _z_ – 1)
( _j_ ) (4 _x_ – 4 _y_ – 3)(3 _x_ – 3 _y_ \+ 4
( _k_ ) _x_ 2(2 _x n_ \+ 1)(2 _x_ _n_ – 3)
**5.15** ( _a_ )(y + 3)(y2 – 3 _y_ \+ 9)
( _b_ ) ( _x_ – 1)( _x_ 2 \+ _x_ \+ 1)
( _c_ ) ( _xy_ \+ 2)( _x_ 2 _y_ 2 – 2 _xy_ \+ 4)
( _d_ ) _z_ 4(2 – 3 _z_ )(4 + 6 _z_ \+ 9 _z_ 2)
( _e_ ) 8 _xy_ ( _x_ – 2 _y_ )( _x_ 2 \+ 2 _xy_ \+ 4 _y_ 2)
( _f_ ) ( _m_ – _n_ )( _m_ 2 \+ _mn_ \+ _n_ 2)( _m_ 6 \+ _m_ 3 _n_ 3 \+ _n_ 6)
( _g_ ) ( _y_ 2 \+ 1)( _y_ 4 – _y_ 2 \+ 1)
( _h_ ) ( _x_ \+ _y_ – 1)( _x_ 2 – _xy_ \+ _y_ 2 – 5 _x_ \+ 4 _y_ \+ 7)
( _i_ ) (2 _x_ – 1)(4 _x_ 2 \+ 2 _x_ \+ 1)( _x_ \+ 1)( _x_ 2 – _x_ \+ 1)
**5.16** ( _a_ )( _x_ \+ 3)( _y_ – 2)
( _b_ ) (2 _r_ – _s_ )( _p_ \+ 3 _q_ )
( _c_ ) ( _ax_ \+ _b_ )( _x_ – 1)
( _d_ ) ( _x_ – _y_ )2( _x_ \+ _y_ )
( _e_ ) _z_ 3( _z_ – 2)( _z_ \+ 1)( _z_ 2 – _z_ \+ 1)
( _f_ ) ( _m_ \+ _n_ )( _m_ – _n_ )( _m_ \+ _n_ \+ 1)
**5.17** ( _a_ ) ( _z_ \+ 1)( _z_ 4 – _z_ 3 \+ _z_ – _z_ \+ 1)
( _b_ ) ( _x_ \+ 2 _y_ )( _x_ 4 – 2 _x_ 3 _y_ \+ 4 _x_ 2 _y_ 2 – 8 _xy_ 3 \+ 16 _y_ 4)
( _c_ ) (2 – _u_ )(16 + 8 _u_ \+ 4 _u_ 2 \+ 2 _u_ 3 \+ _u_ 4)
( _d_ ) ( _m_ \+ 1)( _m_ 4 – _m_ 3 \+ _m_ 2 – _m_ \+ 1)( _m_ – 1)( _m_ 4 \+ _m_ 3 \+ _m_ 2 \+ _m_ \+ 1)
( _e_ ) (1 – _z_ )(1 + _z_ \+ _z_ 2 \+ _z_ 3 \+ _z_ 4 \+ _z_ 5 \+ _z_ 6)
**5.18** ( _a_ ) ( _z_ 2 \+ 4 _z_ \+ 8)( _z_ 2 – 4 _z_ \+ 8)
( _b_ ) (2 _x_ 2 \+ _xy_ \+ _y_ 2)(2 _x_ 2 – _xy_ \+ _y_ 2)
( _c_ ) ( _x_ 4 \+ 2 _x_ 2 – 4)( _x_ 4 – 2 _x_ 2 – 4)
( _d_ ) ( _m_ \+ 2 _n_ \+ 2 _p_ )( _m_ \+ 2 _n_ – 2 _p_ )
( _e_ ) (2 + _a_ – 3 _b_ )(2 – _a_ \+ 3 _b_ )
( _f_ ) (3 _x_ \+ _xy_ \+ 2 _y_ )(3 _x_ – _xy_ \+ 2 _y_ )
( _g_ ) ( _x_ \+ _y_ \+ 4 _z_ )( _x_ \+ _y_ – _z_ )
**5.19**
## CHAPTER 6
Fractions
### **6.1 RATIONAL ALGEBRAIC FRACTIONS**
A rational algebraic fraction is an expression which can be written as the quotient of two polynomials, _P/Q. P_ is called the numerator and _Q_ the denominator of the fraction. Thus
are rational algebraic fractions.
Rules for manipulation of algebraic fractions are the same as for fractions in arithmetic. One such fundamental rule is: The value of a fraction is unchanged if its numerator and denominator are both multiplied by the same quantity or both divided by the same quantity, provided only that this quantity is not zero. In such case we call the fractions _equivalent_.
For example, if we multiply the numerator and denominator of ( _x_ \+ 2)/( _x_ – 3) by ( _x_ – 1) we obtain the equivalent fraction
provided ( _x_ – 1) is not zero, i.e., provided _x_ ≠ 1.
Similarly, given the fraction ( _x_ 2 \+ 3 _x_ \+ 2)/( _x_ 2 \+ 4 _x_ \+ 3) we may write it as
and divide numerator and denominator by ( _x_ \+ 1) to obtain ( _x_ \+ 2)/( _x_ \+ 3) provided ( _x_ \+ 1) is not zero, i.e., provided _x_ ≠ –1. The operation of dividing out common factors of the numerator and denominator is called _cancellation_ and may be indicated by a sloped line thus:
To simplify a given fraction is to convert it into an equivalent form in which numerator and denominator have no common factor (except ±1). In such case we say that the fraction is _reduced to lowest terms_. This reduction is achieved by factoring numerator and denominator and canceling common factors assuming they are not equal to zero.
Three signs are associated with a fraction: the sign of the numerator, of the denominator, and of the entire fraction. Any two of these signs may be changed without changing the value of the fraction. If there is no sign before a fraction, a plus sign is implied.
**EXAMPLES 6.1.**
Change of sign may often be of use in simplification. Thus
### **6.2 OPERATIONS WITH ALGEBRAIC FRACTIONS**
The algebraic sum of fractions having a _common denominator_ is a fraction whose numerator is the algebraic sum of the numerators of the given fractions and whose denominator is the common denominator.
**EXAMPLES 6.2.**
To add and subtract fractions having _different denominators_ , write each of the given fractions as equivalent fractions all having a common denominator.
The _least common denominator_ (LCD) of a given set of fractions is the LCM of the denominators of the fractions.
Thus the LCD of and is the LCM of 4, 5, 10 which is 20, and the LCD of
**EXAMPLES 6.3.**
The product of two or more given fractions produces a fraction whose numerator is the product of the numerators of the given fractions and whose denominator is the product of the denominators of the given fractions.
**EXAMPLES 6.4.**
The quotient of two given fractions is obtained by inverting the divisor and then multiplying.
**EXAMPLES 6.5.**
### **6.3 COMPLEX FRACTIONS**
A complex fraction is one which has one or more fractions in the numerator or denominator, or in both. To simplify a complex fraction:
#### **Method 1**
(1) Reduce the numerator and denominator to simple fractions.
(2) Divide the two resulting fractions.
**EXAMPLE 6.6.**
#### **Method 2**
(1) Multiply the numerator and denominator of the complex fraction by the LCM of all denominators of the fractions in the complex fraction.
(2) Reduce the resulting fraction to lowest terms.
**EXAMPLE 6.7.**
### **Solved Problems**
**Reduction of Fractions to Lowest Terms**
**6.1** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
( _j_ )
**Multiplication of Fractions**
**6.2** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
**Division of Fractions**
**6.3** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
**Addition and Subtraction of Fractions**
**6.4** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
( _j_ )
( _k_ )
( _l_ )
**Complex Fractions**
**6.5** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
### **Supplementary Problems**
Show that:
**6.6** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
**6.7** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
**6.8** ( _a_ )
( _b_ )
( _c_ )
**6.9** ( _a_ )
( _b_ )
( _c_ )
**6.10** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
**6.11** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
## **CHAPTER 7
Exponents**
### **7.1 POSITIVE INTEGRAL EXPONENT**
If _n_ is a positive integer, _a n_ represents the product of _n_ factors each of which is _a_. Thus _a_ 4 _= a_ · _a_ · _a_ · _a_. In _a_ _n_ , _a_ is called the base and _n_ the exponent or index. We may read _a_ _n_ as the " _n_ th power of _a_ " or " _a_ to the _n_ th." If _n_ = 2 we read _a_ 2 as " _a_ squared"; _a_ 3 is read " _a_ cubed."
**EXAMPLES 7.1.**
_x_ 3 = _x_ · _x_ · _x_ , 25 = 2 · 2 · 2 · 2 · 2 = 32, (–3)3 = (–3)(–3)(–3) = –27
### **7.2 NEGATIVE INTEGRAL EXPONENT**
If _n_ is a positive integer, we define
**EXAMPLES 7.2.**
### **7.3 ROOTS**
If _n_ is a positive integer and if _a_ and _b_ are such that _a n = b_, then _a_ is said to be an _n_ th root of _b_.
If _b_ is positive, there is only one positive number _a_ such that _a n = b_. We write this positive number and call it the _principal n_ th root of _b_.
**EXAMPLES 7.3.** is that positive number which when raised to the 4th power yields 16. Clearly this is +2, so we write
**EXAMPLES 7.4.** The number –2 when raised to the 4th power also yields 16. We call –2 a 4th root of 16 but not the principal 4th root of 16.
If _b_ is negative, there is no positive _n_ th root of _b_ , but there is a negative _n_ th root of _b_ if _n_ is odd. We call this negative number the principal _n_ th root of _b_ and we write it
**EXAMPLES 7.5.** is that number which raised to the third power (or cubed) yields –27. Clearly this is –3 and so we write as the principal cube root of –27.
**EXAMPLES 7.6.** If _n_ is even, as in there is no principal _n_ th root in terms of real numbers.
_Note_. In advanced mathematics it can be shown that there are exactly _n_ values of _a_ such that _a_ _n_ = _b, b_ ≠ 0, provided we allow imaginary (or complex) numbers.
### **7.4 RATIONAL EXPONENTS**
If _m_ and _n_ are positive integers we define
**EXAMPLES 7.7.**
If _m_ and _n_ are positive integers we define
**EXAMPLES 7.8.**
We define _a_ 0 = 1 if a ≠ 0.
**EXAMPLES 7.9.**
### **7.5 GENERAL LAWS OF EXPONENTS**
If _p_ and _q_ are real numbers, the following laws hold.
_A_. _a_ _p_ · _a_ _q_ = _a_ _p+q_
**EXAMPLES 7.10.**
_B_. ( _a_ _p_ ) _q_ = _a_ _pq_
**EXAMPLES 7.11.**
_C_.
**EXAMPLES 7.12.**
_D_. ( _ab_ ) _p_ = _a_ _p_ _b_ _p_
**EXAMPLES 7.13.**
_E_.
**EXAMPLES 7.14.**
### **7.6 SCIENTIFIC NOTATION**
Very large and very small numbers are often written in scientific notation when they are used in computation. A number is written in scientific notation by expressing it as a number _N_ times a power of 10 where 1 _≤ N <_ 10 and _N_ contains all of the significant digits on the number.
**EXAMPLES 7.15.** Write each number in scientific notation. ( _a_ ) 5 834 000, ( _b_ ) 0.028 031, ( _c_ ) 45.6.
( _a_ ) 5 834 000 = 5.834 × 106
( _b_ ) 0.028 031 = 2.8031 × 10–2
( _c_ ) 45.6 = 4.56 × 101
We can enter the number 3.1416 × 103 into a calculator by using the EE key or the EXP key. When we enter 3.1416, press the EE key, and then enter 3 followed by pressing the ENTER key or the = key, we get a display of 3141.6. Similarly, we can enter 4.902 × 10–2 by entering 4.902, pressing the EE key, and then entering –2 followed by pressing the ENTER key to obtain a display of 0.049 02. The exponent can usually be any integer from –99 to 99. Depending on the number of digits in the number and the exponent used, a calculator may round off the number and/or leave the result in scientific notation. How many digits you can have in the number N varies from calculator to calculator, as does whether or not the calculator displays a particular result in scientific notation or in standard notation.
Calculators sometimes display the result in scientific notation, such as 3.69E–7 or 3.69–07. In each case the answer is to be interpreted as 3.69 × 10–7. Calculators display the significant digits in the result followed by the power of 10 to be used. When entering a number in scientific notation into your calculator as part of a computation, press the operation sign after each number until you are ready to compute the result.
**EXAMPLES 7.16.** Compute (1.892 × 108) × (5.34 × 10–3) using a calculator.
Enter 1.892, press the EE key, enter 8, press the × sign, enter 5.34, press the EE key, enter –3, and press the ENTER key and we get a display of 1 010 328.
(1.892 × 108) × (5.34 × 10–3) = 1 010 328
### **Solved Problems**
**Positive Integral Exponent**
**7.1** ( _a_ ) 23 = 2 · 2 · 2 = 8
( _b_ ) (–3)4 = (– 3)(– 3)(– 3)(– 3) = 81
( _c_ )
( _d_ ) (3 _y_ )2(2 _y_ )3 = (3 _y_ )(3 _y_ )(2 _y_ )(2 _y_ )(2 _y_ ) = 72 _y_ 5
( _e_ ) (–3 _xy_ 2)3 = (–3 _xy_ 2)(–3 _xy_ 2)(–3 _xy_ 2)
= –27 _x_ 3 _y_ 6
**Negative Integral Exponent**
**7.2** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
( _j_ )
( _k_ )
( _l_ )
( _m_ )
**Rational Exponents**
**7.3** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
**7.4** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
**7.5** ( _a_ ) 70 = 1, (–3)0 = 1, (–2/3)0 = 1
( _b_ ) ( _x_ – _y_ )0 = 1, if _x_ – _y_ ≠ 0
( _c_ ) 3 _x_ 0 = 3 · 1 = 3, if _x_ ≠ 0
( _d_ ) (3 _x_ )0 = 1, if 3 _x_ ≠ 0, i.e. if _x_ ≠ 0
( _e_ ) 4 · 100 = 4 · 1 = 4
( _f_ ) (4 · 10)0 = (40)0 = 1
( _g_ ) –(1)0 = –1
( _h_ ) (–1)0 = 1
( _i_ )
( _j_ )
( _k_ )
( _l_ ) 4( _x_ 2 \+ _y_ 2)( _x_ 2 \+ _y_ 2)0 = 4( _x_ 2 \+ _y_ 2)(1) = 4( _x_ 2 \+ _y_ 2), if _x_ 2 \+ _y_ 2 ≠ 0
**General Laws of Exponents**
**7.6** ( _a_ ) _a p_ · _a q_ = _a p+q_
( _b_ ) _a_ 3 · _a_ 5 = _a_ 3+5 = _a_ 8
( _c_ ) 34 _·_ 35 = 39
( _d_ ) _a_ _n_ +1 · _a_ _n_ –2 = _a_ 2 _n_ –1
( _e_ ) _x_ 1/2 · _x_ 1/3 = _x_ 1/2+1/3 = _x_ 5/6
( _f_ ) _x_ 1/2 · _x_ –1/3 = _x_ 1/2–1/3 = _x_ 1/6
( _g_ ) 107 · 10–3 = 107–3 = 104
( _h_ ) (4 _·_ 10–6)(2 _·_ 104) = 8 · 104–6 = 8 · 10–2
( _i_ ) _a x_ · _a y_ · _a_ – _z_ = _a_ _x_ + _y_ – _z_
( _j_ )
( _k_ ) 101.7 _·_ 102.6 = 104.3
( _l_ ) 10–4.1 _·_ 103.5 · 10–.1 = 10–4.1+3.5–.1 = 10–.7
( _m_ )
( _n_ )
( _o_ )
**7.7** ( _a_ ) ( _a p_) _q_ = _a pq_
( _b_ ) ( _x_ 3)4 = _x_ 3.4 = _x_ 12
( _c_ ) ( _a_ _m_ +2) _n_ = _a_ ( _m_ +2) _n_ = _a_ _mn_ +2 _n_
( _d_ ) (103)2 = 103.2 = 106
( _e_ ) (10–3)2 = 10–3.2 = 10–6
( _f_ ) (49)3/2 = (72)3/2 = 72.3/2 = 73 = 343
( _g_ ) (3–1/2)–2 = 31 = 3
( _h_ ) ( _u_ –2)–3 = _u_ (–2)(–3) = _u_ 6
( _i_ ) (81)3/4 = (34)3/4 = 33 = 27
( _j_ )
( _k_ )
( _l_ )
**7.8** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
( _j_ )
( _k_ )
( _l_ )
( _m_ )
( _n_ )
**7.9** ( _a_ ) ( _ab_ ) _p_ = _a_ _p_ _b_ _b_
( _b_ ) (2 _a_ )4 = 24 _a_ 4 = 16 _a_ 4
( _c_ ) (3 · 102)4 = 34 · 108 = 81 · 108
( _d_ ) (4 _x_ 8 _y_ 4)1/2 = 41/2( _x_ 8)1/2( _y_ 4)1/2 = 2 _x_ 4 _y_ 2
( _e_ )
( _f_ ) ( _x_ 2 _n_ _y_ –1/2 _Z_ _n_ –1)2 = _x_ _4n_ _y_ –1 _z_ 2 _n_ –2
( _g_ ) (27 _x_ 3 _p_ _y_ 6 _q_ z12 _r_ )1/3 = (27)1/3( _x_ 3 _p_ )1/3( _y_ 6 _q_ )1/3( _z_ 12 _r_ )1/3 = 3 _x_ _p_ _y_ 2 _q_ _z_ 4 _r_
**7.10** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
( _j_ )
( _k_ )
( _l_ )
**Miscellaneous Examples**
**7.11** ( _a_ )
( _b_ )
( _c_ )
( _d_ ) 104 \+ 103 \+ 102 \+ 101 \+ 100 \+ 10–1 \+ 10–2 = 10 000 + 1000 + 100 + 10 + 1 + 0.1 + 0.01
= 11 111.11
( _e_ ) 3 · 103 \+ 5 · 102 \+ 2 · 101 \+ 4 · 100 = 3524
( _f_ )
( _g_ )
**7.12** ( _a_ ) Evaluate 4 _x_ –2/3 \+ 3 _x_ 1/3 _+_ 2 _x_ 0 when _x_ = 8.
( _b_ ) Evaluate
when _x_ = 2.
**7.13**
( _b_ )
( _c_ )
( _d_ )
( _e_ )
**7.14** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
**7.15** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ ) 2232 = 229 = 2512
**7.16** ( _a_ ) (0.004)(30 000)2 = (4 × 10–3)(3 × 104)2 = (10–3)(32 × 108) = 4 · 32 × 10–3+8
= 36 × 105 or 3 600 000
( _b_ )
( _c_ )
( _d_ )
( _e_ )
**7.17** For what real values of the variables involved will each of the following operations be valid and yield real numbers?
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
**SOLUTION**
( _a_ ) When _x_ is a real number, must be positive or zero. Assuming were true for all _x_ , then if _x_ = – 1 we would have a contradiction. Thus cannot be true for all values of _x_. We will have If _x_ ≤ 0 we have A result valid both for _x_ ≥ 0 and _x_ ≤ 0 may be (absolute value of _x_ ).
( _b_ ) must be positive or zero and thus will equal _a_ \+ 1 if _a_ \+ 1 ≥ 0, i.e. if _a_ ≥ –1. A result valid for all real values of _a_ is given by
( _c_ ) ( _a_ –2 – _b_ –2)/( _a_ –1 – _b_ –1) is not defined if _a_ or _b_ or both is equal to zero. Similarly it is not defined if the denominator _a_ –1 – _b_ –1 = 0, i.e. if _a_ –1 = _b_ –1 or _a_ = _b_. Hence the result ( _a_ –2 – _b_ –2)/( _a_ –1 – _b_ –1) = _a_ –1 \+ _b_ –1 is valid if and only if _a_ ≠ 0, _b_ ≠ 0, and _a_ ≠ _b_.
( _d_ ) must be positive or zero and will equal _x_ 2 \+ 1 if _x_ 2 \+ 1 _≥_ 0. Since _x_ 2 \+ 1 is greater than zero for all real numbers _x_ , the result is valid for all real values of _x_.
( _e_ ) will not be a real number if _x_ – 1 < 0, i.e. if _x_ < 1. Also, will not be defined if the denominator is zero, i.e. if _x_ = 1. Hence if and only if _x_ > 1.
**7.18** A student was asked to evaluate the expression for _x_ = 2, _y_ = 4. She wrote
and thus obtained the value 2 _x_ = 2(2) = 4 for her answer. Was she correct?
**SOLUTION**
Putting _x_ = 2, _y_ = 4 in the given expression, we obtain
The student made the mistake of writing which is true only if _x_ ≥ 2 _y_. If _x_ ≤ 2 _y_ , In all cases, The required simplification should have been _x_ \+ 2 _y_ \+ 2 _y_ — _x_ = 4 _y_ , which does give 16 when _y_ = 4.
### **Supplementary Problems**
Evaluate each of the following.
**7.19** ( _a_ ) 34
( _b_ ) (–2 _x_ )3
( _c_ )
( _d_ ) 4–3
( _e_ ) (–4 _x_ )–2
( _f_ ) (2 _y_ –1)–1
( _g_ )
( _h_ ) (16)1/4
( _i_ )
( _j_ ) (– _a_ 3 _b_ 3)–2/3
( _k_ ) –3(–1)–1/5(4)–1/2
( _l_ ) (103)0
( _m_ ) ( _x_ – _y_ )0[( _x_ – _y_ )4]–1/2
( _n_ ) _x_ _y_ · _x_ 4 _y_
( _o_ ) 3 _y_ 2/3 · _y_ 4/3
( _p_ ) (4 · 103)(3 · 10–5)(6 · 104)
**7.20** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ ) [( _x_ –1)–2]–3
( _g_ )
( _h_ )
( _i_ )
**7.21** ( _a_ )
( _b_ )
( _c_ )
( _d_ ) 272/3 – 3(3 _x_ )0 _+_ 251/2
( _e_ ) 82/3 · 16–3/4 · 20 – 8–2/3
( _f_ )
( _g_ ) _x_ 3/2 \+ 4 _x_ –1 – 5 _x_ 0 when _x_ = 4
( _h_ ) _y_ 2/3 \+ 3 _y_ –1 – 2 _y_ 0 when _y_ = 1/8
( _i_ )
( _j_ )
( _k_ )
**7.22** ( _a_ ) 250 \+ 0.251/2 – 81/3 · 4–1/2 \+ 0.0271/3
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ ) (64)–2/3 – 3(150)0 \+ 12(2)–2
( _g_ )
( _h_ )
( _i_ )
**7.23** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**7.19** ( _a_ ) 81
( _b_ ) –8 _x_ 3
( _c_ )
( _d_ ) 1/64
( _e_ )
( _f_ ) _y_ /2
( _g_ )
( _h_ ) 2
( _i_ ) 1/2
( _j_ )
( _k_ ) 3/2
( _l_ ) 1
( _m_ )
( _n_ ) _x_ 5 _y_
( _o_ ) 3 _y_ 2
( _p_ ) 7200
**7.20** ( _a_ ) 29
( _b_ ) 1/10
( _c_ ) 1
( _d_ ) 1
( _e_ ) 10–4
( _f_ ) _x_ –6
( _g_ )
( _h_ )
( _i_ )
**7.21** ( _a_ )
( _b_ )
( _c_ ) 4
( _d_ ) 11
( _e_ )
( _f_ )
( _g_ ) 4
( _h_ )
( _i_ ) 18
( _j_ )
( _k_ )
**7.22** ( _a_ ) 0.8
( _b_ )
( _c_ )
( _d_ ) 34
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ ) 150
**7.23** ( _a_ )
( _b_ )
( _c_ )
( _d_ ) 2 _x_ if _x_ ≥ –1, –2 if _x_ ≤ –1
( _e_ ) _x_ – _y_ if 2 _x_ ≥ _y_ , 5 _x_ – 3 _y_ if 2 _x_ ≤ _y_
## **CHAPTER 8
Radicals**
### **8.1 RADICAL EXPRESSIONS**
A radical is an expression of the form which denotes the principal _n_ th root of _a_. The positive integer _n_ is the index, or order, of the radical and the number _a_ is the radicand. The index is omitted if _n_ = 2.
Thus are radicals which have respectively indices 3, 4, and 2 and radicands 5, 7 _x_ 3 – 2 _y_ 2, and _x_ \+ 10.
### **8.2 LAWS FOR RADICALS**
The laws for radicals are the same as the laws for exponents, since The following are the laws most frequently used. _Note:_ If _n_ is even, assume _a, b_ ≥ 0.
_A_.
**EXAMPLES 8.1.**
_B_.
**EXAMPLES 8.2.**
_C_.
**EXAMPLES 8.3.**
_D_.
**EXAMPLE 8.4.**
_E_.
**EXAMPLES 8.5.**
### **8.3 SIMPLIFYING RADICALS**
The form of a radical may be changed in the following ways.
( _a_ ) Removal of perfect _n_ th powers from the radicand.
**EXAMPLES 8.6.**
( _b_ ) Reduction of the index of the radical.
**EXAMPLES 8.7.** where the index is reduced from 4 to 2. where the index is reduced from 6 to 3. _Note_. It is _incorrect_ to write
( _c_ ) Rationalization of the denominator in the radicand.
**EXAMPLE 8.8.** Rationalize the denominator of
Multiply the numerator and denominator of the radicand (9/2) by such a number as will make the denominator a perfect _n_ th power (here _n_ = 3) and then remove the denominator from under the radical sign. The number in this case is 22. Then
**EXAMPLE 8.9.** Rationalize the denominator of
To make 8 _b_ 6 _x_ 3 a perfect 4th power, multiply numerator and denominator by 2 _b_ 2 _x_. Then
A radical is said to be in simplest form if:
( _a_ ) all perfect _n_ th powers have been removed from the radical,
( _b_ ) the index of the radical is as small as possible,
( _c_ ) no fractions are present in the radicand, i.e., the denominator has been rationalized.
### **8.4 OPERATIONS WITH RADICALS**
Two or more radicals are said to be similar if after being reduced to simplest form they have the same index and radicand.
Thus are similar since
Here each radicand is 2 and each index is 2. However, are dissimilar since
To add algebraically two or more radicals, reduce each radical to simplest form and combine terms with similar radicals. Thus:
When multiplying two radicals, we choose the procedure to use based on whether or not the indices of the radicals are the same.
( _a_ ) To multiply two or more radicals having the _same index_ , use Law _B:_
**EXAMPLES 8.10.**
( _b_ ) To multiply radicals with _different indices_ it is convenient to use fractional exponents and the laws of exponents.
**EXAMPLES 8.11.**
When dividing two radicals, we choose the procedure to use based on whether or not the indices of the radicals are the same.
( _a_ ) To divide two radicals having the _same index_ , use Law _C_ ,
and simplify.
**EXAMPLE 8.12.**
We may also rationalize the denominator directly, as follows.
( _b_ ) To divide two radicals with _different indices_ it is convenient to use fractional exponents and the laws of exponents.
**EXAMPLES 8.13.**
### **8.5 RATIONALIZING BINOMIAL DENOMINATORS**
The binomial irrational numbers are called conjugates of each other. Thus are conjugates. The property of these conjugates that makes them useful is the fact that they are the sum and difference of the same two terms, so their product is the difference of the squares of these terms. Hence,
To rationalize a fraction whose denominator is a binomial with radicals of index 2, multiply numerator and denominator by the conjugate.
**EXAMPLE 8.14.**
If the denominator of a fraction is we multiply the numerator and denominator of the fraction by and get a denominator of _a_ \+ _b_. If the original denominator has the form we multiply the numerator and denominator of the fraction by and get a denominator of _a_ – _b_. (See Section 4.2 for the special product rules.)
**EXAMPLE 8.15.**
### **Solved Problems**
**Reduction of a Radical Expression to Simplest Form**
**8.1** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
( _j_ )
( _k_ ) The student is reminded that is a positive number or zero; hence only if _a_ \+ 3 ≥ 0. If values of _a_ such that _a_ \+ 3 < 0 are included, we must write
( _l_ )
( _m_ ) Note that this is valid only if 2 _x_ 2 ≥ 3 _y_ 2. see ( _k_ ) above.
( _n_ )
( _o_ )
( _p_ )
( _q_ )
**Change in Form of a Radical**
**8.2** Express as radicals of the 12th order.
( _a_ )
( _b_ )
( _c_ )
**8.3** Express in terms of radicals of least order.
( _a_ )
( _b_ )
( _c_ )
**8.4** Convert into entire radicals, i.e., radicals having coefficient 1.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
**8.5** Determine which of the following irrational numbers is the greater:
( _a_ )
( _b_ )
( _c_ )
**SOLUTION**
( _a_ )
( _b_ )
( _c_ )
**8.6** Rationalize the denominator.
( _a_ )
( _b_ )
_Another method_ :
( _c_ )
( _d_ )
( _e_ )
**Addition and Subtraction of Similar Radicals**
**8.7** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
**Multiplication of Radicals**
**8.8** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
( _j_ )
( _k_ )
( _l_ )
( _m_ )
( _n_ )
( _o_ )
( _p_ )
( _q_ )
( _r_ )
( _s_ )
**8.9** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
**Division of Radicals. Rationalization of Denominators**
**8.10** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
( _j_ )
( _k_ )
( _l_ )
( _m_ )
( _n_ )
( _o_ )
( _p_ )
### **Supplementary Problems**
Show that:
**8.11** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
( _j_ )
( _k_ )
( _l_ )
( _m_ )
( _n_ )
( _o_ )
**8.12** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
( _j_ )
( _k_ )
( _l_ )
( _m_ )
**8.13** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
( _j_ )
( _k_ )
( _l_ )
( _m_ )
( _n_ )
( _o_ )
( _p_ )
( _q_ )
**8.14** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
( _j_ )
( _k_ )
( _l_ )
( _m_ )
( _n_ )
( _o_ )
( _p_ )
## **CHAPTER 9
Simple Operations with Complex Numbers**
### **9.1 COMPLEX NUMBERS**
The unit of imaginary numbers is and is designated by the letter _i_. Many laws which hold for real numbers hold for imaginary numbers as well.
Thus Also, since we have and similarly for any integral power of _i_.
Note. One must be very careful in applying some of the laws which hold for real numbers. For example, one might be tempted to write
To avoid such difficulties, always express where _m_ is a positive number, as and use _i_ 2 = –1 whenever it arises. Thus:
A complex number is an expression of the form _a_ \+ _bi_ , where _a_ and _b_ are real numbers and In the complex number _a_ \+ _bi, a_ is called the _real part_ and _bi_ the _imaginary part_. When _a_ = 0, the complex number is called a _pure imaginary_. If _b_ = 0, the complex number reduces to the real number _a_. Thus complex numbers include all real numbers and all pure imaginary numbers.
Two complex numbers _a_ \+ _bi_ and _c_ \+ _di_ are _equal_ if and only if _a_ = _c_ and _b_ = _d_. Thus _a_ \+ _bi_ = 0 if and only if _a_ = 0, _b_ = 0. If _c_ \+ _di_ = 3, then _c_ = 3, _d_ = 0.
The conjugate of a complex number _a_ \+ _bi_ is _a_ – _bi_ , and conversely. Thus 5 – 3 _i_ and 5 + 3 _i_ are conjugates.
### **9.2 GRAPHICAL REPRESENTATION OF COMPLEX NUMBERS**
Employing rectangular coordinate axes, the complex number _x_ \+ _yi_ is represented by, or corresponds to, the point whose coordinates are ( _x, y_ ). See Fig. 9-1.
**Fig. 9-1**
To represent the complex number 3 + 4 _i_ , measure off 3 units distance along _X′X_ and to the right of _O_ , and then up 4 units distance.
To represent the number –2 + 3 _i_ , measure off 2 units distance along _X′X_ and to the left of _O_ , and then up 3 units distance.
To represent the number –1 – 4 _i_ , measure off 1 unit distance along _X′X_ and to the left of _O_ , and then down 4 units distance.
To represent the number 2 – 4 _i_ , measure off 2 units distance along _X′X_ and to the right of _O_ , and then down 4 units distance.
Pure imaginary numbers (such as 2 _i_ and – 2 _i)_ are represented by points on the line _Y′Y_. Real numbers (such as 4 and – 3) are represented by points on the line _X′X_.
### **9.3 ALGEBRAIC OPERATIONS WITH COMPLEX NUMBERS**
To _add_ two complex numbers, add the real parts and the imaginary parts separately. Thus
_To subtract_ two complex numbers, subtract the real parts and the imaginary parts separately. Thus
_To multiply_ two complex numbers, treat the numbers as ordinary binomials and replace _i_ 2 by –1. Thus
_To divide_ two complex numbers, multiply the numerator and denominator of the fraction by the conjugate of the denominator, replacing _i_ 2 by –1. Thus
### **Solved Problems**
**9.1** Express each of the following in terms of _i_.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
( _j_ )
( _k_ )
( _l_ )
( _m_ )
**9.2** Perform the indicated operations both algebraically and graphically:
( _a_ ) (2 + 6 _i_ ) + (5 + 3 _i_ ),
( _b_ ) (–4 + 2 _i_ ) – (3 + 5 _i_ ).
**Fig. 9-2**
**Fig. 9-3**
**SOLUTION**
( _a_ ) Algebraically: (2 + 6 _i_ ) + (5 + 3 _i_ ) = 7 + 9 _i_
Graphically: Represent the two complex numbers by the points _P_ 1 and _P_ 2 respectively, as shown in Fig. 9-2. Connect _P_ 1 and _P_ 2 with the origin _O_. Complete the parallelogram having _OP_ 1 and _OP_ 2 as adjacent sides. The vertex _P_ (point 7 + 9 _i_ ) represents the sum of the two given complex numbers.
( _b_ ) Algebraically: (–4 + 2 _i_ ) – (3 + 5 _i_ ) = –7 – 3 _i_
Graphically: (–4 + 2 _i)_ – (3 + 5i) = (–4 + 2 _i_ ) + (–3 – 5 _i_ ). We now proceed to add (–4 + 2 _i_ ) with (–3 – 5i).
Represent the two complex numbers (–4 + 2 _i_ ) and (–3 – 5 _i)_ by the points _P_ 1 and _P_ 2 respectively, as shown in Fig. 9-3. Connect _P_ 1 and _P_ 2 with the origin _O_. Complete the parallelogram having _OP_ 1 and _OP_ 2 as adjacent sides. The vertex _P_ (point –7 –3 _i_ ) represents the subtraction (–4 + 2 _i_ ) – (3 + 5 _i)_.
**9.3** Perform the indicated operations and simplify.
( _a_ ) (5 – 2 _i)_ \+ (6 + 3 _i)_ = 11 + _i_
( _b_ ) (6 + 3 _i)_ – (4 – 2 _i)_ = 6 + 3 _i_ – 4 + _2i_ = 2 + _5i_
( _c_ ) (5 – 3 _i)_ – (–2 + 5 _i)_ = 5 – 3 _i_ \+ 2 – 5 _i_ = 7 – 8 _i_
( _d_ )
( _e_ ) ( _a_ \+ _bi_ ) + ( _a_ – _bi_ ) = _2a_
( _f_ ) ( _a_ \+ _bi_ ) – ( _a_ – _bi_ ) = _a_ \+ _bi_ – _a_ \+ _bi_ = 2 _bi_
( _g_ )
**9.4** ( _a_ )
( _b_ )
( _c_ ) (4 _i_ )(–3 _i_ ) = –12 _i_ 2 = 12
( _d_ ) (6 _i_ )2 = 36 _i_ 2 = –36
( _e_ )
( _f_ ) 3 _i_ ( _i_ \+ 2) = 3 _i_ 2 \+ 6 _i_ = –3 + 6 _i_
( _g_ ) (3 – 2 _i_ )(4 + _i_ ) = 3 · 4 + 3 · _i_ – (2 _i_ ) 4 – (2 _i_ ) _i_ = 12 + 3 _i_ – 8 _i_ \+ 2 = 14 – 5 _i_
( _h_ ) (5 – 3 _i_ )( _i_ \+ 2) = 5 _i_ \+ 10 – 3 _i_ 2 – 6 _i_ = 5 _i_ \+ 10 + 3 – 6 _i_ = 13 – _i_
( _i_ ) (5 + 3 _i_ )2 = 52 \+ 2(5)3 _i_ \+ (3 _i_ )2 = 25 + 30 _i_ \+ 9 _i_ 2 = 16 + 30 _i_
( _j_ )
( _k_ )
( _l_ ) (1 + _i_ )3 = 1 + 3 _i_ \+ 3 _i_ 2 \+ _i_ 3 = 1 + 3 _i_ – 3 – _i_ = –2 + _2i_
( _m_ ) (3 – 2 _i_ )3 = 33 \+ 3(32)(–2 _i_ ) + 3(3)(–2 _i_ ) 2 + (–2 _i_ 3
= 27 + 3(9)(–2 _i_ ) + 3(3)(4 _i_ 2) – 8 _i_ 3 = 27 – 54 _i_ – 36 + _8i_ = –9 – 46 _i_
( _n_ )(3 + 2 _i_ ) 3 = 33 \+ 3(32)(2 _i_ ) + 3(3)(2 _i_ )2 \+ (2 _i_ )3
= 27 + 54 _i_ \+ 36 _i_ 2 \+ 8 _i_ 3 = 27 + 54 _i_ – 36 – 8 _i_ = –9 + 46 _i_
( _o_ ) (1 + 2 _i_ ) 4 = [(1 + 2 _i_ ) 2]2 = (1 + 4 _i_ \+ 4 _i_ 2)2 = (–3 + 4 _i_ )2 = 9 – 24 _i_ \+ 16 _i_ 2 = –7 – 24 _i_
( _p_ ) (–1 + _i_ )8 = [(–1 + _i_ )2]4 = (1 – 2 _i_ \+ _i_ 2)4 = (–2 _i_ )4 = 16 _i_ 4 = 16
**9.5** ( _a_ )
( _b_ )
( _c_ )
### **Supplementary Problems**
**9.6** Express each of the following in terms of _i_.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
( _j_ )
**9.7** Perform the indicated operations both algebraically and graphically.
( _a_ ) (3 + 2 _i_ ) + (2 + 3 _i_ )
( _b_ ) (2 – i) + (–4 + 5 _i_ )
( _c_ ) (4 – 3 _i_ ) – (–2 + _i_ )
( _d_ ) (–2 + 2 _i_ ) – (–2 – _i_ )
**9.8** Perform each of the indicated operations and simplify.
( _a_ ) (3 + 4 _i_ ) + (–1 –6 _i_ )
( _b_ ) (–2 + 5 _i_ ) – (3 –2 _i_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ ) (2 _i_ )4
( _h_ )
( _i_ ) 5 _i_ (2 – _i_ )
( _j_ ) (2 + _i_ )(2 – _i_ )
( _k_ ) (–3 + 4 _i_ )(–3 –4 _i_ )
( _l_ ) (2 – 5 _i_ )(3 + 2 _i_ )
( _m_ ) (3 – 4 _i_ )2
( _n_ ) (1 + _i_ )(2 + 2 _i_ )(3 – _i_ )
( _o_ ) ( _i_ – 1)3
( _p_ ) (2 + 3 _i_ )3
( _q_ ) (1 – _i_ )4
( _r_ ) ( _i_ \+ 2)5
**9.9** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**9.6** ( _a_ ) 14 _i_
( _b_ ) –32 _i_
( _c_ ) 2 _i_
( _d_ )
( _e_ ) –35 _i_
( _f_ )
( _g_ ) –2 + _i_
( _h_ )
( _i_ ) 18 _i_ \+ 20
( _j_ )
**9.7** ( _a_ ) 5 + 5 _i_ and Fig. 9-4
( _b_ ) –2 + 4 _i_ and Fig. 9-5
( _c_ ) 6 – 4 _i_ and Fig. 9-6
( _d_ ) 3 _i_ and Fig. 9-7
**Fig. 9-4**
**Fig. 9-5**
**Fig. 9-6**
**Fig. 9-7**
**9.8** ( _a_ ) 2 – 2 _i_
( _b_ ) –5 + 7 _i_
( _c_ ) 1 – _i_
( _d_ )
( _e_ ) –6
( _f_ ) 2
( _g_ ) 16
( _h_ ) –27/64
( _i_ ) 5 + 10 _i_
( _j_ ) 5
( _k_ ) 25
( _l_ ) 16–11 _i_
( _m_ ) –7 –24 _i_
( _n_ ) 4 + 12 _i_
( _o_ ) 2 + 2 _i_
( _p_ ) –46 + 9 _i_
( _q_ ) –4
( _r_ ) –38 + 41 _i_
**9.9** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ ) 0
( _h_ ) _i_
( _i_ ) 3 + 4 _i_
## **CHAPTER 10
Equations in General**
### **10.1 EQUATIONS**
An equation is a statement of equality between two expressions called members.
An equation which is true for only certain values of the variables (sometimes called unknowns) involved is called a _conditional equation_ or simply an equation.
An equation which is true for all permissible values of the variables (or unknowns) involved is called an _identity_. By permissible values are meant the values for which the members are defined.
**EXAMPLE 10.1.** _x_ \+ 5 = 8 is true only for _x_ = 3; it is a conditional equation.
**EXAMPLE 10.2.** _x_ 2 – _y_ 2 = ( _x_ – _y_ )( _x_ \+ _y_ ) is true for all values of _x_ and _y_ ; it is an identit _y_.
**EXAMPLE 10.3.**
is true for all values except for the nonpermissible values _x_ = 2, _x_ = 3; these excluded values lead to division by zero which is not allowed. Since the equation is true for all permissible values of _x_ , it is an identity.
The symbol ≡ is sometimes used for identities instead of = .
The solutions of a conditional equation are those values of the unknowns which make both members equal. These solutions are said to satisfy the equation. If only one unknown is involved the solutions are also called _roots_. To solve an equation means to find all of the solutions.
Thus _x_ = 2 is a solution or root of 2 _x_ \+ 3 = 7, since if we substitute _x_ = 2 into the equation we obtain 2(2) + 3 = 7 and both members are equal, i.e., the equation is satisfied. Similarly, three (of the many) solutions of 2 _x_ \+ _y_ = 4 are: _x_ = 0, _y_ = 4; _x_ = 1, _y_ = 2; _x_ = 5, _y_ –6.
### **10.2 OPERATIONS USED IN TRANSFORMING EQUATIONS**
_A_. If equals are added to equals, the results are equal.
Thus if _x_ – _y_ = _z_ , we may add _y_ to both members and obtain _x_ = _y_ \+ _z_.
_B_. If equals are subtracted from equals, the results are equal.
Thus if _x_ \+ 2 = 5, we may subtract 2 from both members to obtain _x_ = 3.
_Note_. Because of rules _A_ and _B_ we may transpose a term from one member of an equation to the other member merely by changing the sign of the term. Thus if 3 _x_ \+ 2 _y_ – 5 = _x_ – 3 _y_ \+ 2, then 3 _x_ – _x_ \+ 2 _y_ \+ 3 _y_ = 5 + 2 or 2 _x_ \+ 5 _y_ = 7.
_C_. If equals are multiplied by equals, the results are equal.
Thus if both members of _y_ = 2 _X_ 2 are multiplied by 4 the result is _y_ = 8 _x_ 2.
Similarly, if both members of _C_ = _F_ – 32 are multiplied by the result is _C_ = ( _F_ – 32).
_D_. If equals are divided by equals, the results are equal provided there is no division by zero.
Thus if –4 _x_ = –12, we may divide both members by –4 to obtain _x_ = 3. Similarly, if _E_ = _IR_ we may divide both sides by _R_ = 0 to obtain _I_ = _E_ / _R_.
_E_. The same powers of equals are equal.
Thus if then
_F_. The same roots of equals are equal. Thus:
_G_. Reciprocals of equals are equal provided the reciprocal of zero does not occur.
Thus if 1/ _x_ = 1/3, then _x_ = 3. Similarly,
Operations _A–F_ are sometimes called axioms of equality.
### **10.3 EQUIVALENT EQUATIONS**
Equivalent equations are equations having the same solutions.
Thus _x_ – 2 = 0 and 2 _x_ = 4 have the same solution _x_ = 2 and so are equivalent. However, _x_ – 2 = 0 and _x_ 2 – 2 _x_ = 0 are not equivalent, since _x_ 2 – 2 _x_ = 0 has the additional solution _x_ = 0.
The above operations used in transforming equations may not all yield equations equivalent to the original equations. The use of such operations may yield derived equations with either more or fewer solutions than the original equation.
If the operations yield an equation with more solutions than the original, the extra solutions are called _extraneous_ and the derived equation is said to be _redundant_ with respect to the original equation. If the operations yield an equation with fewer solutions than the original, the derived equation is said to be _defective_ with respect to the original equation.
Operations _A_ and _B_ always yield equivalent equations. Operations _C_ and _E_ may give rise to redundant equations and extraneous solutions. Operations _D_ and _F_ may give rise to defective equations.
### **10.4 FORMULAS**
A formula is an equation which expresses a general fact, rule, or principle.
For example, in geometry the formula _A_ = _πr 2_ gives the area _A_ of a circle in terms of its radius _r_.
In physics the formula _s_ = _gt_ 2, where _g_ is approximately 32.2 ft/s2, gives the relation between the distance s, in feet, which an object will fall freely from rest during a time _t_ , in seconds.
To solve a formula for one of the variables involved is to perform the same operations on both members of the formula until the desired variable appears on one side of the equation but not on the other side.
Thus if _F_ = _ma_ , we may divide by _m_ to obtain _a_ = _F_ / _m_ and the formula is solved for _a_ in terms ofthe other variables _F_ and m. To check the work, substitute _a_ = _F_ / _m_ into the original equation to obtain _F_ = _m_ ( _F_ / _m_ ), an identity.
### **10.5 POLYNOMIAL EQUATIONS**
A monomial in a number of unknowns _x_ , _y_ , _z_ ,... has the form _a x_ _p_ _y_ _q_ _z_ _r_... where the exponents _p_ , _q_ , _r_ ,... are either positive integers or zero and the coefficient _a_ is independent of the unknowns. The sum of the exponents _p_ \+ _q_ \+ _r_ \+... is called the _degree_ of the term in the unknowns _x_ , _y_ , _z_ ,...
**EXAMPLES 10.4.**
When reference is made to degree without specifying the unknowns considered, the degree in all unknowns is implied.
A polynomial in various unknowns consists of terms each of which is rational and integral. The degree of such a polynomial is defined as the degree of the terms of highest degree.
**EXAMPLE 10.5.** 3 _x_ 3 _y_ 4z+ _x y_ 2z5 – 8 _x_ \+ 3 is a polynomial of degree 3 in _x_ , 4in _y_ , 5in z, 7in _x_ and _y_ , 7in _y_ and z, 6in _x_ and z, and 8 in _x_ , _y_ , and z.
A polynomial equation is a statement of equality between two polynomials. The degree of such an equation is the degree of the term of highest degree present in the equation.
**EXAMPLE 10.6.** _xyz_ 2\+ 3 _xz_ = 2 _x_ 3 _y_ \+ 3 _z_ 2 is of degree 3 in _x_ , l in _y_ , 2 in _z_ , 4 in _x_ and _y_ , 3 in _y_ and _z_ , 3 in _x_ and _z_ , and 4 in _x_ , _y_ , and _z_.
It should be understood that like terms in the equation have been combined. Thus 4 _x_ 3 _y_ \+ _x_ 2 _z_ – _xy_ 2 = 4 _x_ 3 _y_ \+ _z_ should be written _x_ 2 _z_ – _xy_ 2 = _z_.
An equation is called _linear_ if it is of degree 1 and _quadratic_ if it is of degree 2. Similarly the words _cubic, quartic_ and _quintic_ refer to equations of degree 3, 4, and 5 respectively.
**EXAMPLES 10.7.**
A polynomial equation of degree _n_ in the unknown _x_ may be written
where _a_ 0, _a_ 1,..., _a n_ are given constants and _n_ is a positive integer.
As special cases we see that
### **Solved Problems**
**10.1** Which of the following are conditional equations and which are identities?
( _a_ ) 3 _x_ – ( _x_ \+ 4) = 2( _x_ – 2), 2 _x_ – 4 = 2 _x_ – 4; identity.
( _b_ ) ( _x_ – 1)( _x_ \+ 1) = ( _x_ – 1)2, _x_ 2 – 1 = _x_ 2 – 2 _x_ \+ 1; conditional equation.
( _c_ ) ( _y_ – 3)2 \+ 3(2 _y_ – 3) = _y_ ( _y_ \+ 1) – _y_ , _y_ 2 – 6 _y_ \+ 9 + 6 _y_ – 9 = _y_ 2 \+ _y_ – _y_ , _y_ 2 = _y_ 2; identity.
( _d_ ) _x_ \+ 3 _y_ – 5 = 2( _x_ \+ 2 _y_ ) + 3, _x_ \+ 3 _y_ – 5 = 2 _x_ \+ 4 _y_ \+ 3; conditional equation.
**10.2** Check each of the following equations for the indicated solution or solutions.
( _a_ )
( _b_ )
( _c_ )
**10.3** Use the axioms of equality to solve each equation.
( _a_ ) 2( _x_ \+ 3) = 3( _x_ – 1), 2 _x_ \+ 6 = 3 _x_ – 3.
Transposing terms: 2 _x_ – 3 _x_ = –6 – 3, – _x_ = –9. Multiplying by –1: _x_ = 9.
Check: 2(9 + 3) = 3(9 – 1), 24 = 24.
( _b_ )
Multiplying by 6: 2 _x_ \+ _x_ = 6, 3 _x_ = 6. Dividing by 3: _x_ = 2.
Check: 2/3 + 2/6 = 1, 1 = 1.
( _c_ ) 3 _y_ – 2( _y_ – 1) = 4( _y_ \+ 2), 3 _y_ – 2 _y_ \+ 2 = 4 _y_ \+ 8, _y_ \+ 2 = 4 _y_ \+ 8.
Transposing: _y_ – 4 _y_ = 8 – 2, –3 _y_ = 6. Dividing by –3: _y_ = 6/(–3) = –2.
Check: 3(–2) – 2(–2 – 1) = 4(–2 + 2), 0 = 0.
( _d_ )
Check: Substituting _x_ = 1 into the given equation, we find –1/0 = –1/0. This is meaningless, since division by zero is an excluded operation, and the given equation has no solution. Note that
are not equivalent equations. When (i) is multiplied by _x_ – 1 an _extraneous solution x_ = 1 is introduced, and equation (ii) is _redundant_ with respect to equation (i).
( _e_ ) _x_ ( _x_ – 3) = 2( _x_ – 3). Dividing each member by _x_ – 3 gives a solution _x_ = 2.
Now _x_ – 3 = 0 or _x_ = 3 is also a solution of the given equation which was lost in the division. The required roots are _x_ = 2 and _x_ = 3.
The equation _x_ = 2 is _defective_ with respect to the given equation.
( _f_ ) Squaring both sides, _x_ \+ 2 = 1 or _x_ = –1.
Check: Substituting _x_ = –1 into the given equation, or 1 = –1 which is false.
Thus _x_ = – 1 is an _extraneous solution_. The given equation has no solution.
(g) Squaring both sides, 2 _x_ – 4 = 36 or _x_ = 20.
Check: If _x_ = 20, which is true.
Hence _x_ = 20 is a solution. In this case no extraneous root was introduced.
**10.4** In each of the following formulas, solve for the indicated letter.
( _a_ ) _E_ = _IR_ , for _R_. Dividing both sides by _I_ = 0, we have _R_ = _E_ / _I_.
( _b_ )
( _c_ )
Taking reciprocals,
( _d_ )
Multiplying by _g_ , _gT_ 2 = 4 _π_ 2 _l_. Dividing by _T_ 2 ≠ 0, _g_ = 4 _π_ 2 _l_ / _T_ 2.
**10.5** In each of the following formulas, find the value of the indicated letter, given the values of the other letters.
( _a_ )
( _b_ )
( _c_ )
**10.6** Determine the degree of each of the following equations in each of the indicated unknowns.
( _a_ )
( _b_ )
( _c_ )
As it stands it is not a polynomial equation. However, it can be transformed into one by multiplying by _y_ \+ _z_ to obtain _x_ 2( _y_ \+ _z_ ) = 3 or _x_ 2 _y_ \+ _x_ 2 _z_ = 3. The derived equation is a polynomial equation of degree 2 in _x_ , 3 in _x_ and _z_ , and 3 in _x_ , _y_ , and _z_.
( _d_ )
As given it is not a polynomial equation, but it can be transformed into one by squaring both sides. Thus we obtain _x_ \+ 3 = _x_ 2 \+ 2 _xy_ \+ y2, which is of degree 2 in _y_ and 2 in _x_ and _y_.
It should be mentioned, however, that the equations are not equivalent, since _x_ 2 \+ 2 _xy_ \+ y2 = _x_ \+ 3 includes both
**10.7** Find all values of _x_ for which ( _a_ ) _x_ 2 = 81, ( _b_ )( _x_ – 1)2 = 4.
**SOLUTION**
( _a_ ) There is nothing to indicate whether _x_ is a positive or negative number, so we must assume either is possible. Taking the square root of both sides of the given equation, we obtain Now represents a positive number (or zero) if _x_ is real. Hence we have = _x_ if _x_ is positive while = – _x_ if _x_ is negative. Thus when writing we must consider that it is either _x_ (if _x_ > 0) or – _x_ (if _x_ < 0). Therefore the equation = 9 may be written as either _x_ = 9 or – _x_ = 9 (i.e. _x_ = – 9). The two solutions may be written _x_ = ±9.
( _b_ ) ( _x_ – 1)2 = 4, ± ( _x_ – 1) = 2 or ( _x_ – 1) = ±2, and the two roots are _x_ = 3 and _x_ = – 1.
**10.8** Explain the fallacy in the following sequence of steps.
**SOLUTION**
There is nothing wrong in steps ( _a_ ), ( _b_ ), ( _c_ ), ( _d_ ).
However, in ( _e_ ) we divide by _x_ – _y_ , which from the original assumption is zero. Since division by zero is not defined, everything we do from ( _e_ ) on is to be looked upon with disfavor.
**10.9** Show that is an irrational number, i.e., it cannot be the quotient of two integers.
**SOLUTION**
Assume that = _p_ / _q_ where _p_ and _q_ are integers having no common factor except ±1, i.e., _p_ / _q_ is in lowest terms. Squaring, we have _p_ 2/ _q_ 2 = 2 or _p_ 2 = 2 _q_ 2. Since 2 _q_ 2 is an even number, _p_ 2 is even and hence _p_ is even (if _p_ were odd, _p_ 2 would be odd); then _p_ = 2 _k_ , where _k_ is an integer. Thus _p_ 2 = 2 _q_ 2 becomes (2 _k_ )2 = 2 _q_ 2 or _q_ 2 = 2 _k_ 2; hence _q_ 2 is even and _q_ is even. But if _p_ and _q_ are both even they would have a common factor 2, thus contradicting the assumption that they have no common factor except ±1. Hence is irrational.
### **Supplementary Problems**
**10.10** State which of the following are conditional equations and which are identities.
( _a_ ) 2 _x_ \+ 3 – (2 – _x_ ) = 4 _x_ – 1
( _b_ ) (2 _y_ – 1)2 \+ (2 _y_ \+ 1)2 = (2 _y_ )2 \+ 6
( _c_ ) 2{ _x_ \+ 4 – 3(2 _x_ – 1)} = 3(4 – 3 _x_ ) + 2 – _x_
( _d_ ) ( _x_ \+ 2 _y_ )( _x_ – 2 _y_ ) – ( _x_ – 2 _y_ )2 \+ 4 _y_ (2 _y_ – _x_ ) = 0
( _e_ )
( _f_ ) ( _x_ – 3) ( _x_ 2 \+ 3 _x_ \+ 9) = _x_ 3 – 27
( _g_ )
( _h_ )( _x_ 2 – _y_ 2)2 \+ (2 _xy_ )2 = ( _x_ 2 \+ _y_ 2)2
**10.11** Check each of the following equations for the indicated solution or solutions.
( _a_ )
( _b_ ) _x_ 2 – 3 _x_ = 4; –1, –4
( _c_ )
( _d_ ) _x_ 3 – 6 _x_ 2 \+ 11 _x_ – 6 = 0; 1, 2, 3
( _e_ )
( _f_ )
( _g_ ) _x_ 2 – 2 _y_ = 3 _y_ 2; _x_ = 4, _y_ = 2; _x_ = 1, _y_ = –1
( _h_ ) ( _x_ \+ _y_ )2 \+ ( _x_ – _y_ )2 = 2( _x_ 2 \+ _y_ 2); any values of _x_ , _y_
**10.12** Use the axioms of equality to solve each equation. Check the solutions obtained.
( _a_ ) 5( _x_ – 4) = 2( _x_ \+ 1) – 7
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ ) ( _y_ \+ 1)2 = 16
( _j_ )(2 _x_ \+ 1)2 \+ (2 _x_ – 1)2 = 34
**10.13** In each of the following formulas, solve for the indicated letter.
( _a_ )
( _b_ )
( _c_ )
( _d_ ) _v_ 2 = _v_ 20 \+ 2 _as_ ; _a_
( _e_ )
( _f_ )
**10.14** In each formula find the value of the indicated letter given the values of the other letters.
( _a_ ) _v_ = _v_ 0 \+ at; find _a_ if _v_ = 20, _v_ 0 = 30, _t_ = 5.
( _b_ )
( _c_ )
( _d_ )
( _e_ )
**10.15** Determine the degree of each equation in each of the indicated unknowns.
( _a_ ) _x_ 3 – 3 _x_ \+ 2 = 0: _x_
( _b_ ) _x_ 2 \+ _xy_ \+ 3 _y_ 4 = 6: _x_ ; _y_ ; _x_ and _y_
( _c_ ) 2 _xy_ 3 – 3 _x_ 2 _y_ 2 \+ 4 _xy_ = 2 _x_ 3: _x_ ; _y_ ; _x_ and _y_
( _d_ ) _xy_ \+ _yz_ \+ _xz_ \+ _z_ 2 _x_ = _y_ 4: _x_ ; _y_ ; _z_ ; _x_ and _z_ ; _y_ and _z_ ; _x_ , _y_ , and _z_
**10.16** Classify each equation according as it is (or can be transformed into) an equation which is linear, quadratic, cubic, quartic or quintic in all of the unknowns present.
( _a_ ) 2 _x_ 4 \+ 3 _x_ 3 – _x_ – 5 = 0
( _b_ ) _x_ – 2 _y_ = 4
( _c_ ) 2 _x_ 2 \+ 3 _xy_ \+ _y_ 2 = 10
( _d_ ) _x_ 2 _y_ 3 – 2 _x yz_ = 4 + _y_ 5
( _e_ )
( _f_ )
( _g_ ) 3 _y_ 2 – 4 _y_ \+ 2 = 2( _y_ – 3)2
( _h_ ) ( _z_ \+ 1)2( _z_ – 2) = 0
**10.17** Is the equation an identity? Explain.
**10.18** Prove that is irrational.
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**10.10** ( _a_ ) Conditional equation
( _b_ ) Conditional equation
( _c_ ) Identity
( _d_ ) Identity
( _e_ ) Conditional equation
( _f_ ) Identity
( _g_ ) Conditional equation
( _h_ ) Identity
**10.11** ( _a_ ) _y_ = 3 is a solution.
( _b_ ) _x_ = –1 is a solution, _x_ = –4 is not.
( _c_ ) _x_ = 34 is a solution, _x_ = 2 is not.
( _d_ ) _x_ = 1, 2, 3 are all solutions.
( _e_ ) _x_ = 3 is a solution.
( _f_ ) _y_ = ± , –1 are all solutions.
( _g_ ) _x_ = 4, _y_ = 2; _x_ = 1, _y_ = – 1 are solutions.
( _h_ ) The equation is an identity; hence any values of _x_ and _y_ are solutions.
**10.12** ( _a_ ) _x_ = 5
( _b_ ) _y_ = 4
( _c_ ) _y_ = 1/2
( _d_ ) _x_ = 3
( _e_ ) no solution
( _f_ ) _x_ = 6
( _g_ ) no solution
( _h_ ) _x_ = 1
( _i_ ) _y_ = 3, –5
( _J_ ) _x_ = ±2
**10.13** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
**10.14** ( _a_ ) a = –2
( _b_ ) d = 17
( _c_ ) q = –15
( _d_ ) v = ±20
( _e_ ) C = 0.0063
**10.15**
( _a_ ) 3
( _b_ ) 2, 4, 4
( _c_ ) 3, 3, 4
( _d_ ) 1, 4, 2, 3, 4, 4
**10.16**
( _a_ ) quartic
( _b_ ) linear
( _c_ ) quadratic
( _d_ ) quintic
( _e_ ) quadratic
( _f_ ) linear
( _g_ ) quadratic
( _h_ ) cubic
**10.17**
The given equation is not an identity.
**10.18** Assume = _p_ / _q_ where _p_ and _q_ are integers having no common factor except ±1. Squaring, we get _p_ 2/ _q_ 2 = 3 or _p_ 2 = 3 _q_ 2. So _p_ 2 is a multiple of 3 or _p_ = 3 _k_ where _k_ is an integer (if _p_ = 3 _k_ \+ 1 or _p_ = 3 _k_ \+ 2, then _p_ 2 is not a multiple of 3). Thus _p_ 2 = 3 _q_ 2 becomes (3 _k_ )2 = 3 _q_ 2 and _q_ 2 = 3 _k_ 2. Since _q_ 2 is a multiple of 3, _q_ is a multiple of 3. But if both _p_ and _q_ are multiples of 3, then they have a common factor of 3. This contradicts the assumption that they have no common factor except ±1. Hence, is irrational.
## **CHAPTER 11
Ratio, Proportion, and Variation**
### **11.1 RATIO**
The ratio of two numbers _a_ and _b_ , written _a_ : _b_ , is the fraction _a/b_ provided _b_ = 0.
Thus _a_ : _b_ = _a/b, b_ = 0. If _a = b_ = 0, the ratio is 1 : 1 or 1/1 = 1.
**EXAMPLES 11.1.**
(1) The ratio of 4 to 6 =
(2)
(3)
### **11.2 PROPORTION**
A proportion is an equality of two ratios. Thus _a_ : b = _c_ : d, or a/b = _c/d_ , is a proportion in which _a_ and d are called the _extremes_ and b and c the _means_ , while d is called the _fourth proportional_ to a, b, and c.
In the proportion _a_ : _b_ = _b_ : c, _c_ is called the _third proportional_ to a and b, and _b_ is called a _mean proportional_ between _a_ and _c_.
Proportions are equations and can be transformed using procedures for equations. Some of the transformed equations are used frequently and are called the laws of proportion. If _a/b c/d_ , then
(1) ad = _bc_
(2)
(3)
(4)
(5)
(6)
### **11.3 VARIATION**
In reading scientific material, it is common to find such statements as "The pressure of an enclosed gas varies directly as the temperature." This and similar statements have precise mathematical meanings and they represent a specific type of function called variation functions. The three general types of variation functions are direct, inverse, and joint.
(1) If _x_ varies _directly_ as _y_ , then x _= ky_ or _x/y = k_ , where _k_ is called the constant of proportionality or the constant of variation.
(2) If x varies _directly_ as y2, then x _= ky 2_.
(3) If x varies _inversely_ as y, then x _= k/y_.
(4) If _x_ varies _inversely_ as y2, then _x = k/y 2_.
(5) If _x_ varies _jointly_ as _y_ and z, then _x = kyz_.
(6) If _x_ varies _directly_ as y2 and _inversely_ as z, then _x = ky 2/z_.
The constant _k_ may be determined if one set of values of the variables is known.
### **11.4 UNIT PRICE**
When shopping we find that many items are sold in different sizes. To compare the prices, we must compute the price per unit of measure for each size of the item.
**EXAMPLES 11.2.** What is the unit price for each item?
( _a_ ) 3 ounce jar of olives costing 87¢
( _b_ ) 12 ounce box of cereal costing $1.32
**EXAMPLES 11.3.** What is the unit price for each item to the nearest tenth of a cent?
( _a_ ) 6.5 ounce can of tuna costing $1.09
( _b_ ) 14 ounce can of salmon costing $1.95
### **11.5 BEST BUY**
To determine the best buy, we compare the unit price for each size of the item and the size with the lowest unit price is the best buy. In doing this, we make two assumptions – a larger size will not result in any waste and the buyer can afford the total price for each of the sizes of the item. The unit price is usually rounded to the nearest tenth of a cent when finding the best buy.
**EXAMPLE 11.4.** Which is the best buy on a bottle of vegetable oil when 1 gallon costs $5.99, 1 pint costs 89¢, and 24 ounces costs $1.29?
The best buy is one gallon of vegetable oil for $5.99.
### **Solved Problems**
**Ratio and Proportion**
**11.1** Express each of the following ratios as a simplified fraction.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
**11.2** Find the ratio of each of the following quantities.
( _a_ ) 6 pounds to 12 ounces.
It is customary to express the quantities in the same units.
Then the ratio of 96 ounces to 12 ounces is 96 : 12 = 8 : 1.
( _b_ ) 3 quarts to 2 gallons.
The required ratio is 3 quarts to 8 quarts or 3 : 8.
( _c_ ) 3 square yards to 6 square feet.
Since 1 square yard = 9 square feet, the required ratio is 27 ft2 : 6 ft2 = 9 : 2.
**11.3** In each of the following proportions determine the value of _x_.
( _a_ )
( _b_ )
( _c_ )
**11.4** Find the fourth proportional to each of the following sets of numbers. In each case let _x_ be the fourth proportional.
( _a_ )
( _b_ )
( _c_ )
**11.5** Find the third proportional to each of the following pairs of numbers. In each case let _x_ be the third proportional.
( _a_ )
( _b_ )
**11.6** Find the mean proportional between 2 and 8.
**SOLUTION**
Let _x_ be the required mean proportional. Then 2 : _x_ = _x_ : 8, x2 = 16 and _x_ = ±4.
**11.7** A line segment 30 inches long is divided into two parts whose lengths have the ratio 2 : 3. Find the lengths of the parts.
**SOLUTION**
Let the required lengths be _x_ and 30 – _x_. Then
**11.8** Two brothers are respectively 5 and 8 years old. In how many years ( _x_ ) will the ratio of their ages be 3 : 4?
**SOLUTION**
In _x_ years their respective ages will be 5 + _x_ and 8 + _x_.
Then (5 + x) : (8 + _x_ ) = 3 : 4, 4(5 + x) = 3(8 + _x_ ), and _x_ = 4.
**11.9** Divide 253 into four parts proportional to 2, 5, 7, 9.
**SOLUTION**
Let the four parts be _2k, 5k, 7k, 9k_.
Then 2k + _5k + 7k + 9k_ = 253 and _k_ = 11. Thus the four parts are 22, 55, 77, 99.
**11.10** If _x_ : _y_ : _z_ = 2 : – 5 : 4 and _x_ – 3 _y_ \+ _z_ = 63, find _x_ , _y_ , _z_.
**SOLUTION**
Let _x_ = 2 _k_ , _y_ = –5 _k_ , _z_ = 4 _k_.
Substitute these values in _x_ – 3 _y_ \+ _z_ = 63 and obtain 2 _k_ – 3(–5 _k_ ) + 4 _k_ = 63 or _k_ = 3.
Hence _x_ = 2 _k_ = 6, _y_ = –5 _k_ =–15, _z_ = 4 _k_ = 12.
**Variation**
**11.11** For each of the following statements write an equation, employing _k_ as the constant of proportionality.
( _a_ ) The circumference _C_ of a circle varies as its diameter _d. Ans. C = kd_
( _b_ ) The period of vibration _T_ of a simple pendulum at a given place is proportional to the square root of its length _l. Ans_. _T_ = _k_
( _c_ ) The rate of emission of radiant energy _E_ per unit area of a perfect radiator is proportional to the fourth power of its absolute temperature _T. Ans. E = kT_ 4
( _d_ ) The heat _H_ in calories developed in a conductor of resistance _R_ ohms when using a current of _I_ amperes, varies jointly as the square of the current, the resistance of the conductor, and the time _t_ during which the conductor draws the current. _Ans. H = kI 2Rt_
( _e_ ) The intensity _I_ of a sound wave varies jointly as the square of its frequency _n_ , the square of its amplitude _r_ , the speed of sound v, and the density _d_ of the undisturbed medium. _Ans. I = kn 2r2vd_
( _f_ ) The force of attraction _F_ between two masses _m_ 1 and _m_ 2 varies directly as the product of the masses and inversely as the square of the distance _r_ between them. _Ans_. _F_ = _km_ 1 _m_ 2/ _r_ 2
( _g_ ) At constant temperature, the volume _V_ of a given mass of an ideal gas varies inversely as the pressure _p_ to which it is subjected. _Ans. V = k/p_
( _h_ ) An unbalanced force _F_ acting on a body produces in it an acceleration _a_ which is directly proportional to the force and inversely proportional to the mass _m_ of the body. _Ans. a = kF/m_
**11.12** The kinetic energy _E_ of a body is proportional to its weight _W_ and to the square of its velocity _v_. An 8 lb body moving at 4 ft/sec has 2 ft-lb of kinetic energy. Find the kinetic energy of a 3 ton (6000 lb) truck speeding at 60 mi/hr (88 ft/sec).
**SOLUTION**
To find
Thus the kinetic energy of the truck is
**11.13** The pressure _p_ of a given mass of ideal gas varies inversely as the volume _V_ and directly as the absolute temperature _T_. To what pressure must 100 cubic feet of helium at 1 atmosphere pressure and 253° temperature be subjected to be compressed to 50 cubic feet when the temperature is 313°?
**SOLUTION**
To find
Thus the required pressure is
_Another method_ : Let the subscripts 1 and 2 refer to the initial and final conditions of the gas, respectively.
Then
**11.14** If 8 men take 12 days to assemble 16 machines, how many days will it take 15 men to assemble 50 machines?
**SOLUTION**
The number of days ( _x_ ) varies directly as the number of machines ( _y_ ) and inversely as the number of men (z).
Then
Hence the required number of days is _x_ =
**Unit Price and Best Buy**
**11.15** What is the unit price for 12 oranges costing 99¢?
**SOLUTION**
**11.16** What is the unit price for trash bags when 20 bags cost $2.50?
**SOLUTION**
**11.17** Which is the best buy when 7 cans of soup cost $2.25 and 3 cans of soup cost 95¢?
**SOLUTION**
The best buy is 3 cans of soup costing 95¢.
**11.18** Which is the best buy when a 3 ounce package of cream cheese costs 43¢ and an 8 ounce package of cream cheese costs 87¢?
**SOLUTION**
The best buy is the 8 ounce package of cream cheese costing 87¢.
### **Supplementary Problems**
**11.19** Express each ratio as a simplified fraction.
( _a_ ) 40 : 64
( _b_ ) 4/5 : 8/3
( _c_ ) x2y3 : 3 _xy_ 4
( _d_ ) (a2b + _ab_ 2) : (a2b3 \+ _a 3 b_2)
**11.20** Find the ratio of the following quantities.
( _a_ ) 20 yards to 40 feet
( _b_ ) 8 pints to 5 quarts
( _c_ ) 2 square feet to 96 square inches
( _d_ ) 6 gallons to 30 pints
**11.21** In each proportion determine the value of _x_.
( _a_ ) ( _x_ \+ 3) : ( _x_ – 2) = 3 : 2
( _b_ ) ( _x_ \+ 4) : 1 = (2 – _x_ ) : 2
( _c_ ) ( _x_ \+ 1) : 4 = ( _x_ \+ 6) : 2 _x_
( _d_ ) (2 _x_ \+ 1) : ( _x_ \+ 1) = 5 _x_ : ( _x_ \+ 4)
**11.22** Find the fourth proportional to each set of numbers.
( _a_ ) 3, 4, 12
( _b_ ) –2, 5, 6
( _c_ ) a, b, _c_
( _d_ ) _m_ \+ 2, _m_ – 2, 3
**11.23** Find the third proportional to each pair of numbers.
( _a_ ) 3, 5
( _b_ ) –2, 4
( _c_ ) _a_ , _b_
( _d_ ) _ab_ ,
**11.24** Find the mean proportional between each pair of numbers.
( _a_ ) 3, 27
( _b_ ) –4, –8
( _c_ ) 3 and 6
( _d_ ) _m_ \+ 2 and _m_ \+ 1
**11.25** If ( _x_ \+ _y_ ) : ( _x_ – _y_ ) = 5 : 2, find _x_ : _y_.
**11.26** Two numbers have the ratio 3 : 4. If 4 is added to each of the numbers the resulting ratio is 4 : 5. Find the numbers.
**11.27** A line segment of length 120 inches is divided into three parts whose lengths are proportional to 3, 4, 5. Find the lengths of the parts.
**11.28** If _x_ : _y_ : _z_ = 4 : –3 : 2 and 2 _x_ \+ 4 _y_ – 3 _z_ = 20, find _x_ , _y_ , _z_.
**11.29**
( _a_ ) If _x_ varies directly as _y_ and if _x_ = 8 when _y_ = 5, find _y_ when _x_ = 20.
( _b_ ) If _x_ varies directly as y2 and if _x_ = 4 when y = 3, find x when y = 6.
( _c_ ) If _x_ varies inversely as y and if _x_ = 8 when _y_ = 3, find y when _x_ = 2.
**11.30** The distance covered by an object falling freely from rest varies directly as the square of the time of falling. If an object falls 144 ft in 3 sec, how far will it fall in 10 sec?
**11.31** The force of wind on a sail varies jointly as the area of the sail and the square of the wind velocity. On a square foot of sail the force is 1 lb when the wind velocity is 15 mi/hr. Find the force of a 45 mi/hr wind on a sail of area 20 square yards.
**11.32** If 2 men can plow 6 acres of land in 4 hours, how many men are needed to plow 18 acres in 8 hours?
**11.33** What is the unit price to the nearest tenth of a cent for each item?
( _a_ ) 1.36 liter can of fruit punch costing $1.09
( _b_ ) 283 gram jar of jelly costing 79¢
( _c_ ) 10.4 ounce jar of face cream costing $3.73
( _d_ ) 1 dozen cans of peas costing $4.20
( _e_ ) 25 pounds of grass seed costing $27.75
( _f_ ) 3 doughnuts costing 49¢
**11.34** Which is the best buy?
( _a_ ) 100 aspirin tablets for $1.75 or 200 aspirin tablets for $2.69
( _b_ ) a 6 ounce jar of peanut butter for 85¢ or a 12 ounce jar of peanut butter for $1.59
( _c_ ) a 14 ounce bottle of mouthwash for $1.15 or a 20 ounce bottle of mouthwash for $1.69
( _d_ ) a 9 ounce jar of mustard for 35¢ or a 24 ounce jar of mustard for 89¢
( _e_ ) a 454 gram box of crackers for $1.05 or a 340 gram box of crackers for 93¢
( _f_ ) a 0.94 liter bottle of fabric softener for 99¢ or a 2.76 liter bottle of fabric softener for $2.65
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**11.19**
( _a_ ) 5/8
( _b_ ) 3/10
( _c_ ) _x_ /3 _y_
( _d_ ) 1/ _ab_
**11.20**
( _a_ ) 3 : 2
( _b_ ) 4 : 5
( _c_ ) 3 : 1
( _d_ ) 8 : 5
**11.21**
( _a_ ) 12
( _b_ ) –2
( _c_ ) 4, –3
( _d_ ) 2, –2/3
**11.22**
( _a_ ) 16
( _b_ ) –15
( _c_ ) bc/a
( _d_ ) 3( _m_ – 2)/( _m_ \+ 2)
**11.23**
( _a_ ) 25/3
( _b_ ) –8
( _c_ ) _b_ 2/ _a_
( _d_ ) 1
**11.24**
( _a_ ) ±9
( _b_ ) ±4
( _c_ ) ±6
( _d_ )
**11.25** 7/3
**11.26** 12, 16
**11.27** 30, 40, 50 in.
**11.28** –8, 6, –4
**11.29** ( _a_ ) 12
( _b_ ) 16
( _c_ ) 12
**11.30** 1600 ft
**11.31** 1620 lb
**11.32** 3 men
**11.33**
( _a_ ) 80.1¢ per liter
( _b_ ) 0.3¢ per gram
( _c_ ) 35.9¢ per ounce
( _d_ ) 35¢ per can
( _e_ ) 111¢ per pound
( _f_ ) 16.3¢ per doughnut
**11.34**
( _a_ ) 200 aspirins for $2.69
( _b_ ) 12 ounce jar for $1.59
( _c_ ) 14 ounce bottle for $1.15
( _d_ ) 24 ounce jar for 89¢
( _e_ ) 454 gram box for $1.05
( _f_ ) 2.76 liter bottle for $2.65
## **CHAPTER 12
Functions and Graphs**
### **12.1 VARIABLES**
A variable is a symbol which may assume any one of a set of values during a discussion. A _constant_ is a symbol which stands for only one particular value during the discussion.
Letters at the end of the alphabet, such as _x_ , _y_ , _z_ , _u_ , _v_ , and _w_ , are usually employed to represent variables, and letters at the beginning of the alphabet, such as _a_ , _b_ , and _c_ , are used as constants.
### **12.2 RELATIONS**
A relation is a set of ordered pairs. The relation may be specified by an equation, a rule, or a table. The set of the first components of the ordered pairs is called the domain of the relation. The set of the second components is called the range of the relation. In this chapter, we shall consider only relations that have sets of real numbers for their domain and range.
**EXAMPLE 12.1.** What is the domain and range of the relation {(1, 3), (2, 6), (3, 9), (4, 12)}?
### **12.3 FUNCTIONS**
A function is a relation such that each element in the domain is paired with exactly one element in the range.
**EXAMPLES 12.2.** Which relations are functions?
( _a_ ) {(1, 2), (2, 3), (3, 4), (4, 5)}
function – each first element is paired with exactly one second element
( _b_ ) {(1, 2), (1, 3), (2, 8), (3, 9)}
not a function – 1 is paired with 2 and with 3
( _c_ ) {(1, 3), (2, 3), (4 3), (9, 3)}
function – each first element is paired with exactly one second element
Often functions and relations are stated as equations. When the domain is not stated, we determine the largest subset of the real numbers for which the equation is defined, and that is the domain. Once the domain has been determined, we determine the range by finding the value of the equation for each value of the domain. The variable associated with the domain is called an independent variable and the variable associated with the range is called the dependent variable. In equations with the variables _x_ and _y_ , we generally assume that _x_ is the independent variable and that _y_ is the dependent variable.
**EXAMPLE 12.3.** What is the domain and range of _y_ = _x_ 2 \+ 2?
The domain is the set of all real numbers since the square of each real number is a real number and that a real number plus 2 is still a real number. Domain = {all real numbers}
The range is the set of all real numbers greater than or equal to 2 since the square of a real number is at least zero and when you add 2 to each value we have the real numbers that are at least 2. Range = {all real numbers ≥ 2}
**EXAMPLE 12.4.** What is the domain and range of _y_ = 1/( _x_ – 3)?
The equation is not defined when _x_ = 3, so the domain is the set of all real numbers not equal to 3. Domain = {real numbers ≠ 3}
A fraction can be zero only when the numerator can be zero. Since the numerator of this fraction is always 1, the fraction can never equal zero. Thus, the range is the set of all real numbers not equal to 0. Range = {real numbers ≠ 0}.
### **12.4 FUNCTION NOTATION**
The notation _y_ = _f_ ( _x_ ), read " _y_ equals _f_ of _x_ ," is used to designate that _y_ is a function of _x_. With this notation _f_ ( _a_ ) represents the value of the dependent variable _y_ when _x_ = _a_ (provided that there is a value).
Thus _y_ = _x_ 2 – 5 _x_ \+ 2 may be written _f_ ( _x_ ) = _x_ 2 – 5 _x_ \+ 2. Then _f_ (2), i.e., the value of _f_ ( _x_ ) or _y_ when _x_ = 2, is _f_ (2) = 22 – 5(2) + 2 = –4. Similarly, _f_ (–1) = (–1)2 – 5(–1) + 2 = 8.
Any letter may be used in the function notation; thus _g_ ( _x_ ), _h_ ( _x_ ), _F_ ( _x_ ), etc., may represent functions of _x_.
### **12.5 RECTANGULAR COORDINATE SYSTEM**
A rectangular coordinate system is used to give a picture of the relationship between two variables.
Consider two mutually perpendicular lines _X′X_ and _Y′Y_ intersecting in the point _O_ , as shown in Fig. 12-1.
**Fig. 12-1**
The line _X′ X_ , called the _x_ axis, is usually horizontal.
The line _Y′ Y_ , called the _y_ axis, is usually vertical.
The point _O_ is called the origin.
Using a convenient unit of length, lay off points on the _x_ axis at successive units to the right and left of the origin _O_ , labeling those points to the right 1, 2, 3, 4,... and those to the left – 1, – 2, – 3, –4,.... Here we have arbitrarily chosen _OX_ to be the positive direction; this is customary but not necessary.
Do the same on the _y_ axis, choosing _OY_ as the positive direction. It is customary (but not necessary) to use the same unit of length on both axes.
The _x_ and _y_ axes divide the plane into 4 parts known as _quadrants_ , which are labeled I, II, III, IV as in Fig. 12-1.
Given a point _P_ in this _xy_ plane, drop perpendiculars from _P_ to the _x_ and _y_ axes. The values of _x_ and _y_ at the points where these perpendiculars meet the _x_ and _y_ axes determine respectively the _x coordinate_ (or abscissa) of the point and the _y coordinate_ (or ordinate) of the point _P_. These coordinates are indicated by the symbol ( _x_ , _y_ ).
Conversely, given the coordinates of a point, we may locate or plot the point in the _xy_ plane.
For example, the point _P_ in Fig. 12-1 has coordinates (3, 2); the point having coordinates (–2, –3) is _Q_.
The graph of a function _y_ = _f_ ( _x_ ) is the set of all points ( _x_ , _y_ ) satisfied by the equation _y_ = _f_ ( _x_ ).
### **12.6 FUNCTION OF TWO VARIABLES**
The variable _z_ is said to be a function of the variables _x_ and _y_ if there exists a relation such that to each pair of values of _x_ and _y_ there corresponds one or more values of _z_. Here _x_ and _y_ are independent variables and _z_ is the dependent variable.
The function notation used in this case is _z_ = _f_ ( _x_ , _y_ ): read " _z_ equals _f_ of _x_ and _y_." Then _f_ ( _a_ , _b_ ) denotes the value of _z_ when _x_ = _a_ and _y_ = _b_ , provided the function is defined for these values.
Thus if _f_ ( _x_ , _y_ ) = _x_ 3 \+ _xy_ 2 – 2 _x_ , then _f_ (2, 3) = 23 \+ 2 · 32 – 2 · 3 = 20.
In like manner we may define functions of more than two independent variables.
### **12.7 SYMMETRY**
When the left half of a graph is a mirror image of the right half, we say the graph is symmetric with respect to the _y_ axis (see Fig. 12-2). This symmetry occurs because for any _x_ value, both _x_ and – _x_ result in the same _y_ value, that is _f_ ( _x_ ) = _f_ (– _x_ ). The equation may or may not be a function for _y_ in terms of _x_.
**Fig. 12-2**
Some graphs have a bottom half that is the mirror image of the top half, and we say these graphs are symmetric with respect to the _x_ axis. Symmetry with respect to the _x_ axis results when for each _y_ , both _y_ and – _y_ result in the same _x_ value (see Fig. 12-3). In these cases, you do not have a function for _y_ in terms of _x_.
**Fig. 12-3**
If substituting for – _x_ for _x_ and – _y_ for _y_ in an equation yields an equivalent equation, we say the graph is symmetric with respect to the origin (see Fig. 12-4). These equations represent relations that are not always functions.
**Fig. 12-4**
Symmetry can be used to make sketching the graphs of relations and functions easier. Once the type of symmetry, if any, and the shape of half of the graph have been determined, the other half of the graph can be drawn by using this symmetry. Most graphs are not symmetric with respect to the _y_ axis, _x_ axis, or the origin. However, many frequently used graphs do display one of these symmetries and using that symmetry in graphing the relation simplifies the graphing process.
**EXAMPLE 12.5.** Test the relation _y_ = 1/ _x_ for symmetry.
Substituting – _x_ for _x_ , we get _y_ = –1/ _x_ , so the graph is not symmetric with respect to the _y_ axis.
Substituting – _y_ for _y_ , we get – _y_ = 1/ _x_ , so the graph is not symmetric with respect to the _x_ axis.
Substituting – _x_ for _x_ and – _y_ for _y_ , we get – _y_ = –1/ _x_ which is equivalent to _y_ = 1/ _x_ , so the graph is symmetric with respect to the origin.
### **12.8 SHIFTS**
The graph of _y_ = _f_ ( _x_ ) is shifted upward by adding a positive constant to each _y_ value in the graph. It is shifted downward by adding a negative constant to each _y_ value in the graph of _y_ = _f_ ( _x_ ). Thus, the graph of _y_ = _f_ ( _x_ ) + _b_ differs from the graph of _y_ = _f_ ( _x_ ) by a vertical shift of | _b_ | units. The shift is up if _b_ > 0 and the shift is down if _b_ < 0.
**EXAMPLES 12.6.** How do the graphs of _y_ = _x_ 2 \+ 2 and _y_ = _x_ 2 – 3 differ from the graph of _y_ = _x_ 2?
The graph of _y_ = _x_ 2 is shifted up 2 units to yield the graph of _y_ = _x_ 2 \+ 2 (see Figs. 12-5( _a_ ) and ( _b_ )).
The graph of _y_ = _x_ 2 is shifted 3 units down to yield the graph of _y_ = _x_ 2 – 3 (see Figs. 12-5( _a_ ) and ( _c_ )).
**Fig. 12-5**
The graph of _y_ = _f_ ( _x_ ) is shifted to the right when a positive number is subtracted from each _x_ value. It is shifted to the left if a negative number is subtracted from each _x_ value. Thus, the graph of _y_ = _f_ ( _x_ – _a_ ) differs from the graph of _y_ = _f_ ( _x_ ) by a horizontal shift of | _a_ | units. The shift is to the right if _a_ > 0 and the shift is to the left if _a_ < 0.
**EXAMPLES 12.7.** How do the graphs of _y_ = ( _x_ \+ 1)2 and _y_ = ( _x_ – 2)2 differ from the graph of _y_ = _x_ 2?
The graph of _y_ = _x_ 2 is shifted 1 unit to the left to yield the graph of _y_ = ( _x_ \+ 1)2 since _x_ \+ 1 = _x_ – (–1) (see Figs. 12-6( _a_ ) and ( _b_ )).
The graph of _y_ = _x_ 2 is shifted 2 units to the right to yield the graph of _y_ = ( _x_ – 2)2 (see Figs. 12-6( _a_ ) and ( _c_ )).
**Fig. 12-6**
### **12.9 SCALING**
If each _y_ value is multiplied by a positive number greater than 1, the rate of change in _y_ is increased from the rate of change in _y_ values for _y_ = _f_ ( _x_ ). However, if each _y_ value is multiplied by a positive number between 0 and 1, the rate of change in _y_ values is decreased from the rate of change in _y_ values for _y_ = _f_ ( _x_ ). Thus, the graph of _y_ = _cf_ ( _x_ ), where _c_ is a positive number, differs from the graph of _y_ = _f_ ( _x_ ) by the rate of increase in _y_. If _c_ > 1 the rate of change in _y_ is increased and if 0 < _c_ < 1 the rate of change in _y_ is decreased.
The graph of _y_ = _f_ ( _x_ ) is reflected across the _x_ axis when each _y_ value is multiplied by a negative number. So the graph of _y_ = _cf_ ( _x_ ), where _c_ < 0, is the reflection of _y_ = | _c_ | _f_ ( _x_ ) across the _x_ axis.
**EXAMPLES 12.8.** How do the graphs of _y_ = –| _x_ |, _y_ = 3| _x_ |, and _y_ = 1/2| _x_ | differ from the graph of _y_ = | _x_ |?
The graph of _y_ = | _x_ | is reflected across the _x_ axis to yield _y_ = –| _x_ | (see Figs. 12-7( _a_ ) and ( _b_ )).
**Fig. 12-7**
The graph of _y_ = | _x_ | has the _y_ value multiplied by 3 for each _x_ value to yield the graph of _y_ = 3| _x_ | (see Figs. 12-7( _a_ ) and ( _c_ )).
The graph of _y_ = | _x_ | has the _y_ value multiplied by 1/2 for each _x_ value to yield the graph of _y_ = 1/2| _x_ | (see Figs. 12-7( _a_ ) and ( _d_ ).
### **12.10 USING A GRAPHING CALCULATOR**
In discussing graphing calculators the information given will be general, but a Texas Instruments TI-84 graphing calculator was used to verify the general procedures. Most graphing calculators operate in a somewhat similar manner to one another, but you need to use the instruction manual for your calculator to see how to do these operations for that particular make and model of graphing calculator.
A graphing calculator allows you to graph functions easily. The key to graphing is setting the graphing window appropriately. To do this, you need to use the domain of the function to set the maximum and minimum _x_ values and the range to set the maximum and minimum _y_ values. When the domain or range is a large interval, it may be necessary to use the scale for _x_ or _y_ to make the graph smaller, increasing the size of the units along either or both axes. Occasionally, it may be necessary to view the graph in parts of its domain or range to see how the graph actually looks.
To compare the graphs of _y_ = _x_ 2, _y_ = _x_ 2 \+ 2, and _y_ = _x_ 2 – 3 on a graphing calculator, you enter each function in the _y_ = menu. Let _y_ 1 = _x_ 2, _y_ 2 = _x_ 2 \+ 2, and _y_ 3 = _x_ 2 – 3. Turn off the functions _y_ 2 and _y_ 3 and set the graphing window at the standard setting. When you press the graph key, you will see a graph as shown in Fig. 12-5( _a_ ). Turn off the function _y_ 1 and turn on the function _y_ 2, then press the graph key. The graph displayed will be Fig. 12-5( _b_ ). Now turn off function _y_ 2 and turn on function _y_ 3. Press the graph key and you will see Fig. 12-5( _c_ ). When you turn on functions _y_ 1, _y_ 2, and _y_ 3 and press the graph key and you will see all three functions graphed on the same set of axes (see Fig. 12-8). The graph of _y_ 2 = _x_ 2 \+ 2 is 2 units above the graph of _y_ 1 = _x_ 2, while the graph of _y_ 3 = _x_ 2 – 3 is 3 units below the graph of _y_ 1.
**Fig. 12-8**
In a similar fashion, you can compare the graph of _y_ 1 = _f_ ( _x_ ) and _y_ 2 = _f_ ( _x_ ) + _b_ for any function _f(x)_. Notice that when _b_ > 0, the graph of _y_ 2 is _b_ units above the graph of _y_ 1. When _b_ < 0, the graph of _y_ 2 is | _b_ | units below the graph of _y_ 1.
Consider the graphs of _y_ = _x_ 2, _y_ = ( _x_ \+ 1)2, and _y_ = ( _x_ – 2)2. To compare these graphs using a calculator, we need to set _y_ 1 = _x_ 2, _y_ 2 = ( _x_ \+ 1)2, and _y_ 3 = ( _x_ – 2)2. Using the standard window and graphing all three functions at once, we see that _y_ 2 = ( _x_ \+ 1)2 is 1 unit to the left of the graph of _y_ 1 = _x_ 2. Also, the graph of _y_ 3 = ( _x_ – 2)2 is 2 units to the right of the graph of _y_ 1 (see Fig. 12-9).
**Fig.12-9**
In general, to compare the graphs of _y_ 1 = _f_ ( _x_ ) and _y_ 2 = _f(x_ – _a_ ) for all functions _f_ ( _x_ ), we note that the graph of _y_ 2 = _f_ ( _x_ – _a_ ) is _a_ units to the right of _y_ 1 = _f_ ( _x_ ) when _a_ > 0. When _a_ < 0, the graph of _y_ 2 is | _a_ | units to the left of _y_ 1.
**EXAMPLE 12.9.** Graph _x_ 2 \+ _y_ 2 = 9.
To graph _x_ 2 \+ _y_ 2 = 9 on a calculator, we first solve the equation for _y_ and get We let and and graph them on the same set of axes. If we use the standard window, we get a distorted view of this graph because the scale on the _y_ axis is not equal to the scale on the _x_ axis. By multiplying by a factor of 0.67 (for the TI-84), we can adjust the _y_ interval and get a more accurate view of the graph. Thus using the domain –10, 10] and a range of [–6.7, 6.7], we get the graph of a circle (see [Fig. 12-10).
**Fig. 12-10**
### **Solved Problems**
**12.1** Express the area _A_ of a square as a function of its ( _a_ ) side _x_ , ( _b_ ) perimeter _P_ , and ( _c_ ) diagonal _D_ (see Fig. 12-11).
**Fig. 12-11**
**SOLUTION**
( _a_ ) _A_ = _x_ 2
( _b_ )
( _c_ )
**12.2** Express the ( _a_ ) area _A_ , ( _b_ ) perimeter _P_ , and ( _c_ ) diagonal _D_ of a rectangle as a function of its sides _x_ and _y_. Refer to Fig. 12-12.
**Fig. 12-12**
**SOLUTION**
( _a_ ) _A_ = _xy_ ,
( _b_ ) _P_ = 2 _x_ \+ 2 _y_ ,
( _c_ )
**12.3** Express the ( _a_ ) altitude _h_ and ( _b_ ) area _A_ of an equilateral triangle as a function of its side _s_. Refer to Fig. 12-13.
**Fig. 12-13**
**SOLUTION**
( _a_ )
( _b_ )
**12.4** The surface area _S_ and volume _V_ of a sphere of radius _r_ are given by _S_ = 4 _πr_ 2 and Express ( _a_ ) _r_ as a function of _S_ and also as a function of _V_ , ( _b_ ) _V_ as a function of _S_ , and ( _c_ ) _S_ as a function of _V_.
**SOLUTION**
( _a_ ) From _S_ = 4 _πr_ 2 obtain
From obtain
( _b_ ) Set
in and obtain
( _c_ ) Set
in _S_ = 4 _πr_ 2 and obtain
**12.5** Given _y_ = 3 _x_ 2 – 4 _x_ \+ 1, find the values of _y_ corresponding to _x_ = –2, –1, 0, 1, 2.
**SOLUTION**
For _x_ = –2, _y_ = 3(–2)2 – 4(–2) + 1 = 21; for _x_ = –1, _y_ = 3(–1)2 – 4(–1) + 1 = 8; for _x_ = 0, _y_ = 3(0)2 – 4(0) + 1 = 1; for _x_ = 1, _y_ = 3(1)2 – 4(1) + 1 = 0; for _x_ = 2, _y_ = 3(2)2 – 4(2) + 1 = 5. These values of _x_ and _y_ are conveniently listed in the following table.
**12.6** Extend the table of values in Problem 12.5 by finding the values of _y_ corresponding to _x_ = –3/2, –1/2, 1/2, 3/2.
**SOLUTION**
For etc. The following table of values summarizes the results.
**12.7** State the domain and range for each relation.
( _a_ ) _y_ = 3 – _x_ 2
( _b_ ) _y_ = _x_ 3 \+ 1
( _c_ )
( _d_ )
**SOLUTION**
( _a_ ) Domain = {all real numbers} Since every real number can be squared, 3 – _x_ 2 is defined for all real numbers.
Range = {all real numbers ≤ 3} Since _x_ 2 is non-negative for all real numbers, 3 – _x_ 2 does not exceed 3.
( _b_ ) Domain = {all real numbers} Since every real number can be cubed, _x_ 3 \+ 1 is defined for all real numbers.
Range = {all real numbers} Since _x_ 3 yields all real numbers, _x_ 3 \+ 1 also yields all real numbers.
( _c_ ) Domain = {all real numbers ≥ –2} Since the square root yields real numbers only for non-negative real numbers, _x_ must be at least –2.
Range = {all real numbers ≥ 0} Since we want the principal square root, the values will be non-negative numbers.
( _d_ ) Domain = {all real numbers} Since the cube root yields a real number for all real numbers, _x_ can be any real number.
Range = {all real numbers} Since any real number can be the cube root of a real number, we get all real numbers.
**12.8** In which of these equations is _y_ a function of _x_?
( _a_ ) _y_ = 3 _x_ 3
( _b_ ) _y_ 2 = _x_
( _c_ ) _xy_ = 1
( _d_ ) _y_ = 2 _x_ \+ 5
( _e_ )
( _d_ ) _y_ 3 = 8 _x_
**SOLUTION**
**12.9** If _f_ ( _x_ ) = _x_ 3 – 5 _x_ – 2, find _f_ (–2), _f_ (–3/2), _f_ (–1), _f_ (0), _f_ (1), _f_ (2).
**SOLUTION**
We may arrange these values in a table.
**12.10** If find _F_ (–2), _F_ ( _x_ ), _F_ (– _x_ ).
**SOLUTION**
**12.11** Given _R_ ( _x_ ) = (3 _x_ – 1)/(4 _x_ \+ 2), find
( _a_ )
( _b_ )
( _c_ ) _R_ [ _R_ ( _x_ )].
**SOLUTION**
( _a_ )
( _b_ )
( _c_ )
**12.12** If _F_ ( _x_ , _y_ ) = _x_ 3 – 3 _xy_ \+ _y_ 2, find
( _a_ ) _F_ (2, 3),
( _b_ ) _F_ (–3, 0),
( _c_ )
**SOLUTION**
( _a_ ) _F_ (2, 3) = 23 – 3(2)(3) + 32 = –1
( _b_ ) _F_ (–3, 0) = (–3)3 – 3(–3)(0) + 02 = –27
( _c_ )
**12.13** Plot the following points on a rectangular coordinate system: (2, 1), (4, 3), (–2, 4), (–4, 2), (–4, –2), (–5/2, –9/2), (4, –3), (2, – ).
**SOLUTION**
See Fig. 12-14.
**Fig. 12-14**
**12.14** Given _y_ = 2 _x_ – 1, obtain the values of _y_ corresponding to _x_ = –3, –2, –1, 0, 1, 2, 3 and plot the points ( _x_ , _y_ ) thus obtained.
**SOLUTION**
The following table lists the values of _y_ corresponding to the given values of _x_.
The points (–3, –7), (–2, –5), (–1, –3), (0, –1), (1, 1), (2, 3), (3, 5) are plotted, as shown in Fig. 12-l5.
**Fig. 12-15**
Note that all points satisfying _y_ = 2 _x_ – 1 lie on a straight line. In general the graph of _y_ = _ax_ \+ _b_ , where _a_ and _b_ are constants, is a straight line; hence _y_ = _ax_ \+ _b_ or _f_ ( _x_ ) = _ax_ \+ _b_ is called a _linear function_. Since two points determine a straight line, only two points need be plotted and the line drawn connecting them.
**12.15** Obtain the graph of the function defined by _y_ = _x_ 2 – 2 _x_ – 8 or _f_ ( _x_ ) = _x_ 2 – 2 _x_ – 8.
**SOLUTION**
The following table gives the values of _y_ or _f_ ( _x_ ) for various values of _x_.
Thus the following points lie on the graph: (–4, 16), (–3, 7), (–2, 0), (–1, –5), etc.
In plotting these points it is convenient to use different scales on the _x_ and _y_ axes, as shown in Fig. 12-16. The points marked × were added to those already obtained in order to get a more accurate picture.
**Fig.12-16**
The curve thus obtained is called a _parabola_. The lowest point _P_ , called a minimum point, is the _vertex_ of the parabola.
**12.16** Graph the function defined by _y_ = 3 – 2 _x_ – _x_ 2.
**SOLUTION**
The curve obtained is a parabola, as shown in Fig. 12-17. The point _Q_ (–1, 4), the vertex of the parabola, is a maximum point. In general, _y_ = _ax_ 2 \+ _bx_ \+ _c_ represents a parabola whose vertex is either a maximum or minimum point depending on whether _a_ is – or +, respectively. The function _f_ ( _x_ ) = _ax_ 2 \+ _bx_ \+ _c_ is sometimes called a quadratic function.
**Fig. 12-17**
**12.17** Obtain the graph of _y_ = _x_ 3 \+ 2 _x_ 2 – 7 _x_ – 3.
**SOLUTION**
The graph is shown in Fig. 12-18. Points marked × are not listed in the table; they were added in order to improve the accuracy of the graph.
**Fig. 12-18**
Point _A_ is called a _relative maximum point_ ; it is not the highest point on the entire curve, but points on either side are lower. Point _B_ is called a _relative minimum point_. The calculus enables us to determine such relative maximum and minimum points.
**12.18** Obtain the graph of _x_ 2 \+ _y_ 2 = 36.
**SOLUTION**
We may write Note that _x_ must have a value between –6 and +6 if _y_ is to be a real number.
The points to be plotted are etc.
Figure 12-19 shows the graph, a circle of radius 6.
**Fig. 12-19**
In general, the graph of _x_ 2 \+ _y_ 2 = _a_ 2 is a circle with center at the origin and radius _a_.
It should be noted that if the units had not been taken as the same on the _x_ and _y_ axes, the graph would not have looked like a circle.
**12.19** Determine whether the graph is symmetric with respect to the _y_ axis, _x_ axis, or the origin.
( _a_ ) _y_ = 4 _x_
( _b_ ) _x_ 2 \+ _y_ 2 = 8
( _c_ ) _xy_ 2 = 1
( _d_ ) _x_ = _y_ 2 \+ 1
( _e_ ) _y_ = _x_ 3
( _f_ )
**SOLUTION**
**12.20** Use the graph of _y_ = _x_ 3 to graph _y_ = _x_ 3 \+ 1.
**SOLUTION**
The graph of _y_ = _x_ 3 is shown in Fig. 12-20. The graph of _y_ = _x_ 3 \+ 1 is the graph of _y_ = _x_ 31 unit up and is shown in Fig. 12-21.
**Fig. 12-20**
**Fig. 12-21**
**12.21** Use the graph of _y_ = | _x_ | to graph _y_ = | _x_ \+ 2|.
**SOLUTION**
The graph of _y_ = | _x_ | is shown in Fig. 12-22. The graph of _y_ = | _x_ \+ 2| is the graph of _y_ = | _x_ | shifted 2 units to the left, since | _x_ \+ 2| = | _x_ – (–2)| and is shown in Fig. 12-23.
**Fig. 12-22**
**Fig. 12-23**
**12.22** Use the graph of _y_ = _x_ 2 to graph _y_ = – _x_ 2.
**SOLUTION**
The graph of _y_ = _x_ 2 is shown in Fig. 12-24. The graph of _y_ = – _x_ 2 is the graph of _y_ = _x_ 2 reflected across the _x_ axis and is shown in Fig. 12-25.
**Fig. 12-24**
**Fig. 12-25**
**12.23** A man has 40 ft of wire fencing with which to form a rectangular garden. The fencing is to be used only on three sides of the garden, his house providing the fourth side. Determine the maximum area which can be enclosed.
**SOLUTION**
Let _x_ = length of each of two of the fenced sides of the rectangle; then 40 – 2 _x_ = length of the third fenced side.
The area _A_ of the garden is _A_ = _x_ (40 – 2 _x_ ) = 40 _x_ – 2 _x_ 2. We wish to find the maximum value of _A_. A table of values and the graph of _A_ plotted against _x_ are shown. It is clear that _x_ must have a value between 0 and 20 ft if _A_ is to be positive.
From the graph in Fig. 12-26 the maximum point _P_ has coordinates (10, 200) so that the dimensions of the garden are 10 ft by 20 ft and the area is 200 ft2.
**Fig. 12-26**
**12.24** A rectangular piece of tin has dimensions 12 in. by 18 in. It is desired to make an open box from this material by cutting out equal squares from the corners and then bending up the sides. What are the dimensions of the squares cut out if the volume of the box is to be as large as possible?
**SOLUTION**
Let _x_ be the length of the side of the square cut out from each corner. The volume _V_ of the box thus obtained is _V_ = _x_ (12 – 2 _x_ )(18 – 2 _x_ ). It is clear that _x_ must be between 0 and 6 in. if there is to be a box (see Fig. 12-27).
**Fig. 12-27**
From the graph, the value of _x_ corresponding to the maximum value of _V_ lies between 2 and 2.5 in. By plotting more points it is seen that _x_ = 2.4 in. approximately.
Problems such as this and Problem 12.23 may often be solved easily and exactly by methods of the calculus.
**12.25** A cylindrical can is to have a volume of 200 cubic inches. Find the dimensions of the can which is made of the least amount of material.
**SOLUTION**
Let _x_ be the radius and _y_ the height of the cylinder.
The area of the top or bottom of the can is _πx_ 2 and the lateral area is 2 _πxy_ ; then the total area _S_ = 2 _πx_ 2 \+ 2 _πxy_.
The volume of the cylinder is _πx_ 2 _y_ , so that _πx_ 2 _y_ = 200 and _y_ = 200/ _πx_ 2. Then
A table of values and the graph of _S_ plotted against _x_ (Fig. 12-28) are shown. We take π = 3.14 approximately.
**Fig. 12-28**
From the graph in Fig. 12-28, minimum _S_ = 189 in2 occurs when _x_ = 3.2 in. approximately; and from _y_ = 200/ _πx_ 2 we have _y_ = 6.2 in. approximately.
**12.26** Find approximately the values of _x_ for which _x_ 3 \+ 2 _x_ 2 – 7 _x_ – 3 = 0.
**SOLUTION**
Consider _y_ = _x_ 3 \+ 2 _x_ 2 – 7 _x_ – 3. We must find values of _x_ for which _y_ = 0.
From the graph of _y_ = _x_ 3 \+ 2 _x_ 2 – 7 _x_ – 3, which is shown in Fig. 12-18, it is clear that there are three real values of _x_ for which _y_ = 0 (the values of _x_ where the curve intersects the _x_ axis). These values are _x_ = –3.7, _x_ = –0.4, and _x_ = 2.1 approximately. More exact values may be obtained by advanced techniques.
**12.27** The following table shows the population of the United States (in millions) for the years 1840, 1850,..., 1950. Graph these data.
**SOLUTION**
See Fig. 12-29.
**Fig. 12-29**
**12.28** The time _T_ (in seconds) required for one complete vibration of a simple pendulum of length _l_ (in centimeters) is given by the following observations obtained in a physics laboratory. Exhibit graphically _T_ as a function of _l_.
**SOLUTION**
The observation points are connected by a smooth curve (Fig. 12-30) as is usually done in science and engineering.
**Fig. 12-30**
### **Supplementary Problems**
**12.29** A rectangle has sides of lengths _x_ and 2 _x_. Express the area _A_ of the rectangle as a function of its ( _a_ ) side _x_ , ( _b_ ) perimeter _P_ , and ( _c_ ) diagonal _D_.
**12.30** Express the area _A_ of a circle in terms of its ( _a_ ) radius _r_ , ( _b_ ) diameter _d_ , and ( _c_ ) circumference, _C_.
**12.31** Express the area _A_ of an isosceles triangle as a function of _x_ and _y_ , where _x_ is the length of the two equal sides and _y_ is the length of the third side.
**12.32** The side of a cube has length _x_. Express ( _a_ ) _x_ as a function of the volume _V_ of the cube, ( _b_ ) the surface area _S_ of the cube as a function of _x_ , and ( _c_ ) the volume _V_ as a function of the surface _S_.
**12.33** Given _y_ = 5 + 3 _x_ – 2 _x_ 2, find the values of _y_ corresponding to _x_ = –3, –2, –1, 0, 1, 2, 3.
**12.34** Extend the table of values in Problem 12.33 by finding the values of _y_ which correspond to _x_ = –5/2, –3/2, –1/2, 1/2, 3/2, 5/2.
**12.35** State the domain and range for each equation.
( _a_ ) _y_ = –2 _x_ \+ 3
( _b_ ) _y_ = _x_ 2 – 5
( _c_ ) _y_ = _x_ 3 – 4
( _d_ ) _y_ = 5 – 2 _x_ 2
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
**12.36** For which of these relations is _y_ a function of _x_?
( _a_ ) _y_ = _x_ 3 \+ 2
( _b_ ) _x_ = _y_ 3 \+ 2
( _c_ ) _x_ = _y_ 2 \+ 4
( _d_ ) _y_ = _x_ 2 – 5
( _e_ ) _x_ 2 \+ _y_ 2 = 5
( _f_ )
( _g_ ) _x_ = | _y_ |
( _h_ )
( _i_ )
**12.37** If _f_ ( _x_ ) = 2 _x_ 2 \+ 6 _x_ – 1, find _f_ (–3), _f_ (–2), _f_ (0), _f_ (1/2), _f_ (3).
**12.38** If _F_ find ( _a_ ) _F_ (1), ( _b_ ) _F_ (2), ( _c_ ) _F_ ( _x)_ , ( _d_ ) _F_ (– _x_ ).
**12.39** If _G_ find
( _a_ )
( _b_ )
( _c_ ) _G_ ( _x_ 2 \+ 1).
**12.40** If _F_ ( _x_ , _y_ ) = 2 _x_ 2 \+ 4 _xy_ – _y_ 2, find ( _a_ ) _F_ (1, 2), ( _b_ ) _F_ (–2, –3), ( _c_ ) _F_ ( _x_ \+ 1, _y_ – 1).
**12.41** Plot the following points on a rectangular coordinate system:
( _a_ ) (1, 3),
( _b_ ) (–2, 1),
( _c_ ) (–1/2, –2),
( _d_ ) (–3, 2/3),
( _e_ )
**12.42** If _y_ = 3 _x_ \+ 2, ( _a_ ) obtain the values of _y_ corresponding to _x_ = –2, –1, 0, 1, 2 and ( _b_ ) plot the points ( _x_ , _y_ ) thus obtained.
**12.43** Determine whether the graph of each relation is symmetric with respect to the _y_ axis, _x_ axis, or origin.
( _a_ ) _y_ = 2 _x_ 4 \+ 3
( _b_ ) _y_ = ( _x_ – 3)3
( _c_ )
( _d_ ) _y_ = 3
( _e_ ) _y_ = –5 _x_ 3
( _f_ ) _y_ = 7 _x_ 2 \+ 4
( _g_ ) _y_ 2 = _x_ \+ 2
( _h_ ) _y_ = 3 _x_ – 1
( _i_ ) _y_ = 5 _x_
**12.44** State how the graph of the first equation relates to the graph of the second equation.
( _a_ ) _y_ = – _x_ 4 and _y_ = _x_ 4
( _b_ ) _y_ = 3 _x_ and _y_ = _x_
( _c_ ) _y_ = _x_ 2 \+ 10 and _y_ = _x_ 2
( _d_ ) _y_ = ( _x_ – 1)3 and _y_ = _x_ 3
( _e_ ) _y_ = _x_ 2 – 7 and _y_ = _x_ 2
( _f_ ) _y_ = | _x_ | + 1 and _y_ = | _x_ |
( _g_ ) _y_ = | _x_ \+ 5| and _y_ = | _x_ |
( _h_ ) _y_ = – _x_ 3 and _y_ = _x_ 3
( _i_ ) _y_ = _x_ 2/6 and _y_ = _x_ 2
( _j_ ) _y_ = ( _x_ \+ 8)2 and _y_ = _x_ 2
**12.45** Graph the functions ( _a_ ) _f_ ( _x_ ) = 1 – 2 _x_ , ( _b_ ) _f_ ( _x_ ) = _x_ 2 – 4 _x_ \+ 3, ( _c_ ) _f_ ( _x_ ) = 4 – 3 _x_ – _x_ 2.
**12.46** Graph _y_ = _x_ 3 – 6 _x_ 2 \+ 11 _x_ – 6.
**12.47** Graph ( _a_ ) _x_ 2 \+ _y_ 2 = 16, ( _b_ ) _x_ 2 \+ 4 _y_ 2 = 16.
**12.48** There is available 120 ft of wire fencing with which to enclose two equal rectangular gardens _A_ and _B_ , as shown in Fig. 12-31. If no wire fencing is used along the sides formed by the house, determine the maximum combined area of the gardens.
**Fig. 12-31**
**12.49** Find the area of the largest rectangle which can be inscribed in a right triangle whose legs are 6 and 8 in. respectively (see Fig. 12-32).
**Fig. 12-32**
**12.50** Obtain the relative maximum and minimum values of the function _f_ ( _x_ ) = 2 _x_ 3 – 15 _x_ 2 \+ 36 _x_ – 23.
**12.51** From the graph of _y_ = _x_ 3 – 7 _x_ \+ 6 obtain the roots of the equation _x_ 3 – 7 _x_ \+ 6 = 0.
**12.52** Show that the equation _x_ 3 – _x_ 2 \+ 2 _x_ – 3 = 0 has only one real root.
**12.53** Show that _x_ 4 – _x_ 2 \+ 1 = 0 cannot have any real roots.
**12.54** The percentage of workers in the USA employed in agriculture during the years 1860, 1870,..., 1950 is given in the following table. Graph the data.
**12.55** The total time required to bring an automobile to a stop after perceiving danger is composed of the _reaction time_ (time between recognition of danger and application of brakes) plus _braking time_ (time for stopping after application of brakes). The following table gives the stopping distances _d_ (feet) of an automobile traveling at speeds _v_ (miles per hour) at the instant danger is sighted. Graph _d_ against _v_.
**12.56** The time _t_ taken for an object to fall freely from rest through various heights _h_ is given in the following table.
( _a_ ) Graph _h_ against _t_.
( _b_ ) How long would it take an object to fall freely from rest through 48 ft? 300 ft?
( _c_ ) Through what distance can an object fall freely from rest in 3.6 seconds?
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**12.29**
**12.30**
**12.31**
**12.32**
**12.33**
**12.34**
**12.35**
**12.36**
( _a_ ) Function
( _b_ ) Function
( _c_ ) Not a function
( _d_ ) Function
( _e_ ) Not a function
( _f_ ) Not a function
( _g_ ) Not a function
( _h_ ) Function
( _i_ ) Function
**12.37** _f_ (–3) = –1, _f_ (–2) = –5, _f_ (0) = –1, _f_ (1/2) = 5/2, _f_ (3) = 35
**12.38**
**12.39**
**12.40**
( _a_ ) 6,
( _b_ ) 23,
( _c_ ) 2 _x_ 2 \+ 4 _xy_ – _y_ 2 \+ 6 _y_ – 3
**12.41** See Fig. 12-33.
**Fig. 12-33**
**12.42**
( _a_ ) –4, –1, 2, 5, 8
( _b_ ) See Fig. 12-34.
**Fig. 12-34**
**12.43**
( _a_ ) _y_ axis
( _b_ ) Origin
( _c_ ) None
( _d_ ) _y_ axis
( _e_ ) Origin
( _f_ ) _y_ axis
( _g_ ) _x_ axis
( _h_ ) None
( _i_ ) Origin
**12.44**
( _a_ ) Reflected across _x_ axis
( _b_ ) _y_ increases 3 times as fast
( _c_ ) Shifted 10 units up
( _d_ ) Shifted 1 unit right
( _e_ ) Shifted 7 units down
( _f_ ) Shifted 1 unit up
( _g_ ) Shifted 5 units left
( _h_ ) Reflected across the _x_ axis
( _i_ ) _y_ increases 1/6 as fast
( _j_ ) Shifted 8 units left
**12.45**
( _a_ ) See Fig. 12-35.
( _b_ ) See Fig. 12-36.
( _c_ ) See Fig. 12-37.
**Fig. 12-35**
**Fig. 12-36**
**Fig. 12-37**
**12.46** See Fig. 12-38.
**Fig. 12-38**
**12.47**
( _a_ ) See Fig. 12-39.
( _b_ ) See Fig. 12-40.
**Fig. 12-39**
**Fig. 12-40**
**12.48** 1200 ft2
**12.49** 12 in2
**12.50** Maximum value of _f_ ( _x_ ) is 5 (at _x_ = 2); minimum value of _f_ ( _x_ ) is 4 (at _x_ = 3).
**12.51** Roots are _x_ = –3, _x_ = 1, _x_ = 2.
**12.52** See Fig. 12-41.
**Fig. 12-41**
**12.53** See Fig. 12-42.
**Fig. 12-42**
**12.54** See Fig. 12-43.
**Fig. 12-43**
**12.55** See Fig. 12-44.
**Fig. 12-44**
**12.56**
( _a_ ) See Fig. 12-45;
( _b_ ) 1.7 sec, 4.3 sec;
( _c_ ) 207 ft
**Fig. 12-45**
## **CHAPTER 13
Linear Equations in One Variable**
### **13.1 LINEAR EQUATIONS**
A linear equation in one variable has the form _ax_ \+ _b_ = 0, where _a_ ≠ 0 and _b_ are constants. The solution of this equation is given by _x_ = – _b/a_.
When a linear equation is not in the form _ax_ \+ _b_ = 0, we simplify the equation by multiplying each term by the LCD for all fractions in the equation, removing any parentheses, or combining like terms. In some equations we do more than one of the procedures.
**EXAMPLE 13.1.** Solve the equation _x_ \+ 8 – 2( _x_ \+ 1) = 3 _x_ – 6 for _x_.
Check:
### **13.2 LITERAL EQUATIONS**
Most literal equations we encounter are formulas. Frequently, we want to use the formula to determine a value other than the standard one. To do this, we consider all variables except the one we are interested in as constants and solve the equation for the desired variable.
**EXAMPLE 13.2.** Solve _p_ = 2( _1_ \+ _w_ ) for _l_.
### **13.3 WORD PROBLEMS**
In solving a word problem, the first step is to decide what is to be found. The next step is to translate the conditions stated in the problem into an equation or to state a formula that expresses the conditions of the problem. The solution of the equation is the next step.
**EXAMPLE 13.3.** If the perimeter of a rectangle is 68 meters and the length is 14 meters more than the width, what are the dimensions of the rectangle?
Let _w_ = the number of meters in the width and _w_ \+ 14 = the number of meters in the length.
The rectangle is 24 meters long by 10 meters wide.
**EXAMPLE 13.4.** The sum of two numbers is -4 and their difference is 6. What are the numbers?
Let _n_ = the smaller number and _n_ \+ 6 = the larger number.
The two numbers are – 5 and 1.
**EXAMPLE 13.5.** If one pump can fill a pool in 16 hours and if two pumps can fill the pool in 6 hours, how fast can the second pump fill the pool?
Let _h_ = the numbers of hours for the second pump to fill the pool.
The second pump takes 9.6 hours (or 9 hours and 36 minutes) to fill the pool.
**EXAMPLE 13.6.** How many liters of pure alcohol must be added to 15 liters of a 60% alcohol solution to obtain an 80% alcohol solution?
Let _n_ = the number of liters of pure alcohol to be added.
Fifteen liters of pure alcohol must be added.
### **Solved Problems**
**13.1** Solve each of the following equations.
( _a) x_ \+ 1 = 5, _x_ = 5 – 1, _x_ = 4.
_Check:_ Put _x_ = 4 in the original equation and obtain 4 + 1 ? 5, 5 = 5.
( _b_ ) 3 _x_ – 7 = 14, 3 _x_ = 14 + 7, 3 _x_ = 21, _x_ = 7.
_Check:_ 3(7) – 7 ? 14, 14 = 14.
( _c_ ) 3 _x_ \+ 2 = 6 _x_ – 4, 3 _x_ – 6 _x_ = –4 – 2, –3 _x_ = –6, _x_ = 2.
( _d_ ) _x_ \+ 3( _x_ – 2) = 2 _x_ – 4, _x_ \+ 3 _x_ – 6 = 2 _x_ – 4, 4 _x_ – 2 _x_ = 6 – 4, 2 _x_ = 2, _x_ = 1.
( _e_ ) 3 _x_ – 2 = 7 – 2 _x_ , 3 _x_ \+ 2 _x_ = 7 + 2, 5 _x_ = 9, _x_ = 9/5.
( _f_ ) 2( _t_ \+ 3) = 5( _t_ – 1) –7( _t_ – 3), 2 _t_ \+ 6 = 5 _t_ – 5 – 7 _t_ \+ 21, 4 _t_ = 10, _t_ = 10/4 = 5/2.
( _g_ ) 3 _x_ \+ 4( _x_ – 2) = _x_ – 5 + 3(2 _x_ – 1), 3 _x_ \+ 4 _x_ – 8 = _x_ – 5 + 6 _x_ – 3, 7 _x_ – 8 = 7 _x_ – 8.
This is an identity and is true for all values of _x_.
( _h_ )
( _i_ ) 3 + 2[ _y_ – (2 _y_ \+ 2)] = 2[ _y_ \+ (3 _y_ – 1)], 3 + 2[ _y_ – 2 _y_ – 2] = 2[ _y_ \+ 3 _y_ – 1], 3 + 2 _y_ – 4 _y_ – 4 = 2 _y_ \+ 6 _y_ – 2, –2 _y_ – 1 = 8 _y_ – 2, –l0 _y_ = –1, _y_ = 1/10.
( _j_ ) ( _s_ \+ 3)2 = ( _s_ – 2)2 – 5, _s_ 2 \+ 6 _s_ \+ 9 = _s_ 2 – 4 _s_ \+ 4 – 5, 6 _s_ \+ 4 _s_ = –9 – 1, _s_ = –l.
( _k_ )
_Check_ :
( _l_ )
There is no value of _x_ which satisfies this equation.
( _m_ ) Multiplying by 2 _x_ , 5(2) + 5 = 12 _x_ , 12 _x_ = 15, _x_ = 5/4.
( _n_ ) Multiplying by 2 _x_ ( _x_ – 1), the LCD of the fractions,
( _x_ \+ 3)( _x_ – 1) + 5(2 _x_ ) = _x_ ( _x_ – 1), _x_ 2 \+ 2 _x_ – 3 + 10 _x_ = _x_ 2 – _x_ , 13 _x_ = 3, _x_ = 3/13.
( _o_ ) Multiplying by ( _x_ – 3)( _x_ \+ 3) or _x_ 2 – 9, _x_ – 9, 2( _x_ \+ 3) – 4( _x_ – 3) = 16, 2 _x_ \+ 6 – 4 _x_ \+ 12 = 16, –2 _x_ = –2, _x_ = 1.
( _p_ )
( _q_ ) Multiplying by _x_ ( _x_ – 4)(2 _x_ \+ 3), the LCD of the fractions,
3(2 _x_ \+ 3) – 2 _x_ = 9( _x_ – 4), 6 _x_ \+ 9 – 2 _x_ = 9 _x_ – 36, 45 = 5 _x_ , _x_ = 9.
**13.2** Solve for _x_.
( _a_ ) 2 _x_ – 4 _p_ = 3 _x_ \+ _2p, 2x_ – 3 _x_ = 2 _p_ \+ 4 _p_ , – _x_ = 6 _p_ , _x_ = –6 _p_.
( _b_ ) provided _a ≠ b_.
If _a_ = _b_ the equation is an identity and is true for all values of _x_.
( _c_ ) provided 3 _a_ ≠ 2 _c_.
If 3 _a_ = 2 _c_ there is no solution unless _d_ = – _b_ , in which case the original equation is an identity.
( _d_ ) (provided 3 _a_ ≠ 4 _b_ ).
**13.3** Express each statement in terms of algebraic symbols.
( _a_ ) One more than twice a certain number.
Let _x_ = the number. Then 2 _x_ = twice the number, and one more than twice the number = 2 _x_ \+ 1.
( _b_ ) Three less than five times a certain number.
Let _x_ = the number. Then three less than five times the number = 5 _x_ – 3.
( _c_ ) Each of two numbers whose sum is 100.
If _x_ = one of the numbers, then 100 – _x_ = the other number.
( _d_ ) Three consecutive integers (for example, _5_ , 6, 7).
If _x_ is the smallest integer, then ( _x_ \+ 1) and ( _x_ \+ 2) are the other two integers.
( _e_ ) Each of two numbers whose difference is 10.
Let _x_ = the smaller number; then ( _x_ \+ 10) = the larger number.
( _f_ ) The amount by which 100 exceeds three times a given number.
Let _x_ = given number. Then the excess of 100 over 3 _x_ is (100 – 3 _x_ ).
( _g_ ) Any odd integer.
Let _x_ = any integer. Then 2 _x_ is always an even integer, and (2 _x_ \+ 1) is an odd integer.
( _h_ ) Four consecutive odd integers (for example, 1, 3, 5, 7; 17, 19, 21, 23).
The difference between two consecutive odd integers is 2.
Let 2 _x_ \+ 1 = smallest odd integer. Then the required numbers are 2 _x_ \+ 1, 2 _x_ \+ 3, 2 _x_ \+ 5, 2 _x_ \+ 7.
( _i_ ) The number of cents in _x_ dollars.
Since 1 dollar = 100 cents, _x_ dollars = 100 _x_ cents.
( _j_ ) John is twice as old as Mary, and Mary is three times as old as Bill. Express each of their ages in terms of a single unknown.
Let _x_ = Bill's age. Then Mary's age is 3 _x_ and John's age is 2(3 _x_ ) = 6 _x_.
Another method. Let _y_ = John's age. Then Mary's age and Bill's age
( _k_ ) The three angles _A, B_ , C of a triangle if angle _A_ has 10° more than twice the number of degrees in angle C.
Let C = _x_ °; then _A_ = (2 _x_ \+ 10)°. Since _A_ \+ _B_ \+ _C_ = 180°, _B_ = 180° – ( _A_ \+ _C_ ) = (170 – 3 _x_ )°.
( _l_ ) The time it takes a boat traveling at a speed of 20 mi/hr to cover a distance of _x_ miles.
Distance = speed × time. Then
( _m_ ) The perimeter and area of a rectangle if one side is 4 ft longer than twice the other side.
Let _x_ ft = length of shorter side; then (2 _x_ \+ 4)ft = length of longer side. The perimeter = 2( _x_ ) + 2(2 _x_ \+ 4) = (6 _x_ \+ 8)ft, and the area = _x_ (2 _x_ \+ 4) ft2.
( _n_ ) The fraction whose numerator is 3 less than 4 times its denominator.
Let _x_ = denominator; then numerator = 4 _x_ – 3. The fraction is (4 _x_ – 3)/ _x_.
( _o_ ) The number of quarts of alcohol contained in a tank holding _x_ gallons of a mixture which is 40% alcohol by volume.
In _x_ gallons of mixture are 0.40 _x_ gallons of alcohol or 4(0.40 _x_ ) = 1.6 _x_ quarts of alcohol.
**13.4** The sum of two numbers is 21, and one number is twice the other. Find the numbers.
**SOLUTION**
Let _x_ and 2 _x_ be the two numbers. Then _x_ \+ 2 _x_ = 21 or _x_ = 7, and the required numbers are _x_ = 7 and 2 _x_ = 14.
_Check_. 7 + 14 = 21 and 14 = 2(7), as required.
**13.5** Ten less than four times a certain number is 14. Determine the number.
**SOLUTION**
Let _x_ = required number. Then 4 _x_ – 10 = 14, 4 _x_ = 24, and _x_ = 6.
_Check_. Ten less than four times 6 is 4(6) –10 = 14, as required.
**13.6** The sum of three consecutive integers is 24. Find the integers.
**SOLUTION**
Let the three consecutive integers be _x_ , _x_ \+ 1, _x_ \+ 2. Then _x_ \+ ( _x_ \+ 1) + ( _x_ \+ 2) = 24 or _x_ = 7, and the required integers are 7, 8, 9.
**13.7** The sum of two numbers is 37. If the larger is divided by the smaller, the quotient is 3 and the remainder is 5. Find the numbers.
**SOLUTION**
Let _x_ = smaller number, 37 – _x_ = larger number.
Then
Solving, 37 – _x_ = 3 _x_ \+ 5, 4 _x_ = 32, _x_ = 8. The required numbers are 8, 29.
**13.8** A man is 41 years old and his son is 9. In how many years will the father be three times as old as the son?
**SOLUTION**
Let _x_ = required number of years.
Father's age in _x_ years = 3(son's age in _x_ years)
41 + _x_ = 3(9 + _x_ ) and _x_ = 7 years.
**13.9** Ten years ago Jane was four times as old as Bianca. Now she is only twice as old as Bianca. Find their present ages.
**SOLUTION**
Let _x_ = Bianca's present age; then 2 _x_ = Jane's present age.
Jane's age ten years ago = 4(Bianca's age ten years ago)
2 _x_ – 10 = 4( _x_ – 10) and _x_ = 15 years.
Hence Bianca's present age is _x_ = 15 years and Jane's present age is 2 _x_ = 30 years.
_Check_. Ten years ago Bianca was 5 and Jane 20, i.e., Jane was four times as old as Bianca.
**13.10** Robert has 50 coins, all in nickels and dimes, amounting to $3.50. How many nickels does he have?
**SOLUTION**
Let _x_ = number of nickels; then 50 – _x_ = number of dimes.
Amount in nickels + amount in dimes = 350¢
5 _x_ ¢ + 10(50 – _x_ )¢ = 350¢ from which _x_ = 30 nickels.
**13.11** In a purse are nickels, dimes, and quarters amounting to $1.85. There are twice as many dimes as quarters, and the number of nickels is two less than twice the number of dimes. Determine the number of coins of each kind.
**SOLUTION**
Let _x_ = number of quarters; then 2 _x_ = no. of dimes, and 2(2 _x_ ) – 2 = 4 _x_ – 2 = no. of nickels.
Amount in quarters + amount in dimes + amount in nickels = 185¢
25( _x_ )¢ + 10(2 _x_ )¢ + 5(4 _x_ – 2)¢ = 185¢ from which _x_ = 3.
Hence there are _x_ = 3 quarters, 2 _x_ = 6 dimes, and 4 _x_ – 2 = 10 nickels.
_Check_. 3 quarters = 75¢, 6 dimes = 60¢. 10 nickels = 50¢, and their sum = $1.85.
**13.12** The tens digit of a certain two-digit number exceeds the units digit by 4 and is 1 less than twice the units digit. Find the two-digit number.
**SOLUTION**
Let _x_ = units digit; then _x_ \+ 4 = tens digit.
Since the tens digit = 2(units digit) – 1, we have _x_ \+ 4 = 2( _x_ ) – 1 or _x_ = 5.
Thus _x_ = 5, _x_ \+ 4 = 9, and the required number is 95.
**13.13** The sum of the digits of a two-digit number is 12. If the digits are reversed, the new number is 4/7 times the original number. Determine the original number.
**SOLUTION**
Let _x_ = units digit; 12 – _x_ = tens digit.
Original number = 10(12 – _x_ ) + _x_ ; reversing digits, the new number = 10 _x_ \+ (12 – _x_ ). Then
Solving, _x_ = 4, 12 – _x_ = 8, and the original number is 84.
**13.14** A man has $4000 invested, part at 5% and the remainder at 3% simple interest. The total income per year from these investments is $168. How much does he have invested at each rate?
**SOLUTION**
Let _x_ = amount invested at 5%; $4000 – _x_ = amount at 3%.
Solving, _x_ = $2400 at 5%, $4000 – _x_ = $1600 at 3%.
**13.15** What amount should an employee receive as bonus so that she would net $500 after deducting 30% for taxes?
**SOLUTION**
Let _x_ = required amount.
**13.16** At what price should a merchant mark a sofa that costs $120 in order that it may be offered at a discount of 20% on the marked price and still make a profit of 25% on the selling price?
**SOLUTION**
Let _x_ = marked price; then sale price = _x_ – 0.20 _x_ = 0.80 _x_.
Since profit = 25% of sale price, cost = 75% of sale price. Then
**13.17** When each side of a given square is increased by 4 feet the area is increased by 64 square feet. Determine the dimensions of the original square.
**SOLUTION**
Let _x_ = side of given square; _x_ \+ 4 = side of new square.
**13.18** One leg of a right triangle is 20 inches and the hypotenuse is 10 inches longer than the other leg. Find the lengths of the unknown sides.
**SOLUTION**
Let _x_ = length of unknown leg; _x_ \+ 10 = length of hypotenuse.
The required sides are _x_ = 15 in. and _x_ \+ 10 = 25 in.
**13.19** Temperature Fahrenheit = (temperature Celsius) + 32. At what temperature have the Fahrenheit and Celsius readings the same value?
**SOLUTION**
Let _x_ = required temperature = temperature Fahrenheit = temperature Celsius.
Then _x_ = _x_ \+ 32 or _x_ = –40°. Thus –40°F = –40°C.
**13.20** A mixture of 40 lb of candy worth 60¢ a pound is to be made up by taking some worth 45¢/lb and some worth 85¢/lb. How many pounds of each should be taken?
**SOLUTION**
Let _x_ = weight of 45¢ candy; 40 – _x_ = weight of 85¢ candy.
Value of 45¢/lb candy + value of 85¢/lb candy = value of mixture
Solving, _x_ = 25 lb of 45¢/lb candy; 40 – _x_ = 15 lb of 85¢/lb candy.
**13.21** A tank contains 20 gallons of a mixture of alcohol and water which is 40% alcohol by volume. How much of the mixture should be removed and replaced by an equal volume of water so that the resulting solution will be 25% alcohol by volume?
**SOLUTION**
Let _x_ = volume of 40% solution to be removed.
Volume of alcohol in final solution = volume of alcohol in 20 gal of 25% solution
or 0.40(20 – _x_ ) = 0.25(20) from which _x_ = 7.5 gallons.
**13.22** What weight of water must be evaporated from 40 lb of a 20% salt solution to produce a 50% solution? All percentages are by weight.
**SOLUTION**
Let _x_ = weight of water to be evaporated.
**13.23** How many quarts of a 60% alcohol solution must be added to 40 quarts of a 20% alcohol solution to obtain a mixture which is 30% alcohol? All percentages are by volume.
**SOLUTION**
Let _x_ = number of quarts of 60% alcohol to be added.
**13.24** Two unblended manganese (Mn) ores contain 40% and 25% of manganese respectively. How many tons of each must be mixed to give 100 tons of blended ore containing 35% of manganese? All percentages are by weight.
**SOLUTION**
Let _x_ = weight of 40% ore required; 100 – _x_ = weight of 25% ore required.
from which _x_ = 66 tons of 40% ore and 100 _– _x_ =_ 33 tons of 25% ore.
**13.25** Two cars _A_ and _B_ having average speeds of 30 and 40 mi/hr respectively are 280 miles apart. They start moving toward each other at 3:00 P.M. At what time and where will they meet?
**SOLUTION**
Let _t_ = time in hours each car travels before they meet. Distance = speed × time.
They meet at 7:00 P.M. at a distance 30t = 120 mi from initial position of A or at a distance 40 _t_ = 160 mi from initial position of _B_.
**13.26** _A_ and _B_ start from a given point and travel on a straight road at average speeds of 30 and 50 mi/hr respectively. If _B_ starts 3 hr after _A_ , find ( _a_ ) the time and ( _b_ ) the distance they travel before meeting.
**SOLUTION**
Let _t_ and ( _t_ – 3) be the number of hours _A_ and _B_ respectively travel before meeting.
( _a_ ) Distance in miles = average speed in mi/hr × time in hours. When they meet,
Hence _A_ travels _t_ = 7 hr and _B_ travels ( _t_ – 3) = 4 hr.
( _b_ ) Distance = 30 _t_ = 30(7 ) = 225 mi, or distance = 50( _t_ – 3) = 50(4 ) = 225 mi.
**13.27** _A_ and _B_ can run around a circular mile track in 6 and 10 minutes respectively. If they start at the same instant from the same place, in how many minutes will they pass each other if they run around the track ( _a_ ) in the same direction, ( _b_ ) in opposite directions?
**SOLUTION**
Let _t_ = required time in minutes.
( _a_ ) They will pass each other when _A_ covers 1 mile more than _B_. The speeds _A_ and _B_ are 1/6 and 1/10 mi/min respectively. Then, since distance = speed × time:
Distance by _A_ – distance by _B_ = 1 mile
( _b_ ) Distance by _A_ \+ distance by _B_ = 1 mile
**13.28** A boat, propelled to move at 25 mi/hr in still water, travels 4.2 mi against the river current in the same time that it can travel 5.8 mi with the current. Find the speed of the current.
**SOLUTION**
Let _v_ = speed of current. Then, since time = distance/speed,
time against the current = time in direction of the current
or
**13.29** _A_ can do a job in 3 days, and _B_ can do the same job in 6 days. How long will it take them if they work together?
**SOLUTION**
Let _n_ = number of days it will take them working together.
In 1 day A does 1/3 of the job and B does 1/6 of the job, thus together completing 1/ _n_ of the job (in 1 day). Then
from which _n_ = 2 days.
_Another method_. In _n_ days _A_ and _B_ together complete
complete job. Solving, _n_ = 2 days.
**13.30** A tank can be filled by three pipes separately in 20, 30, and 60 minutes respectively. In how many minutes can it be filled by the three pipes acting together?
**SOLUTION**
Let _t_ = time required, in minutes.
In 1 minute three pipes together fill of the tank. Then in _t_ minutes they together fill
complete tank. Solving, _t_ = 10 minutes.
**13.31** _A_ and _B_ working together can complete a job in 6 days. _A_ works twice as fast as _B_. How many days would it take each of them, working alone, to complete the job?
**SOLUTION**
Let _n_ , 2 _n_ = number of days required by _A_ and _B_ respectively, working alone, to do the job.
In 1 day _A_ can do l/ _n_ of the job and _B_ can do l/2 _n_ of the job. Then in 6 days they can do
complete job. Solving, _n_ = 9 days, 2 _n_ = 18 days.
**13.32** _A_ 's rate of doing work is three times that of _B_. On a given day _A_ and _B_ work together for 4 hours; then _B_ is called away and _A_ finishes the rest of the job in 2 hours. How long would it take _B_ to do the complete job alone?
**SOLUTION**
Let _t_ , 3 _t_ = time in hours required by _A_ and _B_ respectively, working alone, to do the job.
In 1 hour A does 1/ _t_ of job and B does l/3 _t_ of job. Then
complete job. Solving, 3 _t_ = 22 hours.
**13.33** A man is paid $18 for each day he works and forfeits $3 for each day he is idle. If at the end of 40 days he nets $531, how many days was he idle?
**SOLUTION**
Let _x_ = number of days idle; 40 – _x_ = number of days worked.
Amount earned – amount forfeited = $531
or $18(40 – _x_ ) – 3 _x_ = $531 and _x_ = 9 days idle.
### **Supplementary Problems**
**13.34** Solve each of the following equations.
( _a_ ) 3 _x_ – 2 = 7
( _b_ ) _y +_ 3( _y_ – 4) = 4
( _c_ ) 4 _x_ – 3 = 5 – 2 _x_
( _d_ ) _x_ – 3 – 2(6 – 2 _x_ ) = 2(2 _x_ – 5)
( _e_ )
( _f_ )
( _g_ )( _x_ – 3)2 \+ ( _x_ \+ 1)2 = ( _x_ – 2)2 \+ ( _x_ \+ 3)2
( _h_ ) (2 _x_ \+ 1)2 = ( _x_ – 1)2 \+ 3 _x(x_ \+ 2)
( _i_ )
( _j_ )
( _k_ )
( _l_ )
**13.35** Solve for the indicated letter.
( _a_ ) 2( _x_ – _p_ ) = 3(6 _p_ – _x_ ): _x_
( _b_ ) 2 _by_ – 2 _a_ = _ay_ – 4 _b_ : _y_
( _c_ )
( _d_ )
( _e_ )
**13.36** Express each of the following statements in terms of algebraic symbols.
( _a_ ) Two more than five times a certain number.
( _b_ ) Six less than twice a certain number.
( _c_ ) Each of two numbers whose difference is 25.
( _d_ ) The squares of three consecutive integers.
( _e_ ) The amount by which five times a certain number exceeds 40.
( _f_ ) The square of any odd integer.
( _g_ ) The excess of the square of a number over twice the number.
( _h_ ) The number of pints in _x_ gallons.
( _i_ ) The difference between the squares of two consecutive even integers.
( _j_ ) Bob is six years older than Jane who is half as old as Jack. Express each of their ages in terms of a single unknown.
( _k_ ) The three angles _A_ , _B, C_ of a triangle _ABC_ if angle _A_ exceeds twice angle _B_ by 20°.
( _l_ ) The perimeter and area of a rectangle if one side is 3 ft shorter than three times the other side.
( _m_ ) The fraction whose denominator is 4 more than twice the square of the numerator.
( _n_ ) The amount of salt in a tank holding _x_ quarts of water if the concentration is 2 lb of salt per gallon.
**13.37**
( _a_ ) One half of a certain number is 10 more than one sixth of the number. Find the number.
( _b_ ) The difference between two numbers is 20 and their sum is 48. Find the numbers.
( _c_ ) Find two consecutive even integers such that twice the smaller exceeds the larger by 18.
( _d_ ) The sum of two numbers is 36. If the larger is divided by the smaller, the quotient is 2 and the remainder is 3. Find the numbers.
( _e_ ) Find two consecutive positive odd integers such that the difference of their squares is 64.
( _f_ ) The first of three numbers exceeds twice the second number by 4, while the third number is twice the first. If the sum of the three numbers is 54, find the numbers.
**13.38**
( _a_ ) A father is 24 years older than his son. In 8 years he will be twice as old as his son. Determine their present ages.
( _b_ ) Mary is fifteen years older than her sister Jane. Six years ago Mary was six times as old as Jane. Find their present ages.
( _c_ ) Larry is now twice as old as Bill. Five years ago Larry was three times as old as Bill. Find their present ages.
**13.39**
( _a_ ) In a purse is $3.05 in nickels and dimes, 19 more nickels than dimes. How many coins are there of each kind?
( _b_ ) Richard has twice as many dimes as quarters, amounting to $6.75 in all. How many coins does he have?
( _c_ ) Admission tickets to a theater were 60¢ for adults and 25¢ for children. Receipts for the day showed that 280 persons attended and $140 was collected. How many children attended that day?
**13.40**
( _a_ ) The tens digit of a certain two-digit number exceeds the units digit by 3. The sum of the digits is 1/7 of the number. Find the number.
( _b_ ) The sum of the digits of a certain two-digit number is 10. If the digits are reversed, a new number is formed which is one less than twice the original number. Find the original number.
( _c_ ) The tens digit of a certain two-digit number is 1/3 of the units digit. When the digits are reversed, the new number exceeds twice the original number by 2 more than the sum of the digits. Find the original number.
**13.41**
( _a_ ) Goods cost a merchant $72. At what price should he mark them so that he may sell them at a discount of 10% from his marked price and still make a profit of 20% on the selling price?
( _b_ ) A woman is paid $20 for each day she works and forfeits $5 for each day she is idle. At the end of 25 days she nets $450. How many days did she work?
( _c_ ) A labor report states that in a certain factory a total of 400 men and women are employed. The average daily wage is $16 for a man and $12 for a woman. If the labor cost is $5720 per day, how many women are employed?
( _d_ ) A woman has $450 invested, part at 2% and the remainder at 3% simple interest. How much is invested at each rate if the total annual income from these investments is $11?
( _e_ ) A man has $2000 invested at 7% and $5000 at 4% simple interest. What additional sum must he invest at 6% to give him an overall return of 5%?
**13.42**
( _a_ ) The perimeter of a rectangle is 110 ft. Find the dimensions if the length is 5 ft less than twice the width.
( _b_ ) The length of a rectangular floor is 8 ft greater than its width. If each dimension is increased by 2 ft, the area is increased by 60 ft2. Find the dimensions of the floor.
( _c_ ) The area of a square exceeds the area of a rectangle by 3 in2. The width of the rectangle is 3 in. shorter and the length 4 in. longer than the side of the square. Find the side of the square.
( _d_ ) A piece of wire 40 in. long is bent into the form of a right triangle, one of whose legs is 15 in. long. Determine the lengths of the other two sides.
( _e_ ) The length of a rectangular swimming pool is twice its width. The pool is surrounded by a cement walk 4 ft wide. If the area of the walk is 784 ft2, determine the dimensions of the pool.
**13.43**
( _a_ ) Lubricating oil worth 28 cents/quart is to be mixed with oil worth 33 cents/quart to make up 45 quarts of a mixture to sell at 30¢/qt. What volume of each grade should be taken?
( _b_ ) What weight of water must be added to 50 lb of a 36% sulfuric acid solution to yield a 20% solution? All percentages are by weight.
( _c_ ) How many quarts of pure alcohol must be added to 10 quarts of a 15% alcohol solution to obtain a mixture which is 25% alcohol? All percentages are by volume.
( _d_ ) There is available 60 gallons of a 50% solution of glycerin and water. What volume of water must be added to the solution to reduce the glycerin concentration to 12%? All percentages are by volume.
( _e_ ) The radiator of a jeep has a capacity of 4 gallons. It is filled with an anti-freeze solution of water and glycol which analyzes 10% glycol. What volume of the mixture must be drawn off and replaced with glycol to obtain a 25% glycol solution? All percentages are by volume.
( _f_ ) One thousand quarts of milk testing 4% butterfat are to be reduced to 3%. How many quarts of cream testing 23% butterfat must be separated from the milk to produce the required result? All percentages are by volume.
( _g_ ) There are available 10 tons of coal containing 2.5% sulfur, and also supplies of coal containing 0.80% and 1.10% sulfur respectively. How many tons of each of the latter should be mixed with the original 10 tons to give 20 tons containing 1.7% sulfur?
**13.44**
( _a_ ) Two motorists start toward each other at 4:30 P.M. from towns 255 miles apart. If their respective average speeds are 40 and 45 mi/hr, at what time will they meet?
( _b_ ) Two planes start from Chicago at the same time and fly in opposite directions, one averaging a speed of 40 mi/hr greater than the other. If they are 2000 miles apart after 5 hours, find their average speeds.
( _c_ ) At what rate must motorist _A_ travel to overtake motorist _B_ who is traveling at a rate 20 mi/hr slower if _A_ starts two hours after _B_ and wishes to overtake _B_ in 4 hours?
( _d_ ) A motorist starts from city _A_ at 2:00 P.M. and travels to city _B_ at an average speed of 30 mi/hr. After resting at _B_ for one hour, she returns over the same route at an average speed of 40 mi/hr and arrives at _A_ that evening at 6:30 P.M. Determine the distance between _A_ and _B_.
( _e_ ) Tom traveled a distance of 265 miles. He drove at 40 mi/hr during the first part of the trip and at 35 mi/hr during the remaining part. If he made the trip in 7 hours, how long did he travel at 40 mi/hr?
( _f_ ) A boat can move at 8 mi/hr in still water. If it can travel 20 miles downstream in the same time it can travel 12 miles upstream, determine the rate of the stream.
( _g_ ) The speed of a plane is 120 mi/hr in a calm. With the wind it can cover a certain distance in 4 hours, but against the wind it can cover only 3/5 of that distance in the same time. Find the velocity of the wind.
**13.45**
( _a_ ) A farmer can plow a certain field three times as fast as his son. Working together, it would take them 6 hours to plow the field. How long would it take each to do it alone?
( _b_ ) A painter can do a given job in 6 hours. Her helper can do the same job in 10 hours. The painter begins the work and after 2 hours is joined by the helper. In how many hours will they complete the job?
( _c_ ) One group of workers can do a job in 8 days. After this group has worked 3 days, another group joins it and together they complete the job in 3 more days. In what time could the second group have done the job alone?
( _d_ ) A tank can be filled by two pipes separately in 10 and 15 minutes respectively. When a third pipe is used simultaneously with the first two pipes, the tank can be filled in 4 minutes. How long would it take the third pipe alone to fill the tank?
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**13.34**
( _a_ ) _x_ = 3
( _b_ ) _y_ = 4
( _c_ ) _x_ = 4/3
( _d_ ) _x_ = 5
( _e_ ) _t_ = –6
( _f_ ) _x_ = 1/11
( _g_ ) _x_ = –1/2
( _h_ ) all values of _x_ ( _identity_ )
( _i_ ) _z_ = 22
( _j_ ) _x_ = 1/4
( _k_ ) y = 5
( _l_ ) _x_ = –1
**13.35**
( _a_ ) _x_ = 4 _p_
( _b_ ) y = –2 if _a_ ≠ 2 _b_
( _c_ )
( _d_ )
( _e_ )
**13.36**
( _a_ ) 5 _x_ \+ 2
( _b_ ) 2 _x_ – 6
( _c_ ) _x_ \+ 25, _x_
( _d_ ) _x_ 2,( _x_ \+ 1)2,( _x_ \+ 2)2
( _e_ ) 5 _x_ – 40
( _f_ )(2 _x_ \+ 1)2 where _x_ = integer
( _g_ ) _x_ 2 – 2 _x_
( _h_ ) 8 _x_
( _i_ ) (2 _x_ \+ 2)2 – (2 _x_ )2, _x_ = integer
( _j_ ) Jane's age _x_ , Bob's age _x_ \+ 6, Jack's age 2 _x_
( _k_ ) _B_ = _x_ °, _A_ = (2 _x_ \+ 20)°, _C_ = (160 – 3 _x_ )°
( _l_ ) One side is x, adjacent side is 3 _x_ – 3.
Perimeter = 8 _x_ – 6, area = 3 _x_ 2 – 3 _x_
( _m_ )
( _n_ )
**13.37**
( _a_ )30
( _b_ ) 34, 14
( _c_ ) 20, 22
( _d_ ) 25, 11
( _e_ ) 15, 17
( _f_ ) 16, 6, 32
**13.38**
( _a_ ) Father 40, son 16
( _b_ ) Mary 24, Jane 9
( _c_ ) Larry 20, Bill 10
**13.39**
( _a_ ) 14 dimes, 33 nickels
( _b_ ) 15 quarters, 30 dimes
( _c_ ) 200 adults, 80 children
**13.40**
( _a_ ) 63
( _b_ ) 37
( _c_ ) 26
**13.41**
( _a_ ) $100
( _b_ ) 23 days
( _c_ ) 170 women
( _d_ ) $200 at 3%, $250 at 2%
( _e_ ) $1000
**13.42**
( _a_ ) width 20 ft, length 35 ft
( _b_ ) width 10 ft, length 18 ft
( _c_ ) 9 in.
( _d_ ) other leg 8 ft, hypotenuse 17 ft
( _e_ ) 30 ft by 60 ft
**13.43**
( _a_ ) 18 qt of 33¢, 27 qt of 28¢
( _b_ ) 40lb
( _c_ ) 4/3qt
( _d_ ) 190 gal
( _e_ ) 2/3 gal
( _f_ ) 50 qt
( _g_ ) 6.7 tons of 0.80%, 3.3 tons of 1.10%
**13.44**
( _a_ ) 7:30 P.M.
( _b_ ) 180, 220 mi/hr
( _c_ ) 60 mi/hr
( _d_ ) 60mi
( _e_ ) 4hr
( _f_ ) 2mi/hr
( _g_ ) 30 mi/hr
**13.45**
( _a_ ) Father 8 hr, son 24 hr
( _b_ ) 2 hr
( _c_ ) 12 days
( _d_ ) 12 minutes
## CHAPTER 14
Equations of Lines
### **14.1 SLOPE OF A LINE**
The equation _ax_ \+ _by_ = _c_ , where not both _a_ and _b_ are 0 and _a_ , _b_ , and _c_ are real numbers is the standard (or general) form of the equation of a line.
The slope of a line is the ratio of the change in _y_ compared to the change in _x_.
If ( _x_ 1, _y_ 1) and ( _x_ 2, _y_ 2) are two points on a line and _m_ is the slope of the line, then
**EXAMPLE 14.1**. What is the slope of the line through the points (5, –8) and (6, 2)?
The slope of the line through the two points (5, –28) and (6, 2) is –10.
**EXAMPLE 14.2.** What is the slope of the line 3 _x_ – 4 _y_ = 12?
First we need to find two points that satisfy the equation of the line 3 _x_ – 4 _y_ = 12. If _x_ = 0, then 3(0) – 4 _y_ = 12 and _y_ = –3. Thus, one point is (0, –3). If _x_ = –4, then 3(– 4) – 4 _y_ = 12 and _y_ = –6. So, (– 4, –6) is another point on the line.
The slope of the line 3 _x_ – 4 _y_ = 12 is 3/4.
In Example 14.1, the slope of the line is negative. This means that, as we view the graph of the line from left to right, as _x_ increases _y_ decreases (see Fig. 14-1). In Example 14.2, the slope is positive, which means that as _x_ increases so does _y_ (see Fig. 14-2).
**Fig. 14-1**
**Fig. 14-2**
A horizontal line _y_ = _k_ , where _k_ is a constant, has zero slope. Since all of the _y_ values are the same, _y_ 2 – _y_ 1 = 0.
A vertical line _x_ = _k_ , where _k_ is a constant, does not have a slope, that is, the slope is not defined. Since all of the _x_ values are the same, _x_ 2 – _x_ 1 = 0 and division by zero is not defined.
### **14.2 PARALLEL AND PERPENDICULAR LINES**
Two non-vertical lines are parallel if and only if the slopes of the lines are equal.
**EXAMPLE 14.3.** Show that the figure _PQRS_ with vertices _P_ (0, –2), _Q_ (–2, 3), _R_ (3, 5), and _S_ (5, 0) is a parallelogram. Quadrilateral _PQRS_ is a parallelogram if and are parallel and and are parallel.
Since and have the same slope, they are parallel, and since and have the same slope, they are parallel. Thus, the opposite sides of _PQRS_ are parallel, so _PQRS_ is a parallelogram.
**EXAMPLE 14.4.** Show that the points _A_ (0, 4), _B_ (2, 3), and _C_ (4, 2) are collinear, that is, lie on the same line.
The points _A, B_ , and _C_ are collinear if the slopes of the lines through any two pairings of the points are the same.
The lines _AB_ and _BC_ have the same slope and share a common point, _B_ , so the lines are the same line. Thus, the points _A, B_ , and _C_ are collinear.
Two non-vertical lines are perpendicular if and only if the product of their slopes is –1. The slope of each line is said to be the negative reciprocal of the slope of the other line.
**EXAMPLE 14.5.** Show that the line through the points _A_ (3, 3) and _B_ (6, – 3) is perpendicular to the line through the points _C_ (4, 2) and _D_ (8, 4).
Since (–2)(1/2) = – 1, the lines _AB_ and _CD_ are perpendicular.
### **14.3 SLOPE-INTERCEPT FORM OF EQUATION OF A LINE**
If a line has slope _m_ and _y_ intercept (0, _b_ ), then for any point ( _x, y_ ), where _x_ ≠ 0, on the line we have
The slope-intercept form of the equation of a line with slope _m_ and _y_ intercept _b_ is _y_ = _mx_ \+ _b_.
**EXAMPLE 14.6.** Find the slope and _y_ intercept of the line 3 _x_ \+ 2 _y_ = 12.
We solve the equation 3 _x_ \+ 2 _y_ = 12 for _y_ to get _y_ = – _x_ \+ 6. The slope of the line is – and the _y_ intercept is 6.
**EXAMPLE 14.7.** Find the equation of the line with slope –4 and y intercept 6.
The slope of the line is –4, so _m_ = –4 and the _y_ intercept is 6, so _b_ = 6. Substituting into _y_ = _mx_ \+ _b_ , we get _y_ = – 4 _x_ \+ 6 for the equation of the line.
### **14.4 SLOPE-POINT FORM OF EQUATION OF A LINE**
If a line has slope _m_ and goes through a point ( _x_ 1, _y_ 1), then for any other point ( _x, y_ ) on the line, we have _m_ = ( _y_ – _y_ 1)/( _x_ – _x_ 1) and _y_ – _y_ 1 = _m_ ( _x_ – _x_ 1).
The slope-point form of the equation of a line is _y_ – _y_ 1 = _m_ ( _x_ – _x_ 1).
**EXAMPLE 14.8.** Write the equation of the line passing through the point (1, –2) and having slope —2/3.
Since ( _x_ 1, _y_ 1) = (1, –2) and _m_ = – 2/3, we substitute into _y_ – _y_ 1 = _m_ ( _x_ – _x_ 1) to get _y_ \+ 2 = – 2/3( _x_ – 1). Simplifying we get 3( _y_ \+ 2) = – 2( _x_ – 1), and finally 2 _x_ \+ 3 _y_ = –4.
The equation of the line through (1, –2) with slope – 2/3 is 2 _x_ \+ 3 _y_ = – 4.
### **14.5 TWO-POINT FORM OF EQUATION OF A LINE**
If a line goes through the points ( _x_ 1, _y_ 1) and ( _x_ 2, _y_ 2), it has slope _m_ = ( _y_ 2 – _y_ 1)/( _x_ 2 – _x_ 1) if _x_ 2 ≠ _x_ 1. Substituting into the equation _y_ – _y_ 1 = _m_ ( _x_ – _x_ 1), we get
The two-point form of the equation of a line is
If _x_ 2 = _x_ 1, we get the vertical line _x_ = _x_ 1. If _y_ 2 = _y_ 1, we get the horizontal line _y_ = _y_ 1.
**EXAMPLE 14.9.** Write the equation of the line passing through (3, 6) and (– 4, 4).
Let ( _x_ 1, _y_ 1) = (3, 6) and ( _x_ 2, _y_ 2) = (– 4, 4) and substitute into
The equation of the line through the points (3, 6) and (– 4, 4) is 2 _x_ – 7 _y_ = – 36.
### **14.6 INTERCEPT FORM OF EQUATION OF A LINE**
If a line has _x_ intercept _a_ and _y_ intercept _b_ , it goes through the points ( _a_ , 0) and (0, _b_ ). The equation of the line is
which simplifies to _bx_ \+ _ay_ = _ab_. If both _a_ and _b_ are non-zero we get _x/a_ \+ _y/b_ = 1.
If a line has _x_ intercept _a_ and _y_ intercept _b_ and both _a_ and _b_ are non-zero, the equation of the line is
**EXAMPLE 14.10.** Find the intercepts of the line 4 _x_ – 3 _y_ = 12.
We divide the equation 4 _x_ – 3 _y_ = 12 by 12 to get
The _x_ intercept is 3 and the _y_ intercept is – 4 for the line 4 _x_ – 3 _y_ = 12.
**EXAMPLE 14.11.** Write the equation of the line that has an _x_ intercept of 2 and a _y_ intercept of 5.
We have _a_ = 2 and _b_ = 5 for the equation
Substituting we get
Simplifying we get 5 _x_ \+ 2 _y_ = 10.
The line with _x_ intercept 2 and _y_ intercept 5 is 5 _x_ \+ 2 _y_ = 10.
### **Solved Problems**
**14.1** What is the slope of the line through each pair of points?
( _a_ ) (4, 1) and (7, 6)
( _b_ ) (3, 9) and (7, 4)
( _c_ ) (–4, 1) and (–4, 3)
( _d_ ) (–3, 2) and (2, 2)
**SOLUTION**
**14.2** Determine whether the line containing the points _A_ and _B_ is parallel, perpendicular, or neither to the line containing the points _C_ and _D_.
( _a_ ) _A_ (2, 4), _B_ (3, 8), _C_ (5, 1), and _D_ (4, –3)
( _b_ ) _A_ (2, –3), _B_ (–4, 5), _C_ (0, –1), and _D_ (–4, –4)
( _c_ ) _A_ (1, 9), _B_ (4, 0), _C_ (0, 6), and _D_ (5, 3)
( _d_ ) _A_ (8, – 1), _B_ (2, 3), _C_ (5, 1), and _D_ (2, – 1)
**SOLUTION**
( _a_ ) slope slope
Since the slopes are equal, the lines AB and CD are parallel.
( _b_ ) slope slope
Since (–4/3)(3/4) = – 1, the lines _AB_ and _CD_ are perpendicular.
( _c_ ) slope slope
Since the slopes are not equal and do not have a product of – 1, the lines _AB_ and _CD_ are neither parallel nor perpendicular.
( _d_ ) slope slope
Since the slopes are not equal and do not have a product of – 1, the lines _AB_ and _CD_ are neither parallel nor perpendicular.
**14.3** Determine whether the given three points are collinear or not.
( _a_ ) (0, 3), (1, 1), and (2, – 1)
( _b_ ) (1, 5), (–2, – 1), and (– 3, –4)
**SOLUTION**
( _a_ ) and
Since the line between (0, 3) and (1, 1) and the line between (1, 1) and (2, – 1) have the same slope, the points (0, 3), (1, 1), and (2, – 1) are collinear.
( _b_ ) and
Since the slope of the line between (1, 5) and (– 2, – 1) and the slope of the line between (– 2, –1) and (–3, –4) are different, the points (1, 5), (– 2, – 1), and (– 3, –4) are not collinear.
**14.4** Write the equation of the line with slope _m_ and _y_ intercept _b_.
( _a_ ) _m_ = –2/3, _b_ = 6
( _b_ ) _m_ = – 3, _b_ = – 4
( _c_ ) _m_ = 0, _b_ = 8
( _d_ ) _m_ = 3, _b_ = 0
**SOLUTION**
**14.5** Write the equation of the line that contains point _P_ and has slope _m_.
( _a_ ) _P_ (2, 5), _m_ = 4
( _b_ ) _P_ (1, 4), _m_ = 0
( _c_ ) _P_ (–1, –6), _m_ = 1/4
( _d_ ) _P_ (2, –3), _m_ = –3/7
**SOLUTION**
We use the formula _y_ – _y_ 1 = _m_ ( _x_ – _x_ 1).
**14.6** Write the equation of the line passing through points _P_ and _Q_.
( _a_ ) _P_ (1, – 4), _Q_ (2, 3)
( _b_ ) _P_ (6, – 1), _Q_ (0, 2)
( _c_ ) _P_ (–1, 4), _Q_ (3, 4)
( _d_ ) _P_ (1, 5), _Q_ (–2, 3)
( _e_ ) _P_ (7, 1), _Q_ (8, 3)
( _f_ ) _P_ (4, 21), _Q_ (4, 3)
**SOLUTION**
( _f_ ) Since _P_ and _Q_ have the same _x_ value, slope is not defined. However, the line through _P_ and _Q_ has to have 4 for its _x_ coordinate in all points. Thus, the line is _x_ = 4.
**14.7** Write the equation of the line that has _x_ intercept – 3 and _y_ intercept 4.
**SOLUTION**
**14.8** Write the equation of the line through (–5, 6) that is parallel to the line 3 _x_ – 4 _y_ = 5.
**SOLUTION**
We write the equation 3 _x_ – 4 _y_ = 5 in slope-intercept form to identify its slope, _y_ = 3/4 _x_ – 5/4. Since the form is _y_ = _mx_ \+ _b, m_ = 3/4. Parallel lines have the same slope, so the line we want has a slope of 3/4.
Now that we have the slope and a point the line goes through, so we can write the equation using the point-slope form: _y_ – _y_ 1 = _m_ ( _x_ – _x_ 1). Substituting we get _y_ – 6 = 3/4( _x_ \+ 5). Simplifying we get 4 _y_ – 24 = 3 _x_ \+ 15 and finally 3 _x_ – 4 _y_ = – 39. The equation we want is 3 _x_ – 4 _y_ = – 39.
**14.9** Write the equation of the line through (4, 6) that is perpendicular to the line 2 _x_ – _y_ = 8.
**SOLUTION**
In the slope-intercept form the given line is _y_ = 2 _x_ – 8. The slope of the line is 2, so the slope of a perpendicular line is the negative reciprocal of 2, which is –1/2. We want to write the equation of the line with slope of –1/2 and going through (4, 6). Thus, _y_ – 6 = –l/2( _x_ – 4), so 2 _y_ – 12 = – _x_ \+ 4, and finally _x_ \+ 2 _y_ = 16. The line we want is _x_ \+ 2 _y_ = 16.
### **Supplementary Problems**
**14.10** What is the slope of the line through each pair of points?
( _a_ ) (–1, 2), (4, –3)
( _b_ ) (3, 4), (– 4, –3)
( _c_ ) (5, 4), (5, –2)
( _d_ ) (–5, 3), (2, 3)
( _e_ ) (–1, 5), (–2, 3)
( _f_ ) (7, 3), (8, –3)
**14.11** Determine whether the line containing the points _P_ and _Q_ is parallel, perpendicular, or neither to the line containing the points _R_ and _S_.
( _a_ ) _P_ (4, 2), _Q_ (8, 3), _R_ (–2, 8), and _S_ (1, –4)
( _b_ ) _P_ (0, –5), _Q_ (15, 0), _R_ (1, 2), and _S_ (0, 5)
( _c_ ) _P_ (–7, 8), _Q_ (8, –7), _R_ (–8, 10), and _S_ (6, –4)
( _d_ ) _P_ (8, – 2), _Q_ (2, 8), _R_ (– 2, – 8), and _S_ (– 8, – 2)
**14.12** Determine a constant real number k such that the lines _AB_ and _CD_ are (1) parallel and (2) perpendicular.
( _a_ ) _A_ (2, 1), _B_ (6, 3), _C_ (4, _k_ ), and _D_ (3, 1)
( _b_ ) _A_ (1, _k_ ), _B_ (2, 3), _C_ (1, 7), and _D_ (3, 6)
( _c_ ) _A_ (9, 4), _B_ ( _k_ , 10), _C_ (11, –2), and _D_ (–2, 4)
( _d_ ) _A_ (1, 2), _B_ (4, 0), _C_ ( _k_ , 2), and _D_ (1, –3)
**14.13** Determine whether the given three points are collinear or not.
( _a_ ) (–3, 1), (–11, –1), and (–15, –2)
( _b_ ) (1, 1), (4, 2), and (2, 3)
**14.14** Write the equation of the line with slope m and y intercept _b_.
( _a_ ) _m_ = –3, _b_ = 4
( _b_ ) _m_ = 0, _b_ = – 3
( _c_ ) _m_ = 2/3, _b_ = –2
( _d_ ) _m_ = 4, _b_ = 0
( _e_ ) _m_ = –1/2, _b_ = 3
( _f_ ) _m_ = – 5/6, _b_ = 1/6
**14.15** Write the equation of the line that goes through point _P_ and has slope _m_.
( _a_ ) _P_ (– 5, 2), _m_ = –1
( _b_ ) _P_ (–4, –3), _m_ = 4
( _c_ ) _P_ (4, –1), _m_ = 2/3
( _d_ ) _P_ (0, 4), _m_ = –4/3
( _e_ ) _P_ (2, 6), _m_ = –5
( _f_ ) _P_ (–1, 6), _m_ = 0
**14.16** Write the equation of the line through the points _P_ and _Q_.
( _a_ ) _P_ (1, 2), _Q_ (2, 4)
( _b_ ) _P_ (1.6, 3), _Q_ (0.3, 1.4)
( _c_ ) _P_ (0.7, 3), _Q_ (0.7, – 3)
( _d_ ) _P_ (10, 2), _Q_ (5, 2)
( _e_ ) _P_ (3, 6), _Q_ (– 3, 8)
( _f_ ) _P_ (– 4, 2), _Q_ (2, 4)
( _g_ ) _P_ (– 1, 3), _Q_ (0, 6)
( _h_ ) _P_ (0, 0), _Q_ (– 3, 6)
**14.17** Write the equation of the line that has _x_ intercept _a_ and _y_ intercept _b_.
( _a_ ) _a_ = –2, _b_ = –2
( _b_ ) _a_ = 6, _b_ = –3
( _c_ ) _a_ = –1/2, _b_ = 4
( _d_ ) _a_ = 6, _b_ = 1/3
**14.18** Write the equation of the line through point _P_ and parallel to the line _l_.
( _a_ ) _P_ (2, – 4), line _l_ : _y_ = 4 _x_ – 6
( _b_ ) _P_ (1, 0), line _l_ : _y_ = 3 _x_ \+ 1
( _c_ ) _P_ (– 1, – 1), line _l_ : 4 _x_ \+ 5 _y_ = 5
( _d_ ) _P_ (3, 5), line _l_ : 3 _x_ – 2 _y_ = 18
**14.19** Write the equation of the line through the point _P_ and perpendicular to the line _l_.
( _a_ ) _P_ (2, –1), line _l_ : _x_ = 4 _y_
( _c_ ) P(1, 1), line _l_ : 3 _x_ – 2 _y_ = 4
( _b_ ) _P_ (0, 6), line _l_ : 2 _x_ \+ 3 _y_ = 5
( _d_ ) _P_ (1, –2), line _l_ : 4 _x_ \+ _y_ = 7
**14.20** Determine if the triangle with vertices _A, B_ , and _C_ is a right triangle.
( _a_ ) _A_ (4, 0), _B_ (7, –7), and _C_ (2, –5)
( _b_ ) _A_ (5, 8), _B_ (–2, 1), and _C_ (2, –3)
( _c_ ) _A_ (2, 1), _B_ (3, –1), and _C_ (1, –2)
( _d_ ) _A_ (–6, 3), _B_ (3, –5), and _C_ (–1, 5)
**14.21** Show by using slopes that the diagonals and of quadrilateral _PQRS_ are perpendicular.
( _a_ ) _P_ (0, 0), _Q_ (5, 0), _R_ (8, 4), and _S_ (3, 4)
( _b_ ) _P_ (–3, 0), _Q_ (6, –3), _R_ (7, 5), and _S_ (3, 3)
**14.22** Show that the points _P, Q, R_ , and _S_ are the vertices of a parallelogram _PQRS_.
( _a_ ) _P_ (5, 0), _Q_ (8, 2), _R_ (6, 5), and _S_ (3, 3)
( _b_ ) _P_ (–9, 0), _Q_ (–10, –6), _R_ (4, 8), and _S_ (5, 14)
**14.23** Write the equation of the line through (7, 3) that is parallel to the _x_ axis.
**14.24** Write the equation of the horizontal line through the point (–2, – 3).
**14.25** Write the equation of the vertical line through the point (2, 4).
**14.26** Write the equation of the line through (5, 8) that is perpendicular to the _x_ axis.
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**14.10**
( _a_ ) –1
( _b_ ) 1
( _c_ ) not defined
( _d_ ) 0
( _e_ ) 2
( _f_ ) –6
**14.11**
( _a_ ) perpendicular, slopes 1/4 and –4
( _b_ ) perpendicular, slopes 1/3 and – 3
( _c_ ) parallel, slopes –1 and –1
( _d_ ) neither, slopes –5/3 and –1
**14.12**
( _a_ ) (1) 3/2 (2) –1
( _b_ ) (1) 7/2 (2) 1
( _c_ ) (1) –4 (2) 153/13
( _d_ ) (1) –13/2 (2)13/3
**14.13**
( _a_ ) yes: _m_ = 1/4
( _b_ ) no: slopes are not equal
**14.14**
( _a_ ) _y_ = –3 _x_ \+ 4
( _b_ ) _y_ = –3
( _c_ ) 2 _x_ – 3 _y_ = 6
( _d_ ) _y_ = 4 _x_
( _e_ ) _x_ \+ 2 _y_ = 6
( _f_ ) 5 _x_ \+ 6 _y_ = 1
**14.15**
( _a_ ) _x_ \+ _y_ = –3
( _b_ ) 4 _x_ – _y_ = –13
( _c_ ) 2 _x_ – 3 _y_ = 11
( _d_ ) 4 _x_ \+ 3 _y_ = 12
( _e_ ) 5 _x_ \+ _y_ = 16
( _f_ ) _y_ = 6
**14.16**
( _a_ ) _y_ = 2 _x_
( _b_ ) 80 _x_ – 65 _y_ = – 67
( _c_ ) 10 _x_ = 7
( _d_ ) _y_ = 2
( _e_ ) _x_ \+ 3 _y_ = 21
( _g_ ) 3 _x_ – _y_ = –6
( _f_ ) _x_ – 3 _y_ = –10
( _h_ ) _y_ = –2 _x_
**14.17**
( _a_ ) _x_ \+ _y_ = – 2
( _b_ ) _x_ – 2 _y_ = 6
( _c_ ) 8 _x_ – _y_ = – 4
( _d_ ) _x_ \+ 18 _y_ = 6
**14.18**
( _a_ ) _y_ = 4 _x_ – 12
( _b_ ) _y_ = 3 _x_ – 3
( _c_ ) 4 _x_ \+ 5 _y_ = – 9
( _d_ ) 3 _x_ – 2 _y_ = – 1
**14.19**
( _a_ ) 4 _x_ \+ _y_ = 7
( _b_ ) 3 _x_ – 2 _y_ = –12
( _c_ ) 2 _x_ \+ 3 _y_ = 5
( _d_ ) _x_ – 4 _y_ = 9
**14.20**
( _a_ )
( _b_ )
( _c_ )
( _d_ )
**14.21**
( _a_ ) yes: slopes are 1/2 and – 2
( _b_ ) yes: slopes are 1/2 and – 2
**14.22**
( _a_ )
( _b_ )
**14.23** _y_ = 3
**14.24** _y_ = –3
**14.25** _x_ = 2
**14.26** _x_ = 5
## CHAPTER 15
Simultaneous Linear Equations
### **15.1 SYSTEMS OF TWO LINEAR EQUATIONS**
A linear equation in two variables _x_ and _y_ is of the form _ax_ \+ _by_ = _c_ where _a, b, c_ are constants and _a, b_ are not both zero. If we consider two such equations
_a_ 1 _x_ \+ _b_ 1 _y_ = _c_ 1
_a_ 2 _x_ \+ _b_ 2 _y_ = _c_ 2
we say that we have two simultaneous linear equations in two unknowns or a system of two linear equations in two unknowns. A pair of values for _x_ and _y_ , ( _x, y_ ), which satisfies both equations is called a _simultaneous solution_ of the given equations.
Thus the simultaneous solution of _x_ \+ _y_ = 7 and _x_ – _y_ = 3 is (5, 2).
Three methods of solving such systems of linear equations are illustrated here.
A. Solution by addition or subtraction. If necessary, multiply the given equations by such numbers as will make the coefficients of one unknown in the resulting equations numerically equal. If the signs of the equal coefficients are unlike, add the resulting equations; if like, subtract them. Consider
(1) 2 _x_ – _y_ = 4
(2) _x_ \+ 2 _y_ = –3.
To eliminate _y_ , multiply (1) by 2 and add to (2) to obtain
Substitute _x_ = 1 in (1) and obtain 2 – _y_ = 4 or _y_ = – 2. Thus the simultaneous solution of (1) and (2) is (1, 2).
Check: Put _x_ = 1, _y_ = –2 in (2) and obtain 1 + 2(–2) ? –3, –3, = –3.
B. Solution by substitution. Find the value of one unknown in either of the given equations and substitute this value in the other equation.
For example, consider the system (1), (2) above. From (1) obtain _y_ = 2 _x_ –4 and substitute this value into (2) to get _x_ \+ 2(2 _x_ – 4) = – 3 which reduces to _x_ = 1. Then put _x_ = 1 into either (1) or (2) and obtain _y_ = – 2. The solution is (1, – 2).
C. Graphical solution. Graph both equations, obtaining two straight lines. The simultaneous solution is given by the coordinates ( _x, y_ ) of the point of intersection of these lines. Fig. 15-1 shows that the simultaneous solution of (1) 2 _x_ – _y_ = 4 and (2) _x_ \+ 2 _y_ = – 3 is _x_ = 1, _y_ = – 2, also written (1, – 2).
**Fig. 15-1**
If the lines are parallel, the equations are _inconsistent_ and have no simultaneous solution. For example, (3) _x_ \+ _y_ = 2 and (4) 2 _x_ \+ 2 _y_ = 8 are inconsistent, as indicated in Fig. 15-2. Note that if equation (3) is multiplied by 2 we obtain 2 _x_ \+ 2 _y_ = 4, which is obviously inconsistent with (4).
**Fig. 15-2**
_Dependent_ equations are represented by the same line. Thus every point on the line represents a solution and, since there are an infinite number of points, there are an infinite number of simultaneous solutions. For example, (5) _x_ \+ _y_ = 1 and (6) 4 _x_ \+ 4 _y_ = 4 are dependent equations as indicated in Fig. 15-3. Note that if (5) is multiplied by 4 the result is (6).
**Fig. 15-3**
### **15.2 SYSTEMS OF THREE LINEAR EQUATIONS**
A system of three linear equations in three variables is solved by eliminating one unknown from any two of the equations and then eliminating the same unknown from any other pair of equations.
Linear equations in three variables represent planes and can result in two or more parallel planes, which are thus inconsistent and have no solution. The three planes can coincide or all three can intersect in a common line and be dependent. The three planes can intersect in a single point, like the ceiling and two walls forming a corner in a room, and are consistent.
Linear equations in three variables _x, y_ , and _z_ are of the form _ax_ \+ _by_ \+ _cz_ = _d_ , where _a, b, c_ , and _d_ are real numbers and not all three of _a_ , _b_ , and _c_ are zero. If we consider three such equations
_a_ 1 _x_ \+ _b_ 1 _y_ \+ _c_ 1 _z_ = _d_ 1
_a_ 2 _x_ \+ _b_ 2 _y_ \+ _c_ 2 _z_ = _d_ 2
_a_ 3 _x_ \+ _b_ 3 _y_ \+ _c_ 3 _z_ = _d_ 3
and find a value ( _x, y, z_ ) that satisfies all three equations, we say we have a simultaneous solution to the system of equations.
**EXAMPLE 15.1.** Solve the system of equations 2 _x_ \+ 5 _y_ \+ 4 _z_ = 4, _x_ \+ 4 _y_ \+ 3 _z_ = 1, and _x_ – 3 _y_ – 2 _z_ = 5.
First we will eliminate _x_ from (1) and (2) and from (2) and (3).
Now we eliminate _z_ from equations (4) and (5).
We solve (6) and get _y_ = – 2.
Substituting into (4) or (5), we solve for _z_.
Substituting into (1), (2), or (3), we solve for _x_.
The solution to the system of equations is (3, – 2, 2).
We check the solution by substituting the point (3, – 2, 2) into equations (1), (2), and (3).
Thus (3, – 2, 2) checks in each of the given equations and is the answer to the problem.
### **Solved Problems**
Solve the following systems.
**15.1**
(1) 2 _x_ – _y_ = 4
(2) _x_ \+ _y_ = 5
**SOLUTION**
Add (1) and (2) and obtain 3 _x_ = 9, _x_ = 3.
Now put _x_ = 3 in (1) or (2) and get _y_ = 2. The solution is _x_ = 3, _y_ = 2 or (3, 2).
_Another method_. From (1) obtain _y_ = 2 _x_ – 4 and substitute this value into equation (2) to get _x_ \+ 2 _x_ – 4 = 5, 3 _x_ = 9, _x_ = 3. Now put _x_ = 3 into (1) or (2) and obtain _y_ = 2.
Check: 2 _x_ – _y_ = 2(3) – 2 = 4 and _x_ \+ _y_ = 3 + 2 = 5.
_Graphical solution_. The graph of a linear equation is a straight line. Since a straight line is determined by two points, we need plot only two points for each equation. However, to insure accuracy we shall plot three points for each line.
The simultaneous solution is the point of intersection (3, 2) of the lines (see Fig. 15-4).
**Fig. 15-4**
**15.2**
(1) 5 _x_ \+ 2 _y_ = 3
(2) 2 _x_ \+ 3 _y_ = –1
**SOLUTION**
To eliminate _y_ , multiply (1) by 3 and (2) by 2 and subtract the results.
Now put _x_ = 1 in (1) or (2) and obtain _y_ = – 1. The simultaneous solution is (1, – 1).
**15.3**
(1) 2 _x_ \+ 3 _y_ = 3
(2) 6 _y_ – 6 _x_ = 1
**SOLUTION**
Rearranging (2),
To eliminate _x_ , multiply (1) by 3 and add the result with (2) to get
Now put _y_ = 2/3 into (1) or (2) and obtain _x_ = 1/2. The solution is (1/2, 2/3).
**15.4**
(1) 5 _y_ = 3 – 2 _x_
(2) 3 _x_ = 2 _y_ \+ 1
**SOLUTION**
Solve by substitution.
From (1),
Put this value into (2) and obtain
Then
and the solution is
**15.5** (1)
(2)
**SOLUTION**
To eliminate fractions, multiply (1) by 6 and (2) by 4 and simplify to obtain
Solving, we find
**15.6**
(1) _x_ – 3 _y_ = 2 _a_
(2) 2 _x_ \+ _y_ = 5 _a_
**SOLUTION**
To eliminate _x_ , multiply (1) by 2 and subtract (2); then _y_ = _a_ /7.
To eliminate _y_ , multiply (2) by 3 and add with (1); then _x_ = 17 _a_ /7.
The solution is
**15.7**
(1) 3 _u_ \+ 2 _v_ = 7 _r_ \+ _s_
(2) 2 _u_ – _v_ = 3 _s_
Solve for _u_ and _v_ in terms of _r_ and _s_.
**SOLUTION**
To eliminate v, multiply (2) by 2 and add with (1); then 7 _u_ = 7 _r_ \+ 7 _s_ or _u_ = _r_ \+ _s_.
To eliminate _u_ , multiply (1) by 2, (2) by –3, and add the results; then _v_ = 2 _r_ – _s_.
The solution is ( _r_ \+ _s_ , 2 _r_ – _s_ ).
**15.8**
(1) _ax_ \+ _by_ = 2 _a_ 2 – 3 _b_ 2
(2) _x_ \+ 2 _y_ = 2 _a_ – 6 _b_
**SOLUTION**
Multiply (2) by _a_ and subtract from (1); then _by_ – 2 _ay_ = 6 _ab_ – 3 _b_ 2, _y_ ( _b_ – 2 _a_ ) = 3 _b_ (2 _a_ – _b_ ), and provided _b_ — 2 _a_ ≠ 0.
Similarly, we obtain _x_ = 2 _a_ provided _b_ – 2 _a_ ≠ 0.
Check: (1) _a_ (2 _a_ ) + _b_ (–3 _b_ ) = 2 _a_ 2 — 3 _b_ 2, (2) 2 _a_ \+ 2(–3 _b_ ) = 2 _a_ – 6 _b_.
_Note_. If _b_ – 2 _a_ = 0, or _b_ = 2 _a_ , the given equations become
which are dependent since (11) may be obtained by multiplying (21) by _a_. Thus if _b_ = 2 _a_ the system possesses an infinite number of solutions, i.e., any values of _x_ and _y_ which satisfy _x_ \+ 2 _y_ = – 10 _a_.
**15.9** The sum of two numbers is 28 and their difference is 12. Find the numbers.
**SOLUTION**
Let _x_ and _y_ be the two numbers. Then (1) _x_ \+ _y_ = 28 and (2) _x_ – _y_ = 12.
Add (1) and (2) to obtain 2 _x_ = 40, _x_ = 20. Subtract (2) from (1) to obtain 2 _y_ = 16, _y_ = 8.
_Note_. Of course this problem may also be solved easily by using one unknown. Let the numbers be _n_ and 28 – _n_.
Then _n_ – (28 – _n_ ) = 12 or _n_ = 20, and 28 – _n_ = 8.
**15.10** If the numerator of a certain fraction is increased by 2 and the denominator is increased by 1, the resulting fraction equals 1/2. If, however, the numerator is increased by 1 and the denominator decreased by 2, the resulting fraction equals 3/5. Find the fraction.
**SOLUTION**
Let _x_ = numerator, _y_ = denominator, and _x/y_ = required fraction. Then
Solve (1) and (2) simultaneously and obtain _x_ = 2, _y_ = 7. The required fraction is 2/7.
**15.11** Two years ago a man was six times as old as his daughter. In 18 years he will be twice as old as his daughter. Determine their present ages.
**SOLUTION**
Let _x_ = father's present age in years, _y_ = daughter's present age in years.
Solve (1) and (2) simultaneously and obtain _x_ = 32, _y_ = 7.
**15.12** Find the two-digit number satisfying the following two conditions. (1) Four times the units digit is six less than twice the tens digit. (2) The number is nine less than three times the number obtained by reversing the digits.
**SOLUTION**
Let _t_ = tens digit, _u_ = units digit.
The required number = 10 _t_ \+ _u_ ; reversing digits, the new number = 10 _u_ \+ _t_. Then
Solving (1) and (2) simultaneously, _t_ = 7, _u_ = 2, and the required number is 72.
**15.13** Five tables and eight chairs cost $115; three tables and five chairs cost $70. Determine the cost of each table and of each chair.
**SOLUTION**
Let _x_ = cost of a table, _y_ = cost of a chair. Then
Solve (1) and (2) simultaneously and obtain _x_ = $15, _y_ = $5.
**15.14** A merchant sold his entire stock of shirts and ties for $1000, the shirts being priced at 3 for $10 and the ties at $2 each. If he had sold only 1/2 of the shirts and 2/3 of the ties he would have collected $600. How many of each kind did he sell?
**SOLUTION**
Let _s_ = number of shirts sold, _t_ = number of ties sold. Then
Solving (1) and (2) simultaneously, _s_ = 120, _t_ = 300.
**15.15** An investor has $1100 income from bonds bearing 4% and 5%. If the amounts at 4% and 5% were interchanged she would earn $50 more per year. Find the total sum invested.
**SOLUTION**
Let _x_ = amount invested at 4%, _y_ = amount at 5%. Then
Solving (1) and (2) simultaneously, _x_ = $15 000, _y_ = $10 000, and their sum is $25 000.
**15.16** Tank _A_ contains a mixture of 10 gallons water and 5 gallons pure alcohol. Tank _B_ has 12 gallons water and 3 gallons alcohol. How many gallons should be taken from each tank and combined in order to obtain an 8 gallon solution containing 25% alcohol by volume?
**SOLUTION**
In 8 gal of required mixture are 0.25(8) = 2 gal alcohol.
Let _x, y_ = volumes taken from tanks _A, B_ respectively; then (1) _x_ \+ _y_ = 8.
Thus in _x_ gal of _A_ are _x_ /3 gal alcohol, and in _y_ gal of _B_ are _y_ /5 gal alcohol; then (2) _x_ /3 + _y_ /5 = 2.
Solving (1) and (2) simultaneously, _x_ = 3 gal, _y_ = 5 gal.
_Another method_ , using only one unknown. Let _x_ = volume taken from tank _A_ , 8 – _x_ = volume taken from tank _B_.
Then from which _x_ = 3 gal, 8 – _x_ = 5 gal.
**15.17** A given alloy contains 20% copper and 5% tin. How many pounds of copper and of tin must be melted with 100 lb of the given alloy to produce another alloy analyzing 30% copper and 10% tin? All percentages are by weight.
**SOLUTION**
Let _x, y_ = number of pounds of copper and tin to be added, respectively.
In 100 lb of given alloy are 20 lb copper and 5 lb tin. Then, in the new alloy,
The simultaneous solution of (1) and (2) is _x_ = 17.5 lb copper, _y_ = 7.5 lb tin.
**15.18** Determine the rate of a woman's rowing in still water and the rate of the river current, if it takes her 2 hours to row 9 miles with the current and 6 hours to return against the current.
**SOLUTION**
Let _x_ = rate of rowing in still water, _y_ = rate of current.
Solving (1) and (2) simultaneously, _x_ = 3 mi/hr, _y_ = 3/2 mi/hr.
**15.19** Two particles move at different but constant speeds along a circle of circumference 276 ft. Starting at the same instant and from the same place, when they move in opposite directions they pass each other every 6 sec and when they move in the same direction they pass each other every 23 sec. Determine their rates.
**SOLUTION**
Let _x, y_ = their respective rates in ft/sec.
Solving (1) and (2) simultaneously, _x_ = 29 ft/sec, _y_ = 17 ft/sec.
**15.20** Fahrenheit temperature = _m_ (Celsius temperature) + _n_ , or _F_ = _mC_ \+ _n_ , where _m_ and _n_ are constants. At one atmosphere pressure, the boiling point of water is 212 °F or 100 °C and the freezing point of water is 32 °F or 0 °C. ( _a_ ) Find _m_ and _n_. ( _b_ ) What Fahrenheit temperature corresponds to –273 °C, the lowest temperature obtainable?
**SOLUTION**
Solve the following systems.
**15.21** (1) 2 _x_ – _y_ \+ _z_ = 3
(2) _x_ \+ 3 _y_ – 2 _z_ = 11
(3) 3 _x_ – 2 _y_ \+ 4 _z_ = 1
**SOLUTION**
To eliminate _y_ between (1) and (2) multiply (1) by 3 and add with (2) to obtain
To eliminate _y_ between (2) and (3) multiply (2) by 2, (3) by 3, and add the results to get
Solving (11) and (21) simultaneously, we find _x_ = 3, _z_ = – 1. Substituting these values into any of the given equations, we find _y_ = 2.
Thus the solution is (3, 2, – 1).
**15.22**
**SOLUTION**
To remove fractions, multiply the equations by 12 and obtain the system
To eliminate _x_ between (11) and (21) multiply (11) by 3, (21) by –4, and add the results to obtain
To eliminate _x_ between (21) and (31) multiply (21) by 2 and subtract (31) to obtain
The simultaneous solution of (12) and (22) is _y_ = 2, _z_ = 4. Substituting these values of _y_ and _z_ into any of the given equations, we find _x_ = 6.
Thus the simultaneous solution of the three given equations is (6, 2, 4).
**15.23**
**SOLUTION**
Let
so that the given equations may be written
from which we find _u_ = – 2, _v_ = 3, _w_ = –4.
Thus
The solution is (–1/2, 1/3, –1/4).
Check:
**15.24**
(1) 3 _x_ \+ _y_ – _z_ = 4,
(2) _x_ \+ _y_ \+ 4 _z_ = 3,
(3) 9 _x_ \+ 5 _y_ \+ 10 _z_ = 8.
**SOLUTION**
Subtracting (2) from (1), we obtain (11) 2 _x_ – 5 _z_ = 1.
Multiplying (2) by 5 and subtracting (3), we obtain (21) — 4 _x_ \+ 10 _z_ = 7.
Now (11) and (21) are inconsistent since (11) multiplied by – 2 gives – 4 _x_ \+ 10 _z_ = – 2, thus contradicting (21). This indicates that the original system is inconsistent and hence has no simultaneous solution.
**15.25** _A_ and _B_ working together can do a given job in 4 days, _B_ and _C_ together can do the job in 3 days, and _A_ and _C_ together can do it in 2.4 days. In how many days can each do the job working alone?
**SOLUTION**
Let _a, b, c_ = number of days required by each working alone to do the job, respectively. Then 1/ _a_ , 1/ _b_ , 1/ _c_ = fraction of complete job done by each in 1 day, respectively. Thus
Solving (1), (2), (3) simultaneously, we find _a_ = 6, _b_ = 12, _c_ = 4 days.
### **Supplementary Problems**
**15.26** Solve each of the following pairs of simultaneous equations by the methods indicated.
**15.27** Solve each of the following pairs of simultaneous equations by any method.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
**15.28** Indicate which of the following systems are (1) consistent, (2) dependent, (3) inconsistent.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
**15.29**
( _a_ ) When the first of two numbers is added to twice the second the result is 21, but when the second number is added to twice the first the result is 18. Find the two numbers.
( _b_ ) If the numerator and denominator of a certain fraction are both increased by 3, the resulting fraction equals 2/3. If, however, the numerator and denominator are both decreased by 2, the resulting fraction equals 1/2. Determine the fraction.
( _c_ ) Twice the sum of two numbers exceeds three times their difference by 8, while half the sum is one more than the difference. What are the numbers?
( _d_ ) If three times the larger of two numbers is divided by the smaller, the quotient is 6 and the remainder is 6. If five times the smaller is divided by the larger, the quotient is 2 and the remainder is 3. Find the numbers.
**15.30**
( _a_ ) Six years ago Bob was four times as old as Mary. In four years he will be twice as old as Mary. How old are they now?
( _b_ ) _A_ is eleven times as old as _B_. In a certain number of years _A_ will be five times as old as _B_ , and five years after that she will be three times as old as _B_. How old are they now?
**15.31**
( _a_ ) Three times the tens digit of a certain two-digit number is two more than four times the units digit. The difference between the given number and the number obtained by reversing the digits is two less than twice the sum of the digits. Find the number.
( _b_ ) When a certain two-digit number is divided by the number obtained by reversing the digits, the quotient is 2 and the remainder is 7. If the number is divided by the sum of its digits, the quotient is 7 and the remainder 6. Find the number.
**15.32**
( _a_ ) Two pounds of coffee and 3 lb of butter cost $4.20. A month later the price of coffee advanced 10% and that of butter 20%, making the total cost of a similar order $4.86. Determine the original cost of a pound of each.
( _b_ ) If 3 gallons of Grade A oil are mixed with 7 gal of Grade B oil the resulting mixture is worth 43 ¢/gal. However, if 3 gal of Grade A oil are mixed with 2 gal of Grade B oil the resulting mixture is worth 46 ¢/gal. Find the price per gallon of each grade.
( _c_ ) An investor has $116 annual income from bonds bearing 3% and 5% interest. Then he buys 25% more of the 3% bonds and 40% more of the 5% bonds, thereby increasing his annual income by $41. Find his initial investment in each type of bond.
**15.33**
( _a_ ) Tank A contains 32 gallons of solution which is 25% alcohol by volume. Tank B has 50 gal of solution which is 40% alcohol by volume. What volume should be taken from each tank and combined in order to make up 40 gal of solution containing 30% alcohol by volume?
( _b_ ) Tank A holds 40 gal of a salt solution containing 80 lb of dissolved salt. Tank B has 120 gal of solution containing 60 lb of dissolved salt. What volume should be taken from each tank and combined in order to make up 30 gal of solution having a salt concentration of 1.5 lb/gal?
( _c_ ) A given alloy contains 10% zinc and 20% copper. How many pounds of zinc and of copper must be melted with 1000 lb of the given alloy to produce another alloy analyzing 20% zinc and 24% copper? All percentages are by weight.
( _d_ ) An alloy weighing 600 lb is composed of 100 lb copper and 50 lb tin. Another alloy weighing 1000 lb is composed of 300 lb copper and 150 lb tin. What weights of copper and tin must be melted with the two given alloys to produce a third alloy analyzing 32% copper and 28% tin. All percentages are by weight.
**15.34**
( _a_ ) Determine the speed of a motorboat in still water and the speed of the river current, if it takes 3 hr to travel a distance of 45 mi upstream and 2 hr to travel 50 mi downstream.
( _b_ ) When two cars race around a circular mile track starting from the same place and at the same instant, they pass each other every 18 seconds when traveling in opposite directions and every 90 seconds when traveling in the same direction. Find their speeds in mi/hr.
( _c_ ) A passenger on the front of train A observes that she passes the complete length of train B in 33 seconds when traveling in the same direction as B and in 3 seconds when traveling in the opposite direction. If B is 330 ft long, find the speeds of the two trains.
**15.35** Solve each of the following systems of equations.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
**15.36** Indicate which of the following systems are (1) consistent, (2) dependent, (3) inconsistent.
( _a_ )
( _b_ )
( _c_ )
**15.37** The first of three numbers exceeds the third by one-half of the second. The sum of the second and third numbers is one more than the first. If the second is subtracted from the sum of the first and third numbers the result is 5. Determine the numbers.
**15.38** When a certain three-digit number is divided by the number with digits reversed, the quotient is 2 and the remainder 25. The tens digit is one less than twice the sum of the hundreds digit and units digit. If the units digit is subtracted from the tens digit, the result is twice the hundreds digit. Find the number.
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**15.26**
( _a_ ) _x_ = 2, _y_ = – 1
( _b_ ) _x_ = – 1, _y_ = 3
( _c_ ) _x_ = 1/2, _y_ = 3/2
**15.27**
( _a_ ) _x_ = 5/2, _y_ = –1
( _b_ ) _x_ = 3, _y_ = 2
( _c_ ) _x_ = 6, _y_ = 10
( _d_ ) _x_ = 5, _y_ = 2
( _e_ ) _x_ = 3 _t_ /2, _y_ = – 2 _t_
( _f_ ) _x_ = – 1, _y_ = 1
( _g_ ) _u_ = _r_ – 2 _s_ , _v_ = 2 _r_ \+ _s_
( _h_ ) _x_ = 1/2, _y_ = 1/3
( _i_ ) _x_ = _a_ \+ _b_ , _y_ = _a_ – _b_ if _a_ 2 ≠ 2 _b_ 2
**15.28**
( _a_ ) Consistent,
( _b_ ) Inconsistent,
( _c_ ) Dependent
( _d_ ) Inconsistent,
( _e_ ) Consistent,
( _f_ ) Dependent.
**15.29**
( _a_ ) 5, 8
( _b_ ) 7/12
( _c_ ) 7, 3
( _d_ ) 16, 7
**15.30**
( _a_ ) Mary 11 yr, Bob 26 yr
( _b_ ) _A_ is 22 yr, _B_ is 2 yr
**15.31**
( _a_ ) 64
( _b_ ) 83
**15.32**
( _a_ ) Coffee 90 ¢/lb, butter 80 ¢/lb
( _b_ ) Grade A 50 ¢/gal, Grade B 40 ¢/gal
( _c_ ) $1200 at 3%, $1600 at 5%
**15.33**
( _a_ ) 26 2/3 gal from A 13 1/3 gal from B
( _b_ ) 20 gal from A 10 gal from B
( _c_ ) 150 lb zinc 100 lb copper
( _d_ ) 400 lb copper 500 lb tin
**15.34**
( _a_ ) Boat 20 mi/hr, river 5 mi/hr
( _b_ ) 120 mi/hr, 80 mi/hr
( _c_ ) 60 ft/sec, 50 ft/sec
**15.35**
( _a_ ) _x_ = –1, _y_ = 2, _z_ = –2
( _b_ ) _x_ = 0, _y_ = 2, _z_ = 1
( _c_ ) _x_ = 6, _y_ = 4, _z_ = –3
( _d_ ) _x_ = 1/2, _y_ = – 1/3, _z_ = 1/6
**15.36**
( _a_ ) Dependent
( _b_ ) Inconsistent
( _c_ ) Consistent
**15.37** 4, 2, 3
**15.38** 371
## CHAPTER 16
Quadratic Equations in One Variable
### **16.1 QUADRATIC EQUATIONS**
A quadratic equation in the variable _x_ has the form _ax_ 2 \+ _bx_ \+ _c_ = 0 where _a, b_ , and _c_ are constants and _a_ ≠ 0.
Thus _x_ 2 – 6 _x_ \+ 5 = 0, 2 _x_ 2 \+ _x_ – 6 = 0, _x_ 2 \+ 3 _x_ = 0, and 3 _x_ 2 – 5 = 0 are quadratic equations in one variable. The last two equations may be divided by 2 and 4 respectively to obtain where the coefficient of _x_ 2 in each case is 1.
An incomplete quadratic equation is one which either _b_ = 0 or _c_ = 0, e.g., 4 _x_ 2 – 5 = 0, 7 _x_ 2 – 2 _x_ = 0, and 3 _x_ 2 = 0.
To solve a quadratic equation _ax_ 2 \+ _bx_ \+ _c_ = 0 is to find values of _x_ which satisfy the equation. These values of _x_ are called _zeros_ or _roots_ of the equation.
For example, _x_ 2 – 5 _x_ \+ 6 = 0 is satisfied by _x_ = 2 and _x_ = 3. Then _x_ = 2 and _x_ = 3 are zeros or roots of the equation.
### **16.2 METHODS OF SOLVING QUADRATIC EQUATIONS**
A. Solution by square root
**EXAMPLES 16.1.** Solve each quadratic equation for _x_.
( _a_ ) _x_ 2 – 4 = 0
( _b_ ) 2 _x_ 2 – 21 = 0
( _c_ ) _x_ 2 \+ 9 = 0
( _a_ ) _x_ 2 – 4 = 0. Then _x_ 2 = 4, _x_ = ±2, and the roots are _x_ = 2, –2.
( _b_ ) 2 _x_ 2 – 21 = 0. Then _x_ 2 = 21/2 and the roots are
( _c_ ) _x_ 2 \+ 9 = 0. Then _x_ 2 = – 9 and the roots are
B. Solution by factoring
**EXAMPLES 16.2.** Solve each quadratic equation for _x_.
( _a_ ) 7 _x_ 2 – 5 _x_ = 0
( _b_ ) _x_ 2 – 5 _x_ \+ 6 = 0
( _c_ ) 3 _x_ 2 \+ 2 _x_ – 5 = 0
( _d_ ) _x_ 2 – 4 _x_ \+ 4 = 0
( _a_ ) 7 _x_ 2 – 5 _x_ = 0 may be written as _x_ (7 _x_ – 5) = 0. Since the product of the two factors is zero, we set each factor equal to 0 and solve the resulting linear equations. _x_ = 0 or 7 _x_ – 5 = 0. So _x_ = 0 and _x_ = 5/7 are the roots of the equation.
( _b_ ) _x_ 2 – 5 _x_ \+ 6 = 0 may be written as ( _x_ – 3)( _x_ – 2) = 0. Since the product is equal to 0, we set each factor equal to 0 and solve the resulting linear equations. _x_ – 3 = 0 or _x_ – 2 = 0. So _x_ = 3 and _x_ = 2 are the roots of the equation.
( _c_ ) 3 _x_ 2 \+ 2 _x_ – 5 = 0 may be written as (3 _x_ \+ 5)( _x_ – 1) = 0. Thus, 3 _x_ \+ 5 = 0 or _x_ – 1 = 0 and the roots of the equation are _x_ = – 5/3 and _x_ = 1.
( _d_ ) _x_ 2 – 4 _x_ \+ 4 = 0 may be written as ( _x_ – 2)( _x_ – 2) = 0. Thus, _x_ – 2 = 0 and the equation has a double root _x_ = 2.
C. Solution by completing the square
**EXAMPLE 16.3.** Solve _x_ 2 – 6 _x_ – 2 = 0.
Write the unknowns on one side and the constant term on the other; then
_x_ 2 – 6 _x_ = 2.
Add 9 to both sides, thus making the left-hand side a perfect square; then
Hence and the required roots are
_Note_. In the method of completing the square (1) the coefficient of the _x_ 2 term must be 1 and (2) the number added to both sides is the square of half the coefficient of _x_.
**EXAMPLE 16.4.** Solve 3 _x_ 2 – 5 _x_ \+ 1 = 0.
Dividing by 3,
Adding to both sides,
D. Solution by quadratic formula.
The solutions of the quadratic equation _ax_ 2 \+ _bx_ \+ _c_ = 0 are given by the formula
where _b_ 2 – 4 _ac_ is called the _discriminant_ of the quadratic equation.
For a derivation of the quadratic formula see Problem 16.5.
**EXAMPLE 16.5.** Solve 3 _x_ 2 – 5 _x_ \+ 1 = 0. Here _a_ = 3, _b_ = –5, _c_ = 1 so that
**EXAMPLE 16.6.** Solve 4 _x_ 2 – 6 _x_ \+ 3 = 0.
Here _a_ = 4, _b_ = – 6, and _c_ = 3.
E. Graphical solution
The real roots or zeros of _ax_ 2 \+ _bx_ \+ _c_ = 0 are the values of _x_ corresponding to _y_ = 0 on the graph of the parabola _y_ = _ax_ 2 \+ _bx_ \+ _c_. Thus the solutions are the abscissas of the points where the parabola intersects the _x_ axis. If the graph does not intersect the _x_ axis the roots are not real.
### **16.3 SUM AND PRODUCT OF THE ROOTS**
The sum _S_ and the product _P_ of the roots of the quadratic equation _ax_ 2 \+ _bx_ \+ _c_ = 0 are given by _S_ = – _b_ / _a_ and _P_ = _c/a_.
Thus in 2 _x_ 2 \+ 7 _x_ – 6 = 0 we have _a_ = 2, _b_ = 7, _c_ = – 6 so that _S_ = –7/2 and _P_ = – 6/2 = –3.
It follows that a quadratic equation whose roots are _r_ 1, _r_ 2 is given by _x_ 2 – _Sx_ \+ _P_ = 0 where _S_ = _r_ 1 \+ _r_ 2 and _P_ = _r_ 1 _r_ 2. Thus the quadratic equation whose roots are _x_ = 2 and _x_ = – 5 is _x_ 2 – (2 – 5) _x_ \+ 2(– 5) = 0 or _x_ 2 \+ 3 _x_ – 10 = 0.
### **16.4 NATURE OF THE ROOTS**
The nature of the roots of the quadratic equation _ax_ 2 \+ _bx_ \+ _c_ = 0 is determined by the discriminant _b_ 2 – 4 _ac_. When the roots involve the imaginary unit _i_ , we say the roots are not real.
Assuming _a, b, c_ are _real numbers_ then
(1) if _b_ 2 – 4 _ac_ > 0, the roots are _real_ and _unequal_ ,
(2) if _b_ 2 – 4 _ac_ = 0, the roots are _real_ and _equal_ ,
(3) if _b_ 2 – 4 _ac_ < 0, the roots are _not real_.
Assuming _a, b, c_ are _rational numbers_ then
(1) if _b_ 2 – 4 _ac_ is a perfect square ≠ 0, the roots are real, rational, and unequal,
(2) if _b_ 2 – 4 _ac_ = 0, the roots are real, rational, and equal,
(3) if _b_ 2 – 4 _ac_ > 0 but not a perfect square, the roots are real, irrational, and unequal,
(4) if _b_ 2 – 4 _ac_ < 0, the roots are not real.
Thus 2 _x_ 2 \+ 7 _x_ – 6 = 0, with discriminant _b_ 2 – 4 _ac_ = 72 – 4(2)(–6) = 97, has roots which are real, irrational and unequal.
### **16.5 RADICAL EQUATIONS**
A radical equation is an equation having one or more unknowns under a radical.
Thus are radical equations.
To solve a radical equation, isolate one of the radical terms on one side of the equation and transpose all other terms to the other side. If both members of the equation are then raised to a power equal to the index of the isolated radical, the radical will be removed. This process is continued until radicals are no longer present.
**EXAMPLE 16.7.** Solve
It is very important to check the values obtained, as this method often introduces extraneous roots which are to be rejected.
### **16.6 QUADRATIC-TYPE EQUATIONS**
An equation of quadratic type has the form _az_ 2 _n_ \+ _bz n_ \+ _c_ = 0 where _a_ = 0, _b, c_ , and _n_ = 0 are constants, and where _z_ depends on _x_. Upon letting _z n_ = _u_ this equation becomes _au_ 2 \+ _bu_ \+ _c_ = 0, which may be solved for _u_. These values of _u_ may be used to obtain _z_ from which it may be possible to obtain _x_.
**EXAMPLE 16.8.** Solve _x_ 4 – 3 _x_ 2 – 10 = 0.
**EXAMPLE 16.9.** Solve (2 _x_ – 1)2 \+ 7(2 _x_ – 1) + 12 = 0.
### **Solved Problems**
**16.1** Solve
( _a_ ) _x_ 2 – 16 = 0. Then _x_ 2 = 16, _x_ = ±4.
( _b_ ) 4 _t_ 2 – 9 = 0. Then 4 _t_ 2 = 9, _t_ 2 = 9/4, _t_ = ±3/2.
( _c_ )
( _d_ )
( _e_ )
_Check_. If _x_ = 3 is substituted into the original equation, we have division by zero which is not allowed. Hence _x_ = 3 is not a solution.
**16.2** Solve by factoring
( _a_ ) _x_ 2 \+ 5 _x_ – 6 = 0, ( _x_ \+ 6)( _x_ – 1) = 0, _x_ = –6, 1.
( _b_ ) _t_ 2 = 4 _t_ , _t_ 2 – 4 _t_ = 0, _t_ ( _t_ – 4) = 0, _t_ = 0, 4.
( _c_ ) _x_ 2 \+ 3 _x_ = 28, _x_ 2 \+ 3 _x_ – 28 = 0, ( _x_ \+ 7)( _x_ – 4) = 0, _x_ = –7, 4.
( _d_ ) 5 _x_ – 2 _x_ 2 = 2, 2 _x_ 2 – 5 _x_ \+ 2 = 0, (2 _x_ – 1)( _x_ – 2) = 0, _x_ = 1/2, 2.
( _e_ )
4( _t_ – 4) + 4( _t_ – 1) = 5( _t_ – 1)( _t_ – 4), 5 _t_ 2 – 33 _t_ \+ 40 = 0, ( _t_ – 5)(5 _t_ – 8) = 0, _t_ = 5, 8/5.
( _f_ )
**16.3** What term must be added to each of the following expressions in order to make it a perfect square trinomial?
( _a_ )
( _b_ )
( _c_ )
( _d_ )
**16.4** Solve by completing the square.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
**16.5** Solve the equation _ax_ 2 \+ _bx_ \+ _c_ = 0, _a_ ≠ 0, by the method of completing the square.
**SOLUTION**
**16.6** Solve by the quadratic formula.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
**16.7** Solve graphically: ( _a_ ) 2 _x_ 2 \+ 3 _x_ – 5 = 0, ( _b_ ) 4 _x_ 2 – 12 _x_ \+ 9 = 0, ( _c_ ) 4 _x_ 2 – 4 _x_ \+ 5 = 0.
**SOLUTION**
( _a_ ) _y_ = 2 _x_ 2 \+ 3 _x_ – 5
The graph of _y_ = 2 _x_ 2 \+ 3 _x_ – 5 indicates that when _y_ = 0, _x_ = 1, and –2.5.
Thus the roots of 2 _x_ 2 \+ 3 _x_ – 5 = 0 are _x_ = 1, –2.5 (see Fig. 16-1( _a_ )).
**Fig. 16-1**
( _b_ ) _y_ = 4 _x_ 2 – 12 _x_ \+ 9
The graph of _y_ = 4 _x_ 2 – 12 _x_ \+ 9 is tangent to the _x_ axis at _x_ = 1.5, i.e., when _y_ = 0, _x_ = 1.5.
Thus 4 _x_ 2 – 12 _x_ \+ 9 = 0 has the equal roots _x_ = 1.5 (see Fig. 16-1( _b_ )).
( _c_ ) _y_ = 4 _x_ 2 – 4 _x_ \+ 5
The graph of _y_ = 4 _x_ 2 – 4 _x_ \+ 5 does not intersect the _x_ axis, i.e., there is no real value of _x_ for which _y_ = 0.
Hence the roots of 4 _x_ 2 – 4 _x_ \+ 5 = 0 are not real (see Fig. 16-1( _c_ )).
(By the quadratic formula the roots are _x_ = ± _i_.)
**16.8** Prove that the sum _S_ and product _P_ of the roots of the quadratic equation _ax_ 2 \+ _bx_ \+ _c_ = 0 are _S_ = – _b/a_ and _P_ = _c/a_.
**SOLUTION**
By the quadratic formula the roots are
The sum of the roots is
The product of the roots is
**16.9** Without solving, find the sum _S_ and product _P_ of the roots.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
**16.10** Find the discriminant _b_ 2 – 4 _ac_ of each of the following equations and thus determine the character of their roots.
( _a_ ) _x_ 2 – 8 _x_ \+ 12 = 0. _b_ 2 – 4 _ac_ = (– 8)2 – 4(1)(12) = 16; the roots are real, rational, unequal.
( _b_ ) 3 _y_ 2 \+ 2 _y_ – 4 = 0. _b_ 2 – 4 _ac_ = 52; the roots are real, irrational, unequal.
( _c_ ) 2 _x_ 2 – _x_ \+ 4 = 0. _b_ 2 – 4 _ac_ = – 31; the roots are conjugate not real numbers.
( _d_ ) 4 _z_ 2 – 12 _z_ \+ 9 = 0. _b_ 2 – 4 _ac_ = 0; the roots are real, rational, equal.
( _e_ ) 2 _x_ – 4 _x_ 2 = 1 or 4 _x_ 2 – 2 _x_ \+ 1 = 0. _b_ 2 – 4 _ac_ = – 12; the roots are conjugate not real numbers.
( _f_ ) Here the coefficients are real but not rational numbers.
_b_ 2 – 4 _ac_ = 16; the roots are real and unequal.
**16.11** Find a quadratic equation with integer coefficients having the given pair of roots. ( _S_ = sum of roots, _P_ = product of roots.)
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
**16.12** In each quadratic equation find the value of the constant _p_ subject to the given condition.
( _a_ )
( _b_ )
( _c_ )
( _d_ ) 3 _x_ 2 \+ ( _p_ \+ 1) + 24 = 0 has one root equal to twice the other. Let the roots be _r_ , 2 _r_.
Product of the roots is _r_ (2 _r_ ) = 8; then _r_ 2 = 4 and _r_ = ±2.
Sum of the roots is 3 _r_ = – ( _p_ \+ 1)/3. Substitute _r_ = 2 and _r_ = –2 into this equation and obtain _p_ = –19 and _p_ = 17 respectively.
( _e_ ) 2 _x_ 2 – 12 _x_ \+ _p_ \+ 2 = 0 has the difference between its roots equal to 2.
Let the roots be _r, s_ ; then (1) _r_ – _s_ = 2. The sum of the roots is 6; then (2) _r_ \+ _s_ = 6. The simultaneous solution of (1) and (2) is _r_ = 4, _s_ = 2.
Now put _x_ = 2 or _x_ = 4 into the given equation and obtain _p_ = 14.
**16.13** Find the roots of each quadratic equation subject to the given conditions.
( _a_ ) (2 _k_ \+ 2) _x_ 2 \+ (4 – 4 _k_ ) _x_ \+ _k_ – 2 = 0 has roots which are reciprocals of each other.
Let _r_ and 1/ _r_ be the roots, their product being 1.
Product of roots is from which _k_ = –4.
Put _k_ = – 4 into the given equation; then 3 _x_ 2 – 10 _x_ \+ 3 = 0 and the roots are 1/3, 3.
( _b_ ) _kx_ 2 – (1 + _k_ ) _x_ \+ 3 _k_ \+ 2 = 0 has the sum of its roots equal to twice the product of its roots.
Sum of roots = 2(product of roots); then
Put _k_ = –3/5 into the given equation; then 3 _x_ 2 \+ 2 _x_ – 1 = 0 and the roots are –1, 1/3.
( _c_ ) ( _x_ \+ _k_ )2 = 2 – 3 _k_ has equal roots.
Write the equation as _x_ 2 \+ 2 _kx_ \+ ( _k_ 2 \+ 3 _k_ – 2) = 0 where _a_ = 1, _b_ = 2 _k_ , _c_ = _k_ 2 \+ 3 _k_ – 2.
The roots are equal if the discriminant ( _b_ 2 – 4 _ac_ ) = 0.
Then from _b_ 2 – 4 _ac_ = (2 _k_ )2 – 4(1)( _k_ 2 \+ 3 _k_ – 2) = 0 we get _k_ = 2/3.
Put _k_ = 2/3 into the given equation and solve to obtain the double root –2/3.
**16.14** Solve
( _a_ )
( _b_ )
( _c_ )
( _d_ )
**16.15** Solve
( _a_ )
( _b_ )
( _c_ )
**16.16** Solve
( _a_ )
( _b_ )
**16.17** Solve the equation
**SOLUTION**
Let _x_ 2 – 6 _x_ = _u_ ; then or (1)
Squaring (1), _u_ – 3 = _u_ 2 – 10 _u_ \+ 25, _u_ 2 – 11 _u_ \+ 28 = 0, and _u_ = 7, 4.
Since only _u_ = 7 satisfies (1), substitute _u_ = 7 in _x_ 2 – 6 _x_ = _u_ and obtain
Both _x_ = 7 and _x_ = –1 satisfy the original equation and are thus solutions.
_Note_. If we write the given equation as and square both sides, the resulting fourth degree equation would be difficult to solve.
**16.18** Solve the equation
**SOLUTION**
Squaring,
then 25(16 – 8 _x_ \+ _x_ 2) = 9( _x_ 2 – 8 _x_ \+ 32), _x_ 2 – 8 _x_ \+ 7 = 0, and _x_ = 7, 1. The only solution is _x_ = 1; reject _x_ = 7, an extraneous solution.
**16.19** Solve
( _a_ ) _x_ 4 – 10 _x_ 2 \+ 9 = 0. Let _x_ 2 = _u_ ; then _u_ 2 – 10 _u_ \+ 9 = 0 and _u_ = 1, 9.
For _u_ = 1, _x_ 2 = 1 and _x_ = ±1; for _u_ = 9, _x_ 2 = 9 and _x_ = ±3.
The four solutions are _x_ = ±1, ±3; each satisfies the given equation.
( _b_ ) 2 _x_ 4 \+ _x_ 2 – 1 = 0. Let _x_ 2 = _u_ ; then 2 _u_ 2 \+ _u_ – 1 = 0 and _u_ = , – 1.
If if _u_ = –1, _x_ 2 = –1 and _x_ = ± _i_.
The four solutions are
( _c_ )
( _d_ )
The four solutions are
( _e_ )
**16.20** Find the values of _x_ which satisfy each of the following equations.
The equation
has no solution; the equation
has solution _x_ = –1/2.
If _u_ = 9/16,
The required solutions are _x_ = –1/2, – 3/7, 3.
( _b_ ) ( _x_ 2 \+ 3 _x_ \+ 2)2 – 8( _x_ 2 \+ 3 _x_ ) = 4. Let _x_ 2 \+ 3 _x_ = _u_ ; then ( _u_ \+ 2)2 – 8 _u_ = 4 and _u_ = 0, 4.
If _u_ = 0, _x_ 2 \+ 3 _x_ = 0 and _x_ = 0, – 3; if _u_ = 4, _x_ 2 \+ 3 _x_ = 4 and _x_ = –4, 1.
The solutions are _x_ = – 4, –3, 0, 1.
**16.21** One positive number exceeds three times another positive number by 5. The product of the two numbers is 68. Find the numbers.
**SOLUTION**
Let _x_ = smaller number; then 3 _x_ \+ 5 = larger number.
Then _x_ (3 _x_ \+ 5) = 68, 3 _x_ 2 \+ 5 _x_ – 68 = 0, (3 _x_ \+ 17)( _x_ – 4) = 0, and _x_ = 4, – 17/3.
We exclude – 17/3 since the problem states that the numbers are positive.
The required numbers are _x_ = 4 and 3 _x_ \+ 5 = 17.
**16.22** When three times a certain number is added to twice its reciprocal, the result is 5. Find the number.
**SOLUTION**
Let _x_ = the required number and 1/ _x_ = its reciprocal.
Then 3 _x_ \+ 2(1/ _x_ ) = 5, 3 _x_ 2 – 5 _x_ \+ 2 = 0, (3 _x_ – 2)( _x_ – 1) = 0, and _x_ = 1, 2/3.
_Check_. For _x_ = 1, 3(1) + 2(1/1) = 5; for _x_ = 2/3, 3(2/3) + 2(3/2) = 5.
**16.23** Determine the dimensions of a rectangle having perimeter 50 feet and area 150 square feet.
**SOLUTION**
Sum of four sides = 50 ft; hence, sum of two adjacent sides = 25 ft (see Fig. 16-2). Let _x_ and 25 – _x_ be the lengths of two adjacent sides.
**Fig. 16-2**
The area is _x_ (25 – _x_ ) = 150; then _x_ 2 – 25 _x_ \+ 150 = 0, ( _x_ – 10)( _x_ – 15) = 0, and _x_ = 10, 15.
Then 25 – _x_ = 15, 10; and the rectangle has dimensions 10 ft by 15 ft.
**16.24** The hypotenuse of a right triangle is 34 inches. Find the lengths of the two legs if one leg is 14 inches longer than the other.
**SOLUTION**
Let _x_ and _x_ \+ 14 be the lengths of the legs (see Fig. 16-3).
**Fig. 16-3**
Then _x_ 2 \+ ( _x_ \+ 14)2 = (34)2, _x_ 2 \+ 14 _x_ – 480 = 0, ( _x_ \+ 30)( _x_ – 16) = 0, and _x_ = – 30, 16.
Since _x_ = – 30 has no physical significance, we have _x_ = 16 in. and _x_ \+ 14 = 30 in.
**16.25** A picture frame of uniform width has outer dimensions 12 in. by 15 in. Find the width of the frame ( _a_ ) if 88 square inches of picture show, ( _b_ ) if 100 square inches of picture show.
**SOLUTION**
Let _x_ = width of frame; then the dimensions of the picture are (15 – 2 _x_ ), (12 – 2 _x_ ) (see Fig. 16-4).
**Fig. 16-4**
( _a_ ) Area of picture = (15 – 2 _x_ )(12 – 2 _x_ ) = 88; then 2 _x_ 2 – 27 _x_ \+ 46 = 0, ( _x_ – 2)(2 _x_ – 23) = 0, and _x_ = 2, . Clearly, the width cannot be in. Hence the width of the frame is 2 in.
_Check_. The area of the picture is (15 – 4)(12 – 4) = 88 in.2, as given.
( _b_ ) Here (15 – 2 _x_ )(12 – 2 _x_ ) = 100, 2 _x_ 2 – 27 _x_ \+ 40 = 0 and, by the quadratic formula,
Reject _x_ = 11.8 in., which cannot be the width. The required width is 1.7 in.
**16.26** A pilot flies a distance of 600 miles. He could fly the same distance in 30 minutes less time by increasing his average speed by 40 mi/hr. Find his actual average speed.
**SOLUTION**
Let _x_ = actual average speed in mi/hr.
Time to fly 600 mi at _x_ mi/hr – time to fly 600 mi at ( _x_ \+ 40) mi/hr = hr.
Then
Solving, the required speed is _x_ = 200 mi/hr.
**16.27** A retailer bought a number of shirts for $180 and sold all but 6 at a profit of $2 per shirt. With the total amount received she could buy 30 more shirts than before. Find the cost per shirt.
**SOLUTION**
Let _x_ = cost per shirt in dollars; 180/ _x_ = number of shirts bought.
Then
Solving, _x_ = $3 per shirt.
**16.28** _A_ and _B_ working together can do a job in 10 days. It takes _A_ 5 days longer than _B_ to do the job when each works alone. How many days would it take each of them, working alone, to do the job?
**SOLUTION**
Let _n, n_ – 5 = number of days required by _A_ and _B_ respectively, working alone, to do the job.
In 1 day _A_ does 1/ _n_ of job and _B_ does 1/( _n_ – 5) of job. Thus in 10 days they do together
Then 10(2 _n_ – 5) = _n_ ( _n_ – 5), _n_ 2 – 25 _n_ \+ 50 = 0, and
Rejecting _n_ = 2.2, the required solution is _n_ = 22.8 days, _n_ – 5 = 17.8 days.
**16.29** A ball projected vertically upward with initial speed _v_ 0 ft/sec is at time _t_ sec at a distance _s_ ft from the point of projection as given by the formula _s_ = _v_ 0 _t_ – 16 _t_ 2. If the ball is given an initial upward speed of 128 ft/sec, at what times would it be 100 ft above the point of projection?
**SOLUTION**
At _t_ = 0.88 sec, _s_ = 100 ft and the ball is rising; at _t_ = 7.12 sec, _s_ = 100 ft and the ball is falling. This is seen from the graph of _s_ plotted against _t_ (see Fig. 16-5).
**Fig. 16-5**
### **Supplementary Problems**
**16.30** Solve each equation.
( _a_ ) _x_ 2 – 40 = 9
( _b_ ) 2 _x_ 2 – 400 = 0
( _c_ ) _x_ 2 \+ 36 = 9 – 2 _x_ 2
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
**16.31** Solve each equation by factoring.
( _a_ ) _x_ 2 – 7 _x_ = – 12
( _b_ ) _x_ 2 \+ _x_ = 6
( _c_ ) _x_ 2 = 5 _x_ \+ 24
( _d_ ) 2 _x_ 2 \+ 2 = 5 _x_
( _e_ ) 9 _x_ 2 = 9 _x_ – 2
( _f_ ) 4 _x_ – 5 _x_ 2 = –12
( _g_ )
( _h_ )
( _i_ )
( _j_ )
**16.32** Solve each equation by completing the square.
( _a_ ) _x_ 2 \+ 4 _x_ – 5 = 0
( _b_ ) _x_ ( _x_ – 3) = 4
( _c_ ) 2 _x_ 2 = _x_ \+ 1
( _d_ ) 3 _x_ 2 – 2 = 5 _x_
( _e_ ) 4 _x_ 2 = 12 _x_ – 7
( _f_ ) 6 _y_ 2 = 19 _y_ – 15
( _g_ ) 2 _x_ 2 \+ 3 _a_ 2 = 7 _ax_
( _h_ ) 12 _x_ – 9 _x_ 2 = 5
**16.33** Solve each equation by the quadratic formula.
( _a_ ) _x_ 2 – 5 _x_ = 6
( _b_ ) _x_ 2 – 6 _x_
( _c_ ) 3 _x_ 2 – 2 _x_
( _d_ ) 16 _x_ 2 – 8 _x_ \+ 1 = 0
( _e_ ) _x_ (5 _x_ – 4) = 2
( _f_ ) 9 _x_ 2 \+ 6 _x_ = –4
( _g_ )
( _h_ )
**16.34** Solve each equation graphically.
( _a_ ) 2 _x 2 \+ _x_ – 3 = 0_
( _b_ ) 4 _x_ 2 – 8 _x_ \+ 4 = 0
( _c_ ) _x_ 2 – 2 _x_ = 2
( _d_ ) 2 _x_ 2 \+ 2 = 3 _x_
( _e_ ) 6 _x_ 2 – 7 _x_ – 5 = 0
( _f_ ) 2 _x_ 2 \+ 8 _x_ \+ 3 = 0
**16.35** Without solving, find the sum _S_ and product _P_ of the roots of each equation.
( _a_ ) 2 _x_ 2 \+ 3 _x_ \+ 1 = 0
( _b_ ) _x_ – _x_ 2 = 2
( _c_ ) 2 _x_ ( _x_ \+ 3) = 1
( _d_ ) 2 _x_ 2 \+ 6 _x_ – 5 = 0
( _e_ ) 3 _x_ 2 – 4 = 0
( _f_ ) 4 _x_ 2 \+ 3 _x_ = 0
( _g_ ) 2 _x_ 2 \+ 5 _kx_ \+ 3 _k_ 2 = 0
( _h_ ) 0.2 _x_ 2 – 0.1 _x_ \+ 0.03 = 0
( _i_ )
**16.36** Find the discriminant _b_ 2 – 4 _ac_ and thus determine the character of the roots.
( _a_ ) 2 _x_ 2 – 7 _x_ \+ 4 = 0
( _b_ ) 3 _x_ 2 = 5 _x_ – 2
( _c_ ) 3 _x_ – _x_ 2 = 4
( _d_ ) _x_ (4 _x_ \+ 3) = 5
( _e_ ) 2 _x_ 2 = 5 + 3 _x_
( _f_ )
( _g_ ) 1 + 2 _x_ = 2 _x_ 2 = 0
( _h_ ) 3 _x_ \+ 25/3 _x_ = 10
**16.37** Find a quadratic equation with integer coefficients (if possible) having the given roots.
( _a_ ) 2, –3
( _b_ ) –3, 0
( _c_ ) 8, –4
( _d_ ) –2, –5
( _e_ ) –1/3, 1/2
( _f_ )
( _g_ ) –1 + _i_ , -1 – _i_
( _h_ )
( _i_ )
( _j_ ) – , \+
( _k_ ) _a_ \+ _bi_ , _a_ – _bi a, b_ integers
( _l_ )
**16.38** In each quadratic equation, evaluate the constant _p_ subject to the given condition.
( _a_ ) _px_ 2 – _x_ \+ 5 – 3 _p_ = 0 has one root equal to 2.
( _b_ ) (2 _p_ \+ 1) _x_ 2 \+ _px_ \+ _p_ = 4( _px_ \+ 2) has the sum of its roots equal to the product of its roots.
( _c_ ) 3 _x_ 2 \+ _p_ ( _x_ – 2) + 1 = 0 has roots which are reciprocals.
( _d_ ) 4 _x_ 2 – 8 _x_ \+ 2 _p_ – 1 = 0 has one root equal to three times the other.
( _e_ ) 4 _x_ 2 – 20 _x_ \+ _p_ 2 – 4 = 0 has one root equal to two more than the other.
( _f_ ) _x_ 2 = 5 _x_ – 3 _p_ \+ 3 has the difference between its roots equal to 11.
**16.39** Find the roots of each equation subject to the given condition.
( _a_ ) 2 _px_ 2 – 4 _px_ \+ 5 _p_ = 3 _x_ 2 \+ _x_ – 8 has the product of its roots equal to twice their sum.
( _b_ ) _x_ 2 – 3( _x_ – _p_ ) – 2 = 0 has one root equal to 3 less than twice the other root.
( _c_ ) _p_ ( _x_ 2 \+ 3 _x_ – 9) = _x_ – _x_ 2 has one root equal to the negative of the other.
( _d_ ) ( _m_ \+ 3) _x_ 2 \+ 2 _m_ ( _x_ \+ 1) + 3 = 0 has one root equal to half the reciprocal of the other.
( _e_ ) (2 _m_ \+ 1) _x_ 2 – 4 _mx_ = 1 – 3 _m_ has equal roots.
**16.40** Solve each equation.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
( _j_ )
( _k_ )
( _l_ )
**16.41** Solve the equation.
( _a_ ) _x_ 4 – 13 _x_ 2 \+ 36 = 0
( _b_ ) _x_ 4 – 3 _x_ 2 – 10 = 0
( _c_ ) 4 _x_ –4 – 17 _x_ –2 \+ 4 = 0
( _d_ ) _x_ –4/3 – 5 _x_ –2/3 \+ 4 = 0
( _e_ ) ( _x_ 2 – 6 _x_ )2 – 2( _x_ 2 – 6 _x_ ) = 35
( _f_ )
( _g_ )
( _h_ )
( _i_ ) _x_ 3 – 7 _x_ 3/2 – 8 = 0
( _j_ )
**16.42**
( _a_ ) The sum of the squares of two numbers is 34, the first number being one less than twice the second number. Determine the numbers.
( _b_ ) The sum of the squares of three consecutive integers is 110. Find the numbers.
( _c_ ) The difference between two positive numbers is 3, and the sum of their reciprocals is 1/2. Determine the numbers.
( _d_ ) A number exceeds twice its square root by 3. Find the number.
**16.43**
( _a_ ) The length of a rectangle is three times its width. If the width is diminished by 1 ft and the length increased by 3 ft, the area will be 72 ft2. Find the dimensions of the original rectangle.
( _b_ ) A piece of wire 60 in. long is bent into the form of a right triangle having hypotenuse 25 in. Find the other two sides of the triangle.
( _c_ ) A picture 8 in. by 12 in. is placed in a frame which has uniform width. If the area of the frame equals the area of the picture, find the width of the frame.
( _d_ ) A rectangular piece of tin 9 in. by 12 in. is to be made into an open box with base area 60 in.2 by cutting out equal squares from the four corners and then bending up the edges. Find to the nearest tenth of an inch the length of the side of the square cut from each corner.
**16.44**
( _a_ ) The tens digit of a certain two-digit number is twice the units digit. If the number is multiplied by the sum of its digits, the product is 63. Find the number.
( _b_ ) Find a number consisting of two digits such that the tens digit exceeds the units digit by 3 and the number is 4 less than the sum of the squares of its digits.
**16.45**
( _a_ ) Two men start at the same time from the same place and travel along roads that are at right angles to each other. One man travels 4 mi/hr faster than the other, and at the end of 2 hours they are 40 miles apart. Determine their rates of travel.
( _b_ ) By increasing her average speed by 10 mi/hr a motorist could save 36 minutes in traveling a distance of 120 miles. Find her actual average speed.
( _c_ ) A woman travels 36 miles down a river and back in 8 hours. If the rate of her boat in still water is 12 mi/hr, what is the rate of the river current?
**16.46**
( _a_ ) A merchant purchased a number of coats, each at the same price, for a total of $720. He sold them at $40 each, thus realizing a profit equal to his cost of 8 coats. How many did he buy?
( _b_ ) A grocer bought a number of cans of corn for $14.40. Later the price increased 2 cents a can and as a result she received 24 fewer cans for the same amount of money. How many cans were in his first purchase and what was the cost per can?
**16.47**
( _a_ ) It takes _B_ 6 hours longer than _A_ to assemble a machine. Together they can do it in 4 hr. How long would it take each working alone to do the job?
( _b_ ) Pipe _A_ can fill a given tank in 4 hr. If pipe _B_ works alone, it takes 3 hr longer to fill the tank than if pipes _A_ and _B_ act together. How long will it take pipe _B_ working alone?
**16.48** A ball projected vertically upward is distance _s_ ft from the point of projection after _t_ seconds, where _s_ = 64 _t_ – 16 _t_ 2.
( _a_ ) At what times will the ball be 40 ft above the ground?
( _b_ ) Will the ball ever be 80 ft above the ground?
( _c_ ) What is the maximum height reached?
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**16.30**
( _a_ ) _x_ = ±7
( _b_ ) _x_ = ± 10
( _c_ ) _x_ = ±3 _i_
( _d_ ) _x_ = ±8
( _e_ )
( _f_ ) _x_ = ±1
( _g_ ) _x_ = ±3/2
( _h_ ) _x_ = ±2
**16.31**
( _a_ ) 3, 4
( _b_ ) 2 –3
( _c_ ) 8 –3
( _d_ ) 2, 1/2
( _e_ ) 1/3, 2/3
( _f_ ) 2, –6/5
( _g_ ) 2 _a_ , –4 _a_
( _h_ ) 2, –8
( _i_ ) 1, –7
( _j_ ) 2 _c_ /5, 4 _c_ /5
**16.32**
( _a_ ) 1, –5
( _b_ ) 4, –1
( _c_ ) 1, –1/2
( _d_ ) 2, –1/3
( _e_ )
( _f_ ) 3/2, 5/3
( _g_ ) 3 _a_ , _a_ /2
( _h_ )
**16.33**
( _a_ ) 6, –1
( _b_ ) 3, –2
( _c_ ) 2, –4/3
( _d_ ) 1/4, 1/4
( _e_ )
( _f_ )
( _g_ )
( _h_ )
**16.34**
( _a_ ) _x_ = –3/2 and _x_ = 1 (see Fig. 16-6).
**Fig. 16-6**
( _b_ ) Double root of _x_ = 1 (see Fig. 16-7).
**Fig. 16-7**
( _c_ ) Real zeros between –1 and 0 and between 2 and 3 (see Fig. 16-8).
**Fig. 16-8**
( _d_ ) No real zeros (see Fig. 16-9).
**Fig. 16-9**
( _e_ ) Real zeros between – 1 and 0 and between 1 and 2 (see Fig. 16-10).
**Fig. 16-10**
( _f_ ) Real zeros between – 4 and – 3 and between – 1 and 0 (see Fig. 16-11).
**Fig. 16-11**
**16.35**
( _a_ ) _S_ = –3/2, _P_ = 1/2
( _b_ ) _S_ = 1, _P_ = 2
( _c_ ) _S_ = –3, _P_ = –1/2
( _d_ ) _S_ = –3, _P_ = –5/2
( _e_ ) _S_ = 0, _P_ = –4/3
( _f_ ) _S_ = –3/4, _P_ = 0
( _g_ ) _S_ = –5 _k_ /2, _P_ = 3 _k_ 2/ _2_
( _h_ ) _S_ = 0.5, _P_ = 0.15
( _i_ )
**16.36**
( _a_ ) 17; real, irrational, unequal
( _b_ ) 1; real, rational, unequal
( _c_ ) – 7; not real
( _d_ ) 89; real, irrational, unequal
( _e_ ) 49; real, rational, unequal
( _f_ ) 0; real, equal
( _g_ ) – 4; not real
( _h_ ) 0; real, rational, equal
**16.37**
( _a_ ) _x_ 2 \+ _x_ – 6 = 0
( _b_ ) _x_ 2 \+ 3 _x_ = 0
( _c_ ) _x_ 2 – 4 _x_ – 32 = 0
( _d_ ) _x_ 2 \+ 7 _x_ \+ 10 = 0
( _e_ ) 6 _x_ 2 – _x_ – 1 = 0
( _f_ ) _x_ 2 – 4 _x_ \+ 1 = 0
( _g_ ) _x_ 2 \+ 2 _x_ \+ 2 = 0
( _h_ ) _x_ 2 \+ 4 _x_ – 2 = 0
( _i_ ) 4 _x_ 2 – 16 _x_ \+ 25 = 0
( _j_ )
( _k_ ) _x_ 2 – 2 _ax_ \+ _a_ 2 \+ _b_ 2 = 0
( _l_ ) 4 _x_ 2 – 4 _mx_ \+ _m_ 2 – _n_ = 0
**16.38**
( _a_ ) _p_ = – 3
( _b_ ) _p_ = –4
( _c_ ) _p_ = –1
( _d_ ) _p_ = 2
( _e_ ) _p_ = ±5
( _f_ ) _p_ = –7
**16.39**
( _a_ ) 3, 6
( _b_ ) 1, 2
( _c_ ) ±3/2
( _d_ ) 1/2 ± _i_ /2
( _e_ ) If _m_ = – 1, the roots are 2, 2; if _m_ = 1/2, the roots are 1/2, 1/2.
**16.40**
( _a_ ) 2, – 1
( _b_ ) 1, 3
( _c_ ) 4/9
( _d_ ) – 4, 2
( _e_ ) 9, 1
( _f_ ) 8, – 2
( _g_ ) ±2
( _h_ ) 1
( _i_ ) 3/2
( _j_ ) no solution
( _k_ ) – 2
( _l_ ) 5/4
**16.41**
( _a_ ) ±2, ±3
( _b_ )
( _c_ ) ±2, ±1/2
( _d_ ) ±1, ±1/8
( _e_ ) 7, 5, ±1
( _f_ )
( _g_ )
( _h_ ) 79
( _i_ ) 4
( _j_ )
**16.42**
( _a_ ) 5, 3 or – 27/5, – 11/5
( _b_ ) 5, 6, 7 or – 7, – 6, – 5
( _c_ ) 3, 6
( _d_ ) 9
**16.43**
( _a_ ) 5, 15 ft
( _b_ ) 15, 20 in.
( _c_ ) 2 in.
( _d_ ) 1.3 in.
**16.44**
( _a_ ) 21
( _b_ ) 85
**16.45**
( _a_ ) 12, 16 mi/hr
( _b_ ) 40 mi/hr
( _c_ ) 6 mi/hr
**16.46**
( _a_ ) 24
( _b_ ) 144, 10¢
**16.47**
( _a_ ) _A_ , 6 hr; _B_ , 12 hr
( _b_ ) 5.3 hr approx.
**16.48**
( _a_ ) 0.78 and 3.22 seconds after projection
( _b_ ) No
( _c_ ) 64 ft
## **CHAPTER 17
Conic Sections**
### **17.1 GENERAL QUADRATIC EQUATIONS**
The general quadratic equation in the two variables _x_ and _y_ has the form
where _a, b, c, d, e, f_ are given constants and _a, b, c_ are not all zero.
Thus 3 _x_ 2 \+ 5 _xy_ = 2, _x_ 2 – _xy_ \+ _y_ 2 \+ 2 _x_ \+ 3 _y_ = 0, _y_ 2 = 4 _x_ , _xy_ = 4 are quadratic equations in _x_ and _y_.
The graph of equation (1), if _a, b, c, d, e, f_ are real, depends on the value of _b_ 2 – 4 _ac_.
(1) If _b_ 2 – 4 _ac_ < 0, the graph is in general an ellipse. However, if _b_ = 0 and _a_ = _c_ the graph may be a circle, a point, or non-existent. The point and non-existent situations are called the degenerate cases.
(2) If _b_ 2 – 4 _ac_ = 0, the graph is a parabola, two parallel or coincident lines, or non-existent. The parallel or coincident lines and non-existent situations are called the degenerate cases.
(3) If _b_ 2 – 4 _ac_ > 0, the graph is a hyperbola or two intersecting lines. The two intersecting lines situation is called the degenerate case.
These graphs are the intersections of a plane and a right circular cone, and for this reason are called conic sections.
**EXAMPLES 17.1.** Identify the type of conic section described by each equation.
( _a_ ) _x_ 2 \+ _xy_ = 6
( _b_ ) _x_ 2 \+ 5 _xy_ – 4 _y_ 2 = 10
( _c_ ) 2 _x_ 2 – _y_ 2 = 7
( _d_ ) 3 _x_ 2 \+ 2 _y_ 2 = 14
( _e_ ) 3 _x_ 2 \+ 3 _y_ 2 – 4 _x_ \+ 3 _y_ \+ 10 = 0
( _f_ ) y2 \+ 4 _x_ \+ 3 _y_ \+ 4 = 0
### **17.2 CONIC SECTIONS**
Each conic section is the locus (set) of all points in a plane meeting a given set of conditions. The set of points can be described by an equation. When the locus is positioned at the origin, the figure is called a central conic section. A general equation used to describe a conic section is called the standard equation, which may have more than one form for a conic section. The conic sections are the circle, parabola, ellipse, and hyperbola. We will consider only conic sections in which _b_ = 0, thus having the general quadratic equation _Ax_ 2 \+ _Cy_ 2 \+ _Dx_ \+ _Ey_ \+ _F_ = 0. Trigonometry is needed to discuss fully the general quadratic equations in which _b_ ≠ 0.
### **17.3 CIRCLES**
A circle is the locus of all points in a plane which are at a fixed distance from a fixed point in the plane. The fixed point is the center of the circle and the fixed distance is the radius of the circle.
When the center of the circle is the origin, (0, 0), and the radius is _r_ , the standard form of the equation of a circle is _x_ 2 \+ _y_ 2 = _r_ 2. If the center of the circle is the point ( _h, k_ ) and the radius is _r_ , the standard form of the equation of a circle is ( _x_ – _h_ )2 \+ ( _y_ – _k_ )2 = _r_ 2. If _r_ 2 = 0, we have the degenerate case of a single point which is sometimes called the point circle. If _r_ 2 < 0, we have the non-existent degenerate case, which is sometimes called the imaginary circle, since the radius would have to be an imaginary number.
The graph of the circle ( _x_ – 2)2 \+ ( _y_ \+ 3)2 = 9 has its center at (2, – 3) and a radius of 3 (see Fig. 17-1).
**Fig. 17-1**
**EXAMPLES 17.2.** For each circle state the center and radius.
( _a_ ) _x_ 2 \+ _y_ 2 = 5
( _b_ ) _x_ 2 \+ _y_ 2 = 28
( _c_ ) ( _x_ \+ 2)2 \+ ( _y_ – 4)2 = 81
**EXAMPLES 17.3.** Write the equation of each circle in standard form.
( _a_ ) _x_ 2 \+ _y_ 2 – 8 _x_ \+ 12 _y_ – 48 = 0
( _b_ ) _x_ 2 \+ _y_ 2 – 4 _x_ \+ 6 _y_ \+ 100 = 0
_Note:_ In (1) _r_ 2 = 100, so we have a circle, but in (2) _r_ 2 = –87 so we have the degenerate case.
**EXAMPLE 17.4.** Write the equation of the circle going through the points _P_ (2, –1), _Q_ (–3, 0), and _R_ (1, 4).
By substituting the points _P, Q_ , and _R_ into the general form of a circle, _x_ 2 \+ _y_ 2 \+ _Dx_ \+ _Ey_ \+ _F_ = 0, we get a system of three linear equations.
Eliminating _F_ from (1) and (2) and from (1) and (3), we get
Solving (4) and (5) we get _D_ = 1/3 and _E_ = –7/3, and by substituting _D_ and _E_ in (1) we get _F_ = –8.
The equation of the circle is _x_ 2 \+ _y_ 2 \+ 1/3 _x_ – 7/3 _y_ – 8 = 0 or 3 _x_ 2 \+ 3 _y_ 2 \+ _x_ – 7 _y_ – 24 = 0.
### **17.4 PARABOLAS**
A parabola is the locus of all points in a plane equidistant from a fixed line, the directrix, and a fixed point, the focus.
Central parabolas have their vertex at the origin, focus on one axis, and directrix parallel to the other axis. We denote the distance from the focus to the vertex by | _p_ |. The distance from the directrix to the vertex is also | _p_ |. The equations of the central parabolas are (1) and (2) below.
In (1) the focus is on the _x_ axis and the directrix is parallel to the other axis. If _p_ is positive, the curve opens to the right and if _p_ is negative, the curve opens to the left (see Fig. 17-2). In (2) the focus is on the _y_ axis and the directrix is parallel to the _x_ axis. If _p_ is positive, the curve opens up and if _p_ is negative, the curve opens down (see Fig. 17-3).
**Fig. 17-2**
**Fig. 17-3**
The line through the vertex and the focus is the axis of the parabola and the graph is symmetric with respect to this line.
The parabolas with vertex at the point ( _h, k_ ) and with the axis and the directrix parallel to the _x_ axis and the _y_ axis have the standard forms listed in (3) and (4) below.
In (3) the focus is _F_ ( _h_ \+ _p_ , _k_ ), the directrix is _x_ = _h_ – _p_ , and the axis is _y_ = _k_ (see Fig. 17-4). However, in (4) the focus is _F_ ( _h_ , _k_ \+ _p_ ), the directrix is _y_ = _k_ – _p_ , and the axis is _x_ = _h_ (see Fig. 17-5).
**Fig. 17-4**
**Fig. 17-5**
**EXAMPLES 17.5.** Determine the vertex, focus, directrix, and axis for each parabola.
( _a_ ) _y_ 2 = –8 _x_
( _b_ ) _x_ 2 = 6 _y_
( _c_ ) ( _y_ – 3)2 = 5( _x_ \+ 7)
( _d_ ) ( _x_ – 1)2 = –4( _y_ \+ 4)
( _a_ ) _y_ 2 = – 8 _x_ : vertex ( _h, k_ ) = (0, 0), 4 _p_ = –8, so _p_ = –2, focus ( _p_ , 0) = (–2, 0), and directrix is _x_ = – _p_ , so _x_ = –(–2) = 2, axis is _y_ = 0
( _b_ ) _x_ 2 = 6 _y_ : vertex ( _h, k_ ) = (0, 0), 4 _p_ = 6, so _p_ = 3/2, focus (0, _p_ ) = (0, 3/2), and directrix is _y_ = – _p_ , so _y_ = – 3/2, axis is _y_ = 0
( _c_ ) ( _y_ – 3)2 = 5( _x_ \+ 7): vertex ( _h, k_ ) = (–7, 3), 4 _p_ = 5, so _p_ = 5/4, focus ( _h_ \+ _p_ , _k_ ) = (–7 + 5/4, 3) = (–23/4, 3), and directrix is _x_ = _h_ – _p_ , so _x_ = –7 – 5/4 = –33/4, axis _y_ = _k_ , so _y_ = 3
( _d_ ) ( _x_ – 1)2 = –4( _y_ \+ 4): vertex ( _h_ , _k_ ) = (1, –4), 4 _p_ = –4, so _p_ = –1, focus ( _h_ , _k_ \+ _p_ ) = (1, –4 + (–1)) = (1, –5), and directrix is _y_ = _k_ – _p_ , so _y_ = –4 – (–1) = –3, axis is _x_ = _h_ , so _x_ = 1
**EXAMPLES 17.6.** Write the equation of the parabola with the given characteristics.
( _a_ ) vertex (4, 6) and focus (4, 8)
( _b_ ) focus (3, 5) and directrix _y_ = 3
( _a_ ) Since the vertex (4, 6) and the focus (4, 8) lie on the line _x_ = 4 (see Fig. 17-4), we have a parabola of the form ( _x_ – _h_ )2 = 4 _p_ ( _y_ – _k_ ).
Since the vertex is (4, 6), we have _h_ = 4 and _k_ = 6.
The focus is ( _h_ , _k_ \+ _p_ ), so _k_ \+ _p_ = 8 and 6 + _p_ = 8, so _p_ = 2.
The equation of the parabola is ( _x_ – 4)2 = 8( _y_ – 6).
( _b_ ) Since the directrix is _y_ = 3 (see Fig. 17-5), the parabola has the form ( _x_ – _h_ )2 = 4 _p_ ( _y_ – _k_ ).
The focus (3, 5) is 2 units above the directrix _y_ = 3, so _p_ > 0. The distance from the focus to the directrix is 2| _p_ |, so 2 _p_ = 2 and _p_ = 1.
The focus is ( _h_ , _p_ \+ _k_ ), so _h_ = 3 and _k_ \+ _p_ = 5. Since _p_ = 1, _k_ = 4.
The equation of the parabola is ( _x_ – 3)2 = 4( _y_ – 4).
**EXAMPLES 17.7.** Write the equation of each parabola in standard form.
( _a_ ) _x_ 2 – 4 _x_ – 12 _y_ – 32 = 0
( _b_ ) _y_ 2 \+ 3 _x_ – 6 _y_ = 0
### **17.5 ELLIPSES**
An ellipse is the locus of all points in a plane such that the sum of the distances from two fixed points, the foci, to any point on the locus is a constant.
Central ellipses have their center at the origin, vertices and foci lie on one axis, and the covertices lie on the other axis. We will denote the distance from a vertex to the center by _a_ , the distance from a covertex to the center by _b_ , and the distance from a focus to the center by _c_. For an ellipse, the values _a, b_ , and _c_ are related by _a_ 2 = _b_ 2 \+ _c_ 2 and _a_ > _b_. We call the line segment between the vertices the major axis and the line segment between the covertices the minor axis.
The standard forms for the central ellipses are:
The larger denominator is always _a_ 2 for an ellipse. If the numerator for _a_ 2 is _x_ 2, then the major axis lies on the _x_ axis. In (1) the vertices have coordinates _V_ ( _a_ , 0) and _V_ ′(– _a_ , 0), the foci have coordinates _F_ ( _c_ , 0) and _F_ ′(– _c_ , 0), and the covertices have coordinates _B_ (0, _b_ ) and _B_ ′(0, – _b_ ) (see Fig. 17-6). If the numerator for _a_ 2 is _y_ 2, then the major axis lies on the _y_ axis. In (2) the vertices are at _V_ (0, _a_ ) and _V_ ′(0 – _a_ ), the foci are at _F_ (0, _c_ ) and _F_ ′(0, – _c_ ), and the covertices are at _B_ ( _b_ , 0) and _B_ ′(– _b_ , 0) (see Fig. 17-7).
**Fig. 17-6**
**Fig. 17-7**
If the center of an ellipse is _C_ ( _h_ , _k_ ) then the standard forms for the ellipses are:
In (3) the major axis is parallel to the _x_ axis and the minor axis is parallel to the _y_ axis. The foci have coordinates _F_ ( _h_ \+ _c_ , _k_ ) and _F_ ′( _h_ – _c_ , _k_ ), the vertices are at _V_ ( _h_ \+ _a_ , _k_ ) and _V_ ′( _h_ – _a_ , _k_ ), and the covertices are at _B_ ( _h_ , _k_ \+ _b_ ) and _B_ ′( _h_ , _k_ – _b_ ) (see Fig. 17-8). In (4) the major axis is parallel to the _y_ axis and the minor axis is parallel to the _x_ axis. The foci are at _F_ ( _h_ , _k_ \+ _c_ ) and _F_ ′( _h_ , _k_ – _c_ ), the vertices have coordinates _V_ ( _h_ , _k_ \+ _a_ ) and _V_ ′( _h_ , _k_ – _a_ ), and the covertices are at _B_ ( _h_ \+ _b_ , _k_ ) and _B_ ′( _h_ – _b_ , _k_ ) (see Fig. 17-9).
**Fig. 17-8**
**Fig. 17-9**
**EXAMPLES 17.8.** Determine the center, foci, vertices, and covertices for each ellipse.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _a_ )
Since _a_ 2 is the greater denominator, _a_ 2 = 25 and _b_ 2 = 9, so _a_ = 5 and _b_ = 3. From _a_ 2 = _b_ 2 \+ _c_ 2, we get 25 = 9 + _c_ 2 and _c_ = 4. The center is at (0, 0). The vertices are at ( _a_ , 0) and (– _a_ , 0), so _V_ (5, 0) and _V_ ′(–5, 0). The foci are at ( _c_ , 0) and (– _c_ , 0), so _F_ (4, 0) and _F_ (–4, 0). The covertices are at (0, _b_ ) and (0, – _b_ ), so _B_ (0, 3) and _B_ ′(0, –3).
( _b_ )
( _c_ )
( _d_ )
**EXAMPLES 17.9.** Write the equation of the ellipse having the given characteristics.
( _a_ ) central ellipse, foci at (±4, 0), and vertices at (±5, 0)
( _b_ ) center at (0, 3), major axis of length 12, foci at (0, 6) and (0, 0)
( _a_ ) A central ellipse has its center at the origin, so ( _h, k_ ) = (0, 0). Since the vertices are on the _x_ axis and the center is at (0, 0), the form of the ellipse is
( _b_ ) Since the center is at (0, 3), _h_ = 0 and _k_ = 3. Since the foci are on the _y_ axis, the form of the equation of the ellipse is
The foci are ( _h_ , _k_ \+ _c_ ) and ( _h_ , _k_ – _c_ ), so (0, 6) = ( _h_ , _k_ \+ _c_ ) and 3 + _c_ = 6 and _c_ = 3.
The major axis length is 12, so we know 2 _a_ = 12 and _a_ = 6.
**EXAMPLE 17.10.** Write the equation of the ellipse 18 _x_ 2 \+ 12 _y_ 2 – 144 _x_ \+ 48 _y_ \+ 120 = 0 in standard form.
### **17.6 HYPERBOLAS**
The hyperbola is the locus of all points in a plane such that for any point of the locus the difference of the distances from two fixed points, the foci, is a constant.
Central hyperbolas have their center at the origin and their vertices and foci on one axis, and are symmetric with respect to the other axis. The standard form equations for central hyperbolas are:
The distance from the center to a vertex is denoted by _a_ and the distance from the center to a focus is _c_. For a hyperbola, _c_ 2 = _a_ 2 \+ _b_ 2 and _b_ is a positive number. The line segment between the vertices is called the transverse axis. The denominator of the positive fraction for the standard form is always _a_ 2.
In (1) the transverse axis lies on the _x_ axis, the vertices are _V_ ( _a_ , 0) and _V_ ′(– _a_ , 0), and the foci are at _F_ ( _c_ , 0) and _F_ ′(– _c_ , 0) (see Fig. 17-10). In (2) the transverse axis lies on the _y_ axis, the vertices are at _V_ (0, _a_ ) and _V_ (0, – _a_ ), and the foci are at _F_ (0, _c_ ) and _F_ ′(0, – _c_ ) (see Fig. 17-11). When lines are drawn through the points _R_ and _C_ and the points _S_ and _C_ , we have the asymptotes of the hyperbola. The asymptote is a line that the graph of the hyperbola approaches but does not reach.
If the center of the hyperbola is at ( _h, k_ ) the standard forms are (3) and (4):
In (3) the transverse axis is parallel to the _x_ axis, the vertices have coordinates _V_ ( _h_ \+ _a_ , _k_ ) and _V_ ′( _h_ – _a_ , _k_ ), the foci have coordinates _F_ ( _h_ \+ _c_ , _k_ ) and _F_ ′( _h_ – _c_ , _k_ ), and the points _R_ and _S_ have coordinates _R_ ( _h_ \+ _a_ , _k_ \+ _b_ ) and _S_ ( _h_ \+ _a_ , _k_ – _b_ ). The lines through _R_ and _C_ and _S_ and _C_ are the asymptotes of the hyperbola (see Fig. 17-12). In equation (4) the transverse axis is parallel to the _y_ axis, the vertices are at _V_ ( _h_ , _k_ \+ _a_ ) and _V_ ′( _h_ , _k_ – _a_ ), the foci are at _F_ ( _h_ , _k_ \+ _c_ ) and _F_ ′( _h_ , _k_ – _c_ ), and the points _R_ and _S_ have coordinates _R_ ( _h_ \+ _b_ , _k_ \+ _a_ ) and _S_ ( _h_ – _b_ , _k_ \+ _a_ ) (see Fig. 17-13).
**EXAMPLES 17.11.** Find the coordinates of the center, vertices, and foci for each hyperbola.
( _a_ )
( _b_ )
( _c_ )
**EXAMPLES 17.12.** Write the equation of the hyperbola that has the given characteristics.
( _a_ ) Foci are at (2, 5) and (–4, 5) and transverse axis has length 4.
( _b_ ) Center at (1, –3), a focus is at (1, 2) and a vertex is at (1, 1).
( _a_ ) The foci are on a line parallel to the _x_ axis, so the form is
The center is half-way between the foci, so _c_ = 3 and the center is at _C_ (–1, 5).
The transverse axis joins the vertices, so its length is 2 _a_ , so 2 _a_ = 4 and _a_ = 2.
Since _c_ 2 = _a_ 2 \+ _b_ 2, _c_ = 3 and _a_ = 2, so _b_ 2 = 5.
The equation of the hyperbola is
( _b_ ) The distance from the vertex (1, 1) to the center (1, –3) is _a_ , so _a_ = 4.
The distance from the focus (1, 2) to the center (1, –3) is _c_ , so _c_ = 5.
Since _c_ 2 = _a_ 2 \+ _b_ 2, _a_ = 4, and _c_ = 5, _b_ 2 = 9.
Since the center, vertex, and focus lie on a line parallel to the _y_ axis, the hyperbola has the form
The center is (1, –3), so _h_ = 1, and _k_ = –3.
The equation of the hyperbola is
**EXAMPLES 17.13.** Write the equation of each hyperbola in standard form.
( _a_ ) 25 _x_ 2 – 9 _y_ 2 – 100 _x_ – 72 _y_ – 269 = 0
( _b_ ) 4 _x_ 2 – 9 _y_ 2 – 24 _x_ – 90 _y_ – 153 = 0
### **17.7 GRAPHING CONIC SECTIONS WITH A CALCULATOR**
Since most conic sections are not functions, an important step is to solve the standard form equation for _y_. If _y_ is equal to an expression in _x_ that contains a ± quantity, we need to separate the expression into two parts: _y_ 1 = the expression using the + quantity and _y_ 2 = the expression using the – expression. Otherwise, set _y_ 1 = the expression. Graph either _y_ 1 or _y_ 1 and _y_ 2 simultaneously. The window may need to be adjusted to correct for the distortion caused by unequal scales used on the _x_ axis and the _y_ axis in many graphing calculators' standard windows. Setting the _y_ scale to 0.67 often corrects for this distortion.
For the circle, ellipse, and hyperbola, it is usually necessary to center the graphing window at the point ( _h, k_ ), the center of the conic section. However, the parabola is viewed better if the vertex ( _h, k_ ) is at one end of the viewing window.
### **Solved Problems**
**17.1** Draw the graph of each of the following equations:
( _a_ ) 4 _x_ 2 \+ 9 _y_ 2 = 36,
( _b_ ) 4 _x_ 2 – 9 _y_ 2 = 36,
( _c_ ) 4 _x_ \+ 9 _y_ 2 = 36.
**SOLUTION**
( _a_ )
Note that _y_ is real when 9 – _x_ 2 ≥ 0, i.e., when –3 ≤ _x_ ≤ 3. Hence values of _x_ greater than 3 or less than –3 are excluded.
The graph is an ellipse with center at the origin (see Fig. 17-14(a)).
**Fig. 17-14**
( _b_ )
Note that _x_ cannot have a value between –3 and 3 if _y_ is to be real.
The graph consists of two branches and is called a hyperbola (see Fig. 17-14( _b_ )).
( _c_ )
Note that if _x_ is greater than 9, _y_ is imaginary.
The graph is a parabola (see Fig. 17-14(c)).
**17.2** Plot the graph of each of the following equations:
( _a_ ) _xy_ = 8,
( _b_ ) 2 _x_ 2 – 3 _xy_ \+ _y_ 2 \+ _x_ – 2 _y_ – 3 = 0,
( _c_ ) _x_ 2 \+ _y_ 2 – 4 _x_ \+ 8 _y_ \+ 25 = 0.
**SOLUTION**
( _a_ ) _xy_ = 8, _y_ = 8/ _x_. Note that if _x_ is any real number except zero, _y_ is real. The graph is a hyperbola (see Fig. 17-15( _a_ )).
**Fig. 17-15**
( _b_ ) 2 _x_ 2 – 3 _xy_ \+ _y_ 2 \+ _x_ – 2 _y_ – 3 = 0. Write as _y_ 2 – (3 _x_ \+ 2) _y_ \+ (2 _x_ 2 \+ _x_ – 3) = 0 and solve by the quadratic formula to obtain
The given equation is equivalent to two linear equations, as can be seen by writing the given equation as (2 _x_ – _y_ \+ 3)( _x_ – _y_ – 1) = 0. The graph consists of two intersecting lines (see Fig. 17-15( _b_ )).
( _c_ ) Write as _y_ 2 \+ 8 _y_ \+ ( _x_ 2 – 4 _x_ \+ 25) = 0; solving,
Since _x_ 2 – 4 _x_ \+ 9 = _x_ 2 – 4 _x_ \+ 4 + 5 = ( _x_ – 2)2 \+ 5 is always positive, the quantity under the radical sign is negative. Thus _y_ is imaginary for all real values of _x_ and the graph does not exist.
**17.3** For each equation of a circle, write it in standard form and determine the center and radius.
( _a_ ) _x_ 2 \+ _y_ 2 – 8 _x_ \+ 10 _y_ – 4 = 0
( _b_ ) 4 _x_ 2 \+ 4 _y_ 2 \+ 28 _y_ \+ 13 = 0
**SOLUTION**
( _a_ )
( _b_ )
**17.4** Write the equation of the following circles.
( _a_ ) center at the origin and goes through (2, 6)
( _b_ ) ends of diameter at (–7, 2) and (5, 4)
**SOLUTION**
( _a_ ) The standard form of a circle with center at the origin is _x_ 2 \+ _y_ 2 = _r_ 2. Since the circle goes through (2, 6), we substitute _x_ = 2 and _y_ = 6 to determine _r_ 2. Thus, _r_ 2 = 22 \+ 62 = 40. The standard form of the circle is _x_ 2 \+ _y_ 2 = 40.
( _b_ ) The center of a circle is the midpoint of the diameter. The midpoint _M_ of the line segment having endpoints ( _x_ 1, _y_ 1) and ( _x_ 2, _y_ 2) is
Thus, the center is
The radius of a circle is the distance from the center to the endpoint of the diameter. The distance, _d_ , between two points ( _x_ 1, _y_ 1) and ( _x_ 2, _y_ 2) is Thus, the distance from the center _C_ (–1, 3) to (5, 4) is
The equation of the circle is ( _x_ \+ 1)2 \+ ( _y_ – 3)2 = 37.
**17.5** Write the equation of the circle passing through three points (3, 2), (–1, 4), and (2, 3).
**SOLUTION**
The general form of the equation of a circle is _x_ 2 \+ _y_ 2 \+ _Dx_ \+ _Ey_ \+ _F_ = 0, so we must substitute the given points into this equation to get a system of equations in _D_ , _E_ , and _F_.
We eliminate _F_ from (1) and (2) and from (1) and (3) to get
We solve the system of (4) and (5) to get _D_ = 2 and _E_ = 2 and substituting into (1) we get _F_ = –23.
The equation of the circle is _x_ 2 \+ _y_ 2 \+ 2 _x_ \+ 2 _y_ \- 23 = 0.
**17.6** Write the equation of the parabola in standard form and determine the vertex, focus, directrix, and axis.
( _a_ ) _y_ 2 \- 4 _x_ \+ 10 _y_ \+ 13 = 0
( _b_ ) 3 _x_ 2 \+ 18 _x_ \+ 11 _y_ \+ 5 = 0.
**SOLUTION**
( _a_ )
( _b_ )
**17.7** Write the equation of the parabola with the given characteristics.
( _a_ ) vertex at origin and directrix _y_ = 2
( _b_ ) vertex (–1, –3) and focus (–3, –3)
**SOLUTION**
( _a_ ) Since the vertex is at the origin, we have the form _y_ 2 = 4 _px_ or _x_ 2 = 4 _py_. However, since the directrix is _y_ = 2, the form is _x_ 2 = 4 _py_.
The vertex is (0, 0) and the directrix is _y_ = _k_ – _p_. Since _y_ = 2 and _k_ = 0, we have _p_ = –2.
The equation of the parabola is _x_ 2 = –8 _y_.
( _b_ ) The vertex is (–1, –3) and the focus is (–3, –3) and since they lie on a line parallel to the _x_ axis, the standard form is ( _y_ – _k_ )2 = 4 _p_ ( _x_ – _h_ ).
From the vertex we get _h_ = –1 and _k_ = –3, and since the focus is ( _h_ \+ _p_ , _k_ ), _h_ \+ _p_ = –3 and –1 + _p_ = –3, we get _p_ = –2.
Thus, the standard form of the parabola is ( _y_ \+ 3)2 = –8( _x_ \+ 1).
**17.8** Write the equation of the ellipse in standard form and determine its center, vertices, foci, and covertices.
( _a_ ) 64 _x_ 2 \+ 81 _y_ 2 = 64
( _b_ ) 9 _x_ 2 \+ 5 _y_ 2 \+ 36 _x_ \+ 10 _y_ – 4 = 0
**SOLUTION**
( _a_ )
( _b_ )
**17.9** Write the equation of the ellipse that has these characteristics.
( _a_ ) foci are (1, 0) and (–1, 0) and length of minor axis is 2
( _b_ ) vertices are at (5, – 1) and (–3, –1) and _c_ = 3.
**SOLUTION**
( _a_ ) The midpoint of the line segment between the foci is the center, so the center is _C_ (0, 0) and we have a central ellipse. The standard form is
The foci are ( _c_ , 0) and (– _c_ , 0) so ( _c_ , 0) = (1, 0) and _c_ = 1.
The minor axis has length 2 , so 2 _b_ = 2 and _b_ = and _b_ 2 = 2.
For the ellipse, _a_ 2 = _b_ 2 \+ _c_ 2 and _a_ 2 = 1 + 2 = 3.
Since the foci are on the _x_ axis, the standard form is
The equation of the ellipse is
( _b_ ) The midpoint of the line segment between the vertices is the center, so the center is
We have an ellipse with center at ( _h_ , _k_ ) where _h_ = 1 and _k_ = –1.
The standard form of the ellipse is
The vertices are ( _h_ \+ _a_ , _k_ ) and ( _h_ – _a_ , _k_ ), so ( _h_ \+ _a_ , _k_ ) = (1 + _a_ , – 1) = (5, – 1). Thus, 1 + _a_ = 5 and _a_ = 4.
For the ellipse, _a_ 2 = _b_ 2 \+ _c_ 2, _c_ is given to be 3, and we found _a_ to be 4. Thus, _a_ 2 = 42 = 16 and _c_ 2 = 32 = 9. Therefore, _a_ 2 = _b_ 2 \+ _c_ 2 yields 16 = _b_ 2 \+ 9 and _b_ 2 = 7.
Since the vertices are on a line parallel to the x axis, the standard form is
The equation of the ellipse is
**17.10** For each hyperbola, write the equation in standard form and determine the center, vertices, and foci.
( _a_ ) 16 _x_ 2 – 9 _y_ 2 \+ 144 = 0
( _b_ ) 9 _x_ 2 – 16 _y_ 2 \+ 90 _x_ \+ 64 _y_ \+ 17 = 0
**SOLUTION**
( _a_ )
center ( _h, k_ ) = (0, 0) _a_ 2 = 16 and _b_ 2 = 9, so _a_ = 4 and _b_ = 3
Since _c_ 2 = _a_ 2 \+ _b_ 2 for a hyperbola, _c_ 2 = 16 + 9 = 25 and _c_ = 5.
The foci are (0, _c_ ) and (0, – _c_ ), so _F_ (0, 5) and _F_ '(0, –5).
The vertices are (0, _a_ ) and (0, – _a_ ), so _V_ (0, 4) and _V'_ (0, –4)
( _b_ )
center ( _h_ , k) = (–5, 2) _a_ 2 = 16 and _b_ 2 = 9, so _a_ = 4 and _b_ = 3
Since _c_ 2 = _a_ 2 \+ _b_ 2, _c_ 2 = 16 + 9 = 25 and _c_ = 5.
The foci are ( _h_ \+ _c_ , _k_ ) and ( _h_ – _c_ , _k_ ), so _F_ (0, 2) and _F'_ (–10, 2).
The vertices are ( _h_ \+ _a_ , _k_ ), and ( _h_ – _a_ , _k_ ) so _V_ (–1, 2), and _V_ '(–9, 2).
**17.11** Write the equation of the hyperbola with the given characteristics.
( _a_ ) vertices are (0, ± 2) and foci are (0, ± 3)
( _b_ ) foci (1, 2) and (–11, 2) and the transverse axis has length 4
**SOLUTION**
( _a_ ) Since the vertices are (0, ±2), the center is at (0, 0), and since they are on a vertical line the standard form is
The vertices are at (0, + _a_ ) so _a_ = 2 and the foci are at (0, + 3) so _c_ = 3.
Since _c_ 2 = _a_ 2 \+ _b_ 2, 9 = 4 + _b_ 2 so _b_ 2 = 5.
The equation of the hyperbola is
( _b_ ) Since the foci are (1, 2) and (–11, 2), they are on a line parallel to the _x_ axis, so the form is
The midpoint of the line segment between the foci (1, 2) and (– 11, 2) is the center, so _C_ ( _h_ , _k_ ) = (–5, 2). The foci are at ( _h_ \+ _c_ , _k_ ) and ( _h_ – _c_ , _k_ ), so ( _h_ \+ _c_ , _k_ ) = (1, 2) and –5 + _c_ = 1, with _c_ = 6. The transverse axis has length 4 so 2 _a_ = 4 and _a_ = 2. From _c_ 2 = _a_ 2 \+ _b_ 2, we get 36 = 4 + _b_ 2 and _b_ 2 = 32.
The equation of the hyperbola is
### **Supplementary Problems**
**17.12** Graph each of the following equations.
( _a_ ) _x_ 2 \+ _y_ 2 = 9
( _b_ ) _xy_ = – 4
( _c_ ) 4 _x_ 2 \+ _y_ 2 = 16
( _d_ ) _x_ 2 – 4 _y_ 2 = 36
( _e_ ) _y_ 2 = 4 _x_
( _f_ ) _x_ 2 \+ 3 _y_ 2 – 1 = 0
( _g_ ) _x_ 2 \+ 3 _xy_ \+ _y_ 2 = 16
( _h_ ) _x_ 2 \+ 4 _y_ = 4
( _i_ ) _x_ 2 \+ _y_ 2 – 2 _x_ \+ 2 _y_ \+ 2 = 0
( _j_ ) 2 _x_ 2 – _xy_ – _y_ 2 – 7 _x_ – 2 _y_ \+ 3 = 0
**17.13** Write the equation of the circle that has the given characteristics.
( _a_ ) center (4, 1) and radius 3
( _b_ ) center (5, –3) and radius 6
( _c_ ) goes through (0, 0), (– 4, 0), and (0, 6)
( _d_ ) goes through (2, 3), (–1, 7), and (1, 5)
**17.14** Write the equation of the circle in standard form and state the center and radius.
( _a_ ) _x_ 2 \+ _y_ 2 \+ 6 _x_ – 12 _y_ – 20 = 0
( _b_ ) _x_ 2 \+ _y_ 2 \+ 12 _x_ – 4 _y_ – 5 = 0
( _c_ ) _x_ 2 \+ _y_ 2 \+ 7 _x_ \+ 3 _y_ – 10 = 0
( _d_ ) 2 _x_ 2 \+ 2 _y_ 2 – 5 _x_ – 9 _y_ \+ 11 = 0
**17.15** Write the equation of the parabola that has the given characteristics.
( _a_ ) vertex (3, –2) and directrix x = –5
( _b_ ) vertex (3, 5) and focus (3, 10)
( _c_ ) passes through (5, 10), vertex is at the origin, and axis is the x axis
( _d_ ) vertex (5, 4) and focus (2, 4)
**17.16** Write the equation of the parabola in standard form and determine its vertex, focus, directrix, and axis.
( _a_ ) _y_ 2 \+ 4 _x_ – 8 _y_ \+ 28 = 0
( _b_ ) _x_ 2 – 4 _x_ \+ 8 _y_ \+ 36 = 0
( _c_ ) _y_ 2 – 24 _x_ \+ 6 _y_ – 15 = 0
( _d_ ) 5 _x_ 2 \+ 20 _x_ – 9 _y_ \+ 47 = 0
**17.17** Write the equation of the ellipse that has these characteristics.
( _a_ )
( _b_ ) covertices (±3, 0), major axis length 10
( _c_ ) center (–3, 2), vertex (2, 2), _c_ = 4
( _d_ ) vertices (3, 2) and (3, –6), covertices (1, –2) and (5, –2)
**17.18** Write the equation of the ellipse in standard form and determine the center, vertices, foci, and covertices.
( _a_ ) 3 _x_ 2 \+ 4 _y_ 2 – 30 _x_ – 8 _y_ \+ 67 = 0
( _b_ ) 16 _x_ 2 \+ 7 _y_ 2 – 64 _x_ \+ 28 _y_ – 20 = 0
( _c_ ) 9 _x_ 2 \+ 8 _y_ 2 \+ 54 _x_ \+ 80 _y_ \+ 209 = 0
( _d_ ) 4 _x_ 2 \+ 5 _y_ 2 – 24 _x_ – 10 _y_ \+ 21 = 0
**17.19** Write the equations of the hyperbola that has the given characteristics.
( _a_ ) vertices (± 3, 0), foci (±5, 0)
( _b_ ) vertices (0, ±8), foci (0, ±10)
( _c_ ) foci (4, –1) and (4, 5), transverse axis length is 2
( _d_ ) vertices (–1, –1) and (–1, 5), b = 5
**17.20** Write the equation of the hyperbola in standard form and determine the center, vertices, and foci.
( _a_ ) 4 _x_ 2 – 5 _y_ 2 – 8 _x_ – 30 _y_ – 21 = 0
( _b_ ) 5 _x_ 2 – 4 _y_ 2 – 10 _x_ – 24 _y_ – 51 = 0
( _c_ ) 3 _x_ 2 – _y_ 2 – 18 _x_ \+ 10 _y_ – 10 = 0
( _d_ ) 4 _x_ 2 – _y_ 2 \+ 8 _x_ \+ 6 _y_ \+ 11 = 0
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**17.12**
( _a_ ) circle, Fig. 17-16
( _b_ ) hyperbola, Fig. 17-17
( _c_ ) ellipse, Fig. 17-18
( _d_ ) hyperbola, Fig. 17-19
( _e_ ) parabola, Fig. 17-20
( _f_ ) ellipse, Fig. 17-21
( _g_ ) hyperbola, Fig. 17-22
( _h_ ) parabola, Fig. 17-23
( _i_ ) single point, (1, –1)
( _j_ ) two intersecting lines, Fig. 17-24 ( _y_ = _x_ – 3 and _y_ = –2 _x_ \+ 1)
**Fig. 17-16**
**Fig. 17-17**
**Fig. 17-18**
**Fig. 17-19**
**Fig. 17-20**
**Fig. 17-21**
**Fig. 17-22**
**Fig. 17-23**
**Fig. 17-24**
**17.13**
( _a_ ) ( _x_ – 4)2 \+ ( _y_ –1)2 = 9
( _b_ ) ( _x_ – 5)2 \+ ( _y_ \+ 3)2 = 9
( _c_ ) _x_ 2 \+ _y_ 2 \+ 4 _x_ – 6 _y_ = 0
( _d_ ) _x_ 2 \+ _y_ 2 \+ 11 _y_ – _y_ – 32 = 0
**17.14**
**17.15**
( _a_ ) ( _y_ \+ 2)2 = 32( _x_ – 3)
( _b_ ) ( _x_ – 3)2 = 20 ( _y_ – 5)
( _c_ ) _y_ 2 = 20 _x_
( _d_ ) ( _x_ – 5)2 = –12( _y_ – 4)
**17.16**
**17.17**
**17.18**
**17.19**
( _a_ )
( _b_ )
( _c_ )
( _d_ )
**17.20**
( _a_ ) center (1, –3), vertices (1, –1) and (1, –5), foci (1, 0) and (1, –8)
( _b_ ) center (1, 3), vertices (–1, 3) and (3, 3), foci (4, 3) and (–2, 3)
( _c_ ) center (3, –5), vertices (5, –5) and (1, –5), foci (7, –5) and (–1, –5)
( _d_ ) center (–1, 3), vertices (–1, 7) and (–1, –1), foci (1, 3 + 2 ) and (1, 3 –2 )
## **CHAPTER 18
Systems of Equations Involving Quadratics**
### **18.1 GRAPHICAL SOLUTION**
The real simultaneous solutions of two quadratic equations in _x_ and _y_ are the values of _x_ and _y_ corresponding to the points of intersection of the graphs of the two equations. If the graphs do not intersect, the simultaneous solutions are imaginary.
### **18.2 ALGEBRAIC SOLUTION**
_A_. One linear and one quadratic equation Solve the linear equation for one of the unknowns and substitute in the quadratic equation.
**EXAMPLE 18.1.** Solve the system
(1) _x_ \+ _y_ = 7
(2) _x_ 2 \+ _y_ 2 = 25
Solving (1) for _y_ , _y_ = 7 – _x_. Substitute in (2) and obtain _x_ 2 \+ (7 – _x_ )2 = 25, _x_ 2 – 7 _x_ \+ 12 = 0, ( _x_ – 3)( _x_ – 4) = 0, and _x_ = 3, 4. When _x_ = 3, _y_ = 7 – _x_ = 4; when _x_ = 4, _y_ = 7 – _x_ = 3. Thus the simultaneous solutions are (3, 4) and (4, 3).
_B_. Two equations of the form _ax_ 2 \+ _by_ 2 = _c_ Use the method of addition or subtraction.
**EXAMPLE 18.2.** Solve the system
(1) 2 _x_ 2 – _y_ 2 = 7
(2) 3 _x_ 2 \+ 2 _y_ 2 = 14
To eliminate y, multiply (1) by 2 and add to (2); then
7 _x_ 2 = 28, _x_ 2 = 4 and _x_ = ±2.
Now put _x_ = 2 or _x_ = – 2 in (1) and obtain _y_ = ±1.
The four solutions are:
(2, 1); (–2, 1); (2, –1); (–2, –1)
_C_. Two equations of the form _ax 2 \+ bxy + cy2 = d_
**EXAMPLE 18.3**. Solve the system
(1) _x_ 2 \+ _xy_ = 6
(2) _x_ 2 \+ 5 _xy_ – 4 _y_ 2 = 10
_Method 1_.
Eliminate the constant term between both equations. Multiply (1) by 5, (2) by 3, and subtract; then
_x_ 2 – 5 _xy_ \+ 6 _y_ 2 = 0, ( _x_ \- 2 _y_ )( _x_ \- 3 _y_ ) = 0, _x_ = 2 _y_ and _x_ = 3 _y_.
Now put _x_ = 2 _y_ in (1) or (2) and obtain _y_ 2 = 1, _y_ = ± 1.
When _y_ = 1, _x_ = 2 _y_ = 2; when _y_ = – 1, _x_ = 2 _y_ = –2. Thus two solutions are: _x_ = 2, _y_ = 1; _x_ = –2, _y_ = –1.
Then put _x_ = 3 _y_ in (1) or (2) and get
When
when
Thus the four solutions are:
_Method 2_.
Let _y_ = _mx_ in both equations.
From
From
Then
from which hence _y_ = _x_ /2, _y_ = _x_ /3. The solution proceeds as in Method 1.
_D_. Miscellaneous methods
(1) Some systems of equations may be solved by replacing them by equivalent and simpler systems (see Problems 18.8-18.10).
(2) An equation is called symmetric in _x_ and _y_ if interchange of _x_ and _y_ does not change the equation. Thus _x_ 2 \+ _y_ 2 – 3 _xy_ \+ 4 _x_ \+ 4 _y_ = 8 is symmetric in _x_ and _y_. Systems of symmetric equations may often be solved by the substitutions _x_ = _u_ \+ _v_ , _y_ = _u_ – _v_ (see Problems 18.11-18.12).
### **Solved Problems**
**18.1** Solve graphically the following systems:
_(a)_
_(b)_
_(C)_
**SOLUTION**
See Fig. 18-1.
**Fig. 18-1**
**18.2** Solve the following systems:
_(a)_
_(b)_
**SOLUTION**
_(a)_ Solving the linear equation for _x_ , _x_ = 4 – 2 _y_. Substituting in the quadratic equation,
_y_ 2 – _y_ (4 – 2 _y_ ) = 7, 3 _y_ 2 – 4 _y_ – 7 = 0, ( _y_ \+ l)(3 _y_ – 7) = 0 and _y_ = –1, 7/3.
If _y_ = –1, _x_ = 4 – 2 _y_ = 6; if _y_ = 7/3, _x_ = 4 – 2 _y_ = – 2/3.
The solutions are (6, – 1) and (–2/3, 7/3).
_(b)_ Solving the linear equations for Substituting in the quadratic equation,
**18.3** Solve the system: (1) 2 _x_ 2 – 3 _y_ 2 = 6, (2) 3 _x_ 2 \+ 2 _y_ 2 = 35.
**SOLUTION**
To eliminate _y_ , multiply (1) by 2, (2) by 3 and add; then 13 _x_ 2= 117, _x_ 2 = 9, _x_ = ±3.
Now put _x_ = 3 or _x_ = – 3 in (1) and obtain _y_ = ±2.
The solutions are: (3, 2); (–3, 2); (3, –2); (–3, –2).
**18.4** Solve the system:
(1)
(2)
**SOLUTION**
The equations are quadratic in Substituting we obtain
8 _u_ 2 \- 3 _v_ 2 = 5 and 5 _u_ 2 \+ 2 _v_ 2 = 38.
Solving simultaneously, _u_ 2 = 4, _v_ 2 = 9 or _x_ 2 = 1/4, _y_ 2 = 1/9; then _x_ = ± 1/2, _y_ = ± 1/3. The solutions are:
**18.5** Solve the system
(1) 5 _x_ 2 \+ 4 _y_ 2 = 48
(2) _x_ 2 \+ 2 _xy_ = 16
by eliminating the constant terms.
**SOLUTION**
Multiply (2) by 3 and subtract from (1) to obtain
2 _x_ 2 \- 6 _xy_ \+ 4 _y_ 2 = 0, _x_ 2 – 3 _xy_ \+ 2 _y_ 2 = 0, ( _x_ \- _y_ )( _x_ \- 2 _y_ ) = 0 and _x_ = _y_ , _x_ = 2 _y_.
Substituting _x_ = _y_ in (1) or (2), we have
Substituting _x_ = 2 _y_ in (1) or (2), we have _y_ 2 = 2 and
The four solutions are:
**18.6** Solve the system
by using the substitution _y_ = _mx_.
**SOLUTION**
Put _y_ = _mx_ in (1); then 3 _x_ 2 – 4 _mx_ 2 = 4 and
Put _y_ = _mx_ in (2); then _x_ 2 – 2 _m_ 2 _x_ 2 = 2 and
Thus
Now substitute in (1) or (2) and obtain _x_ 2 = 4, _x_ = ± 2.
The solutions are (2, 1) and (–2, –1).
**18.7** Solve the system:
(1) _x_ 2 \+ _y_ 2 = 40,
(2) _xy =_ 12.
**SOLUTION**
From (2), _y_ = 12/ _x_ ; substituting in (1), we have
For _x_ = ±6, _y_ = 12/ _x_ = ±2; for _x_ = ±2, _y_ = ±6.
The four solutions are: (6, 2); (– 6, –2); (2, 6); (– 2, –6).
_Note_. Equation (2) indicates that those solutions in which the product _xy_ is negative (e.g. _x_ = 2, _y_ = –6) are extraneous.
**18.8** Solve the system:
(1) _x_ 2 \+ _y_ 2 \+ 2 _x_ – _y_ = 14,
(2) _x_ 2 \+ _y_ 2 \+ _x_ – 2 _y_ = 9.
**SOLUTION**
Subtract (2) from (1): _x_ \+ _y_ = 5 or _y_ = 5 – _x_.
Substitute _y_ = 5 – _x_ in (1) or (2): 2 _x_ 2 – 7 _x_ \+ 6 = 0, (2 _x_ – 3)( _x_ – 2) = 0 and _x_ = 3/2, 2.
The solutions are and (2, 3).
**18.9** Solve the system:
(1) _x_ 3 \+ _y_ 3 = 35,
(2) _x_ \+ _y_ = 5.
**SOLUTION**
Dividing (1) by (2),
From (2), _y_ = 5 – _x_ ; substituting in (3), we have
The solutions are (3, 2) and (2, 3).
**18.10** Solve the system:
(1) _x_ 2 \+ 3 _xy_ \+ 2 _y_ 2 = 3,
(2) _x_ 2 \+ 5 _xy_ \+ 6 _y_ 2 = 15.
**SOLUTION**
Dividing (1) by (2),
From Substituting _y_ = –2 _x_ in (1) or (2), _x_ 2 = 1 and _x_ = ± 1.
The solutions are (1, –2) and (– 1, 2).
**18.11** Solve the system:
(1) _x_ 2 \+ _y_ 2 \+ 2 _x_ \+ 2 _y_ = 32,
(2) _x_ \+ _y_ \+ 2 _xy_ = 22.
**SOLUTION**
The equations are symmetric in _x_ and _y_ since interchange of _x_ and _y_ yields the same equation. Substituting _x_ = _u_ \+ _v_ , _y_ = _u_ – _v_ in (1) and (2), we obtain
Adding (3) and (4), we get 2 _u_ 2 \+ 3 _u_ – 27 = 0, ( _u_ – 3)(2 _u_ \+ 9) = 0 and _u_ = 3, –9/2.
When _u_ = 3, _v_ 2 = 1 and _v_ = ± 1; when _u_ = –9/2, _v_ 2 = 19/4 and Thus the solutions of (3) and (4) are: _u_ = 3, _v_ = 1; _u_ = 3, _v_ = –1; _u_ = –9/2,
Then, since _x_ = _u_ \+ _v_ , _y_ = _u_ – _v_ , the four solutions of (1) and (2) are:
**18.12** Solve the system:
(1) **x 2** _\+ y 2 =_ 180,
(2)
**SOLUTION**
From (2) obtain (3) 4 _x_ \+ 4 _y_ – _xy_ = 0. Since (1) and (3) are symmetric in _x_ and _y_ , substitute _x_ = _u_ \+ _v_ , _y_ = _u_ – _v_ in (1) and (3) and obtain
Subtracting (5) from (4), we have _u_ 2 – 4 _u_ – 45 = 0, ( _u_ – 9)( _u_ \+ 5) = 0 and _u_ = 9, –5.
When _u =_ 9, _v_ = ±3; when Thus the solutions of (4) and (5) are: _u_ = 9, v = 3; _u_ = 9, _v_ = –3; _u_ = –5,
Hence the four solutions of (1) and (2) are:
**18.13** The sum of two numbers is 25 and their product is 144. What are the numbers?
**SOLUTION**
Let the numbers be _x_ , _y_. Then (1) _x_ \+ _y_ = 25 and (2) _xy_ = 144.
The simultaneous solutions of (1) and (2) are _x_ = 9, _y_ = 16 and _x_ = 16, _y_ = 9. Hence the required numbers are 9, 16.
**18.14** The difference of two positive numbers is 3 and the sum of their squares is 65. Find the numbers.
**SOLUTION**
Let the numbers be _p_ , _q_. Then (1) _p_ – _q_ = 3 and (2) _p_ 2 \+ _q_ 2 = 65.
The simultaneous solutions of (1) and (2) are _p_ = 7, _q_ = 4 and _p_ = –4, _q_ = –7. Hence the required (positive) numbers are 7, 4.
**18.15** A rectangle has perimeter 60 ft and area 216 ft2. Find its dimensions.
**SOLUTION**
Let the rectangle have sides of lengths _x_ , _y_. Then (1) 2 _x_ \+ 2 _y_ = 60 and (2) _xy_ = 216.
Solving (1) and (2) simultaneously, the required sides are 12 and 18 ft.
**18.16** The hypotenuse of a right triangle is 41 ft long and the area of the triangle is 180 ft2. Find the lengths of the two legs.
**SOLUTION**
Let the legs have lengths _x_ , _y_. Then (1) _x_ 2 \+ _y_ 2 = (41)2 and (2) ( _xy_ ) = 180.
Solving (1) and (2) simultaneously, we find the legs have lengths 9 and 40 ft.
### Supplementary Problems
**18.17** Solve the following systems graphically.
( _a_ ) _x_ 2 \+ _y_ 2 = 20, 3 _x_ – _y_ = 2
( _b_ ) _x_ 2 \+ 4 _y_ 2 = 25, _x_ 2 – _y_ 2 = 5
( _c_ ) _y_ 2 = _x_ , _x_ 2 \+ 2 _y_ 2 = 24
( _d_ ) _x_ 2 \+ 1 = 4 _y_ , 3 _x_ – 2 _y_ = 2
**18.18** Solve the following systems algebraically.
( _a_ ) 2 _x_ 2 – _y_ 2 = 14, _x_ – _y_ = 1
( _b_ ) _xy_ \+ _x_ 2 = 24, _y_ – 3 _x_ \+ 4 = 0
( _c_ ) 3 _xy_ – 10 _x_ = _y_ , 2 – _y_ \+ _x_ = 0
( _d_ ) 4 _x_ \+ 5 _y_ = 6, _xy_ = –2
( _e_ ) 2 _x_ 2 – _y_ 2 = 5, 3 _x_ 2 \+ 4 _y_ 2 = 57
( _f_ ) 9/ _x_ 2 \+ 16/ _y_ 2 = 5, 18/ _x_ 2 – 12/ _y_ 2 = –1
( _g_ ) _x_ 2 – _xy_ = 12, _xy_ – _y_ 2 = 3
( _h_ ) _x_ 2 \+ 3 _xy_ = 18, _x_ 2 – 5 _y_ 2 = 4
( _i_ ) _x_ 2 \+ 2 _xy_ = 16, 3 _x_ 2 – 4 _xy_ \+ 2 _y_ 2 = 6
( _j_ ) _x_ 2 – _xy_ \+ _y_ 2 = 7, _x_ 2 \+ _y_ 2 = 10
( _k_ ) _x_ 2 – 3 _y_ 2 \+ 10 _y_ = 19, _x_ 2 – 3 _y_ 2 \+ 5 _x_ = 9
( _l_ ) _x_ 3 – _y_ 3 = 9, _x_ – _y_ = 3
( _m_ ) _x_ 3 – _y_ 3 = 19, _x_ 2 _y_ – _xy_ 2 = 6
( _n_ ) 1/ _x_ 3 \+ 1/ _y_ 3 = 35, 1/ _x_ 2 – 1/ _xy_ \+ 1/ _y_ 2 = 7
**18.19** The square of a certain number exceeds twice the square of another number by 16. Find the numbers if the sum of their squares is 208.
**18.20** The diagonal of a rectangle is 85 ft. If the short side is increased by 11 ft and the long side decreased by 7 ft, the length of the diagonal remains the same. Find the dimensions of the original rectangle.
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**18.17** ( _a_ ) (2, 4), (–0.8, –4.4) See Fig. 18-2
( _b_ ) (3, 2), (–3, 2), (3, –2), (–3, –2) See Fig. 18-3
( _c_ ) (4, 2), (4, –2) See Fig. 18-4
( _d_ ) (1, 0.5), (5, 6.5) See Fig. 18-5
**18.18** ( _a_ ) (3, 2), (–5, – 6)
( _b_ ) (3, 5), (–2, –10)
( _c_ ) (2, 4), (–1/3, 5/3)
( _d_ ) (–1, 2), (5/2, –4/5)
( _e_ )
( _f_ ) (3, 2), (3, –2), (–3, 2), (–3, –2)
( _g_ ) (4, 1),(–4, –1)
( _h_ )
( _i_ ) (2, 3), (–2, –3)
( _j_ ) (1, 3), (–1, –3), (3, 1),(–3, –1)
( _k_ ) (–12, –5), (4, 3)
( _l_ ) (1, –2), (2, –1)
( _m_ ) (–2, –3), (3, 2)
( _n_ ) (1/2, 1/3), (1/3, 1/2)
**18.19** 12, 8; –12, –8; 12, –8; –12, 8
**18.20** 40 ft, 75 ft
**Fig. 18-2**
**Fig. 18-3**
**Fig. 18-4**
**Fig. 18-5**
## CHAPTER 19
**Inequalities**
### **19.1 DEFINITIONS**
An _inequality_ is a statement that one real quantity or expression is greater or less than another real quantity or expression.
The following indicate the meaning of inequality signs.
(1) _a > b_ means " _a_ is greater than _b_ " (or _a – b_ is a positive number).
(2) _a < b_ means " _a_ is less than _b_ " (or _a – b_ is a negative number).
(3) _a ≥ b_ means " _a_ is greater than or equal to _b_."
(4) _a ≤ b_ means " _a_ is less than or equal to _b_."
(5) 0 < _a_ < 2 means " _a_ is greater than zero but less than 2."
(6) –2 ≤ _x_ < 2 means " _x_ is greater than or equal to – 2 but less than 2."
An _absolute inequality_ is true for all real values of the letters involved. For example, ( _a_ – _b_ )2 > – 1 holds for all real values of _a_ and _b_ , since the square of any real number is positive or zero.
A _conditional inequality_ holds only for particular values of the letters involved. Thus _x –_ 5 > 3 is true only when _x_ is greater than 8.
The inequalities _a > b_ and _c > d_ have the _same sense_. The inequalities _a > b_ and _x < y_ have _opposite sense_.
### **19.2 PRINCIPLES OF INEQUALITIES**
(1) The sense of an inequality is unchanged if each side is increased or decreased by the same real number. It follows that any term may be transposed from one side of an inequality to the other, provided the sign of the term is changed.
Thus if _a > b_, then _a + c > b + c_, and _a – c > b – c_, and _a – b >_ 0.
(2) The sense of an inequality is unchanged if each side is multiplied or divided by the same positive number.
Thus if _a > b_ and _k >_ 0, then
(3) The sense of an inequality is reversed if each side is multiplied or divided by the same negative number.
Thus if _a > b_ and _k <_ 0, then
(4) If _a > b_ and _a, b, n_ are positive, then _a n > bn_ but _a –n < b–n_.
**EXAMPLES 19.1.**
(5) If _a > b_ and _c > d_, then (a _\+ c) > (b + d_ ).
(6) If _a > b >_ 0 and c > d > 0, then _ac > bd_.
### **19.3 ABSOLUTE VALUE INEQUALITIES**
The absolute value of a quantity represents the distance that the value of the expression is from zero on a number line. So _|x – a| = b_ , where _b_ > 0, says that the quantity _x – a_ is _b_ units from 0, _x – a_ is _b_ units to the right of 0, or x _– a_ is _b_ units to the left of 0. When we say _|x – a| > b, b >_ 0, then _x – a_ is at a distance from 0 that is greater than _b_. Thus, _x – a > b_ or _x – a < –b_. Similarly, if _|x – a| < b, b >_ 0, then _x – a_ is at a distance from 0 that is less than _b_. Hence, _x – a_ is between _b_ units below 0, _–b_ , and _b_ units above 0.
**EXAMPLES 19.2.** Solve each of these inequalities for _x_.
( _a_ ) | _x_ – 3| > 4
( _b_ ) | _x_ \+ 4| < 7
( _c_ ) | _x_ – 5| < –3
( _d_ ) | _x_ – 5| > –5
( _a_ ) | _x_ – 3| > 4, then _x_ – 3 > 4 or _x_ – 3 < –4. Thus, _x_ > 7 or _x_ < –1. The solution interval is (–∞, –1) ∪ (7, ∞), (where ∪ represents the union of the two intervals).
( _b_ ) | _x_ \+ 4| < 7 then –7 < _x_ \+ 4 < 7. Thus, –11 < _x_ < 3. The solution interval is (–11, 3).
( _c_ ) | _x_ – 5| < –3 Since the absolute value of a number is always greater than or equal to zero, there are no values for which the absolute value will be less than –3. Thus, there is no solution and we may write ∅ for the solution interval.
( _d_ ) | _x_ \+ 3| > –5 Since the absolute value of a number is always at least zero, it is always greater than –5. Thus the solution is all real numbers, and for the solution interval we write (– ∞, ∞).
### **19.4 HIGHER DEGREE INEQUALITIES**
Solving higher degree inequalities is similar to solving higher degree equations: we must always compare the expression to zero. If _f(x)_ > 0, then we are interested in the values of _x_ that will produce a product and/or quotient of factors that is positive, while if _f(x)_ < 0, we wish to find the values of _x_ that will produce a product and/or quotient that is negative.
If _f(x)_ is a quadratic expression we have just two factors to consider, and we can do this by examining cases based on the possible signs of the two factors that will produce the desired sign for the expression (see Problems 19.3 _(c)_ and 19.14). When the number of factors in _f(x)_ increases by one, the number of cases to consider doubles. Thus, for an expression with 2 factors there are 4 cases, with 3 factors there are 8 cases, and with 4 factors there are 16 cases. In each instance half of the cases will produce a positive expression and half a negative one. Thus, the case procedure gets to be a very long one quite quickly. An alternative procedure to the case method is the sign chart.
**EXAMPLE 19.3.** Solve the inequality _x_ 2 \+ 15 < 8 _x_.
The inequality _x_ 2 \+ 15 < 8 _x_ is equivalent to _x_ 2 – 8 _x_ \+ 15 < 0 and to ( _x_ – 3)( _x_ – 5) < 0 and is true when the product of _x_ – 3 and _x_ – 5 is negative. The critical values of the product are the values that make these factors 0, because they represent where the product may change signs.
The critical values of _x_ , 3 and 5, are placed on a number line and divide it into three intervals. We need to find the sign of the product of _x_ – 3 and _x_ – 5 on each of these intervals to find the solution (see Fig. 19-1). Vertical lines are drawn through each critical value. A dashed line indicates that the critical value is not in the solution and a solid line indicates that the critical value is in the solution.
**Fig. 19-1**
The signs above the number line are the signs for the factors and are found by selecting an arbitrary value in the interval as a test value and determining whether each factor is positive or negative for the test value. For the interval to the left of 3, we choose a test value of 1 and substitute it into _x_ – 3 and see that the value is –2, so we record a – sign, and for _x_ – 5 the value is –4 and again we record a – sign. For the interval between 3 and 5 we choose any value, such as 3.5, and determine that _x_ – 3 is positive and _x_ – 5 is negative. Finally, for the interval to the right of 5, we choose a value of 12 and see that both _x_ – 3 and _x_ – 5 are positive. The sign for the problem, written below the line, in each interval is determined by the signs of the factors in that interval. If an even number of factors in a product or quotient are negative, the product or quotient is positive. If an odd number of factors are negative, the product or quotient is negative.
We select the intervals that satisfy our problem ( _x_ – 3)( _x_ – 5) < 0, so we select the intervals that are negative in the sign chart. In the interval between 3 and 5 the problem is negative (see Fig. 19-1), so the solution is the interval (3, 5). The parentheses mean that the 3 and 5 are not included in the interval, and we know this since the boundary lines are dashed. If they had been in the solution, we would have used a bracket instead of a parenthesis at the end of the interval next to the 3.
The solution for _x_ 2 \+ 15 < 8 _x_ is the interval (3, 5).
**EXAMPLE 19.4.** Solve the inequality
The inequality is compared to 0 and the numerator and denominator are factored, so we can see that the critical values for the problem are the solution of _x_ = 0, _x_ – 3 = 0, and _x_ \+ 4 = 0. Thus, the critical values are _x_ = 0, _x_ = 3 and _x_ = –4. Since there are three critical values, the number line is divided into four distinct intervals, as shown in Fig. 19-2.
**Fig. 19-2**
The signs above the line are the signs of each factor in each interval. The sign below is the sign for the problem and it is + when an even number of factors are negative and – when an odd number of factors are negative. Since the problem uses the > sign, values that make the numerator zero are solutions, so a solid line is drawn through 3. Since 0 and –4 make the denominator of the fraction 0, they are not solutions and dashed lines were drawn through 0 and –4 (see Fig. 19-2).
Since the problem
indicates that a positive or zero value is wanted, we want the regions with a + sign in the sign chart. Thus, the solutions are the intervals, (–4, 0) and [3, ∞), and the solution is written (–4, 0) ∪ [3, ∞). The ∪ indicates that we want the union of the two intervals. Note that the bracket, [, is used because the critical value 3 is in the solution and a parenthesis, ), is always used for the infinite, ∞, side of an interval.
### **19.5 LINEAR INEQUALITIES IN TWO VARIABLES**
The solution of linear inequalities in two variables _x_ and _y_ consists of all points ( _x, y_ ) that satisfy the inequality. Since a linear equation represents a line, a linear inequality is the points on one side of a line. The points on the line are included when the sign ≥ or ≤ is used in the statement of the inequality. The solutions of linear inequalities are usually found by graphical methods.
**EXAMPLE 19.5.**
Find the solution for 2 _x_ – _y_ ≤ 3.
We graph the line related to the inequality 2 _x_ – _y_ ≤ 3, which is 2 _x_ – _y_ = 3. Since the symbol ≤ is used, the line is part of the solution and a solid line is used to indicate this (see Fig. 19-3). If the line is not part of the solution, we use a dashed line to indicate that fact. We shade the region on the side of the line where the points are solutions of the inequality. The solution region is determined by selecting a test point that is not on the line. If the test point satisfies the inequality, then all points on that side of the line are in the solution. If the test point does not satisfy the inequality, no points on that side of the line are in the solution. Hence the solution points are on the opposite side of the line from the test point.
**Fig. 19-3**
The point _P_ (2, 4) is not on the line 2 _x_ – _y_ = 3, so it can be used as a test point. When we substitute (2, 4) into the inequality 2 _x_ – _y_ ≤ 3, we get 2(2) –4 ≤ 3, which is true, since 0 ≤ 3. We shade on the side of the line that contains the test point (2, 4) to indicate the solution region. If we had selected _Q_ (5, –2) and substituted into 2 _x_ – _y_ ≤ 3, we would have obtained 12 ≤ 3, which is false, and would have shaded on the opposite side of the line from _Q_. This is the same region we found using the test point _P_.
The solution for 2 _x_ – _y_ ≤ 3 is shown in Fig. 19-3 and consists of the shaded region and the line.
### **19.6 SYSTEMS OF LINEAR INEQUALITIES**
If we have two or more linear inequalities in two variables, we say we have a system of linear inequalities and the solution of the system is the intersection, or common region, of the solution regions for the inequalities.
A system with two inequalities whose related equations intersect always has a solution region. If the related equations are parallel, the system may or may not have a solution. Systems with three or more inequalities may or may not have a solution.
**EXAMPLE 19.6.**
Solve the system of inequalities 2 _x_ \+ _y_ > 3 and _x_ –2 _y_ ≤ –1.
We graph the related equations 2 _x_ \+ _y_ = 3 and _x_ – 2 _y_ = –1 on the same set of axes. The line 2 _x_ \+ _y_ = 3 is dashed, since it is not included in 2 _x_ \+ _y_ > 3, but the line _x_ – 2 _y_ = – 1 is solid, since it is included in _x_ –2 _y_ ≤ –1.
Now we select a test such as (0, 5) that is not on either line, determine which side of each line to shade and shade only the common region. Since 2(0) + 5 > 3 is true, the solution region is to the right and above the line 2 _x_ \+ _y_ = 3. Since 0 – 2(5) ≤ –1 is true, the solution region is to the left and above the line _x_ – 2 _y_ = – 1.
The solution region of 2 _x_ \+ _y_ > 3 and _x_ – 2 _y_ ≤ –1 is the shaded region of Fig. 19-4, which includes the part of the solid line bordering the shaded region.
**Fig. 19-4**
### **19.7 LINEAR PROGRAMMING**
Many practical problems from business involve a function (objective) that is to be either maximized or minimized subject to a set of conditions (constraints). If the objective is a linear function and the constraints are linear inequalities, the values, if any, that maximize or minimize the objective occur at the corners of the region determined by the constraints.
**EXAMPLE 19.7.** The Green Company uses three grades of recycled paper, called grades A, B, and C, produced from scrap paper it collects. Companies that produce these grades of recycled paper do so as the result of a single operation, so the proportion of each grade of paper is fixed for each company. The Ecology Company process produces 1 unit of grade A, 2 units of grade B, and 3 units of grade C for each ton of paper processed and charges $300 for the processing. The Environment Company process produces 1 unit of grade A, 5 units of grade B, and 1 unit of grade C for each ton of paper processed and charges $500 for processing. The Green Company needs at least 100 units of grade A paper, 260 units of grade B paper, and 180 units of grade C paper. How should the company place its order so that costs are minimized?
If _x_ represents the number of tons of paper to be recycled by the Ecology Company and y represents the number of tons of paper to be processed by the Environment Company, then the objective function is _C_ ( _x, y_ ) = 300 _x_ \+ 500 _y_ , and we want to minimize _C_ ( _x, y_ ).
The constraints stated in terms of _x_ and _y_ are for grade A: 1 _x_ \+ 1 _y_ ≥ 100; for grade B: 2 _x_ \+ 5 _y_ ≥ 260; and for grade C: 3 _x_ \+ 1 _y_ ≥ 180. Since you can not have a company process a negative number of tons of paper, _x_ ≥ 0 and y ≥ 0. These last two constraints are called natural or implied constraints, because these conditions are true as a matter of fact and need not be stated in the problem.
We graph the inequalities determined from the constraints (see Fig. 19-5). The vertices of the region are _A_ (0, 180), _B_ (40, 60), _C_ (80, 20), and _D_ (130, 0).
**Fig. 19-5**
The minimum for _C(x, y)_ , if it exists, will occur at point _A, B, C_ , or _D_ , so we evaluate the objective function at these points.
The Green Printing Company can minimize the cost of recycled paper to $34 000 by having the Ecology Company process 80 tons of paper and the Environment Company process 20 tons of paper.
### **Solved Problems**
**19.1** If _a > b_ and _c > d_, prove that _a + c > b + d_.
**SOLUTION**
Since ( _a – b_ ) and ( _c – d_ ) are positive, ( _a – b_ ) + ( _c – d_ ) is positive.
Hence ( _a–b_ ) + ( _c–d_ ) > 0, ( _a + c_ )– ( _b + d_ ) > 0 and ( _a + c_ ) > ( _b + d_ ).
**19.2** Find the fallacy.
**SOLUTION**
There is nothing wrong with steps ( _a_ ), ( _b_ ), ( _c_ ), ( _d_ ). The error is made in step ( _e_ ) where the inequality is divided by _a-b_ , a negative number, without reversing the inequality sign.
**19.3** Find the values of x for which each of the following inequalities holds.
_(a)_ 4 _x_ \+ 5 > 2 _x_ \+ 9. We have 4 _x_ – 2 _x_ > 9 – 5, 2 _x_ > 4 and _x_ > 2.
_(b)_ Multiplying by 6, we obtain
3 _x_ – 2 < 4 _x_ \+ 3, 3 _x_ – 4 _x_ < 2 + 3, – _x_ < 5, _x_ > –5.
(c) _x_ 2 < 16.
_Method 1_. _x_ 2 – 16 < 0, ( _x_ – 4)( _x_ \+ 4) < 0. The product of the factors ( _x_ – 4) and ( _x_ \+ 4) is negative. Two cases are possible.
(1) _x_ – 4 > 0 and _x_ \+ 4 < 0 simultaneously. Thus _x_ > 4 and _x_ < –4. This is impossible, as x cannot be both greater than 4 and less than –4 simultaneously.
(2) _x_ – 4 < 0 and _x_ \+ 4 > 0 simultaneously. Thus _x_ < 4 and _x_ > –4. This is possible if and only if –4 < x < 4. Hence –4 < x < 4.
_Method 2_. ( _x_ 2)1/2 < (16)1/2. Now ( _x_ 2)1/2 = _x_ if _x_ ≥ 0, and ( _x_ 2)1/2 = – _x_ if _x_ ≤ 0. If _x_ ≥ 0, ( _x_ 2)1/2 < (16)1/2 may be written _x_ < 4. Hence 0 ≤ _x_ < 4.
If _x_ ≤ 0, ( _x_ 2)1/2 < (16)1/2 may be written – _x_ < 4 or _x_ > –4. Hence –4 < _x_ ≤ 0. Thus 0 ≤ _x_ < 4 and –4 < _x_ ≤ 0, or –4 < _x_ < 4.
**19.4** Prove that _a_ 2 \+ _b_ 2 > 2 _ab_ if _a_ and _b_ are real and unequal numbers.
**SOLUTION**
If _a_ 2 \+ _b_ 2 > 2 _ab_ , then _a_ 2 – 2 _ab_ \+ _b_ 2 > 0 or ( _a_ – _b_ )2 > 0. This last statement is true since the square of any real number different from zero is positive.
The above provides a clue as to the method of proof. Starting with ( _a – b_ )2 > 0, which we know to be true if _a ≠ b_ , we obtain _a_ 2 – 2 _ab_ \+ _b_ 2 > 0 or _a_ 2 \+ _b_ 2 > 2 _ab_.
Note that the proof is essentially a reversal of the steps in the first paragraph.
**19.5** Prove that the sum of any positive number and its reciprocal is never less than 2.
**SOLUTION**
We must prove that ( _a_ \+ 1/ _a_ ) ≥ 2 if _a_ > 0.
If (a + 1/a) ≥ 2, then _a_ 2 \+ 1 ≥ 2a, a2 – 2a + 1 ≥ 0, and ( _a – 1_ )2 ≥ 0 which is true.
To prove the theorem we start with ( _a_ – 1)2 ≥ 0, which is known to be true.
Then _a_ 2 – 2 _a_ \+ 1 ≥ 0, _a_ 2 \+ 1 > 2 _a_ and _a_ \+ 1/ _a_ > 2 upon division by _a_.
**19.6** Show that _a_ 2 \+ _b_ 2 \+ _c_ 2 > _ab_ \+ _bc_ \+ _ca_ for all real values of _a_ , _b_ , _c_ unless _a_ = _b_ = _c_.
**SOLUTION**
Since _a_ 2 \+ _b_ 2 > 2 _ab, b_ 2 \+ _c_ 2 > 2 _bc_ , _c_ 2 \+ _a_ 2 > 2 _ca_ (see Problem 19.4), we have by addition
2( _a_ 2 \+ _b_ 2 \+ _c_ 2) > 2( _ab_ + _bc_ \+ _ca_ ) or _a_ 2 \+ _b_ 2 \+ _c_ 2 > _ab_ \+ _bc_ \+ _ca_.
(If _a_ = _b_ = _c_ , then _a_ 2 \+ _b_ 2 \+ _c_ 2 = _ab_ \+ _bc_ \+ _ca_.)
**19.7** If _a_ 2 \+ _b_ 2 = 1 and _c_ 2 \+ _d_ 2 = 1, show that _ac + bd_ < 1.
**SOLUTION**
_a 2 \+ c2 >_ 2 _ac_ and _b 2 \+ d2 >_ 2 _bd;_ hence by addition
( _a_ 2 \+ _b_ 2) + ( _c_ 2 \+ _d_ 2) > 2 _ac_ \+ 2 _bd_ or 2 > 2 _ac_ \+ 2 _bd_ , i.e., 1 > _ac_ \+ _bd_.
**19.8** Prove that _x_ 3 \+ _y_ 3 > _x_ 2 _y_ \+ _y_ 2 _x_ , if _x_ and _y_ are real, positive and unequal numbers.
**SOLUTION**
If _x_ 3 \+ _y_ 3 > x2 _y_ \+ _y_ 2 _x_ , then ( _x_ \+ _y_ )( _x_ 2 – _xy_ \+ _y_ 2) > _xy_ ( _x_ \+ _y_ ). Dividing by _x + y_ , which is positive.
_x_ 2 – _xy_ \+ _y_ 2 > _xy_ or _x_ 2 – 2 _xy_ \+ _y_ 2 > 0, i.e., ( _x_ – _y_ )2 > 0 which is true if _x_ ≠ _y_.
The steps are reversible and supply the proof. Starting with (x – y)2 > 0, _x ≠ y_ , obtain
_x_ 2 – _xy_ \+ y2 > _xy_.
Multiplying both sides by _x + y_ , we have _(x + y_ )(x2 – xy + y2) > _xy(x + y_ ) or _x 3 \+ y3 > x2y + y2x_.
**19.9** Prove that _a n \+ bn > an–1b + abn–1_, provided _a_ and _b_ are positive and unequal, and _n >_ l.
**SOLUTION**
If _a_ _n_ \+ _b_ _n_ > _a_ _n_ –1 _b_ \+ _ab_ _n_ –1, then ( _a_ _n_ – _a_ _n_ –1 _b_ ) – ( _ab_ _n_ –1 – _b_ _n_ ) > 0 or
_a_ _n_ –1( _a_ – _b_ ) – _b_ _n_ –1( _a_ – _b_ ) > 0, i.e., ( _a_ _n_ –1 – _b_ _n_ –1)( _a_ – _b_ ) > 0.
This is true since the factors are both positive or both negative.
Reversing the steps, which are reversible, provides the proof.
**19.10** Prove that
if _a_ > 0 and _a_ ≠ 1.
**SOLUTION**
Multiplying both sides of the inequality by _a 3_ (which is positive since _a >_ 0), we have
_a_ 6 \+ 1 > _a_ 5 \+ _a_ , _a_ 6 – _a_ 5 – _a_ \+ 1 > 0 and ( _a_ 5 – 1)( _a_ – 1) > 0.
If _a >_ 1 both factors are positive, while if 0 < _a <_ 1 both factors are negative. In either case the product is positive.
(If _a_ = 1 the product is zero.)
Reversal of the steps provides the proof.
**19.11** If _a, b, c, d_ are positive numbers and
prove that
**SOLUTION**
_Method_ 1. If
then multiplying by _d(b + d)_ we obtain
and, dividing by _bd_ ,
which is given as true. Reversing the steps provides the proof. _Method 2_. Since
then
**19.12** Prove:
**SOLUTION**
_(a)_ Since _x > y, x – y >_ 0. Multiplying both sides of _x + y >_ 1 by the positive number _x – y_ ,
_(x + y)(x – y) > (x – y)_ or _x 2 – y2 > x – y_.
(b) Since _x < y, x – y_ < 0. Multiplying both sides of _x + y_ > 1 by the negative number _x – y_ reverses the sense of the inequality; thus
( _x_ \+ _y_ )( _x_ – _y_ ) < ( _x_ – _y_ ) or _x_ 2 – _y_ 2 < _x_ – _y_.
**19.13** The arithmetic mean of two numbers _a_ and _b_ is ( _a_ \+ _b_ )/2, the geometric mean is and the harmonic mean is _2ab_ /( _a_ \+ _b_ ). Prove that
if _a_ and _b_ are positive and unequal.
**SOLUTION**
( _a_ ) If (a + b)/2 > then (a + b)2 > a2 \+ 2 _ab_ \+ b2 > 4 _ab_ , a2 – 2 _ab_ \+ b2 > 0 and (a – b)2 > 0 which is true if _a ≠ b_. Reversing the steps, we have (a + b)/2 >
( _b_ ) If
then
( _a_ \+ _b_ )2 > 4 _ab_ and ( _a_ – _b_ )2 > 0
which is true if _a ≠ b_. Reversing the steps, we have 2 _ab_ /(a + b). From ( _a_ ) and ( _b_ ),
.
**19.14** Find the values of _x_ for which ( _a_ ) _x_ 2 – 7 _x_ \+ 12 = 0, ( _b_ ) x2 – 7 _x_ \+ 12 > 0, ( _c_ ) _x_ 2 – 7 _x_ \+ 12 < 0.
**SOLUTION**
(a) _x_ 2 – 7 _x_ \+ 12 _= ( _x_ –_ 3)( _x_ – 4) = 0 when _x_ = 3 or 4.
(b) _x_ 2 – 7 _x_ \+ 12 > 0 or ( _x_ – 3)( _x_ – 4) > 0 when ( _x_ – 3) > 0 and ( _x_ – 4) > 0 simultaneously, or when ( _x_ – 3) < 0 and _( _x_ –_ 4) < 0 simultaneously.
( _x_ – 3) > 0 and ( _x_ – 4) > 0 simultaneously when _x_ > 3 and _x_ > 4, i.e., when _x_ > 4.
( _x_ –3) < 0 and ( _x_ – 4) < 0 simultaneously when _x_ < 3 and _x_ < 4, i.e., when _x_ < 3.
Hence _x_ 2 – 7 _x_ \+ 12 > 0 is satisfied when _x_ > 4 or _x_ < 3.
(c) _x_ 2 – 7 _x_ \+ 12 < 0 or ( _x_ – 3)( _x_ – 4) < 0 when ( _x_ – 3) > 0 and ( _x_ – 4) < 0 simultaneously, or when ( _x_ – 3) < 0 and ( _x_ – 4) > 0 simultaneously.
( _x_ – 3) > 0 and ( _x_ – 4) < 0 simultaneously when _x_ > 3 and _x_ < 4, i.e., when 3 < _x_ < 4.
( _x_ – 3) < 0 and ( _x_ – 4) > 0 simultaneously when _x_ < 3 and _x_ > 4, which is absurd.
Hence _x_ 2 – 7 _x_ \+ 12 < 0 is satisfied when 3 < _x_ < 4.
**19.15** Determine graphically the range of values of _x_ defined by
( _a_ ) _x_ 2 \+ 2 _x_ – 3 = 0
( _b_ ) _x_ 2 \+ 2 _x_ – 3 > 0
( _c_ ) _x_ 2 \+ 2 _x_ – 3 < 0.
**SOLUTION**
Figure 19-6 shows the graph of the function defined by _y_ = _x_ 2 \+ 2 _x_ – 3. From the graph it is clear that
**Fig. 19-6**
( _a_ ) _y_ = 0 when _x_ = 1, _x_ = –3
( _b_ ) _y_ > 0 when _x_ > 1 or _x_ < –3
( _c_ ) _y_ < 0 when –3 < _x_ < 1.
**19.16** Solve for _x_ : ( _a_ ) |3 _x_ – 6| + 2 > 9 ( _b_ ) |7 _x_ – 1 |– 6 < 2.
( _a_ ) |3 _x_ – 6| + 2 > 9
The solution of |3 _x_ – 6| + 2 > 9 is the interval (–∞, –1/3) ∪ (13/3, ∞).
( _b_ ) |7 _x_ – 1| – 6 < 2
The solution of |7 _x_ – 1| –6 < 2 is the interval (–1, 9/7).
**19.17** Solve for _x_ :
**SOLUTION**
The critical values are _x_ = –1 and _x_ = 2. We make a sign chart (see Fig. 19-7), with a solid line through _x_ = 2, since it makes the fraction 0, and 0 is included in the solution, and a dashed line through _x_ = –1, since it makes the fraction undefined. Next, we determine the sign of each factor in the three intervals. Finally, in intervals where an even number of factors are negative the problem is positive and in those where an odd number of factors are negative the problem is negative. The solution of
**Fig. 19-7**
is the interval (–1, 2].
The critical values are _x_ = 2, _x_ = 3, and _x_ = 7. We make a sign chart (see Fig. 19-8), with dashed lines through _x = 2_ and _x_ = 3 and a solid line through _x_ = 7. Since _x_ = 3 makes the denominator of the fraction zero it is excluded, even though it also makes the numerator 0. The signs for the factors are determined for each interval and then used to determine the sign for the problem in each interval. The factor _x_ – 3 is used an even number of times in the problem and could be omitted from the sign chart, since any factor raised to an even power is always non-negative.
**Fig. 19-8**
The solution of
is the interval (2, 3) ∪ (3, 7].
_Note 1_ : If we had canceled the common factor _x_ – 3, we might have overlooked the fact that the problem is not defined when _x_ = 3 and it cannot be in the solution set.
_Note 2:_ When a factor appears in the problem an even number of times, it may be excluded from the sign chart and is usually omitted. When a factor appears in the problem an odd number of times it must be included in the sign chart an odd number of times, and is usually included exactly once.
**19.18** Find the solution for the system of inequalities –2 _x_ \+ _y_ ≥ 2 and 2 _x_ – _y_ ≤ 6.
**SOLUTION**
Graph the related equations – _2x_ \+ _y = 2_ and _2x – y =_ 6. Both lines are solid, since they are included in the solution.
Using (0, 0) as the test point, we get –2(0) + 0 ≥ 2, which is false, and 2(0) – 0 ≤ 6, which is true.
Since the test point (0, 0) makes – _2x_ \+ _y_ ≤ 2 false, the solution lies on the opposite side of the line – _2x_ \+ _y = 2_ from the point (0, 0). So we shade above and to the left of the line – _2x_ \+ _y =_ 2.
Since the test point (0, 0) makes _2x – y ≤_ 6 true, the solution is on the same side of the line _2x – y =_ 6 as the point (0, 0). So we shade above and to the left of the line _2x – y =_ 6.
The common solution is the region above and to the left of – _2x_ \+ _y =_ 2, and is the shaded region shown in Fig. 19-9.
**Fig. 19-9**
**19.19** The Close Shave Company manufactures two types of electric shaver. One shaver is cordless, requires 4 hours to make, and sells for $40. The other shaver is a cord type, takes 2 hours to make, and sells for $30. The Company has only 800 work hours to use in manufacturing each day and the shipping department can pack and ship only 300 shavers per day. How many of each type of shaver should the Close Shave Company produce per day to maximize its sales revenue?
**SOLUTION**
Let _x_ be the number of cordless shavers made per day and y be the number of cord-type shavers made per day.
The objective function is _R_ ( _x_ , _y_ ) = 40 _x_ \+ 30 _y_.
The stated constraints are 4 _x_ \+ 2 _y_ ≤ 800 and _x_ \+ _y_ ≤ 300.
The natural constraints are _x_ ≥ 0 and _y_ ≥ 0.
From Fig. 19-10, we see that the vertices of the region formed by the constraints are A(0, 0), B(200, 0), C(100, 200), and D(0, 300).
**Fig. 19-10**
R(0, 0) = 40(0) + 30(0) = 0 + 0 = 0
R(200, 0) = 40(200) + 30(0) = 8000 + 0 = 8000
R(100, 200) = 40(100) + 30(200) = 4000 + 6000 = 10000
R(0, 300) = 40(0) + 30(300) = 0 + 9000 = 9000.
The Close Shave Company achieves the maximum sales revenue of $10 000 per day by producing 100 cordless shavers and 200 cord-type shavers per day.
### **Supplementary Problems**
**19.20** If _a > b_, prove that _a – c > b – c_ where _c_ is any real number.
**19.21** If _a > b_ and _k >_ 0, prove that _ka > kb_.
**19.22** Find the values of _x_ for which of the following inequalities holds.
( _a_ ) 2 ( _x_ \+ 3) > 3( _x_ – 1) + 6
( _d_ ) x2 > 9
**19.23** For what values of _a_ will (a + 3) < 2(2a + 1)?
**19.24** Prove that ( _a_ 2 \+ _b_ 2) ≥ _ab_ for all real values of _a_ and b, the equality holding if and only if _a = b_.
**19.25** Prove that
if _x_ and _y_ are positive and _x ≠ y_.
**19.26** Prove that
if _x_ > 0, _y_ > 0.
**19.27** Prove that _xy_ \+ 1 ≥ _x_ \+ _y_ if _x_ ≥ 1 and _y_ ≥ 1 or if _x_ ≤ 1 and _y_ ≥ 1.
**19.28** If _a_ > 0, _a_ ≠ 1 and n is any positive integer, prove that
**19.29** Show that
**19.30** Determine the values of _x_ for which each of the following inequalities holds.
( _a_ ) _x_ 2 \+ 2 _x_ – 24 > 0
( _b_ ) _x_ 2 – 6 < _x_
( _c_ ) 3 _x_ 2 – 2 _x_ < 1
**19.31** Determine graphically the range of values of _x_ for which ( _a_ ) _x_ 2 – 3 _x_ – 4 > 0, ( _b_ ) 2 _x_ 2 – 5 _x_ \+ 2 < 0.
**19.32** Write the solution for each inequality in interval notation.
( _a_ ) |3 _x_ \+ 3| – 15 ≥ –6 ( _b_ ) |2 _x_ – 3| < 7
**19.33** Write the solution for each inequality in the interval notation.
( _a_ ) _x_ 2 ≥ 10 _x_ – 21
( _c_ ) ( _x_ – 1)( _x_ – 2)( _x_ \+ 3) > 0
( _f_ )
**19.34** Graph each inequality and shade the solution region.
( _a_ ) 4 _x_ – _y_ ≤ 5
( _b_ ) _y_ – 3 _x_ > 2
**19.35** Graph each system of inequalities and shade the solution region.
( _a_ ) _x_ \+ 2 _y_ ≤ 20 and 3 _x_ \+ 10 _y_ ≤ 80
( _b_ ) 3 _x_ \+ _y_ ≥ 4, _x_ \+ _y_ ≥ 2, –x + _y_ ≤ 4, and _x_ ≤ 5
**19.36** Use linear programming to solve each problem.
( _a_ ) Ramone builds portable storage buildings. He uses 10 sheets of plywood and 15 studs in a small building and 15 sheets of plywood and 45 studs in a large building. Ramone has 60 sheets of plywood and 135 studs available for use. If Ramone makes a profit of $400 on a small building and $500 on a large building, how many of each type of building should he make to maximize his profit?
( _b_ ) Jean and Wesley make wind chimes and bird houses in their craft shop. Each wind chime requires 3 hours of work from Jean and 1 hour of work from Wesley. Each bird house requires 4 hours of work from Jean and 2 hours of work from Wesley. Jean cannot work more than 48 hours per week and Wesley cannot work more than 20 hours per week. If each wind chime sells for $12 and each bird house sells for $20, how many of each item should they make to maximize their revenue?
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**19.22** ( _a_ ) _x_ < 3
( _b_ ) _x_ > 2
( _c_ ) 0 < _x_ < 2
( _d_ ) _x_ < -3 or _x_ > 3
**19.23**
**19.30** ( _a_ ) _x_ > 4 or _x_ < -6
( _b_ ) -2 < _x_ < 3
( _c_ )
( _d_ )
**19.31** ( _a_ ) _x_ > 4 or _x_ < –1
( _b_ )
**19.32** ( _a_ ) (–∞, –4) ∪ [2, ∞)
( _b_ ) (–2, 5)
**19.33** ( _a_ ) (–∞, 3) ∪ [7, ∞)
( _b_ ) (–∞, –1) ∪ (0, 1)
( _c_ ) (–3, 1) ∪ (2, ∞)
( _d_ ) (–2, 1]
( _e_ ) (–∞, –4] ∪ (–1, ∞)
( _f_ ) (–2, 3) ∪ [6, ∞)
**19.34** ( _a_ ) Figure 19-11 ( _b_ ) Figure 19-12
**Fig. 19-11**
**Fig. 19-12**
**19.35** ( _a_ ) Figure 19-13 ( _b_ ) Figure 19-14
**Fig. 19-13**
**Fig. 19-14**
**19.36** ( _a_ ) Ramone maximizes his profit by making 6 small buildings and 0 large buildings.
( _b_ ) Jean and Wesley will maximize their revenue by making 6 wind chimes and 8 bird houses.
## CHAPTER 20
**Polynomial Functions**
### **20.1 POLYNOMIAL EQUATIONS**
A rational integral equation of degree _n_ in the variable _x_ is an equation which can be written in the form
_a nx_ _n_ \+ _a_ _n_ – 1 _x_ _n_ – 1 \+ _a_ _n_ –2 _x_ _n_ – 2 \+... + _a_ 1 _x_ \+ _a_ 0 = 0, _a_ _n_ ≠ 0
where _n_ is a positive integer and _a_ 0, _a_ 1, _a_ 2,..., _a_ n – 1, _a_ n are constants.
Thus 4 _x_ 3 – 2 _x_ 2 \+ 3 _x_ – 5 = 0, _x_ 2 – and _x_ 4 \+ are rational integral equations in _x_ of degree 3, 2 and 4 respectively. Note that in each equation the exponents of _x_ are positive and integral, and the coefficients are constants (real or complex numbers).
The coefficient of the highest degree term is called the lead coefficient and a0 is called the constant term.
In this chapter, only rational integral equations are considered.
A polynomial of degree _n_ in the variable _x_ is a function of _x_ which can be written in the form
_P_ ( _x_ ) = _a_ n _x_ _n_ \+ _a_ _n_ – 1 _x_ n – 1 \+ _a_ n – 2 _x_ _n_ – 2 \+... + _a_ 1 _x_ \+ _a_ 0, _a_ n ≠ 0
where _n_ is a positive integer and _a_ 0, _a_ 1, _a_ 2,..., _a_ _n_ –1, _a_ _n_ are constants. Then _P_ ( _x_ ) = 0 is a rational integral equation of degree _n_ in _x_.
If _P_ ( _x_ ) = 3 _x_ 3 \+ _x_ 2 \+ 5 _x_ – 6, then _P_ (–2) = 3(–2)3 \+ (–2)2 \+ 5(–2) – 6 = –36.
If _P_ ( _x_ ) = _x_ 2 \+ 2 _x_ – 8, then .
Any value of _x_ which makes _P(x)_ vanish is called a root of the equation _P_ ( _x_ ) = 0. Thus 2 is a root of the equation _P_ ( _x_ ) = 3 _x_ 3 – 2 _x_ 2 – 5 _x_ – 6 = 0, since _P_ (2) = 24 – 8 – 10 – 6 = 0.
### **20.2 ZEROS OF POLYNOMIAL EQUATIONS**
A. Remainder theorem If _r_ is any constant and if a polynomial _P_ ( _x_ ) is divided by ( _x_ – _r_ ), the remainder is _P_ ( _r_ ).
For example, if _P_ ( _x_ ) = 2 _x_ 3 – 3 _x_ 2 – _x_ \+ 8 is divided by _x_ \+ 1, then _r_ = – 1 and the remainder = _P_ (–1) = –2 – 3 + 1 + 8 = 4. That is,
where _Q(x)_ is a polynomial in _x_.
B. Factor theorem If _r_ is a root of the equation _P_ ( _x_ ) = 0, i.e. if _P_ ( _r_ ) = 0, then ( _x_ – _r_ ) is a factor of _P_ ( _x_ ). Conversely, if ( _x_ – _r_ ) is a factor of _P(x)_ , then _r_ is a root of _P(x)_ = 0, or _P_ ( _r_ ) = 0.
Thus 1, – 2, –3 are the three roots of the equation _P_ ( _x_ ) = _x_ 3 \+ 4 _x_ 2 \+ _x_ – 6 = 0, since _P_ (1) = _P_ (– 2) = _P_ (– 3) = 0. Then ( _x_ – 1), ( _x_ \+ 2) and ( _x_ \+ 3) are factors of _x_ 3 \+ 4 _x_ 2 \+ _x_ – 6.
C. Synthetic division Synthetic division is a simplified method of dividing a polynomial _P_ ( _x_ ) by _x_ – _r_ , where _r_ is any assigned number. By this method we determine values of the coefficients of the quotient and the value of the remainder can readily be determined.
**EXAMPLE 20.1.** Divide (5 _x_ \+ _x_ 4 – 14 _x_ 2) by ( _x_ \+ 4) using synthetic division.
Write the terms of the dividend in descending powers of the variable and fill in missing terms using zero for the coefficients; write the divisor in the form _x_ – _a_.
( _x_ 4 \+ 0 _x_ 3 – 14 _x_ 2 \+ 5 _x_ \+ 0) ÷ ( _x_ – (–4))
Write the constant term a from the divisor on the left in a and write the coefficients from the dividend to the right of the symbol
Bring down the first term in the divisor to the third row, leaving a blank row for now.
Multiply the term in the quotient row (third row) by the divisor and write the product in the second row under the second term in the first row, add the numbers in the column formed, and write the sum as the second term in the quotient row.
Multiply the last term on the right in the quotient row by the divisor, write it under the next term in the top row, add, and write the sum in the quotient row. Continue this process until all of the terms in the top row have a number under them.
The third row is the quotient row with the last term being the remainder. The degree of the quotient polynomial is one less than the degree of the dividend because we are dividing by a linear factor. The terms of the quotient row are the coefficients of the terms in the quotient polynomial. The degree of the quotient polynomial here is 3. The quotient with remainder for (5 _x_ \+ _x_ 4 – 14 _x_ 2) ÷ ( _x_ \+ 4) is
D. Fundamental theorem of algebra Every polynomial equation _P_ ( _x_ ) = 0 has at least one root, real or complex.
Thus _x_ 7 – 3 _x_ 5 \+ 2 = 0 has at least one root.
But has no root, since no number _r_ exists such that _f_ ( _r_ ) = 0. Since this equation is not rational, the fundamental theorem does not apply.
E. Number of roots of an equation Every rational integral equation _P_ ( _x_ ) = 0 of the _n_ th degree has exactly _n_ roots.
Thus 2 _x_ 3 \+ 5 _x_ 2 – 14 _x_ –8 = 0 has exactly 3 roots, namely 2, – , –4.
Some of the _n_ roots may be equal. Thus the equation of the sixth degree ( _x_ – 2)3 ( _x_ – 5)2 ( _x_ \+ 4) = 0 has 2 as a triple root, 5 as a double root, and –4 as a single root; i.e., the six roots are 2, 2, 2, 5, 5, -4.
### **20.3 SOLVING POLYNOMIAL EQUATIONS**
A. Complex and irrational roots
(1) If a complex number _a + bi_ is a root of the rational integral equation _P(x)_ = 0 with _real coefficients_ , then the conjugate complex number _a – bi_ is also a root.
It follows that every rational integral equation of odd degree with real coefficients has at least one real root.
(2) If the rational integral equation _P_ ( _x_ ) = 0 with _rational coefficients_ has _a_ \+ as a root, where _a_ and _b_ are rational and is irrational, then _a_ – is also a root.
B. Rational root theorem
If _b/c_ , a rational fraction in lowest terms, is a root of the equation
_a nx_ _n_ \+ _a_ n – 1 _x_ _n_ – 1 \+ _a_ _n_ – 2 _x_ _n_ – 2 +... + _a_ 1 _x_ \+ _a_ 0 = 0, _a_ n ≠ 0
with integral coefficients, then _b_ is a factor of a0 and _c_ is a factor of _a n_.
Thus if _b/c_ is a rational root of 6 _x_ 3 \+ 5 _x_ 2 – 3 _x_ – 2 = 0, the values of _b_ are limited to the factors of 2, which are ± 1, ±2; and the values of _c_ are limited to the factors of 6, which are ±1, ±2, ±3, ±6. Hence the only possible rational roots are ±1, ±2, ±1/2, ±1/3, ±1/6, ±2/3.
C. Integral root theorem
It follows that if an equation _P_ ( _x_ ) = 0 has integral coefficients and the lead coefficient is 1 :
_x_ n \+ _a_ _n_ – 1 _x_ _n_ – 1 \+ _a_ n – 2 _x_ _n_ – 2 +... + _a_ 1 _x_ \+ _a_ 0 = 0,
then any rational root of _P_ ( _x_ ) = 0 is an integer and a factor of _a_ 0.
Thus the rational roots, if any, of _x_ 3 \+ 2 _x_ 2 – 11 _x_ – 12 = 0 are limited to the integral factors of 12, which are ±1, ±2, ±3, ±4, ±6, ±12.
D. Intermediate value theorem
If _P_ ( _x_ ) = 0 is a polynomial equation with real coefficients, then approximate values of the real roots of _P_ ( _x_ ) = 0 may be found by obtaining the graph of _y_ = _P_ ( _x_ ) and determining the values of _x_ at the points where the graph intersects the _x_ -axis ( _y_ = 0). Fundamental in this procedure is the fact that if _P_ ( _a_ ) and _P(b)_ have opposite signs then _P_ ( _x_ ) = 0 has at least one root between _x_ = _a_ and _x_ = _b_. This fact is based on the continuity of the graph of _y_ = _P_ ( _x_ ) when _P_ ( _x_ ) is a polynomial with real coefficients.
**EXAMPLE 20.2.** For each real zero of _P_ ( _x_ ) = 2 _x_ 3 – 5 _x_ 2 – 6 _x_ \+ 4 isolate the zero between two consecutive integers.
Since _P_ ( _x_ ) = 2 _x_ 3 – 5 _x_ 2 – 6 _x_ \+ 4 has degree 3, there are at most 3 real zeros. We will look for the real zeros in the interval –5 to 5. The interval is arbitrary and may need to be expanded if the real zeros are not found here. By synthetic division, we will find the value of _P_ ( _x_ ) for each integer in the interval selected. The remainders from the synthetic division are the values of _P_ ( _x_ ) and are summarized in the table below.
Note that _P_ (– 2) = – 20 and _P_ (– 1) = 3 have opposite signs, so from the Intermediate Value Theorem there is a real zero between –2 and –1. Similarly, since _P_ (0) = 4 and _P_ (1) = –5 there is a real zero between 0 and 1, and since _P_ (3) = –5 and _P_ (4) = 28 there is a real zero between 3 and 4. Three real zeros have been isolated, so we have located all the real zeros of _P_ ( _x_ ).
It is not always possible to locate all the real zeros this way because there could be more than one zero between two consecutive integers. When there is an even number of zeros between two consecutive integers the Intermediate Value Theorem will not reveal them when we use just integers for _x_. The Intermediate Value Theorem does not tell you how many real zeros are in the interval, just that there is at least one real zero in the interval.
E. Upper and lower limits for the real roots
A number _a_ is called an _upper limit_ or _upper bound_ for the real roots of _P_ ( _x_ ) = 0 if no root is greater than _a_. A number _b_ is called a _lower limit_ or _lower bound_ for the real roots of _P_ ( _x_ ) = 0 if no root is less than _b_. The following theorem is useful in determining upper and lower limits.
Let _P_ ( _x_ ) = _a_ _n_ _x_ _n_ \+ _a_ _n_ – 1 _x_ _n_ – 1 \+ _a_ _n_ – 2x _n_ – 2 +... + _a_ 0 = 0, where _a_ 0, _a_ 1,..., _a_ _n_ are real and _a_ _n_ > 0.
Then:
(1) If upon synthetic division of _P_ ( _x_ ) by _x_ – _a_ , where _a_ ≥ 0, all of the numbers obtained in the third row are positive or zero, then _a_ is an upper limit for all the real roots of _P_ ( _x_ ) = 0.
(2) If upon synthetic division of _P_ ( _x_ ) by _x_ – _b_ , where _b_ ≤ 0, all of the numbers obtained in the third row are alternately positive and negative (or zero), then _b_ is a lower limit for all the real roots of _P_ ( _x_ ) = 0.
**EXAMPLE 20.3.** Find an interval that contains all the real zeros of _P_ ( _x_ ) = 2 _x_ 3 – 5 _x_ 2 \+ 6.
We will find the integer, b, that is the least upper limit of the real zeros of _P_ ( _x_ ) and the integer, _a_ , that is the lower limit on the real zeros of _P_ ( _x_ ). All real zeros will be in the interval [ _a_ , _b_ ]. To find _a_ and _b_ we use synthetic division on _P_ ( _x_ ) = 2 _x_ 3 – 5 _x_ 2 \+ 6.
When we divide using 3, the quotient row is all positive, so 3 is the smallest integer that is an upper limit for the real zeros of _P_ ( _x_ ). Thus _b_ = 3.
When we divide using –1, the quotient row alternates in sign, so –1 is the greatest integer that is a lower limit for the real zeros of _P_ ( _x_ ). Thus, _a_ = –1.
The real zeros of _P_ ( _x_ ) = 2 _x_ 3 – 5 _x_ 2 \+ 6 are in the interval (–1, 3) or –1 < _x_ < 3. Since _P_ (– 1) ≠ 0 and _P_ (3) = 0, we used interval notation that indicates that neither endpoint is a zero.
F. Descartes' Rule of Signs
If the terms of a polynomial _P_ ( _x_ ) with real coefficients are arranged in order of descending powers of _x_ , a _variation of sign_ occurs when two consecutive terms differ in sign. For example, _x_ 3 – 2 _x_ 2 \+ 3 _x_ – 12 has 3 variations of sign, and 2 _x_ 7 – 6 _x_ 5 – 4 _x_ 4 \+ _x_ 2 – 2 _x_ \+ 4 has 4 variations of sign.
Descartes' Rule of Signs says that the number of positive roots of _P_ ( _x_ ) = 0 is either equal to the number of variations of sign of _P_ ( _x_ ) or is less than that number by an even integer. The number of negative roots of _P_ ( _x_ ) = 0 is either equal to the number of variations of sign of _P_ (– _x_ ) or is less than that number by an even integer.
Thus in _P_ ( _x_ ) = _x_ 9 – 2 _x_ 5 \+ 2 _x_ 2 – 3 _x_ \+ 12 = 0 there are 4 variations of sign of _P_ ( _x_ ); hence the number of positive roots of _P_ ( _x_ ) = 0 is 4, (4 – 2) or (4 – 4). Since _P_ (- _x_ ) = (- _xf_ – 2(- _xf_ \+ 2(- _x_ )2 – 3(- _x_ ) + 12 = – _x_ 9 \+ 2 _x_ 5 \+ 2 _x_ 2 \+ 3 _x_ \+ 12 = 0 has one variation of sign, then _P_ ( _x_ ) = 0 has exactly one negative root. Hence there are 4, 2, or 0 positive roots, 1 negative root, and at least 9 – (4 + 1) = 4 nonreal roots. (There are 4, 6, or 8 nonreal roots. Why?)
### **20.4 APPROXIMATING REAL ZEROS**
In solving a polynomial equation _P_ ( _x_ ) = 0, it is not always possible to find all the zeros by the previous methods. We have been able to determine the irrational and imaginary zeros when we were able to find quadratic factors that we could solve using the quadratic formula. If we cannot find the quadratic factors of _P_ ( _x_ ) = 0, we will not be able to solve for the imaginary zeros, but we can often find an approximation for some of the real zeros.
To approximate a real zero of _P_ ( _x_ ) = 0, we must first find an interval that contains a real zero of _P_ ( _x_ ) = 0. We can do this using the Intermediate Value Theorem to locate to numbers _a_ and _b_ such that _P_ ( _a_ ) and _P_ ( _b_ ) have opposite signs. We keep using the Intermediate Value Theorem until we have isolated the real zero in an interval small enough that it will be known to the desired degree of accuracy.
**EXAMPLE 20.4.** Find a real zero of _x_ 3 \+ 3 _x_ \+ 8 = 0 correct to two decimal places.
By Descartes' Rule of Signs, _P_ ( _x_ ) = _x_ 3 \+ 3 _x_ \+ 8 has no positive real zeros and 1 negative real zero.
Using synthetic division, we find _P_ (–2) = –6 and _P_ (– 1) = 4, so by the Intermediate Value Theorem _P_ ( _x_ ) = _x_ 3 \+ 3 _x_ \+ 8 has a real zero between –2 and –1.
We now use synthetic division and the Intermediate Value Theorem to determine the tenths interval containing the zero. The results are summarized in the table below.
We can see that _P_ (–1.5) is positive and _P_ (–1.6) is negative so the zero is between – 1.6 and – 1.5.
Now we check for the hundredths digit by using synthetic on the interval between –1.6 and –1.5. We do not have to find all the hundredths values, just a sign change between two consecutive values.
We see that _P_ (–1.51) is positive and _P_ (–1.52) is negative, so by the Intermediate Value Theorem there is a real zero between –1.51 and –1.52.
Since the real zero is located between –1.51 and –1.52, we just need to determine whether it rounds off to –1.51 or – 1.52. To do this we find _P(–_ 1.515), which is about –0.02. This value of _P(–_ 1.515) is negative and _P_ (–1.51) is positive, so we know that the zero is between –1.515 and –1.510, and all the numbers in this interval rounded to two decimal places are –1.51.
Thus, to two decimal places the only real zero of _x_ 3 \+ 3 _x_ \+ 8 = 0 is –1.51.
To use a graphing calculator to approximate the real zeros of a polynomial we graph the function and use the trace and zoom features of the calculator. After we graph the function, we use the trace feature to locate an interval that contains a real zero using the Intermediate Value Theorem. We then use the zoom feature to focus in on this interval. We continue using the trace and zoom feature until we find two _x_ values that round to the desired degree of accuracy and have function values that are opposite in sign.
### **Solved Problems**
**20.1** Prove the remainder theorem: If a polynomial _P_ ( _x_ ) is divided by ( _x_ – _r_ ) the remainder is _P_ ( _r_ ).
**SOLUTION**
In the division of _P_ ( _x_ ) by ( _x_ – _r_ ), let _Q_ ( _x_ ) be the quotient and _R_ , a constant, the remainder. By definition _P_ ( _x_ ) = ( _x_ – _r_ ) _Q_ ( _x_ ) + _R_ , an identity for all values of _x_. Letting _x_ = _r_ , _P_ ( _r_ ) = _R_.
**20.2** Determine the remainder _R_ after each of the following divisions.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
**20.3** Prove the factor theorem: If _r_ is a root of the equation _P_ ( _x_ ) = 0, then ( _x_ – _r_ ) is a factor of _P_ ( _x_ ); and conversely if ( _x_ – _r_ ) is a factor of _P_ ( _x_ ), then _r_ is a root of _P_ ( _x_ ) = 0.
**SOLUTION**
In the division of _P_ ( _x_ ) by ( _x_ – _r_ ), let _Q_ ( _x_ ) be the quotient and _R_ , a constant, the remainder. Then _P_ ( _x_ ) = ( _x_ – _r_ ) _Q_ ( _x_ ) + _R_ or _P_ ( _x_ ) = ( _x_ – _r_ ) _Q_ ( _x_ ) + _P_ ( _r_ ) by the remainder theorem.
If _r_ is a root of _P_ ( _x_ ) = 0, then _P_ ( _r_ ) = 0. Hence _P_ ( _x_ ) = ( _x_ – _r_ ) _Q_ ( _x_ ), or ( _x_ – _r_ ) is a factor of _P_ ( _x_ ).
Conversely if ( _x_ – _r_ ) is a factor of _P_ ( _x_ ), then the remainder in the division of _P_ ( _x_ ) by ( _x_ – _r_ ) is zero. Hence _P_ ( _r_ ) = 0, i.e., _r_ is a root of _P_ ( _x_ ) = 0.
**20.4** Show that ( _x_ – 3) is a factor of the polynomial _P_ ( _x_ ) = _x_ 4 – 4 _x_ 3 – 7 _x_ 2 \+ 22 _x_ \+ 24.
**SOLUTION**
_P_ (3) = 81 – 108 – 63 + 66 + 24 = 0. Hence ( _x_ – 3) is a factor of _P_ ( _x_ ), 3 is a _zero_ of the polynomial _P_ ( _x_ ), and 3 is a _root_ of the equation _P_ ( _x_ ) = 0.
**20.5**
( _a_ ) Is – 1 a root of the equation _P_ ( _x_ ) = _x_ 3 – 7 _x_ – 6 = 0?
( _b_ ) Is 2 a root of the equation _P_ ( _y_ ) = _y_ 4 – 2y2 – _y_ \+ 7 = 0?
( _c_ ) Is 2 _i_ a root of the equation _P_ ( _z_ ) = 2 _z_ 3 \+ 3z2 \+ 8 _z_ \+ 12 = 0?
**SOLUTION**
( _a_ ) _P_ (–1) = –1 + 7 – 6 = 0. Hence – 1 is a root of the equation _P_ ( _x_ ) = 0, and [ _x_ – (– 1)] = _x_ \+ 1 is a factor of the polynomial _P_ ( _x_ ).
( _b_ ) _P_ (2) = 16 – 8 – 2 + 7 = 13. Hence 2 is not a root of _P_ ( _y_ ) = 0, and ( _y_ – 2) is not a factor of _y_ 4 – 2 _y_ 2 – _y_ \+ 7.
( _c_ ) _P_ (2 _i_ ) = 2(2 _i_ )3 \+ 3(2 _i_ )2 \+ 8(2 _i_ ) + 12 = – 16 _i_ – 12 + 16 _i_ \+ 12 = 0. Hence 2 _i_ is a root of _P_ ( _z_ ) = 0, and ( _z_ – 2 _i_ ) is a factor of the polynomial _P_ ( _z_ ).
**20.6** Prove that _x_ – _a_ is a factor of _x_ _n_ – _a_ _n_ , if _n_ is any positive integer.
**SOLUTION**
_P_ ( _x_ ) = _x_ _n_ – _a_ _n_ ; then _P_ ( _a_ ) = _a_ _n_ – _a_ _n_ = 0. Since _P_ ( _a_ ) = 0, _x_ – _a_ is a factor of _x_ _n_ – _a_ _n_.
**20.7** _(a)_ Show that _x_ 5 \+ _a_ 5 is exactly divisible by _x_ \+ _a_.
( _b_ ) What is the remainder when _y_ 6 \+ _a_ 6 is divided by _y_ \+ _a_?
**SOLUTION**
( _a_ ) _P_ ( _x_ ) = _x_ 5 \+ _a_ 5; then _P_ (– _a_ ) = (– _a_ )5 \+ _a_ 5 = – _a_ 5 \+ _a_ 5 = 0. Since _P_ (– _a_ ) = 0, _x_ 5 \+ _a_ 5 is exactly divisible by _x_ \+ _a_.
( _b_ ) _P_ ( _y_ ) = _y_ 6 \+ _a_ 6. Remainder = _P_ (– _a_ ) = (– _a_ )6 \+ _a_ 6 = _a_ 6 \+ _a_ 6 = 2 _a_ 6.
**20.8** Show that _x_ \+ _a_ is a factor of _x_ _n_ – _a_ _n_ when _n_ is an even positive integer, but not a factor when _n_ is an odd positive integer. Assume _a_ ≠ 0.
**SOLUTION**
_P(x)_ = _x_ _n_ – _d_ _n_.
When _n_ is even, _P_ (– _a_ ) = (– _a_ ) _n_ – _a_ _n_ = _a_ _n_ – _a_ _n_ = 0. Since _P_ (– _a_ ) = 0, _x_ \+ _a_ is a factor of _x_ _n_ – _a_ _n_ when _n_ is even.
When _n_ is odd, _P_ (– _a_ ) = (– _a_ ) _n_ – _a_ _n_ = – _a_ _n_ – _a_ _n_ = –2 _a_ _n_. Since _P_ (– _a_ ) ≠ 0, _x_ _n_ – _a_ _n_ is not exactly divisible by _x_ \+ _a_ when _n_ is odd (the remainder being –2 _a_ _n_ ).
**20.9** Find the values of _p_ for which
( _a_ ) 2 _x_ 3 – _px_ 2 \+ 6 _x_ – 3 _p_ is exactly divisible by _x_ \+ 2,
( _b_ ) ( _x_ 4 – _p_ 2 _x_ \+ 3 – _p_ ) ÷ ( _x_ – 3) has a remainder of 4.
**SOLUTION**
( _a_ ) The remainder is 2(– 2)3 – _p_ (– 2)2 \+ 6(–2) – 3 _p_ = –16 – 4 _p_ – 12 – 3 _p_ = –28 – 7 _p_ = 0. Then _p_ = –4.
( _b_ ) The remainder is 34 – _p_ 2(3) + 3 – _p_ = 84 – 3 _p_ 2 – _p_ = 4. Then 3 _p_ 2 \+ _p_ – 80 = 0, ( _p_ – 5)(3 _p_ \+ 16) = 0 and _p_ = 5, –16/3.
**20.10** By synthetic division, determine the quotient and remainder in the following.
(3 _x_ 5 – 4 _x_ 4 – 5 _x_ 3 – 8 _x_ \+ 25) ÷ ( _x_ – 2)
**SOLUTION**
The top row of figures gives the coefficients of the dividend, with zero as the coefficient of the missing power of _x_ (0 _x_ 2). The 2 at the extreme left is the second term of the divisor with the sign changed (since the coefficient of _x_ in the divisor is 1).
The first coefficient of the top row, 3, is written first in the third row and then multiplied by the 2 of the divisor. The product 6 is placed first in the second row and added to the –4 above it to give 2, which is the next figure in the third row. This 2 is then multiplied by the 2 of the divisor. The product 4 is placed in the second row and added to the –5 above it to give the – 1 in the third row; etc. The last figure in the third row is the Remainder, while all the figures to its left constitute the coefficients of the Quotient.
Since the dividend is of the 5th degree and the divisor of the 1st degree, the quotient is of the 4th degree.
The answer may be written:
**20.11** ( _x_ 4 – 2 _x_ 3 – 24 _x_ 2 \+ 15 _x_ \+ 50) ÷ ( _x_ \+ 4)
**SOLUTION**
**20.12** (2 _x_ 4 – 17 _x_ 2 – 4) ÷ ( _x_ \+ 3)
**SOLUTION**
**20.13** (4 _x_ 3 – 10 _x_ 2 \+ _x_ – 1) ÷ ( _x_ \+ 1/2)
**SOLUTION**
**20.14** Given _P_ ( _x_ ) = _x_ 3 – 6 _x_ 2 – 2 _x_ \+ 40, compute ( _a_ ) _p_ (–5) and ( _b_ ) _P_ (4) using synthetic division.
**SOLUTION**
( _a_ )
( _b_ )
**20.15** Given that one root of _x_ 3 \+ 2 _x_ 2 – 23 _x_ \- 60 = 0 is 5, solve the equation.
**SOLUTION**
The depressed equation is _x_ 2 \+ 7 _x_ \+ 12 = 0, whose roots are –3, –4. The three roots are 5, –3, –4.
**20.16** Two roots of _x_ 4 – 2 _x_ 2 – 3 _x_ – 2 = 0 are – 1 and 2. Solve the equation.
**SOLUTION**
The first depressed equation is _x_ 3 – _x_ 2 – _x_ – 2 = 0.
The second depressed equation is _x_ 2 \+ _x_ \+ 1 = 0, whose roots are
The four roots are –1, 2,
**20.17** Determine the roots of each of the following equations.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
**20.18** Write the equation having only the following roots.
( _a_ ) 5, 1, –3;
( _b_ ) 2, –1/4, –1/2;
( _c_ )
( _d_ ) 0, 1 ± 5 _i_.
**SOLUTION**
( _a_ ) ( _x_ – 5)( _x_ – 1)( _x_ \+ 3) = 0 or _x_ 3 – 3 _x_ 2 – 13 _x_ \+ 15 = 0.
( _b_ ) which has integral coefficients.
( _c_ )
( _d_ )
**20.19** Form the equation with integral coefficients having only the following roots.
( _a_ )
( _b_ )
( _c_ )
( _d_ ) 2 as a triple root, –1.
**SOLUTION**
( _a_ ) ( _x_ – 1)(2 _x_ – 1)(3 _x_ \+ 1) = 0 or 6 _x_ 3 – 7 _x_ 2 \+ 1 = 0
( _b_ ) _x_ (4 _x_ – 3)(3 _x_ – 2)( _x_ \+ 1) = 0 or 12 _x_ 4 – 5 _x_ 3 – 11 _x_ 2 \+ 6 _x_ = 0
( _c_ )
or 2 _x_ 4 \+ 17 _x_ 2 – 9 = 0
( _d_ ) ( _x_ – 2)3( _x_ \+ 1) = 0 or _x_ 4 – 5 _x_ 3 \+ 6 _x_ 2 \+ 4 _x_ – 8 = 0
**20.20** Each given number is a root of a polynomial equation with _real coefficients_. What other number is a root?
( _a_ ) 2 _i_ ,
( _b_ ) –3 + 2 _i_ ,
( _c_ )
**SOLUTION**
( _a_ ) –2 _i_ ,
( _b_ ) –3 – 2 _i_ ,
( _c_ )
**20.21** Each given number is a root of a polynomial equation with _rational coefficients_. What other number is a root?
( _a_ )
( _b_ )
( _c_ )
**SOLUTION**
( _a_ )
( _b_ )
( _c_ )
**20.22** Criticize the validity of each of the following conclusions.
( _a_ ) _x_ 3 \+ 7 _x_ – 6 _i_ = 0 has _x_ = _i_ as a root; hence _x_ = – _i_ is a root.
( _b_ )
( _c_ )
**SOLUTION**
( _a_ ) _x_ = – _i_ is not necessarily a root, since not all the coefficients of the given equation are real. By substitution it is, in fact, found that _x_ = – _i_ is not a root.
( _b_ ) The conclusion is valid, since the given equation has real coefficients.
( _c_ ) is not necessarily a root, since not all the coefficients of the given equation are _rational. By_ substitution it is found that is not a root.
**20.23** Write the polynomial equation of lowest degree with real coefficients having 2 and 1 – 3 _i_ as two of its roots.
**SOLUTION**
( _x_ – 2)[ _x_ – (1 – 3 _i_ )][ _x_ – (1 + 3 _i_ )] = ( _x_ – 2)( _x_ 2 – 2 _x_ \+ 10) = 0 or _x_ 3 – 4 _x_ 2 \+ 14 _x_ – 20 = 0
**20.24** Form the polynomial equation of lowest degree with rational coefficients having and –6 as two of its roots.
**SOLUTION**
**20.25** Form the quartic polynomial equation with rational coefficients having as two of its roots
( _a_ )
( _b_ )
**SOLUTION**
( _a_ )
( _b_ ) = ( _x_ 2 – 4 _x_ \+ 5)( _x_ 2 – 2 _x_ – 2) = 0 or _x_ 4 – 6 _x_ 3 \+ 11 _x_ 2 – 2 _x_ – 10 = 0
**20.26** Find the four roots of _x_ 4 \+ 2 _x_ 2 \+ 1 = 0.
**SOLUTION**
_x_ 4 \+ 2 _x_ 2 \+ 1 = ( _x_ 2 \+ 1)2 = [( _x_ \+ _i_ )( _x_ – _i_ )]2 = 0. The roots are _i, i, –i, –i_.
**20.27** Solve _x_ 4 – 3 _x_ 3 \+ 5 _x_ 2 – 27 _x_ – 36 = 0, given that one root is a pure imaginary number of the form _bi_ where _b_ is real.
**SOLUTION**
Substituting _bi_ for _x, b_ 4 \+ 3 _b_ 3 _i_ – 5 _b_ 2 – 27 _bi_ – 36 = 0.
Equating real and imaginary parts to zero:
_b_ 4 – 5 _b_ 2 – 36 = 0, _(b_ 2 – 9)( _b_ 2 \+ 4) = 0 and _b_ = ±3 since _b_ is real;
3 _b_ 3 – 27 _b_ = 0, 3 _b(b_ 2 – 9) = 0 and _b_ = 0, ±3.
The common solution is _b_ = ±3; hence two roots are ± _3i_ and ( _x_ – 3 _i_ )( _x_ \+ 3 _i_ ) = _x_ 2 \+ 9 is a factor of _x_ 4 – 3 _x_ 3 \+ 5 _x_ 2 – 27 _x_ – 36. By division the other factor is _x_ 2 – 3 _x_ – 4 = ( _x_ – 4)( _x_ \+ 1), and the other two roots are 4, – 1.
The four roots are ±3 _i_ , 4, –1.
**20.28** Form the polynomial equation of lowest degree with _rational coefficients_ , one of whose roots is
( _a_ )
( _b_ )
**SOLUTION**
( _a_ )
Squaring both sides,
Squaring again, _x_ 4 – 10 _x_ 2 \+ 25 = 24 and _x_ 4 – 10 _x_ 2 \+ 1 = 0.
( _b_ )
Squaring both sides,
Squaring again, _x_ 4 – 2 _x_ 2 \+ 1 = –8 and _x_ 4 – 2 _x_ 2 \+ 9 = 0.
**20.29** ( _a_ ) Write the polynomial equation of lowest degree with _constant_ (real or complex) coefficients having the roots 2 and 1 – 3 _i_. Compare with Problem 20.23 above.
( _b_ ) Write the polynomial equation of lowest degree with _real_ coefficients having the roots –6 and – Compare with Problem 20.24 above.
**SOLUTION**
( _a_ )
( _b_ )
**20.30** Obtain the rational roots, if any, of each of the following polynomial equations.
( _a_ ) _x_ 4 – 2 _x_ 2 – 3 _x_ – 2 = 0
The rational roots are limited to the integral factors of 2, which are ±1, ±2.
Testing these values for _x_ in order, +1, –1, +2, –2, by synthetic division or by substitution, we find that the only rational roots are – 1 and 2.
( _b_ ) _x_ 3 – _x_ – 6 = 0
The rational roots are limited to the integral factors of 6, which are ±1, ±2, ±3, ±6.
Testing these values for _x_ in order, +1, –1, +2, –2, +3, –3, +6, –6, the only rational root obtained is 2.
( _c_ ) 2 _x_ 3 \+ _x_ 2 – 7 _x_ – 6 = 0
If _b/c_ (in lowest terms) is a rational root, the only possible values of _b_ are ±1, ±2, ±3, ±6; and the only possible values of _c_ are ±1, ±2. Hence the possible rational roots are limited to the following numbers: ±1, ±2, ±3, ±6, ±1/2, ±3/2.
Testing these values for _x_ , we obtain –1, 2, –3/2 as the rational roots.
_(d_ ) 2 _x_ 4 \+ _x_ 2 \+ 2 _x_ – 4 = 0
If _b/c_ is a rational root, the values of _b_ are limited to ±1, ±2, ±4; and the values of _c_ are limited to ±1, ±2. Hence the possible rational roots are limited to the numbers ±1, ±2, ±4, ±1/2.
Testing these values for _x_ , we find that there are no rational roots.
**20.31** Solve the polynomial equation _x_ 3 – 2 _x_ 2 – 31 _x_ \+ 20 = 0.
**SOLUTION**
Any rational root of this equation is an integral factor of 20. Then the possibilities for rational roots are: ±1, ±2, ±4, ±5, ±10, ±20.
Testing these values for _x_ by synthetic division, we find that –5 is a root.
The depressed equation _x_ 2 – 7 _x_ \+ 4 = 0 has irrational roots
Hence the three roots of the given equation are
**20.32** Solve the polynomial equation 2 _x_ 4 – 3 _x_ 3 – 7 _x_ 2 – 8 _x_ \+ 6 = 0.
**SOLUTION**
If _b/c_ is a rational root, the only possible values of _b_ are ±1, ±2, ±3, ±6; and the only possible values of _c_ are ±1, ±2. Hence the possibilities for rational roots are ±1, ±2, ±3, ±6, ±1/2, ±3/2.
Testing these values of _x_ by synthetic division, we find that 3 is a root.
The first depressed 2 _x_ 3 \+ 3 _x_ 2 \+ 2 _x_ – 2 = 0 is tested and 1/2 is obtained as a root.
The second depressed equation 2 _x_ 2 \+ 4 _x_ \+ 4 = 0 or _x_ 2 \+ 2 _x_ \+ 2 = 0 has the nonreal roots –1 ± _i_. The four roots are 3, 1/2, –1 ± _i_.
**20.33** Prove that is an irrational number.
**SOLUTION**
Squaring again, _x_ 4 – 10 _x_ 2 \+ 25 = 24 or _x_ 4 – 10 _x_ 2 \+ 1 = 0. The only possible rational roots of this equation are + 1. Testing these values, we find that there is no rational root. Hence _x_ = is irrational.
**20.34** Graph _P(x_ ) = _x_ 3 \+ _x_ – 3. From the graph determine the number of positive, negative and nonreal roots of _x_ 3 \+ _x_ – 3 = 0.
**SOLUTION**
From the graph it is seen that there is one positive and no negative real root (see Fig. 20-1). Hence there are two conjugate nonreal roots.
**20.35** Find upper and lower limits of the real roots of ( _a_ ) _x_ 3 – 3 _x_ 2 \+ 5 _x_ \+ 4 = 0, ( _b_ ) _x_ 3 \+ _x_ 2 – 6 = 0.
**Fig. 20-1**
**SOLUTION**
( _a_ ) The possible rational roots are ±1, ±2, ±4.
_Testing for upper limit_.
Since all the numbers in the third row of the synthetic division of _P(x_ ) by _x_ – 3 are positive (or zero), an upper limit of the roots is 3, i.e., no root is greater than 3.
_Testing for lower limit_.
Since the numbers in the third row are alternately positive and negative, – 1 is a lower limit of the roots, i.e., no root is less than – 1.
( _b_ ) The possible rational roots are ±1, ±2, ±3, ±6.
_Testing for upper limit_.
Hence 2 is an upper limit of the roots.
_Testing for lower limit_.
Since all the numbers of the third row are alternately positive and negative (or zero), a lower limit to the roots is –1.
**20.36** Determine the rational roots of 4 _x_ 3 \+ 15 _x_ – 36 = 0 and thus solve the equation completely.
**SOLUTION**
The possible rational roots are ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36, ±1/2, ±3/2, ±9/2, ±1/4, ±3/4, ±9/4. To avoid testing all of these possibilities, find upper and lower limits of the roots.
_Testing for upper limit_.
Hence no ( _real_ ) root is greater than or equal to 2.
_Testing for lower limit_.
Hence no real root is less than or equal to –1.
The only possible rational roots greater than –1 and less than 2 are +1, ±1/2, ±3/2, ±1/4, ±3/4. Testing these we find that 3/2 is the only rational root.
The other roots are solutions of 4 _x_ 2 \+ 6 _x_ \+ 24 = 0 or 2 _x_ 2 \+ _3x +_ 12 = 0, i.e.,
**20.37** Employing Descartes' Rule of Signs, what may be inferred as to the number of positive, negative and nonzero roots of the following equations?
( _a_ ) 2 _x_ 3 \+ 3 _x_ 2 – 13 _x_ \+ 6 = 0
( _b_ ) _x_ 4 – 2 _x_ 2 – 3 _x_ – 2 = 0
( _c_ ) _x_ 2 – 2 _x_ \+ 7 = 0
( _d_ ) 2x4 \+ 7 _x_ 2 \+ 6 = 0
( _e_ ) _x_ 4 – 3 _x_ 2 – 4 = 0
( _f_ ) _x_ 3 \+ 3 _x_ – 14 = 0
( _g_ ) _x_ 6 \+ _x_ 3 – 1 = 0
( _h_ ) _x_ 6 – 3 _x_ 2 – 4 _x_ \+ 1 = 0
**SOLUTION**
( _a_ ) There are 2 variations of sign in P( _x_ ) = 2 _x_ 3 \+ 3 _x_ 2 – 13 _x_ \+ 6. There is 1 variation of sign in _P_ (– _x_ ) = –2 _x_ 3 \+ 3 _x_ 2 \+ 13 _x_ \+ 6. Hence there are at most 2 positive roots and 1 negative root.
The roots may be: (1)2 positive, 1 negative, 0 nonreal; or (2) 0 positive, 1 negative, 2 nonreal. (Nonreal roots occur in conjugate pairs.)
( _b_ ) There is 1 variation of sign in _P_ ( _x_ ) = _x_ 4 – 2 _x_ 2 – 3 _x_ – 2, and 3 variations of sign in _P_ (– _x_ ) = _x_ 4 – 2 _x_ 2 \+ 3 _x_ – 2. Hence there are at most 1 positive root and 3 negative roots.
( _c_ ) There are 2 variations of sign in _P_ ( _x_ ) = _x_ 2 – 2 _x_ \+ 7, and no variation of sign in _P_ (– _x_ ) = _x_ 2 \+ 2 _x_ \+ 7.
( _d_ ) Neither _P_ ( _x_ ) = 2 _x_ 4 \+ 7 _x_ 2 \+ 6 nor _P_ (– _x_ ) = 2 _x_ 4 \+ 7 _x_ 2 \+ 6 has a variation of sign. Hence all 4 roots are nonreal, since _P_ (0) ≠ 0.
( _e_ ) There is 1 variation of sign in _P_ ( _x_ ) = _x_ 4 – 3 _x_ 2 – 4 = 0, and 1 variation of sign in _P_ (– x) = _x_ 4 – 3 _x_ 2 – 4.
Hence the roots are: 1 positive, 1 negative, 2 nonreal.
( _f_ ) There is 1 variation of sign in _P_ ( _x_ ) = _x_ 3 \+ 3 _x_ – 14, and no variation in _P_ (– _x_ ) = –x3 – 3 _x_ – 14.
Hence the roots are: 1 positive, 2 nonreal.
( _g_ ) There is 1 variation of sign in _P_ ( _x_ ) = _x_ 6 \+ _x_ 3 – 1, and 1 variation in _P_ (– _x_ ) = _x_ 6 – _x_ 3 – 1.
Hence the roots are: 1 positive, 1 negative, 4 nonreal.
( _h_ ) There are 2 variations of sign in _P_ ( _x_ ) = _x_ 6 – 3 _x_ 2 – 4 _x_ \+ 1, and 2 variations of sign in _P_ (– _x_ ) = _x_ 6 – 3 _x_ 2 \+ 4 _x_ \+ 1.
Hence the roots may be:
(1) 2 positive, 2 negative, 2 nonreal;
(2) 2 positive, 0 negative, 4 nonreal;
(3) 0 positive, 2 negative, 4 nonreal;
(4) 0 positive, 0 negative, 6 nonreal.
**20.38** Determine the nature of the roots of x _n_ – 1 = 0 when _n_ is a positive integer and ( _a_ ) _n_ is even, ( _b_ ) _n_ is odd.
**SOLUTION**
( _a_ ) P( _x_ ) = x _n_ – 1 has 1 variation of sign, and _P_ (– _x_ ) = _x_ n\- 1 has 1 variation of sign. Hence the roots are: 1 positive, 1 negative, ( _n_ – 2) nonreal.
( _b_ ) P( _x_ ) = x _n_ 1 has 1 variation of sign, and _P_ (– _x_ ) = – _x_ _n_ – 1 has no variation of sign. Hence the roots are: 1 positive, 0 negative, ( _n_ – 1) nonreal.
**20.39** Obtain the rational roots, if any, of each equation, making use of Descartes' Rule of Signs.
( _a_ ) _x_ 3 – _x_ 2 \+ 3 _x_ – 27 = 0,
( _b_ ) _x_ 3 \+ 2 _x_ \+ 12 = 0,
( _c_ )2 _x_ 5 \+ _x_ – 66 = 0,
( _d_ )3 _x_ 4 \+ 7 _x_ 2 \+ 6 = 0.
**SOLUTION**
( _a_ ) By Descartes' Rule of Signs, the equation has 3 or 1 positive roots and no negative root. Hence the rational roots are limited to positive integral factors of 27, which are 1, 3, 9, 27.
Testing these values for _x_ , the only rational root obtained is 3.
( _b_ ) By the rule of signs, the equation has no positive root and 1 negative root. Hence the rational roots are limited to the negative integral factors of 12, i.e., to –1, –2, –3, –4, –6, –12.
Testing these values for _x_ , we obtain –2 as the only rational root.
( _c_ ) By the rule of signs, the equation has 1 positive root and no negative root. Hence the rational roots are limited to positive rational numbers of the form _b/c_ where _b_ is limited to integral factors of 66 and c is limited to integral factors of 2. The possible rational roots are then 1, 2, 3, 6, 11, 22, 33, 66, 1/2, 3/2, 11/2, 33/2.
Testing these values for _x_ , we obtain 2 as the only rational root.
( _d_ ) The equation has no real root, since neither _P_ ( _x_ ) = 3 _x_ 4 \+ 7 _x_ 2 \+ 6 nor _P_ (– _x_ ) = 3 _x_ 4 \+ 7 _x_ 2 \+ 6 has a variation of sign and _P_ (0) ≠ 0.
Hence all of its four roots are nonreal.
**20.40** Use the Intermediate Value Theorem to isolate each of the real zeros of P( _x_ ) between two consecutive integers.
( _a_ ) P( _x_ ) = 3 _x_ 3 – 8 _x_ 2 – 8 _x_ \+ 8
( _b_ ) P( _x_ ) = 5 _x_ 3 – 4 _x_ 2 – 10 _x_ \+ 8
**SOLUTION**
( _a_ ) For _P_ ( _x_ ) = 3 _x_ 3 – 8 _x_ 2 – 8 _x_ \+ 8, we find the upper and lower limits of the real zeros.
Since the quotient row when synthetic division is done with 4 is all positive, the upper limit of the real zeros of _P_ ( _x_ ) is 4.
Since the quotient row when synthetic division is done with –2 alternates in sign, the lower limit of the real zeros of _P_ ( _x_ ) is –2.
We now examine the interval from –2 to 4 to isolate the real zeros of _P_ ( _x_ ) between consecutive integers. Since _P_ (0) = 8 and _P_ (1) = – 5 there is a real zero between 0 and 1. Since P(3) = – 7 and _P_ (4) = 40 there is a real zero between 3 and 4. Since _P_ (– 1) = 5 and _P_ (– 2) = –32, there is a real zero between –2 and – 1.
The real zeros of _P_ ( _x_ ) = 3 _x_ 3 – 8 _x_ 2 – 8 _x_ \+ 8 are between –2 and –1, 0 and 1, and 3 and 4.
( _b_ ) For _P_ ( _x_ ) = 5 _x_ 3 – 4 _x_ 2 – 10 _x_ \+ 8, we find the upper and lower limits of the real zeros.
The upper limit for the zeros of _P_ ( _x_ ) is 2.
The lower limit of the real zeros of _P_ ( _x_ ) is –2.
We now examine the interval from –2 to 2 to isolate the real zeros of _P_ ( _x_ ). Since P( _0_ ) = 8 and P( _1_ ) = – 1, there is a real zero between 0 and 1. Since P( _1_ ) = – 1 and P( _2_ ) = 12, there is a real zero between 1 and 2. Since _P_ (– 1) = 9 and _P_ (– 2) = –28, there is a real zero between –2 and –1.
Thus, the real zeros of _P_ ( _x_ ) = 5 _x_ 3 – 4 _x_ 2 – 10 _x_ \+ 8 are located between – 2 and –1, 0 and 1, and 1 and 2.
**20.41** Approximate a real zero of _P_ ( _x_ ) = _x_ 3 – _x_ – 5 to two decimal places.
**SOLUTION**
By Descartes' Rule of Signs, _P_ ( _x_ ) = _x_ 3 – _x_ – 5 has 1 positive real zero and 2 or 0 negative real zeros.
Now we determine the upper limit of real zeros of _P_ ( _x_ ).
The upper limit on the real zeros of _P_ ( _x_ ) is 2.
Since _P_ ( _1_ ) = –5 and _P_ ( _2_ ) = 1, the positive real zero is between 1 and 2.
We now determine the tenths interval of the zero. Using synthetic division, we determine the tenths values until we find two values with different signs. Since _P_ ( _2_ ) = 1 is closer to 0 than _P_ ( _1_ ) = – 5, we start with _x_ = 1.9. Since _P_ (1.9) = –0.041 and _P_ (2.0) = 1, the real zero is between 1.9 and 2.0.
Since _P_ (1.90) = –0.41 is closer to 0 than _P_ (2.0) = 1, we look for the hundredths interval starting with _x_ = 1.91. _P_ (1.91) = 0.579. Since _P_ (1.90) = –0.041 and _P_ (1.91) = 0.058, there is a real zero between 1.90 and 1.91.
We now determine _P_ (1.905) to decide whether the zero rounds off to 1.90 or 1.91. _P_ (1.905) = 0.008. Since _P_ (1.900) is negative and _P_ (1.905) is positive, the zero is between 1.900 and 1.905. When rounded to two decimal places, all numbers in this interval round to 1.90.
Thus, rounded to two decimal places, a real zero of _P_ ( _x_ ) = _x_ 3 – _x_ – 5 is 1.90.
**20.42** Approximate to three decimal places.
**SOLUTION**
= 3 and _P_ ( _x_ ) = _x_ 3 – 3 = 0.
By Descartes' Rule of Signs, _P_ ( _x_ ) has one positive real zero and no negative real zeros.
Since _P_ ( _1_ ) = –2 and _P_ ( _2_ ) = 5, the zero is between 1 and 2.
Since _P_ ( _1_ ) is closer to 0 than _P_ ( _2_ ), locate the tenths interval by evaluating _P_ ( _x_ ) from _x_ = 1 to _x_ = 2 starting with _x_ = 1.1. Once a sign change in _P_ ( _x_ ) is found we stop.
Since _P_ (1.4) is negative and _P_ (1.5) is positive, the real zero is between 1.4 and 1.5.
Now determine the hundredths interval by exploring the values of _P_ ( _x_ ) in the interval from x= 1.40 to x= 1.50.
Since _P_ (1.4) is negative and _P_ (1.5) is positive, the zero is between 1.4 and 1.5.
The next step is to determine the thousandths interval for the zero by exploring the values of _P_ ( _x_ ) in the interval between _x_ = 1.440 and _x_ = 1.450.
Since _P_ (1.442) is negative and _P_ (1.443) is positive, the zero is between 1.442 and 1.443.
Since _P_ (1.4425) = 0.002, the real zero is between 1.4420 and 1.4425. All values in the interval from 1.4420 to 1.4425 round to 1.442 to three decimal places.
Therefore, = 1.442 to three decimal places.
### **Supplementary Problems**
**20.43**
**20.44** Determine the remainder in each of the following.
( _a_ ) (2 _x_ 5 – 7) ÷ ( _x_ \+ 1)
( _b_ )( _x_ 3 \+ 3 _x_ 2 – 4 _x_ \+ 2) ÷ ( _x_ – 2)
( _c_ ) (3 _x_ 3 \+ 4 _x_ – 4) ÷ ( _x_ – )
( _d_ ) (4 _y_ 3 \+ _y_ \+ 27) ÷ (2 _y_ \+ 3)
( _e_ )
( _f_ ) (2 _x_ 33 \+ 35) ÷ ( _x_ \+ 1)
**20.45** Prove that _x_ \+ 3 is a factor of _x_ 3 \+ 7 _x_ 2 \+ 10 _x_ – 6 and that _x_ = –3 is a root of the equation _x_ 3 \+ 7 _x_ 2 \+ 10 _x_ – 6 = 0.
**20.46** Determine which of the following numbers are roots of the equation _y_ 4 \+ 3 _y_ 3 \+ 12 _y_ – 16 = 0:
( _a_ ) 2,
( _b_ ) -4,
( _c_ ) 3,
( _d_ ) 1,
( _e_ ) _2i_.
**20.47** Find the values of _k_ for which
( _a_ ) 4 _x_ 3 \+ 3 _x_ 2 – k _x_ \+ 6 _k_ is exactly divisible by _x_ \+ 3,
( _b_ ) _x_ 5 \+ 4 _kx_ – 4 _k_ 2 = 0 has the root _x_ = 2.
**20.48** By synthetic division determine the quotient and remainder in each of the following.
( _a_ ) (2 _x_ 3 \+ 3 _x_ 2 – 4 _x_ – 2) ÷ ( _x_ \+ 1)
( _b_ ) (3 _x_ 5 \+ _x_ 3 – 4) ÷ ( _x_ – 2)
( _c_ ) ( _y_ 6 – 3 _y_ 5 \+ 4 _y_ – 5) ÷ ( _y_ \+ 2)
( _d_ ) (4 _x_ 3 \+ 6 _x_ 2 – 2 _x_ \+ 3) ÷ (2 _x_ \+ 1)
**20.49** If _P_ ( _x_ ) = 2 _x_ 4 – 3 _x_ 3 \+ 4 _x_ – 4, compute _P_ (2) and _P_ (–3) using synthetic division.
**20.50** Given that one root of _x_ 3 – 7 _x_ – 6 = 0 is –1, find the other two roots.
**20.51** Show that 2 _x_ 4 – _x_ 3 – 3 _x_ 2 – 31 _x_ – 15 = 0 has roots 3, . Find the other roots.
**20.52** Find the roots of each equation.
( _a_ ) ( _x_ \+ 3)2( _x_ – 2)3( _x_ \+ 1) = 0
( _b_ ) 4 _x_ 4( _x_ \+ 2)4( _x_ – 1) = 0
( _c_ ) ( _x_ 2 \+ 3 _x_ \+ 2)( _x_ 2 – 4 _x_ \+ 5) = 0
( _d_ ) ( _y_ 2 \+ 4)2( _y_ \+ 1)2 = 0
**20.53** Form equations with integral coefficients having only the following roots.
( _a_ ) 2, – 3, –
( _b_ ) 0, -4, 2/3, 1
( _c_ ) ± 3 _i_ , double root 2
( _d_ ) -1 ± 2 _i_ , 2 ± _i_
**20.54** Form an equation whose only roots are
**20.55** Write the equation of lowest possible degree with integral coefficients having the given roots.
( _a_ ) 1, 0, _i_
( _b_ ) 2 + _i_
( _c_ )
( _d_ )
( _e_ )
( _f_ ) _i_ /2, 6/5
**20.56** In the equation _x_ 3 \+ _ax_ 2 \+ _bx_ \+ _a_ = 0, _a_ and _b_ are real numbers. If _x_ = 2 + _i_ is a root of the equation, find _a_ and _b_.
**20.57** Write an equation of lowest degree with integral coefficients having as a double root.
**20.58** Write an equation of lowest degree with integral coefficients having as a root.
**20.59** Solve each equation, given the indicated root.
( _a_ ) _x_ 4 \+ _x_ 3 – 12 _x_ 2 \+ 32 _x_ – 40 =
( _b_ )6 _x_ 4 – 11 _x_ 3 \+ _x_ 2 \+ 33 _x_ – 45 =
( _c_ ) _x_ 3 – 5 _x_ 2 \+ 6 =
( _d_ ) _x_ 4 – 4 _x_ 3 \+ 6 _x_ 2 – 16 _x_ \+ 8 = 0; 2 _i_
**20.60** Obtain the rational roots, if any, of each equation.
( _a_ ) _x_ 4 \+ 2 _x_ 3 – 4 _x_ 2 – 5 _x_ – 6 = 0
( _b_ ) 4 _x_ 3 – 3 _x_ \+ 1 = 0
( _c_ ) 2 _x_ 4 – _x_ 3 \+ 2 _x_ 2 – 2 _x_ – 4 = 0
( _d_ ) 3 _x_ 3 \+ _x_ 2 – 12 _x_ – 4 = 0
**20.61** Solve each equation.
( _a_ ) _x_ 3 – _x_ 2 – 9 _x_ \+ 9 = 0
( _b_ ) 2 _x_ 3 – 3 _x_ 2 – 11 _x_ \+ 6 = 0
( _c_ ) 3 _x_ 3 \+ 2 _x_ 2 \+ 2 _x_ – 1 = 0
( _d_ ) 4 _x_ 4 \+ 8 _x_ 3 – 5 _x_ 2 – 2 _x_ \+ 1 = 0
( _e_ ) 5 _x_ 4 \+ 3 _x_ 3 \+ 8 _x_ 2 \+ 6 _x_ – 4 = 0
( _f_ ) 3 _x_ 5 \+ 2 _x_ 4 – 15 _x_ 3 – 10 _x_ 2 \+ 12 _x_ \+ 8 = 0
**20.62** Prove that ( _a_ ) and ( _b_ ) are irrational numbers.
**20.63** If _P_ ( _x_ ) = 2 _x_ 3 – 3 _x_ 2 \+ 12 _x_ – 16, determine the number of positive, negative and nonreal roots.
**20.64** Locate between two successive integers the real roots of _x_ 4 – 3 _x_ 2 – 6 _x_ – 2 = 0. Find the least positive root of the equation accurate to two decimal places.
**20.65** Find upper and lower limits for the real roots of each equation.
( _a_ ) _x_ 3 – 3 _x_ 2 \+ 2 _x_ – 4 = 0
( _b_ ) 2 _x_ 4 \+ 5 _x_ 2 – 6 _x_ – 14 = 0
**20.66** Find the rational roots of 2 _x_ 3 – 5 _x_ 2 \+ 4 _x_ \+ 24 = 0 and thus solve the equation completely.
**20.67** Using Descartes' Rule of Signs, what may be inferred as to the number of positive, negative and nonreal roots of the following equations?
( _a_ ) 2 _x_ 3 \+ 3 _x_ 2 \+ 7 = 0
( _b_ ) 3 _x_ 3 – _x_ 2 \+ 2 _x_ – 1 = 0
( _c_ ) _x_ 5 \+ 4 _x_ 3 – 3 _x_ 2 – _x_ \+ 12 = 0
( _d_ ) _x_ 5 – 3 _x_ – 2 = 0
**20.68** Given the equation 3 _x_ 4 – _x_ 3 \+ _x_ 2 – 5 = 0, determine ( _a_ ) the maximum number of positive roots, ( _b_ ) the minimum number of positive roots, ( _c_ ) the exact number of negative roots, and ( _d_ ) the maximum number of nonreal roots.
**20.69** Given the equation 5 _x_ 3 \+ 2 _x_ – 4 = 0, how many roots are ( _a_ ) negative, ( _b_ ) real?
**20.70** Does the equation _x_ 6 \+ 4 _x_ 4 \+ 3 _x_ 2 \+ 16 = 0 have ( _a_ ) 4 nonreal and 2 real roots, ( _b_ ) 4 real and 2 nonreal roots, ( _c_ ) 6 nonreal roots, or ( _d_ ) 6 real roots?
**20.71** ( _a_ ) How many positive roots has the equation _x_ 6 – 7 _x_ 2 – 11 = 0?
( _b_ ) How many complex roots has the equation _x_ 7 \+ _x_ 4 – _x_ 2 – 3 = 0?
( _c_ ) Show that _x_ 6 \+ 2 _x_ 3 \+ 3 _x_ – 4 = 0 has exactly 4 nonreal roots.
( _d_ ) Show that _x_ 4 \+ _x_ 3 – _x_ 2 – 1 = 0 has only one negative root.
**20.72** Solve completely each equation.
( _a_ ) 8 _x_ 3 – 20 _x_ 2 \+ 14 _x_ – 3 = 0
( _b_ ) 8 _x_ 4 – 14 _x_ 3 – 9 _x_ 2 \+ 11 _x_ – 2 = 0
( _c_ ) 4 _x_ 3 \+ 5 _x_ 2 \+ 2 _x_ – 6 = 0
( _d_ ) 2 _x_ 4 – _x_ 3 – 23 _x_ 2 \+ 18 _x_ \+ 18 = 0
**20.73** Approximate the indicated root of each equation to the specified accuracy.
( _a_ ) 2 _x_ 3 \+ 3 _x_ 2 – 9 _x_ – 7 = 0; positive root, to the nearest tenth
( _b_ ) _x_ 3 \+ 9 _x_ 2 \+ 27 _x_ – 50 = 0; positive root, to the nearest hundredth
( _c_ ) _x_ 3 – 3 _x_ 2 – 3 _x_ \+ 18 = 0; negative root, to the nearest tenth
( _d_ ) _x_ 3 \+ 6 _x_ 2 \+ 9 _x_ \+ 17 = 0; negative root, to the nearest tenth
( _e_ ) _x_ 5 \+ _x_ 4 – 27 _x_ 3 – 83 _x_ 2 \+ 50 _x_ \+ 162 = 0; root between 5 and 6, to the nearest hundredth
( _f_ ) _x_ 4 – 3 _x_ 3 \+ _x_ 2 – 7 _x_ \+ 12 = 0; root between 1 and 2, to the nearest hundredth
**20.74** In finding the maximum deflection of a beam of given length loaded in a certain way, it is necessary to solve the equation 4 _x_ 3 – 150 _x_ 2 \+ 1500 _x_ – 2871 = 0. Find correct to the nearest tenth the root of the equation lying between 2 and 3.
**20.75** The length of a rectangular box is twice its width, and its depth is one foot greater than its width. If its volume is 64 cubic feet, find its width to the nearest tenth of a foot.
**20.76** Find correct to the nearest hundredth.
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**20.43** ( _a_ ) 2
( _b_ ) 12
( _c_ ) 0
( _d_ ) 3/2
( _e_ )
**20.44** ( _a_ ) –9
( _b_ ) 14
( _c_ ) –13/8
( _d_ ) 12
( _e_ ) 1
( _f_ ) 33
**20.46** –4, 1 and 2 _i_ are roots
**20.47** ( _a_ ) k = 9
( _b_ ) _k_ = 4, –2
**20.48** ( _a_ )
( _b_ )
( _c_ )
( _d_ )
**20.49** 12, 227
**20.50** 3, –2
**20.51** –1 ±2 _i_
**20.52** ( _a_ ) double root –3, triple root 2, –1
( _b_ ) quadruple root 0, quadruple root –2, 1
( _c_ ) –1, –2, 2 ± _i_
( _d_ ) double roots ±2 _i_ , double root – 1
**20.53** ( _a_ ) 2 _x_ 3 \+ 3 _x_ 2 – 11 _x_ – 6 = 0
( _b_ ) 3 _x_ 4 \+ 7 _x 3_ – 18 _x_ 2 \+ 8 _x_ = 0
( _c_ ) _x_ 4 – 4 _x_ 3 \+ 13 _x_ 2 – 36 _x_ \+ 36 = 0
( _d_ ) _x_ 4 –2 _x_ 3 +2 _x_ 2 – 10 _x_ \+ 25 = 0
**20.54** _x_ 4 – _x_ 2 – 10 _x_ – 4 = 0
**20.55** ( _a_ ) _x_ 4 – _x_ 3 \+ _x_ 2 – _x_ = 0
( _b_ ) _x_ 2 – 4 _x_ \+ 5 = 0
( _c_ ) 3 _x_ 3 \+ 5 _x_ 2 – 8 _x_ \+ 2 = 0
( _d_ ) _x_ 3 \+ 2 _x_ 2 \+ 3 _x_ \+ 6 = 0
( _e_ ) _x_ 4 – _x_ 2 – 2 = 0
( _f_ ) 20 _x_ 3 – 24 _x_ 2 \+ 5 _x_ – 6 = 0
**20.56** _a_ = –5, _b_ = 9
**20.57** _x_ 4 \+ 4 _x_ 3 \+ 2 _x_ 2 – 4 _x_ \+ 1 = 0
**20.58** _x_ 4 \+ 2 _x_ 2 \+ 49 = 0
**20.59** ( _a_ )
( _b_ )
( _c_ )
( _d_ ) ±2 _i_ , 2 ±
**20.60** ( _a_ ) –3, 2
( _b_ )1/2, 1/2, –1
( _c_ ) no rational root
( _d_ ) –1/3, ±2
**20.61** ( _a_ )1, ± 3
( _b_ ) 3, – 2, 1/2
( _c_ )
( _d_ )
( _e_ )
( _f_ )
**20.63** 1 positive root, 0 negative roots, 2 nonreal roots.
**20.64** Positive root between 2 and 3; negative root between –1 and 0; positive root = 2.41 approx.
**20.65** ( _a_ ) Upper limit 3, lower limit –1
( _b_ ) Upper limit 2, lower limit –2
**20.66** –3/2, 2 _±_ 2 _i_
**20.67** ( _a_ ) 1 negative, 2 nonreal
( _b_ ) 3 positive or 1 positive, 2 nonreal
( _c_ ) 1 negative, 2 positive, 2 nonreal or 1 negative, 4 nonreal
( _d_ ) 1 positive, 2 negative, 2 nonreal or 1 positive, 4 nonreal
**20.68** ( _a_ ) 3
( _b_ ) 1
( _c_ ) 1
( _d_ ) 2
**20.69** ( _a_ ) none
( _b_ ) one
**20.70** ( _c_ )
**20.71** ( _a_ ) one
( _b_ ) four or six
**20.72** ( _a_ )1/2, 1/2, 3/2
( _b_ )2, –1, 1/4, 1/2
( _c_ )3/4, –1 ± _i_
( _d_ )
**20.73** ( _a_ ) 1.9
( _b_ ) 1.25
( _c_ ) –2.2
( _d_ ) –4.9
( _e_ ) 5.77
( _f_ ) 1.38
**20.74** 2.5
**20.75** 2.9 ft
**20.76** 2.71
## **CHAPTER 21
Rational Functions**
### **21.1 RATIONAL FUNCTIONS**
A rational function is the ratio of two polynomial functions. If _P_ ( _x_ ) and _Q_ ( _x_ ) are polynomials, then a function of the form _R_ ( _x_ ) = _P_ ( _x_ )/ _Q_ ( _x_ ) is a rational function where _Q_ ( _x_ ) ≠ 0.
The domain of _R_ ( _x_ ) is the intersection of the domains of _P_ ( _x_ ) and _Q_ ( _x_ ).
### **21.2 VERTICAL ASYMPTOTES**
If _R_ ( _x_ ) = _P_ ( _x_ )/ _Q_ ( _x_ ), then values of _x_ that make _Q_ ( _x_ ) = 0 result in vertical asymptotes if _P_ ( _x_ ) ≠ 0. However, if for some value _x_ = _a_ , _P_ ( _a_ ) = 0 and _Q_ ( _a_ ) = 0, then _P_ ( _x_ ) and _Q_ ( _x_ ) have a common factor of _x_ – _a_. If _R_ ( _x_ ) is then reduced to lowest terms, the graph of _R_ ( _x_ ) has a hole in it where _x_ = _a_.
A vertical asymptote for _R_ ( _x_ ) is a vertical line _x_ = _k, k_ is a constant, that the graph of _R_ ( _x_ ) approaches but does not touch. _R_ ( _k_ ) is not defined because _Q_ ( _k_ ) = 0 and _P_ ( _k_ ) ≠ 0. The domain of _R_ ( _x_ ) is separated in distinct intervals by the vertical asymptotes of _R_ ( _x_ ).
**EXAMPLE 21.1.** What are the vertical asymptotes of
Since
is undefined when _x_ 2 – 4 = 0, _x_ = 2 and _x_ = – 2 could result in vertical asymptotes. When _x_ = 2, 2 _x_ – 3 ≠ 0 and when _x_ = – 2, 2 _x_ – 3 ≠ 0. Thus, the graph of _R_ ( _x_ ) has vertical asymptotes of _x_ = 2 and _x_ = – 2.
### **21.3 HORIZONTAL ASYMPTOTES**
A rational function _R_ ( _x_ ) = _P_ ( _x_ )/ _Q_ ( _x_ ) has a horizontal asymptote _y_ = _a_ if, as | _x_ | increases without limit, _R_ ( _x_ ) approaches _a_. _R_ ( _x_ ) has at most one horizontal asymptote. The horizontal asymptote of _R_ ( _x_ ) may be found from a comparison of the degree of _P_ ( _x_ ) and the degree of _Q_ ( _x_ ).
(1) If the degree of _P_ ( _x_ ) is less than the degree of _Q_ ( _x_ ), then _R_ ( _x_ ) has a horizontal asymptote of _y_ = 0.
(2) If the degree of _P_ ( _x_ ) is equal to the degree of _Q_ ( _x_ ), then _R_ ( _x_ ) has a horizontal asymptote of _y_ = _a n_/ _b n_, where _a n_ is the lead coefficient (coefficient of the highest degree term) of _P_ ( _x_ ) and _b n_ is the lead coefficient of _Q_ ( _x_ ).
(3) If the degree of _P_ ( _x_ ) is greater than the degree of _Q_ ( _x_ ), then _R_ ( _x_ ) does not have a horizontal asymptote.
The graph of _R_ ( _x_ ) may cross a horizontal asymptote in the interior of its domain. This is possible since we are only concerned with how _R_ ( _x_ ) behaves as | _x_ | increases without limit in determining the horizontal asymptote.
**EXAMPLE 21.2.** What are the horizontal asymptotes of each rational function _R_ ( _x_ )?
( _a_ )
( _b_ )
( _c_ )
the degree ofthe numerator3 _x_ 3 is 3 and the degree of the denominatoris 2. Since the numeratorexceeds the degree of the denominator, _R_ ( _x_ ) does not have a horizontal asymptote.
(b) The degree of the numerator of
is 1 and the degree of the denominator is 2, so _R_ ( _x_ ) has a horizontal asymptote of _y_ = 0.
(c) The numerator and denominator of
each have degree 1. Since the lead coefficient of the numerator is 2 and the lead coefficient of the denominator is 5, _R_ ( _x_ ) has a horizontal asymptote of
### **21.4 GRAPHING RATIONAL FUNCTIONS**
To graph a rational function _R_ ( _x_ ) = _P_ ( _x_ )/ _Q_ ( _x_ ), we first determine the holes: values of _x_ for which both _P_ ( _x_ ) and _Q_ ( _x_ ) are zero. After any holes are located, we reduce _R_ ( _x_ ) to lowest terms. The value of the reduced form of _R_ ( _x_ ) for an _x_ that yields a hole is the _y_ coordinate of the point that corresponds to the hole.
Once _R_ ( _x_ ) is in lowest terms, we determine the asymptotes, symmetry, zeros, and _y_ intercept if they exist. We graph the asymptotes as dashed lines, plot the zeros and _y_ intercept, and plot several other points to determine how the graph approaches the asymptotes. Finally, we sketch the graph through the plotted points and approaching the asymptotes.
**EXAMPLE 21.3.** Sketch a graph of each rational function _R_ ( _x_ ).
( _a_ )
( _b_ )
has vertical asymptotes at _x_ = 1, and at _x_ = – 1, a horizontal asymptote of _y_ = 0, and no holes.
Since the numerator of _R_ ( _x_ ) is a constant, it does not have any zeros. Since _R_ (0) = –3, _R_ ( _x_ ) has a _y_ intercept of (0, –3).
Plot the _y_ intercept and graph the asymptotes as dashed lines. We determine some values of _R_ ( _x_ ) in each interval of the domain (– ∞, – 1), and (– 1, 1), and (1, ∞). _R_ (– _x_ ) = _R_ ( _x_ ), so _R_ ( _x_ ) is symmetric with respect to the _y_ axis.
Plot (2, 1), (–2, 1), (0.5, –4), and (–0.5, – 4). Using the asymptotes as a boundary, we sketch the graph. The graph of
is shown in Fig. 21-1.
**Fig. 21-1**
( _b_ )
has vertical asymptotes at _x_ = 2 and _x_ = – 2, a horizontal asymptote of _y_ = – 1, and no holes.
The zeros of _R_ ( _x_ ) are for _x_ = 0. Since when _x_ = 0, _R_ (0) = 0 and (0, 0) is the zero and the _y_ intercept.
Plot the point (0, 0) and the vertical and horizontal asymptotes.
We determine some values of _R_ ( _x_ ) in each interval of its domain (– ∞, – 2), (– 2, 2), and (2, ∞). Since _R_ (– _x_ ) = _R_ ( _x_ ), the graph is symmetric with respect to the _y_ axis.
Plot (3, – 9/5), (–3, – 9/5), (1, 1/3), and (1, –1/3).
Using the asymptotes as boundary lines, sketch the graph of _R_ ( _x_ ).
The graph of
is shown in Fig. 21-2.
**Fig. 21-2**
### **21.5 GRAPHING RATIONAL FUNCTIONS USING A GRAPHING CALCULATOR**
The graphing features of a graphing calculator allow for easy graphing of rational functions. However, unless the graphing calculator specifically plots the _x_ values for the asymptotes, it connects the distinct branches of the rational function. You need to determine the vertical asymptotes, then set the scale on the _x_ axis of the graphing window so that the values of the vertical asymptotes are used.
Horizontal asymptotes must be read from the graph itself, since they are not drawn or labeled and only appear as a characteristic of the graph in the calculator display window.
Holes are based on factors that can be canceled out from the rational function. These are difficult to locate from the display on a graphing calculator.
In general, when using a graphing calculator to aid in the production of a graph on paper, it is good practice not to depend on the graphing calculator to determine the vertical asymptotes, horizontal asymptotes, or holes for a rational function. Determine these values for yourself and place them on the graph you are constructing. Use the display on the graphing calculator to indicate the location and shape of the graph and to guide your sketch of the graph.
### **Solved Problems**
**21.1** State the domain of each rational function _R_ ( _x_ ).
( _a_ )
( _b_ )
( _c_ )
**SOLUTION**
( _a_ ) For _R_ ( _x_ ) = 3 _x_ /( _x_ \+ 2), set _x_ \+ 2 = 0 and see that _R_ ( _x_ ) is not defined for _x_ = – 2. The domain of _R_ ( _x_ ) is {all real numbers except –2} or domain = (– ∞, –2) ∪ (– 2, ∞).
( _b_ ) For _R_ ( _x_ ) = ( _x_ 3 – 2 _x_ 2 – 3 _x_ )/ _x_ , we see that _R_ ( _x_ ) is not defined for _x_ = 0. Thus, the domain of _R_ ( _x_ ) is {all real numbers except 0} or domain = (– ∞, 0) ∪ (0, ∞).
( _c_ ) For _R_ ( _x_ ) = (3 _x_ 2 – 1)/( _x_ 3 – _x_ ), we set _x_ 3 – _x_ = 0 and determine that for _x_ = 0, _x_ = 1, and _x_ = –1 _R_ ( _x_ ) is undefined. The domain of _R_ ( _x_ ) is {all real numbers except –1, 0, 1} or domain = (– ∞, – 1) ∪ (– 1, 0) ∪ (0, 1) ∪ (1, ∞).
**21.2** Determine the vertical asymptotes, horizontal asymptotes, and holes for each rational function _R_ ( _x_ ).
( _a_ )
( _b_ )
( _c_ )
**SOLUTION**
( _a_ ) Values that make the denominator zero but do not make the numerator zero yield asymptotes. In _R_ ( _x_ ) = 3 _x_ /( _x_ \+ 2), _x_ = – 2 makes the denominator _x_ \+ 2 = 0 but does not make the numerator 3 _x_ = 0. Thus, _x_ = – 2 is a vertical asymptote.
Since the degree of the numerator 3 _x_ is 1 and the degree of the denominator _x_ \+ 2 is 1, _R_ ( _x_ ) has a horizontal asymptote of _y_ = 3/1 = 3, where 3 is the lead coefficient of the numerator and 1 is the lead coefficient of the denominator.
Since _x_ = 0 is the only value that makes the numerator 0 and _x_ = 0 does not make the denominator 0, _R_ ( _x_ ) has no holes in its graph.
( _b_ ) Only _x_ = 0 makes the denominator of _R_ ( _x_ ) = ( _x_ 3 – 2 _x_ 2 – 3 _x_ )/ _x_ zero. Since _x_ = 0 also makes the numerator zero, _R_ ( _x_ ) has no vertical asymptotes.
Since the degree of the numerator of _R_ ( _x_ ) is 3 and the degree of the denominator is 1, the degree of the numerator exceeds the degree of the denominator and there is no horizontal asymptote.
Since _x_ = 0 makes both the numerator and denominator of _R_ ( _x_ ) zero, there is a hole in the graph of _R_ ( _x_ ) when _x_ = 0. We reduce _R_ ( _x_ ) to lowest terms and get _R_ ( _x_ ) = _x_ 2 – 2 _x_ – 3 when _x_ ≠ 0. The graph of this reduced form would have the value of – 3 if _x_ were 0, so the graph of _R_ ( _x_ ) has a hole at (0, –3).
( _c_ ) Since _x_ 3 – _x_ = 0 has solutions of _x_ = 0, _x_ = 1, and _x_ = –1, the vertical asymptotes of _R_ ( _x_ ) = (3 _x_ 2 – 1)/( _x_ 3 – _x_ ) are _x_ = 0, _x_ = 1, and _x_ = – 1.
The degree of the numerator of _R_ ( _x_ ) is less than the degree of the denominator, so _y_ = 0 is the horizontal asymptote of _R_ ( _x_ ).
The numerator is not zero for any values, _x_ = – 1, _x_ = 0, and _x_ = 1, that make the denominator zero, so the graph of _R_ ( _x_ ) does not have any holes in it.
**21.3** What are the zeros and _y_ intercept of each rational function _R_ ( _x_ )?
( _a_ )
( _b_ )
( _c_ )
**SOLUTION**
( _a_ ) For _R_ ( _x_ ) = 3 _x_ /( _x_ \+ 2), the numerator 3 _x_ is zero if _x_ = 0. Since _x_ = 0 does not make the denominator zero, there is a zero when _x_ = 0. Thus, (0, 0) is the zero of _R_ ( _x_ ). The _y_ intercept is the value of _y_ when _x_ = 0. Thus, (0, 0) is the point for the _y_ intercept.
( _b_ ) For _R_ ( _x_ ) = ( _x_ 3 – 2 _x_ 2 – 3 _x_ )/ _x_ , the numerator _x_ 3 – 2 _x_ 2 – 3 _x_ is zero when _x_ = 0, _x_ = –1, and _x_ = 3. However, _x_ = 0 makes the denominator zero, so it will not yield a zero of _R_ ( _x_ ). The zeros of _R_ ( _x_ ) are (3, 0) and (–1, 0).
From Problem 21.1( _b_ ), we know that _x_ = 0 is not in the domain of _R_ ( _x_ ). Thus, _R_ ( _x_ ) does not have a _y_ intercept.
( _c_ ) For _R_ ( _x_ ) = (3 _x_ 2 – 1)/( _x_ 3 – _x_ ), the numerator 3 _x_ 2 – 1 = 0 has solutions and Thus the zeros of _R_ ( _x_ ) are
From Problem 21.1 ( _c_ ), we know that the domain of _R_ ( _x_ ) does not contain _x_ = 0, so _R_ ( _x_ ) does not have a _y_ intercept.
**21.4** Sketch a graph of each rational function _R_ ( _x_ ).
( _a_ )
( _b_ )
( _c_ )
**SOLUTION**
From Problems 21.1, 21.2, and 21.3, we know the domains, vertical asymptotes, horizontal asymptotes, holes, zeros, and _y_ intercepts for these three rational functions. We will use this information in graphing each of these rational functions.
( _a_ ) _R_ ( _x_ ) = 3 _x_ /( _x_ \+ 2) has a vertical asymptote of _x_ = –2, a horizontal asymptote of _y_ = 3, and (0, 0) is the point for the zero and _y_ intercept. We draw dashed lines for the asymptotes and plot (0, 0). We select some _x_ values and determine the points, then plot them. For _x_ = – 4, –3, –1, and 2, we get the points (–4, 6), (– 3, 9), (– 1, –3), and (2, 1.5). Now we sketch the graph of _R_ ( _x_ ) going through the plotted points and approaching the asymptotes. Since the domain of _R_ ( _x_ ) is separated into two parts, we will have two parts to the graph of _R_ ( _x_ ). See Fig. 21-3.
**Fig. 21-3**
( _b_ ) _R_ ( _x_ ) = ( _x_ 3 – 2 _x_ 2 – 3 _x_ )/ _x_ = _x_ 2 – 2 _x_ – 3 when _x_ ≠ 0, and there is a hole at (0, – 3). There are no asymptotes for the graph of _R_ ( _x_ ), there is no _y_ intercept, but there are zeros at (3, 0) and (–1, 0). We plot the zeros and place an open circle O around the point (0, –3) to indicate the hole in the graph. Now we select values of _x_ , determine the corresponding points, and plot them. For _x_ = –2, 1, 2, and 4, we get the points (–2, 5), (1, –4), (2, 3), and (4, 5). Since the domain of _R_ ( _x_ ) is separated into two parts by the _x_ value for the hole of _R_ ( _x_ ), the graph of _R_ ( _x_ ) is separated into two parts by the hole at (0, –3). See Fig. 21-4.
**Fig. 21-4**
( _c_ ) _R_ ( _x_ ) = (3 _x_ 2 – 1)/( _x_ 3 – _x_ ) has vertical asymptotes of _x_ = –1, _x_ = 0, and _x_ = 1, a horizontal asymptote of _y_ = 0, and zeros of We approximate the zeros to be (0.6, 0) and (–0.6, 0) and we plot these points and graph the asymptotes. To be sure that we graph all the parts of _R_ ( _x_ ), we select _x_ values from each interval in the domain. For _x_ = –2, –1.5, –0.75, –0.25, 0.25, 0.75, 1.5, and 2, we get the points (–2, –1.8), (–1.5, –3.1), (–0.75, 2.1), (–0.25, –3.5), (0.25, 3.5), (0.75, –2.1), (1.5, 3.1), and (2, 1.8). Since the domain of _R_ ( _x_ ) is separated into four parts the graph of _R_ ( _x_ ) is in four separate parts. See Fig. 21-5.
**Fig. 21-5**
### **Supplementary Problems**
**21.5** State the domain of each rational function.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
**21.6** Determine the asymptotes of each rational function.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
**21.7** Determine the zeros and _y_ intercept for each rational function.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
**21.8** Graph each rational function.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
**21.9** Graph each rational function.
( _a_ )
( _b_ )
**21.10** Graph each rational function.
( _a_ )
( _b_ )
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**21.5** ( _a_ ) (–∞, – 2) ∪ (–2, ∞)
( _b_ ) (–∞, 2) ∪ (2, ∞)
( _c_ ) (–∞, – 2) ∪ (– 2, 2) ∪ (2, ∞)
( _d_ ) (–∞, – 2) ∪ (– 2, 1) ∪ (1, ∞)
( _e_ ) (–∞, –3) ∪ (–3, ∞)
( _f_ ) (–∞, –4) ∪ (–4, ∞)
( _g_ ) (–∞, 0) ∪ (0, ∞)
( _h_ ) (–∞, –1/3) ∪ (– 1/3, 0) ∪ (0, 1/3) ∪ (1/3, ∞)
( _i_ ) (– ∞, 2) ∪ (2, 3) ∪ (3, ∞)
**21.6**
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
**21.7**
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
**21.8** ( _a_ ) Fig. 21-6
**Fig. 21-6**
( _b_ ) Fig. 21-7
**Fig. 21-7**
( _c_ ) Fig. 21-8
**Fig. 21-8**
( _d_ ) Fig. 21-9
**Fig. 21-9**
( _e_ ) Fig. 21-10
**Fig. 21-10**
( _f_ ) Fig. 21-11
**Fig. 21-11**
( _g_ ) Fig. 21-12
**Fig. 21-12**
( _h_ ) Fig. 21-13
**Fig. 21-13**
( _i_ ) Fig. 21-14
**Fig. 21-14**
**21.9** ( _a_ ) Reduced to lowest terms _R_ ( _x_ ) = 1/( _x_ – 2) when _x_ ≠ –2. Graph is Fig. 21-15.
**Fig. 21-15**
( _b_ ) Reduced to lowest terms _R_ ( _x_ ) ≠ 3/( _x_ – 2) when _x_ ≠ 0. Graph is Fig. 21-16.
**Fig. 21-16**
**21.10** ( _a_ ) Fig. 21-17
**Fig. 21-17**
( _b_ ) Fig. 21-18
**Fig. 21-18**
## **CHAPTER 22
Sequences and Series**
### **22.1 SEQUENCES**
A sequence of numbers is a function defined on the set of positive integers. The numbers in the sequence are called _terms_. A series is the sum of the terms of a sequence.
### 22.2 ARITHMETIC SEQUENCES
A. _An arithmetic sequence_ is a sequence of numbers each of which, after the first, is obtained by adding to the preceding number a constant number called the _common difference_.
Thus 3, 7, 11, 15, 19,... is an arithmetic sequence because each term is obtained by adding 4 to the preceding number. In the arithmetic sequence 50, 45, 40,... the common difference is 45 – 50 = 40-45 = -5.
B. _Formulas for arithmetic sequences_
(1) The _n_ th term, or last term: _l_ = _a_ \+ ( _n_ – 1) _d_
(2) The sum of the first _n_ terms: _ _
where _a_ = first term of the sequence; _d_ = common difference;
_n_ = number of terms; _l_ = _n_ th term, or last term;
_S_ = sum of first _n_ terms.
**EXAMPLE 22.1.** Consider the arithmetic sequence 3, 7, 11,... where _a_ = 3 and _d_ = 7 – 3 = 11 – 7 = 4. The sixth term is _l_ = _a_ \+ ( _n_ \- 1) _d_ = 3 + (6 - 1)4 = 23.
The sum of the first six terms is
or
### **22.3 GEOMETRIC SEQUENCES**
A. _A geometric sequence_ is a sequence of numbers each of which, after the first, is obtained by multiplying the preceding number by a constant number called the _common ratio_.
Thus 5, 10, 20, 40, 80,... is a geometric sequence because each number is obtained by multiplying the preceding number by 2. In the geometric sequence 9, –3, 1 ,... the common ratio is
B. _Formulas for geometric sequences_.
(1) The nth term, or last term: _l = ar_ n–1
(2) The sum of the first _n_ terms:
where _a_ = first term; _r_ = common ratio; _n_ = number of terms;
_l = nth_ term, or last term; _S_ = sum of first _n_ terms.
**EXAMPLE 22.2.** Consider the geometric sequence 5, 10, 20,... where _a_ = 5 and
The seventh term is _l = ar n–1_ = 5(27–1) = 5(26) = 320.
The sum of the first seven terms is
### **22.4 INFINITE GEOMETRIC SERIES**
The sum to infinity (S∞) of any geometric sequence in which the common ratio _r_ is numerically less than 1 is given by
where | _r_ | < 1
**EXAMPLE 22.3.** Consider the infinite geometric series
where _a_ = 1 and _r_ = — . Its sum to infinity is
### **22.5 HARMONIC SEQUENCES**
A _harmonic sequence_ is a sequence of numbers whose reciprocals form an arithmetic sequence. Thus
is a harmonic sequence because 2, 4, 6, 8, 10,... is an arithmetic sequence.
### **22.6 MEANS**
The terms between any two given terms of sequence are called the _means_ between these _two_ terms.
Thus in the arithmetic sequence 3, 5, 7, 9, 11,... the arithmetic mean between 3 and 7 is 5, and _four_ arithmetic means between 3 and 13 are 5, 7, 9, 11.
In the geometric sequence 2, –4, 8, –16,... _two_ geometric means between 2 and –16 are –4, 8.
In the harmonic sequence
the harmonic mean between and is and _three_ harmonic means between and are ,
### **Solved Problems**
**22.1** Which of the following sequences are arithmetic sequences?
**22.2** Prove the formula _S_ = ( _n_ /2)( _a_ \+ _l_ ) for the sum of the first _n_ terms of an arithmetic sequence.
**SOLUTION**
The sum of the first _n_ terms of an arithmetic sequence may be written
where the sum is written in reversed order.
Adding, 2S _= (a + l_ ) _\+ (a + l_ ) _\+ (a + l_ ) _h_...h (a _\+ l_ ) to _n_ terms.
Hence 2S _= n(a + l_ ) and
**22.3** Find the 16th term of the arithmetic sequence: 4, 7, 10,...
**SOLUTION**
Here _a_ = 4, _n_ = 16, _d_ = 7 _–_ 4 = 10 _–_ 7 = 3, and _l = a + (n –_ 1) _d_ = 4 + (16 _–_ 1)3 = 49.
**22.4** Determine the sum of the first 12 terms of the arithmetic sequence: 3, 8, 13,...
**SOLUTION**
Here _a_ = 3, _d_ = 8 _–_ 3 = 13 _–_ 8 = 5, _n_ = 12, and
_Otherwise_ : _l_ = _a_ \+ ( _n_ – 1) _d_ = 3 + (12 _–_ 1)5 = 58 and
**22.5** Find the 40th term and the sum of the first 40 terms of the arithmetic sequence: 10, 8, 6,...
**SOLUTION**
Here _d_ = 8 _–_ 10 = 6 _–_ 8 _= –2, a_ = 10, _n_ = 40.
Then _l_ = _a_ \+ ( _n_ – 1) _d_ = 10 + (40 _–_ 1)(–2) = –68 and
**22.6** Which term of the sequence 5, 14, 23,... is 239?
**SOLUTION**
_l_ = _a_ \+ ( _n_ – 1) _d_ , 239 = 5 + ( _n_ – 1)9, 9 _n_ = 243 and the required term is _n_ = 27.
**22.7** Compute the sum of the first 100 positive integers exactly divisible by 7.
**SOLUTION**
The sequence is 7, 14, 21,...an arithmetic sequence in which _a_ = 7, _d_ = 7, _n_ = 100.
Hence
**22.8** How many consecutive integers, beginning with 10, must be taken for their sum to equal 2035?
**SOLUTION**
The sequence is 10, 11, 12,... an arithmetic sequence in which _a_ = 10, _d_ = 1, _S_ = 2035.
Using
we obtain _n_ 2 \+ 19 _n_ – 4070 = 0,
( _n_ – 55)( _n_ \+ 74) = 0, _n_ = 55, –74.
Hence 55 integers must be taken.
**22.9** How long will it take to pay off a debt of $880 if $25 is paid the first month, $27 the second month, $29 the third month, etc.?
**SOLUTION**
From
we obtain 880
The debt will be paid off in 20 months.
**22.10** How many terms of the arithmetic sequence 24, 22, 20,... are needed to give the sum of 150? Write the terms.
**SOLUTION**
For _n_ = 10: 24, 22, 20, 18, 16, 14, 12, 10, 8, 6.
For _n_ = 15: 24, 22, 20, 18, 16, 14, 12, 10, 8, 6, 4, 2, 0, –2, –4.
**22.11** Determine the arithmetic sequence whose sum to _n_ terms is _n_ 2 \+ 2 _n_.
**SOLUTION**
The _n_ th term = sum to n terms – sum to _n_ – 1 terms
= _n_ 2 \+ 2 _n_ – [( _n_ — 1)2 \+ 2( _n_ – 1)] = 2 _n_ \+ 1.
Thus the arithmetic sequence is 3, 5, 7, 9,...
**22.12** Show that the sum of _n_ consecutive odd integers beginning with 1 equals _n_ 2.
**SOLUTION**
We are to find the sum of the arithmetic sequence 1, 3, 5,... to n terms.
Then _a_ = 1, _d_ = 2, _n_ = _n_ and
**22.13** Find three numbers in an arithmetic sequence such that the sum of the first and third is 12 and the product of the first and second is 24.
**SOLUTION**
Let the numbers in the arithmetic sequence be ( _a_ – _d_ ), _a_ ,( _a_ \+ _d_ ). Then ( _a_ – _d_ ) + ( _a_ \+ _d_ ) = 12 or _a_ = 6.
Since ( _a_ – _d_ ) a = 24, (6 – _d_ )6 = 24 or _d_ = 2. Hence the numbers are 4, 6, 8.
**22.14** Find three numbers in an arithmetic sequence whose sum is 21 and whose product is 280.
**SOLUTION**
Let the numbers be ( _a_ – _d_ ), _a_ ,( _a_ \+ _d_ ). Then ( _a_ – _d_ ) + _a_ \+ ( _a_ \+ _d_ ) = 21 or _a_ = 7.
Since ( _a_ – _d_ )( _a_ )( _a_ \+ _d_ ) = 280, _a_ ( _a_ 2 – d2) = 7(49 – _d_ 2) = 280 and _d_ = ±3.
The required numbers are 4, 7, 10 or 10, 7, 4.
**22.15** Three numbers are in the ratio of 2 : 5 : 7. If 7 is subtracted from the second, the resulting numbers form an arithmetic sequence. Determine the original numbers.
**SOLUTION**
Let the original numbers be 2x, 5x, 7x. The resulting numbers in the arithmetic sequence are 2 _x_ ,(5 _x_ –7), 7 _x_.
Then (5 _x_ –7) –2 _x = 7x_ –(5 _x_ –7) or _x_ = 14. Hence the original numbers are 28, 70, 98.
**22.16** Compute the sum of all integers between 100 and 800 that are divisible by 3.
**SOLUTION**
The arithmetic sequence is 102, 105, 108,..., 798. Then _I = a + (n —_ 1) d, 798 = 102 _\+ (n —_ 1)3, n = 233, and
**22.17** A slide of uniform grade is to be built on a level surface and is to have 10 supports equidistant from each other. The heights of the longest and shortest supports will be 42 feet and 2 feet respectively. Determine the required height of each support.
**SOLUTION**
From _l_ = _a_ \+ ( _n_ –1) _d_ we have 42 = 2 + (10 – 1) _d_ and _d_ = 4 ft.
Thus the heights are 2, 6 , 11, 15 , 20, 24 , 29, 33 , 38, 42 feet respectively.
**22.18** A freely falling body, starting from rest, falls 16 ft during the first second, 48 ft during the second second, 80 ft during the third second, etc. Calculate the distance it falls during the fifteenth second and the total distance it falls in 15 seconds from rest.
**SOLUTION**
Here _d_ = 48 –16 = 80 –48 = 32.
During the 15th second it falls a distance _l_ = _a_ \+ ( _n_ –1) _d_ = 16 +(15 –1)32 = 464 ft.
Total distance covered during 15 sec is
**22.19** In a potato race, 8 potatoes are placed 6 ft apart on a straight line, the first being 6 ft from the basket. A contestant starts from the basket and puts one potato at a time into the basket. Find the total distance she must run in order to finish the race.
**SOLUTION**
Here _a_ = 2 · 6 = 12ft and _l_ = 2(6 · 8) = 96ft. Then
**22.20** Show that if the sides of a right triangle are in an arithmetic sequence, their ratio is 3 : 4 : 5.
**SOLUTION**
Let the sides be ( _a – d), a, (a + d_ ), where the hypotenuse is ( _a + d_ ).
Then ( _a + d) 2 = a2 \+ (a – d)2_ or _a = 4d_. Hence ( _a – d_ ) : _a_ : ( _a + d_ ) = 3 _d_ : 4 _d_ : 5 _d_ = 3 : 4 : 5.
**22.21** Derive the formula for the arithmetic mean ( _x_ ) between two numbers _p_ and _q_.
**SOLUTION**
Since _p, x, q_ are in an arithmetic sequence, we have _x_ – _p_ = _q_ – _x_ or _x_ = ( _p_ \+ _q_ ).
**22.22** Find the arithmetic mean between each of the following pairs of numbers.
**22.23** Insert 5 arithmetic means between 8 and 26.
**SOLUTION**
We require an arithmetic sequence of the form 8, —, —, —, —, —, 26; thus _a_ = 8, _l_ = 26 and _n_ = 7.
Then _l_ = _a_ \+ ( _n_ – 1) _d_ , 26 = 8 + (7 –1) _d, d = 3_.
The five arithmetic means are 11, 14, 17, 20, 23.
**22.24** Insert between 1 and 36 a number of arithmetic means so that the sum of the resulting arithmetic sequence will be 148.
**SOLUTION**
_S = n(a +_ l), 148 = _n_ (1 + 36), 37 _n_ = 296 and _n_ = 8.
_l_ = _a_ \+ ( _n –1_ ) _d, 36 = 1 +_ (8 _–1) d, 7d_ = 35 and _d_ = 5.
The complete arithmetic sequence is 1, 6, 11, 16, 21, 26, 31, 36.
**22.25** Which of the following sequences are geometric sequences?
**22.26** Prove the formula
for the sum of the first n terms of a geometric sequence.
**SOLUTION**
The sum of the first n terms of a geometric sequence may be written
Multiplying (1) by r, we obtain
Subtracting (1) from (2),
**22.27** Find the 8th term and the sum of the first eight terms of the sequence 4, 8, 16,...
**SOLUTION**
Here _a_ = 4, _r_ = 8/4 = 16/8 = 2, _n_ = 8.
The 8th term is _l_ = _ar_ _n_ –1 = 4(2)8–1 = 4(27) = 4(128) = 512.
The sum of the first eight terms is
**22.28** Find the 7th term and the sum of the first seven terms of the sequence 9, –6, 4,...
**SOLUTION**
Here _a_ = 9,
Then the 7th term is
**22.29** The second term of a geometric sequence is 3 and the fifth term is 81/8. Find the eighth term.
**SOLUTION**
5th term 2nd term = _ar_ = 3. Then and
Hence the 8th term
**22.30** Find three numbers in a geometric sequence whose sum is 26 and whose product is 216.
**SOLUTION**
Let the numbers in geometric sequence be _a/r, a, ar_. Then ( _a/r)(a)(ar_ ) = 216, _a 3_ = 216 and _a_ = 6.
Also _a/r + a + ar_ = 26, 6/r + 6 _\+ 6r_ = 26, 6r2 – 20r + 6 = 0 and r = 1/3, 3.
For _r_ = 1/3, the numbers are 18, 6, 2; for _r_ = 3, the numbers are 2, 6, 18.
**22.31** The first term of a geometric sequence is 375 and the fourth term is 192. Find the common ratio and the sum of the first four terms.
**SOLUTION**
1st term _= a_ = 375, 4th term _= ar 3_ = 192. Then 375r3 = 192, r3 = 64/125 and r = 4/5.
The sum of the first four terms is
**22.32** The first term of a geometric sequence is 160 and the common ratio is 3/2. How many consecutive terms must be taken to give a sum of 2110?
**SOLUTION**
The five consecutive terms are 160, 240, 360, 540, 810.
**22.33** In a geometric sequence consisting of four terms in which the ratio is positive, the sum of the first two terms is 8 and the sum of the last two terms is 72. Find the sequence.
**SOLUTION**
The four terms are _a, ar, ar_ 2, _ar_ 3. Then _a + ar_ = 8 and _ar_ 2 \+ _ar_ 3 = 72.
Hence so that _r_ = 3.
Since _a + ar_ = 8, _a = 2_ and the sequence is 2, 6, 18, 54.
**22.34** Prove that _x_ , _x_ \+ 3, _x_ \+ 6 cannot be a geometric sequence.
**SOLUTION**
If _x, x +_ 3, _x +_ 6 is a geometric sequence then
or 9 = 0
Since this equality can never be true, _x, x +_ 3, _x +_ 6 cannot be a geometric sequence.
**22.35** A boy agrees to work at the rate of one cent the first day, two cents the second day, four cents the third day, eight cents the fourth day, etc. How much would he receive at the end of 12 days?
**SOLUTION**
Here _a_ = 1, _r_ = 2, _n_ = 12.
**22.36** It is estimated that the population of a certain town will increase 10% each year for four years. What is the percentage increase in population after four years?
**SOLUTION**
Let _p_ denote the initial population. After one year the population is 1.10 _p_ , after two years (1.10)2 _p_ , after three years (1.10)3p, after four years (1.10)4 _p_ = 1.46 _p_. Thus the population increases 46%.
**22.37** From a tank filled with 240 gallons of alcohol, 60 gallons are drawn off and the tank is filled up with water. Then 60 gallons of the mixture are removed and replaced with water, etc. How many gallons of alcohol remain in the tank after 5 drawings of 60 gallons each are made?
**SOLUTION**
After the first drawing, 240 – 60 = 180 gal of alcohol remain in the tank.
After the second drawing,
of alcohol remain, etc.
The geometric sequence for the number of gallons of alcohol remaining in the tank after successive drawings is
After the fifth drawing ( _n_ = 5):
of alcohol remain.
**22.38** A sum of $400 is invested today at 6% per year. To what amount will it accumulate in five years if interest is compounded (a) annually, (b) semiannually, (c) quarterly?
**SOLUTION**
Let _P_ = initial principal, _i_ = interest rate per period, _S_ = compound amount after _n_ periods.
At end of 1st period: interest = _Pi_ , new amount = _P_ \+ _Pi_ = _P_ (1 + _i_ ).
At end of 2nd period: interest = _P_ (l + _i_ ) _i_ , new amount = _P_ (l + _i_ ) + _P_ (l + _i_ ) _i_ = _P_ (l + _i_ )2.
Compound amount at end of _n_ periods is _S_ = _P_ (1 + _i_ ) _n_.
( _a_ ) Since there is 1 interest period per year, _n_ = 5 and _i_ = 0.06.
_S_ = _P_ (1 + _i_ ) _n_ = 400(1 + 0.06)5 = 400(1.3382) = $535.28
( _b_ ) Since there are 2 interest periods per year, _n_ = 2(5) = 10 and
_S_ = _P_ (1 + _i_ ) _n_ = 400(1 + 0.03)10 = 400(1.3439) = $537.56.
( _c_ ) Since there are 4 interest periods per year, _n_ = 4(5) = 20 and
_S_ = _P_ (1 + _i_ ) _n_ = 400(1 + 0.015)20 = 400(1.3469) = $538.76.
**22.39** What sum ( _P_ ) should be invested in a loan association at 4% per annum compounded semiannually, so that the compound amount ( _S_ ) will be $500 at the end of 3 years?
**SOLUTION**
Since there are 2 interest periods per year, (periods) and the interest rate per period is
Then _S_ = _P_ (1 + _i_ ) _n_ or _P_ = _S_ (1 + _i_ )– _n_ = 500(1 + 0.02)–7 = 500(0.87056) = $435.28.
**22.40** Derive the formula for the geometric mean, _G_ , between two numbers _p_ and _q_.
**SOLUTION**
Since _p_ , _G_ , _q_ are in geometric sequence, we have _G_ / _p_ = _q_ / _G_ , _G_ 2 = _pq_ and
It is customary to take if _p_ and _q_ are positive.
and if _p_ and _q_ are negative.
**22.41** Find the geometric mean between each of the following pairs of numbers.
( _a_ ) 4 and 9.
( _b_ ) –2 and –8.
( _c_ )
**22.42** Show that the arithmetic mean _A_ of two positive numbers _p_ and _q_ is greater than or equal to their geometric mean _G_.
**SOLUTION**
Arithmetic mean of _p_ and _q_ is Geometric mean of _p_ and _q_ is
Then
Now is always positive or zero; hence A ≥ G. (A _= G_ if and only if p = q.)
**22.43** Insert two geometric means between 686 and 2.
**SOLUTION**
We require a geometric sequence of the form 686, —, —, 2 where _a_ = 686, _l_ = 2, _n_ = 4.
Then _l_ = _ar_ _n_ -1, 2 = 686 _r_ 3, _r 3_ = 1/343 and _r_ = 1/7.
Thus the geometric sequence is 686, 98, 14, 2 and the means are 98, 14.
_Note_. Actually, _r_ 3 = 1/343 is satisfied by three different values of _r_ , one of the roots being real and two imaginary. It is customary to exclude geometric sequences with imaginary numbers.
**22.44** Insert five geometric means between 9 and 576.
**SOLUTION**
We require a geometric sequence of the form 9,—, —, —, —, —, 576 where _a_ = 9, _l_ = 576, _n_ = 7.
Then _l_ = _ar_ _n_ –1, 576 = 9 _r_ 6, _r_ 6 = 64, r3 = ±8 and _r_ = ±2.
Thus the sequences are 9, 18, 36, 72, 144, 288, 576 and 9, –18, 36, –72, 144, –288, 576; and the corresponding means are 18, 36, 72, 144, 288 and –18, 36, –72, 144, –288.
**22.45** Find the sum of the infinite geometric series.
( _a_ )
( _b_ )
( _c_ )
**22.46** Express each of the following repeating decimals as a rational fraction.
( _a_ ) 0.444...
( _b_ ) 0.4272727...
( _c_ ) 6.305305...
( _d_ ) 0.78367836...
**SOLUTION**
( _a_ ) 0.444... = 0.4 + 0.04 + 0.004 +..., where _a_ = 0.4, r = 0.1.
( _b_ )
( _c_ )
( _d_ ) 0.78367836... = 0.7836 + 0.00007836 +..., where _a_ = 0.7836, _r_ = 0.0001.
**22.47** The distances passed over by a certain pendulum bob in succeeding swings form the geometric sequence 16, 12, 9,... inches respectively. Calculate the total distance traversed by the bob before coming to rest.
**SOLUTION**
**22.48** Find the least number of terms of the series that should be taken so that their sum will differ from their sum to infinity by less than 1/1000.
**SOLUTION**
Let _S ∞_ = sum to infinity, _S n_ = sum to _n_ terms. Then
It is required that
Then
When Thus at least 10 terms should be taken.
**22.49** Which of the following sequences are harmonic sequences?
( _a_ ) is a harmonic sequence since 3, 5, 7,... is an arithmetic sequence.
( _b_ ) 2, 4, 6,... is not a harmonic sequence since . is not an arithmetic sequence.
( _c_ ) is a harmonic sequence since is an arithmetic sequence.
**22.50** Compute the 15th term of the harmonic sequence
**SOLUTION**
The corresponding arithmetic sequence is 4, 7, 10,...; its 15th term is _l_ = _a_ \+ ( _n_ – 1) _d_ = 4 + (15 – 1)3 = 46.
Hence the 15th term of the harmonic progression is
**22.51** Derive the formula for the harmonic mean, _H_ , between two numbers _p_ and _q_.
**SOLUTION**
Since _p, H, q_ is a harmonic sequence, is an arithmetic sequence.
Then
_Another method_.
Harmonic mean between _p_ and _q_ = reciprocal of the arithmetic mean between
Arithmetic mean between
Hence the harmonic mean between _p_ and
**22.52** What is the harmonic mean between 3/8 and 4?
**SOLUTION**
Arithmetic mean between
Hence the harmonic mean between and 4 = 24/35.
Or, by formula, harmonic mean
**22.53** Insert four harmonic means between 1/4 and 1/64.
**SOLUTION**
To insert four arithmetic means between 4 and 64: _l_ = _a_ \+ ( _n_ — 1) _d_ , 64 = 4 + (6 – 1) _d, d_ = 12.
Thus the four arithmetic means between 4 and 64 are 16, 28, 40, 52.
Hence the four harmonic means between
**22.54** Insert three harmonic means between 10 and 20.
**SOLUTION**
To insert three arithmetic means between
Thus the three arithmetic means between
Hence the three harmonic means between 10 and 20 are
**22.55** Determine whether the sequence — 1, –4, 2 is in arithmetic, geometric or harmonic sequence.
**SOLUTION**
Since –4 – (— 1) ≠ 2 – (–4), it is not in arithmetic sequence.
Since it is not in geometric sequence.
Since are in arithmetic sequence, i.e., the given sequence is in harmonic sequence.
### **Supplementary Problems**
**22.56** Find the nth term and the sum of the first n terms of each arithmetic sequence for the indicated value of _n_.
( _a_ ) 1, 7, 13,... _n_ = 100
( _b_ ) 2, 9,... _n_ = 23
( _c_ ) –26, –24, –22,... _n_ = 40
( _d_ ) 2, 6, 10,... _n_ = 16
( _e_ )
( _f_ ) x –y, x, _x + y,... n_ = 30
**22.57** Find the sum of the first _n_ terms of each arithmetic sequence.
( _a_ ) 1, 2, 3,...
( _b_ ) 2, 8, 14,...
( _c_ ) 5, ...
**22.58** An arithmetic sequence has first term 4 and last term 34. If the sum of its terms is 247, find the number of terms and their common difference.
**22.59** An arithmetic sequence consisting of 49 terms has last term 28. If the common difference of its terms is 1/2, find the first term and the sum of the terms.
**22.60** Find the sum of all even integers between 17 and 99.
**22.61** Find the sum of all integers between 84 and 719 which are exactly divisible by 5.
**22.62** How many terms of the arithmetic sequence 3, 7, 11,... are needed to yield the sum 1275?
**22.63** Find three numbers in an arithmetic sequence whose sum is 48 and such that the sum of their squares is 800.
**22.64** A ball starting from rest rolls down an inclined plane and passes over 3 in. during the 1st second, 5 in. during the 2nd second, 7 in. during the 3rd second, etc. In what time from rest will it cover 120 inches?
**22.65** If 1¢ is saved the 1st day, 2¢ the 2nd day, 3¢ the third day, etc., find the sum that will accumulate at the end of 365 days.
**22.66** The sum of 40 terms of a certain arithmetic sequence is 430, while the sum of 60 terms is 945. Determine the nth term of the arithmetic sequence.
**22.67** Find an arithmetic sequence which has the sum of its first n terms equal to 2n2 \+ 3n.
**22.68** Determine the arithmetic mean between ( _a_ ) 15 and 41, ( _b_ ) – 16 and 23, ( _c_ ) ( _d_ ) _x_ – _3y_ and 5x + 2y.
**22.69** (a) Insert 4 arithmetic means between 9 and 24.
(b) Insert 2 arithmetic means between –1 and 11.
(c) Insert 3 arithmetic means between x _\+ 2y_ and x + 10y.
(d) Insert between 5 and 26 a number of arithmetic means such that the sum of the resulting arithmetic sequence will be 124.
**22.70** Find the _n_ th term and the sum of the first n terms of each geometric sequence for the indicated value of n.
(a) 2, 3, 9/2,... n = 5
(b) 6, –12, 24,... n = 9
(c) 1, 1/2, 1/4,... n = 10
(d) 1, 3, 9,... n = 8
(e) 8, 4, 2,... n = 12
(f) 3, 3 ... n = 8
**22.71** Find the sum of the first n terms of each geometric sequence.
( _a_ ) 1, 1/3, 1/9,...
( _b_ ) 4/3, 2, 3,...
( _c_ ) 1, –2, 4,...
**22.72** A geometric sequence has first term 3 and last term 48. If each term is twice the previous term, find the number of terms and the sum of the geometric sequence.
**22.73** Prove that the sum S of the terms of a geometric sequence in which the first term is a, the last term is l and the common ratio is r is given by
**22.74** In a geometric sequence the second term exceeds the first term by 4, and the sum of the second and third terms is 24. Show that there are two possible geometric sequences satisfying these conditions and find the sum of the first 5 terms of each geometric sequence.
**22.75** In a geometric sequence consisting of four terms, in which the ratio is positive, the sum of the first two terms is 10 and the sum of the last two terms is Find the sequence.
**22.76** The first two terms of a geometric sequence are _b_ /(1 + _c_ ) and _b_ /(1 + _c_ )2. Show that the sum of n terms of this sequence is given by the formula
**22.77** Find the sum of the first n terms of the geometric sequence:
**22.78** The third term of a geometric sequence is 6 and the fifth term is 81 times the first term. Write the first five terms of the sequence, assuming the terms are positive.
**22.79** Find three numbers in a geometric sequence whose sum is 42 and whose product is 512.
**22.80** The third term of a geometric sequence is 144 and the sixth term is 486. Find the sum of the first five terms of the geometric sequence.
**22.81** A tank contains a salt water solution in which is dissolved 972 lb of salt. One third of the solution is drawn off and the tank is filled with pure water. After stirring so that the solution is uniform, one third of the mixture is again drawn off and the tank is again filled with water. If this process is performed four times, what weight of salt remains in the tank?
**22.82** The sum of the first three terms of a geometric sequence is 26 and the sum of the first six terms is 728. What is the nth term of the geometric sequence?
**22.83** The sum of three numbers in geometric sequence is 14. If the first two terms are each increased by 1 and the third term decreased by 1, the resulting numbers are in an arithmetic sequence. Find the geometric sequence.
**22.84** Determine the geometric mean between:
(a) 2 and 18,
(b) 4 and 6,
(c) –4 and –16,
(d) _a + b_ and 4 _a_ \+ 4 _b_.
**22.85** (a) Insert two geometric means between 3 and 192.
(b) Insert four geometric means between and 8.
(c) The geometric mean of two numbers is 8. If one of the numbers is 6, find the other.
**22.86** The first term of an arithmetic sequence is 2; and the first, third and eleventh terms are also the first three terms of a geometric sequence. Find the sum of the first eleven terms of the arithmetic sequence.
**22.87** How many terms of the arithmetic sequence 9, 11, 13,... must be added in order that the sum should equal the sum of nine terms of the geometric sequence 3, –6, 12, –24,...?
**22.88** In a set of four numbers, the first three are in a geometric sequence and the last three are in an arithmetic sequence with a common difference of 6. If the first number is the same as the fourth, find the four numbers.
**22.89** Find two numbers whose difference is 32 and whose arithmetic mean exceeds the geometric mean by 4.
**22.90** Find the sum of the infinite geometric series.
( _a_ ) 3 + 1 + 1/3 + ···
( _b_ ) 4 + 2 + 1 + ···
( _c_ ) 1 + l/22 \+ 1/24 \+ ···
( _d_ ) 6 – 2 + 2/3 –···
( _e_ ) 4 – 8/3 + 16/9–···
( _f_ ) 1 + 0.1 + 0.01 + ···
**22.91** The sum of the first two terms of a decreasing geometric series is 5/4 and the sum to infinity is 9/4. Write the first three terms of the geometric series.
**22.92** The sum of an infinite number of terms of a decreasing geometric series is 3, and the sum of their squares is also 3. Write the first three terms of the series.
**22.93** The successive distances traversed by a swinging pendulum bob are respectively 36, 24, 16,... in. Find the distance which the bob will travel before coming to rest.
**22.94** Express each repeating decimal as a rational fraction.
( _a_ ) 0.121212···
( _b_ ) 0.090909···
( _c_ ) 0.270270···
( _d_ ) 1.424242···
( _e_ ) 0.1363636···
( _f_ ) 0.428571428571428···
**22.95** ( _a_ ) Find the 8th term of the harmonic sequence 2/3, 1/2, 2/5,...
( _b_ ) Find the 10th term of the harmonic sequence 5, 30/7, 15/4,...
( _c_ ) What is the _n_ th term of the harmonic sequence 10/3, 2, 10/7,...?
**22.96** Find the harmonic mean between each pair of numbers.
( _a_ ) 3 and 6
( _b_ ) 1/2 and 1/3
( _c_ )
( _d_ ) _a_ \+ _b_ and _a_ – _b_
**22.97** ( _a_ ) Insert two harmonic means between 5 and 10.
( _b_ ) Insert four harmonic means between 3/2 and 3/7.
**22.98** An object moves at uniform speed _a_ from _A_ to _B_ and then travels at uniform speed _b_ from _B_ to _A_. Show that the average speed in making the round trip is _2ab/(a + b_ ), the harmonic mean between _a_ and _b_. Calculate the average speed if _a_ = 30 ft/sec and _b_ = 60 ft/sec.
#### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**22.56** ( _a_ ) _l_ = 595, _S_ = 29 800
( _b_ ) _l_ = 19, _S_ = 93l
( _c_ ) _l_ = 52, _S_ = 520
( _d_ ) _l_ = 62, _S_ = 512
( _e_ ) _l_ = 57, _S_ = 1110
( _f_ ) _l_ = _x_ \+ 28 _y_ , _S_ = 30 _x_ \+ 405 _y_
**22.57** ( _a_ )
( _b_ ) _n_ (3 _n_ – 1)
(c)
**22.58** _n = 13, d = 5/2_
**22.59** _a_ = 4, _S_ = 784
**22.60** 2378
**22.61** 50 800
**22.62** 25
**22.63** 12, 16, 20
**22.64** 10 sec
**22.65** $667.95
**22.66**
**22.67** 5, 9, 13, 17,... _n_ th term = 4 _n_ \+ 1
**22.68** ( _a_ ) 28,
( _b_ )7/2,
( _c_ )
( _d_ ) 3 _x_ – _y_ /2
**22.69** ( _a_ ) 12, 15, 18, 21
( _b_ ) 3, 7
( _c_ ) _x_ \+ 4 _y_ , _x_ \+ 6 _y_ , _x_ \+ 8 _y_
( _d_ ) The arithmetic sequence is 5, 8, 11, 14, 17, 20, 23, 26
**22.70** ( _a_ ) _l_ = 81/8, _S_ = 211/8
( _b_ ) _l_ = 1536, _S_ = 1026
( _c_ ) _l_ = 1/512, _S_ = 1023/512
( _d_ ) _l_ = 2187, _S_ = 3280
( _e_ ) _l_ = 1/256, _S_ = 4095/256
( _f_ ) _l_ = 81, S = 120 + 40
**22.71** ( _a_ )
( _b_ )
( _c_ )
**22.72** _n_ = 5, _S_ = 93
**22.74** 2, 6, 18,... and _S_ = 242; 4, 8, 16,... and _S_ = 124
**22.75** 4, 6, 9, 27/2
**22.77**
**22.78** 2/3, 2, 6, 18, 54
**22.79** 2, 8, 32
**22.80** 844
**22.81** 192 lb
**22.82** 2·3 _n_ -1
**22.83** 2, 4, 8
**22.84** ( _a_ )6
( _b_ )
( _c_ ) –8
( _d_ ) 2 _a_ \+ 2 _b_
**22.85** ( _a_ ) 12, 48
( _b_ )2, 4,
( _c_ )32/3
**22.86** 187 or 22
**22.87** 19
**22.88** 8, –4, 2, 8
**22.89** 18, 50
**22.90** ( _a_ ) 9/2
( _b_ ) 8
( _c_ ) 4/3
( _d_ ) 9/2
( _e_ ) 12/5
( _f_ ) 10/9
**22.91** 3/4, 1/2, 1/3
**22.92** 3/2, 3/4, 3/8
**22.93** 108 in.
**22.94** ( _a_ ) 4/33
( _b_ ) 1/11
( _c_ ) 10/37
( _d_ ) 47/33
( _e_ ) 3/22
( _f_ ) 3/7
**22.95** ( _a_ ) 1/5
( _b_ ) 2
( _c_ )
**22.96** ( _a_ ) 4
( _b_ ) 2/5
( _c_ )
( _d_ )
**22.97** ( _a_ ) 6, 15/2
( _b_ ) 1, 3/4, 3/5, 1/2
**22.98** 40 ft/sec
## CHAPTER 23
Logarithms
### **23.1 DEFINITION OF A LOGARITHM**
If _b x = N_, where _N_ is a positive number and _b_ is a positive number different from 1, then the exponent _x_ is the logarithm of _N_ to the base _b_ and is written _x_ = log _b_ _N_.
**EXAMPLE 23.1.** Write 32 = 9 using logarithmic notation.
Since 32 = 9, then 2 is the logarithm of 9 to the base 3, i.e., 2 = log3 9.
**EXAMPLE 23.2.** Evaluate log2 8.
log2 8 is that number _x_ to which the base 2 must be raised in order to yield 8, i.e., 2 _x_ = 8, _x_ = 3. Hence log2 8 = 3.
Both _b_ _x_ = _N_ and _x_ = log _ _b_ N_ are equivalent relationships; _b_ _x_ = _N_ is called the _exponential form_ and _x_ = log _b_ _N_ the _logarithmic form_ of the relationship. As a consequence, corresponding to _laws of exponents_ there are _laws of logarithms_.
### **23.2** LAWS OF LOGARITHMS
I. The logarithm of the product of two positive numbers _M_ and _N_ is equal to the sum of the logarithms of the numbers, i.e.,
II. The logarithm of the quotient of two positive numbers _M_ and _N_ is equal to the difference of the logarithms of the numbers, i.e.,
III. The logarithm of the _p_ th power of a positive number _M_ is equal to _p_ mutiplied by the logarithm of the number, i.e.,
**EXAMPLES 23.3.** Apply the laws of logarithms to each expression.
( _a_ ) log2 3(5)
( _b_ )
( _c_ ) log753
( _d_ )
( _a_ ) log23(5) = log2 3 + log2 5
( _b_ )
( _c_ ) log753 = 3log7 5
( _d_ )
### **23.3 COMMON LOGARITHMS**
The system of logarithms whose base is 10 is called the common logarithm system. When the base is omitted, it is understood that base 10 is to be used. Thus log 25 = log10 25.
Consider the following table.
It is obvious that 101.5377 will give some number greater than 10 (which is 101) but smaller than 100 (which is 102). Actually, 101.5377 = 34.49; hence log 34.49 = 1.5377.
The digit before the decimal point is the _characteristic_ of the log, and the decimal fraction part is the _mantissa_ of the log. In the above example, the characteristic is 1 and the mantissa is. 5377.
The mantissa of the log of a number is found in tables, ignoring the decimal point of the number. Each mantissa in the tables is understood to have a decimal point preceding it, and the mantissa is always considered positive.
The characteristic is determined by inspection from the number itself according to the following rules.
(1) For a number greater than 1, the characteristic is positive and is one _less_ than the number of digits before the decimal point. For example:
(2) For a number less than 1, the characteristic is negative and is one _more_ than the number of zeros immediately following the decimal point. The negative sign of the characteristic is written in either of these two ways: ( _a_ ) above the characteristic as etc.; ( _b_ ) as 9 – 10, 8 – 10, etc. Thus the characteristic of 0.3485 is or 9 – 10, of 0.0513 is or 8 – 10, and of 0.0024 is or 7 – 10.
### **23.4 USING A COMMON LOGARITHM TABLE**
To find the common logarithm of a positive number use the table of common logarithms in Appendix A.
Suppose it is required to find the log of the number 728. In the table of common logarithms glance down the _N_ column to 72, then horizontally to the right to column 8 and note the figure 8621, which is the required mantissa. Since the characteristic is 2, log 728 = 2.8621. (This means that 728 = 102.86212.)
The mantissa for log 72.8, for log 7.28, for log 0.728, for log 0.0728, etc., is. 8621, but the characteristics differ. Thus:
When the number contains four digits, interpolate using the method of proportional parts.
**EXAMPLE 23.4.** Find log 4.638.
The characteristic is 0. The mantissa is found as follows.
.8 × tabular difference = .000 72 or.0007 to four decimal places.
Mantissa of log 4638 = .6656 +.0007 = .6663 to four digits.
Hence log 4.638 = 0.6663.
The mantissa for log 4638, for log 463.8, for log 46.38, etc., is. 6663, but the characteristics differ. Thus:
The antilogarithm is the number corresponding to a given logarithm. "The antilog of 3" means "the number whose log is 3"; that number is obviously 1000.
**EXAMPLES 23.5.** Find the value of _N_.
( _a_ ) log _N_ = 1.9058
( _b_ ) log _N_ = 7.8657 – 10
( _c_ ) log _N_ = 9.3842 – 10.
( _a_ ) In the table the mantissa.9058 corresponds to the number 805. Since the characteristic of log _N_ is 1, the number must have two digits before the decimal point; thus _N_ = 80.5 (or antilog 1.9058 = 80.5).
( _b_ ) In the table the mantissa.8657 corresponds to the number 734. Since the characteristic is 7 – 10, the number must have two zeros immediately following the decimal point; thus _N_ = 0.007 34 (or antilog 7.8657 – 10 = 0.007 34).
( _c_ ) Since the mantissa.3842 is not found in the tables, interpolation must be used.
### **23.5 NATURAL LOGARITHMS**
The system of logarithms whose base is the constant e is called the natural logarithm system. When we want to indicate the base of a logarithm is e we write ln. Thus, ln 25 = loge 25.
The exponential form of ln _a_ = _b_ is e _b_ = _a_. The number e is an irrational number that has a decimal expansion e = 2.718 281 828 450 45....
### **23.6 USING A NATURAL LOGARITHM TABLE**
To find the natural logarithm of a positive number use the table of natural logarithms in Appendix B.
To find the natural logarithm of a number from 1 to 10, such as 5.26, we go down the _N_ column to 5.2, then across to the right to the column headed by.06 to get the value 1.6601. Thus, ln 5.26 = 1.6601. This means that 5.26 = e1.6601.
If we want to find the natural logarithm of a number greater than 10 or less than one, we write the number in scientific notation, apply the laws of logarithms, and use the natural logarithm table and the fact that ln 10 = 2.3026.
**EXAMPLES 23.6.** Find the natural logarithm of each number.
( _a_ ) 346
( _b_ ) 0.0217
( _b_ )
The value of ln 4.638 cannot be found directly from the natural logarithm table since it has four significant digits, but we can interpolate to find it.
0.8 × tabular difference = 0.8 × 0.0021 = 0.001 68 or 0.0017 to four decimal places.
Thus, ln 4.638 = ln 4.630 + 0.0017 = 1.5326 + 0.0017 = 1.5343.
The antilogarithm of a natural logarithm is the number that has the given logarithm. The procedure for finding the antilogarithm of a natural logarithm less than 0 or greater than 2.3026, requires us to add or subtract multiples of ln 10 = 2.3026 to bring the natural logarithm into the range 0 to 2.3026 that can be found from the table in Appendix B.
**EXAMPLES 23.7.** Find the value of _N_.
( _a_ ) ln _N_ = 2.1564
( _b_ ) ln _N_ = –4.9705
( _c_ ) ln _N_ = 1.8869
( _a_ ) ln _N_ = 2.1564 is between 0 and 2.3026, so we look in the natural logarithm table for 2.1564. It is in the table, so we get _N_ from the sum of the numbers that head the row and column for 1.1564. Thus, _N_ = antilogarithm 2.1564 = 8.64.
( _b_ ) Since ln _N_ = – 4.9705 is less than 0, we must rewrite it as a number between 0 and 2.3026 minus a multiple of 2.3026 = ln 10. Since if we add 3 times 2.3026 to – 4.9705 we get a positive number between 0 and 2.3026, we will rewrite –4.9705 as 1.9373 – 3(2.3026).
( _c_ ) Since ln _N_ = 1.8869 is between 0 and 2.3026, we look for 1.8869 in the natural logarithm table, but it is not there. We will have to use interpolation to find _N_.
### **23.7 FINDING LOGARITHMS USING A CALCULATOR**
If the number we want to find the logarithm of a number with four or more significant digits, we can round the number to four significant digits and use the logarithm tables and interpolation or we can use a scientific or graphing calculator to find the logarithm for the given number. The use of the calculator will yield a more accurate result.
A scientific calculator can be used to find logarithms and antilogarithms to base 10 or base e. Scientific calculators have keys for the log and ln functions, and the inverse of these yields the antilogarithms.
Much of the computation once done using logarithms can be done directly on a scientific calculator. The advantages of doing a problem on the calculator are that the numbers rarely have to be rounded and that the problem can be worked quickly and accurately.
### **Solved Problems**
**23.1** Express each of the following exponential forms in logarithmic form:
_(a) p q_ = _r_ ,
( _b_ ) 23 = 8,
( _c_ ) 42 = 16,
( _d_ )
( _e_ )
**SOLUTION**
_(a) q_ = log _p_ _r_ ,
( _b_ ) 3 = log2 8,
( _c_ ) 2 = log4 16,
( _d_ )
( _e_ )
**23.2** Express each of the following logarithmic forms in exponential form:
( _a_ ) log5 25 = 2,
( _b_ ) log2 64 = 6,
( _c_ )
( _d_ ) log _a_ a3 = 3,
( _e_ ) log _r_ 1 = 0.
**SOLUTION**
( _a_ ) 52 = 25,
( _b_ ) 26 = 64,
( _c_ )
_(d) a_ 3 = _a_ 3,
_(e) r_ 0 = 1
**23.3** Determine the value of each of the following.
( _a_ ) log4 64. Let log4 64 = _x_ ; then 4 _x_ = 64 = 43 and _x_ = 3.
( _b_ ) log3 81. Let log3 81 = _x_ ; then 3 _x_ = 81 = 34 and _x_ = 4.
( _c_ ) log1/2 8. Let log1/2 8 = _x_ ; then = 8, (2–1) _x_ = 23, 2– _x_ = 23 and _x_ = – 3.
( _d_ )
( _e_ )
**23.4** Solve each of the following equations.
( _a_ ) log3 _x_ = 2, 32 = _x_ , _x_ = 9
( _b_ )
( _c_ ) log _x_ 25 = 2, _x_ 2 = 25, _x_ = ±5. Since bases are positive, the solution is _x_ = 5.
( _d_ )
( _e_ ) log(3 _x_ 2 \+ 2 _x_ – 4) = 0, 100 = 3 _x_ 2 \+ 2 _x_ – 4, 3 _x_ 2 \+ 2 _x_ – 5 = 0, _x_ = 1, –5/3.
**23.5** Prove the laws of logarithms.
**SOLUTION**
Let _M_ = _b x_ and _N = b y_; then _x_ = log _b_ _M_ and _y_ = log _b_ _N_.
I. Since _MN_ = _b x_. _b y_ = _b x+y_, then log _b_ _MN_ = _x + y_ = log _b_ _M_ \+ log _b_ _N_.
II.
III. Since _M p_ = ( _b x_) _p_ = _b px_, then log _b Mp_ = _px_ = _p_ log _b M_.
**23.6** Express each of the following as an algebraic sum of logarithms, using the laws I, II, III.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
**23.7** Given that log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990, log 7 = 0.8451 (all base 10) accurate to four decimal places, evaluate the following.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
_Note_. ln exponential form this means 10–2.0916 = 0.0081.
**23.8** Express each of the following as a single logarithm (base is 10 unless otherwise indicated).
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
**23.9** In each of the following equations, solve for the indicated letter in terms of the other quantities.
( _a_ )
( _b_ )
( _c_ )
( _d_ ) 2 log _x_ \+ 3 log _y_ = 4 log _z_ – 2 _: y_.
Solving for log _y_ , 3 log _y_ = 4 log _z_ – 2 – 2 log _x_ and
Hence _y_ = 10–2/3 _x_ –2/3 _z_ 4/3.
( _e_ )
**23.10** Determine the characteristic of the common logarithm of each of the following numbers.
( _a_ ) 57
( _b_ ) 57.4
( _c_ ) 5.63
( _d_ ) 35.63
( _e_ ) 982.5
( _f_ ) 7824
( _g_ ) 186 000
( _h_ ) 0.71
( _i_ ) 0.7314
( _j_ ) 0.0325
( _k_ ) 0.0071
( _l_ ) 0.0003
**SOLUTION**
( _a_ ) 1
( _b_ ) 1
( _c_ ) 0
( _d_ ) 1
( _e_ ) 2
( _f_ ) 3
( _g_ ) 5
( _h_ ) 9 – 10
( _i_ ) 9 – 10
( _j_ ) 8 – 10
( _k_ ) 7 – 10
( _l_ ) 6 – 10
**23.11** Verify each of the following common logarithms.
**23.12** Verify each of the following.
**23.13** Write each of the following numbers as a power of 10: ( _a_ ) 893, ( _b_ ) 0.358.
**SOLUTION**
( _a_ ) We require _x_ such that 10 _x_ = 893. Then _x_ = log 893 = 2.9509 and 893 = 102.9509.
( _b_ ) We require _x_ such that 10 _x_ = 0.358.
Then _x_ = log 0.358 = 9.5539 – 10 = –0.4461 and 0.358 = 10–0.4461.
Calculate each of the following using logarithms.
**23.14** _P_ = 3.81 × 43.4
**SOLUTION**
Hence _P_ = antilog 2.2184 = 165.3.
Note the exponential significance of the computation. Thus
**23.15** _P_ = 73.42 × 0.004 62 × 0.5143
**SOLUTION**
Hence _P_ = 0.1744.
**23.16**
**SOLUTION**
log _P_ = log 784.6 + log 0.0431 – log 28.23
**23.17** _P_ = (7.284)5
**SOLUTION**
log _P_ = 5 log 7.284 = 5(0.8623) = 4.3115 and _P_ = 20490.
**23.18**
**SOLUTION**
**23.19**
**SOLUTION**
**23.20** The period _T_ of a simple pendulum of length _l_ is given by the formula where _g_ is the acceleration due to gravity. Find _T_ (in seconds) if _l_ = 281.3 cm and _g_ = 981.0 cm/sec2. Take 2π = 6.283.
**SOLUTION**
**23.21** Solve for _x_ : 52 _x_ +2 = 35 _x_ –1.
**SOLUTION**
**23.22** Find the value of each of these natural logarithms.
( _a_ ) ln 5.78
( _b_ ) ln 8.62
( _c_ ) ln 3.456
( _d_ ) ln 4.643
( _e_ ) ln 190
( _f_ ) ln 0.0084
( _g_ ) ln 2839
( _h_ ) ln 0.014 85
**SOLUTION**
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
**23.23** Find the value of _N_.
( _a_ ) ln _N_ = 2.4146
( _b_ ) ln _N_ = 0.9847
( _c_ ) ln _N_ = –1.7654
**SOLUTION**
( _a_ )
( _b_ )
( _c_ )
### **Supplementary Problems**
**23.24** Evaluate:
( _a_ ) log2 32,
( _b_ )
( _c_ ) log31/9,
( _d_ ) log1/416,
( _e_ ) log _e_ _e_ _x_ ,
( _f_ ) log84.
**23.25** Solve each equation for the unknown.
( _a_ ) log2 _x_ = 3
( _b_ ) log _y_ = –2
( _c_ ) log _x_ 8 = –3
( _d_ ) log3 (2 _x_ \+ 1) = 1
( _e_ ) log4 _x_ 3 = 3/2
( _f_ ) log( _x_ –1) (4 _x_ – 4) = 2
**23.26** Express as an algebraic sum of logarithms.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
**23.27** Solve each equation for the indicated letter in terms of the other quantities.
( _a_ ) 2 log _x_ = log 16; _x_
( _b_ ) 3 log _y_ \+ 2 log 2 = log 32; _y_
( _c_ ) log3 _F_ = log3 4 – 2log3 _x_ ; _F_
( _d_ ) ln (30 – _U_ ) = ln 30 – 2 _t_ ; _U_
**23.28** Prove that if _a_ and _b_ are positive and ≠ 1, (log _a_ _b_ )(log _b_ _a_ ) = 1.
**23.29** Prove that 10log _N_ = _N_ where _N_ > 0.
**23.30** Determine the characteristic of the common logarithm of each number.
( _a_ ) 248
( _b_ ) 2.48
( _c_ ) 0.024
( _d_ ) 0.162
( _e_ ) 0.0006
( _f_ ) 18.36
( _g_ ) 1.06
( _h_ ) 6000
( _i_ ) 4
( _j_ ) 40.60
( _k_ ) 237.63
( _l_ ) 146.203
( _m_ ) 7 000 000
( _n_ ) 0.000 007
**23.31** Find the common logarithm of each number.
( _a_ ) 237
( _b_ ) 28.7
( _c_ ) 1.26
( _d_ ) 0.263
( _e_ ) 0.086
( _f_ ) 0.007
( _g_ ) 10 400
( _h_ ) 0.00 607
( _i_ ) 0.000 000 728
( _j_ ) 6 000 000
( _k_ ) 23.70
( _l_ ) 6.03
( _m_ ) 1
( _n_ ) 1000
**23.32** Find the antilogarithm of each of the following.
( _a_ ) 2.8802
( _b_ ) 1.6590
( _c_ ) 0.6946
( _d_ ) .9042
( _e_ ) 8.3160 – 10
( _f_ ) 7.8549 – 10
( _g_ ) 4.6618
( _h_ ) 0.4216
( _i_ ) .9484
( _j_ ) 9.8344 – 10
**23.33** Find the common logarithm of each number by interpolation.
( _a_ ) 1463
( _b_ ) 810.6
( _c_ ) 86.27
( _d_ ) 8.106
( _e_ ) 0.6041
( _f_ ) 0.046 22
( _g_ ) 1.006
( _h_ ) 300.6
( _i_ ) 460.3
( _j_ ) 0.003 001
**23.34** Find the antilogarithm of each of the following by interpolation.
( _a_ ) 2.9060
( _b_ ) .4860
( _c_ ) 1.6600
( _d_ ) .9840
( _e_ ) 3.7045
( _f_ ) 8.9266 – 10
( _g_ ) 2.2500
( _h_ ) 0.8003
( _i_ ) .4700
( _j_ ) 1.2925
**23.35** Write each number as a power of 10:
( _a_ ) 45.4,
( _b_ ) 0.005 278.
**23.36** Evaluate.
( _a_ ) (42.8)(3.26)(8.10)
( _b_ )
( _c_ )
( _d_ )
( _e_ )
( _f_ )
( _g_ )
( _h_ )
( _i_ )
( _j_ )
**23.37** Solve the following hydraulics equation:
**23.38** Solve for _x_.
( _a_ ) 3 _x_ = 243
( _b_ ) 5 _x_ = 1/125
( _c_ ) 2 _x_ +2 = 64
_(d) x_ –2 = 16
_(e) x_ –3/4 = 8
_(f) x_ –2/3 = 1/9
( _g_ ) 7 _x_ –1/2 = 4
( _h_ ) 3 _x_ = 1
( _i_ ) 5 _x_ –2 = 1
( _j_ ) 22 _x_ +3 = 1
**23.39** Solve each exponential equation:
( _a_ ) 42 _x_ – 1 = 5 _x_ \+ 2,
( _b_ ) 3 _x_ –1 = 4 · 51–3 _x_.
**23.40** Find the natural logarithms.
( _a_ ) ln 2.367
( _b_ ) ln 8.532
( _c_ ) ln 4875
( _d_ ) ln 0.000 189 4
**23.41** Find _N_ , the antilogarithm of the given number.
( _a_ ) ln _N_ = 0.7642
( _b_ ) ln _N_ = 1.8540
( _c_ ) ln _N_ = 8.4731
( _d_ ) ln _N_ = –6.2691
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**23.24**
( _a_ ) 5
( _b_ ) 1/4
( _c_ ) –2
( _d_ ) –2
_(e) x_
( _f_ ) 2/3
**23.25**
( _a_ ) 8
( _b_ ) 0.01
( _c_ ) 1/2
( _d_ ) 1
( _e_ ) 2
( _f_ ) 5
**23.26**
( _a_ ) 3 log _U +_ 2 log _V_ – 5 log _W_
( _b_ )
( _c_ )
( _d_ )
**23.27**
( _a_ ) 4
( _b_ ) 2
_(c) F_ = 4/ _x_ 2
_(d) U_ = 30(1 – _e_ –2 _t_ )
**23.30**
( _a_ ) 2
( _b_ ) 0
( _c_ )
( _d_ )
( _e_ )
( _f_ ) 1
( _g_ ) 0
( _h_ ) 3
( _i_ ) 0
( _j_ ) 1
( _k_ ) 2
( _l_ ) 2
( _m_ ) 6
( _n_ )
**23.31**
( _a_ ) 2.3747
( _b_ ) 1.4579
( _c_ ) 0.1004
( _d_ ) .4200
( _e_ ) .9345
( _f_ ) 7.8451 – 10
( _g_ ) 4.0170
( _h_ ) .7832
( _i_ ) .8621
( _j_ ) 6.7782
( _k_ ) 1.3747
( _l_ ) 0.7803
( _m_ ) 0.0000
( _n_ ) 3.0000
**23.32**
( _a_ ) 759
( _b_ ) 45.6
( _c_ ) 4.95
( _d_ ) 0.0802
( _e_ ) 0.0207
( _f_ ) 0.00716
( _g_ ) 45 900
( _h_ ) 2.64
( _i_ ) 0.888
( _j_ ) 0.683
**23.33**
( _a_ ) 3.1653
( _b_ ) 2.9088
( _c_ ) 1.9359
( _d_ ) 0.9088
( _e_ ) .7811
( _f_ ) 8.6648 – 10
( _g_ ) 0.0026
( _h_ ) 2.4780
( _i_ ) 2.6631
( _j_ ) 7.4773 – 10
**23.34**
( _a_ ) 805.4
( _b_ ) 0.3062
( _c_ ) 45.71
( _d_ ) 0.9638
( _e_ ) 5064
( _f_ ) 0.08445
( _g_ ) 177.8
( _h_ ) 6.314
( _i_ ) 0.2951
( _j_ ) 19.61
**23.35**
( _a_ ) 101.6571
( _b_ ) 10–2.2776
**23.36**
( _a_ ) 1130
( _b_ ) 0.0248
( _c_ ) 29.9
( _d_ ) 4.27
( _e_ ) 1.124
( _f_ ) 860
( _g_ ) 1.90
( _h_ ) 4.44
( _i_ ) 145.5
( _j_ ) 8.54
**23.37**
0.0486
**23.38**
( _a_ ) 5
( _b_ ) –3
( _c_ ) 4
( _d_ ) ± 1/4
( _e_ ) 1/16
( _f_ ) ± 27
( _g_ ) 49/16
( _h_ ) 0
( _i_ ) 2
( _j_ ) –3/2
**23.39**
( _a_ ) 3.958
( _b_ ) 0.6907
**23.40**
( _a_ ) 0.8616
( _b_ ) 2.1438
( _c_ ) 8.4919
( _d_ ) – 8.5717
**23.41**
( _a_ ) 2.147
( _b_ ) 6.385
( _c_ ) 4784
( _d_ ) 0.001 894
## **CHAPTER 24
Applications of Logarithms and Exponents**
### **24.1 INTRODUCTION**
Logarithms have their major use in solving exponential equations and solving equations in which the variables are related logarithmically. To solve equations in which the variable is in the exponent, we generally start by changing the expression from exponential form to logarithmic form.
### **24.2 SIMPLE INTEREST**
Interest is money paid for the use of a sum of money called the principal. The interest is usually paid at the ends of specified equal time intervals, such as monthly, quarterly, semiannually, or annually. The sum of the principal and the interest is called the amount.
The simple interest, _I_ , on the principal, _P_ , for a time in years, _t_ , at an interest rate per year, _r_ , is given by the formula _I = Prt_ , and the amount, _A_ , is found by _A = P + Prt_ or _A = P_ (1 + _rt_ ).
**EXAMPLE 24.1.** If an individual borrows $800 at 8% per year for two and one-half years, how much interest must be paid on the loan?
**EXAMPLE 24.2.** If a person invests $3000 at 6% per year for five years, how much will the investment be worth at the end of the five years?
### **24.3 COMPOUND INTEREST**
Compound interest means that the interest is paid periodically over the term of the loan which results in a new principal at the end of each interval of time.
If a principal, _P_ , is invested for _t_ years at an annual interest rate, _r_ , compounded _n_ times per year, then the amount, _A_ , or ending balance is given by:
**EXAMPLE 24.3.** Find the amount of an investment if $20 000 is invested at 6% compounded monthly for three years.
When the interest is compounded more and more frequently, we get to a situation of continuously compounded interest. If a principal, _P_ , is invested for _t_ years at an annual interest rate, _r_ , compounded continuously, then the amount, _A_ , or ending balance, is given by:
_A = Pe_ _rt_
**EXAMPLE 24.4.** Find the amount of an investment if $20 000 is invested at 6% compounded continuously for three years.
_A_ = _Pe_ _rt_
_A_ = 20 000e0.06(3)
_A_ = 20 000e0.18
ln _A_ = ln 20 000e0.18
ln _A_ = ln 20 000 + ln e0.18
ln _A_ = ln(2 00 × 104) + 0.18 ln e
ln _A_ = ln 2.00 + 4 ln 10 + 0.18(1) ln e = 1
ln _A_ = 0.6931 + 4(2.3026) + 0.18 ln 2.00 = 0.6931 and ln 10 = 2.3026
ln _A_ = 10.0835
ln _A_ = 0.8731 + 4(2.3026)
ln _A_ = ln (2.39 + 0.004) + ln 104
ln _A_ = ln 2.394 + ln 104
ln _A_ = ln (2.394 × 104)
ln _A_ = ln 23 940
_A_ = $23 940
In doing Examples 24.3 and 24.4 we found the answers to four significant digits. However, using the logarithm tables and doing interpolation results in some error. Also, we may have a problem if the interest is compounded daily, because when we divide _r_ by _n_ the result could be zero when rounded to thousandths. To deal with this problem and to get greater accuracy, we can use five-place logarithm tables, calculators, or computers. Generally, banks and other businesses use computers or calculators to get the accuracy they need.
**EXAMPLE 24.5.** Use a scientific or graphing calculator to find the amount of an investment if $20 000 is invested at 6% compounded monthly for three years.
To the nearest cent, the amount has been increased by $33.61 from the amount found in Example 24.3. It is possible to compute the answer to the nearest cent here, while we were able to compute the result to the nearest ten dollars in Example 24.3.
**EXAMPLE 24.6.** Use a scientific or graphing calculator to find the amount of an investment if $20 000 is invested at 6% compounded continuously for three years.
_A_ = _Pe_ _rt_
_A_ = $20 000e0.06(3)
_A_ = $20 000e0.18 use the inverse of ln _x_ to compute e0.18
_A_ = $23 944.35
To the nearest cent, the amount has increased by $4.35 from the amount found in Example 24.4. The greater accuracy was possible because the calculator computes with more decimal places in each operation and then the answer is rounded. In our examples, we rounded to hundredths because cents are the smallest units of money that have a general usefulness. Most calculators compute with 8, 10, or 12 significant digits in doing the operations.
### **24.4 APPLICATIONS OF LOGARITHMS**
The loudness, L, of a sound (in decibels) perceived by the human ear depends on the ratio of the intensity, _I_ , of the sound to the threshold, _I_ 0, of hearing for the average human ear.
**EXAMPLE 24.7.** Find the loudness of a sound that has an intensity 10 000 times the threshold of hearing for the average human ear.
Chemists use the hydrogen potential, pH, of a solution to measure its acidity or basicity. The pH of distilled water is about 7. If the pH of a solution exceeds 7, it is called a base, but if its pH is less than 7 it is called an acid. If [H+] is the concentration of hydrogen ions in moles per liter, the pH is given by the formula:
**EXAMPLE 24.8.** Find the pH of the solution whose concentration of hydrogen ions is 5.32 _x_ 10-5 moles per liter.
pH = –log[H+]
pH = –log (5.32 × 10–5)
pH = –[log 5.32 + log 10–5]
pH = –log 5.32 × (–5)log 10 log 10 = 1
pH = –log 5.32 + 5(1)
pH = –0.7259 + 5
pH = 4.2741
pH = 4.3
Seismologists use the Richter scale to measure and report the magnitude of earthquakes. The magnitude or Richter number of an earthquake depends on the ratio of the intensity, _I_ , of an earthquake to the reference intensity, _I_ 0, which is the smallest earth movement that can be recorded on a seismograph. Richter numbers are usually rounded to the nearest tenth or hundredth. The Richter number is given by the formula:
**EXAMPLE 24.9.** If the intensity of an earthquake is determined to be 50 000 times the reference intensity, what is its reading on the Richter scale?
### **24.5 APPLICATIONS OF EXPONENTS**
The number e is involved in many functions occurring in nature. The growth curve of many materials can be described by the exponential growth equation:
_A_ = _A_ 0e _rt_
where A0 is the initial amount of the material, r is the annual rate of growth, t is the time in years, and A is the amount of the material at the ending time.
**EXAMPLE 24.10.** The population of a country was 2 400 000 in 1990 and it has an annual growth rate of 3%. If the growth is exponential, what will its population be in 2000?
The decay or decline equation is similar to the growth except the exponent is negative.
where _A_ 0 is the initial amount, _r_ is the annual rate of decay or decline, _t_ is the time in years, and _A_ is the ending amount.
**EXAMPLE 24.11.** A piece of wood is found to contain 100 grams of carbon-14 when it is removed from a tree. If the rate of decay of carbon-14 is 0.0124% per year, how much carbon-14 will be left in the wood after 200 years?
### **Solved Problems**
**24.1** A woman borrows $400 for 2 years at a simple interest rate of 3%. Find the amount required to repay the loan at the end of 2 years.
**SOLUTION**
Interest _I = Prt_ = 400(0.03)(2) = $24. Amount _A_ = principal _P +_ interest _I_ = $424.
**24.2** Find the interest _I_ and amount _A_ for
( _a_ ) $600 for 8 months (2/3 yr) at 4%,
( _b_ ) $1562.60 for 3 years, 4 months (10/3 yr) at 3 %.
**SOLUTION**
( _a_ ) _I_ = _Prt_ = 600(0.04)(2/3) = $16. _A_ = _P + I_ = $616.
( _b_ ) _I_ = _Prt_ = 1562.60(0.035)(10/3) = $182.30. _A = P + I_ = $1744.90.
**24.3** What principal invested at 4% for 5 years will amount to $1200?
**SOLUTION**
The principal of $1000 is called the present value of $1200. This means that $1200 to be paid 5 _years from now_ is worth $1000 _now_ (the interest rate being 4%).
**24.4** What rate of interest will yield $1000 on a principal of $800 in 5 years?
**SOLUTION**
**24.5** A man wishes to borrow $200. He goes to the bank where he is told that the interest rate is 5%, interest payable in advance, and that the $200 is to be paid back at the end of one year. What interest rate is he actually paying?
**SOLUTION**
The simple interest on $200 for 1 year at 5% is _I_ = 200(0.05)(1) = $10. Thus he receives $200 – $10 = $190. Since he must pay back $200 after a year, _P_ = $190, _A_ = $200, _t_ = 1 year. Thus
i.e., the effective interest rate is 5.26%.
**24.6** A merchant borrows $4000 under the condition that she pay at the end of every 3 months $200 on the principal plus the simple interest of 6% on the principal outstanding at the time. Find the total amount she must pay.
**SOLUTION**
Since $4000 is to be paid (excluding interest) at the rate of $200 every 3 months, it will take 4000/200(4) = 5 years, i.e., 20 payments.
The total interest is 60 + 57 + 54 +... + 9 + 6 + 3: an arithmetic sequence with the sum given by _S_ = ( _n_ /2)( _a_ \+ _l_ ), where _a_ = 1st term, _l_ = last term, _n_ = number of terms.
Then _S_ = (20/2)(60 + 3) = $630, and the total amount she must pay is $4630.
**24.7** What will $500 deposited in a bank amount to in 2 years if interest is compounded semiannually at 2%?
**SOLUTION**
_Method 1_. Without formula.
_Method 2_. Using formula.
_P_ = $500, _i_ = rate per period = 0.02/2 = 0.01. _n_ = number of periods = 4.
_A_ = _P_ (1 + _i_ ) _n_ = 500(1.01)4 = 500(1.0406) = $520.30.
_Note_. (1.01)4 may be evaluated by the binomial formula, logarithms or tables.
**24.8** Find the compound interest and amount of $2800 in 8 years at 5% compounded quarterly.
**SOLUTION**
_A_ = _P_ (1 + _i_ ) _n_ = 2800(1 + 0.05/4)32 = 2800(1.0125)32 = 2800(1.4881) = $4166.68.
Interest _= A – P_ = $4166.68 – $2800 = $1366.68.
**24.9** What rate of interest compounded annually is the same as the rate of interest of 6% compounded semiannually?
**SOLUTION**
Amount from principal _P_ in 1 year at rate _r_ = _P_ (l + _r_ ).
Amount from principal _P_ in 1 year at rate 6% compounded semiannually = _P_ (1 + 0.03)2.
The amounts are equal if _P_ (1 + _r_ ) = _P_ (1.03)2, 1 + _r_ = (1.03)2, _r_ = 0.0609 or 6.09%.
The rate of interest _i_ per year compounded a given number of times per year is called the _nominal rate_. The rate of interest _r_ which, if compounded annually, would result in the same amount of interest is called the _effective rate_. In this example, 6% compounded semiannually is the nominal rate and 6.09% is the effective rate.
**24.10** Derive a formula for the effective rate in terms of the nominal rate.
**SOLUTION**
Amount from principal _P_ in 1 year at rate _r = P_ (1 + _r_ ).
Amount from principal _P_ in 1 year at rate _i_ compounded _k_ times per year = _P_ (1 + _i_ / _k_ ) _k_.
The amounts are equal if _P_ (1 + _r_ ) = _P_ (1 + _i_ / _k_ ) _k_.
Hence _r_ = (1 + _i_ / _k_ ) _k_ – 1.
**24.11** The population in a country grows at a rate of 4% compounded annually. At this rate, how long will it take the population to double?
**SOLUTION**
**24.12** If $1000 is invested at 10% compounded continuously, how long will it take the investment to triple?
**SOLUTION**
**24.13** Find the pH of blood if the concentration of hydrogen ions is 3.98 × 10–8.
**SOLUTION**
pH = – log[H+]
pH = – log (3.98 × 10–8)
pH = – log 3.98 – (–8) log 10
pH = –0.5999 + 8
pH = 7.4001
pH = 7.40
**24.14** An earthquake in San Francisco in 1989 was reported to have a Richter number of 6.90. How does the intensity of the earthquake compare with the reference intensity?
**SOLUTION**
**24.15** The population of the world increased from 2.5 billion in 1950 to 5.0 billion in 1987. If the growth was exponential, what was the annual growth rate?
**SOLUTION**
**24.16** In Nigeria the rate of deforestation is 5.25% per year. If the decrease in forests in Nigeria is exponential, how long will it take until only 25% of the current forests are left?
**SOLUTION**
### **Supplementary Problems**
**24.17** If $5.13 interest is earned in two years on a deposit of $95, then what is the simple annual interest rate?
**24.18** If $500 is borrowed for one month and $525 must be paid back at the end of the month, what is the simple annual interest?
**24.19** If $4000 is invested at a bank that pays 8% interest compounded quarterly, how much will the investment be worth in 6 years?
**24.20** If $8000 is invested in an account that pays 12% interest compounded monthly, how much will the investment be worth after 10 years?
**24.21** A bank tried to attract new, large, long-term investments by paying 9.75% interest compounded continuously if at least $30 000 was invested for at least 5 years. If $30 000 is invested for 5 years at this bank, how much would the investment be worth at the end of the 5 years?
**24.22** What interest will be earned if $8000 is invested for 4 years at 10% compounded semiannually?
**24.23** What interest will be earned if $3500 is invested for 5 years at 8% compounded quarterly?
**24.24** What interest will be earned if $4000 is invested for 6 years at 8% compounded continuously?
**24.25** Find the amount that will result if $9000 is invested for 2 years at 12% compounded monthly.
**24.26** Find the amount that will result if $9000 is invested for 2 years at 12% compounded continuously.
**24 27** In 1990 an earthquake in Iran was said to have about 6 times the intensity of the 1989 San Francisco earthquake, which had a Richter number of 6.90. What is the Richter number of the Iranian earthquake?
**24.28** Find the Richter number of an earthquake if its intensity is 3 160 000 times as great as the reference intensity.
**24.29** An earthquake in Alaska in 1964 measured 8.50 on the Richter scale. What is the intensity of this earthquake compared with the reference intensity?
**24.30** Find the intensity as compared with the reference intensity of the 1906 San Francisco earthquake if it has a Richter number of 8.25.
**24.31** Find the Richter number of an earthquake with an intensity 20 000 times greater than the reference intensity.
**24.32** Find the pH of each substance with the given concentration of hydrogen ions.
( _a_ ) beer: [H+] = 6.31 × 10–5
( _b_ ) orange juice: [H+] = 1.99 × 10–4
( _c_ ) vinegar: [H+] = 6.3 × 10–3
( _d_ ) tomato juice: [H+] = 7.94 × 10–5
**24.33** Find the approximate hydrogen ions concentration, [H+], for the substances with the given pH.
( _a_ ) apples: pH = 3.0
( _b_ ) eggs: pH = 7.8
**24.34** If gastric juices in your stomach have a hydrogen ion concentration of 1.01 × 10–1 moles per liter, what is the pH of the gastric juices?
**24.35** A relatively quiet room has a background noise level of 32 decibels. How many times the hearing threshold intensity is the intensity of a relatively quiet room?
**24.36** If the intensity of an argument is about 3 980 000 times the hearing threshold intensity, what is the decibel level of the argument?
**24.37** The population of the world compounds continuously. If in 1987 the growth rate was 1.63% annually and an initial population of 5 billion people, what will the world population be in the year 2000?
**24.38** During the Black Plague the world population declined by about 1 million from 4.7 million to about 3.7 million during the 50-year period from 1350 to 1400. If world population decline was exponential, what was the annual rate of decline?
**24.39** If the world population grew exponentially from 1.6 billion in 1900 to 5.0 billion in 1987, what was the annual rate of population growth?
**24.40** If the deforestation of El Salvador continues at the current rate for 20 more years only 53% of the present forests will be left. If the decline of the forests is exponential, what is the annual rate of deforestation for El Salvador?
**24.41** A bone is found to contain 40% of the carbon-14 that it contained when it was part of a living animal. If the decay of carbon-14 is exponential with an annual rate of decay of 0.0124%, how long ago did the animal die?
**24.42** Radioactive strontium-90 is used in nuclear reactors and decays exponentially with an annual rate of decay of 2.48%. How much of 50 grams of strontium-90 will be left after 100 years?
**24.43** How long does it take 12 grams of carbon-14 to decay to 10 grams when the decay is exponential with an annual rate of decay of 0.0124%?
**24.44** How long does it take for 10 grams of strontium-90 to decay to 8 grams if the decay is exponential and the annual rate of decay is 2.48%?
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
_Note_ : The tables in Appendices A and B were used in computing these answers. If a calculator is used your answers may vary.
**24.17** 2.7%
**24.18** 60%
**24.19** $6437
**24.20** $26 250
**24.21** $48 850
**24.22** $3820
**24.23** $1701
**24.24** $2464
**24.25** $11 410
**24.26** $11 440
**24.27** 7.68
**24.28** 6.50
**24.29** 316 200 000 _I_ 0
**24.30** 177 800 000 _I_ 0
**24.31** 4.30
**24.32**
( _a_ ) pH = 4.2
( _b_ ) pH = 3.7
( _c_ ) pH = 2.2
( _d_ ) pH = 4.1
**24.33**
( _a_ ) [H+] = 0.001 or 1.00 × 10–3
( _b_ ) [H+] = 1.585 × 108
**24.34** 1.0
**24.35** 1585 _I_ 0
**24.36** 66 decibels
**24.37** 6.18 billion
**24.38** 0.48% per year
**24.39** 1.31% per year
**24.40** 3.17% per year
**24.41** 7390 years
**24.42** 4.2 grams
**24.43** 1471 years
**24.44** 8.998 years
## **CHAPTER 25
Permutations and Combinations**
### **25.1 FUNDAMENTAL COUNTING PRINCIPLE**
If one thing can be done in _m_ different ways and, when it is done in any one of these ways, a secan thing can be done in _n_ different ways, then the two things in succession can be done in _mn_ different ways.
For example, if there are 3 candidates for governor and 5 for mayor, then the two offices may be filled in 3·5 = 15 ways.
In general, if _a_ 1 can be done in _x_ 1 ways, _a_ 2 can be done in _x_ 2 ways, _a_ 3 can be done in _x_ 3 ways,..., and _a n_ can be done in _x n_ ways, then the event _a_ 1 _a_ 2 _a_ 3... _a n_ can be done in _x_ 1 · _x_ 2 · _x_ 3 ··· _x n_ ways.
**EXAMPLE 25.1.** A man has 3 jackets, 10 shirts, and 5 pairs of slacks. If an outfit consists of a jacket, a shirt, and a pair of slacks, how many different outfits can the man make?
### **25.2 PERMUTATIONS**
A permutation is an arrangement of all or part of a number of things in a definite order.
For example, the permutations of the three letters _a, b, c_ taken all at a time are _abc, acb, bca, bac, cba, cab_.
The permutations of the three letters _a, b, c_ taken two at a time are _ab, ac, ba, bc, ca, cb_.
For a natural number _n, n_ factorial, denoted by _n_!, is the product of the first _n_ natural numbers. That is, _n_! = _n_ · ( _n_ – 1) · ( _n_ – 2)... 2·1. Also, _n_! = _n_ · ( _n_ – 1)!
Zero factorial is defined to be 1 : 0! = 1.
**EXAMPLES 25.2.** Evaluate each factorial.
( _a_ ) 7!
( _b_ ) 5!
( _c_ ) 1!
( _d_ ) 2!
( _e_ ) 4!
( _a_ ) 7! = 7·6·5·4·3·2·1 = 5040
( _b_ ) 5! = 5·4·3·2·1= 120
( _c_ ) 1! = 1
( _d_ ) 2! = 2·1 = 2
( _e_ ) 4! = 4· 3·2·1 = 24
The symbol _n Pr_ represents the number of permutations (arrangements, orders) of _n_ things taken _r_ at a time.
Thus 8 _P_ 3 denotes the number of permutations of 8 things taken 3 at a time, and 5 _P_ 5 denotes the number of permutations of 5 things taken 5 at a time.
_Note_. The symbol _P_ ( _n, r_ ) having the same meaning as _n Pr_ is sometimes used.
A. Permutations of _n_ different things taken _r_ at a time
When
#### **EXAMPLES 25.3.**
5 _P_ 1 = 5, 5 _P_ 2 = 5 ·4 = 20, 5 _P_ 3 = 5 ·4·3 = 60, 5 _P_ 4 = 5 ·4·3 ·2= 120, 5 _P_ 5 = 5! = 5 ·4·3 ·2· 1 = 120, 10 _P_ 7 = 10 · 9 · 8 · 7 · 6 · 5 ·4 = 604 800.
The number of ways in which 4 persons can take their places in a cab having 6 seats is 6 _P_ 4 = 6 · 5 ·4· 3 = 360.
B. Permutations with some things alike, taken all at a time
The number of permutations _P_ of _n_ things taken all at a time, of which _n_ 1 are alike, _n_ 2 others are alike, _n_ 3 others are alike, etc., is
For example, the number of ways 3 dimes and 7 quarters can be distributed among 10 boys, each to receive one coin, is
C. Circular permutations
The number of ways of arranging _n_ different objects around a circle is ( _n_ \- 1)! ways.
Thus 10 persons may be seated at a round table in (10 – 1)! = 9! ways.
### **25.3 COMBINATIONS**
A combination is a grouping or selection of all or part of a number of things without reference to the arrangement of the things selected.
Thus the combinations of the three letters _a, b, c_ taken 2 at a time are _ab, ac, bc_. Note that _ab_ and _ba_ are 1 combination but 2 permutations of the letters _a, b_.
The symbol _n Cr_ represents the number of combinations (selections, groups) of _n_ things taken _r_ at a time.
Thus 9 _C_ 4 denotes the number of combinations of 9 things taken 4 at a time.
_Note_. The symbol _C(n, r_ ) having the same meaning as _n Cr_ is sometimes used.
A. Combinations of _n_ different things taken _r_ at a time
For example, the number of handshakes that may be exchanged among a party of 12 students if each student shakes hands once with each other student is
The following formula is very useful in simplifying calculations:
This formula indicates that the number of selections of _r_ out of _n_ things is the same as the number of selections of _n_ – _r_ out of _n_ things.
#### **EXAMPLES 25.4.**
Note that in each case the numerator and denominator have the same number of factors.
B. Combinations of different things taken any number at a time
The total number of combinations _C_ of _n_ different things taken 1, 2, 3,..., _n_ at a time is
For example, a woman has in her pocket a quarter, a dime, a nickel, and a penny. The total number of ways she can draw a sum of money from her pocket is 24 – 1 = 15.
### **25.4 USING A CALCULATOR**
Scientific and graphing calculators have keys for factorials, _n_!; permutations, _n Pr_, and combinations, _n Cr_. As factorials get larger, the results are displayed in scientific notation. Many calculators have only two digits available for the exponent, which limits the size of the factorial that can be displayed. Thus, 69! can be displayed and 70! cannot, because 70! needs more than two digits for the exponent in scientific notation. When the calculator can perform an operation, but it cannot display the result an error message is displayed instead of an answer.
The values of _n Pr_ and _n Cr_ can often be computed on the calculator when _n_! cannot be displayed. This can be done because the internal procedure does not require the result to be displayed, just used.
### **Solved Problems**
**25.1** Evaluate 20 _P_ 2, 8 _P_ 5, 7 _P_ 5, 7 _P_ 7.
**SOLUTION**
**25.2** Find _n_ if ( _a_ ) 7 · _n P_3 = 6 · _n_ +l _P_ 3, ( _b_ ) 3· _n P_4 = _n_ –1 _P_ 5.
**SOLUTION**
( _a_ ) 7 _n_ ( _n_ – 1)( _n_ – 2) = 6( _n_ \+ 1)( _n_ )( _n_ – 1).
Since _n_ ≠ 0, 1 we may divide by _n_ ( _n_ – 1) to obtain 7( _n_ – 2) = 6( _n_ \+ 1), _n_ = 20.
( _b_ ) 3 _n_ ( _n_ – 1)( _n_ – 2)( _n_ – 3) = ( _n_ – 1)( _n_ – 2)( _n_ – 3)( _n_ – 4)( _n_ – 5).
Since _n_ ≠ 1, 2, 3 we may divide by ( _n_ – 1)( _n_ – 2)( _n_ – 3) to obtain
Thus _n_ = 10.
**25.3** A student has a choice of 5 foreign languages and 4 sciences. In how many ways can he choose 1 language and 1 science?
**SOLUTION**
He can choose a language in 5 ways, and with each of these choices there are 4 ways of choosing a science.
Hence the required number of ways = 5 · 4 = 20 ways.
**25.4** In how many ways can 2 different prizes be awarded among 10 contestants if both prizes ( _a_ ) may not be given to the same person, ( _b_ ) may be given to the same person?
**SOLUTION**
( _a_ ) The first prize can be awarded in 10 different ways and, when it is awarded, the second prize can be given in 9 ways, since both prizes may not be given to the same contestant.
Hence the required number of ways = 10 · 9 = 90 ways.
( _b_ ) The first prize can be awarded in 10 ways, and the second prize also in 10 ways, since both prizes may be given to the same contestant.
Hence the required number of ways = 10 · 10 = 100 ways.
**25.5** In how many ways can 5 letters be mailed if there are 3 mailboxes available?
**SOLUTION**
Each of the 5 letters may be mailed in any of the 3 mailboxes.
Hence the required number of ways = 3 · 3 · 3 · 3 · 3 = 35 = 243 ways.
**25.6** There are 4 candidates for president of a club, 6 for vice-president and 2 for secretary. In how many ways can these three positions be filled?
**SOLUTION**
A president may be selected in 4 ways, a vice-president in 6 ways, and a secretary in 2 ways. Hence the required number of ways = 4· 6 · 2 = 48 ways.
**25.7** In how many different orders may 5 persons be seated in a row?
**SOLUTION**
The first person may take any one of 5 seats, and after the first person is seated, the second person may take any one of the remaining 4 seats, etc. Hence the required number of orders = 5 · 4 · 3 · 2 · 1 = 120 orders.
_Otherwise_. Number of orders = number of arrangements of 5 persons taken all at a time
= 5 _P_ 5 = 5! = 5 ·4·3 ·2· 1 = 120 orders.
**25.8** In how many ways can 7 books be arranged on a shelf?
**SOLUTION**
Number of ways = number of arrangements of 7 books taken all at a time
= 7 _P_ 7 = 7! = 7 · 6 · 5 ·4 · 3 · 2 · 1 = 5040 ways.
**25.9** Twelve different pictures are available, of which 4 are to be hung in a row. **In** how many ways can this be done?
**SOLUTION**
The first place may be occupied by any one of 12 pictures, the second place by anyone of 11, the third place by any one of 10, and the fourth place by any one of 9.
Hence the required number of ways = 12 · 11 · 10 · 9 = 11 880 ways.
_Otherwise_. Number of ways = number of arrangements of 12 pictures taken 4 at a time
= 12 _P_ 4 = 12·11 · 10·9 = 11880 ways.
**25.10** It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?
**SOLUTION**
The men may be seated in _5 P5_ ways, and the women in _4 P4_ ways. Each arrangement of the men may be associated with each arrangement of the women.
Hence the required number of arrangements = _5 P5_ · _4 P4_ = 5! 4! = 120·24 = 2880.
**25.11** In how many orders can 7 different pictures be hung in a row so that 1 specified picture is ( _a_ ) at the center, ( _b_ ) at either end?
**SOLUTION**
( _a_ ) Since 1 given picture is to be at the center, 6 pictures remain to be arranged in a row. Hence the number of orders = _6 P6_ = 6! = 720 orders.
( _b_ ) After the specified picture is hung in any one of 2 ways, the remaining 6 can be arranged in _6 P6_ ways. Hence the number of orders = 2· _6 P6_ = 1440 orders.
**25.12** In how many ways can 9 different books be arranged on a shelf so that ( _a_ ) 3 of the books are always together, ( _b_ ) 3 of the books are never all 3 together?
**SOLUTION**
( _a_ ) The specified 3 books can be arranged among themselves in _3 P3_ ways. Since the specified 3 books, are always together, they may be considered as 1 thing. Then together with the other 6 books (things) we have a total of 7 things which can be arranged in _7 P7_ ways.
Total number of ways = _3 P3_ · _7 P7_ = 3!7! = 6 · 5040 = 30240 ways.
( _b_ ) Number of ways in which 9 books can be arranged on a shelf if there are no restrictions = 9! = 362 880 ways.
Number of ways in which 9 books can be arranged on a shelf when 3 specified books are always together (from ( _a_ ) above)= 3!7! = 30240 ways.
Hence the number of ways in which 9 books can be arranged on a shelf so that 3 specified books are never all 3 together = 362 880 – 30240 = 332 640 ways.
**25.13** In how many ways can _n_ women be seated in a row so that 2 particular women will not be next to each other?
**SOLUTION**
With no restrictions, _n_ women may be seated in a row in _n Pn_ ways. If 2 of the _n_ women must always sit next to each other, the number of arrangements = 2!(n–1 _P_ n–1).
Hence the number of ways _n_ women can be seated in a row if 2 particular women may never sit together= _n_ _P_ n – 2( _n_ –1 _P_ n–l) = _n_! – 2( _n_ – 1)! = _n_ ( _n_ – 1)! – 2( _n_ – 1)! = ( _n_ – 2)· ( _n_ – 1)!
**25.14** Six different biology books, 5 different chemistry books and 2 different physics books are to be arranged on a shelf so that the biology books stand together, the chemistry books stand together, and the physics books stand together. How many such arrangements are possible?
**SOLUTION**
The biology books can be arranged among themselves in 6! ways, the chemistry books in 5! ways, the physics books in 2! ways, and the three groups in 3! ways.
Required number of arrangements = 6!5!2!3! = 1 036 800.
**25.15** Determine the number of different words of 5 letters each that can be formed with the letters of the word _chromate (a_ ) if each letter is used not more than once, ( _b_ ) if each letter may be repeated in any arrangement. (These words need not have meaning.)
**SOLUTION**
( _a_ ) Number of words = arrangements of 8 different letters taken 5 at a time
_= 8P5_ = 8 · 7 · 6 · 5 ·4 = 6720 words.
( _b_ ) Number of words = 8 · 8 · 8 · 8 · 8 = 85 = 32 768 words.
**25.16** How many numbers may be formed by using 4 out of the 5 digits **1, 2, 3, 4, 5** ( _a_ ) if the digits must not be repeated in any number, ( _b_ ) if they may be repeated? If the digits must not be repeated, how many of the 4-digit numbers ( _c_ ) begin with 2, ( _d_ ) end with 25?
**SOLUTION**
( _a_ ) Numbers formed = 5 _P_ 4 = 5 ·4·3·2 = 120 numbers.
( _b_ ) Numbers formed = 5 ·5·5·5 = 54 = 625 numbers.
( _c_ ) Since the first digit of each number is specified, there remain 4 digits to be arranged in 3 places.
Numbers formed = 4 _P_ 3 = 4 · 3 · 2 = 24 numbers.
( _d_ ) Since the last two digits of every number are specified, there remain 3 digits to be arranged in 2 places.
Numbers formed = 3 _P_ 2 = 3 · 2 = 6 numbers.
**25.17** How many 4-digit numbers may be formed with the 10 digits **0, 1, 2, 3,..., 9** ( _a_ ) if each digit is used only once in each number? ( _b_ ) How many of these numbers are odd?
**SOLUTION**
( _a_ ) The first place may be filled by any one of the 10 digits except 0, i.e., by any one of 9 digits. The 9 digits remaining may be arranged in the 3 other places in _9 P3_ ways.
Numbers formed = 9 · 9 _P_ 3 = 9(9 · 8 · 7) = 4536 numbers.
( _b_ ) The last place may be filled by any one of the 5 odd digits, 1, 3, 5, 7, 9. The first place may be filled by any one of the 8 digits, i.e., by the remaining 4 odd digits and the even digits, 2, 4, 6, 8. The 8 remaining digits may be arranged in the 2 middle positions in _8 P2_ ways.
Numbers formed = 5 · 8 · 8 _P_ 2 = 5 · 8 · 8 · 7 = 2240 odd numbers.
**25.18** ( _a_ ) How many 5-digit numbers can be formed from the 10 digits 0, **1, 2, 3,**..., 9, repetitions allowed? How many of these numbers ( _b_ ) begin with 40, ( _c_ ) are even, ( _d_ ) are divisible by 5?
**SOLUTION**
( _a_ ) The first place may be filled by any one of 9 digits (any of the 10 except 0). Each of the other 4 places may be filled by any one of the 10 digits whatever.
Numbers formed = 9 · 10 · 10 · 10 · 10 = 9 · 104 = 90000 numbers.
( _b_ ) The first 2 places may be filled in 1 way, by 40. The other 3 places may be filled by any one of the 10 digits whatever.
Numbers formed = 1 · 10 · 10 · 10 = 103 = 1000 numbers.
( _c_ ) The first place may be filled in 9 ways, and the last place in 5 ways (0, 2, 4, 6, 8). Each of the other 3 places may be filled by any one of the 10 digits whatever.
Even numbers = 9 · 10 · 10 · 10 · 5 = 45 000 numbers.
( _d_ ) The first place may be filled in 9 ways, the last place in 2 ways (0, 5), and the other 3 places in 10 ways each.
Numbers divisible by 5 = 9 · 10 · 10 · 10 · 2 = 18 000 numbers.
**25.19** How many numbers between 3000 and 5000 can be formed by using the 7 digits 0, 1, 2, 3, 4, 5, 6 if each digit must not be repeated in any number?
**SOLUTION**
Since the numbers are between 3000 and 5000, they consist of 4 digits. The first place may be filled in 2 ways, i.e., by digits 3, 4. Then the remaining 6 digits may be arranged in the 3 other places in 6 _P_ 3 ways.
Numbers formed = 2· 6 _P_ 3 = 2(6 · 5·4) = 240 numbers.
**25.20** From 11 novels and 3 dictionaries, 4 novels and 1 dictionary are to be selected and arranged on a shelf so that the dictionary is always in the middle. How many such arrangements are possible?
**SOLUTION**
The dictionary may be chosen in 3 ways. The number of arrangements of 11 novels taken 4 at a time is 11 _P_ 4.
Required number of arrangements = 3· ll _P_ 4 = 3(11 · 10 · 9 · 8) = 23 760.
**25.21** How many signals can be made with _5_ different flags by raising them any number at a time?
**SOLUTION**
Signals may be made by raising the flags 1, 2, 3, 4, and 5 at a time. Hence the total number of signals is
5 _P_ 1 \+ 5 _P_ 2 \+ 5 _P_ 3 \+ 5 _P_ 4 \+ 5 _P_ 5 = 5 + 20 + 60 + 120 + 120 = 325 signals.
**25.22** Compute the sum of the 4-digit numbers which can be formed with the four digits 2, 5, 3, 8 if each digit is used only once in each arrangement.
**SOLUTION**
The number of arrangements, or numbers, is 4 _P_ 4 = 4! = 4·3·2·1 = 24.
The sum of the digits = 2 + 5 + 3 + 8 = 18, and each digit will occur 24/4 = 6 times each in the units, tens, hundreds, and thousands positions. Hence the sum of all the numbers formed is
1(6· 18) + 10(6 · 18) + 100(6 · 18) + 1000(6· 18) = 119988.
**25.23** ( _a_ ) How many arrangements can be made from the letters of the word _cooperator_ when all are taken at a time? How many of such arrangements ( _b_ ) have the three os together, ( _c_ ) begin with the two _rs_?
**SOLUTION**
( _a_ ) The word _cooperator_ consists of 10 letters: 3 _o_ s, 2 _r_ s, and 5 different letters.
Number of arrangements =
( _b_ ) Consider the 3 os as 1 letter. Then we have 8 letters of which 2 rs are alike.
Number of arrangements =
( _c_ ) The number of arrangements of the remaining 8 letters, of which 3 os are alike, = 8!/3! = 6720.
**25.24** There are 3 copies each of 4 different books. In how many different ways can they be arranged on a shelf?
**SOLUTION**
There are 3 · 4 = 12 books of which 3 are alike, 3 others alike, etc.
Number of arrangements =
**25.25** ( _a_ ) In how many ways can 5 persons be seated at a round table?
( _b_ ) In how many ways can 8 persons be seated at a round table if 2 particular persons must always sit together?
**SOLUTION**
( _a_ ) Let 1 of them be seated anywhere. Then the 4 persons remaining can be seated in 4! ways. Hence there are 4! = 24 ways of arranging 5 persons in a circle.
( _b_ ) Consider the two particular persons as one person. Since there are 2! ways of arranging 2 persons among themselves and 6! ways of arranging 7 persons in a circle, the required number of ways = 2!6! = 2 ·720 = 1440 ways.
**25.26** In how many ways can 4 men and 4 women be seated at a round table if each woman is to be between two men?
**SOLUTION**
Consider that the men are seated first. Then the men can be arranged in 3! ways, and the women in 4! ways.
Required number of circular arrangements = 3!4! = 144.
**25.27** By stringing together 9 differently colored beads, how many different bracelets can be made?
**SOLUTION**
There are 8! arrangements of the beads on the bracelet, but half of these can be obtained from the other half simply by turning the bracelet over.
Hence there are (8!) = 20 160 different bracelets.
**25.28** In each case, find _n_ : ( _a_ ) _n Cn_–2 = 10, ( _b_ ) _n C_15 = _n C_11, ( _c_ ) _n P_4 = 30· _n C_5.
**SOLUTION**
( _a_ )
( _b_ )
( _c_ )
Then
**25.29** Given _n Pr_ = 3024 and _n Cr_ = 126, find _r_.
**SOLUTION**
**25.30** How many different sets of 4 students can be chosen out of 17 qualified students to represent a school in a mathematics contest?
**SOLUTION**
Number of sets = number of combinations of 4 out of 17 students
**25.31** In how many ways can 5 styles be selected out of 8 styles?
**SOLUTION**
Number of ways = number of combinations of 5 out of 8 styles
**25.32** In how many ways can 12 books be divided between _A_ and _B_ so that one may get 9 and the other 3 books?
**SOLUTION**
In each separation of 12 books into 9 and 3, _A_ may get the 9 and _B_ the 3, or _A_ may get the 3 and _B_ the 9.
Hence the number of ways
**25.33** Determine the number of different triangles which can be formed by joining the six vertices of a hexagon, the vertices of each triangle being on the hexagon.
**SOLUTION**
Number of triangles = number of combinations of 3 out of 6 points
**25.34** How many angles less than 180° are formed by 12 straight lines which terminate in a point, if no two of them are in the same straight line?
**SOLUTION**
Number of angles = number of combinations of 2 out of 12 lines
**25.35** How many diagonals has an octagon?
**SOLUTION**
Lines formed = number of combinations of 2 out of 8 corners (points)
Since 8 of these 28 lines are the sides of the octagon, the number of diagonals = 20.
**25.36** How many parallelograms are formed by a set of 4 parallel lines intersecting another set of 7 parallel lines?
**SOLUTION**
Each combination of 2 lines out of 4 can intersect each combination of 2 lines out of 7 to form a parallelogram.
Number of parallelograms = 4 _C_ 2 · 7 _C_ 2 = 6 · 21 = 126 parallelograms.
**25.37** There are 10 points in a plane. No three of these points are in a straight line, except 4 points which are all in the same straight line. How many straight lines can be formed by joining the 10 points?
**SOLUTION**
Number of lines formed if no 3 of the 10 points were in a straight line
Number of lines formed by 4 points, no 3 of which are collinear
Since the 4 points are collinear, they form 1 line instead of 6 lines.
Required number of lines = 45 – 6 + 1 = 40 lines.
**25.38** In how many ways can 3 women be selected out of 15 women?
( _a_ ) if 1 of the women is to be included in every selection,
( _b_ ) if 2 of the women are to be excluded from every selection,
( _c_ ) if 1 is always included and 2 are always excluded?
**SOLUTION**
( _a_ ) Since 1 is always included, we must select 2 out of 14 women.
Hence the number of ways
( _b_ ) Since 2 are always excluded, we must select 3 out of 13 women.
Hence the number of ways
( _c_ ) Number of ways
**25.39** An organization has 25 members, 4 of whom are doctors. In how many ways can a committee of 3 members be selected so as to include at least 1 doctor?
**SOLUTION**
Total number of ways in which 3 can be selected out of 25 = 25C3.
Number of ways in which 3 can be selected so that no doctor is included = 25–4 _C_ 3 = 21 _C_ 3.
Then the number of ways in which 3 members can be selected so that at least 1 doctor is included is
**25.40** From 6 chemists and 5 biologists, a committee of 7 is to be chosen so as to include 4 chemists. In how many ways can this be done?
**SOLUTION**
Each selection of 4 out of 6 chemists can be associated with each selection of 3 out of 5 biologists.
Hence the number of ways = 6 _C_ 4 · 5 _C_ 3 = 6 _C_ 2 · 5 _C_ 2 = 15 · 10 = 150 ways.
**25.41** Given 8 consonants and 4 vowels, how many 5-letter words can be formed, each word consisting of 3 different consonants and 2 different vowels?
**SOLUTION**
The 3 different consonants can be selected in 8 _C_ 3 ways, the 2 different vowels in 4 _C_ 2 ways, and the 5 different letters (3 consonants, 2 vowels) can be arranged among themselves in 5 _P_ 5 = 5! ways.
Hence the number of words = 8 _C_ 3 · 4 _C_ 2 · 5! = 56 · 6 · 120 = 40320.
**25.42** From 7 capitals, 3 vowels and 5 consonants, how many words of 4 letters each can be formed if each word begins with a capital and contains at least 1 vowel, all the letters of each word being different?
**SOLUTION**
The first letter, or capital, may be selected in 7 ways.
The remaining 3 letters may be
( _a_ ) 1 vowel and 2 consonants, which may be selected in 3 _C_ 1 · 5 _C_ 2 ways,
( _b_ ) 2 vowels and 1 consonant, which may be selected in 3 _C_ 2 · 5 _C_ 1 ways, and
( _c_ ) 3 vowels, which may be selected in 3 _C_ 3 = 1 way.
Each of these selections of 3 letters may be arranged among themselves in 3 _P_ 3 = 3! ways.
Hence the number of words = 7 · 3!(3C1 · 5C2 \+ _3 C2_ · 5C1 \+ 1)
= 7 · 6(3 · 10 + 3 ·5+ 1) = 1932 words.
**25.43** _A_ has 3 maps and _B_ has 9 maps. Determine the number of ways in which they can exchange maps if each keeps his initial number of maps.
**SOLUTION**
_A_ can exchange 1 map with _B_ in 3C1 ·9C1 = 3 ·9 = 27 ways.
_A_ can exchange 2 maps with _B_ in _3 C2_ · _9 C2_ = 3 · 36 = 108 ways.
_A_ can exchange 3 maps with _B_ in _3 C3_ · _9 C3_ = 1 · 84 = 84 ways.
Total number of ways = 27 + 108 + 84 = 219 ways.
_Another method_. Consider that _A_ and _B_ put their maps together. Then the problem is to find the number of ways _A_ can select 3 maps out of 12, not including the selection by _A_ of his original three maps.
Hence,
**25.44** ( _a_ ) In how many ways can 12 books be divided among 3 students so that each receives 4 books?
( _b_ ) In how many ways can 12 books be divided into 3 groups of 4 each?
**SOLUTION**
( _a_ ) The first student can select 4 out of 12 books in 12 _C_ 4 ways.
The second student can select 4 of the remaining 8 books in 8C4 ways.
The third student can select 4 of the remaining 4 books in 1 way.
Number of ways = 12C4 · 8C4 · 1 = 495 · 70 · 1 = 34 650 ways.
( _b_ ) The 3 groups could be distributed among the students in 3! = 6 ways.
Hence the number of groups = 34650/3! = 5775 groups.
**25.45** In how many ways can a person choose 1 or more of 4 electrical appliances?
**SOLUTION**
Each appliance may be dealt with in 2 ways, as it can be chosen or not chosen. Since each of the 2 ways of dealing with an appliance is associated with 2 ways of dealing with each of the other appliances, the number of ways of dealing with the 4 appliances = 2 · 2 · 2 · 2 = 24 ways. But 24 ways includes the case in which no appliance is chosen.
Hence the required number of ways = 24 – 1 = 16 – 1 = 15 ways.
_Another method_. The appliances may be chosen singly, in twos, etc. Hence the required number of ways = 4 _C_ 1 \+ 4 _C_ 2 \+ 4 _C_ 3 \+ 4 _C_ 4 = 4 + 6 + 4 + 1 = 15 ways.
**25.46** How many different sums of money can be drawn from a wallet containing one bill each of 1, 2, 5, 10, 20 and 50 dollars?
**SOLUTION**
Number of sums = 26 – 1 = 63 sums.
**25.47** In how many ways can 2 or more ties be selected out of 8 ties?
**SOLUTION**
One or more ties may be selected in (28 – 1) ways. But since 2 or more must be chosen, the required number of ways = 28 – 1 – 8 = 247 ways.
_Another method_. 2, 3, 4, 5, 6, 7, or 8 ties may be selected in
**25.48** There are available 5 different green dyes, 4 different blue dyes, and 3 different red dyes. How many selections of dyes can be made, taking at least 1 green and 1 blue dye?
**SOLUTION**
The green dyes can be chosen in (25 – 1) ways, the blue dyes in (24 – 1) ways, and the red dyes in 23 ways.
Number of selections = (25 – 1)(24 – 1)(23) = 31 · 15 · 8 = 3720 selections.
### **Supplementary Problems**
**25.49** Evaluate: 16 _P_ 3, 7 _P_ 4, 5 _P_ 5, 12 _P_ 1.
**25.50** Find _n_ if ( _a_ ) 10 · _n_ _P_ 2 = _n_ +1 _P_ 4, ( _b_ ) 3 · 2 _n_ +4 _P_ 3 = 2 · _n_ +4 _P_ 4.
**25.51** In how many ways can six people be seated on a bench?
**25.52** With four signal flags of different colors, how many different signals can be made by displaying two flags one above the other?
**25.53** With six signal flags of different colors, how many different signals can be made by displaying three flags one above the other?
**25.54** In how many ways can a club consisting of 12 members choose a president, a secretary, and a treasurer?
**25.55** If no two books are alike, in how many ways can 2 red, 3 green, and 4 blue books be arranged on a shelf so that all the books of the same color are together?
**25.56** There are 4 hooks on a wall. In how many ways can 3 coats be hung on them, one coat on a hook?
**25.57** How many two-digit numbers can be formed with the digits 0, 3, 5, 7 if no repetition in any of the numbers is allowed?
**25.58** How many even numbers of two different digits can be formed from the digits 3, 4, 5, 6, 8?
**25.59** How many three-digit numbers can be formed from the digits 1, 2, 3, 4, 5 if no digit is repeated in any number?
**25.60** How many numbers of three digits each can be written with the digits 1, 2,..., 9 if no digit is repeated in any number?
**25.61** How many three-digit numbers can be formed from the digits 3, 4, 5, 6, 7 if digits are allowed to be repeated?
**25.62** How many odd numbers of three digits each can be formed, without the repetition of any digit in a number, from the digits ( _a_ ) 1, 2, 3, 4, ( _b_ ) 1, 2, 4, 6, 8?
**25.63** How many even numbers of four different digits each can be formed from the digits 3, 5, 6, 7, 9?
**25.64** How many different numbers of 5 digits each can be formed from the digits 2, 3, 5, 7, 9 if no digit is repeated?
**25.65** How many integers are there between 100 and 1000 in which no digit is repeated?
**25.66** How many integers greater than 300 and less than 1000 can be made with the digits 1, 2, 3, 4, 5 if no digit is repeated in any number?
**25.67** How many numbers between 100 and 1000 can be written with the digits 0, 1, 2, 3, 4 if no digit is repeated in any number?
**25.68** How many four-digit numbers greater than 2000 can be formed with the digits 1, 2, 3, 4 if repetitions ( _a_ ) are not allowed, ( _b_ ) are allowed?
**25.69** How many of the arrangements of the letters of the word _logarithm_ begin with a vowel and end with a consonant?
**25.70** In a telephone system four different letters _P_ , _R_ , _S_ , _T_ and the four digits 3, 5, 7, 8 are used. Find the maximum number of "telephone numbers" the system can have if each consists of a letter followed by a four-digit number in which the digits may be repeated.
**25.71** In how many ways can 3 girls and 3 boys be seated in a row, if no two girls and no two boys are to occupy adjacent seats?
**25.72** How many four character codes can be made by using three dots and two dashes?
**25.73** In how many ways can three dice fall?
**25.74** How many fraternities can be named with the 24 letters of the Greek alphabet if each has three letters and none is repeated in any name?
**25.75** How many signals can be shown with 8 flags of which 2 are red, 3 white and 3 blue, if they are all strung up on a vertical pole at once?
**25.76** In how many ways can 4 men and 4 women sit at a round table so that no two men are adjacent?
**25.77** How many different arrangements are possible with the factors of the term _a_ 2 _b_ 4 _c_ 5 written at full length?
**25.78** In how many ways can 9 different prizes be awarded to two students so that one receives 3 and the other 6?
**25.79** How many different radio stations can be named with 3 different letters of the alphabet? How many with 4 different letters in which _W_ must come first?
**25.80** In each case find _n_ : ( _a_ ) 4 · _n_ _C_ 2 = _n_ +2 _C_ 3, ( _b_ ) _n_ +2 _C_ _n_ = 45, ( _c_ ) _n_ _C_ 12 = _n_ _C_ 8.
**25.81** If 5 · _n P_3 = 24 · _n_ _C_ 4, find _n_.
**25.82** Evaluate ( _a_ ) 7 _C_ 7, ( _b_ ) 5 _C_ 3, ( _c_ ) 7 _C_ 2, ( _d_ ) 7 _C_ 5, ( _e_ ) 7 _C_ 6, ( _f_ ) 8 _C_ 7, ( _g_ ) 8 _C_ 5, ( _h_ ) 100 _C_ 98.
**25.83** How many straight lines are determined by ( _a_ ) 6, ( _b_ ) _n_ points, no three of which lie in the same straight line?
**25.84** How many chords are determined by seven points on a circle?
**25.85** A student is allowed to choose 5 questions out of 9. In how many ways can she choose them?
**25.86** How many different sums of money can be formed by taking two of the following: a cent, a nickel, a dime, a quarter, a half-dollar?
**25.87** How many different sums of money can be formed from the coins of Problem 25.86?
**25.88** A baseball league is made up of 6 teams. If each team is to play each of the other teams ( _a_ ) twice, ( _b_ ) three times, how many games will be played?
**25.89** How many different committees of two men and one woman can be formed from ( _a_ ) 7 men and 4 women, ( _b_ ) 5 men and 3 women?
**25.90** In how many ways can 5 colors be selected out of 8 different colors including red, blue, and green
( _a_ ) if blue and green are always to be included,
( _b_ ) if red is always excluded,
( _c_ ) if red and blue are always included but green excluded?
**25.91** From 5 physicists, 4 chemists, and 3 mathematicians a committee of 6 is to be chosen so as to include 3 physicists, 2 chemists, and 1 mathematician. In how many ways can this be done?
**25.92** In Problem 25.91, in how many ways can the committee of 6 be chosen so that
( _a_ ) 2 members of the committee are mathematicians.
( _b_ ) at least 3 members of the committee are physicists?
**25.93** How many words of 2 vowels and 3 consonants may be formed (considering any set a word) from the letters of the word ( _a_ ) _stenographic_ , ( _b_ ) _facetious_?
**25.94** In how many ways can a picture be colored if 7 different colors are available for use?
**25.95** In how many ways can 8 women form a committee if at least 3 women are to be on the committee?
**25.96** A box contains 7 red cards, 6 white cards and 4 blue cards. How many selections of three cards can be made so that ( _a_ ) all three are red, ( _b_ ) none is red?
**25.97** How many teams of nine players can be chosen from 13 candidates if _A_ , _B_ , _C_ , _D_ are the only candidates for two positions and can play no other position?
**25.98** How many different committees including 3 Democrats and 2 Republicans can be chosen from 8 Republicans and 10 Democrats?
**25.99** At a meeting, after everyone had shaken hands once with everyone else, it was found that 45 handshakes were exchanged. How many were at the meeting?
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**25.49** 3360, 840, 120, 12
**25.50** ( _a_ ) 4, ( _b_ ) 6
**25.51** 720
**25.52** 12
**25.53** 120
**25.54** 1320
**25.55** 1728
**25.56** 24
**25.57** 9
**25.58** 12
**25.59** 60
**25.60** 504
**25.61** 125
**25.62** ( _a_ ) 12, ( _b_ ) 12
**25.63** 24
**25.64** 120
**25.65** 648
**25.66** 36
**25.67** 48
**25.68** ( _a_ ) 18, ( _b_ ) 192
**25.69** 90 720
**25.70** 1024
**25.71** 72
**25.72** 10
**25.73** 216
**25.74** 12, 144
**25.75** 560
**25.76** 144
**25.77** 6930
**25.78** 168
**25.79** 15 600; 13 800
**25.80** ( _a_ ) 2, 7, ( _b_ ) 8, ( _c_ ) 20
**25.81** 8
**25.82** ( _a_ ) 1, ( _b_ ) 10, ( _c_ ) 21, ( _d_ ) 21, ( _e_ ) 7, ( _f_ ) 8, ( _g_ ) 56, ( _h_ ) 4950
**25.83** ( _a_ )
**25.84** 21
**25.85** 126
**25.86** 10
**25.87** 31
**25.88** ( _a_ ) 30, ( _b_ ) 45
**25.89** ( _a_ ) 84, ( _b_ ) 30
**25.90** ( _a_ ) 20, ( _b_ ) 21, ( _c_ ) 10
**25.91** 180
**25.92** ( _a_ ) 378, ( _b_ ) 462
**25.93** ( _a_ ) 40 320, ( _b_ ) 4800
**25.94** 127
**25.95** 219
**25.96** ( _a_ ) 35, ( _b_ ) 120
**25.97** 216
**25.98** 3360
**25.99** 10
## **CHAPTER 26
The Binomial Theorem**
### **26.1 COMBINATORIAL NOTATION**
The number of combinations of _n_ objects selected _r_ at a time, _n Cr_, can be written in the form
which is called combinatorial notation.
where _n_ and _r_ are integers and _r_ ≤ _n_.
**EXAMPLES 26.1.** Evaluate each expression.
( _a_ )
( _a_ )
( _b_ )
( _c_ )
( _d_ )
### **26.2 EXPANSION OF ( _a_ \+ _x_ ) _n_**
If _n_ is a positive integer we expand ( _a_ \+ _x_ ) _n_ as shown below:
This equation is called the binomial theorem, or binomial formula
Other forms of the binomial theorem exist and some use combinations to express the coefficients. The relationship between the coefficients and combinations are shown below.
So
and
The _r_ th term of the expansion of ( _a_ \+ _x_ ) _n_ is
The _r_ th term formula for the expansion of ( _a_ \+ _x_ ) _n_ can be expressed in terms of combinations.
### **Solved Problems**
**26.1** Evaluate each expression.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
**SOLUTION**
( _a_ )
( _b_ )
( _c_ )
( _d_ )
Expand by the binomial formula.
**26.2**
**26.3**
**26.4**
Note that in the expansion of ( _a_ \+ _x_ ) _n_ :
(1) The exponent of _a_ \+ the exponent of _x_ = _n_ (i.e., the degree of each term is _n_ ).
(2) The number of terms is _n_ \+ 1, when _n_ is a positive integer.
(3) There are _two_ middle terms when _n_ is an odd positive integer.
(4) There is only _one_ middle term when _n_ is an even positive integer.
(5) The coefficients of the terms which are equidistant from the ends are the same. It is interesting to note that these coefficients may be arranged as follows.
This array of numbers is known as _Pascal's Triangle_. The first and last number _n_ in each row are 1, while any other number in the array can be obtained by adding the two numbers to the right and left of it in the preceding row.
**26.5**
In the expansion of a binomial of the form ( _a_ – _b_ ) _n_ , where _n_ is a positive integer, the terms are alternately + and –.
**26.6**
**26.7**
**26.8**
**26.9**
**26.10**
**26.11**
**26.12**
**26.13**
In Problems 26.14-26.18, write the indicated term of each expansion, using the formula
**26.14** Sixth term of ( _x_ \+ _y_ )15.
**SOLUTION**
**26.15** Fifth term of
### **SOLUTION**
**26.16** Fourth term of ( _x_ 2 – _y_ 2)11.
**SOLUTION**
**26.17** Ninth term of
**SOLUTION**
**26.18** Eighteenth term of
**SOLUTION**
**26.19** Find the term involving _x_ 2 in the expansion of
**SOLUTION**
**26.20** Find the term independent of _x_ in the expansion of
**SOLUTION**
From ( _x_ 2)9- _r_ +1( _x_ -1) _r_ -1 = _x_ 0 we obtain 2(9 – _r_ \+ 1) – 1( _r_ – 1) = 0 or _r_ = 7.
For the 7th term: _n_ = 9, _r_ = 7, _n_ – _r_ \+ 2 = 4, _r_ – 1 = 6, _n_ – _r_ \+ 1 = 3.
**26.21** Evaluate (1.03)10 to five significant figures.
**SOLUTION**
Note that all 11 terms of the expansion of (0.03 + 1)10 would be required in order to evaluate (1.03)10.
**26.22** Evaluate (0.99)15 to four decimal places.
**SOLUTION**
**26.23** Find the sum of the coefficients in the expansion of ( _a_ ) (1 + _x_ )10, ( _b_ )(1 – _x_ )10.
**SOLUTION**
( _a_ ) If, 1, _c_ 1, _c_ 2,..., _c_ 10 are the coefficients, we have the identity
Then (1 + 1)10 = 1 + _c_ 1 \+ _c_ 2 \+... + _c_ 10 = sum of coefficients = 210 = 1024
( _b_ ) Let _x_ = 1. Then (1 – _x_ )10 = (1 – 1)10 = 0 = sum of coefficients.
### **Supplementary Problems**
**26.24** Expand by the binomial formula.
( _a_ )
( _b_ ) ( _x_ – 2)5
( _c_ ) ( _y_ \+ 3)4
( _d_ )
( _e_ )( _x_ 2 – _y_ 3)4
( _f_ ) ( _a_ – 2 _b_ )6
( _g_ )
( _h_ ) ( _y_ 1/2 \+ _y_ –1/2)6
**26.25** Write the indicated term in the expansion of each of the following.
( _a_ ) Fifth term of (a – b)7
( _b_ ) Seventh term of
( _c_ ) Middle term of
( _d_ ) Seventh term of
( _e_ ) Sixteenth term of (2 – 1/ _x_ )18
( _f_ ) Sixth term of ( _x_ 2 – 2 _y_ )11
**26.26** Find the term independent of _x_ in the expansion of
**26.27** Find the term involving _x_ 3 in the expansion of
**26.28** Evaluate (0.98)6 correct to five decimal places.
**26.29** Evaluate (1.1)10 correct to the nearest hundredth.
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**26.24** ( _a_ )
( _b_ ) _x_ 5 – 10 _x_ 4 \+ 40 _x_ 3 – 80 _x_ 2 \+ 80 _x_ – 32
( _c_ ) _y_ 4 \+ 12 _y_ 3\+ 54 _y_ 2 \+ 108 _y_ \+ 81
( _d_ )
( _e_ ) _x_ 8 – 4 _x_ 6 _y_ 3 \+ 6 _x_ 4 _y_ 6 – 4 _x_ 2 _y_ 9 \+ _y_ 12
( _f_ ) _a_ 6 – 12 _a_ 5 _b_ \+ 60 _a_ 4 _b_ 2 – 160 _a_ 3 _b_ 3 \+ 240 _a_ 2 _b_ 4 – 192 _ab_ 5 \+ 64 _b_ 6 _x_ 4 3 _x_ 3 27 _x_ 2 54 _x_ 81
( _g_ )
( _h_ ) _y_ 3 \+ 6 _y_ 2 \+ 15 _y_ \+ 20 + 15 _y_ -1 \+ 6 _y_ -2 \+ _y_ -3
**26.25**
( _a_ ) 35 _a_ 3 _b_ 4
( _b_ ) 84
( _c_ ) 70
( _d_ ) 210 _a_
( _e_ )
( _f_ ) –14 784 _x_ 12 _y_ 5
**26.26** 5
**26.27** 792 _x_ 3
**26.28** 0.885 84
**26.29** 2.59
## **CHAPTER 27
Probability**
### **27.1 SIMPLE PROBABILITY**
Suppose that an event can happen in _h_ ways and fail to happen in _f_ ways, all these _h_ \+ _f_ ways suppose equally likely. Then the probability of the occurrence of the event (called its success) is
and the probability of the non-occurrence of the event (called its failure) is
where _n_ = _h + f_
It follows that _p_ \+ _q_ = 1, _p_ = 1 - _q_ , and _q_ = 1 - _p_.
The odds in favor of the occurrence of the event are _h:f_ or _h/f;_ the odds against its happening are _f:h_ or _f/h_.
If _p_ is the probability that an event will occur, the odds in favor of its happening are _p:q_ = _p_ :(1 - _p_ ) or _p/_ (1 - _p_ ); the odds against its happening are _q:p_ = (1 - _p_ ): _p_ or (1 - _p_ )/ _p_.
### **27.2 COMPOUND PROBABILITY**
Two or more events are said to be independent if the occurrence or non-occurrence of any one of them does not affect the probabilities of occurrence of any of the others.
Thus if a coin is tossed four times and it turns up a head each time, the fifth toss may be head or tail and is not influenced by the previous tosses.
The probability that two or more independent events will happen is equal to the product of their separate probabilities.
Thus the probability of getting a head on both the fifth and sixth tosses is ( ) =
Two or more events are said to be dependent if the occurrence or non-occurrence of one of the events affects the probabilities of occurrence of any of the others.
Consider that two or more events are dependent. If _p_ 1 is the probability of a first event, _p_ 2 the probability that after the first has happened the second will occur, _p_ 3 the probability that after the first and second have happened the third will occur, etc., then the probability that all events will happen in the give order is the product _p_ 1. _p_ 2. _p_ 3....
For example, a box contains 3 white balls and 2 black balls. If a ball is drawn at random, the probability that it is black is If this ball is not replaced and a second ball is drawn, the probability that it also is black is Thus the probability that both will be black is
Two or more events are said to be mutually exclusive if the occurrence of any one of them excludes the occurrence of the others.
The probability of occurrence of some one of two or more mutually exclusive events is the _sum_ of the probabilities of the individual events.
**EXAMPLE 27.1.** If a die is thrown, what is the probability of getting a 5 or a 6? Getting a 5 and getting a 6 are mutually exclusive so
Two events are said to be overlapping if the events have at least one outcome in common, hence they can happen at the same time.
The probability of occurrence of some one of two overlapping events is the sum of the probabilities of the two individual events minus the probability of their common outcomes.
**EXAMPLE 27.2.** If a die is thrown, what is the probability of getting a number less than 4 or an even number?
The numbers less than 4 on a die are 1, 2, and 3. The even numbers on a die are 2, 4, and 6. Since these two events have a common outcome, 2, they are overlapping events.
_P_ (less than 4 or even) = _P_ (less than 4) + _P_ (even) – _P_ (less than 4 and even)
### 27.3 MATHEMATICAL EXPECTATION
If _p_ is the probability that a person will receive a sum of money _m_ , the value of his expectation is _p_ · _m_.
Thus if the probability of your winning a $10 prize is 1/5, your expectation is ($10) = $2.
### 27.4 BINOMIAL PROBABILITY
If _p_ is the probability that an event will happen in any single trial and _q_ = 1 - _p_ is the probability that it will fail to happen in any single trial, then the probability of its happening exactly _r_ times in _n_ trials is _n Crprqn-r_. (See Problems 27.22 and 27.23.)
The probability that an event will happen at least _r_ times in _n_ trials is
This expression is the sum of the first _n_ \- _r_ \+ 1 terms of the binomial expansion of ( _p_ \+ _q_ ) _n_. (See Problems 27.24-27.26.)
### 27.5 CONDITIONAL PROBABILITY
The probability that a second event will occur given that the first event has occurred is called conditional probability. To find the probability that the second event will occur given that the first event occurred, divide the probability that both events occurred by the probability of the first event. The probability of event _B_ given that event _A_ has occurred is denoted by _P_ ( _B_ | _A_ ).
**EXAMPLE 27.3.** A box contains black chips and red chips. A person draws two chips without replacement. If the probability of selecting a black chip and a red chip is 15/56 and the probability of drawing a black chip on the first draw is 3/4, what is the probability of drawing a red chip on the second draw, if you know the first chip drawn was black?
If _B_ is the event drawing a black chip and _R_ is the event drawing a red chip, then _P_ ( _R|B_ ) is the probability of drawing a red chip on the second draw given that a black chip was drawn on the first draw.
Thus, the probability of drawing a red chip on the second draw given that a black chip was drawn on the first draw is 5/14.
### **Solved Problems**
**27.1** One ball is drawn at random from a box containing 3 red balls, 2 white balls, and 4 blue balls. Determine the probability _p_ that it is ( _a_ ) red, ( _b_ ) not red, ( _c_ ) white, _(d)_ red or blue.
**SOLUTION**
( _a_ )
( _b_ )
( _c_ )
( _d_ )
**27.2** One bag contains 4 white balls and 2 black balls; another bag contains 3 white balls and 5 black balls. If one ball is drawn from each bag, determine the probability _p_ that ( _a_ ) both are white, ( _b_ ) both are black, ( _c_ ) 1 is white and 1 is black.
**SOLUTION**
( _a_ )
( _b_ )
( _c_ ) Probability that first ball is white and second black
Probability that first ball is black and second white
These are mutually exclusive; hence the required probability
_Another method_.
**27.3** Determine the probability of throwing a total of 8 in a single throw with two dice, each of whose faces is numbered from 1 to 6.
**SOLUTION**
Each of the faces of one die can be associated with any of the 6 faces of the other die; thus the total number of possible cases = 6 · 6 = 36 cases.
There are 5 ways of throwing an 8: 2, 6; 3, 5; 4, 4; 5, 3; 6, 2.
**27.4** What is the probability of getting at least 1 one in 2 throws of a die?
**SOLUTION**
The probability of not getting a one in any single throw = 1 - 1/6 = 5/6.
The probability of not getting a one in 2 throws = (5/6)(5/6) = 25/36.
Hence the probability of getting at least 1 one in 2 throws = 1 - 25/36 = 11/36.
**27.5** The probability of _A_ 's winning a game of chess against _B_ is 1/3. What is the probability that _A_ will win at least 1 of a total of 3 games?
**SOLUTION**
The probability of _A's_ losing any single game = 1 - 1/3 = 2/3, and the probability of _A_ losing all 3 games = (2/3)3 = 8/27.
Hence the probability of _A_ winning at least 1 game = 1 - 8/27 = 19/27.
**27.6** Three cards are drawn from a pack of 52, each card being replaced before the next one is drawn. Compute the probability _p_ that all are ( _a_ ) spades, ( _b_ ) aces, ( _c_ ) red cards.
**SOLUTION**
A pack of 52 cards includes 13 spades, 4 aces, and 26 red cards.
( _a_ )
( _b_ )
( _c_ )
**27.7** The odds are 23 to 2 against a person winning a $500 prize. What is her mathematical expectation?
**SOLUTION**
Expectation = probability of winning x sum of money
**27.8** Nine tickets, numbered from 1 to 9, are in a box. If 2 tickets are drawn at random, determine the probability _p_ that ( _a_ ) both are odd, ( _b_ ) both are even, ( _c_ ) one is odd and one is even, _(d)_ they are numbered 2, 5.
**SOLUTION**
There are 5 odd and 4 even number tickets.
( _a_ )
( _b_ )
**27.9** A bag contains 6 red, 4 white, and 8 blue balls. If 3 balls are drawn at random, determine the probability _p_ that ( _a_ ) all 3 are red, ( _b_ ) all 3 are blue, ( _c_ ) 2 are white and 1 is red, _(d)_ at least 1 is red, _(e)_ 1 of each color is drawn, ( _f_ ) the balls are drawn in the order red, white, blue.
**SOLUTION**
( _a_ )
( _b_ )
( _c_ )
( _d_ ) Probability that none is red =
Hence the probability that at least 1 is red =
( _e_ )
( _f_ )
**27.10** Three cards are drawn from a pack of 52 cards. Determine the probability _p_ that ( _a_ ) all are aces, ( _b_ ) all are aces and drawn in the order spade, club, diamond, ( _c_ ) all are spades, _(d)_ all are of the same suit, _(e)_ no two are of the same suit.
**SOLUTION**
( _a_ ) There are 52 _C_ 3 selections of 3 out of 52 cards, and 4 _C_ 3 selections of 3 out of 4 aces.
Hence
( _b_ ) There are 52 _P_ 3 orders of drawing 3 out of 52 cards, one of which is the given order.
Hence
( _c_ ) There are 13 _C_ 3 selections of 3 out of 13 spades.
Hence
_(d)_ There are 4 suits, each consisting of 13 cards. Hence there are 4 ways of selecting a suit, and 13 _C_ 3 ways of selecting 3 cards from a given suit.
Hence
_(e)_ There are 4 _C_ 3 = 4 _C_ 1 = 4 ways of selecting 3 out of 4 suits, and 13 · 13 · 13 ways of selecting 1 card from each of 3 given suits.
Hence
**27.11** What is the probability that any two different cards of a well-shuffled deck of 52 cards will be together in the deck, if their suit is not considered?
**SOLUTION**
Consider the probability that, for example, an ace and a king are together. There are 4 aces and 4 kings in a deck. Hence an ace can be chosen in 4 ways, and when that is done a king can be chosen in 4 ways. Thus an ace and then a king can be selected in 4 · 4 = 16 ways. Similarly, a king and then an ace can be selected in 16 ways. Then an ace and a king can be together in 2 · 16 = 32 ways.
For everyone way the combination (ace, king) occurs, the remaining 50 cards and the (ace, king) combination can be permuted in 51! ways. The number of favorable arrangements is thus 32(51!). Since the total number of arrangements of all the cards in the deck is 52!, the required probability is
**27.12** A man holds 2 of a total of 20 tickets in a lottery. If there are 2 winning tickets, determine the probability that he has ( _a_ ) both, ( _b_ ) neither, ( _c_ ) exactly one.
**SOLUTION**
( _a_ ) There are 20 _C_ 2 ways of selecting 2 out of 20 tickets.
Hence the probability of his winning both prizes
_Another method_. The probability of winning the first prize = 2/20 = 1/10. After winning the first prize (he has 1 ticket left and there remain 19 tickets from which to choose the second prize) the probability of winning the second prize is 1/19.
Hence the probability of winning both prizes
( _b_ ) There are 20 tickets, 18 of which are losers.
Hence the probability of winning neither prize
_Another method_. The probability of not winning the first prize = 1 - 2/20 = 9/10. If he does not win the first prize (he still has 2 tickets), the probability of not winning the second prize = 1 - 2/19 = 17/19.
Hence the probability of winning neither prize
( _c_ ) Probability of winning exactly one prize
= 1 – probability of winning neither – probability of winning both
_Another method_.
Probability of winning first but not second prize
Probability of not winning first but winning second
Hence the probability of winning exactly 1 prize
**27.13** A box contains 7 tickets, numbered from 1 to 7 inclusive. If 3 tickets are drawn from the box, one at a time, determine the probability that they are alternately either odd, even, odd or even, odd, even.
**SOLUTION**
The probability that the first drawn is odd (4/7), then the second even (3/6) and then the third odd _(3/5)_ is
The probability that the first drawn is even (3/7), then the second odd (4/6), and then the third even (2/5) is
Hence the required probability
_Another method_. Possible orders of 7 numbers taken 3 at a time = 7 _P_ 3 = 7 · 6 · 5 = 210.
Orders where numbers are alternately odd, even, odd = 4 · 3 · 3 = 36.
Orders where numbers are alternately even, odd, even = 3 · 4 · 2 = 24.
Hence the required probability
**27.14** The probability that _A_ can solve a given problem is 4/5, that _B_ can solve it is 2/3, and that C can solve it is 3/7. If all three try, compute the probability that the problem will be solved.
**SOLUTION**
The probability that _A_ will fail to solve it = 1 - 4/5 = 1/5, that _B_ will fail = 1 - 2/3 = 1/3, and that C will fail = 1 - 3/7 = 4/7.
The probability that all three fail
Hence the probability that all three will not fail, i.e., that at least one will solve it, is
**27.15** The probability that a certain man will be alive 25 years hence is 3/7, and the probability that his wife will be alive 25 years hence is 4/5. Determine the probability that, 25 years hence, ( _a_ ) both will be alive, ( _b_ ) at least one of them will be alive, ( _c_ ) only the man will be alive.
**SOLUTION**
( _a_ ) The probability that both will be alive
( _b_ ) The probability that both will die within 25 years
Hence the probability that at least one will be alive
( _c_ ) The probability that the man will be alive = 3/7, and the probability that his wife will not be alive = 1 - 4/5 = 1/5.
Hence the probability that only the man will be alive
**27.16** There are three candidates, _A, B_ , and C, for an office. The odds that _A_ will win are 7 to 5, and the odds that _B_ will win are 1 to 3. ( _a_ ) What is the probability that either _A_ or _B_ will win? ( _b_ ) What are the odds in favor of C?
**SOLUTION**
( _a_ ) Probability that _A_ will win:
that _B_ will win:
Hence the probability that either _A_ or _B_ will win
( _b_ ) Probability that _C_ will win:
Hence the odds in favor of _C_ are 1 to 5.
**27.17** One purse contains 5 dimes and 2 quarters, and a second purse contains 1 dime and 3 quarters. If a coin is taken from one of the two purses at random, what is the probability that it is a quarter?
**SOLUTION**
The probability of selecting the first purse (1/2) and of then drawing a quarter from it (2/7) is (1/2) (2/7) = 1/7.
The probability of selecting the second purse (1/2) and of then drawing a quarter from it (3/4) is (1/2) (3/4) = 3/8.
Hence the required probability
**27.18** A bag contains 2 white balls and 3 black balls. Four persons, _A, B, C, D_ , in the order named, each draw one ball and do not replace it. The first to draw a white ball receives $10. Determine their expectations.
**SOLUTION**
_A's_ probability of winning = , and the expectation =
To find _B's_ expectation: The probability that _A_ fails = 1 – 2/5 = 3/5. If _A_ fails, the bag contains 2 white and 2 black balls. Thus the probability that if _A_ fails _B_ will win = 2/4 = 1/2. Hence _B's_ probability of winning = (3/5)(1/2) = 3/10, and the expectation is $3.
To find _C_ 's expectation: The probability that _A_ fails = 3/5, and the probability that _B_ fails = 1 – 1/2 = 1/2. If _A_ and _B_ both fail, the bag contains 2 white balls and 1 black ball. Thus the probability that, if _A_ and _B_ both fail, _C_ will win = 2/3. Hence _C_ 's probability of winning and the expectation
If _A, B_ , and C fail, only white balls remain and D must win. Hence D's probability of winning and the expectation
**27.19** Eleven books, consisting of 5 engineering books, 4 mathematics books, and 2 chemistry books, are placed on a shelf at random. What is the probability _p_ that the books of each kind are all together?
**SOLUTION**
When the books of each kind are all together, the engineering books could be arranged in 5! ways, the mathematics books in 4! ways, the chemistry books in 2! ways, and the 3 groups in 3! ways.
**27.20** Five red blocks and 4 white blocks are placed at random in a row. What is the probability _p_ that the extreme blocks are both red?
**SOLUTION**
Total possible arrangements of 5 red and 4 white blocks
Arrangements where extreme blocks are both red
Hence the required probability
**27.21** One purse contains 6 copper coins and 1 silver coin; a second purse contains 4 copper coins. Five coins are drawn from the first purse and put into the second, and then 2 coins are drawn from the second and put into the first. Determine the probability that the silver coin is in ( _a_ ) the second purse, ( _b_ ) the first purse.
**SOLUTION**
Initially, the first purse contains 7 coins. When 5 coins are drawn from the first purse and put into the second, the probability that the silver coin is put into the second purse is 5/7, and the probability that it remains in the first purse is 2/7.
The second purse now contains 5 + 4 = 9 coins. Finally, after 2 of these 9 coins are put into the first purse, the probability that the silver coin is in the second purse and the probability that it is in the first purse
**27.22** Compute the probability that a single throw with 9 dice will result in exactly 2 ones.
**SOLUTION**
The probability that a certain pair of the 9 dice thrown will yield ones The probability that the other 7 dice will not yield ones Since _9 C2_ different pairs may be selected from the 9 dice, the probability that exactly 1 pair will be aces is
_Or, by formula:_ Probability
**27.23** What is the probability of getting a 9 exactly once in 3 throws with a pair of dice?
**SOLUTION**
A 9 can occur in 4 ways: 3, 6; 4, 5; 5, 4; 6, 3.
In any throw with a pair of dice, the probability of getting a 9 = 4/(6·6) and the probability of not getting The probability that any given throw with a pair of dice is a 9 and that the other two throws are not Since there are 3Cl = 3 different ways in which one throw is a 9 and the other two throws are not, the probability of throwing a 9 exactly once in 3 throws
_Or, by formula:_ Probability
**27.24** If the probability that the average freshman will not complete four years of college is 1/3, what is the probability _p_ that of 4 freshmen at least 3 will complete four years of college?
**SOLUTION**
Probability that 3 will complete and 1 will not
Probability that 4 will complete
_Or, by formula: p_ = first 2 _(n_ – _r_ \+ 1 = 4 – 3 + 1) terms of the expansion of
**27.25** A coin is tossed 6 times. What is the probability _p_ of getting at least 3 heads? What are the odds in favor of getting at least 3 heads?
**SOLUTION**
On each toss, probability of a head = probability of a tail = 1/2.
The probability that certain 3 of the 6 tosses will give heads = (1/2)3. The probability that none of the other 3 tosses will be heads = (1/2)3. Since 6 _C_ 3, different selections of 3 can be made from the 6 tosses, the probability that exactly 3 will be heads is
Similarly, the probability of exactly 4 heads = 6 _C_ 4(1/2)6 = 6 _C_ 2(1/2)6,
the probability of exactly 5 heads = 6 _C_ 5(1/2)6 = 6 _C_ (1/2)6,
the probability of exactly 6 heads = (1/2)6.
Hence
The odds in favor of getting at least 3 heads is 21: 11 or 21/11.
_Or, by formula: p_ = first 4 _(n_ – _r_ \+ 1 = 6 – 3 + 1) terms of the expansion of
**27.26** Determine the probability _p_ that in a family of 5 children there will be at least 2 boys and 1 girl. Assume that the probability of a male birth is 1/2.
**SOLUTION**
The three favorable cases are: 2 boys, 3 girls; 3 boys, 2 girls; 4 boys, 1 girl.
**27.27** The probability that a student takes chemistry and is on the honor rolls is 0.042. The probability that a student is on the honor roll is 0.21. What is the probability that the student is taking chemistry, given that the student is on the honor roll?
**SOLUTION**
**27.28** At the Pine Valley Country Club, 32% of the members play golf and are female. Also, 80% of the members play golf. If a member of the club is selected at random, find the probability that the member is female given that the member plays golf.
**SOLUTION**
### **Supplementary Problems**
**27.29** Determine the probability that a digit chosen at random from the digits 1, 2, 3,..., 9 will be ( _a_ ) odd, ( _b_ ) even, ( _c_ ) a multiple of 3.
**27.30** A coin is tossed three times. If H = head and T = tail, what is the probability of the tosses coming up in the order ( _a_ ) HTH, ( _b_ ) THH, ( _c_ ) HHH?
**27.31** If three coins are tossed, what is the probability of obtaining ( _a_ ) three heads, ( _b_ ) two heads and a tail?
**27.32** Find the probability of throwing a total of 7 in a single throw with two dice.
**27.33** What is the probability of throwing a total of 8 or 11 in a single throw with two dice?
**27.34** A die is thrown twice. What is the probability of getting a 4 or 5 on the first throw and a 2 or 3 on the second throw? What is the probability of not getting a one on either throw?
**27.35** What is the probability that a coin will turn up heads at least once in six tosses of a coin?
**27.36** Five discs in a bag are numbered 1, 2, 3, 4, 5. What is the probability that the sum of the numbers on three discs chosen at random is greater than 10?
**27.37** Three balls are drawn at random from a box containing 5 red, 8 black, and 4 white balls. Determine the probability that ( _a_ ) all three are white, ( _b_ ) two are black and one red, ( _c_ ) one of each color is selected.
**27.38** Four cards are drawn from a pack of 52 cards. Find the probability that ( _a_ ) all are kings, ( _b_ ) two are kings and two are aces, ( _c_ ) all are of the same suit, _(d)_ all are clubs.
**27.39** A woman will win $3.20 if in 5 tosses of a coin she gets either of the sequences HTHTH or THTHT where H = head and T = tail. Determine her expectation.
**27.40** In a plane crash it was reported that three persons out of the total of twenty passengers were injured. Three newspapermen were in this plane. What is the probability that the three reported injured were the newspapermen?
**27.41** A committee of three is to be chosen from a group consisting of 5 men and 4 women. If the selection is made at random, find the probability that ( _a_ ) all three are women, ( _b_ ) two are men.
**27.42** Six persons seat themselves at a round table. What is the probability that two given persons are adjacent?
**27.43** _A_ and _B_ alternately toss a coin. The first one to turn up a head wins. If no more than five tosses each are allowed for a single game, find the probability that the person who tosses first will win the game. What are the odds against _A's_ losing if she goes first?
**27.44** Six red blocks and 4 white blocks are placed at random in a row. Find the probability that the two blocks in the middle are of the same color.
**27.45** In 8 tosses of a coin determine the probability of ( _a_ ) exactly 4 heads, ( _b_ ) at least 2 tails, ( _c_ ) at most 5 heads, _(d)_ exactly 3 tails.
**27.46** In 2 throws with a pair of dice determine the probability of getting ( _a_ ) an 11 exactly once, ( _b_ ) a 10 twice.
**27.47** What is the probability of getting at least one 11 in 3 throws with a pair of dice?
**27.48** In ten tosses of a coin, what is the probability of getting not less than 3 heads and not more than 6 heads?
**27.49** The probability that an automobile will be stolen and found within one week is 0.0006. The probability that an automobile will be stolen is 0.0015. What is the probability that a stolen automobile will be found in one week?
**27.50** In the Pizza Palace, 95% of the customers order pizza. If 65% of the customers who order pizza also order breadsticks, find the probability that a customer who orders a pizza will also order breadsticks.
**27.51** In a large shopping mall, a marketing agency conducted a survey of 100 people about a ban on smoking in the mall. Of the 60 non-smokers surveyed, 48 preferred a smoking ban. Of the 40 smokers surveyed, 32 preferred a smoking ban. What is the probability that a person selected at random from the group surveyed prefers a smoking ban given that the person is a non-smoker?
**27.52** In a new subdivision, 35% of the houses have a family room and a fireplace, while 70% have family rooms. What is the probability that a house selected at random in this subdivision has a fireplace given that it has a family room?
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**27.29** ( _a_ ) 5/9 ( _b_ ) 4/9 ( _c_ ) 1/3
**27.30** ( _a_ ) 1/8 ( _b_ ) 1/8 ( _c_ ) 1/8
**27.31** ( _a_ ) 1/8 ( _b_ ) 3/8
**27.32** 1/6
**27.33** 7/36
**27.34** 1/9, 25/36
**27.35** 63/64
**27.36** 1/5
**27.37** (a)
(b)
(c)
**27.38** ( _a_ )
( _b_ )
( _c_ ) _(d)_
**27.39** 20 cents
**27.40** 1/1140
**27.41** ( _a_ ) 1/21 ( _b_ ) 10/21
**27.42** 2/5
**27.43**
**27.44**
**27.45** ( _a_ ) ( _b_ ) ( _c_ ) _(d)_
**27.46** ( _a_ ) ( _b_ )
**27.47**
**27.48**
**27.49** 0.4
**27.50**
**27.51** 0.8
**27.52** 0.5
## **CHAPTER 28
Determinants**
### **28.1 DETERMINANTS OF SECOND ORDER**
The symbol
consisting of the four numbers _a_ 1, _b_ 1, _a_ 2, _b_ 2 arranged in two rows and two columns, is called a _determinant of second order_ or _determinant of order two_. The four numbers are called _elements_ of the determinant.
By definition,
Thus
Here the elements 2 and 3 are in the first row, the elements –1 and –2 are in the second row. Elements 2 and – 1 are in the first column, and elements 3 and – 2 are in the second column. A determinant is a number. A determinant of order one is the number itself.
### **28.2 CRAMER'S RULE**
Systems of two linear equations in two unknowns may be solved by use of second-order determinants. Given the system of equations
it is possible by any of the methods of Chapter 15 to obtain
These values for _x_ and _y_ may be written in terms of second-order determinants as follows:
The form involving determinants is easy to remember if one keeps in mind the following:
( _a_ ) The denominators in (2) are given by the determinant
in which the elements are the coefficients of _x_ and _y_ arranged as in the given equations (1). This determinant, usually denoted by D, is called the _determinant of the coefficients_.
( _b_ ) The numerator in the solution for either unknown is the same as the determinant of the coefficients D with the exception that the column of coefficients of the unknown to be determined is replaced by the column of constants on the right side of equations (1). When the column of coefficients for the variable _x_ in determinant D is replaced with the column of constants, we call the new determinant _D x_. When the column of _y_ coefficients in determinant D is replaced with the column of constants, we call the new determinant _D y_.
**EXAMPLE 28.1.** Solve the system
The denominator for both _x_ and _y_ is
Thus, the solution of the system is (1, 2).
The method of solution of linear equations by determinants is called _Cramer's Rule_. If the determinant _D_ = 0, then Cramer's Rule can not be used to solve the system.
### **28.3 DETERMINANTS OF THIRD ORDER**
The symbol
consisting of nine numbers arranged in three rows and three columns is called a _determinant of third order_. By definition, the value of this determinant is given by
and is called the expansion of the determinant.
In order to remember this definition, the following scheme is given. Rewrite the first two columns on the right of the determinant as follows:
( _a_ ) Form the products of the elements in each of the 3 diagonals shown which run down from left to right, and precede each of these 3 terms by a positive sign.
( _b_ ) Form the products of the elements in each of the 3 diagonals shown which run down from right to left, and precede each of these 3 terms by a negative sign.
( _c_ ) The algebraic sum of the six products of (1) and (2) is the required expansion of the determinant.
**EXAMPLE 28.2.**
Evaluate
Rewriting,
The value of the determinant is
Cramer's Rule for linear equations in 3 unknowns is a method of solving the following equations for _x, y, z_
by determinants. It is an extension of Cramer's Rule for linear equations in two unknowns. If we solve equations (3) by the methods of Chapter 12, we obtain
These may be written in terms of determinants as follows
_D_ is the determinant of the coefficients of _x, y, z_ in equations (3) and is assumed not equal to zero. If D is zero, Cramer's Rule cannot be used to solve the system of equations.
The form involving determinants is easy to remember if one keeps in mind the following:
( _a_ ) The denominators in (4) are given by the determinant D in which the elements are the coefficients of _x, y_ , and _z_ arranged as in the given equations (3).
( _b_ ) The numerator in the solution for any unknown is the same as the determinant of the coefficients D with the exception that the column of coefficients of the unknown to be determined is replaced by the column of constants on the right side of equations (3).
( _c_ ) The solution of the system is ( _x, y, z_ ) where
**EXAMPLE 28.3.** Solve the system
The solution of the system is (1, –1, 2).
### **28.4 DETERMINANTS OF ORDER _n_**
An _n_ th order determinant is written
In this notation each element is characterized by two subscripts, the first indicating the _row_ in which the element appears, the second indicating the _column_ in which the element appears. Thus _a_ 23 is the element in the 2nd row and 3rd column whereas _a_ 32 is the element in the 3rd row and 2nd column.
The _principal diagonal_ of a determinant consists of the elements in the determinant which lie in a straight line from the upper left-hand corner to the lower right-hand corner.
### **28.5 PROPERTIES OF DETERMINANTS**
I. Interchanging corresponding rows and columns of a determinant does not change the value of the determinant. Thus any theorem proved true for rows holds for columns, and conversely.
**EXAMPLE 28.4.**
II. If each element in a row (or column) is zero, the value of the determinant is zero.
**EXAMPLE 28.5.**
III. Interchanging any two rows (or columns) reverses the sign of the determinant.
**EXAMPLE 28.6.**
IV. If two rows (or columns) of a determinant are identical, the value of the determinant is zero.
**EXAMPLE 28.7.**
V. If each of the elements in a row (or column) of a determinant is multiplied by the same number _p_ , the value of the determinant is multiplied by _p_.
**EXAMPLE 28.8.**
VI. If each element of a row (or column) of a determinant is expressed as the sum of two (or more) terms, the determinant can be expressed as the sum of two (or more) determinants.
**EXAMPLE 28.9.**
VII. If to each element of a row (or column) of a determinant is added _m_ times the corresponding element of any other row (or column), the value of the determinant is not changed.
**EXAMPLE 28.10.**
These properties may be proved for the special cases of second and third order determinants by using the methods of expansion of Sections 28.2 and 28.3. For proofs of the general cases see the solved problems below.
### **28.6 MINORS**
The minor of an element in a determinant of order _n_ is the determinant of order _n_ – 1 obtained by removing the row and the column which contain the given element.
For example, the minor of _a 32_ in the 4th order determinant
is obtained by crossing out the row and column containing _a32_ as shown, and writing the resulting determinant of order 3, namely
The minor of an element is denoted by capital letters. Thus the minor corresponding to the element _a 32_ is denoted by _A 32_.
### **28.7 VALUE OF A DETERMINANT OF ORDER _n_**
The value of a determinant may be obtained in terms of minors as follows:
(1) Choose any row (or column).
(2) Multiply each element in the row (or column) by its corresponding minor preceded by (–1) _i_ \+ _j_ where _i_ \+ _j_ is the sum of the row number _i_ and column number _j_. The minor of an element with the attached sign is called the _cofactor_ of the element.
(3) Add algebraically the products obtained in (2).
For example, let us expand the determinant
by the elements in the third row. The minors of _a_ 31, _a_ 32, _a_ 33, _a_ 34 are _A_ 31, _A_ 32, _A_ 33, _A_ 34, respectively. The sign corresponding to the element _a_ 31 is + since (–1)3 + 1 = (–1)4 = +1. Similarly, the signs associated with the elements _a_ 32, _a_ 33, _a_ 34 are –, +, – respectively. Thus the value of the determinant is
Property VII is useful in producing zeros in a given row or column. This property coupled with the expansion in terms of minors makes for easy determination of the value of a determinant.
### **28.8 CRAMER'S RULE FOR DETERMINANTS OF ORDER _n_**
Cramer's Rule for the solution of _n_ simultaneous linear equations in _n_ unknowns is exactly analogous to the rule given in Section 28.2 for the case _n_ = 2 and in Section 28.3 for _n_ = 3.
Given _n_ linear equations in _n_ unknowns _x 1, x2, x3,..., xn_
Let _D_ be the determinant of the coefficients of _x_ 1, _x_ 2, _x_ 3,..., _x n_, i.e.,
Denote by _D k_ the determinant _D_ with the _k_ th column (which corresponds to the coefficients of the unknown _x k_) replaced by the column of the coefficients on the right-hand side of (1). Then
If _D_ ≠ 0 there is one and only one solution.
If _D_ = 0 the system of equations may or may not have solutions.
Equations having no simultaneous solution are called _inconsistent_ , otherwise they are consistent.
If _D_ = 0 and at least one of the determinants _D_ 1, _D_ 2,..., _D n_ ≠ 0, the given system is inconsistent. If _D_ = _D_ 1 = _D_ 2 =... = _D n_ = 0, the system may or may not be consistent.
Equations having an infinite number of simultaneous solutions are called _dependent_. If a system of equations is dependent then _D_ 1 = 0 and all of the determinants _D_ 1, _D_ 2,..., _D n_ = 0. The converse, however, is not always true.
### **28.9 HOMOGENEOUS LINEAR EQUATIONS**
If _r_ 1, _r_ 2,..., _r n_ in equations (1) are all zero, the system is said to be _homogeneous_. In this case _D_ 1 = _D_ 2 = _D_ 3 =... = _D_ n = 0 and the following theorem is true.
**Theorem:** A necessary and sufficient condition that _n_ homogeneous linear equations in _n_ unknowns have solutions other than the trivial solution (where all the unknowns equal zero) is that the determinant of the coefficients, _D_ = 0.
A system of _m_ equations in _n_ unknowns may or may not have simultaneous solutions.
(1) If _m_ > _n_ , the unknowns in _n_ of the given equations may be obtained. If these values satisfy the remaining _m_ – _n_ equations the system is consistent, otherwise it is inconsistent.
(2) If _m_ < _n_ , then _m_ of the unknowns may be determined in terms of the remaining _n_ – _m_ unknowns.
### **Solved Problems**
**28.1** Evaluate the following determinants.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
**28.2** ( _a_ ) Show that if the rows and columns of a determinant of order two are interchanged the value of the determinant is the same.
( _b_ ) Show that if the elements of one row (or column) are proportional respectively to the elements of the other row (or column), the determinant is equal to zero.
**SOLUTION**
( _a_ ) Let the determinant be
The determinant with rows and columns interchanged so that 1st row becomes 1st column and 2nd row becomes 2nd column is
( _b_ ) The determinant with proportional rows is
**28.3** Find the values of _x_ for which
**SOLUTION**
Then _2x 2_ – x – 1 = ( _x_ – 1)(2 _x_ \+ 1) = 0 so that _x_ = 1, –1/2.
**28.4** Solve for the unknowns in each of the following systems.
( _a_ )
**SOLUTION**
The solution of the system is (1, 1/2).
( _b_ )
**SOLUTION**
The solution of the system is (–6, 18).
( _c_ )
**SOLUTION**
Rewrite as
The solution of the system is (36/19, –43/19).
**28.5** Solve for _x_ and _y_.
The solution of the system is (9, 7).
The solution of the system is (5, 3).
**28.6** Solve the following systems of equations.
The solution of the system is (6, 18).
The solution of the system is (3/2, 4/5).
**28.7** Evaluate each of the following determinants.
( _a_ )
Repeat the first two columns:
( _b_ )
( _c_ )
( _d_ )
( _e_ )
**28.8**
( _a_ ) Show that if two rows (or two columns) of a third-order determinant have their corresponding elements proportional, the value of the determinant is zero.
( _b_ ) Show that if the elements of any row (or column) are multiplied by any given constant and added to the corresponding elements of any other row (or column), the value of the determinant is unchanged.
**SOLUTION**
( _a_ ) We must show that
where the elements in the first and second rows are proportional. This is shown by expansion of the determinant.
( _b_ ) Let the given determinant be
We must show that if _k_ is any constant
where we have multiplied each of the elements in the second row of the given determinant by _k_ and added to the corresponding elements in the third row. The result is proved by expanding each of the determinants and showing that they are equal.
**28.9** Solve the following systems of equations.
( _a_ )
Here
The solution of the system is (1, 2, –1).
( _b_ )
Write as
Then
The solution of the system is (1, 2, 3).
**28.10** The equations for the currents _i_ 1, _i_ 2, _i_ 3 in a given electrical network are
Find _i_ 3.
**SOLUTION**
**28.11** Write the expansion of the determinant by using minors for row one.
**SOLUTION**
The expansion is
The required expansion is
**28.12** Prove Property III: If two rows (or columns) are interchanged, the sign of the determinant is changed.
**SOLUTION**
For the case of third-order determinants, we must show that
The expansion of each side of the equation yields the same expression. Thus the property holds for third-order determinants. The methods of proof holds in the general case.
**28.13** Prove Property IV: If two rows (or columns) are identical, the determinant has value zero.
**SOLUTION**
Let _D_ be the value of the determinant. By Property III, interchange of the two identical rows should change the value to – _D_. Since the determinants are the same, _D_ = – _D_ or _D_ = 0.
**28.14** Prove Property V: If each of the elements of a row (or column) are multiplied by the same number _p_ , the value of the determinant is multiplied by _p_.
**SOLUTION**
Each term of the determinant contains one and only one element from the row multiplied by _p_ and thus each term has factor _p_. This factor is therefore common to all the terms of the expansion and so the determinant is multiplied by _p_.
**28.15** Prove Property VI: If each element of a row (or column) of a determinant is expressed as the sum of two (or more) terms, the determinant can be expressed as the sum of two (or more) determinants.
**SOLUTION**
For the case of third-order determinants we must show that
We will expand each determinant by minors using the first column.
The expansion for each side is identical. Thus the property holds for third-order determinants. The method of proof holds in the general case.
**28.16** Prove Property VII: If to each element of a row (or column) of a determinant is added _m_ times the corresponding element of any other row (or column), the value of the determinant is not changed.
**SOLUTION**
For the case of a third-order determinant we must show that
By property VI the left-hand side may be written
This last determinant may be written
which is zero by Property IV.
**28.17** Show that
**SOLUTION**
The number 3 may be factored from each element in the first column and 2 may be factored from each element in the third column to yield
which equals zero since the first and third columns are identical.
**28.18** Use Property VII to transform the determinant
into a determinant of equal value with zeros in the first row, second and third columns.
**SOLUTION**
Multiply each element in the first column by 2 and add to the corresponding elements in the second column, thus obtaining
Multiply each element in the first column of the new determinant by –3 and add to the corresponding elements in the third column to obtain
The result could have been obtained in one step by writing
The choice of the numbers 2 and –3 was made in order to obtain zeros in the desired places.
**28.19** Use Property VII to transform the determinant
into an equal determinant having three zeros in the 4th row.
**SOLUTION**
Multiply each element in the 1st column (the _basic_ column) by –3, –4, +2 and add respectively to the corresponding elements in the 2nd, 3rd, 4th columns. The result is
Note that it is useful to choose a basic row or column containing the element 1.
**28.20** Obtain 4 zeros in a row or column of the 5th order determinant
**SOLUTION**
We shall produce zeros in the 2nd column by use of the basic row shown shaded. Multiply the elements in this basic row by –5, –3, 3, –2 and add respectively to the corresponding elements in the 1st, 2nd, 4th, 5th rows to obtain
**28.21** Obtain 3 zeros in a row or column of the determinant
without changing its value.
**SOLUTION**
It is convenient to use Property VII to obtain an element 1 in a row or column. For example, by multiplying each of the elements in column 2 by –1 and adding to the corresponding elements in column 3, we obtain
Using the 3rd column as the basic column, multiply its elements by 2, –2, 2 and add respectively to the 1st, 2nd, 4th columns to obtain
which equals the given determinant.
**28.22** Write the minor and corresponding cofactor of the element in the second row, third column for the determinant
**SOLUTION**
Crossing out the row and column containing the element, the minor is given by
Since the element is in the 2nd row, 3rd column and 2 + 3 = 5 is an odd number, the associated sign is minus. Thus the cofactor corresponding to the given element is
**28.23** Write the minors and cofactors of the elements in the 4th row of the determinant
**SOLUTION**
The elements in the 4th row are –3, –2, –4, 1.
**28.24** Express the value of the determinant of Problem 28.23 in terms of minors or cofactors.
**SOLUTION**
Value of determinant = sum of elements each multiplied by associated cofactor
Upon evaluating each of the 3rd order determinants the result –53 is obtained.
The method of evaluation here indicated is tedious. However, the labor involved may be considerably reduced by first transforming a given determinant into an equivalent one having zeros in a row or column by use of Property VII as shown in the following problem.
**28.25** Evaluate the determinant in Problem 28.16 by first transforming it into one having three zeros in a row or column and then expanding by minors.
**SOLUTION**
Choosing the basic column indicated,
multiply its elements by –2, –5, 3 and add respectively to the corresponding elements of the 1st, 3rd, 4th colums to obtain
Expand according to the cofactors of the elements in the second row and obtain
(0)(its cofactor) + (1)(its cofactor) + (0)(its cofactor) + (0)(its cofactor)
Expanding this determinant, we obtain the value –53 which agrees with the result of Problem 28.23.
Note that the method of this problem may be employed to evaluate 3rd order determinants in terms of 2nd order determinants.
**28.26** Evaluate each of the following determinants.
( _a_ )
Multiply the elements in the indicated basic row by –2, 1, –3 and add respectively to the corresponding elements in the 2nd, 3rd, 4th rows to obtain
Multiply the elements in the indicated basic column by –7 and add to the corresponding elements in the 1st and 3rd columns to obtain
( _b_ )
Multiply the elements in the indicated basic column by 3, 2, 1, –4 respectively and add to the corresponding elements in the 1st, 3rd, 4th, 5th columns to obtain
In the last determinant, multiply the elements in the indicated basic row by 6, –4 and add respectively to the elements in the 1st and 2nd rows to obtain
Multiply the elements in the indicated row of the last determinant by 2 and add to the 2nd row to obtain
Multiply the elements in the indicated row of the last determinant 22, 8 and add respectively to the elements in the 1st and 3rd rows to obtain
**28.27** Factor the following determinant.
**28.28** Solve the system
**SOLUTION**
Then
**28.29** The currents _i_ 1, _i_ 2, _i_ 3, _i_ 4, _i_ 5 (measured in amperes) can be determined from the following set of equations. Find _i_ 3.
**SOLUTION**
**29.30** Determine whether the system
is consistent.
**SOLUTION**
However,
Hence at least one of the determinants _D_ 1, _D_ 2, _D_ 3 ≠ 0 so that the equations are inconsistent.
This could be seen in another way by multiplying the first equation by 2 and adding to the second equation to obtain _4x_ – _5y_ \+ _z_ = 6 which is not consistent with the last equation.
**28.31** Determine whether the system
is consistent.
**SOLUTION**
Nothing can be said about the consistency from these facts. On closer examination of the system it is noticed that the second and third equations are inconsistent. Hence the system is inconsistent.
**28.32** Determine whether the system
is consistent.
**SOLUTION**
_D_ = _D_ 1 = _D_ 2 = _D_ 3 = 0. Hence nothing can be concluded from these facts.
Solving the first two equations for _x_ and _y_ (in terms of _z_ ), These values are found by substitution to satisfy the third equation. (If they did not satisfy the third equation the system would be inconsistent.)
Hence the values satisfy the system and there are infinite sets of solutions, obtained by assigning various values to _z_. Thus if _z_ = 3, then _x_ = 3, _y_ = 4; if _z_ = –2, then _x_ = 0, _y_ = 0; etc.
It follows that the given equations are _dependent_. This may be seen in another way by multiplying the second equation by 3 and adding to the first equation to obtain _5x_ – _5y_ \+ _z_ = –2 which is the third equation.
**28.33** Does the system
possess only the trivial solution _x_ = _y_ = _z_ = 0?
**SOLUTION**
Since _D_ ≠ 0 and _D_ 1 = _D_ 2 = _D_ 3 = 0, the system has only the trivial solution.
**28.34** Find nontrivial solutions for the system
if they exist.
**SOLUTION**
Hence there are nontrivial solutions.
To determine these nontrivial solutions solve for _x_ and _y_ (in terms of _z_ ) from the first two equations (this may not always be possible). We find _x_ = _z/2, y_ = _z/2_. These satisfy the third equation. An infinite number of solutions is obtained by assigning various values to z. For example, if _z_ = 6, then _x_ = 3, _y_ = 3; if _z_ = –4, then _x_ = –2, _y_ = –2; etc.
**28.35** For what values of _k_ will the system
have nontrivial solutions?
**SOLUTION**
Nontrivial solutions are obtained when
Hence D = –3 _k_ 2 \+ 3 _k_ \+ 6 = 0 or _k_ = –1, 2.
### **Supplementary Problems**
**28.36** Evaluate each of the following determinants.
**28.37** Show that if the elements of one row (or column) of a second-order determinant are multiplied by the same number, the determinant is multiplied by the number.
**28.38** Solve the unknowns in each of the following systems.
**28.39** Evaluate each determinant.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
( _e_ )
**28.40** For what value of _k_ does
**28.41** Show that if the elements of one row (or column) of a third-order determinant are multiplied by the same number, the determinant is multiplied by the number.
**28.42** Solve for the unknowns in each of the following systems.
( _a_ )
( _b_ )
( _c_ )
**28.43** Solve for the indicated unknown.
**28.44** ( _a_ ) Prove Property I: If the rows and columns of a determinant are interchanged, the value of the determinant is the same.
( _b_ ) Prove Property II: If each element in a row (or column) is zero, the value of the determinant is zero.
**28.45** Show that the determinant
equals zero.
**28.46** Transform the determinant
into an equal determinant having three zeros in the 3rd column.
**28.47** Without changing the value of the determinant
obtain four zeros in the 4th column.
**28.48** For the determinant
( _a_ ) write the minors and cofactors of the elements in the 3rd row,
( _b_ ) express the value of the determinant in terms of minors or cofactors,
( _c_ ) find the value of the determinant.
**28.49** Transform the determinant
into a determinant having three zeros in a row and then evaluate the determinant by use of expansion by minors.
**28.50** Evaluate each determinant.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
**28.51** Factor each determinant:
( _a_ )
( _b_ )
**28.52** Solve each system:
( _a_ )
( _b_ )
**28.53** Find _i_ 1 and _i_ 4 for the system
**28.54** Determine whether each system is consistent.
( _a_ )
( _b_ )
( _c_ )
( _d_ )
**28.55** Find non-trivial solutions, if they exist, for the system
**28.56** For what value of _k_ will the system
possess non-trivial solutions?
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**28.36**
( _a_ ) 5
( _b_ ) –2
( _c_ ) 4
( _d) 14xy_
( _e_ ) – _a_ 2 – _b_ 2
( _f_ ) _x_ 2 – 8 _x_
**28.38**
( _a_ ) _x_ = 2, _y_ = –3; (2, –3)
( _b_ ) _r_ = 1/2, _s_ = 3/2; (1/2, 3/2)
( _c_ ) _x_ = –2, _y_ = –4; (–2, –4)
( _d_ ) _x_ = 8/23, _y_ = –82/23; (8/23, –82/23)
( _e_ ) _x_ = 12, _y_ = 16; (12, 16)
( _f_ ) _x_ = 5, _y_ = –2; (5, –2)
( _g_ ) _x_ = 12, _y_ = 15; (12, 15)
( _h_ ) _u_ = 2/3, _v_ = 3/5; (2/3, 3/5)
**28.39**
( _a_ ) 43
( _b_ ) 19
( _c_ ) 0
( _d_ ) 5 _x_ \+ 8 _y_ – 14 _z_
( _e_ ) _bc_ 2 – _cb_ 2 \+ _a_ 2 _c_ – _ac 2 \+ _ab_ 2 – _ba_ 2_
**28.40** All values of _k_.
**28.42**
( _a_ ) _x_ = –2, _y_ = 1, _z_ = –3; (–2, 1, –3)
( _b_ ) _u_ = 1, _v_ = –1, w = 2; (1, –1, 2)
( _c_ ) _x_ = –4, _y_ = 2, _Z_ = 3; (–4, 2, 3)
**28.43**
( _a_ ) i2 = 0.8
( _b_ ) _x_ = 6
**28.48**
( _c_ ) –38
**28.49** 28
**28.50**
( _a_ ) 38
( _b_ ) –143
( _c_ ) –108
( _d_ ) 88
**28.51**
( _a_ ) _abc_ ( _a_ – _b_ )( _b – c_ )( _c_ – _a_ )
( _b_ ) ( _x_ – 1)( _y_ – 1)( _z_ – 1)( _x_ – _y_ )( _y_ – _z_ )( _z_ – _x_ )
**28.52**
( _a) x_ = 2, _y_ = –1, _z_ = 3, _w_ = 1
( _b_ ) _x_ = 1, _y_ = –1, _z_ = 2, _w_ = 0
**28.53**
_i 1_ = 3, _i 4_ = –2
**28.54**
( _a_ ) consistent
( _b_ ) dependent
( _c_ ) inconsistent
( _d_ ) inconsistent
**28.55** Only the trivial solution _x_ = _y_ = _z_ = 0.
**28.56** _k_ = –1
## **CHAPTER 29
Matrices**
### **29.1 DEFINITION OF A MATRIX**
A matrix is a rectangular array of numbers. The numbers are the entries or elements of the matrix. The following are examples of matrices.
Matrices are classified by the number of rows and columns. The matrices above are 2 × 2, 3 × 2, 3 × 1, and 2 × 3 with the first number indicating the number of rows and the second number indicating the number of columns. When a matrix has the same number of rows as columns, it is a square matrix.
The matrix **A** is an _m_ × _n_ matrix. The entries in matrix **A** are double subscripted with the first number indicating the row of the entry and the second number indicating the column of the entry. The general entry for the matrix is denoted by _a ij_ The matrix **A** can be denoted by [ _a ij_].
### **29.2 OPERATIONS WITH MATRICES**
If matrix **A** and matrix **B** have the same size, same number of rows and same number of columns, and the general entries are of the form _a ij_ and _b ij_ respectively, then the sum **A + B =** _[a ij]_ \+ _[b ij]_ = _[a ij_ \+ _b ij]_ = _[c ij]_ = **C** for all _i_ and _j_.
**EXAMPLE 29.1.** Find the sum of A
The matrix **—A** is called the opposite of matrix **A** and each entry in **—A** is the opposite of the corresponding entry in **A.**
Thus, for
Multiplying a matrix by a scalar (real number) results in every entry in the matrix being multiplied by the scalar.
**EXAMPLE 29.2.** Multiply the matrix
The product **AB** where **A** is an _m_ × _p_ matrix and **B** is a _p_ × _n_ matrix is **C** , an _m_ × _n_ matrix. The entries _c ij_ in matrix **C** are found by the formula _c ij_ = _a_ _i_ 1 _b_ 1 _j_ \+ _a_ _i_ 2 _b_ 2 _j_ \+ _a_ _i_ 3 _b_ 3 _j_ \+... + _a_ _ip_ _b_ _pj_.
**EXAMPLE 29.3.** Find the product **AB** when
**EXAMPLE 29.4.** Find the products **CD** and **DC** when
In Example 29.4, note that although both products **CD** and **DC** exist, **CD ≠ DC.** Thus, multiplication of matrices is not commutative.
An identity matrix is an _n × n_ matrix with entries of 1 when the row and column numbers are equal and 0 everywhere else. We denote the _n_ × _n_ identity matrix by **I** _n_.
For example,
If **A** is a square matrix and **I** is the identity matrix the same size as **A,** then **AI = IA** = **A.**
For
and
### **29.3 ELEMENTARY ROW OPERATIONS**
Two matrices are said to be row equivalent if one can be obtained from the other by a sequence of elementary row operations.
**Elementary Row Operations**
(1) Interchange two rows.
(2) Multiplying a row by a nonzero constant.
(3) Add a multiple of a row to another row.
A matrix is said to be in reduced row-echelon form if it has the following properties:
(1) All rows consisting of all zeros occur at the bottom of the matrix.
(2) A row that is not all zeros has a 1 as its first non-zero entry, which is called the leading 1.
(3) For two successive non-zero rows, the leading 1 in the higher row is further to the left than the leading 1 in the lower row.
(4) Every column that contains a leading 1 has zeros in every other position in the column.
**EXAMPLE 29.5.** Use elementary row operations to put the matrix A in reduced row-echelon form when
∼ is the symbol used between two matrices to indicate that the two matrices are row equivalent.
R2 in front of a matrix means that the row following it was row 2 in the previous matrix.
R3 — 3R1 in front of a matrix means that the row following it was obtained from the previous matrix by subtracting 3 times row 1 from row 3.
The reduced row-echelon form of matrix
### **29.4 INVERSE OF A MATRIX**
**A** square matrix **A** has an inverse if there is a matrix **A –1** such that **AA –1 = A–1A = I.**
To find the inverse, if it exists, of a square matrix **A** we complete the following procedure.
(1) Form the partitioned matrix **[A|I],** where **A** is the given _n_ × _n_ matrix and **I** is the _n_ × _n_ identity matrix.
(2) Perform elementary row operations on **[A|I]** until the partitioned matrix has the form **[I|B],** that is, until the matrix **A** on the left is transformed into the identity matrix. If **A** cannot be transformed into the identity matrix, matrix **A** does not have an inverse.
(3) The matrix **B** is **A –1,** the inverse of matrix **A.**
**EXAMPLE 29.6.** Find the inverse of matrix
If the matrix **A** is row equivalent to **I,** then the matrix **A** has an inverse and is said to be invertible. **A** does not have an inverse if it is not row equivalent to **I.**
**EXAMPLE 29.7.** Find the inverse, if it exists, for matrix
The matrix **A** is row equivalent to the matrix on the left. Since the matrix on the left has a row of all zeros, it is not row equivalent to **I.** Thus, the matrix **A** does not have an inverse.
Another way to determine whether the inverse of a matrix **A** exists is that the determinant associated with an invertible matrix is non-zero, that is, det **A ≠** 0 if **A –1** exists.
For 2 _×_ 2 matrices, the inverse can be found by a special procedure:
(1) Find the value of det **A.** If det **A** ≠ 0, then the inverse exists.
(2) Exchange the entries on the main diagonal, swap _a_ 11 and _a_ 22.
(3) Change the signs of the entries on the off diagonal, replace _a_ 21 by _–a 21_ and _a_ 12 by – _a_ 12.
(4) Multiply the new matrix by 1/det **A.** This product is **A –1.**
### **29.5 MATRIX EQUATIONS**
A matrix equation **AX** = **B** has a solution if and only if the matrix **A –1** exists and the solution is **X** = **A –1B.**
**EXAMPLE 29.8.** Solve the matrix equation
### **29.6 MATRIX SOLUTION OF A SYSTEM OF EQUATIONS**
To solve a system of equations using matrices, we write a partitioned matrix which is the coefficient matrix on the left augmented by the constants matrix on the right.
The augmented matrix associated with the system
**EXAMPLE 29.9.** Use matrices to solve the system of equations:
Write the augmented matrix for the system.
Put the matrix on the left in reduced row-echelon form.
From the reduced row-echelon form of the augmented matrix, we write the equations:
_x_ 1 = –1, _x_ 2 = 2, _x_ 3 = 1, and _x_ 4 = 3.
Thus, the solution of the system is (–1, 2, 1, 3).
**EXAMPLE 29.10.** Solve the system of equations:
The system has infinitely many solutions of the form (2 _—_ 5 _x_ 3, —1 _+_ 3 _x_ 3, _x_ 3), where **x 3** is a real number.
### **Solved Problems**
**29.1** Find _(a)_ **A + B** , _(b)_ **A** – **B** , _(c)_ 3 **A,** and _(d_ ) 5 **A** – 2 **B** when
**SOLUTION**
**29.2** Find, if they exist, _(a)_ **AB** , _(b)_ **BA,** and _(c)_ **A** 2 when
**SOLUTION**
_(a)_
_(b)_
_(c)_ **A** 2 = [3 2 1][3 2 1]; not possible. **A** _n_ , _n_ >1, exists for square matrices only.
**29.3** Find **AB,** if possible.
_(a)_
_(b)_
**SOLUTION**
_(a)_
**A** is a 3 × 2 matrix and **B** is a 3 × 3. Since **A** has only two columns it can only multiply matrices having two rows, 2 × _k_ matrices.
_(b)_
**29.4** Write each matrix in reduced row-echelon form.
_(a)_
_(b)_
**SOLUTION**
_(a)_
The reduced row-echelon form of
The reduced row-echelon form of
**29.5** Find the inverse, if it exists, for each matrix.
_(a)_
_(b)_
_(c)_
_(d)_
**SOLUTION**
_(a)_
The first form of the matrix is frequently used because it reduces the amount of computation with fractions that needs to be done. Also, it makes it easier to work with matrices on a graphing calculator.
_(b)_ Since det **B =** 0, **B -1** does not exist.
_(c)_
_(d)_
Since the left matrix in the last form is not row equivalent to the identity matrix I, D does not have an inverse.
**29.6** If solve each equation for **X.**
_(a)_ 2 **X** \+ 3 **A = B** _(b)_ 3 **A** \+ 6 **B** = –3 **X**
**SOLUTION**
_(a)_
_(b)_ 3 **A** \+ 6 **B** = –3 **X.** So –3 **X** = 3 **A** \+ 6 **B** and **X = –A** – 2 **B.**
**29.7** Write the matrix equation **AX = B** and use it to solve the system
**SOLUTION**
The solution to the system is (4, 8).
**29.8** Solve each system of equations using matrices.
_(a)_
_(b)_
**SOLUTION**
From the reduced row echelon form of the matrix, we write the equations:
_x_ = 1, _y_ = –1, and _z_ = 2.
The system has the solution (1, –1, 2).
_(b)_
Since the last row results in the equation 0 _z_ = 1, which has no solution, the system of equations has no solutions.
### **Supplementary Problems**
**29.9**
Perform the indicated operations, if possible.
_(a)_ **B** \+ **C**
_(b)_ 5 **A**
_(c)_ 2 **C** –6 **B**
_(d)_ –6 **B**
_(e)_ 3 **B** \+ 2 **C**
_(f)_ **D A**
_(g)_ **A D**
_(h)_ **C** – **B**
_(i)_ **C** –5 **A**
_(j)_ **BC**
_(k)_ ( **DA** ) **B**
_(l)_ **A** 2
_(m)_ **B** 2
_(n)_ **D** ( **AB** )
_(o)_ **A** 3
_(p)_ **DB** \+ **DC**
**29.10** Find the product **AB** , if possible.
_(a)_
_(b)_
_(c)_
_(d)_
**29.11** Solve each system of equations using a matrix equation of the form AX = B.
_(a)_
_(b)_
_(c)_
_(d)_
**29.12** Write each matrix in reduced row echelon form.
_(a)_
_(b)_
_(c)_
_(d)_
_(e)_
_(f)_
**29.13** Find the inverse, if it exists, of each matrix.
_(a)_
_(b)_
_(c)_
_(d)_
_(e)_
_(f)_
_(g)_
_(h)_
**29.14** Solve each system of equations using matrices.
_(a)_
_(b)_
_(c)_
_(d)_
_(e)_
_(f)_
_(g)_
_(h)_
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**29.9**
_(a)_
_(b)_
_(c)_
_(d)_
_(e)_
_(f)_
_(g)_ not possible
_(h)_
_(i)_ not possible
_(j)_ not possible
_(k)_
_(l)_
_(m)_ not possible
_(n)_
_(o)_
_(p)_
**29.10** _(a)_
_(b)_
_(c)_
_(d)_
**29.11** _(a)_ (5, 5) _(b)_ (-3, 2) _(c)_ (1, 1) _(d)_ (8, 8)
**29.12** _(a)_
_(b)_
_(c)_
_(d)_
_(e)_
_(f)_
**29.13** _(a)_
_(b)_
_(c)_
_(d)_
_(e)_
_(f)_
_(g)_
_(h)_
**29.14** _(a)_ (–1, 3, 2)
_(b)_ no solution
_(c)_ (2 _z_ – 1, _z, z)_ where _z_ is a real number
_(d)_ (– _z_ \+ 3, _z_ \+ 1, _z_ ) where _z_ is a real number
_(e)_ (–4, 8, 5)
_(f)_ (0, 0, 0)
_(g)_ (1, 0, 3, 2)
_(h)_ no solution
## **CHAPTER 30
Mathematical Induction**
### **30.1 PRINCIPLE OF MATHEMATICAL INDUCTION**
Some statements are defined on the set of positive integers. To establish the truth of such a statement, we could prove it for each positive integer of interest separately. However, since there are infinitely many positive integers, this case-by-case procedure can never prove that the statement is always true. A procedure called mathematical induction can be used to establish the truth of the state for all positive integers.
**Principle of Mathematical Induction**
Let _P(n)_ be a statement that is either true or false for each positive integer _n_. If the following two conditions are satisfied:
(1) _P_ (1) is true, and
(2) Whenever for _n_ = _k, P_ ( _k_ ) is true implies _P(k_ \+ 1) is true.
Then _P(n)_ is true for all positive integers _n_.
### **30.2 PROOF BY MATHEMATICAL INDUCTION**
To prove a theorem or formula by mathematical induction there are two distinct steps in the proof.
(1) Show by actual substitution that the proposed theorem or formula is true for some one positive integer _n_ , as _n_ = 1, or _n_ = 2, etc.
(2) Assume that the theorem or formula is true for _n_ = _k_. Then prove that it is true for _n_ = _k_ \+ 1.
Once steps (1) and (2) have been completed, then you can conclude the theorem or formula is true for all positive integers greater than or equal to _a_ , the positive integer from step (1).
### **Solved Problems**
**30.1** Prove by mathematical induction that, for all positive integers _n_ ,
**SOLUTION**
_Step 1_. The formula is true for _n =_ 1, since
_Step 2_. Assume that the formula is true for _n_ = _k_. Then, adding ( _k_ \+ 1) to both sides,
which is the value of _n(n_ \+ 1)/2 when ( _k_ \+ 1) is substituted for _n_.
Hence if the formula is true for _n_ = _k_ , we have proved it to be true for _n_ = _k_ \+ 1. But the formula holds for _n_ = 1; hence it holds for _n_ = 1 + 1 = 2. Then, since it holds for _n_ = 2, it holds for _n_ = 2 + 1 = 3, and so on. Thus the formula is true for all positive integers _n_.
**30.2** Prove by mathematical induction that the sum of _n_ terms of an arithmetic sequence _a, a + d_ ,
**SOLUTION**
_Step 1_. The formula holds for _n_ = 1, since
_Step 2_. Assume that the formula holds for _n_ = _k_. Then
Add the ( _k_ \+ 1)th term, which is ( _a_ \+ _kd_ ), to both sides of the latter equation. Then
The right-hand side of this equation
which is the value of ( _n_ /2)[2 _a_ \+ ( _n_ — 1) _d_ ] when _n_ is replaced by ( _k_ \+ 1).
Hence if the formula is true for _n_ = _k_ , we have proved it to be true for _n_ = _k_ \+ 1. But the formula holds for _n =_ 1; hence it holds for _n_ = 1 +1 = 2. Then, since it holds for _n_ = 2, it holds for _n_ = 2 + 1 = 3, and so on. Thus the formula is true for all positive integers _n_.
**30.3** Prove by mathematical induction that, for all positive integers _n_ ,
**SOLUTION**
_Step 1_. The formula is true for _n =_ 1, since
_Step 2_. Assume that the formula is true for _n_ = _k_. Then
Add the ( _k_ \+ 1)th term, which is ( _k_ \+ 1)2, to both sides of this equation. Then
The right-hand side of this equation
which is the value of _n_ ( _n_ \+ 1)(2 _n_ \+ 1)/6 when n is replaced by ( _k_ \+ 1).
Hence if the formula is true for _n_ = _k_ , it is true for _n_ = _k_ \+ 1. But the formula holds for _n_ = 1; hence it holds for _n_ = 1 + 1 = 2. Then, since it holds for _n_ =2, it holds for _n_ = 2 + 1 = 3, and so on. Thus the formula is true for all positive integers.
**30.4** Prove by mathematical induction that, for all positive integers _n_ ,
**SOLUTION**
_Step_ 1. The formula is true for n = 1, since
_Step_ 2. Assume that the formula is true for n = k. Then
Add the ( _k_ \+ 1)th term, which is
to both sides of the above equation. Then
The right-hand side of this equation is
which is the value of _n_ /(2 _n_ \+ 1) when n is replaced by ( _k_ \+ 1).
Hence if the formula is true for _n_ = _k_ , it is true for _n_ = _k_ \+ 1. But the formula holds for _n_ = 1; henceit holds for _n_ = 1 + 1 = 2. Then since it holds for _n_ = 2 it holds for _n_ = 2 + 1 = 3 and so on. Thus the formula is true for all positive integers _n_.
**30.5** Prove by mathematical induction that a2n – b 2n is divisible by _a + b_ when n is any positive integer.
**SOLUTION**
_Step_ 1. The theorem is true for _n_ = 1, since _a_ 2 _– b_ 2 = ( _a_ \+ _b_ )( _a_ – _b_ ).
_Step_ 2. Assume that the theorem is true for _n_ = _k_. Then
_a_ 2 _k_ – _b_ 2 _k_ is divisible by _a_ \+ _b_
We must show that _a_ 2 _k_ +2 – _b_ 2 _k_ +2 is divisible by _a + b_. From the identity
it follows that _a_ 2 _k_ +2 – _b_ 2 _k_ +2 is divisible by _a_ \+ _b_ if _a_ 2 _k_ – _b_ 2 _k_ is.
Hence if the theorem is true for _n_ = _k_ , we have proved it to be true for _n_ = _k_ \+ 1. But the theorem holds for _n_ = 1; hence it holds for _n_ = 1 + 1 = 2. Then, since it holds for _n_ = 2, it holds for _n_ = 2 + 1 = 3, and so on. Thus the theorem is true for all positive integers _n_.
**30.6** Prove the binomial formula
for positive integers _n_.
**SOLUTION**
_Step 1_. The formula is true for _n_ = 1.
_Step 2_. Assume the formula is true for _n_ = _k_. Then
Multiply both sides by _a_ \+ _x_. The multiplication on the right may be written
Since
the product may be written
which is the binomial formula with n replaced by _k_ \+ 1.
Hence if the formula is true for _n_ = _k_ , it is true for _n_ = _k_ \+ 1. But the formula holds for _n_ = 1; hence it holds for _n_ = 1 = 2, and so on. Thus the formula is true for all positive integers _n_.
**30.7** Prove by mathematical induction that the sumof the interior angles, _S(n)_ , of a convex polygon is _S(n)_ = ( _n_ – 2)180°, where _n_ is the number of sides on the polygon.
**SOLUTION**
_Step_ 1. Since a polygon has at least 3 sides, we start with _n_ = 3. For _n_ = 3, _S_ (3) = (3 – 2)180° = (1)180° = 180°. This is true since the sum of the interior angles of a triangle is 180°.
_Step_ 2. Assume that for _n_ = _k_ , the formula is true. Then _S_ ( _k_ ) = ( _k_ – 2)180° is true. Now consider a convex polygon with _k_ \+ 1 sides. We can draw in a diagonal that forms a triangle with two of the sides of the polygon. The diagonal also forms a _k_ -sided polygon with the other sides of the original polygon. The sum of the interior angles of the ( _k_ \+ 1)-sided polygon, _S_ ( _k_ \+ 1), is equal to the sum of the interior angles of the triangle, _S_ (3), and the sum of the interior angles of the _k_ -sided polygon, _S_ ( _k_ ).
Hence, if the formula is true for _n_ = _k_ , it is true for _n_ = _k_ \+ 1.
Since the formula is true for _n_ = 3, and whenever it is true for _n_ = _k_ it is true for _n_ = _k_ \+ 1, the formula is true for all positive integers _n_ ≥ 3.
**30.8** Prove by mathematical induction that _n_ 3 \+ 1 ≥ _n_ 2 \+ _n_ for all positive integers.
**SOLUTION**
_Step 1_. For _n =_ 1, _n 3 +_1 = l3 \+ 1 = 1 + 1 = 2 and _n_ 2 \+ _n_ = 12\+ 1 = 1 + 1 = 2. So _n 3 +_1 ≥ _n_ 2 \+ _n_ is true when _n_ = 1.
_Step 2_. Assume the statement is true for _n_ = _k_. So _k 3_\+ 1 ≥ _k_ 2 \+ _k_ is true.
We know _n_ ≥ 1, so _k_ ≥ 1 and _k_ 3 \+ 2 _k_ 2 ≥ 3. Thus ( _k_ \+ 1)3 \+ 1 ≥ ( _k_ \+ 1)2 \+ ( _k_ \+ 1). Hence, when the statement is true for _n_ = _k_ , it is true for _n_ = _k_ \+ 1.
Since the statement is true for _n_ = 1, and whenever it is true for _n_ = k, is true for _n_ = _k_ \+ 1, the statement is true for all positive integers _n_.
### **Supplementary Problems**
Prove each of the following by mathematical induction. In each case _n_ is a positive integer.
**30.9**
**30.10**
**30.11**
**30.12**
**30.13**
**30.14**
**30.15**
**30.16**
**30.17** _a_ _n_ – _b_ _n_ is divisible by _a_ – _b_ , for _n_ = positive integer.
**30.18** _a_ 2 _n_ –1 \+ _b_ 2 _n_ –1 is divisible by _a + b_ , for _n_ = positive integer.
**30.19**
**30.20** 1 + 2 + 22 \+... + 2 _n_ –1 = 2 _n_ –1
**30.21** ( _ab_ ) _n_ = _a_ _n_ _b_ _n_ , for _n_ = _a_ positive integer
**30.22**
**30.23** _n_ 2 \+ _n_ is even
**30.24** _n_ 3 \+ 5 _n_ is divisible by 3
**30.25** 5 _n_ – 1 is divisible by 4
**30.26** 4 _n_ – 1 is divisible by 3
**30.27** _n_ ( _n_ \+ 1)( _n_ \+ 2) is divisible by 6
**30.28** _n_ ( _n_ \+ 1)( _n_ \+ 2)( _n_ \+ 3) is divisible by 24
**30.29** _n_ 2 \+ 1 > _n_
**30.30** 2 _n_ ≥ _n_ \+ 1
## **CHAPTER 31
Partial Fractions**
### **31.1 RATIONAL FRACTIONS**
A rational fraction in _x_ is the quotient of two polynomials in _x_.
Thus is a rational fraction.
### **31.2 PROPER FRACTIONS**
A proper fraction is one in which the degree of the numerator is less than the degree of the denominator.
Thus are proper fractions:
An improper fraction is one in which the degree of the numerator is greater than or equal to the degree of the denominator.
Thus is an improper fraction.
By division, an improper fraction may always be written as the sum of a polynomial and a proper fraction.
Thus
### **31.3 PARTIAL FRACTIONS**
A given proper fraction may often be written as the sum of other fractions (called partial fractions) whose denominators are of lower degree than the denominator of the given fraction.
**EXAMPLE 31.1.**
### **31.4 IDENTICALLY EQUAL POLYNOMIALS**
If two polynomials of degree _n_ in the same variable _x_ are equal for more than _n_ values of _x_ , the coefficients of like powers of _x_ are equal and the two polynomials are identically equal. If a term is missing in either of the polynomials, it can be written in with a coefficient of 0.
### **31.5 FUNDAMENTAL THEOREM**
A proper fraction may be written as the sum of partial fractions according to the following rules.
(1) Linear factors none of which are repeated
If a linear factor _ax_ \+ _b_ occurs once as a factor of the denominator of the given fraction, then corresponding to this factor associate a partial fraction _A/(ax_ \+ _b_ ), where _A_ is a constant ≠ 0.
**EXAMPLE 31.2.**
(2) Linear factors some of which are repeated
If a linear factor _ax_ \+ _b_ occurs _p_ times as a factor of the denominator of the given fraction, then corresponding to this factor associate the _p_ partial fractions
where _A_ 1, _A_ 2,..., _A_ p are constants and _A_ p ≠ 0.
**EXAMPLES 31.3.**
(a)
(b)
(3) Quadratic factors none of which are repeated
If a quadratic factor _ax 2_ \+ _bx_ \+ _c_ occurs once as a factor of the denominator of the given fraction, then corresponding to this factor associate a partial fraction
where _A_ and _B_ are constants which are not both zero.
_Note_. It is assumed that _ax 2_ \+ _bx_ \+ _c_ cannot be factored into two real linear factors with integer coefficients.
**EXAMPLES 31.4.**
(a)
(b)
(4) Quadratic factors some of which are repeated
If a quadratic factor _ax 2_ \+ _bx_ \+ _c_ occurs _p_ times as a factor of the denominator of the given fraction, then corresponding to this factor associate the _p_ partial fractions
where _A_ 1, _B_ 1, _A_ 2, _B_ 2,..., _A_ _p_ , _B_ _p_ are constants and _A_ _p_ , _B_ _p_ are not both zero.
**EXAMPLE 31.5.**
### **31.6 FINDING THE PARTIAL FRACTION DECOMPOSITION**
Once the form of the partial fraction decomposition of a rational fraction has been determined, the next step is to find the system of equations to be solved to get the values of the constants needed in the partial fraction decomposition. The solution of the system of equations can be aided by the use of a graphing calculator, especially when using the matrix methods discussed in Chapter 29.
Although the system of equations usually involves more than three equations, it is often quite easy to determine the value of one or two variables or relationships among the variables that allow the system to be reduced to a size small enough to be solved conveniently by any method. The methods discussed in Chapter 15 and Chapter 28 are the basic procedures used.
**EXAMPLE 31.6.** Find the partial fraction decomposition of
Using Rules (2) and (3) in Section 31.5, the form of the decomposition is:
Equating the coefficients of the corresponding terms in the two polynomials and setting the others equal to 0, we get the system of equations to solve.
Solving the system, we get _A_ = –1, _B_ = 5, _C_ = 1, and _D_ = 0.
Thus, the partial fraction decomposition is:
### **Solved Problems**
**31.1** Resolve into partial fractions
**SOLUTION**
Let
We must find the constants _A_ and _B_ such that
or
_x_ \+ 2 = ( _A_ \+ 2 _B_ ) _x_ \+ 3 _B_ – 5 _A_.
Equating coefficients of like powers of _x_ , we have 1 = _A_ \+ 2 _B_ and 2 = 3 _B_ – 5 _A_ which when solved simultaneously give _A_ = –1/13, _B_ = 7/13.
Hence
_Another method_. _x_ \+ 2 = _A_ ( _x_ – 5)+ _B_ (2 _x_ \+ 3)
**31.2**
**SOLUTION**
Hence
**31.3**
**SOLUTION**
Equating coefficients of like powers of _x_ , _A_ \+ _C_ = 2, 6 _A_ \+ _B_ \+ 2 _C_ = 7 and 9 _A_ – _B_ – 3 _C_ = 23. Solving simultaneously, _A_ = 2, _B_ = –5, _C_ = 0.
Hence
_Another method_. 2 _x_ 2 \+ 7 _x_ \+ 23 = _A_ ( _x_ \+ 3)2 \+ _B_ ( _x_ – 1) + _C_ ( _x_ – 1)( _x_ \+ 3)
**31.4**
**SOLUTION**
Equating coefficients of like powers of _x_ , _B_ \+ _D_ = 0, _A_ – 4 _B_ \+ _C_ – 2 _D_ = 1, –4 _A_ \+ 4 _B_ = 26, 4 _A_ = 2. The simultaneous solution of these four equations is _A_ = 1/2, _B_ = –1, _C_ = –3/2, _D_ = 1.
Hence
_Another method_. _x_ 2 – 6 _x_ \+ 2 = _A_ ( _x_ – 2)2 – _Bx_ ( _x_ – 2)2 \+ _Cx_ 2 \+ _Dx_ 2( _x_ – 2)
To find _A_ , let _x_ = 0: 2 = 4 _A_ , _A_ = 1/2. To find _C_ , let _x_ = 2: 4 – 12 + 2 = 4 _C_ , _C_ = –3/2.
To find _B_ and _D_ , let _x_ = any values except 0 and 2 (for example, let _x_ = 1, _x_ = –1).
The simultaneous solution of equations (1) and (2) is _B_ = –1, _D_ = 1.
**31.5**
**SOLUTION**
**31.6**
**SOLUTION**
Equating coefficients of like powers of _x_ , 3 _A_ \+ _C_ = 7, 3 _B_ – 2 _A_ – 2 _C_ = –25, –2 _B – C_ = 6. The simultaneous solution of these three equations is _A_ = 1, _B_ = –5, _C_ = 4
Hence
**31.7**
**SOLUTION**
Equating coefficients of like powers of _x_ ,
_A_ \+ _C_ = 0, B + _D_ = 4, 3 _C_ – 2 _A_ = 0, –2 _B_ \+ 3 _D_ = –28.
Solving simultaneously, _A_ = 0, _B_ = 8, _C_ = 0, _D_ = – 4.
Hence
### **Supplementary Problems**
Find the partial fraction decomposition of each rational fraction.
**31.8**
**31.9**
**31.10**
**31.11**
**31.12**
**31.13**
**31.14**
**31.15**
**31.16**
**31.17**
**31.18**
**31.19**
**31.20**
**31.21**
**31.22**
**31.23**
**31.24**
**31.25**
**31.26**
**31.27**
**31.28**
**31.29**
**31.30**
### **ANSWERS TO SUPPLEMENTARY PROBLEMS**
**31.8**
**31.9**
**31.10**
**31.11**
**31.12**
**31.13**
**31.14**
**31.15**
**31.16**
**31.17**
**31.18**
**31.19**
**31.20**
**31.21**
**31.22**
**31.23**
**31.24**
**31.25**
**31.26**
**31.27**
**31.28**
**31.29**
**31.30**
## **APPENDIX A
Table of Common Logarithms**
## **APPENDIX B
Table of Natural Logarithms**
If _N_ ≥ 10, ln 10 = 2.3026 and write _N_ in scientific notation; then use ln _N_ = ln[ _k_ · (10 _m_ )] = ln _k_ \+ _m_ ln 10 = ln _k_ \+ _m_ (2.3026), where 1 ≤ _k_ , < 10 and _m_ is an integer.
## **INDEX**
_Please note that index links point to page beginnings from the print edition. Locations are approximate in e-readers, and you may need to page down one or more times after clicking a link to get to the indexed material_.
Abscissa,
Absolute value,
Absolute value inequality,
Addition,
associative property for,
commutative property for,
of algebraic expressions,
of complex numbers,
of fractions,
of radicals,
rules of signs for,
Algebra
fundamental operations of,
fundamental theorem of,
Algebraic expressions,
Antilogarithm, ,
Arithmetic mean,
Arithmetic means,
Arithmetic sequence,
Associative properties,
Asymptotes,
horizontal,
vertical,
Augmented matrix,
Axioms of equality, ,
Base of logarithms,
Base of powers,
Best buy,
Binomial,
Binomial coefficients,
Binomial expansion,
formula or theorem,
proof of, for positive integral powers,
Bounds, lower and upper, for roots,
Braces,
Brackets,
Cancellation,
Characteristic of a logarithm,
Circle,
Circular permutations,
Closure property,
Coefficients,
in binomial formula,
lead,
relation between roots and,
Cofactor,
Combinations,
Common difference,
Common logarithms,
Common ratio,
Commutative properties,
Completeness property,
Completing the square,
Complex fractions,
Complex numbers,
algebraic operations with,
conjugate of,
equal,
graphical addition and subtraction of, –70
imaginary part of,
pure imaginary,
real part of,
Compound interest,
Conditional equation,
Conditional inequality,
Conic sections, –180
circle,
ellipse,
hyperbola,
parabola,
Conjugate complex numbers,
Conjugate irrational numbers,
Consistent equations,
Constant,
of proportionality or variation,
Continuity,
Coordinate system, rectangular,
Cramer's Rule, –326
Cube of a binomial,
Cubic equation,
Decimal, repeating,
Defective equations,
Degree,
of a monomial,
of a polynomial,
Denominator, ,
Density property,
Dependent equations,
Dependent events,
Dependent variable,
Depressed equations,
Descartes' Rules of Signs, –218
Determinants,
expansion or value of, , ,
of order _n_ ,
of second order,
of third order,
properties of,
solution of linear equations by, , , –329
Difference,
common,
of two cubes,
of two squares,
tabular,
Discriminant,
Distributive law for multiplication,
Dividend,
Division,
by zero,
of algebraic expressions,
of complex numbers,
of fractions, , –5
of radicals,
synthetic,
Divisor,
Double root, ,
Domain, –90
_e_ , base of natural logarithms,
Effective rate of interest,
Element of a determinant,
Elementary row operations,
Ellipse, –174
Equations,
complex roots of,
conditional,
cubic,
defective,
degree of,
depressed,
equivalent,
graphs of, (see Graphs)
identity,
irrational roots of,
limits for roots of,
linear,
literal,
number of roots of,
quadratic, ,
quadratic type,
quartic,
quintic,
radical,
redundant,
roots of,
simultaneous,
solutions of,
systems of,
transformation of, –74
with given roots,
Equivalent equations,
Equivalent fractions,
Expectations, mathematical,
Exponential equations,
Exponential form,
Exponents, ,
applications,
fractional,
laws of, , –50
zero,
Extraneous roots, solutions,
Extremes,
Factor,
greatest common,
prime,
Factor theorem,
Factorial notation,
Factoring,
Failure, probability of,
Formulas,
Fourth proportional,
Fractional exponents,
Fractions, –5, –43
complex,
equivalent,
improper,
operations with, –5
partial,
proper,
rational algebraic,
reduced to lowest terms,
signs of,
Function,
graph of, –95
linear, , –131
notation for,
polynomial,
quadratic, , –152
Fundamental Theorem of Algebra,
Fundamental Counting Principle,
General or _n_ th term,
Geometric mean,
Geometric sequence, –246
infinite,
Geometric series, infinite,
Graphical solution of equations, ,
Graphs, –95
of equations, –95, , –178
of functions, –95
of linear equations in two variables,
of quadratic equations in two variables,
with holes,
Greater than,
Greatest common factor,
Grouping, symbols of,
Grouping of terms, factoring by,
Harmonic mean,
Harmonic sequence,
Holes, in graph,
Homogenous linear equations,
Hyperbola,
_i_ ,
Identically equal polynomials,
Identity,
property,
matrix,
Imaginary numbers, ,
Imaginary part of a complex number,
Imaginary unit, ,
Improper fraction,
Inconsistent equations,
Independent events,
Independent variable,
Index, ,
Index of a radical,
reduction of,
Induction, mathematical,
Inequalities,
absolute,
conditional,
graphical solution of,
higher order,
principles of,
sense of,
signs of,
Infinite geometric sequence or series,
Infinity,
Integers,
Integral roots theorem,
Interest, –277
compound,
simple,
Intermediate Value Theorem,
Interpolation in logarithms,
Interpolation, linear,
Inverse property,
Irrationality, proofs of, –79,
Irrational number,
Irrational roots,
approximating,
Inverse matrix,
Inverse property,
Least common denominator,
Least common multiple,
Less than,
Like terms,
Linear equations,
consistent,
dependent,
determinant solution of system of, –329
graphical solution of systems of,
homogenous,
inconsistent,
in one variable,
simultaneous, systems of,
Linear function,
Linear interpolation,
Linear programming,
Lines,
intercept form,
slope–intercept form,
slope–point form,
two-point form,
Literals,
Logarithms,
applications of, –279
base of,
common system of,
characteristic of common,
laws of,
mantissa of common,
natural base of,
natural system of,
tables of common,
tables of natural,
Lower bound or limit for roots,
Mantissa,
Mathematical expectation,
Mathematical induction,
Matrix,
addition,
identity,
inverse,
multiplication,
scalar multiplication,
Maximum point, relative,
applications, –106
Mean proportional,
Means of a proportion,
Minimum point, relative,
applications, –106
Minors,
Minuend,
Monomial,
Monomial factor,
Multinomial,
Multiplication, , –15
associative property for,
by zero,
commutative property for,
distributive property for,
of algebraic expressions,
of complex numbers,
of fractions,
of radicals,
rules of signs for,
Mutually exclusive events,
Natural logarithms,
Natural numbers, ,
Negative numbers,
Nonreal roots of an equations,
Numbers,
absolute value of,
complex,
counting,
graphical representation of real,
imaginary,
integers,
irrational,
literal,
natural, ,
negative,
operations with real, –5
positive,
prime,
rational,
real,
whole,
Number system, real,
Numerator, ,
Numerical coefficient,
Odds,
Operations, fundamental,
Order of a determinant, , ,
Order of real numbers, ,
Order property,
Ordinate,
Origin,
of a rectangular coordinate system,
Parabola, –172
vertex of,
Parentheses,
Partial fractions,
Pascal's triangle,
Perfect square trinomial,
Perfect _n_ th powers,
Permutations,
Point, coordinates of a,
Polynomials,
degree of,
factors of, –35
identically equal,
operations with, –15
prime,
relatively prime,
Polynomial functions, –234
solving,
zeros, –215
Positive numbers,
Powers, ,
logarithms of,
of binomials, , –309
Prime factor,
number,
polynomial,
Principal,
Principal root,
Probability,
binomial,
conditional, –312
of dependent events,
of independent events,
of mutually exclusive events,
Product, , ,
of roots of quadratic equation,
Products, special,
Proper fraction,
Proportion,
Proportional,
fourth,
mean,
third,
Proportionality, constant of,
Pure imaginary number,
Quadrants,
Quadratic equations,
discriminant of,
formation of, from given roots,
in one variable,
in two variables,
nature of roots of,
product of roots of,
simultaneous, –193
sum of roots of,
Quadratic equations in one variable, –153
solutions of, –152
by completing the square,
by factoring, –151
by formula, –152
by graphical methods,
by square root method,
Quadratic equations in two variables,
circle,
discriminant,
ellipse,
hyperbola,
parabola,
Quadratic formula,
proof of, –155
Quadratic type equations,
Quartic equation,
Quintic equation,
Quotient, , ,
Radical equations, –153
Radicals,
algebraic addition of,
changing the form of, –59
equations involving, –153
index or order of,
multiplication and division of, –61
rationalization of denominator of, –61
reduction of index of,
removal of perfect nth powers of,
similar,
simplest form of,
Radicand,
Range of a function,
Rate of interest,
Ratio,
common,
Rational function,
graphing, –237
Rational number, ,
Rational root theorem,
Rationalization of denominator,
Real numbers, ,
graphical representation of,
Real part of a complex number,
Reciprocal,
Rectangular coordinates,
Redundant equations,
Relation,
Remainder, ,
Remainder theorem,
Repeating decimal,
Roots, , ,
nature of, for quadratic equation,
double, ,
extraneous,
integral,
irrational,
_n_ th,
number of,
of an equation,
of quadratic equations,
principal _n_ th,
rational,
Row-echelon form,
Row equivalent matrices,
Scaling,
Scientific notation,
Sense of an inequality,
Sequence,
arithmetic,
geometric, –246
harmonic,
infinite,
_n_ th or general term of a, ,
Series,
infinite geometric,
Shifts,
horizontal,
vertical,
Signs,
Descartes' Rule of,
in a fraction,
rules of,
Simple interest,
Simultaneous equations, ,
Simultaneous linear equations,
Simultaneous quadratic equations,
Slope,
horizontal lines,
parallel lines,
perpendicular lines,
vertical lines,
Solutions,
extraneous,
graphical, ,
of systems of equations, ,
trivial, non-trivial,
Special products,
Square,
of a binomial,
of a trinomial,
Straight line, –131
Subtraction, , ,
of algebraic expressions,
of complex numbers,
of fractions, ,
of radicals,
Subtrahend,
Sum,
of arithmetic sequence,
of geometric sequence, –246
of infinite geometric sequence,
of roots of a quadratic equation,
of two cubes,
Symbols of grouping,
Symmetry,
Synthetic division,
Systems of equations, ,
Systems of inequalities,
Systems of _m_ equations in _n_ unknowns,
Tables, ,
of common logarithms,
of natural logarithms,
Tabular difference,
Term,
degree of,
integral and rational,
like or similar,
of sequences,
of series,
Trinomial,
factors of a,
square of a,
Trivial solutions,
Unique Factorization Theorem,
Unit price,
Variable,
dependent,
independent,
Variation, –82
direct,
inverse,
joint,
Zero,
degree,
division by,
exponent,
multiplication by,
Zeros, ,
* * *
Downloadable videos may be obtained from McGraw-Hill Professional's MediaCenter at http://mhprofessional.com/mediacenter.
Some material may require a desktop or laptop computer for full access.
Enter this eBook's ISBN and your e-mail address at the MediaCenter to receive an e-mail message with a download link.
This eBook's ISBN is 978-0-07-182181-0.
* * *
Back
|
{
"redpajama_set_name": "RedPajamaBook"
}
| 318
|
Q: how to make a postgres backup on servers which run on linux/windows i am creating a web application in asp.net with database on postgres. i have successfuly created a database backup by putting the destination of database backup on the webconfig file. the web application succesfully creates backups when published on windows server, however i was wondering if the server is changed to linux the destination folder would be invalid.what could be the solution for it to work on different OS??
another problem is that the postgres installation could be on any drive in linux whereas rightnow in my webapplication i open the system diagnostics to run the pgdump command using the info.filename="C:\Program Files\PostgreSQL\9.3\bin\" + commandType + ".exe ";
anyone has solution for multiOS database backup for postgres using asp.net?? is there any solution to this??
A: You can continue to use your backup scripts on Windows. pg_dump will happily connect to the remote database on the Linux box. E.g.
pg_dump -h my_linux_server ....
I strongly suggest reading the PostgreSQL docs on backup/restore. Consider setting up PITR in addition to periodic dumps. I'd recommend running PgBarman on the Linux server (archiving to an external drive, NAS, or network share) for point-in-time recovery, plus periodic dumps to the Linux box.
See: http://www.postgresql.org/docs/current/static/backup.html
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 1,330
|
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Northwest Quadrant Mockups</title>
</head>
<body>
<p>Northwest Quadrant mockups</p>
<ul>
<!-- TODO: Add links to new mockup pages here -->
<li><a href="html/homepage_1.html">Homepage 1</a></li>
<li><a href="html/edit_organization_1.html">Edit Organization 1</a></li>
</ul>
</body>
</html>
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 5,660
|
KB, kB or kb may stand for:
Businesses and organizations
Banks
KB Kookmin Bank, South Korea
Kaupthing Bank, Iceland
Komerční banka, Czech Republic
Kasikornbank, Thailand
Karafarin Bank, Iran
Libraries
National Library of Sweden ()
National Library of the Netherlands ()
Sport
Kalix BF, a Swedish bandy club
Kjøbenhavns Boldklub, a sports club, Copenhagen, Denmark
Other businesses and organizations
KB Home, a US house builder
KB Lager, Australia
KB Toys, US
K&B, a New Orleans, Louisiana, US drugstore
Druk Air (IATA code: KB), Bhutan airline
People
Kevin Bartlett (Australian rules footballer) (born 1947)
KB (rapper) (born 1988), Kevin Elijah Burgess
KB Killa Beats (born 1983), Zambian record producer
Science and technology
Biology
Kilo-base pair (kb or kbp), length of D/RNA molecule
Computing
Kilobit (kb), 1,000 bits
Kilobyte (kB), 1,000 bytes
Knowledge base
Microsoft Knowledge Base ID prefix
Vehicles
KB series, International Harvester trucks
Isuzu Faster or KB, a pickup truck
Isuzu D-Max or KB, a pickup truck
NZR KB class, a New Zealand steam locomotive
Other uses in science and technology
Boltzmann constant, k or kB
Base dissociation constant Kb
Ebullioscopic constant Kb, relating molality to boiling point elevation
Kauri-butanol value Kb, a measure of solvent performance
Other uses
King's Buildings, a University of Edinburgh campus
Knowledge Bowl, a quiz competition
Former Knight Companion of the Order of the Bath
Kuala Belait, a town in Brunei
WWKB, or KB Radio 1520
A slang name for marijuana
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 5,623
|
Q: Automated testing in CD What type of tests should be included in continuous integration?
Is it only unit tests or also integration/API, UI?
Or should integration/API and UI tests be included only in continuous deployment/delivery pipeline?
A: This answer will largely depend on what you are trying to get out of your CI system and how long these test take to execute. If it brings your CI agent down for an extended period you loose the benefits of having a CI system in place.
This leaves you with 3 options:
*
*Run when launching a deployment. If your organization has multiple deployment environments (e.g., dev, test, prod) to deploy to you can set these tests to only execute when being promoted to one of the more release ready environments. This will allow you to continue to deploy to a dev or test environment for local testing.
*Run these tests at a scheduled interval (e.g. over night).
*Create a complex parallel build system where you compile and run your unit tests in one stage, and have a dedicated agent run the tests. However, you could run into a situation where your tests are either queued for days or you have to pay several agents.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 8,791
|
Q: How to display dialog after current dialog in react native I hava an array like this
export const activities = [
{
title: "Take the Stairs",
progressOverview: {
notSatisfied: 5,
neutral: 0,
satisfied: 2,
},
},
{
title: "5 Set of Sit-Ups",
progressOverview: {
notSatisfied: 3,
neutral: 2,
satisfied: 5,
},
},
{
title: "3 Sets of Push-Ups",
progressOverview: {
notSatisfied: 7,
neutral: 3,
satisfied: 1,
},
},
{
title: "Weight Lifting",
progressOverview: {
notSatisfied: 3,
neutral: 2,
satisfied: 0,
},
},
]
And. I have a dialog/modal like this
Now I want to display every title in array to show in the activities that dialog (in this pic is abc), it mean when I click save. (It mean next) it will display all activities one by one until last item, and in the last item I want to have other dialog, How can I do that? All item is the same, except the last one
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 8,224
|
The Mother. 2003. Directed by Roger Michell
7:00 PM - 11:59 PM Museum of Modern Art
Where: Museum of Modern Art, 11 W 53rd St, New York, New York, United States, 10019
The Mother. 2003. Directed by Roger Michell Add to my calendar
map-The Mother. 2003. Directed by Roger Michell Get Directions
The Mother. 2003. Great Britain. Directed by Roger Michell. 112 min.
Daniel Craig plays Darren, a charming yet loutish (and married) carpenter who has simultaneous affairs with Paula, a client, and her recently widowed mother, May. Anne Reid is admirably vulnerable as May, a woman who knowingly risks relationships with her children to pursue Darren. Craig brings a unique mixture of arrogance and sensitivity to the film. John Waters named the film one of the 10 best of 2004.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 5,703
|
Primnoella australasiae är en korallart som först beskrevs av Gray 1849. Primnoella australasiae ingår i släktet Primnoella och familjen Primnoidae. Inga underarter finns listade i Catalogue of Life.
Källor
Koralldjur
australasiae
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 9,140
|
Klas-Göran Karlsson (born 1955) is a professor of history at Lund University. Karlsson is an author of books on conflictology and Holocaust studies such as "The Holocaust as a Problem of Historical Culture". One of the focus areas of Karlsson's research is the topic of genocide.
In Crimes Against Humanity under Communist Regimes, Karlsson prefers using the term crimes against humanity to include "the direct mass killings of politically undesirable elements, as well as forced deportations and forced labour." Karlsson acknowledges that the term may be misleading in the sense that the Communist regimes targeted groups of their own citizens, but he considers it useful as a broad legal term which emphasizes attacks on civilian populations and because the offenses demean humanity as a whole.
In 2019, Karlsson published The Modern Thirty-Year War (Det moderna trettioåriga kriget – Europa 1914–1945). The book's comparative perspective with developments outside Europe was praised by reviewers.
Works in English
Whereas Karlsson has authored over 40 books in his native Swedish, only part of his oeuvre has been written in or translated into English. Among these:
Lessons at the limits, in Remembering the Holocaust in Educational Settings; Routledge, 2018.
Holocaust heritage: inquiries into European historical cultures; with Ulf Zander; Sekel, 2004.
Perspectives on the Entangled History of Communism and Nazism: A Comnaz Analysis; with Johan Stenfeldt and Ulf Zander; Lexington Books (Rowman & Littlefield), 2015
Crimes Against Humanity under Communist Regimes; Forum for Living History, 2008
References
20th-century Swedish historians
Historians of the Holocaust
Living people
1955 births
Academic staff of Lund University
21st-century Swedish historians
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 1,446
|
Pleurothallis grobyi es una especie de orquídea de la tribu Epidendreae que pertenece a la familia Orchidaceae.
Distribución y hábitat
Encontrado en el centro y oeste de América en México, Belice, Guatemala, El Salvador, Nicaragua, Costa Rica, Venezuela, Colombia, Ecuador, Perú, Brasil, Guayana Francesa, Surinam y Guyana.
Descripción
Es una planta diminuta con flores de 4 mm que crece con temperaturas frías y cada vez más cálidas, son cespitosas y epífitas que se encuentran en la parte inferior del bosque nuboso, sotobosques y bosques de galería a alturas de entre 60 y 3200 metros de altura sobre el nivel del mar, con una sola hoja apical, sub-oval-orbicular a elíptica, canalizada sobre una base peciolada, tridentada que florece en un racimo apical de 2 a 8 cm de largo, más largo que la hoja, inflorescencia con unas brácteas triangulares y flores que abren sucesivamente y que puede variar en flores y colores.
Taxonomía
Pleurothallis grobyi fue descrita por Bateman ex Lindl. y publicado en Edwards's Botanical Register 21: t. 1797. 1835.
Etimología
Pleurothallis: nombre genérico que deriva de la palabra griega 'pleurothallos', que significa "ramas parecidas a costillas". Esto se refiere a la similitud de las costillas de los tallos de muchas de sus especies.
grobyi: epíteto otorgado en honor de Lord Grey de Groby, orquídologo inglés.
Sinonimia
Epidendrum marginatum Rich. 1792;
Humboldtia crepidophylla (Rchb.f.) Kuntze 1891;
Humboldtia grobyi (Bateman ex Lindl.) Kuntze 1891;
Humboldtia trilineata (Barb.Rodr.) Kuntze 1891;
Humboltia crepidophylla (Rchb. f.) Kuntze 1891;
Humboltia grobyi (Bateman ex Lindl.) Kuntze 1891;
Humboldtia marginalis (Rchb. f.) Kuntze 1891;
Humboltia marginata (Lindl.) Kuntze 1891;
Lepanthes marmorata Barb.Rodr. 1881;
Lepanthes trilineata (Barb.Rodr.) Barb.Rodr 1881;
Pleurothallis arevaloi Schltr. 1924;
Pleurothallis barbosae Schltr. 1921;
Pleurothallis biglandulosa Schltr. 1922;
Pleurothallis choconiana' S. Wats. 1888;
Pleurothallis crepidophylla Rchb. f. 1878;
Pleurothallis ezechiasi Hoehne 1946;
Pleurothallis grobyi var. trilineata (Barb.Rodr.) Cogn. 1896;
Pleurothallis lindleyana Cogn. 1896;
Pleurothallis marginalis Rchb. f. 1855;
Pleurothallis marginata (Rich.)Cogn 1896;
Pleurothallis marmorata (Barb.Rodr.) Cogn. 1896;
Pleurothallis panamensis Schltr. 1921;
Pleurothallis pergracilis Rolfe 1893;
Pleurothallis surinamensis H.Focke 1849;
Pleurothallis trilineata' Barb.Rodr. 1877;
Specklinia biglandulosa (Schltr.) Pridgeon & M.W.Chase 2001;
Specklinia ezechiasi (Hoehne) Luer 2004;
Specklinia marginalis (Rchb.f.) F.Barros 1983 publ. 1984
Specklinia grobyi (Bateman ex Lindl.) F. Barros (1983).
Referencias
Enlaces externos
http://www.orchidspecies.com/pleurgrobyi.htm
grobyi
Flora de México
Flora de América Central
Flora del norte de Sudamérica
Flora de América del Sur continental
Flora de Sudamérica occidental
Flora de Brasil
Flora de la Guayana
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 794
|
Inmate No. NN7687: Inside the Violent, Deadly, State-of-the-Art Prison Bill Cosby Now Calls Home
Filed to:Bill Cosby
Photo: Pennsylvania Department of Corrections (SCI Phoenix)
Less than an hour after his publicist stood on the steps and compared him to Brett Kavanaugh, Tupac and Jesus, comedy legend and convicted rapist (no, the two are not mutually exclusive) Bill Cosby was put in a single cell inside the brand-spanking new correctional facility where staff have been accused of hate crimes and abuse, and at least one inmate has died since the prison opened in July.
Located on 164 acres 35 miles northwest of Philadelphia in Collegeville, Penn., the Phoenix State Correctional Institution (SCI Phoenix) is a sprawling new maximum security facility that costs taxpayers $400 million dollars (coincidentally, not enough to purchase NBC). The new facility boasts a basketball court, a laundry room, chapel and 3,830 beds, 192 of which are dedicated to a female transitional unit focused on reentry and family unification, according to the Pennsylvania Department of Corrections.
There are no reports of prisoners welcoming Cosby by singing: "It's a different woooorld than where you come from!" and prison officials say their long-term goal is to acclimate the 81-year-old to the general population. Although the vast majority of SCI Phoenix cells are double occupancy, the Chicago Sun-Times reports that Cosby received a single cell in the prison where he will likely spend the entirety of the minimum three years he must serve before he is eligible for parole.
While officials have curiously remained mum on whether they will allow Cosby to use his medical degree from Hillman when he practiced as an OB/GYN under the pseudonym Heathcliff Huxtable, the loathsome bartender/pharmacist will be assigned a job inside the prison. According to court documents, Cosby's punishment includes court costs of $43,611.83. Here is a tour of Cosby's new mansion:
Since opening, SCI Phoenix guards have subjected the inmates to abuses including racial slurs, graphic images drawn on photos of loved ones and the desecration of religious items. The Inquirer reported in April that the correctional officers committed a series of "hate crimes:"
Henry Goodelman said his uncle, inmate Jonathan Margoles, found his tefillin — an item used by some observant Jews during prayer — confiscated. Danny Liptrot, a Muslim inmate, said in a letter that his prayer rug and kufi cap were stolen. "They replaced it with a nude picture of a woman, and an athletic supporter," he wrote.
Anthony Wright said in a letter that his watch and several other possessions were missing, and that "swastikas and derogatory images" were drawn on his photos.
"I just think," he wrote, "even an animal deserves to be treated with dignity and respect."
The vast majority of SCI Phoenix's inmates were transferred from the state prison at Graterford, about a mile down the road from the new facility. Between Graterford and SCI Phoenix, at least six inmates have committed suicide inside the walls of the two prisons in 2018 alone.
But suicides aren't the only thing that made Graterford deadly. It is Cosby's new housemates, too. In January, 68-year-old Arthur Phillips was found dead in his Graterford cell after he was convicted of deviate sexual intercourse with a person less than 16 years old. Although Phillips attorney said he did not seem "overly depressed," prison officials ruled the death as a suicide.
In February, 31-year-old Christopher Stephon Gilchrist was mysteriously found dead in his cell at Graterford; Gilchrist was incarcerated after he was convicted of helping two corrections officers plan an attack on another inmate. The officers planned to leave the cell open so Gilchrist could attack an "unruly" prisoner, according to the Daily Local.
Just three days later, another sex offender, Isaac De Jesus, 37, was found "unresponsive" in his cell at Graterford. DeJesus was serving a sentence for indecent assault on a juvenile, according to Pennsylvania's DOC.
Prison guard Mark Baserman was beaten to death by 22-year-old inmate Paul Kendrick in February, according to Fox News, the first correctional officer death in a Pennsylvania state prison since 1979, which also happened at Graterford.
In 2016, Christopher Ganues, a guard at Graterford, was convicted of institutional sexual assault after he forced an inmate at Graterford to have sex with him, according to the Daily News.
All of Graterford's staff now works at SCI Phoenix.
The Root cannot confirm whether or not Cosby asked some inmates to pull their pants up.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 6,410
|
\section{Introduction}
Aerosol, in the form of pollution from human emissions, enters the atmosphere and eventually interacts with a cloud leading to aerosol-cloud interactions (ACI).
As aerosol enters the cloud, a causal chain of events catalyzes.
It begins with aerosol particles activating as cloud droplet nuclei, which increases the number of droplets within the cloud, reducing the mean radius of cloud droplets to redistribute the water vapor, and eventually increasing the cloud's brightness (\cref{fig:causal}) \cite{twomey1984assessment}.
\replaced[id=MS]{Overall, an increase in atmospheric aerosol leads to larger, brighter, longer-lasting clouds that reflect more incoming sunlight.}{An increase in brightness leads to the cloud reflecting more incoming sunlight.}
ACI are thus a net cooling process and offset some fraction of warming due to rising levels of CO\(_{2}\).
\replaced[id=MS]{The strength of the effect is however dependent on the local environment surrounding the cloud. ACI remain one of the most uncertain effects in our current climate models, as current models are limited in their ability to simulate ACI with such environmental heterogeneity \cite{ipcc-ar5-wg1-ch7,ipcc-ar6-wg1-spm}.}{ACI are one of the most uncertain effects in our current climate models due in part to the strength of their effects being dependent on the local environment surrounding the cloud. Our current climate models are limited in their ability to simulate ACI with such environmental heterogeneity \cite{ipcc-ar5-wg1-ch7,ipcc-ar6-wg1-spm}.}
Climate models can only approximate ACI given their low spatial resolution and limited parameterizations, often dependent on only a few environmental parameters, such as the relative humidity within a grid cell.
These factors lead to increased uncertainty in future projections.
Currently, state-of-the-art climate models estimate that the range of cooling due to ACI may offset 0\%-50\% of the warming due to greenhouse gas emissions.
\replaced[id=MS]{This work uses causal machine learning to estimate ACI from satellite observations, by reframing the problem as a treatment (aerosol) and outcome (change in droplet radius). We predict the causal effect of aerosol on clouds and provide uncertainty bounds that we compare to the parameterizations of climate model ACI.}{This work uses causal machine learning to estimate ACI from satellite observations. We provide uncertainty bounds on these estimates and compare them to the parameterizations of climate model ACI.}
\replaced[id=MS]{We consider uncertainty arising from violations of two assumptions: positivity (or overlap) and unconfoundedness (or no hidden confounding). Positivity violations are due to insufficient representation within the data for all treatment levels, such as "treating" cloud with aerosol. Unmeasured confounding are unobserved factors which influence both the treatment and outcomes, such as humidity causing aerosol swelling and altering cloud properties.}{Uncertainty in causal effect estimates can arise from violations of the positivity assumption, insufficient representation within the data for all treatment levels (such as "treating" cloud with aerosol), and from unmeasured confounding, when unobserved factors influence both the treatment and outcomes (such as humidity causing aerosol swelling and altering cloud properties).}
To better understand these individual sources of uncertainty, we use Overcast \cite{jesson2022scalable}, a prime example of the needs of a community such as ACI leading to methodological contributions in machine learning.
\deleted[id=MS]{Overcast estimates the effects of a continuous valued process and can account for the influence of heterogeneity, positivity violations, and confounding on the outcome.
In the real world, rarely are processes either `on' or `off', therefore creating a model that can encompass a range of effects depending on the magnitude of the treatment is vital to accurately model these processes.}
Compared to prior work such as \cite{jesson2021using}, we consider aerosol optical depth (AOD), our proxy for aerosol concentration, as a continuous treatment rather than discrete and perform an uncertainty-aware sensitivity analysis to study the consequences of possible violations of positivity and unconfoundedness.
\begin{figure}[h]
\centering
\subfigure[Simple Causal Graph of ACI]{
\includegraphics[width=.33\linewidth]{simple_causal.png}
\label{fig:causal}
}
\subfigure[The Southeast Pacific]{
\includegraphics[width=.3\linewidth]{SEP.jpg}
\label{fig:SEP}
}
\subfigure[The South Atlantic]{
\includegraphics[width=.3\linewidth]{SA.jpg}
\label{fig:SA}
}
\caption{The causal graph underlining our knowledge of ACI and satellite imagery of the two regions analyzed, chosen due to their unique aerosol-cloud interactions and breadth of past studies to pull knowledge from.}
\label{fig:causal_regions}
\end{figure}
\section{Methods}
\added[id=MS]{Following \cite{jesson2022scalable},} we use the potential outcomes framework to estimate the effect of a continuous treatment $\mathrm{T} \in \mathcal{T}$ (aerosol), on outcomes of interest $\mathrm{Y} \in \mathcal{Y}$ (cloud property), for a unit described by covariates $\mathbf{X} \in \mathcal{X}$ (environmental information) as shown in \cref{fig:causal} \cite{rubinEstimatingcausaleffects1974, rubinRandomizationAnalysisExperimental1980, sekhonNeymanRubinModel2009, splawa-neymanApplicationProbabilityTheory1990}.
We call a potential outcome and denote by $\mathrm{Y}_{\mathrm{t}}$ what the outcome would have been if the treatment were $\mathrm{t}$.
\added[id=MS]{The covariates considered are relative humidity at 900, 850 and 700 millibar, sea surface temperature, vertical motion at 500 millibars, lower tropospheric stability, and effective inversion strength. The treatment is aerosol optical depth (AOD), a proxy for aerosol concentration. The outcome considered is the cloud droplet size ($r_e$).}
To estimate the treatment-effect, we study the conditional average potential outcome (CAPO) and the average potential outcome (APO)
\[
\text{CAPO} = \mu(\mathbf{x}, \mathrm{t}) \coloneqq \mathbb{E}\left[\mathrm{Y}_{\mathrm{t}} \mid \mathbf{X} = \mathbf{x}\right],
\quad \text{and} \quad
\text{APO} = \mu(\mathrm{t}) \coloneqq \mathbb{E}\left[\mu(\mathbf{X}, \mathrm{t})\right],
\]
which can be identified from the observational distribution $P(\mathbf{X}, \mathrm{T}, \mathrm{Y})$ using
\[
\tilde{\mu}(\mathbf{x}, \mathrm{t}) \coloneqq \mathbb{E}\left[\mathrm{Y} \mid \mathrm{T}=\mathrm{t}, \mathbf{X} = \mathbf{x}\right]
\quad \text{and} \quad
\tilde{\mu}(\mathrm{t}) \coloneqq \mathbb{E}\left[\tilde{\mu}(\mathbf{X}, \mathrm{t})\right],
\]
and further assumptions (unconfoundedness, positivity, no-interference and consistency).
\deleted[id=MS]{Identifying the CAPO using finite observational satellite data depends on assumptions untestable from the data and challenged by the continuous treatment regime.}
\replaced[id=MS]{Here, we study the robustness of treatment-effect estimates to positivity and unconfoundedness violations (see \cref{app:assumptions} for more detail).}{Here, we study the robustness of treatment-effect estimates to violations of the unconfoundedness and the positivity assumptions.}
We compute uncertainty bounds corresponding to user-specified relaxations of these assumptions.
The parameter $\Lambda$, for example, is set by the user to explain an assumed level of unmeasured confounding \cite{jesson2022scalable, jessonQuantifyingIgnoranceIndividualLevel2021, jessonIdentifyingcausaleffectinference2020}.
Some confounding influences are impossible to measure directly with satellites, such as humidity causing aerosol swelling and altering cloud properties, and the parameter $\Lambda$ can be used to encode an expert's belief in the influence of such confounders.
\deleted[id=MS]{Moreover, we rely on a proxy for aerosol, the aerosol optical depth (AOD), which has consequences distinct from both unmeasured confounding and positivity.}
We use daily mean, 1\(^{\circ}\) x 1\(^{\circ}\) of satellite observations in order to homogenize the data from the southeast Pacific and south Atlantic (Figures \ref{fig:SEP} and \ref{fig:SA}).
Mean droplet radius ($r_e$) from the MODIS instrument is used as our outcome for all experiments shown within.
We employ aerosol optical depth from MERRA-2 to approximate the concentration of aerosol.
Our environmental confounders are the relative humidity at 900, 850 and 700 millibars, the stability of the atmosphere, the sea surface temperature, and the vertical motion at 500 mb, all also from MERRA-2. \added[id=MS]{For more detail about data and implementation, please refer to \cref{app:data} and \cref{app:implementation}.}
\deleted[id=MS]{The models are neural-network architectures with two components: a feature extractor $\phi(\mathrm{x}; \mathbf{\theta})$ and a density estimator $f(\phi, \mathrm{t}; {\theta})$, \added[id=MS]{represented in \cref{app:model-arch}.}
The covariates $\mathbf{x}$ are given as input to the feature extractor, whose output is concatenated with the treatment $\mathrm{t}$ and given as input to the density estimator which outputs a Gaussian mixture density, $p(\mathrm{y} \mid \mathrm{t}, \mathbf{x}, {\theta})$, from which we can sample.
The feature extractor uses attention mechanisms to model the spatio-temporal correlations between the covariates on a given day using the geographical coordinates of the observations.
\added[id=MS]{Further implementation detail are given in \cref{app:implementation}.}}
\section{Results}
\subsection{Deducing reasonable treatment-effect bounds using domain knowledge}
Unlike past studies which only crudely estimate an uncertainty range due to quantifiable effects, we are able to derive confidence intervals dependent on the influence of confounding by varying $\Lambda$.
Since it is impossible to know the strength of the confounding effect from observed data alone, we propose a method to select a reasonable $\Lambda$ by contrasting two geographical regions.
We contrast the South-East Pacific and the South Atlantic because these regions have different environmental confounders of ACI, for example aerosol type, aerosol hygroscopicity, aerosol size.
These are important confounders, but are unfortunately not included in the available data.
So we select the parameter $\Lambda$ for the Pacific region such that the treatment effect bounds for the Pacific region cover the effect bounds for the Atlantic region under the assumption of no hidden confounding ($\Lambda \to 1$) for the Atlantic region, as shown in \cref{fig:lambda}.
Setting $\Lambda$ to 1.07 gives bounds for the pacific region that reasonably account for the potential bias induced by the unmodeled confounders.
While a larger $\Lambda$ could still be sensible due to other drivers of confounding, domain knowledge informs us that these are the main missing physical mechanisms.
\subsection{Evaluating climate models using machine learning}
As we now have a possible range of ACI derived using the real, observed outcomes, we can judge how well climate models recreate this observed trend by seeing if their responses lie within our derived interval (\cref{fig:esm}).
We find that the Canadian model CanESM5 simulates ACI better than the UK models HadGEM3-GC3-LL and UKESM1-0-LL.
Our trained machine learning model not only uses the real, observed relationships to derive the magnitude of the effect, but can consider the environmental context and confounding influences to derive real, quantifiable bounds of uncertainty.
Therefore, by using the curves found by Overcast as the true response, we know those models which lie outside of our bounds found by contrasting different regions are likely unphysical and highly unlikely to occur in the real-world.
CanESM5 currently estimates the total cooling effect due to ACI to offset approximately half of the warming due to greenhouse gases; based on our results, we would say it is likely that this estimate is closest to the true value observed on Earth \cite{smith2020effective}.
\begin{figure}[htbp]
\centering
\subfigure[Choosing $\Lambda$ using two geographical regions]{
\includegraphics[width=.4\linewidth]{apo_tr_lrp-vs-lra_lrp_re.pdf}
\label{fig:lambda}
}
\subfigure[Comparison with ESMs in the Pacific]{
\includegraphics[width=.4\linewidth]{ESM.pdf}
\label{fig:esm}
}
\caption{Plausible range of effects of aerosol (AOD) on mean droplet radius ($r_e$).}
\label{fig:results}
\end{figure}
\section{Discussion and conclusions}
\subsection{Machine learning's place in climate model verification}
In this work, we show that machine learning methods offer viable ways to objectively judge how well global climate models reproduce climate processes such as the effect of aerosol on mean droplet radius.
A drawback of historical studies which utilize satellite observations is their inability to quantify how the surrounding environment may affect the magnitude of the aerosol-cloud interactions.
Overcast accounts for such contextual confounding and communicates bounds on the treatment effect due to an expert-informed influence of hidden-confounding.
Utilizing this method gives us insight into whether climate models reproduce the observed relationships between AOD and $r_e$.
Climate models currently only reasonably recreate large scale processes that can be explicitly calculated, leaving processes like aerosol-cloud interactions, which occur on scales smaller than the grid scale, poorly parameterized and approximated.
In order to improve our climate models, we must understand in more relatable terms how well they are doing, such as by comparing their outcomes to those from observations.
Machine learning provides not only a way to judge these outcomes, but the relationships learned by Overcast and similar models could in the future be fine-tuned to replace our current parameterizations \cite{gettelman2021machine}.
\subsection{Collaboration across domains vs. purely data driven}
While different sources of confounding due to regional differences alter the outcomes, the choice of which environmental factors are the main sources of confounding can also be investigated using Overcast.
We perform two experiments, with and without relative humidity at 900, 850 and 700 millibars, to derive varying outcome shapes and fit $\Lambda$ to both dose-response curves.
When $\Lambda$ is set to 1.04, both curves are captured by the bounds of uncertainty, allowing us to view how the response may vary within those bounds due to meteorological uncertainty rather than regional uncertainty, where $\Lambda = 1.07$ was required (\cref{subfig:lrprh_scaled_fitted}).
In the absence of ground truth, purely data-driven techniques cannot decide between the model with and without relative humidity, but as domain knowledge is brought in, it is known that the curve with humidity included is the true response curve (\cref{subfig:lrprh_unscaled}).
Purely data-driven approaches may not be the most appropriate for studying climatological processes such as aerosol-cloud interactions as domain knowledge is essential to select the correct inputs and verify the outcomes.
The most robust model arises from combining data and theory, bringing together experts in machine learning and climate processes.
\begin{figure}[htbp]
\centering
\subfigure[Scaled, with appropriate $\Lambda$]{
\includegraphics[width=.4\linewidth]{apo_tr_lrp-vs-lrprh_lrprh_re.pdf}
\label{subfig:lrprh_scaled_fitted}
}
\subfigure[Unscaled]{
\includegraphics[width=.4\linewidth]{apo_tr_lrp-vs-lrprh_re.pdf}
\label{subfig:lrprh_unscaled}
}
\caption{Plausible range of effects when omitting relative humidity from the covariates.}
\label{fig:lrprh}
\end{figure}
\subsection{Limitations and future works}
This work uses aerosol optical depth as a proxy for aerosol concentration, which could bias treatment-effect estimates. For example, it is known that bias can arise from measurement-error in the treatment \cite{measurement-error}. Further, we rely on low resolution data that does not perfectly capture the microphysical processes. Future work could consider different assumptions on the underlying causal model, and attempt to include other aerosol properties like size, hygroscopicity and type.
\section{Acknowledgements}
This project was supported by the European Union's Horizon 2020 research and innovation program under grant agreement No 821205 (FORCeS), the European Research Council (ERC) project constRaining the EffeCts of Aerosols on Precipitation (RECAP) under the European Union's Horizon 2020 research and innovation programme with grant agreement no. 724602, and the Turing Institute post-doctoral enrichment award.
\printbibliography
\newpage
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 3,768
|
This entry was posted in Oil Painting and tagged Art, Culture, Education, Inspiration, Oil Painting Instruction, Pictures. Bookmark the permalink.
I was thinking of you just last week with all that Ophelia talk. I hope everyone is safe.
So much with so little. Amazing.
Clouds have a lot of personality!
We look up at the clouds and associate them with images.
You think perhaps the clouds do the same with us?
It appears you have turned the comments off on your latest YT video… Good for you! There are other things to do than to reply to people like me! Today's video is a good old Liam painting. Thanks again Liam for all that you have done on this blog and your video channel. It has really helped me over the years to stay grounded and apply good painterly concepts. You Rawk Mate!
You are always such a gentleman Liam! I understand how much work it takes to maintain the many websites you maintain. Family, work and everyday responsibilities do take their toll on us; taking us away from our passion for painting. You do provide us with so much by showing us your style of painting on YT. We all learn tricks and get ideas watching you paint.
I am really excited and happy about your son's success! You must be so proud of him. He is an artist too. Someday, I bet he will follow in your footsteps. Musicians usually go from music to art. It seems like it is a natural progression. Thanks again Mate for all that you do for us. It is deeply appreciated.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 7,306
|
Showing posts by Moira Feil.
Beyond Supply Risks: The Conflict Potential of Natural Resources
Guest Contributor // August 8, 2011 // By Lukas Rüttinger & Moira Feil
While the public debate about resource conflicts focuses on the risk of supply disruptions for developed countries, the potentially more risky types of resource conflict are usually ignored. As part of a two-year research project on behalf of the German Federal Environment Agency, adelphi and the Wuppertal Institute for Climate, Energy, and Environment have analyzed the risks of international conflict linked to natural resources in a series of reports titled Beyond Supply Risks – The Conflict Potential of Natural Resources.
Resource extraction, transportation, and processing can create considerable crises and increase the risk of conflicts in producing and transit countries. This phenomenon – widely referred to as the "resource curse" – impacts consuming countries only if it leads to shortages and higher prices. However, in the producing and transit countries it can have much wider destabilizing effects – from increasing corruption to large-scale violent conflict. In addition, the extraction, processing, and transportation of resources often create serious environmental risks. Overexploitation, pollution, and the degradation of ecosystems often directly affect the livelihoods of local communities, which can increase the potential for conflict.
The eight reports that comprise Beyond Supply Risks explore plausible scenarios over the next two decades, focusing on four case studies: copper and cobalt in the Democratic Republic of Congo; the Nabucco natural gas pipeline project across Southern Europe and Turkey; lithium in Bolivia; and rare earth minerals in China.
Lithium in Bolivia
Bolivia possesses the world's largest known lithium deposits, a potentially important resource for the development of electric vehicles. While the development of Bolivia's lithium reserves could provide major economic benefits for one of the poorest countries in Latin America, our analysis identifies two main potential risks of conflict.
First, the environmental consequences of developing industrial-scale lithium production might have negative effects on the livelihoods of the local population. The local population in the lithium-rich department of Potosí has shown that it is capable of organizing itself effectively in defense of its interests, and past resource conflicts have turned violent, making a conflict-sensitive approach all the more important.
Second, the Bolivian economy is largely dependent on natural resources, and consequently is susceptible to price shocks. At present, this risk is primarily associated with natural gas. But lithium production, if developed, might be subject to the same dynamics, which could potentially destabilize the political system.
For consuming countries, these conflicts threaten supplies of lithium only if local protests or broader destabilization were produce bottlenecks in the supply chain.
Rare Earths and China
Like lithium, rare earths are likewise essential for some new technologies. China's well publicized monopoly on 97 percent of the global production spurred a heated debate on the security of supply of strategic minerals. While our case study identifies supply risks for consuming countries, it also outlines some of the conflict risks China might face internally.
First, local populations could protest against the severe ecological impact of rare earth mining and production. In addition, conflicts might arise if those who profit from economic development (entrepreneurs or regional power-holders) undermine the traditional centralized party structures and expand their own influence.
International conflicts over access to Chinese rare earth resources, while they dominate the headlines, do not appear to be the dominant risk. Instead, internal political tensions could result in a weakened China that is not able to exploit its monopoly position for foreign policy gains. Or the government could enter into multilateral agreements and thus avoid a confrontational approach towards consumer nations.
Ultimately, the actual rate of diffusion of environmental technologies and the development of new technologies remain the key factors in determining whether relative shortages in global supply of rare earths will in fact occur. If industrialized nations and emerging economies commit to the same technologies to attain climate policy goals, international resource governance and coordinated promotion of (environmental) technology will also play a role in preventing conflict and crisis over rare earths.
The series concludes with five recommendations to mitigate the risks of future resource conflicts:
Introduce systematic policy impact assessments to understand how policy goals and strategies, especially in regard to climate and environmental policy, interact with resource conflict risks.
Increase the transparency of raw material markets and value creation chains to prevent extreme fluctuations in prices and improve information on markets, origins, and individual players.
Improve the coherence of raw material policy by linking raw material policies with security, environmental, and development policies.
Demand and promote corporate social responsibility along the whole value chain.
Increase environmental and social sustainability as a means of strengthening crisis and conflict prevention by systematically taking into account social and conflict-related aspects in the resource sector.
However, none of these strategies alone would be capable of mitigating all the risks of future resource conflicts. But together they represent a methodology that, with intense coordination among the key players, could make a far-reaching contribution to reducing risk and preventing international conflict over the long term.
The individual reports from the project can be downloaded here:
Conflict Risks (GERMAN only)
Supply and demand (GERMAN only)
Case Study: Nabucco Pipeline (GERMAN only)
Case Study: Congo
Case Study: Bolivia
Case Study: China
Conflict Resolution Strategies (GERMAN only)
Summary and Recommendations
Lukas Ruettinger is a project manager for adelphi, mainly focusing on the fields of conflict analysis and peacebuilding as well as resources and governance. Moira Feil is a senior project manager for adelphi and has participated in more than 30 projects with various partners and clients on natural resource links to crises, conflicts and peacebuilding, and corporate responsibility.
Sources: Government Accounting Office.
Photo Credit: "Potosí: miners in darkness," courtesy of flickr user Olmovich.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 6,832
|
Columbus Blue Jackets News: ACQUIRE D ADAM BOQVIST, 2021 FIRST AND SECOND ROUND PICKS (12TH, 44TH), 2022 FIRST ROUND PICK FROM CHICAGO FOR D SETH JONES, 2021 FIRST ROUND PICK (32ND) AND 2022 SIXTH ROUND PICK
Posted on 6 months ago by Gary Kid Gulliver
Adam Boqvist
COLUMBUS, OHIO – The Columbus Blue Jackets have acquired defenseman Adam Boqvist, first round picks in the 2021 (12th overall) and 2022 NHL Drafts and a second round pick in the 2021 Draft (44th overall) from the Chicago Blackhawks in exchange for defenseman Seth Jones, Columbus' first round pick acquired from Tampa Bay (32nd overall) in the 2021 NHL Draft and Columbus' sixth round pick in the 2022 NHL Draft, club General Manager Jarmo Kekalainen announced tonight.
Boqvist, 20, was selected by Chicago in the first round, eighth overall, in the 2018 NHL Draft. He has registered six goals and 23 assists for 29 points with 20 penalty minutes in 76 career NHL games with the Blackhawks. In 2020-21, he notched 2-14-16 and 14 penalty minutes in 35 outings. He made his NHL debut in 2019-20, tallying 4-9-13 and six penalty minutes in 41 games with the Blackhawks. He also added 1-5-6 and 10 penalty minutes in 15 outings with the American Hockey League's Rockford IceHogs.
The 6-0, 179-pound native of Falun, Sweden registered 20-40-60, 12 penalty minutes and nine power play tallies in 54 games with the Ontario Hockey League's London Knights in 2018-19. He was named to the OHL's Second All-Star Team after finishing second among leauge defensemen in goals, third-T in power play goals, sixth in points and 11th in assists.
The blueliner also played in the Swedish Hockey League, collecting 0-1-1 in 15 games with Brynas IF in 2017-18. He racked up 18-22-40 in 43 contests with Brynas' junior team during that campaign.
Boqvist has represented Sweden at several international tournaments. He collected 1-3-4 in five games with the country at the 2019 IIHF World Juniors. He was also named Best Defenseman at the 2018 IIHF U18 World Championships and helped the country capture bronze after leading tournament blueliners in goals and points with 3-3-6 in six outings.
FULL ADAM BOQVIST STATS
Jones, 26, has posted 65-221-286 and 194 penalty minutes in 580 career games with the Blue Jackets and Nashville Predators since making his NHL debut in 2013-14. The Arlington, Texas native, who joined the club in a trade from the Predators on Jan. 6, 2016, notched 50-173-223 in 381 career contests with the Blue Jackets from 2016-21. He owns the franchise records for defensemen in career assists and points and ranks second in goals. The 6-4, 213-pound blueliner posted 5-23-28 while playing in all 56 games for the Blue Jackets in 2020-21. He was originally selected by Nashville fourth overall at the 2013 NHL Draft.
Seth Jones (megasportsnews.com photo)
FULL SETH JONES STATS
‐ CBJ ‐
Full and partial season tickets for the Blue Jackets' 2021-22 campaign are currently available for purchase. Information on all ticketing options can be obtained by calling (800) NHL‐COLS or by visiting BlueJackets.com.
COURTESY BLUE JACKETS COMMUNICATIONS
Gary Kid Gulliver
View all posts byGary Kid Gulliver | Website
Posted in Columbus Blue Jackets
Previous article 2021 Big Ten Football Media Days Recap – July 23
Next article Columbus Blue Jackets News: ACQUIRE D JAKE BEAN FROM CAROLINA FOR 2021 SECOND ROUND PICK
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 6,434
|
An Apple a Day with Guest DT Ted!
Ted is back today, guest designing for So Suzy Stamps so a big warm welcome to him!
I stamped the Basket of Apples image and had my buddy, Scan-N-Cut, cut it out. I wanted it to take center stage, so I popped it up on dimensionals. I used the Apple Branch image to be the background, and then topped off the entire center panel with the Apple a Day sentiment before surrounding it with some blue Copic (to emulate the sky).
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 7,008
|
You yearn so much
you could be a yacht.
Your mind has already
set sail. It takes a few days
to arrive
at island pace,
but soon you are barefoot
on the sand,
the slim waves testing
your feet
like health professionals.
You toe shells, sea glass, and odd things
that have drifted for years
washed up here.
You drop your towel
and step out of
your togs, ungainly,
your right foot, then
the sand
There is no discernible
in temperature.
You breaststroke in
the lazy blue.
A guy passing in a rowboat
says, 'Beautiful, isn't it?'
And it is. Your body
afloat in salt
as if cured.
Photo by Robert Cross
James Brown's poetry books are Go Round Power Please (winner of the Jessie Mackay Best First Book of Poetry Award), Lemon, Favourite Monsters, The Year of the Bicycle, Warm Auditorium, and Floods Another Chamber, all published by Victoria University Press. He is also the author behind the useful, non-fiction booklet Instructions for Poetry Readings, and in 2005 edited The Nature of Things: Poems from the New Zealand Landscape. He has been a finalist in the New Zealand Book Awards three times.
James convenes the Poetry workshop at the International Institute of Modern Letters at Victoria University of Wellington.
Brown comments: 'The poem is an attempt to capture the awkwardness and pleasure of skinny dipping. I used the second person to encourage the reader to identify as the protagonist, who also has a hidden medical condition. So the poem is kind of about our bodies, how we relate to them and what they do and don't show about us.
'I put a lot of thought into the line-breaks. I wanted to slow the reader down to island pace. At times I wanted the reading to be smooth, even languid, at other times awkward.
'Waiheke Island is a 45 minute ferry trip from downtown Auckland. It is home to alternative life-stylers, retirees, and wealthy people. Waiheke (cascading waters) was the name of a stream where sailors would stop for fresh water. When I wrote the poem, I also read that the stream was thought to have healing properties, but I can't find anywhere saying that now.'
James's Victoria University Press author page
New Zealand Book Council profile
New Zealand Electronic Poetry Centre author page
'Hindenburg': a transcribed poem
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 3,529
|
Dropbox - articles.txt - archive.fo - Eine wunderbare Wahl!
Working at Larry J Rappoport in Philadelphia, PA; find Larry J Rappoport jobs and employment on Jobs2Careers Local - your Philadelphia, PA career guide.
65 EINE ALTERNATIVE ANNÄHERUNGSWEISE: GEDANKEN ZUM PROBLEM DES HANDELSRECHTS IN DER RÖMISCHEN WELT * András FÖLDI (Université Eötvös Loránd de Budapest) 1.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 9,902
|
Die Polish International 1994 im Badminton fanden vom 1. bis zum 4. April 1994 in Kielce statt.
Medaillengewinner
Weblinks
http://tournamentsoftware.com/sport/tournament.aspx?id=59AE97B1-0C92-4040-96DA-1B6DF5AF1CA0
1994
Badminton 1994
Sportveranstaltung in Kielce
EBU Circuit 1993/94
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 7,881
|
François Poncher, né en 1480 à Tours et mort le , est un ecclésiastique français, évêque de Paris.
Biographie
François Poncher est le fils de Louis Poncher et de Robine Le Gendre.
Il fut abbé commendataire de Bourgueil et de Saint-Pierre-le-Fosseux, conseiller clerc au Parlement le , abbé de Saint-Maur. Il succède à son oncle, Étienne Poncher, à l'évêché de Paris en 1519. En 1523, il consacre l'église de Lésigny, bâtie à l'instigation de son père.
Il s'était d'abord fait connaître comme un simoniaque en employant jusqu'à des falsifications de titres pour se procurer la commende de l'abbaye de Saint-Benoît-sur-Loire, qu'il n'eut point parce que le chancelier Duprat était son concurrent. On l'accusa d'avoir intrigué en Espagne, où il aurait cherché à prolonger la prison du roi ; et par ses cabales il aurait tenté de faire ôter la régence à la duchesse d'Angoulême. Ses trames odieuses ne furent découvertes pleinement qu'en 1529. Il fut enfermé au château de Vincennes, où il mourut en 1532, pendant que la cour se disputait avec Rome sur la qualité de ceux qui devaient le juger. On a de lui des Commentaires sur le droit civil.
Références
Source
Paul Belouino, Dictionnaire général et complet des persécutions souffertes par l'église
Évêque de Paris
Évêque catholique du XVIe siècle
Abbé commendataire du XVIe siècle
Naissance en 1480
Naissance à Tours
Décès en septembre 1532
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 8,730
|
daily mood tracker a personality tests galore chart.
available on activity mood monitoring chart.
left daily mood tracking chart printable tracker.
now mood tracker chart pdf.
printable mental health tracker for your bullet journal mood chart.
printable mood tracker chart beam charts.
with depression mood tracker chart pdf.
click to enlarge mood monitoring chart bipolar.
daily mood chart and emotions simple monitoring.
sample mood chart simple tracker.
configure mood captions thumb daily monitoring chart.
mood tracking app paves way for pocket therapy tracker chart pdf.
like this item mood tracker chart pdf.
mood tracker one of the most popular and highest rated apps daily chart.
the bullet journal mood tracker charts depression chart.
laboratory monitoring when prescribing psychotropics psych central professional depression mood tracker chart.
arthritis 2 simple mood tracker chart.
screenshot of the mood tracker simple chart.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 4,147
|
\section{Introduction}
Let $G$ be a connected reductive algebraic group defined over a number field $F$.
In \cite{Arthur84} and \cite{Arthur89}, Arthur
gives a conjectural description of the discrete spectrum of $G$ by introducing at each place $v$
of $F$ a set of parameters $\Psi_v(G)$, that should parameterize all the unitary representations of $G(F_v)$ that are of interest for global applications.
More precisely, Arthur conjectured that attached to every parameter $\psi_v\in \Psi_v(G)$
we should have a finite set $\Pi_{\psi_v}(G(F_v))$, called an $A$-packet,
of irreducible representations of $G(F_v)$,
uniquely
characterized
by the following properties:
\begin{itemize}
\item $\Pi_{\psi_v}(G(F_v))$ consists of unitary representations.
\item The parameter $\psi_v$ corresponds to a unique $L$-parameter $\varphi_{\psi_v}$ and $\Pi_{\psi_v}(G(F_v))$ contains the $L$-packet associated to
$\varphi_{\psi_v}$.
\item $\Pi_{\psi_v}(G(F_v))$ is the support of a stable virtual character distribution on $G(F_v)$.
\item $\Pi_{\psi_v}(G(F_v))$ verifies the ordinary and twisted spectral transfer identities predicted by the theory of endoscopy.
\end{itemize}
Furthermore, any representation occurring in the discrete spectrum of square integrable automorphic representations of $G$, should be a restricted product over all places of representations in the corresponding $A$-packets
In the case when $G$ is a real reductive algebraic group, Adams, Barbasch and Vogan proposed in \cite{ABV} a candidate for an $A$-packet, proving in the process
all of the predicted properties with the exception of the twisted endoscopic identity and unitarity.
The packets in \cite{ABV}, that we call micro-packets or ABV-packets,
are defined by means of sophisticated geometrical methods.
As explained in the introduction of \cite{ABV}, the
inspiration behind their construction
comes from the combination of ideas of Langlands and Shelstad (concerning dual groups and endoscopy) with those of Kazhdan and
Lusztig (concerning the fine structure of irreducible representations), to describe the representations of $G(\mathbb{R})$ in terms of an appropriate geometry on an $L$-group.
The geometric methods are
remarkable, but they have the constraint
of being extremely difficult to calculate in practice.
Without considering some exceptions, like for example ABV-packets attached to tempered Arthur parameters (see Section 7.1 below) or to principal unipotent Arthur parameters (See Chapter 7 \cite{ABV} and Section 7.2 below),
we cannot identify the members of an ABV-packet in any known classification
(in the Langlands classification for example).
The difficulty
comes from the central role played by characteristic cycles in their construction.
This cycles are geometric invariants
that can be understood as a way to measure how far a constructible
sheaf is from a local system.
In the present article, relying on work of Kashiwara-Schapira
and Schmid-Vilonen, we describe
the behaviour of characteristic cycles with respect to the operation of geometric induction, the
geometric counterpart of taking parabolic or cohomological induction in representation theory.
By doing this, we are able to describe the characteristic cycle associated to an induced representation, in terms of the characteristic cycle of the representation being induced
(see Proposition \ref{prop:ccLG} below). Before continuing with a more detailed description
on the behaviour of characteristic cycles under induction,
let us mention some consequences of it.
As a first application we have
the proof that the cohomology packets defined by Adams and Johnson in \cite{Adams-Johnson} are micro-packets.
In more detail, Adams and Johnson proposed in \cite{Adams-Johnson} a candidate for an $A$-packet
by attaching to any member in a particular family of Arthur parameters (see points (AJ1), (AJ2) and (AJ3)
Section 7.3),
a packet consisting of
representations cohomologically induced from unitary characters.
Now,
from the behaviour of characteristic cycles under induction,
the description of the ABV-packets corresponding to any Arthur parameter
in the family studied in \cite{Adams-Johnson},
reduces to the description of ABV-packets corresponding to essentially unipotent Arthur parameters (see Section 7.2),
and from this reduction
we prove in Theorem \ref{theo:ABV-AJ}, that the cohomological constructions of \cite{Adams-Johnson} are particular cases of the ones in \cite{ABV}. It is important to point out that the equality between Adams-Johnson and ABV-packets is known to experts, but to my knowledge no proof of it can be found in the literature. Let us also say that from this equality
and the proof in \cite{AMR} that for classical groups
the packets defined in \cite{Adams-Johnson} are $A$-packets (\cite{Arthur}),
we conclude that in the
framework of \cite{Adams-Johnson} and for classical groups, the three constructions of $A$-packets coincide.
As a second application we can mention that, an important step in the proof that for classical groups the $A$-packets introduced in \cite{Arthur} are ABV-packets (work in progress with Paul Mezo),
is the description of the ABV-packets for the general linear group.
The understanding of the behaviour of characteristic cycles under induction
will show to be important in the proof that for the general linear group,
ABV-packets are Langlands packets, that is, they consist of a single representation
Let us give now a quick overview on
how geometric induction affects characteristic cycles.
We begin by introducing the geometric induction functor.
Suppose $G$ is a connected reductive complex algebraic group defined over $\mathbb{R}$
with Lie algebra $\mathfrak{g}$.
Denote by $K$ the complexification in $G$ of some maximal
compact subgroup of $G(\mathbb{R})$.
Write $X_G$ for the flag variety of $G$,
and suppose $Q$ is a parabolic subgroup of $G$ with
Levi decomposition $Q=LN$.
Consider the fibration $X_G\rightarrow G/Q$. Its fiber over $Q$
can be identified with the flag variety $X_L$ of $L$. We denote the inclusion of that fiber in
$X_G$ by
\begin{align}\label{eq:mapvarieties}
\iota:X_L&\longrightarrow X_G.
\end{align}
Let $D_{c}^{b}(X_G,K)$ be the
$K$-equivariant bounded derived category of sheaves of complex vector spaces on $X_G$ having cohomology sheaves constructible with respect to an algebraic stratification of $X_G$. Living inside this category we have the subcategory
$\mathcal{P}(X_G,K)$ of $K$-equivariant perverse sheaves on $X_G$.
Set $\mathcal{D}_{X_G}$ to be the sheaf of algebraic differential operators on $X_G$. The Riemann-Hilbert correspondence (see Theorem 7.2.1\cite{Hotta} and Theorem 7.2.5 \cite{Hotta}) defines an equivalence of categories between $\mathcal{P}(X_G,K)$ and the category
$\mathcal{D}(X_G,K)$
of $K$-equivariant
$\mathcal{D}_{X_{G}}$-modules on $X_G$.
Now, write $\mathcal{M}(\mathfrak{g},K)$ for the category of $(\mathfrak{g}, K)$-modules of $G$, and $\mathcal{M}(\mathfrak{g},K, I_{X_G})$ for the subcategory of $(\mathfrak{g}, K)$-modules of $G$ annihilated by the kernel $I_{X_G}$ of the operator representation (see equations (\ref{eq:operatorrepresentation}) and (\ref{eq:keroperatorrepresentation}) below).
The categories $\mathcal{M}(\mathfrak{g},K, I_{X_G})$ and $\mathcal{D}(X_G,K)$ are identified through the Beilinson-Bernstein correspondence (\cite{BB}),
and
composing this functor with the Riemann-Hilbert correspondence we obtain the equivalence of categories:
\begin{align}\label{eq:rhbb}
\Phi_{X_G}:\mathcal{M}(\mathfrak{g},K,I_{X_G}) \xrightarrow{\sim} \mathcal{P}(X_G,K)
\end{align}
and consequently a bijection between the corresponding Grothendieck groups
$K\mathcal{M}(\mathfrak{g},K,I_{X_G})$ and $K\mathcal{P}(X_G,K)$.
Similarly, by denoting
$\mathfrak{l}$ the lie algebra of $L$ and $K_L=K\cap L$ we define as for $X_G$, $\mathcal{P}(X_L,K_L)$,
$\mathcal{M}(\mathfrak{l},K_L,I_{X_L})$ and $\Phi_{X_L}:\mathcal{M}(\mathfrak{l},K_L,I_{X_L}) \xrightarrow{\sim} \mathcal{P}(X_L,K_L)$.
Now, let
\begin{align}
\mathscr{R}_{(\mathfrak{l},K_{L})}^{(\mathfrak{g},K)}:\mathcal{M}(\mathfrak{l},K_L)\rightarrow \mathcal{M}(\mathfrak{g},K)
\end{align}
be the cohomological induction functor of Vogan-Zuckerman.
The induction functor in representation theory has
a geometric analogue
$$I_{L}^{G}:D_{c}^{b}(X_L,K_L)\rightarrow D_{c}^{b}(X_G,K)$$
defined through the Bernstein induction functor (see \cite{ViMir}).
The Bernstein induction functor
$$\Gamma_{L}^{G}:D_{c}^{b}(X_G,K_L)\rightarrow D_{c}^{b}(X_G,K)$$
is
the right adjoint of the forgetful functor $\text{Forget}_{K}^{K_L}:D_{c}^{b}(X_G,K)\rightarrow D_{c}^{b}(X_G,K_L)$. The geometric induction functor
is defined then
as the composition:
$$I_{L}^{G}=\Gamma_{L}^{G}\circ R\iota_{\ast}.$$
It satisfies the identity
$$I_{L}^{G}\circ \Phi_{X_L}=\Phi_{X_G}\circ \mathscr{R}_{(\mathfrak{l},K_{L})}^{(\mathfrak{g},K)},$$
which induces the following commutative diagram between the corresponding Grothendieck groups
\begin{align}\label{eq:diagramintroduction}
\xymatrix{
K\mathcal{M}(\mathfrak{g},K,I_{X_{G}}) \ar[r]^{\Phi_{X_G}} & K\mathcal{P}(X_G,K)\\
K\mathcal{M}(\mathfrak{l},K_{L},I_{X_{L}}) \ar[u]^{\mathscr{R}_{(\mathfrak{l},K_{L})}^{(\mathfrak{g},K)}}\ar[r]^{\Phi_{X_L}} & K\mathcal{P}(X_L,K_{L})\ar[u]^{I_{L}^{G}}. }
\end{align}
Now that the geometric induction functor has been introduced, we
turn to the description of
how geometric induction affects characteristic cycles.
The characteristic cycle of a perverse sheaf can be constructed
through Morse theory, or via the Riemann-Hilbert correspondence as the characteristic cycle of the
associated $\mathcal{D}$-module. In any case, the characteristic cycle
can be seen as a map $CC:K\mathcal{P}(X_G,K)\rightarrow \mathscr{L}(X_G,K)$
from $K\mathcal{P}(X_G,K)$ to the set of formal sums
$$\mathscr{L}(X_G,K)=\left\{\sum_{{K-\mathrm{orbits }~S~\mathrm{in}~X_G}}m_S[\overline{T^{\ast}_{S}X_G}]:m_S\in\mathbb{Z}_{\geq 0}\right\}.$$
Schmid and Vilonen, combining the work of Kashiwara-Schapira on proper direct images, and
their own work on direct open embeddings,
describe in \cite{SV} the effect on characteristic cycles
of taking direct images by an arbitrary algebraic morphism. This, applied to the Bernstein induction functor,
leads to a map $\left(I_{L}^{G}\right)_{\ast}:\mathscr{L}(X_L,K_L)\rightarrow\mathscr{L}(X_G,K)$ that extends (\ref{eq:diagramintroduction}) into the commutative diagram:
\begin{align*}
\xymatrix{
K\mathcal{M}(\mathfrak{g},K,I_{X_{G}}) \ar[r]^{\Phi_{X_G}} & K\mathcal{P}(X_{G},K)\ar[r]^{CC} & \mathscr{L}(X_G,K)\\
K\mathcal{M}(\mathfrak{l},K_{L},I_{X_{L}}) \ar[u]^{\mathscr{R}_{(\mathfrak{l},K_{L})}^{(\mathfrak{g},K)}}\ar[r]^{\Phi_{X_L}} & K\mathcal{P}(X_{L},K_{L})\ar[u]^{I_{L}^{G}}\ar[r]^{CC} & \mathscr{L}(X_{L},K_L).\ar[u]^{\left(I_{L}^{G}\right)_{\ast}}
}
\end{align*}
Consequently, for every $K_L$-equivariant perverse sheaf $\mathcal{F}_L$ on $X_{L}$ we have
$$CC(I_{L}^{G}\mathcal{F}_L)=\left(I_{L}^{G}\right)_{\ast}CC(\mathcal{F}_L).$$
The description of the image of $\left(I_{L}^{G}\right)_{\ast}$
is given in Proposition \ref{prop:ccLG} through a formula for
$CC(I_{L}^{G}\mathcal{F}_L)$
in terms of the characteristic cycle of $\mathcal{F}_L$.
More explicitly, writing
$$CC(\mathcal{F}_{L})=\sum_{K_{L}-\mathrm{orbits }~S_L~\mathrm{in}~X_{L}} m_{S_L}[\overline{T_{S_L}^{\ast}X_{L}}]$$
for the characteristic cycle of $\mathcal{F}_{L}$,
we prove
that the image of $\mathcal{F}_L$ under
$\left(I_{L}^{G}\right)_{\ast}$ is equal to
\begin{align}\label{eq:dernierintro}
CC(\Gamma_{L}^{G}\mathcal{F}_L)&=\left(I_{L}^{G}\right)_{\ast}CC(\mathcal{F}_L)\\
&=\sum_{K_L-\mathrm{orbits }~S_L~\mathrm{in}~X_L} m_{S_{L}}[\overline{T_{K\cdot\iota(S_L)}^{\ast}X_G}]+
\sum_{G-orbits~S\text{~in~}\partial(G\cdot \iota(X_{L}))} m_{S}[\overline{T_{S}^{\ast}X_G}].
\end{align}
From this equality we are able to deduce that
the cohomology packets
introduced by Adams and Johnson in
\cite{Adams-Johnson}, are examples of micro-packets.
The comparison between both types of packets is done in Section 7 where we also describe
the family of Arthur parameters considered in the work of Adams and Johnson and
give a description of the Adams-Johnson packets.
We continue by outlining the contents of the paper.
In Section 2 we describe the context in which we will do representation theory. We recall the notion of real form and introduce the concepts of extended group and representation of a strong real form.
This is just a short review of Chapters 2 \cite{ABV}.
Section 3 is devoted to the introduction of the geometric objects required
in the definition of micro-packets, namely: the category of perverse sheaves, the category of $\mathcal{D}$-modules, and
the notion of characteristic variety and characteristic cycle. We follow mainly Chapter 2 \cite{Hotta} and Chapter 7 and 19 \cite{ABV}.
In Section 4 we introduce the geometric induction functor as a geometric counterpart of the
induction functor in representation theory. We review the work of Kashiwara-Schapira \cite{Kashiwara-Schapira} on proper
direct images and the work of Schmid-Vilonen \cite{SV} on open direct images.
We end the section by explaining how characteristic cycles behave under geometric induction.
In Section 5 we recall the extended Langlands correspondence
and in Section 6 we express the Langlands classification in a more geometric manner. More precisely, following Chapter 6
of \cite{ABV} we introduce a topological space whose set of ${}^{\vee}{G}$-orbits, where
${}^{\vee}{G}$ denotes the dual group of $G$,
is in bijection with the set of ${}^{\vee}{G}$-conjugacy classes of Langlands parameters and express Langlands classification in this new setting.
The interest of this new space resides in its
richer geometry when
compared with the geometry of the usual space of Langlands parameters.
We start Section 7 by introducing micro-packets. Then we describe the micro-packets corresponding to
tempered parameters (Section 7.1) and to essentially unipotent Arthur parameters (Section 7.2).
We end the section by studying the case of cohomologically induced packets or Adams-Johnson packets. Finally, relying on the work done in Section 4 we prove that Adams-Johnson packets are micro-packets (Section 7.3).
It is in sections 7.2 and 7.3,
and in the study of Section 4 of the geometric induction functor,
that most of the original work of the present article is done.\\
~\\
To end with this introduction, let us mention that in \cite{Trappa}
(see Proposition 2.4 \cite{Trappa} and Corollary 2.5 \cite{Trappa}) the authors
obtain a similar result to the one of Proposition \ref{prop:ccLG} (see Equation (\ref{eq:dernierintro}) above) by using the definition of characteristic cycles in terms of normal slices (see Chapter II.6.A \cite{GM}).\\
\textbf{Acknowledgements:} The author wishes to thank Paul Mezo for useful discussions and
enlightening remarks during the preparation of this document.
\begin{comment}
comparison of
of this, is given by the packets of representations with cohomology, the Adams-Johnson-packets. Using cohomological induction it's possible to reduce the description of the packet to the description of an A-B-V-packet attached to the Arthur parameter of an unitary representation.
Another example is given by the general linear group, with Paul Mezo we where able, using parabolic and cohomological induction, to reduce the characterization of A-B-V-packets of the general linear group to the characterization of A-B-V-packets attached to essentially unipotent Arthur parameters of the general linear group.
(i.e. such that its restriction to ). And the description of this last type of A-B-V-packet follows from techniques on nilpotent orbits and associated varieties. In section (6.3) we describe, in the context of classical groups, another cases where the use cohomological and parabolic induction should simplify the description of A-B-V-packets.
\end{comment}
\section{Structure theory: real forms and extended groups
In this section we describe the context in which we will do representation theory. We begin
with a short review on some basics facts about real forms of reductive groups, to recall
next the notion of extended group, strong real forms, and of representations of a strong real form. Following the philosophy of \cite{ABV}, in this
article we are not going to fix a real form and study the corresponding set of representations,
instead, we fix an inner class of real forms and
consider at the same time the set of representations of each real form in the inner class.
Extended groups have been introduced in \cite{ABV},
as a manner
to study and describe
in an organized and uniform way, the representation theory corresponding
to an inner class of real forms.
We follow Chapter 2 \cite{ABV}.\\
Let ${G}$ be a connected reductive complex algebraic group
with Lie algebra $\mathfrak{g}$. A \textbf{real form} of ${G}$ is an antiholomorphic involutive automorphism
$$\sigma:G\rightarrow G$$
with group of real points given by
$$G(\mathbb{R},\sigma):=G^{\sigma}=\{g\in G:\sigma(g)=g\}.$$
The group $G(\mathbb{R},\sigma)$ is a real Lie group with Lie algebra
$$\mathfrak{g}(\mathbb{R},\sigma):=\mathfrak{g}^{d\sigma}.$$
Among all the real forms of $G$, of particular interest
is the \textbf{compact real form} $\sigma_{c}$.
It is characterized up to conjugation by $G$ by the requirement that
$G(\mathbb{R},\sigma_{c})$ is compact. Moreover, Cartan showed that $\sigma_{c}$ may be chosen to commute with $\sigma$.
The commutativity implies that the composition
\begin{align}\label{eq:Cinvolution}
\theta_{\sigma}=\sigma\circ\sigma_{c}=\sigma_{c}\circ\sigma
\end{align}
defines an algebraic involution of $G$ of order two, called the Cartan involution;
it is determined by $\sigma$ up to conjugation by $G(\mathbb{R},\sigma)$.
The group of fixed points
\begin{align}\label{eq:complexification}
K_{\sigma}=G^{\theta_{\sigma}}
\end{align}
is a (possibly disconnected) complex reductive algebraic subgroup of $G$.
The corresponding group of real points
$$K_{\sigma}(\mathbb{R})=G(\mathbb{R},\sigma)\cap K_{\sigma}=G(\mathbb{R},\sigma)\cap G(\mathbb{R},\sigma_{c})=
G(\mathbb{R},\sigma)\cap K_{\sigma}$$
is a maximal compact subgroup of $G(\mathbb{R},\sigma)$ and a maximal compact subgroup
of $K_{\sigma}$ (see (5d)-(5g) \cite{AVParameters}).\\
Let us introduce now the notion of an extended group.
We start by recalling the notion of
inner real forms.
\begin{deftn}
Two real forms $\sigma$ and $\sigma'$ are said to be \textbf{inner to each other} if there is an element $g\in G$
such that
$$\sigma'=\text{Ad}(g)\circ \sigma.$$
\end{deftn}
\begin{deftn}[Definition 2.13 \cite{ABV}]\label{deftn:extendedgroup}
An \textbf{extended group containing} $G$ is a real Lie group $G^{\Gamma}$
subject to the following conditions.
\begin{enumerate}
\item $G^{\Gamma}$ contains $G$ as a subgroup of index two. That is, there is a short exact sequence
\begin{align*}
1\rightarrow {G}\rightarrow {}G^{\Gamma}\rightarrow \Gamma\rightarrow 1,
\end{align*}
where $\Gamma=\mathrm{Gal}(\mathbb{C}/\mathbb{R})$.
\item Every element of $G^{\Gamma}-G$ acts on $G$ as an antiholomorphic automorphism.
\end{enumerate}
A \textbf{strong real form} of $G^{\Gamma}$ is an element
$\delta\in G^{\Gamma}$ such that $\delta^{2}\in Z(G)$ has finite order. To each strong
real form $\delta\in G^{\Gamma}$ we associate a real form $\sigma_{\delta}$ of $G$ defined by
conjugation by $\delta$:
$$\sigma_{\delta}(g):=\delta g\delta^{-1}.$$
The group of real points of $\delta$ is defined to be the group of real points of $\sigma_{\delta}$:
$$G(\mathbb{R},\delta):=G(\mathbb{R},\sigma_{\delta})=\{g\in G:\sigma_{\delta}(g)=\delta g\delta^{-1}=g \}. $$
Two strong real forms of $G^{\Gamma}$ are called equivalent if they are conjugate by $G$.
\end{deftn}
Notice that what we call here an extended group is called a weak extended group in \cite{ABV}. The next result gives a classification of extended groups.
\begin{prop}[Corollary 2.16 \cite{ABV}]\label{prop:classificationextended}
Suppose $G$ is a connected reductive complex algebraic group,
and write $\Psi_0(G)$ for the based root datum of $G$.\begin{enumerate}[i.]
\item Fix a weak extended group $G^{\Gamma}$ for $G$. Let $\sigma_Z$ be the antiholomorphic
involution of $Z(G)$ defined by the conjugation action of any element of $G^{\Gamma}-G$. We can attach to $G^{\Gamma}$ two invariants. The first of these
is an involutive automorphism
$$ a\in \mathrm{Aut}(\Psi_0(G))$$
of the based root datum of $G$. The second is a class
$$\overline{z}\in Z(G)^{\sigma_Z}/(1+\sigma_Z)Z(G),$$
where
$$(1+\sigma_Z)Z(G)=\{z\sigma_Z (z):z\in Z(G)\}.$$
\item Suppose $G^{\Gamma}$ and $(G^{\Gamma})'$ are weak extended groups for G with the
same invariants $(a,\overline{z})$. Then the identity map on $G$ extends to an
isomorphism from $G^{\Gamma}$ to $(G^{\Gamma})'$.
\item Suppose $a\in \mathrm{Aut}(\Psi_0(G))$ is an involutive automorphism
and let $\sigma$ be any real form in the inner class corresponding to $a$ by Proposition 2.12 \cite{ABV}.
Write $\sigma_Z$ for the antiholomorphic involution of $Z(G)$ defined by the action of
$\sigma$;
and suppose
$\overline{z}\in Z(G)^{\sigma_Z}/(1+\sigma_Z)Z(G).$
Then there is a weak extended group $G^{\Gamma}$ with invariants $(a,\overline{z})$.
\end{enumerate}
\end{prop}
The next result states
the relation between extended groups and inner classes of real forms.
\begin{prop}[Proposition 2.14 \cite{ABV}]
Suppose $G^{\Gamma}$ is a weak extended group. Then the
set of real forms of $G$ associated to strong real forms of $G^{\Gamma}$ constitutes
exactly one inner class of real forms.
\end{prop}
We end this section with the definition of a representation of a strong real form.
\begin{deftn}[Definition 2.13 \cite{ABV}]
\label{deftn:representationstrongrealform}
A \textbf{representation of a strong real form} of $G^{\Gamma}$ is a pair $(\pi,\delta)$ subject to
\begin{enumerate}
\item $\delta$ is a strong real form of $G^{\Gamma}$.
\item $\pi$ is an admissible representation of $G(\mathbb{R},\delta)$.
\end{enumerate}
Two such representations $(\pi,\delta)$ and $(\pi',\delta')$ are said to be
equivalent if there is an element $g\in G$ such that
$g\delta g^{-1}=\delta'$ and $\pi\circ Ad(g^{-1})$ is (infinitesimally) equivalent to
$\pi'$.
Finally, define
$$\Pi(G/\mathbb{R})$$
to be the set of infinitesimal equivalence classes of irreducible representations of strong real
forms of $G^{\Gamma}$.
\end{deftn}
\section{$\mathcal{D}$-modules, Perverse sheaf and Characteristic Cycles}
In this section we introduce the geometric objects required
to the definition in Section 7 of the micro-packets. We begin with the definition of the categories that are going to be involved in their construction, to recall
next
the concepts of characteristic variety and characteristic cycle, and
describe some of their properties.
We follow mainly Chapter 2 \cite{Hotta} and Chapter 19 \cite{ABV}.\\
Suppose $X$ is a smooth complex algebraic variety
on which an algebraic group $H$ acts with finitely many orbits. Define (see Appendix B \cite{ViMir} and Definition 7.7 \cite{ABV}):\\
$\bullet$ $D_{c}^{b}(X)$ to be the bounded derived category of sheaves of complex vector spaces on $X$ having cohomology sheaves constructible with respect to an algebraic stratification
on $X$.\\
$\bullet$ $D_{c}^{b}(X,H)$ to be the subcategory of $D_{c}^{b}(X)$ consisting of
$H$-equivariant sheaves of complex vector spaces on $X$ having cohomology sheaves constructible with respect to the algebraic stratification
defined by the
$H$-orbits
on $X$.\\
~\\
Living inside this
last category we have the category of $H$-equivariant perverse sheaves on $X$. We write\\
$\bullet$ $\mathcal{P}(X,H)$ for be the category of $H$-equivariant perverse sheaves on $X$.\\
~\\
Next, set $\mathcal{D}_{X}$ to be the sheaf of algebraic differential operators on $X$ and define:\\
$\bullet$ $\mathcal{D}(X,H)$ to be the category of $H$-equivariant coherent sheaves of
$\mathcal{D}_{X}$-modules on $X$.\\
$\bullet$ ${D}^{b}(\mathcal{D}_X,H)$ to be the $H$-equivariant bounded derived category of sheaves of $\mathcal{D}_{X}$-modules on $X$ having cohomology sheaves coherent.\\
~\\
The categories $\mathcal{P}(X,H)$ and $\mathcal{D}(X,H)$
are abelian, and every object has finite length. To
each of them correspond then a Grothendieck group that we denote respectively by
\begin{equation}\label{eq:ggroups}
K\mathcal{P}(X,H)\quad \text{and}\quad K\mathcal{D}(X,H).
\end{equation}
The four previous categories
are related through the Riemann-Hilbert correspondence.
\begin{theo}[Riemann-Hilbert Correspondence, see Theorem 7.2.1 \cite{Hotta}, Theorem 7.2.5 \cite{Hotta}
and Theorem 7.9 \cite{ABV}]\label{theo:rhcorrespondence}
The de Rham functor induces an equivalence of categories
\begin{align*}
DR:{D}^{b}(\mathcal{D}_X,H)\rightarrow D_{c}^{b}(X,H)
\end{align*}
such that if we restrict $DR$ to the full subcategory $\mathcal{D}(X,H)$ of
${D}^{b}(\mathcal{D}_X,H)$ we obtain an equivalence of categories
\begin{align*}
DR:\mathcal{D}(X,H)\rightarrow\mathcal{P}(X,H).
\end{align*}
This induces an isomorphism of Grothendieck groups
\begin{align*}
DR:K\mathcal{D}(X,H)\rightarrow K\mathcal{P}(X,H)
\end{align*}
\end{theo}
We use the previous isomorphism to identify the Grothendieck groups of (\ref{eq:ggroups})
writing simply $K(X,H)$ instead of $K\mathcal{P}(X,H)$ and $K\mathcal{D}(X,H).$\\
In this paper we are principally interested in the case of $X$ being a flag variety.
More precisely, let $G$ be a connected reductive complex algebraic group
with Lie algebra $\mathfrak{g}$. Fix a real form $\sigma$ of $G$ and as in
(\ref{eq:complexification}), write $K_{\sigma}$
for the group of fixed points of the corresponding Cartan involution.
The flag variety $X_G$ of $G$ is defined as the set of all Borel subgroups of $G$ (or equivalently as the set of all Borel subalgebras of $\mathfrak{g}$). The group $G$ acts on $X_G$ by conjugation,
this action is transitive and if we restrict it to
$K_\sigma$, then the number of $K_{\sigma}$-orbits is finite.
Moreover, for any fixed Borel subgroup $B\in X_{G}$ the normalizer of $B$ in $G$ is $B$ itself, thus we obtain a bijection
$$X_G\cong G/B.$$
The flag variety has then a natural structure of an algebraic variety.\\
We turn now to a short discussion on characteristic varieties and characteristic cycles. These two
objects are going to play a central role in the definition of micro-packets.
We begin with the definition of the characteristic variety
of a $\mathcal{D}_{X}$-module.
\begin{deftn}[Section 2.2 \cite{Hotta}]
Let $M$ be a coherent $\mathcal{D}_{X}$-module of $X$. Choose a good filtration $F$ on $X$ (we invite
the reader to see definition (2.1.2) of \cite{Hotta} for the definition of a good filtration and Theorem (2.1.3) of the same book for the proof that every coherent $\mathcal{D}_{X}$-module admits one).
Write $\mathrm{gr}^{F} M$ for the corresponding graded module.
Let $\pi:T^{\ast}X\rightarrow X$ be the cotangent bundle of $X$.
Since we have $\mathrm{gr}^{F} \mathcal{D}_{X}\cong \pi_{\ast}\mathcal{O}_{T^{\ast}X}$,
the graded module $\mathrm{gr}^{F} M$ is a coherent module over $\pi_{\ast}\mathcal{O}_{T^{\ast}X}$
(Proposition 2.2.1 \cite{Hotta}),
and we can define the coherent $\mathcal{O}_{T^{\ast}X}$-module
$$\widetilde{\mathrm{gr}^{F} M}:=\mathcal{O}_{T^{\ast}X}\otimes_{\pi^{-1}\pi_{\ast}\mathcal{O}_{T^{\ast}X}}\pi^{-1}(\mathrm{gr}M),$$
where $\pi^{-1}$ is the inverse image functor of $\pi$.
The support of $\widetilde{\mathrm{gr}^{F} M}$ is independent of the
choice of a good filtration (see for example Theorem (2.2.1) of \cite{Hotta}).
It is called the \textbf{characteristic variety} of $M$ and is denoted by
$$\mathrm{Ch}(M):=\mathrm{supp}({\widetilde{\mathrm{gr}^{F} M}}).$$
Since $\widetilde{\mathrm{gr}^{F} M}$ is a graded module over the graded ring $\mathcal{O}_{T^{\ast}X}$,
the characteristic variety $\mathrm{Ch}(M)$ is a closed conic algebraic subset in $T^{\ast}X$.
\end{deftn}
To define the characteristic cycle of a $\mathcal{D}_{X}$-module,
we need to introduce first
the more general notion of an associated cycle.
\begin{deftn}[Section 1.5 \cite{fulton}]
Let $X$ be any variety, and let $X_{1},\cdots, X_{n}$ be the irreducible components of $X$. The geometric multiplicity of $X_{i}$ on $X$ is defined to be the length of the local ring $\mathcal{O}_{X_{i},X}$:
$$m_{X_{i}}(X)=l_{\mathcal{O}_{X_{i},X}}(\mathcal{O}_{X_{i},X})$$
(Since the local rings $\mathcal{O}_{X_{i},X}$ are all zero-dimensional, their length is well defined).
We define the \textbf{associated cycle} $Cyc(X)$ of $X$ to be the formal sum
$$Cyc(X)=\sum_{i=1}^{n}m_{X_{i}}(X)[X_{i}].$$
\end{deftn}
\begin{deftn}[Definition 2.2.2 \cite{Hotta}]\label{deftn:defPCycle}
Let $M$ be a coherent $\mathcal{D}_{X}$-module. Denote by $I(\mathrm{Ch}(M))$
the set of the irreducible components of $\mathrm{Ch}(M)$. We define the \textbf{characteristic cycle} of
$M$ by the formal sum
$$CC(M)=Cyc(\mathrm{Ch}(M))=\sum_{C\in I(\mathrm{Ch}(M))}m_{C}(\mathrm{Ch}(M))[C].$$
For $d\in \mathbb{N}$ we denote its degree $d$ part by
$$CC_{d}(M)=\sum_{\substack{C\in I(\mathrm{Ch}(M))\\
\dim C=d}}m_{C}(\mathrm{Ch}(M))[C].$$
\end{deftn}
Let $M$ be a coherent $\mathcal{D}_{X}$-module. Following the notation of \cite{ABV} (Definition 1.30 \cite{ABV} and Proposition 19.12 \cite{ABV})
for each irreducible subvariety $C$ of $T^{\ast}X$ we denote
\begin{align}\label{eq:microlocalmultiplicity}
\chi_{C}^{mic}(M):=\left\{
\begin{array}{cl}
m_{C}(\mathrm{Ch}(M))& \text{if }C\in I(\mathrm{Ch}(M)),\\
0 &\text{otherwise}.
\end{array}
\right.
\end{align}
and call $\chi_{C}^{mic}(M)$ the microlocal multiplicity along $C$.
\begin{prop}[Theorem 2.2.3 \cite{Hotta}]\label{prop:additive}
Let
$$0\rightarrow M\rightarrow N\rightarrow L\rightarrow 0$$
be an exact sequence of coherent $\mathcal{D}_{X}$-modules. Then for any irreducible subvariety
$C$ of $T^{\ast}X$ such that $C\in I(\mathrm{Ch}(N))$ we have
$$\chi_{C}^{mic}(N)=\chi_{C}^{mic}(M)+\chi_{C}^{mic}(L).$$
In particular, for $d=\mathrm{dim}(\mathrm{Ch}(N))$ we have
$$CC_{d}(N)=CC_{d}(M)+CC_{d}(L).$$
\end{prop}
From the previous result, for each irreducible subvariety $C$ of $T^{\ast}X$
the microlocal multiplicity along $C$ defines and additive function for short exact sequences:
$$\chi_{C}^{mic}:\mathcal{D}(X,H)\rightarrow \mathbb{Z}.$$
Therefore,
the microlocal multiplicities
define $\mathbb{Z}$-linear functionals:
$$\chi_{C}^{mic}:K(X,H)\rightarrow \mathbb{Z}.$$
A coherent $\mathcal{D}_{X}$-module $M$ is called a holonomic $\mathcal{D}_{X}$-module (or a
holonomic system) if it satisfies
$\mathrm{dim}(\mathrm{Ch}(M))=\mathrm{dim}(X).$ From Theorem (11.6.1) \cite{Hotta}
every $H$-equivariant coherent $\mathcal{D}_{X}$-module is holonomic. Furthermore, for any $H$-equivariant coherent $\mathcal{D}_{X}$-module, the stratification of $X$ defined by the
$H$-orbits induces a stratification of $\mathrm{Ch}(M)$ by the closure of the conormal bundle to the $H$-orbits, and a more explicit description of the characteristic cycles
is therefore possible. More precisely, denote
at each point $x\in X$
the differential of the $H$-action by
\begin{align}\label{eq:actionmap}
\mathcal{A}_{x}:\mathfrak{h}\rightarrow T_{x}X.
\end{align}
Regarding $\mathfrak{h}\times X$ as a trivial bundle over $X$, we get a bundle map
\begin{align}\label{eq:actionmap2}
\mathcal{A}:\mathfrak{h}\times Y\rightarrow TX.
\end{align}
We define the conormal bundle to the $H$-action as the annihilator of the image of $\mathcal{A}$:
$$T_{H}^{\ast}X=\{(\lambda,x):\lambda\in T_{y}^{\ast}X,\lambda(a_{x}(\mathfrak{h}))=0\}.$$
If $x$ belongs to the $H$-orbit $S$, then
$$a_{x}(\mathfrak{h})=T_{x}S$$
and the fiber of $T_{H}^{\ast}X$ at $x$ is the conormal bundle to $S$ at $x$:
$$T_{H,x}^{\ast}X=T_{S,x}^{\ast}X.$$
Therefore
$$T_{H}^{\ast}X=\bigcup_{H-\text{orbits }S~\mathrm{in}~X}T_{S}^{\ast}X.$$
We have the following result.
\begin{theo}[Proposition 19.12 \cite{ABV}]\label{theo:defcc}
Let $\mathcal{M}$ be a $H$-equivariant coherent $\mathcal{D}_{X}$-module. We have:
\begin{enumerate}[i.]
\item The characteristic variety of $\mathcal{M}$
is contained in the conormal bundle to the $H$-action
$T_{H}^{\ast}X$.
\item The $H$-components of $\mathrm{Ch}(\mathcal{M})$ are closures of conormal bundles
of $H$-orbits on $X$. Consequently
$$CC(\mathcal{M})=\sum_{{H-\mathrm{orbits }~S~\mathrm{in}~X}}\chi_{S}^{mic}(\mathcal{M})[\overline{T^{\ast}_{S}X}],$$
where $\chi_{S}^{mic}(\mathcal{M})$ denotes the microlocal multiplicity along $\overline{T^{\ast}_{S}X}$ (i.e. $\chi_{S}^{mic}(\mathcal{M})=\chi_{\overline{T^{\ast}_{S}X}}^{mic}(\mathcal{M})$).
\item The support of $\mathcal{M}$ is given by
$$\mathrm{Supp}(\mathcal{M})=\bigcup_{\chi_{S}^{mic}(\mathcal{M})\neq 0} \overline{S}.$$
\end{enumerate}
\end{theo}
\begin{deftn}[Section 10.1 \cite{Kashiwara-Schapira85}, Apendix B \cite{ViMir}, Equation (2.4)-(2.5) \cite{SV}]\label{deftn:ccycleperverse}
Let $P$ be an $H$-equivariant perverse sheaf on $X$ and write $M$
for the corresponding $\mathcal{D}_{X}$-module under the Riemann-Hilbert correspondence.
We define the characteristic cycle of $P$ as
$$CC(P):=CC(M).$$
More generally, let $\mathcal{F}\in D_{c}^{b}(X,H)$ and write $\mathcal{M}$
for the corresponding element in $D_{c}^{b}(\mathcal{D}_X,H)$ under the Riemann-Hilbert correspondence. We define the characteristic variety of $\mathcal{F}$ as
$$\mathrm{Ch}(\mathcal{F}):=\bigcup_i \mathrm{Ch}(H^{i}\mathcal{M}),$$
and the characteristic cycle of $\mathcal{F}$ to be the formal sum
$$CC(\mathcal{F}):=\sum_{C\in I(\mathrm{Ch}(\mathcal{F}))}m_{C}[C],$$
where $I(\mathrm{Ch}(\mathcal{F}))$ denotes
the set of the irreducible components of $\mathrm{Ch}(M)$ and for each $C\in I(\mathrm{Ch}(\mathcal{F}))$, $m_{C}$ is an integer defined as in Equation (2.5) \cite{SV}.
Now, if we denote $\mathscr{L}(X)$ to be the set of formal linear combination with $\mathbb{Z}$-coefficients of irreducible \textbf{analytic Lagrangian conic subvarieties} in the
cotangent bundle $T^{\ast}X$ of the variety $X$, then $CC(\mathcal{F})\in \mathscr{L}(X)$.
In particular, if we define $\mathscr{L}(X,H)$
to be the set of formal sums
$$\mathscr{L}(X,H)\subset\left\{\sum_{{H-\mathrm{orbits }~S~\mathrm{in}~X}}m_S[\overline{T^{\ast}_{S}X}]:m_S\in\mathbb{Z}\right\},$$
taking characteristic cycles defines a
map
$$CC:D_{c}^{b}(X,H)\rightarrow \mathscr{L}(X,H)$$
and from the remark following Proposition (\ref{prop:additive})
we obtain a $\mathbb{Z}$-linear map
$$CC:K(X,H)\rightarrow \mathscr{L}(X,H).$$
Finally, for each formal sum
$C=\sum_{i}m_{C_{i}}[C_{i}]\in \mathscr{L}(X)$
we define the support $|C|$ of $C$, as
$$|C|=\overline{\bigcup_{m_{i}\neq 0}C_{i}}.$$
Notice that by definition, for every $\mathcal{F}\in \mathcal{P}(X,H)$,
$|CC(\mathcal{F})|=\mathrm{Ch}(\mathcal{F})$.
\end{deftn}
\section{Geometric induction}
In this section we recall Bernstein's induction functor
and use it to define a geometric analogue of the induction functor in representation theory.
Once the geometric induction functor has been introduced,
we turn to explain how it affects characteristic cycles. By doing this, we are able to describe the characteristic cycle associated to an induced representation (through Beilinson-Bernstein Correspondence), in terms of the characteristic cycle of the representation being induced. As we will see in Section 7.3, this will have as a consequence the possibility to reduce in some particular
cases the computation of the micro-packets associated to a group, to the computation of the micro-packets associated to a Levi subgroup.\\
Suppose $G$ is a connected reductive complex algebraic group with Lie algebra $\mathfrak{g}$,
and let $\sigma$ be a real form of $G$.
As in Equation (\ref{eq:complexification}), denote by $K$ the group of fixed points of the Cartan involution associated to $\sigma$. Define
\begin{align*}
\mathcal{M}(\mathfrak{g},K) \text{ to be the category of }(\mathfrak{g}, K)\text{-modules of }G.
\end{align*}
We have an equivalence of categories between $\mathcal{M}(\mathfrak{g},K)$
and the category of (infinitesimal equivalence classes of) admissible representations of $G(\mathbb{R},\sigma)$.
Following this equivalence, in this article we shall blur the distinction between these two categories by referring to their objects indiscriminately as representations of $G(\mathbb{R},\sigma)$.
We begin by recalling how the Beilinson-Bernstein correspondence
relates the categories introduced in the previous section with the category
of $(\mathfrak{g}, K)$-modules of $G$. We follow Chapter 8 \cite{ABV}.
Let $X_{G}$ be the flag variety of $G$, and as in the previous section write
$\mathcal{D}_{X_{G}}$ for the sheaf of algebraic differential operators on $X_{G}$. We define
$$D_{X_{G}}=\Gamma \mathcal{D}_{X_{G}}$$
to be the algebra of global sections of $\mathcal{D}_{X_{G}}$. We recall that every
element of $\mathfrak{g}$ defines a global vector field on $X_{G}$
and that this identification extends to an algebra homomorphism
\begin{align}\label{eq:operatorrepresentation}
\psi_{X_{G}}:U(\mathfrak{g})\longrightarrow D_{X_{G}}
\end{align}
called the \textbf{operator representation} of $U(\mathfrak{g})$.
The kernel of $\psi_{X_{G}}$ is a two-sided ideal denoted by
\begin{align}\label{eq:keroperatorrepresentation}
I_{X_{G}}=\text{ker}\psi_{X_{G}}.
\end{align}
Now, if $\mathcal{M}$ is any sheaf of $\mathcal{D}_{X_{G}}$-modules,
then the vector space $M=\Gamma \mathcal{M}$ obtained by taking global sections
is in a natural way a $D_{X_{G}}$-module and therefore, via $\psi_{X_{G}}$, a module for $U(\mathfrak{g})/I_{X_{G}}$.
The functor sending the $\mathcal{D}_{X_{G}}$-module $\mathcal{M}$ to the
$U(\mathfrak{g})/I_{X_{G}}$-module $M$ is called the \textbf{global sections functor}.
In the other direction, if $M$ is any module for $U(\mathfrak{g})/I_{X_{G}}$
then we may form the tensor product
$$\mathcal{M}=\mathcal{D}_{X_{G}}\otimes_{\psi_{X_{G}}(U(\mathfrak{g})/I_{X_{G}})} M.$$
This is a sheaf of $\mathcal{D}_{X_{G}}$-modules on $X_{G}$.
The functor sending $M$ to $\mathcal{M}$ is called \textbf{localization}.
\begin{theo}[Beilinson-Bernstein localization theorem,
see \cite{BB}, Theorem 3.8 \cite{BoBrI}, Theorem 1.9 \cite{BoBrIII}
and Theorem 8.3 \cite{ABV}]\label{theo:bbcorespondance}
We have:
\begin{enumerate}[i.]
\item The operator representation $\psi_{X_{G}}:U(\mathfrak{g})\longrightarrow D_{X_{G}}$ is surjective.
\item The global sections and localization functors provide an equivalence of categories between
quasicoherent sheaves of $\mathcal{D}_{X_{G}}$-modules on $X_G$ and modules for $U(\mathfrak{g})/I_{X_{G}}$.
\item Let
\begin{align*}
\mathcal{M}(\mathfrak{g},K,I_{X_{G}})\text{ be the category of }(\mathfrak{g}, K)\text{-modules of }G\text{ annihilated by }I_{X_G}.
\end{align*}
Then the global sections functor and localization functor provide an equivalence of categories between:
$$\mathcal{D}(X_G,K)\quad\text{ and }\quad \mathcal{M}(\mathfrak{g},K,I_{X_{G}}).$$
\end{enumerate}
\end{theo}
Suppose $Q$ is a parabolic subgroup of $G$ with
Levi decomposition $Q=LN$, and such that $L$ is stable under $\sigma$.
Consider the fibration $X_G\rightarrow G/Q$. Its fiber over $Q$
can be identified with the flag variety $X_L$ of $L$. We denote the inclusion of that fiber in
$X_G$ by
\begin{align}\label{eq:mapvarieties}
\iota:X_L&\longrightarrow X_G
\end{align}
Finally, denote $K_{L}=K\cap L$ and let $\mathfrak{l}$ be the lie algebra of $L$. We define
\begin{align}\label{eq:cohomologicalinduction}
\mathscr{R}_{(\mathfrak{l},K_{L})}^{(\mathfrak{g},K)}(\cdot):\mathcal{M}(\mathfrak{l},K_L)\rightarrow \mathcal{M}(\mathfrak{g},K)
\end{align}
to be the cohomological induction functor (see (5.3a)-(5.3b) \cite{Knapp-Vogan} and (11.54a)-(11.54b) \cite{Knapp-Vogan}).
Since we
are identifying representations with the underlying
$(\mathfrak{g},K)$-module, following Proposition 11.57 \cite{Knapp-Vogan}, we are not going to make any distinction between parabolic and cohomological induction, and use the functor in (\ref{eq:cohomologicalinduction}) to express both types of induction.
\begin{comment}
\item $\sigma(P)=P$ we denote
\begin{align}
\text{Ind}_{P(\sigma,\mathbb{R})}^{G(\sigma,\mathbb{R})}(\cdot):\Pi(L(\sigma,\mathbb{R}))\rightarrow \Pi(G(\sigma,\mathbb{R}))
\end{align}
to be the normalized parabolic induction functor.
\end{itemize}
Finally define the representation theory induction functor as
\begin{align*}
I_{L}^{G}:\Pi(L(\sigma,\mathbb{R}))\rightarrow \Pi(G(\sigma,\mathbb{R}))
\pi_{L}\mapsto \left\{
\begin{array}{rl}
\text{Ind}_{} & \text{if } x < 0,\\
0 & \text{if } x = 0,\\
x & \text{if } x > 0.
\end{array} \right.
\end{align*}
\end{comment}
We now begin with the description of the geometric induction functor. The objective is to define a functor
$$I_{L}^{G}:D_{c}^{b}(X_L,K_{L})\rightarrow D_{c}^{b}(X_G,K),$$
that makes the following diagram commutative
\begin{align}\label{eq:cdiagramme1}
\xymatrix{
K\mathcal{M}(\mathfrak{g},K,I_{X_{G}}) \ar[r] & K(X_G,K)\\
K\mathcal{M}(\mathfrak{l},K_{L},I_{X_{L}}) \ar[u]^{\mathscr{R}_{(\mathfrak{l},K_{L})}^{(\mathfrak{g},K)}}\ar[r] & K(X_L,K_{L})\ar[u]^{I_{L}^{G}}, }.
\end{align}
Here the horizontal arrows are given by Theorem \ref{theo:bbcorespondance}. The construction of $I_{L}^{G}$ is based on Bernstein's geometric functor.
\begin{deftn}[Section 1.1 \cite{ViMir}]\label{deftn:geominduction0}
Suppose $Y$ is a smooth complex algebraic variety
on which the algebraic group $G$ acts with finitely many orbits.
For any subgroup $H$ of $G$ we define the Bernstein induction functor
$$\Gamma_{H}^{G}:D_{c}^{b}(Y,H)\rightarrow D_{c}^{b}(Y,G)$$
as the right adjoint of the forgetful functor from $D_{c}^{b}(Y,G)$
to $D_{c}^{b}(Y,H)$ (see \cite{Bi} for the proof of its existence). More precisely,
consider the diagram
\begin{align}\label{eq:diagraminduction1}
\xymatrix{
G\times Y \ar[r]^\mu \ar[d]_p & G\times_{H} Y \ar[d]^a \\
Y & Y
}
\end{align}
given by
$$ \xymatrix{
(g,y) \ar[r] \ar[d] & \overline{(g,y)} \ar[d] \\
y & g\cdot y
}$$
where $G\times_{H} Y$ is the quotient of $G\times Y$ by the $H$-action $h\cdot(g,y)=(gh^{-1},hy)$.
From Theorem (A.2) (iii) \cite{ViMir}, for $\mathcal{F}\in D_{c}^{b}(Y,H)$ there is an unique $\widetilde{\mathcal{F}}\in D_{c}^{b}(G\times_{H}Y,G)$ such that
$p^{\ast}\mathcal{F}=\mu^{\ast}\widetilde{\mathcal{F}}$. Bernstein's induction functor is defined as
\begin{align*}
\Gamma_{H}^{G}\mathcal{F}=Ra_{\ast}\widetilde{\mathcal{F}},\quad \mathcal{F}\in D_{c}^{b}(Y,B),
\end{align*}
where $R{a_{\ast}}:D_{c}^{b}(G\times_{H}Y,G)\rightarrow D_{c}^{b}(Y,G)$
is the right derived functor of the direct image functor defined by $a$.
Equivalently, one can also define $\Gamma_{H}^{G}:D_{c}^{b}(Y,H)\rightarrow D_{c}^{b}(Y,G)$
via the diagram
\begin{align}\label{eq:diagraminduction2}
\xymatrix{
G\times Y \ar[r]^\nu \ar[d]_b & G/H\times Y \ar[d]^p \\
Y & Y
}
\end{align}
given by
$$ \xymatrix{
(g,y) \ar[r] \ar[d] & {(gH,y)} \ar[d] \\
g^{-1}\cdot y & y
}$$
Then for $\mathcal{F}\in D_{c}^{b}(Y,H)$ we have
\begin{align*}
\Gamma_{H}^{G}\mathcal{F}=Rp_{\ast}{\mathcal{F}'},
\end{align*}
where $\mathcal{F}'$ is the unique element in $D_{c}^{b}(G/H\times Y,G)$ such that $b^{\ast}\mathcal{F}=\nu^{\ast}{\mathcal{F}'}$ (Theorem (A.2) (iii) \cite{ViMir}) and $R{p_{\ast}}:D_{c}^{b}(G/{H}\times Y,G)\rightarrow D_{c}^{b}(Y,G)$ is the right derived functor of the direct image functor defined by $p$.
\end{deftn}
The next lemma relates the characteristic variety of $\mathcal{F}$ with
the characteristic variety of $\Gamma_{H}^{G}(\mathcal{F})$.
\begin{lem}\label{lem:contention}
In the setting of Definition \ref{deftn:geominduction0}, let $\mathcal{F}\in D_{c}^{b}(G,H)$. Then
\begin{align}\label{eq:contention}
G\cdot \mathrm{Ch}(\mathcal{F}) \subset \mathrm{Ch}(\Gamma_{H}^{G}\mathcal{F})
\subset \overline{G\cdot \mathrm{Ch}(\mathcal{F})}
\end{align}
\end{lem}
\begin{proof}
The right inclusion in (\ref{eq:contention})
is Lemma (1.2) \cite{ViMir}.
For the left inclusion we consider the definition of $\Gamma_{H}^{G}$ via Diagram (\ref{eq:diagraminduction2}). Let $\mathcal{F}'\in D_{c}^{b}(G/H\times Y,G)$ be such that
$b^{\ast}\mathcal{F}=\nu^{\ast}{\mathcal{F}'}$.
We use the description of Ch$(Rp_{\ast}\mathcal{F}')$
given in Proposition B2 \cite{ViMir}.
Let $\overline{G/H}$ be a smooth compactification of $G/H$.
Then $p:G/H\times Y\rightarrow Y$ factors as
$p:G/H \times Y\xrightarrow{i} \overline{G/H}\times Y\xrightarrow{q} Y$,
and as explained in the proof of Lemma B2 \cite{ViMir} we have
$$\mathrm{Ch}(Rp_{\ast}\mathcal{F}')=pr(\mathrm{Ch}(R{i}_{\ast}\mathcal{F}')),$$ where
$pr$ denote the projection $pr:T^{\ast}(G/H\times Y)\rightarrow T^{\ast}Y$.
Since $i:G/H \times Y\rightarrow \overline{G/H}\times Y$ is an open embedding
$$\mathrm{Ch}(\mathcal{F}')\subset \mathrm{Ch}(R{i}_{\ast}\mathcal{F}').$$
Therefore
$$pr( \mathrm{Ch}(\mathcal{F}'))\subset \mathrm{Ch}(Rp_{\ast}\mathcal{F}')=\mathrm{Ch}(\Gamma_{H}^{G}\mathcal{F}).$$
Finally, by Proposition B1 \cite{ViMir}, $pr( \mathrm{Ch}(\mathcal{F}'))=pr( \mathrm{Ch}(b^{\ast}\mathcal{F}))$ and
from the proof of Lemma (1.2) \cite{ViMir} we obtain $pr( \mathrm{Ch}(b^{\ast}\mathcal{F}))=
G\cdot \mathrm{Ch}(\mathcal{F})$. Equation (\ref{eq:contention}) follows.
\end{proof}
\begin{deftn}\label{deftn:geominduction}
Let $$\iota:X_L\longrightarrow X_G$$ be the inclusion defined in Equation (\ref{eq:mapvarieties}) and write
$R{\iota_{\ast}}:D_{c}^{b}(X_L,K_L)\rightarrow D_{c}^{b}(X_G,K_L)$
for the right derived functor of the direct image functor defined by $\iota$.
The geometric induction functor $I_{L}^{G}:D_{c}^{b}(X_L,K_L)\rightarrow D_{c}^{b}(X_G,K)$ is defined as
\begin{align}\label{eq:induction}
I_{L}^{G}(\mathcal{F})=\Gamma_{K_L}^{K}(R\iota_{\ast}\mathcal{F})=Ra_{\ast}(\widetilde{R\iota_{\ast}\mathcal{F}}),\quad \mathcal{F}\in D_{c}^{b}(X_L,K_L),
\end{align}
where $\widetilde{R\iota_{\ast}\mathcal{F}}$ is the unique element in
$D_{c}^{b}(K\times_{K_L}X_G,K)$ satisfying
$p^{\ast}(R{\iota}_{\ast}\mathcal{F})=\mu^{\ast}(\widetilde{R\iota_{\ast}\mathcal{F}})$.
Equivalently, the geometric induction functor can also be defined as
\begin{align*}
I_{L}^{G}(\mathcal{F})=\Gamma_{K_L}^{K}(R\iota_{\ast}\mathcal{F})=Rp_{\ast}(({R\iota_{\ast}\mathcal{F}})'),\quad \mathcal{F}\in D_{c}^{b}(X_L,K_L),
\end{align*}
where $({R\iota_{\ast}\mathcal{F}})'$ is the unique element in
$D_{c}^{b}(K/{K_L}\times X_G,K)$ satisfying
$b^{\ast}(R{\iota}_{\ast}\mathcal{F})=\nu^{\ast}(({R\iota_{\ast}\mathcal{F}})')$.
\end{deftn}
The objective of this section is to explain how $I_{L}^{G}$
affects characteristic cycles. We do this by giving a formula for
$CC(I_{L}^{G}(\mathcal{F}))$
in terms of the characteristic cycle
of $\mathcal{F}$.
From Definition \ref{deftn:geominduction}
it is clear that in order to compute $CC(I_{L}^{G}(\mathcal{F}))$
it will be necessary first to describe the behaviour of
characteristic cycles under
taking the direct images
$R\iota_{\ast}\text{ and }Ra_{\ast},$ and second
to reduce the characterization of $CC(\widetilde{R\iota_{\ast}\mathcal{F}})$
to the one of $CC({\mathcal{F}})$. The second step will be done in Proposition
\ref{prop:reduction2} and Corollary \ref{cor:reductionstep2}
below. Another option to compute $CC(I_{L}^{G}(\mathcal{F}))$ is
to make use of $Rp_{\ast}$ instead of $Ra_{\ast}$,
and to reduce the characterization of $CC(({R\iota_{\ast}\mathcal{F}})')$
to the one of $CC({\mathcal{F}})$, but since
Proposition 7.14 \cite{ABV} give us a description of $CC(\widetilde{R\iota_{\ast}\mathcal{F}})$, in this article we work principally with the definition of $I_{L}^{G}$ via Diagram (\ref{eq:diagraminduction1})
To deal with the direct images
$R\iota_{\ast}$
and $Ra_{\ast}$, we describe the pushforward of cycles
$$\iota_{\ast}:\mathscr{L}(X_{L},K_L)\rightarrow \mathscr{L}(X_{G},K_L)\quad\text{ and }\quad a_{\ast}:\mathscr{L}(K\times_{K_L}X_{G},K)\rightarrow \mathscr{L}(X_{G},K)$$
that make the diagrams
\begin{align}\label{eq:diagramcycle1}
\xymatrix{
{D}_{c}^{b}(X_{G},K_L)\ar[r]^{CC} & \mathscr{L}(X_{G},K_L)\\
{D}_{c}^{b}(X_{L},K_L)\ar[u]^{R\iota_{\ast}}\ar[r]^{CC} &
\mathscr{L}(X_{L},K_L)\ar[u]^{\iota_{\ast}}
}
~\quad\text{ and }
\xymatrix{
{D}_{c}^{b}(X_{G},K)\ar[r]^{CC} & \mathscr{L}(X_{G},K)\\
{D}_{c}^{b}(K\times_{K_L}X_{G},K)\ar[u]^{Ra_{\ast}}\ar[r]^{CC} & \mathscr{L}(K\times_{K_L}X_{G},K)\ar[u]^{a_{\ast}}
}
\end{align}
commutative.
This will be done in a more general context than the one of the functions $\iota:X_L\rightarrow X_G$
and $a:K\times_{K_L}X_{G}\rightarrow X_{G}$.
Consider a morphism $F:X\rightarrow Y$ between two
smooth algebraic varieties. We work initially in the derived categories
and restrict our attention to the equivariant subcategories when
working with the equivariants maps $a$ and $\iota$.
The definition of $F_{\ast}:\mathscr{L}(X)\rightarrow \mathscr{L}(Y)$ and proof
of the commutativity of the diagram
\begin{align*}
\xymatrix{
D_{c}^{b}(Y)\ar[r]^{CC} & \mathscr{L}(Y)\\
D_{c}^{b}(X)\ar[u]^{RF_{\ast}}\ar[r]^{CC} & \mathscr{L}(X)\ar[u]^{F_{\ast}}
}
\end{align*}
is due principally to the work of Kashiwara-Schapira \cite{Kashiwara-Schapira} and Schmid-Vilonen \cite{SV}.
We give a short review of their work.
We begin by noticing that Schmid-Vilonen work in the derived category of sheaves having cohomology sheaves constructible with respect to a \textbf{\textit{semi-algebraic}} stratification.
We consider then the derived categories introducted at Section 3 as subcategories of
this larger category, and restrict their result to our framework when working with an algebraic map
$F:X\rightarrow Y$. By abuse of notation
we write, as in definition \ref{deftn:ccycleperverse}, $\mathscr{L}(X)$, respectively $\mathscr{L}(Y)$, for the set of \textbf{\textit{semi-algebraic Lagrangian cycles}} in $T^{\ast}X$, respectively $T^{\ast}Y$.
The definition of $F_{\ast}$ for an arbitrary algebraic map reduces to the case of proper maps and open embeddings. We begin by describing $F_{\ast}$ in the
case that $F$ is a proper map.
Consider the diagram
\begin{align}\label{eq:dF}
T^{\ast}X\xleftarrow{dF} X\times_{Y}T^{\ast} Y\xrightarrow{\tau} T^{\ast}Y,
\end{align}
where $\tau:X\times_{Y}T^{\ast} Y\rightarrow T^{\ast}Y$ is the projection on the second coordinate and for all $(x,(F(x),\lambda))\in X\times_Y T^{\ast}Y$ we have $dF(x,(F(x),\lambda))=(x,\lambda\circ dF_{x})$.
The assumption of $F$ being proper implies that $\tau$ is proper. Hence
we can as in Section 1.4 \cite{fulton} define a pushforward
of cycles $\tau_{\ast}:\mathscr{L}(X\times_{Y}T^{\ast}Y)\rightarrow\mathscr{L}(T^{\ast}Y)$.
Moreover, intersection theory (see Section 6.1, 6.2 and 8.1 \cite{fulton})
allows us to construct a
pullback of cycles $dF^{\ast}:\mathscr{L}(T^{\ast}X)\rightarrow\mathscr{L}(X\times_{Y}T^{\ast}Y)$.
The map $F_{\ast}:\mathscr{L}(X)\rightarrow \mathscr{L}(Y)$
is then defined as the composition of these two functions (see Equation 2.16 \cite{SV})
\begin{align}\label{eq:Gysin}
F_{\ast}:=\tau_{\ast}\circ dF^{\ast}.
\end{align}
The following result due to Kashiwara-Schapira
relates $F_{\ast}$ to the right derived functor $RF_{\ast}:D_{c}^{b}(X)\rightarrow D_{c}^{b}(Y)$
and proves the commutativity of (\ref{eq:diagramcycle1}) in the case of proper maps.
\begin{prop}[Proposition 9.4.2 \cite{Kashiwara-Schapira}]\label{prop:KS}
Let $F:X\rightarrow Y$ be a proper map. Then for all $\mathcal{F}\in D_{c}^{b}(X)$
$$CC(RF_{\ast}\mathcal{F})=F_{\ast}CC(\mathcal{F}).$$
\end{prop}
As explained at the end of Section 3 \cite{SV},
we can give a more explicit description of $F_{\ast}CC(\mathcal{F})$
by choosing a transverse family of cycles with limit equal to
$CC(\mathcal{F})$.
More precisely, suppose $C\in \mathscr{L}(X)$ is transverse to the map
$dF: X\times_{Y} T^{\ast}Y \rightarrow T^{\ast}X$. Then the geometric inverse image $dF^{-1}(C)$ of $C$ is well-defined as a cycle in $X\times_{Y} T^{\ast}Y$ and we have
$$dF^{\ast}(C)=dF^{-1}(C).$$
Consequently, $\tau_{\ast}dF^{-1}(C)$ is a well-defined cycle in $\mathscr{L}(Y)$.
Now, by Lemma 3.26 \cite{SV} we can choose
for every cycle $C_{0}\in \mathscr{L}(X)$
a family $\{C_{s}\}_{s\in (0,b)}\subset \mathscr{L}(X)$
such that the map $dF: X\times_{Y} T^{\ast}Y \rightarrow T^{\ast}X$ is transverse to supp$(C_{s})\subset T^{\ast}X$; for every $s\in (0,b)$ and
$$C_0=\lim_{s\rightarrow 0}C_{s}.$$
For more details about the construction of this family of cycles,
see Equation (3.10) and (3.11) \cite{SV}. Equations (3.12-3.16) \cite{SV} provide
the notion of limit of a family of cycles.
Schmid and Vilonen prove:
\begin{prop}[Proposition 3.27 of \cite{SV}]\label{prop:familylimit}
Suppose $F:X\rightarrow Y$ is proper. Let $C_{0}\in \mathscr{L}(X)$. Choose a family $\{C_{s}\}_{s\in (0,b)}\subset \mathscr{L}(X)$
with limit $C_{0}$ and such that
$dF: X\times_{Y} T^{\ast}Y \rightarrow T^{\ast}X$ is transverse to the support $|C_{s}|\subset T^{\ast}X$; for every $s\in (0,b)$.
Then
\begin{align}\label{eq:familylimit}
F_{\ast}(C_{0})=\lim_{s\rightarrow 0}\tau_{\ast}dF^{-1}(C_{s}).
\end{align}
\end{prop}
Having described $F_{\ast}$ when $F$ is proper, we explain now how to define
$F_{\ast}:\mathscr{L}(X)\rightarrow \mathscr{L}(Y)$
in the case when $F:X\rightarrow Y$ is an open embedding.
We follow
Chapter 4 \cite{SV}.
We start by choosing a real valued, semialgebraic $C^{1}$-function $f:X\rightarrow \mathbb{R}$, such that:
\begin{enumerate}
\item the boundary $\partial X$ is the zero set of $f$,
\item $f$ is positive on $X$.
\end{enumerate}
For more details on the existence of this map, see Equation (4.1) \cite{SV} and
Proposition I.4.5 \cite{Shiota}.
Suppose $C\in \mathscr{L}(X)$, and for each $s>0$ define
$C + sd\log f$ as the cycle of $X$ equal to the image of $C$ under the automorphism of
$T^{\ast}X$ defined by
\begin{align}\label{eq:open1}
(x,\xi)\mapsto \left(x,\xi+s\frac{df_{x}}{f(x)}\right).
\end{align}
Theorem 4.2 \cite{SV} relates the limit of the family of cycles $\{C+s d\log f\}_{s>0}$
(see page 468 \cite{SV} for the proof of why this family defines a family of cycles)
to the direct image $RF_{\ast}: D_c^{b}(X)\rightarrow D_c^{b}(Y)$.
We state it here as:
\begin{prop}\label{prop:openembeding}
Suppose $F:X\rightarrow Y$ is an open embedding. Let $\mathcal{F}\in D_{c}^{b}(X)$, then
$$CC(RF_{\ast}\mathcal{F})=\lim_{s\rightarrow 0} CC(\mathcal{F})+s d\log f.$$
\end{prop}
We notice that, while the family of cycles $\{CC(\mathcal{F})+s d\log f\}_{s>0}$
not necessarily lives in the set of characteristic cycles for the derived category of
sheaves whose cohomology is constructible with respect to an algebraic stratification, the limit does.
Following Proposition \ref{prop:openembeding}
we define for each $\mathcal{F}\in D_{c}^{b}(X)$
\begin{align}\label{eq:openpush}
F_{\ast}CC(\mathcal{F}):=\lim_{s\rightarrow 0} CC(\mathcal{F})+s d\log f.
\end{align}
\begin{comment}
Set,
$$C_{0}=\lim_{s\rightarrow 0} (\mathcal{F}+\frac{df}{ds}).$$
and,
$$\partial C=C_{0}-C$$
then the support $|\partial C|$ of the chain $\partial C$ is contained
in the inverse image of $\partial U$ in $T^{\ast}X$.
$|C|$ is contained in the union of the conormal bundles of a locally
finite family of smooth, locally closed, subanalytic subsets of $\partial U$.
\end{comment}
Finally, to treat the case of an arbitrary algebraic map $F:X\rightarrow Y$
we follow Chapter 6 \cite{SV}. We embed $X$ as an open subset of a compact algebraic manifold $\overline{X}$, and we factor $F$ into a product of three mappings: the closed embedding
\begin{align*}
i:X &\rightarrow X\times Y\\
x&\mapsto(x,F(x))
\end{align*}
which is a simple case of a proper direct image, the open inclusion
$$j : X\times Y \rightarrow \overline{X} \times Y ,$$
and the projection
$$\overline{p} : \overline{X}\times Y \rightarrow Y$$
which is also a proper map.
Then we can factor the derived functor $RF_{\ast}:D_{c}^{b}(X)\rightarrow D_{c}^{b}(Y)$ into the product
$$RF_{\ast}=R\overline{p}_{\ast}\circ Rj_{\ast}\circ Ri_{\ast}.$$
From Theorem \ref{prop:KS} and Theorem \ref{prop:openembeding}
for each $\mathcal{F}\in D_{c}^{b}(X)$ we have
\begin{align}\label{eq:generalpushforward}
CC(RF_{\ast}\mathcal{F})&=CC(R\overline{p}_{\ast}\circ Rj_{\ast}\circ Ri_{\ast}(\mathcal{F}))\\
&=\overline{p}_{\ast}CC(Rj_{\ast}\circ Ri_{\ast}(\mathcal{F}))\nonumber\\
&=(\overline{p}_{\ast}\circ j_{\ast})CC(Ri_{\ast}\mathcal{F})\nonumber\\
&=(\overline{p}_{\ast}\circ j_{\ast}\circ i_{\ast})CC(\mathcal{F}).\nonumber
\end{align}
Consequently, we define
\begin{align}\label{eq:svfunctor}
F_{\ast}&:\mathscr{L}(X)\rightarrow \mathscr{L}(Y)\\
F_{\ast}&:=\overline{p}_{\ast}\circ j_{\ast}\circ i_{\ast}.\nonumber
\end{align}
Let us return to our map $a:K\times_{K_L}X_G\rightarrow X_G$ of Definition \ref{deftn:geominduction0}.
Suppose $\mathcal{F}\in D_{c}^{b}(X_{L},K_L)$
and consider the sheaf
$I_{L}^{G}\mathcal{F}=Ra_{\ast}(\widetilde{R\iota_{\ast}\mathcal{F}})$
of Definition \ref{deftn:geominduction}.
Our objective is to give a more explicit description of
$CC(I_{L}^{G}\mathcal{F})$
by reducing its computation to the one of
the cycle $CC(\mathcal{F})$.
From (\ref{eq:generalpushforward})
we can write
\begin{align}\label{eq:reduction}
CC(I_{L}^{G}\mathcal{F})=CC(Ra_{\ast}(\widetilde{R\iota_{\ast}\mathcal{F}}))=a_{\ast}CC
(\widetilde{R\iota_{\ast}\mathcal{F}}).
\end{align}
Thus, to be able to compute
$CC(I_{L}^{G}\mathcal{F})$
the first step is to relate the characteristic cycle of $\widetilde{R\iota_{\ast}\mathcal{F}}$
to the characteristic cycle of $\mathcal{F}$.
This is done in the two following results. The first is a reformulation of Proposition 7.14 \cite{ABV}, Proposition 20.2 \cite{ABV} and Lemma 1.4 \cite{ViMir}.
\begin{prop}\label{prop:reduction2}
Suppose $X$ is a smooth complex algebraic variety
on which an algebraic group $H$ acts with finitely many orbits.
Suppose $G$ is an algebraic group containing $H$.
Consider the bundle
\begin{align*}
Y=G\times_{H}X,
\end{align*}
on which the group G acts by
$$g\cdot (g',x)=(gg',x).$$
Then:
\begin{enumerate}[1.]
\item The inclusion
\begin{align*}
i:X&\rightarrow Y\\
x& \mapsto \text{ equivalence class of }(e,x)
\end{align*}
induces a bijection
from $H$-orbits on $X$
to $G$-orbits on $Y$. Furthermore, this bijection preserves the closure relations
of closures.
\item There are natural equivalences of categories,
\begin{align*}
D_{c}^{b}(X,H)\cong D_{c}^{b}(G\times_{H}X,G),\quad
\mathcal{P}(X,H)\cong \mathcal{P}(G\times_{H}X,G),\quad
\mathcal{D}(X,H)\cong \mathcal{D}(G\times_{H}X,G).
\end{align*}
\item Write $j:\mathfrak{h}\times X\rightarrow \mathfrak{g}\times X$
for the inclusion, and consider the bundle map
$$j\times \mathcal{A}:\mathfrak{h}\times X\rightarrow \mathfrak{g}\times TX,$$
with $\mathcal{A}:\mathfrak{h}\times X\rightarrow TX$ defined as in (\ref{eq:actionmap2}). Write $Q$
for the quotient bundle: the fiber at $x$ is
$$Q_x=(\mathfrak{g}\times T_x X)/\{(X,\mathcal{A}_{x}(X)):X\in\mathfrak{h}\},$$
with $\mathcal{A}_{x}:\mathfrak{h}\times Y\rightarrow TY$ defined as in (\ref{eq:actionmap}). Then the tangent bundle of $Y$ is naturally isomorphic to the bundle on $Y$ induced by $Q$
$$TY\cong G\times_{H}Q.$$
\item The action mapping $\mathcal{A}_{y}:\mathfrak{h}\times Y\rightarrow TY$ (Equation (\ref{eq:actionmap})) may be computed as
follows. Fix a representative $(g,x)$ for the point $y$ of $G\times_{H}X$, and
an element $Z\in\mathfrak{g}$. Then
$$\mathcal{A}_{y}(Z)=\mathrm{~class~of~}(g,(\mathrm{Ad}(g^{-1})Z,(x,0))).$$
Here $(x,0)$ is the zero element of $T_x X$, so the term paired with $g$ on
the right side represents a class in $Q_x$.
\item The conormal bundle to the $G$-action on the induced bundle $G\times_{H}X$ is naturally induced by
the conormal bundle to the $H$-action on $X$:
\begin{align*}
T_{G}^{\ast}(G\times_{H}X)\cong G\times_{H}T_{H}^{\ast}X.
\end{align*}
\item
Suppose $\mathcal{F}$ and $\widetilde{\mathcal{F}}$ correspond
through any of the equivalences of categories of (2), above.
Then
$$CC(\widetilde{\mathcal{F}})=G\times_{H} CC({\mathcal{F}}).$$
In particular, the microlocal multiplicities (see Equation (\ref{eq:microlocalmultiplicity}) and
Theorem \ref{theo:defcc}$(ii)$) are given by
$$\chi_{G\times_{H}S}^{mic}(\widetilde{\mathcal{F}})=\chi_{S}^{mic}(\mathcal{F})$$
and
$$CC(\widetilde{\mathcal{F}})=\sum_{H-\mathrm{orbits }~S~\mathrm{in}~X}\chi_{S}^{mic}(\mathcal{F})[\overline{T_{G\times_{H}S}^{\ast}(G\times_{H}X)}].$$
\end{enumerate}
\end{prop}
\begin{cor}\label{cor:reductionstep2}
Suppose $\mathcal{F}\in D_{c}^{b}(X_{L},K_{L})$. Then
\begin{align}\label{eq:reductionstep2}
CC(\widetilde{R\iota_{\ast}\mathcal{F}})=\sum_{K_L-\mathrm{orbits }~S_L~\mathrm{in}~X_{L}} \chi_{S_L}^{mic}(\mathcal{F})[\overline{T_{K\times_{K_L} S_L}^{\ast}(K\times_{K_L}X_{G}})].
\end{align}
where $\widetilde{R\iota_{\ast}\mathcal{F}}$ is the unique element in
$D_{c}^{b}(K\times_{K_L}X_G,K)$ satisfying
$p^{\ast}(R{\iota}_{\ast}\mathcal{F})=\mu^{\ast}(\widetilde{R\iota_{\ast}\mathcal{F}})$.
\end{cor}
Before giving the proof of the Corollary, notice that from the definition of $\widetilde{R\iota_{\ast}\mathcal{F}}$
and the proof of Proposition \ref{prop:reduction2}(2) (see pages 93-94 \cite{ABV}),
$R\iota_{\ast}\mathcal{F}$ and $\widetilde{R\iota_{\ast}\mathcal{F}}$ correspond through the
equivalence of categories of Proposition \ref{prop:reduction2}(2).
\begin{proof}
Suppose $\mathcal{F}\in D_{c}^{b}(X_{L},K_{L})$ and write
$$CC(\mathcal{F})=\sum_{K_{L}-\mathrm{orbits }~S_L~\mathrm{in}~X_{L}} \chi_{S_L}^{mic}(\mathcal{F})[\overline{T_{S}^{\ast}X_{L}}]$$
for the corresponding characteristic cycle.
From Proposition \ref{prop:reduction2}(6),
the characteristic cycle of
$\widetilde{R\iota_{\ast}\mathcal{F}}$
may be identified as a cycle in $T^{\ast}_{K}(K\times_{K_L}X)$ as
\begin{align*}
CC(\widetilde{R\iota_{\ast}\mathcal{F}})=K\times_{K_L} CC(R\iota_{\ast}\mathcal{F}).
\end{align*}
From Proposition 6.21 \cite{ABV} the inclusion
$\iota:X_L \longrightarrow X_G$
is a closed immersion and in consequence proper.
Consider the diagram
\begin{align}
T^{\ast}X_{L}\xleftarrow{d\iota} X_{L}\times_{X_G}T^{\ast}X_G\xrightarrow{\tau} T^{\ast}X_{L}.
\end{align}
By Proposition \ref{prop:KS} we have
\begin{align*}
CC(R\iota_{\ast}\mathcal{F})&=\iota_{\ast}CC(\mathcal{F})\\
&=\tau_{\ast}\circ d\iota^{\ast}CC(\mathcal{F}),
\end{align*}
where $\tau_{\ast}$ and $d\iota^{\ast}$ are defined as in (\ref{eq:Gysin}).
Writing $CC(\mathcal{F})$ as the limit of a family of cycles transverse to
$d\iota$, by Proposition \ref{prop:familylimit} we
conclude that $\iota_{\ast}CC(\mathcal{F})$ is
the inverse image of $CC(\mathcal{F})$ under
$$d\iota:T^{\ast}{X_G}|_{\iota(X_{L})}\rightarrow T^{\ast}X_{L}.$$
Since the inverse image of the conormal bundle of each $K_L$-orbit
$S_{L}$ in $X_{L}$ is given by
$$d\iota^{-1}(T_{S_{L}}^{\ast}X_{L})=T_{\iota(S_{L})}^{\ast}X_G$$
we obtain
\begin{align*}
CC(R\iota_{\ast}\mathcal{F})&=\iota_{\ast}CC(\mathcal{F})\\
&=\sum_{K_L-\mathrm{orbits }~S_L~\mathrm{in}~X_L} \chi_{S_{L}}^{mic}(\mathcal{F})[\overline{T_{\iota(S_{L})}^{\ast}X_G}].
\end{align*}
Consequently
\begin{align*}
CC(\widetilde{R\iota_{\ast}\mathcal{F}})&=K\times_{K_L} \sum_{K_L-\mathrm{orbits }~S_L~\mathrm{in}~X_L} \chi_{S_{L}}^{mic}(\mathcal{F})[\overline{T_{\iota(S_{L})}^{\ast}X_G}]\\
&=\sum_{K_L-\mathrm{orbits }~S~\mathrm{in}~X_{G}} \chi_{S}^{mic}({R\iota_{\ast}\mathcal{F}})[\overline{T_{K\times_{K_L} \iota_{\ast}(S)}^{\ast}(K\times_{K_L}X_G})]\\
&=\sum_{K_L-\mathrm{orbits }~S_L~\mathrm{in}~X_L} \chi_{S_{L}}^{mic}(\mathcal{F})[\overline{T_{K\times_{K_L} \iota(S_{L})}^{\ast}(K\times_{K_L}X_G})].
\end{align*}
\end{proof}
The remaining step in the characterization of $CC(I_{L}^{G}\mathcal{F})$
is to describe the effect of taking $a_{\ast}$
on $CC(\widetilde{R\iota_{\ast}\mathcal{F}}).$
This is done in the proof of the following proposition.
\begin{prop}\label{prop:ccLG}
Suppose $\mathcal{F}\in D_{c}^{b}(X_{L},K_{L})$. Then
$$CC\left(I_{L}^{G}\mathcal{F}\right)=\sum_{K_L-\mathrm{orbits }~S_L~\mathrm{in}~X_L} \chi_{S_{L}}^{mic}(\mathcal{F})[\overline{T_{K\cdot\iota(S_L)}^{\ast}X_G}]+
\sum_{K-orbits~S\text{~in~}\partial(G\cdot \iota(X_{L}))} \chi_{S}^{mic}(I_{L}^{G}\mathcal{F})[\overline{T_{S}^{\ast}X_G}].$$
\end{prop}
\begin{proof}
Suppose $\mathcal{F}\in D_{c}^{b}(X_{L},K_{L})$ and write
$$CC(\mathcal{F})=\sum_{K_{L}-\mathrm{orbits }~S_L~\mathrm{in}~X_{L}} \chi_{S_L}^{mic}(\mathcal{F})[\overline{T_{S_L}^{\ast}X_{L}}]$$
for the corresponding characteristic cycle.
From Equation (\ref{eq:reduction}) and Corollary \ref{cor:reductionstep2}
\begin{comment} equation (\ref{eq:reduction})
we can divide the computation of (\ref{eq:objective}) into to steps: we start by using proposition (\ref{prop:reduction2})
to compute $CC(\widetilde{R\iota_{\ast}P_{L}}))$
then we apply $a_{\ast}:\mathscr{L}(X_{L})\rightarrow \mathscr{L}(X)$ (equation (\ref{eq:svfunctor}))
to this cycle to obtain $CC(P)$.
Let's start by describing
$CC(\widetilde{R\iota_{\ast}P_{L}})$.
From point (2) of proposition (\ref{prop:reduction2})
we can identify the characteristic cycle of
$\Gamma_{L}^{G}(X(\iota_{\ast})P_{L})$ as a cycle in $T^{\ast}_{G}(G\times_{L}X)$ as
\begin{align*}
CC(\widetilde{R\iota_{\ast}P_{L}}))=G\times_{L} CC(R\iota_{\ast}P_{L}),
\end{align*}
More precisely
\begin{align*}
CC(\widetilde{R\iota_{\ast}P_{L}})=\sum_{S_{L}\in \mathrm{cv}(M_{L})} \chi_{S_{L}}^{mic}({R\iota_{\ast}P_{L}}))[\overline{T_{G\times_{L} S_{L}}^{\ast}(G\times_{L}X})].
\end{align*}
and we can reduce the description of $CC(\widetilde{R\iota_{\ast}P_{L}}))$ to compute $CC({R\iota_{\ast}P_{L}}))$.
The map $\iota:{}^{L}L\rightarrow{}^{L}G$ is injective,
we know then from proposition (6.21) of \cite{ABV}) that $X(\iota):X(L)\rightarrow X(G)$ is
a closed immersion and in consequence proper.
Therefore if we consider the diagram
\begin{align}
T^{\ast}X_{L}\xleftarrow{dX(\iota)} X_{L}\times_{X}T^{\ast}X\xrightarrow{\tau} T^{\ast}X_{L}
\end{align}
from proposition (\ref{prop:KS}) we have
\begin{align*}
CC(R\iota_{\ast}P_{L})&=\iota_{\ast}CC(P_{L})\\
&=\tau_{\ast}\circ d\iota^{\ast}CC(P_{L}),
\end{align*}
where $\tau_{\ast}$ and $d\iota^{\ast}$ are defined as in (\ref{eq:Gysin}).
Writing $CC(P_{L})$ as the limit of a family of cycles transverse to
$d\iota$ it is not difficult to prove that $\iota_{\ast}CC(P_{L})$ it is
just the inverse image of $CC(P_{L})$ under $d\iota:T^{\ast}X|_{\iota(X_{L})}\rightarrow T^{\ast}X_{L}$ and because the inverse image of the conormal bundle of each $L$-orbit
$S_{L}$ in $X_{L}$ is given by
$$d\iota^{-1}(T_{S_{L}}^{\ast}X_{L})=T_{S_{L}}^{\ast}X$$
we obtain
\begin{align*}
CC(R\iota_{\ast}P_{L})&=\iota_{\ast}CC(P_{L})\\
&=\sum_{S_{L}\in \mathrm{cv}(P_{L})} \chi_{S_{L}}^{mic}(P_{L})[\overline{T_{X(\iota)(S_{L})}^{\ast}X}].
\end{align*}
Therefore
\begin{align*}
CC(\widetilde{R\iota_{\ast}P_{L}})&=G\times_{L} \sum_{S_{L}\in \mathrm{cv}(P_{L})} \chi_{S_{L}}^{mic}(P_{L})[\overline{T_{X(\iota)(S_{L})}^{\ast}X}]\\
&=\sum_{S_{L}\in \mathrm{cv}(P_{L})} \chi_{S_{L}}^{mic}(P_{L})[\overline{T_{G\times_{L} S_{L}}^{\ast}(G\times_{L}X})].
\end{align*}
and\end{comment}
we can write
\begin{align*}
CC\left(I_{L}^{G}\mathcal{F}\right)&=CC(Ra_{\ast}(\widetilde{R\iota_{\ast}\mathcal{F}}))\\
&=a_{\ast}(CC(\widetilde{R\iota_{\ast}\mathcal{F}}))\\
&=a_{\ast}\left( \sum_{K_L-\mathrm{orbits }~S_L~\mathrm{in}~X_L} \chi_{S_{L}}^{mic}(\mathcal{F})[\overline{T_{K\times_{K_L} \iota(S_{L})}^{\ast}(K\times_{K_L}X_G})]\right)
\end{align*}
We recall how $a_{\ast}$ is defined. As explained after Proposition \ref{prop:openembeding},
we embed ${K\times_{K_L} X_G}$ as an open subset of a compact algebraic manifold
$\overline{K\times_{K_L} X_G}$ and factorize $a$ into a product of three maps:
the closed embedding
\begin{align*}
i : K\times_{K_L} X_G &\rightarrow (K\times_{K_L} X_G)\times X_G,\\
(k,x)&\mapsto ((k,x),a(k,x))=((k,x),kx)
\end{align*}
the open inclusion
$$j : (K\times_{K_L} X_G)\times X_G \rightarrow \overline{K\times_{K_L} X_G}\times X_G,$$
and the projection
$$\overline{p} : \overline{K\times_{K_L} X_G}\times X_G\rightarrow X_G.$$
For later use we also denote the restriction of $\overline{p}$ to ${K\times_{K_L} X_G}\times X_G$ as
$$p : {K\times_{K_L} X_G}\times X_G\rightarrow X_G.$$
The map $a_{\ast}$ is defined by
$$a_{\ast}:=\overline{p}_{\ast}\circ j_{\ast}\circ i_{\ast}$$
with $\overline{p}_{\ast}$ and $i_{\ast}$ defined by (\ref{eq:Gysin})
and $j_{\ast}$ as in (\ref{eq:openpush}).
We have
\begin{comment}
\begin{align*}
CC(P)=CC(Ra_{\ast}X(\iota)P_{L})&=CC(\overline{p}_{\ast}(j_{\ast}(i_{\ast}(X(\iota)P_{L}))))\\
\end{align*}
because $\overline{p}$ is a proper map by Kashiwara-Schapira's proper push-forward we have
\begin{align*}
CC(P)=\overline{p}_{\ast}CC(j_{\ast}(i_{\ast}(X(\iota)P_{L}))).
\end{align*}
To deal with the open embedding $j$ we use Schmid-Vilonen open push-forward.
Defining $j_{\ast}CC(i_{\ast}(X(\iota)P_{L})))$ as in ()
from theorem () we obtain
\begin{align*}
CC(P)=\overline{p}_{\ast}(j_{\ast}CC(i_{\ast}(X(\iota)P_{L})))
\end{align*}
Finally $i$
is a proper direct image thus by using once again
Kashiwara-Schapira's proper push-forward
\end{comment}
\begin{align*}
CC\left(I_{L}^{G}\mathcal{F}\right)&=\overline{p}_{\ast}(j_{\ast}(i_{\ast}(CC(\widetilde{R\iota_{\ast}\mathcal{F}}))))\\
&=\overline{p}_{\ast}\left(j_{\ast}\left(i_{\ast}\left( \sum_{K_L-\mathrm{orbits }~S_L~\mathrm{in}~X_L} \chi_{S_{L}}^{mic}(\mathcal{F})[\overline{T_{K\times_{K_L} \iota(S_{L})}^{\ast}(K\times_{K_L}X_G})]\right)\right)\right).
\end{align*}
We start by describing
$i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}})$.
Notice that
the same argument used to compute $\iota_{\ast}CC(\mathcal{F})$ in Corollary \ref{cor:reductionstep2} allows us to conclude that
$i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}})$ is the inverse image of $CC(\widetilde{R\iota_{\ast}\mathcal{F}})$ under
$$di:T^{\ast}((K\times_{K_L}X_G)\times X_G)|_{i(K\times_{K_L}X_G)}\rightarrow T^{\ast} (K\times_{K_L}X_G).$$
Since the map $a$ occurs in the definition of $i$, to compute the inverse image
of $di$ we need first to compute the inverse image under
$$da: (K\times_{K_L}X_G)\times_{X_G}T^{\ast}X_G\rightarrow T^{\ast}(K\times_{K_L}X_G)$$
of the conormal bundle of each $K$-orbit in $K\times_{K_L}X_G$. To do that
we begin by noticing that, since by Proposition \ref{prop:reduction2}$(3)$ for each $(k,x)\in K\times_{K_L}X_G$ we have
$$T_{(k,x)}(K\times_{K_L}X_G)\cong (\mathfrak{k}\times T_{x}X_G)/\{(X,\mathcal{A}_{x}(X)):X\in \mathfrak{k}\cap \mathfrak{l}\},$$
with $\mathcal{A}_{x}:\mathfrak{k}\rightarrow T_{x}X_G$ defined as in (\ref{eq:actionmap}),
the map $da_{(k,x)}:T_{(k,x)}(K\times_{K_L}X_G)\rightarrow T_{kx}X_G$
can be represented as
$$da_{(k,x)}=\text{Ad}(k)\circ \mathcal{A}_{x}+\text{Ad}(k).$$
Then from Proposition \ref{prop:reduction2}$(4)$, is an easy
exercise
to verify that
for each $K$-orbit
$K\times_{K_L} \iota(S_{L})$ in $K\times_{K_L}X_G$, we have
$$d{a}^{-1}({T_{(k,x),K\times_{K_L} \iota(S_{L})}^{\ast}(K\times_{K_L}X_G)})={T_{kx,K\cdot\iota(S_L)}^{\ast}X_G}.$$
Now, for each $((k,x),x')\in (K\times_{K_L}\times X_G)\times X_G$ we have
$$T^{\ast}_{((k,x),x')}((K\times_{K_L}X_G)\times X_G)\cong
T^{\ast}_{(k,x)}(K\times_{K_L}X_G)\times T^{\ast}_{x'}X_G,$$
and the image of each element $(\lambda,\lambda')\in T^{\ast}_{(k,x)}(K\times_{K_L}X_G)\times T^{\ast}_{kx}X_G$ under the map $di$ is
$$(\lambda,\lambda')\mapsto \lambda+\lambda'\circ da_{(k,x)}.$$
Consequently, each element of $T^{\ast}_{(k,x)}(K\times_{K_L}X_G)\times T^{\ast}_{kx}X_G$
in the preimage under $di$ of the annihilator of $T_{(k,x)} K\times_{K_L} \iota(S_{L})$ in
$T^{\ast}_{(k,x)} (K\times_{K_L}X_G)$
must be a linear combination of elements of the form
\begin{itemize}
\item $(\lambda,0)\in T^{\ast}_{(k,x)}(K\times_{K_L}X_G)\times T^{\ast}_{kx}X_G$,\text{ with } $\lambda|_{T_{(k,x)} K\times_{K_L} \iota(S_{L})}=0,$
\item $(0,\lambda')\in T^{\ast}_{(k,x)}(K\times_{K_L}X_G)\times T^{\ast}_{kx}X_G$,\text{ with } $\lambda'|_{T_{kx} K\cdot\iota(S_{L})}=0,$
\item $(\lambda,\lambda')\in T^{\ast}_{(k,x)}(K\times_{K_L}X_G)\times T^{\ast}_{kx}X_G$,\text{ with } $(\lambda+\lambda'\circ da_{(k,x)})|_{T_{(k,x)} K\times_{K_L} \iota(S_{L})}=0,$
\end{itemize}
and one may verify that this space is ${T_{((k,x),kx),i(K\times_{K_L}\iota(S_{L}))}^{\ast}((K\times_{K_L}X_G)\times X_G)}$.
Therefore,
for the conormal bundle of each $K$-orbit
$K\times_{K_L} \iota(S_{L})$ in $K\times_{K_L}X_G$ we obtain
\begin{align}\label{eq:before}
di^{-1}({T_{K\times_{K_L} S_{L}}^{\ast}(K\times_{K_L}X_G)})
={T_{i(K\times_{K_L}\iota(S_{L}))}^{\ast}((K\times_{K_L}X_G)\times X_G)}
\end{align}
and so
\begin{align*}
i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}})
&=i_{\ast}\left( \sum_{K_L-\mathrm{orbits }~S_L~\mathrm{in}~X_L} \chi_{S_{L}}^{mic}(\mathcal{F})[\overline{T_{K\times_{K_L} \iota(S_{L})}^{\ast}(K\times_{K_L}X_G})]\right)\\
&=\sum_{K_L-\mathrm{orbits }~S_L~\mathrm{in}~X_L} \chi_{S_{L}}^{mic}(\mathcal{F})[\overline{T_{i(K\times_{K_L}\iota(S_{L}))}^{\ast}((K\times_{K_L}X_G)\times X_G)}].
\end{align*}
Next we compute $j_{\ast}(i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}}))$.
In order to do this we fix as in the paragraph previous to (\ref{eq:open1}) a
function
$$f:\overline{K\times_{K_L}X_G}\rightarrow \mathbb{R},$$
which takes strictly positive values
on $K\times_{K_L}X_G$ and vanishes on the boundary $\partial({K\times_{K_L}X_G})$.
Define the family of cycles
$\{i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}})+sd\log f\}_{s>0}$ as in (\ref{eq:open1}).
From Proposition \ref{prop:openembeding} we have
\begin{align}\label{eq:intermediate}
j_{\ast}(i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}}))&=\lim_{s\rightarrow 0}
i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}})+sd\log f\\
&=\lim_{s\rightarrow 0} \sum_{K_L-\mathrm{orbits }~S_L~\mathrm{in}~X_L} \chi_{S_{L}}^{mic}(\mathcal{F})[\overline{T_{i(K\times_{K_L}\iota(S_{L}))}^{\ast}((K\times_{K_L}X_G)\times X_G)}]+sd\log f.\nonumber
\end{align}
It only remains to compute the image under $\overline{p}_{\ast}$ of $j_{\ast}(i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}}))$.
Consider the diagrams
\begin{align*}
T^{\ast}(\overline{K\times_{K_L}X_G}\times X_G)\xleftarrow{d\overline{p}}(\overline{K\times_{K_L}X_G}\times X_G)\times_{X_G} T^{\ast}X_G\xrightarrow{\bar{\tau}}T^{\ast}X_G\\
T^{\ast}({K\times_{K_L}X_G}\times X_G)\xleftarrow{d{p}}({K\times_{K_L}X_G}\times X_G)\times_{X_G} T^{\ast}X_G\xrightarrow{{\tau}}T^{\ast}X_G
\end{align*}
Then by (\ref{eq:familylimit}) and (\ref{eq:intermediate})
\begin{align*}
\overline{p}_{\ast}(j_{\ast}(i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}})))=\overline{\tau}_{\ast} d\overline{p}^{\ast}\left(\lim_{s\rightarrow 0}
i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}})+sd\log f\right).
\end{align*}
By Lemma 6.4 \cite{SV}, the function $f$ can be chosen
in such a way that for every sufficiently small $s>0$,
$T^{\ast}_{K}(K\times_{K_L}X_G\times X_G)+sd\log f$
is transverse to
$(\overline{K\times_{K_L}X_G}\times X_G)\times_{X_G} T^{\ast}X_G$.
The transversality condition implies that
the geometric inverse image
$d\overline{p}^{-1}(i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}})+sd\log f)$
is well-defined as a cycle and
$$d\overline{p}^{\ast}(i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}})+sd\log f)=
d\overline{p}^{-1}(i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}})+sd\log f).$$
Moreover, by Proposition \ref{prop:familylimit}
\begin{align}\label{eq:dercor}
\overline{\tau}_{\ast} d\overline{p}^{\ast}\left(\lim_{s\rightarrow 0}
i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}})+sd\log f\right)
=\lim_{s\rightarrow 0}\overline{\tau}_{\ast}d\overline{p}^{-1}(i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}})+sd\log f).
\end{align}
Next, from Equation (\ref{eq:open1}) for each $s>0$, $i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}})+s d\log f$
is a cycle of $K\times_{K_L}X_G\times X_G$. Consequently, the family of cycles $\{i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}})+s d\log f\}_{s>0}$ lies entirely in $T^{\ast}(K\times_{K_L}X_G\times X_G)$;
this permits us to use the map $\tau_{\ast}$ and $dp$ instead of $\overline{\tau}_{\ast}$ and $d\overline{p}$
on the right hand side of (\ref{eq:dercor}) and write
\begin{align}\label{eq:eqcorfinal}
\overline{p}_{\ast}(j_{\ast}(i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}})))&=
\lim_{s\rightarrow 0}\tau_{\ast}d{p}^{-1}(i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}})+s d\log f)\\
&=\lim_{s\rightarrow 0}\tau_{\ast}d{p}^{-1}\left(\sum_{K_L-\mathrm{orbits }~S_L~\mathrm{in}~X_L} \chi_{S_{L}}^{mic}(\mathcal{F})[\overline{T_{i(K\times_{K_L}\iota(S_{L}))}^{\ast}((K\times_{K_L}X_G)\times X_G)}]+s d\log f\right).\nonumber
\end{align}
To compute the right hand side of (\ref{eq:eqcorfinal}) we begin by noticing that
$$(K\times_{K_L}X_G)\times X_G\times_{X_G} T^{\ast}X_G=(K\times_{K_L}X_G)\times T^{\ast}X_G=T^{\ast}_{K\times_{K_L}X_G}(K\times_{K_L}X_G)\times T^{\ast}X_G.$$
Hence $dp$ can be written as
$$dp: T^{\ast}_{K\times_{K_L}X_G}(K\times_{K_L}X_G)\times T^{\ast}X_G \rightarrow T^{\ast}(K\times_{K_L}X_G\times X_G),$$
with the image of each point $(0,\lambda)\in T^{\ast}_{(k,x),K\times_{K_L}X_G}(K\times_{K_L}X_G)\times T_{x'}^{\ast}X_G$ given by
\begin{align}
(0,\lambda) \mapsto \lambda\circ dp_{((k,x),x')}=(0,\lambda)
\end{align}
Therefore $dp$ defines an embedding, and we conclude that $dp^{-1}(i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}})+
s d\log f)$ is simply the intersection
between $T_{K\times_{K_L}X_G}^{\ast}(K\times_{K_L}X_G)\times T^{\ast}X_G$ and
$i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}})+s d\log f$.
Now, by Lemma \ref{lem:contention} the
space $K\cdot \mathrm{Ch}(R{\iota}_{\ast}\mathcal{F})$ is contained
in the characteristic variety of $I_{L}^{G}(\mathcal{F})$. Moreover, for each $s>0$, $dp^{-1}(i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}})+
s d\log f)$ defines a non-zero cycle. Hence
from the description of the elements in
$T_{((k,x),kx),i(K\times_{K_L}\iota(S_{L}))}^{\ast}((K\times_{K_L}X_G)\times X_G)$
given before Equation (\ref{eq:before}),
we obtain that for each point $(k,x)\in K\times_{K_L}\iota(S_{L})$ there exists
$\lambda(f)_{kx}\in T_{kx}^{\ast}X_G$ such that
$$\frac{df_{(k,x)}}{f(k,x)}=(\lambda(f)_{kx}\circ da_{(k,x)}).$$
Each point at the intersection of $T_{(k,x,kx),K\times_{K_L}X_G}(K\times_{K_L}X_G)\times T^{\ast}X_G$ and
$i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}})+s d\log f$ is then of the form
$$(-s\lambda(f)_{kx}\circ da_{(k,x)},\lambda+s\lambda(f)_{kx})+\left(s\frac{df_{(k,x)}}{f(k,x)},0\right)=(0,\lambda+s\lambda(f)_{kx}),\text{ where }\lambda\in T_{kx,K\cdot\iota(S_L)}^{\ast}X_G.$$
Consequently
$$\tau_{\ast}dp^{-1}\left({T_{((k,x),kx),i(K\times_{K_L}\iota(S_{L}))}^{\ast}(K\times_{K_L} X_G)\times X_G}+s \frac{df_{(k,x)}}{f(k,x)}\right)=
{T_{kx,K\cdot\iota(S_L)}^{\ast}X_G}+s\lambda(f)_{kx}.$$
Thus, if for each cycle $C\in \mathscr{L}(X_G,K)$
with $|C|\subset \overline{K\cdot \iota(X_L)}$ (See Definition \ref{deftn:ccycleperverse}), and for each $s>0$ we define
$C + s\lambda(f)$ as the cycle of $X_G$ equal to the image of $C$ under the automorphism
\begin{align*}
(x,\xi)\mapsto (x,\xi+s\lambda (f)_{x}),\quad x\in \overline{K\cdot \iota(X_L)},~\xi \in T_{x}^{\ast}X_G,
\end{align*}
we obtain
\begin{align*}
\tau_{\ast}d{p}^{-1}(i_{\ast}CC(\widetilde{R\iota_{\ast}\mathcal{F}})+s d\log f)
&=\tau_{\ast}dp^{-1}\left(\sum_{K_L-\mathrm{orbits }~S_L~\mathrm{in}~X_L} \chi_{S_{L}}^{mic}(\mathcal{F})[\overline{T_{i(K\times_{K_L}\iota(S_{L}))}^{\ast}((K\times_{K_L}X_G)\times X_G)}]\right)\\
&=\sum_{K_L-\mathrm{orbits }~S_L~\mathrm{in}~X_L} \chi_{S_{L}}^{mic}(\mathcal{F})[\overline{T_{K\cdot\iota(S_L)}^{\ast}X_G}]+s\lambda(f).
\end{align*}
Therefore
\begin{align*}
CC\left(I_{L}^{G}\mathcal{F}\right)&=a_{\ast}(CC(\widetilde{R\iota_{\ast}\mathcal{F}}))\\
&=\overline{p}_{\ast}(j_{\ast}(i_{\ast}(CC(\widetilde{R\iota_{\ast}\mathcal{F}}))))\\
&=\lim_{s\rightarrow 0}\sum_{K_L-\mathrm{orbits }~S_L~\mathrm{in}~X_L} \chi_{S_{L}}^{mic}(\mathcal{F})[\overline{T_{K\cdot\iota(S_L)}^{\ast}X_G}]+s\lambda(f).
\end{align*}
Next, since
for each $x\in a(K\times_{K_L}\iota(X_{L}))=K\cdot \iota(X_{L}) $ and every element $\lambda\in T_{x,K\cdot \iota(S_L)}^{\ast}X_G$,
we have $\lim_{s\rightarrow 0} \lambda+ s\lambda(f)|_{x}=\lambda,$
the restriction of $CC(I_{L}^{G}\mathcal{F})$ to
$a(K\times_{K_L}\iota(X_{L}))$
must coincide with the cycle
$$\sum_{K_L-\mathrm{orbits }~S_L~\mathrm{in}~X_L} \chi_{S_{L}}^{mic}(\mathcal{F})[\overline{T_{K\cdot\iota(S_L)}^{\ast}X_G}].$$ Consequently, if we
write
$$CC_{\partial} =CC\left(I_{L}^{G}\mathcal{F}\right)-\sum_{K_L-\mathrm{orbits }~S_L~\mathrm{in}~X_L} \chi_{S_{L}}^{mic}(\mathcal{F})[\overline{T_{K\cdot\iota(S_L)}^{\ast}X_G}],$$
then the support $|CC_{\partial}|$ of the cycle $CC_{\partial}$ will be
contained in the inverse image of the boundary
$\partial(K\cdot \iota(X_{L}))$ in $T^{\ast}X_G$. By a similar argument to the one of (4.6c)\cite{SV}, we can moreover conclude
that $|CC_{\partial}|$ is the union of the conormal bundles of a family of
$K$-orbits in the boundary of $K\cdot \iota(X_{L})$.
This leads us to the desired equality
$$CC\left(I_{L}^{G}\mathcal{F}\right)=\sum_{K_L-\mathrm{orbits }~S_L~\mathrm{in}~X_L} \chi_{S_{L}}^{mic}(\mathcal{F})[\overline{T_{K\cdot\iota(S_L)}^{\ast}X_G}]+
\sum_{K-orbits~S\text{~in~}\partial(K\cdot \iota(X_{L}))} \chi_{S}^{mic}(I_{L}^{G}\mathcal{F})[\overline{T_{S}^{\ast}X_G}].$$
\begin{comment}
it's a subset of $G\times \epsilon(\iota(X(L)))$
$CC$ its a cycle of and we can
restrict ourselves to the set and consider the map,
$\rightarrow$
To compute we factor into a product of two maps,
$$\rightarrow$$
and the inclusion
$$\rightarrow$$
Then,
$$p_{\ast}CC().$$
Let's choose a family of cycles,
$$=.$$
From (),
$$=$$
moreover,
$$dp^{-1}(T_{S_{L}}X)=T_{G\times \epsilon(\iota)(S_{L})}^{\ast}X$$
Therefore,
$$=.$$
To treat the inclusion we use proposition.
Let f be a such that,
We extend into the inclusion
and define the family.
By proposition we have
\begin{align*}
&=\\
&= open embedding
\end{align*}
\end{comment}
\end{proof}
To end with this section notice that from (\ref{eq:diagramcycle1})
and Proposition \ref{prop:reduction2}(6),
we can by defining
\begin{align}
\left(I_{L}^{G}\right)_{\ast}:\mathscr{L}(X_{L},K_L)&\rightarrow \mathscr{L}(X_{G},K)\\
\sum_{K_{L}-\text{orbits }S\text{ in }X_L}m_S[\overline{T_{S}^{\ast}X_L}]& \mapsto a_{\ast}\left(
\sum_{K_{L}-\text{orbits }S\text{ in }X_L}m_S[\overline{T_{K\times_{K_L}\iota(S)}^{\ast}(K\times_{K_L} X_G)}]\right)\nonumber
\end{align}
extend (\ref{eq:cdiagramme1}) to obtain the commutative diagram
\begin{align*}
\xymatrix{
K\mathcal{M}(\mathfrak{g},K,I_{X_{G}}) \ar[r] & K(X_{G},K)\ar[r]^{CC} & \mathscr{L}(X_G,K)\\
K\mathcal{M}(\mathfrak{l},K_{L},I_{X_{L}}) \ar[u]^{\mathscr{R}_{(\mathfrak{l},K_{L})}^{(\mathfrak{g},K)}}\ar[r] & K(X_{L},K_{L})\ar[u]^{I_{L}^{G}}\ar[r]^{CC} & \mathscr{L}(X_{L},K_L).\ar[u]^{\left(I_{L}^{G}\right)_{\ast}}
}
\end{align*}
\section{The Langlands Correspondence}
In this section we give a quick review on the Langlands classification as explained in
\cite{ABV}.
We follow Chapters 4, 5 and 10 \cite{ABV}.
The Local Langlands Correspondence gives a classification of representations of strong real forms of
$G$ in terms of a set of parameters of an $L$-group of $G$.
In \cite{ABV}, the authors generalize the Langlands Correspondence to
include in the classification, representations of
a special type of covering group of $G$.
In this more general setting
the role of the $L$-groups
in the descriptions of the representations
is played by the more general notion of $E$-group.
The $E$-groups
are introduced for the first time by Adams and Vogan in \cite{AV}.
$E$-groups will also be necessary
to describe in Section 7.3 the Adams-Johnson packets.
The section is divided as follows. We begin
by introducing the notions of
$L$-group
and $E$-group. Next, we recall the definition of an $L$-parameter. The section ends with the formulation of the Local Langlands
Correspondence.\\
Let $\Psi_{0}(G)=\left(X^{\ast},\Delta,X_{\ast},\Delta^{\vee}\right)$ be the based root datum of $G$ and
write ${}^{\vee}\Psi_{0}(G)=\left(X_{\ast},\Delta^{\vee},X^{\ast},\Delta\right)$ for the dual based root datum to $\Psi_{0}(G)$. Then
\begin{align}\label{eq:dualautomorphism1}
\text{Aut}(\Psi_{0}({G}))\cong\text{Aut}\left({}^{\vee}\Psi_{0}({G})\right).
\end{align}
\begin{deftn}[Definition 4.2 \cite{AV} and Definition 4.3 \cite{ABV}]
Suppose $G$ is a complex connected reductive algebraic group. A \textbf{dual
group} for $G$ is a complex connected reductive algebraic group ${}^{\vee}{G}$ whose
based root datum
is dual to the based root
datum of $G$, i.e.
\begin{align}\label{eq:dualautomorphism2}
\Psi_{0}\left({}^{\vee}{G}\right)\cong{}^{\vee}\Psi_{0}({G}).
\end{align}
A \textbf{weak $E$-group} for $G$ is an algebraic group ${}^{\vee}G^{\Gamma}$ containing the dual group ${}^{\vee}{G}$ for $G$
as a subgroup of index two. That is, there is a short exact sequence
\begin{align*}
1\rightarrow {}^{\vee}{G}\rightarrow {}^{\vee}G^{\Gamma}\rightarrow \Gamma\rightarrow 1,
\end{align*}
where $\Gamma=\mathrm{Gal}(\mathbb{C}/\mathbb{R})$.
\end{deftn}
As explained in Chapter 4 of \cite{ABV}, we can give a simple classification of weak $E$-groups
${}^{\vee}G^{\Gamma}$ for $G$. The classification is similar to that
given in Proposition \ref{prop:classificationextended} for extended groups.
In order to give this description in more details we begin by recalling (see for example Proposition 2.11 of
\cite{ABV})
that there is a natural short exact sequence
\begin{align}\label{eq:sequenceexactmorphism}
1\rightarrow \text{Int}\left({}^{\vee}G\right)\rightarrow \text{Aut}\left({}^{\vee}G\right)\xrightarrow{\Psi_0} \text{Aut}\left(\Psi_{0}\left({}^{\vee}{G}\right)\right)\rightarrow 1,
\end{align}
and that this sequence splits (not canonically), as follows;
Choose a Borel subgroup ${}^{\vee}B$ of ${}^{\vee}G$, a maximal torus ${}^{\vee}T\subset {}^{\vee}B$, and a set of basis
vectors $\{X_{\alpha}\}$ for the simple root spaces of ${}^{\vee}T$ in the Lie algebra ${}^{\vee}\mathfrak{b}$ of ${}^{\vee}B$; and define $\text{Aut}\left({}^{\vee}G,^{\vee}B,^{\vee}T,\{X_{\alpha}\}\right)$ to be the set of algebraic automorphisms of ${}^{\vee}G$ preserving ${}^{\vee}B$, ${}^{\vee}T$, and $\{X_{\alpha}\}$ as sets.
Then the restriction of $\Psi_0$ to $\text{Aut}\left({}^{\vee}G,^{\vee}B,^{\vee}T,\{X_{\alpha}\}\right)$ is an isomorphism.
An automorphism belonging to one of the sets $\text{Aut}(^{\vee}G,^{\vee}B,^{\vee}T,\{X_{\alpha}\})$
is called \textit{\textbf{distinguished}}.
Now, let ${}^{\vee}\delta$ be any element in ${}^{\vee}G^{\Gamma}-{}^{\vee}G$
and write $\sigma_{{}^{\vee}\delta}$
for the automorphism of ${}^{\vee}G$ defined by the conjugation action of ${}^{\vee}\delta$ in ${}^{\vee}G$.
From (\ref{eq:sequenceexactmorphism}), we see that $\sigma_{{}^{\vee}\delta}$
induces an involutive automorphism of the based root datum
$$a=\Psi_0(\sigma_{{}^{\vee}\delta})\in \text{Aut}\left(\Psi_0({}^{\vee}G)\right)$$
which is independent of the choice of ${}^{\vee}\delta$.
Suppose moreover that
$\sigma_{{}^{\vee}\delta}$
is a distinguished automorphism of ${}^{\vee}G$.
Set $\theta_Z$ to be the restriction of $\sigma_{{}^{\vee}\delta}$ to $Z({}^{\vee}G)$, which is independent of the choice of ${}^{\vee}\delta$,
and consider the class
$$\overline{z}\in Z\left({}^{\vee}G\right)^{\theta_Z}/(1+\theta_Z)Z({}^{\vee}G)$$
of the element $z={}^{\vee}\delta^{2}\in Z({}^{\vee}G)^{\theta_Z}$,
where $$(1+\theta_Z)Z({}^{\vee}G)=\{z\theta_Z (z):z\in Z({}^{\vee}G)\}.$$
We notice that $\overline{z}$ is independent of the choice of the
distinguished automorphism
$\sigma_{{}^{\vee}\delta}$, indeed, from (\ref{eq:sequenceexactmorphism}),
any other element ${}^{\vee}\delta$
defining a distinguished automorphism of ${}^{\vee}G$
will be of the form ${}^{\vee}\delta'=z_1g{}^{\vee}\delta g^{-1}$
for some $z_1\in Z({}^{\vee}G)$ and $g\in {}^{\vee}G$.
Consequently $({}^{\vee}\delta')^{2}=z_1\theta_Z(z_1)^{\vee}\delta^{2}$, thus ${}^{\vee}\delta^{2}$ and $({}^{\vee}\delta')^{2}$
belong to the same class.
Therefore, to any week $E$-group
${}^{\vee}G^{\Gamma}$ for $G$ we can attach two invariants:
$${}^{\vee}G^{\Gamma}\longrightarrow (a,\overline{z}).$$
Furthermore, from point $b)$ of Proposition 4.4 \cite{ABV}, any two $E$-groups with the same couple of invariants are isomorphic. Conversely, suppose $a\in \text{Aut}(\Psi({}^{\vee}G))$
is an involutive automorphism. Write $\theta_Z$ for the involutive automorphism of $Z({}^{\vee}G)$ defined by the action of any automorphism of ${}^{\vee}G$ corresponding to $a$ under
$\Psi_{0}$, and suppose $$\overline{z}\in Z({}^{\vee}G)^{\theta_Z}/(1+\theta_Z)Z({}^{\vee}G).$$
Then from point $c)$ of Proposition 4.4 of \cite{ABV}, there is a weak $E$-group
with invariants $(a,\overline{z})$.
\begin{deftn}[Definition 4.6, 4.12 and 4.14 \cite{ABV}]\label{deftn:egroup}
Suppose $G$ is a complex connected reductive algebraic group. An \textbf{$E$-group} for $G$ is a pair $\left({}^{\vee}G^{\Gamma},\mathcal{S}\right)$, subject to the following conditions.
\begin{enumerate}
\item ${}^{\vee}G^{\Gamma}$ is a weak $E$-group for $G$.
\item $\mathcal{S}$ is a conjugacy class of pairs $({}^{\vee}\delta,{}^{d}B)$
with ${}^{\vee}\delta$ an element of finite order in ${}^{\vee}G^{\Gamma}-{}^{\vee}{G}$ and
${}^{d}B$ a Borel subgroup of ${}^{\vee}G$.
\item Suppose $\left({}^{\vee}\delta,{}^{d}B\right)\in \mathcal{S}$. Then conjugation by
${}^{\vee}{\delta}$ is a distinguished involutive automorphism
${}^{\vee}\sigma$ of ${}^{\vee}G$
preserving ${}^{d}B$.
\end{enumerate}
The invariants of the $E$-group are the automorphism $a$ attached to ${}^{\vee}G^{\Gamma}$ as before and the element
$$z={}^{\vee}\delta^{2}\in Z({}^{\vee}G)^{\theta_Z}$$
with ${}^{\vee}\delta$ any element in $\left({}^{\vee}\delta,{}^{d}B\right)$.
An \textbf{$L$-group} for $G$ is an $E$-group whose second invariant is equal to $1$, that is to say $z={}^{\vee}\delta^{2}=1$.
\end{deftn}
Just as for weak $E$-groups, $E$-groups can be completely classified by the couple of
invariants $(a,z)$ described above. From point $a)$ of Proposition 4.7 of \cite{ABV},
two $E$-groups with the same couple of parameters are isomorphic.
Furthermore, if we fix a weak $E$-group for $G$
with invariants $(a,\overline{z})$ and if $\left({}^{\vee}G^{\Gamma},\mathcal{S}\right)$ is
an $E$-group, then from point $c)$ of the same proposition,
its second invariant is a representative for the class of $\overline{z}$.
Conversely, if $z\in Z({}^{\vee}G)^{\theta_Z}$ is an element
of finite order representing the class of $\overline{z}$, then there is an $E$-group
structure on ${}^{\vee}G^{\Gamma}$ with second invariant $z$.
\\
Suppose now that $G^{\Gamma}$ is an extended group for $G$. As mentioned earlier, to each extended group
correspond an inner class of real forms. Furthermore,
from proposition (2.12) \cite{ABV}, inner classes of real forms
are in one-to-one correspondence with involutive automorphisms of $\Psi_{0}({G})$,
and from (\ref{eq:dualautomorphism1}) and (\ref{eq:dualautomorphism2})
with involutive automorphisms of $\Psi_{0}({}^{\vee}{G})$.
Let $a$ be the involutive automorphism of $\Psi_{0}({}^{\vee}{G})$
corresponding to the inner class of real form defined by $G^{\Gamma}$.
Then a (weak) $E$-group ${}^{\vee}G^{\Gamma}$ with first invariant $a$
will be called
\textit{\textbf{a (weak)} $E$\textbf{-group for} $G^{\Gamma}$}
or \textit{\textbf{a (weak)} $E$\textbf{-group for} $G$ \textbf{and the specified inner class of real forms}}.\\
As explained at the beginning of this section, to describe the Langlands classification in the more general setting of \cite{ABV},
we need to introduce a family of covering groups of $G$.
We do this in the next definition.
\begin{deftn}[Definition 10.1 and 10.3 \cite{ABV}]\label{deftn:canonicalcover}
Suppose $G^{\Gamma}$ is an extended group for $G$. A connected finite covering group
$$1\rightarrow F\rightarrow \widetilde{G}\rightarrow G\rightarrow 1$$
is said to be \textbf{distinguished} if the following two conditions are satisfied.
\begin{enumerate}
\item For every $x\in G^{\Gamma}-G$, the conjugation action $\sigma_x$ of $x$ lifts to an automorphism of $\widetilde{G}$.
\item The
restriction of $\sigma_x$ to $Z(\widetilde{G})$, which is independent of the choice of $x$,
sends every element of $F$ to its inverse.
\end{enumerate}
We define the \textbf{canonical covering} $G^{can}$ of $G$ as the projective limit of all
the distinguished coverings of $G$ and we write
$$1\rightarrow \pi_1(G)^{can}\rightarrow {G}^{can}\rightarrow G\rightarrow 1.$$
Now, if $\delta$ is a strong real form of $G^{\Gamma}$, let ${G}(\mathbb{R},\delta)^{can}$ be the preimage of ${G}(\mathbb{R},\delta)$ in
${G}^{can}$. Then there is a short exact sequence
$$1\rightarrow \pi(G)^{can}\rightarrow {G}(\mathbb{R},\delta)^{can}\rightarrow
{G}(\mathbb{R},\delta)\rightarrow 1.$$
A \textbf{canonical projective representation of a strong real form} of $G^{\Gamma}$
is a pair $(\pi,\delta)$ subject to
\begin{enumerate}
\item $\delta$ is a strong real form of $G^{\Gamma}$.
\item $\pi$ is an admissible representation of $G^{can}(\mathbb{R},\delta)$.
\end{enumerate}
With equivalence defined as in definition \ref{deftn:representationstrongrealform}.
Let $a$ be the involutive automorphism of $\Psi_{0}({}^{\vee}{G})$
corresponding to the inner class of real form attached to $G^{\Gamma}$
and write $\theta_Z$ for the involutive automorphism of $Z({}^{\vee}G)$ defined by the action of any automorphism of ${}^{\vee}G$ corresponding to $a$ under
$\Psi_{0}$.
From Lemma 10.2$(d)$ \cite{ABV}
we have an isomorphism
\begin{align*}
\text{Elements of finite order in }Z({}^{\vee}G)^{\theta_Z}&\rightarrow \mathrm{Hom}_{\mathrm{cont}}(\pi_{1}(G)^{can},C^{\times})\\
z&\mapsto \chi_z.
\end{align*}
Suppose $z\in Z({}^{\vee}G)^{\theta_Z}$. We say that $(\pi,\delta)$ is of \textbf{type z} if the restriction of $\pi$ to
$\pi_{1}(G)^{can}$ is a multiple of $\chi_z$. Finally define
$$\Pi^{z}(G/\mathbb{R})$$
to be the set of infinitesimal equivalence classes of irreducible canonical representations
of type $z$.
\end{deftn}
We turn now to the definition of the Langlands parameters.
Fix an extended group $G^{\Gamma}$ for $G$, and an $E$-group $\left({}^{\vee}{G}^{\Gamma},\mathcal{S}\right)$
for the corresponding inner class of real forms. We begin by recalling the definition of the Weil group of $\mathbb{R}$
\begin{comment}
\begin{deftn}
A (weak) extended group containing $G$ is a real Lie group $G^{\Gamma}$
subject to the following conditions.
\begin{enumerate}
\item $G^{\Gamma}$ contains $G$ as a subgroup of index two.
\item Every element of $G^{\Gamma}-G$ acts on $G$ as an antiholomorphic automorphism.
\end{enumerate}
A strong real form $\delta$ is an element of
$G^{\Gamma}$ such that $\delta^{2}\in Z(G)$ has finite order.
The group of real points of $\delta$ is defined to be:
$$G(\mathbb{R},\delta)=\{g\in G:\delta g\delta^{-1}=g \} $$
Two strong real forms of $G^{\Gamma}$ are called equivalent if they are conjugated by $G$.
\end{deftn}
\begin{deftn}
A representation of a strong real form of is a pair $(\pi,\delta)$ subject to
\begin{enumerate}
\item $\delta$ is a strong real form of $G^{\Gamma}$.
\item $\pi$ is an admissible representation of $G(\mathbb{R},\delta)$.
\end{enumerate}
Two such representations $(\pi,\delta)$, $(\pi',\delta')$ are said to be
equivalent if there is an element $g\in G$ such that
$g\delta g^{-1}=\delta'$ and $\pi\circ Ad(g^{-1})$ is (infinitesimally) equivalent to
$\pi'$.
We define,
$$\Pi(G/\mathbb{R})$$
to be the set of infinitesimal equivalence classes of irreducible representations of strong real
forms of $G^{\Gamma}$.
\end{deftn}
\end{comment}
\begin{deftn}
The Weil group of $\mathbb{R}$ it's a non-split extension of the Galois group of $\mathbb{C}/\mathbb{R}$
by the group of non-zero complex numbers $\mathbb{C}^{\times}$:
$$1\rightarrow \mathbb{C}^{\times}\rightarrow W_{\mathbb{R}}\rightarrow \mathrm{Gal}(\mathbb{C}/\mathbb{R})\rightarrow 1$$
By identifying $\mathbb{C}^{\times}$ with his image in $W_{\mathbb{R}}$
and $\mathrm{Gal}(\mathbb{C}/\mathbb{R})$ with $\{\pm 1\}$, we can see that is generated by
$\mathbb{C}^{\times}$ and a distinguished element $j$, subject to the relations
$$j^{2}=-1\in\mathbb{C}^{\times},\qquad jzj^{-1}=\bar{z},~z\in\mathbb{C}^{\times}.$$
\end{deftn}
\begin{deftn}(see \cite{Langlands})
A \textbf{Langlands parameter} $\varphi$ for
the weak $E$-group ${}^{\vee}{G}^{\Gamma}$ is a continuous group homomorphism,
$$\varphi:W_{\mathbb{R}}\longrightarrow {}^{\vee}{G}^{\Gamma},$$
such that:
\begin{itemize}
\item the diagram
\begin{align*}
\xymatrix{ W_{\mathbb{R}} \ar[rr]^{\varphi} \ar[rd] && {}^{\vee}{G}^{\Gamma}\ar[ld] \\ & \Gamma }
\end{align*}
is commutative
\item for every $w\in W_{\mathbb{R}}$ the projection of $\varphi(w)$ in ${}^{\vee}{{G}}$ is semisimple.
\end{itemize}
The set of Langlands parameters will be denoted $P\left({}^{\vee}{G}^{\Gamma}\right)$.
We make ${}^{\vee}{G}$ act on $P\left({}^{\vee}{G}^{\Gamma}\right)$ by conjugation
and denote the set of conjugacy classes of Langlands parameters
by $\Phi\left({}^{\vee}{G}^{\Gamma}\right)$.
\end{deftn}
\begin{deftn}(Definition 5.3 \cite{ABV})
Suppose
$\left({}^{\vee}G^{\Gamma},\mathcal{S}\right)$ is an $L$-group
Then we write
$$\Phi\left(G/\mathbb{R}\right)=\Phi\left({}^{\vee}G^{\Gamma}\right).$$
More generally, if
$\left({}^{\vee}G^{\Gamma},\mathcal{S}\right)$
is an $E$-group for $G$
with second invariant $z$, then we write
$$\Phi^{z}\left(G/\mathbb{R}\right)=\Phi\left({}^{\vee}G^{\Gamma}\right).$$
\end{deftn}
\begin{deftn}(Definition 5.11 \cite{ABV})\label{deftn:completeLanglandsparameter}
Write ${}^{\vee}G^{alg}$
for the algebraic universal covering of ${}^{\vee}G$, i.e.
the projective limit of all the finite covers of ${}^{\vee}G$.
Suppose $\varphi\in P\left({}^{\vee}{G}^{\Gamma}\right)$. Set
\begin{align*}
{}^{\vee}{G}_{\varphi}=\text{centralizer~in~} {}^{\vee}{G}\text{~of~} \varphi(W_{\mathbb{R}})
\end{align*}
and write
\begin{align*}
{}^{\vee}{G}_{\varphi}^{alg}=\text{preimage~of~}{}^{\vee}{G}_{\varphi} \text{~in~the~ algebraic~universal~cover~of~}{}^{\vee}{G}.
\end{align*}
Then we define the Langlands component group for $\varphi$
to be the quotient
$$A_{\varphi}={}^{\vee}{{G}}_{\varphi}/({}^{\vee}{{G}}_{\varphi})_{0},$$
where $({}^{\vee}{G}_{\varphi})_0$ denotes the identity component of ${}^{\vee}{G}_{\varphi}$.
Similarly, the universal component group for $\varphi$ is defined as
$$A_{\varphi}^{alg}={}^{\vee}{{G}}_{\varphi}^{alg}/({}^{\vee}{{G}}_{\varphi}^{alg})_{0}.$$
A \textbf{complete Langlands parameter} for ${}^{\vee}{G}^{\Gamma}$
is a pair $(\varphi,\tau)$ with
$\varphi\in P\left({}^{\vee}{G}^{\Gamma}\right)$
and $\tau$ an irreducible representation of $A_{\varphi}^{alg}$.
We make ${}^{\vee}{G}$ act on the set of complete Langlands parameter by conjugation and write
$$\Xi\left({}^{\vee}{G}^{\Gamma}\right):=\text{Set~of~conjugacy~classes~of~complete~Langlands~parameters~for~} {}^{\vee}{G}^{\Gamma}.$$
Finally,
let $\left({}^{\vee}G^{\Gamma},\mathcal{S}\right)$ be an $E$-group for $G$
with second invariant $z$, then
we write
$$\Xi^{z}({G}/\mathbb{R})=\Xi\left({}^{\vee}{G}^{\Gamma}\right)$$
or simply $\Xi({G}/\mathbb{R})$ if ${}^{\vee}{G}^{\Gamma}$ is an $L$-group.
\end{deftn}
We can now state the Local Langlands Correspondence (Theorem 10.4 \cite{ABV}).
\begin{theo}[\textbf{Local Langlands Correspondence for $E$-groups}]\label{theo:4.1}
Suppose $G^{\Gamma}$
is an extended group for $G$, and $({}^{\vee}{G}^{\Gamma},\mathcal{S})$ is a $E$-group for the corresponding inner class of real forms. Write $z$ for the second invariant of
the $E$-group. Then there is a natural bijection between the set $\Pi^{z}({G}/\mathbb{R})$
of equivalence classes of canonical irreducible projective representations of strong real forms of ${G}$ of type $z$
and the set $\Xi^{z}({G}/\mathbb{R})$ of complete Langlands parameters
for ${}^{\vee}{G}^{\Gamma}$.
\end{theo}
We notice that, when $\left({}^{\vee}{G}^{\Gamma},\mathcal{S}\right)$ is a $L$-group for $G$
the projective representations in the Correspondence
are actual representations of strong real forms of ${G}$.
Moreover, in this parameterization, the set of representations of a fixed real form ${G}(\mathbb{R},\delta)$ corresponding to complete Langlands parameter supported on a single orbit is precisely
the $L$-packet for ${G}(\mathbb{R},\delta)$ attached to that orbit.
\section{Geometric parameters}
Following Chapter 6 \cite{ABV}, in this section we introduce a new set of parameters
that is going to replace the set of Langlands parameters
in the Local Langlands Correspondence.
The section is divided as follows. We begin with a short motivation for the definition by Adams, Barbasch and Vogan of this new set of parameters.
Next, we describe
some of its properties, to end with the reformulation in this new setting
of the Local Langlands Correspondence.
The Langlands classification stated in the previous section
can be expressed in a more geometric manner.
We can think of the space
$P\left({}^{\vee}{G}^{\Gamma}\right)$ as a variety with an action of
${}^{\vee}{G}$ by conjugation. The ${}^{\vee}G$-conjugacy class of any $L$-parameter $\varphi\in P\left({}^{\vee}{G}^{\Gamma}\right)$
is thus a ${}^{\vee}{G}$-orbit $S_{\varphi}$ on $P\left({}^{\vee}{G}^{\Gamma}\right)$ and
$$S_{\varphi}\cong {}^{\vee}{G}/{}^{\vee}{G}_{\varphi}.$$
Furthermore, a complete Langlands parameter $(\varphi,\tau)$ for ${}^{\vee}{G}^{\Gamma}$
corresponds to a ${}^{\vee}G$-equivariant local system $\mathcal{V}_{\varphi}$ on $S_{\varphi}$
(i.e. a ${}^{\vee}G$-equivariant vector bundle with a flat connection):
\begin{align}\label{eq:localsystem}
\mathcal{V}_{\varphi}\cong {}^{\vee}{{G}}\times_{{}^{\vee}{{G}}_{\varphi}}V_{\tau}\rightarrow S_{\varphi}.
\end{align}
As
explained in the introduction of {\cite{ABV}},
with this more geometric viewpoint one might hope that,
by analogy
with the theory created by Kazhdan-Lusztig and Beilinson-Bernstein in {\cite{KL}} and {\cite{BB}},
information about irreducible characters should be encoded
by perverse sheaves on the closure of ${}^{\vee}{{G}}$-orbits on $P\left({}^{\vee}{G}^{\Gamma}\right)$.
Unfortunately, the orbits
on $P\left({}^{\vee}{G}^{\Gamma}\right)$ are already closed, hence these
perverse sheaves are nothing but the local systems of (\ref{eq:localsystem}).
To remedy to this situation Adams, Barbasch and Vogan
introduce a new space
with a ${}^{\vee}{{G}}$-action
having the same set of orbits as $P\left({}^{\vee}{G}^{\Gamma}\right)$,
but with a more interesting geometry in which orbits are not necessary closed.
This new set will be called, set of geometric parameters.
\begin{deftn}[Definition 1.7 \cite{ABV}] Let $\lambda\in {}^{\vee}{\mathfrak{g}}$ be a semisimple element. Set
\begin{align}\label{eqt:nilpotentpart}
{}^{\vee}{\mathfrak{g}}(\lambda)_{n}&=\{\mu\in {}^{\vee}{\mathfrak{g}}:[\lambda,\mu]=n\mu\},\quad n\in\mathbb{Z}\nonumber\\
\mathfrak{n}(\lambda)&=\sum_{n=1}^{\infty}{}^{\vee}{\mathfrak{g}}(\lambda)_{n}.
\end{align}
The \textbf{canonical flat through} $\lambda$ is the affine subspace
\begin{align*}
\Lambda=\lambda+\mathfrak{n}(\lambda).
\end{align*}
\end{deftn}
\begin{deftn}[Definition 6.9 \cite{ABV}]\label{deft:geomP}
Suppose ${}^{\vee}{G}^{\Gamma}$ is a weak $E$-group for $G$.
A \textbf{geometric parameter} for
${}^{\vee}{G}^{\Gamma}$ is a pair $(y,\Lambda)$ satisfying
\begin{enumerate}[i.]
\item $y\in{}^{\vee}{G}^{\Gamma}-{}^{\vee}{{G}}$.
\item $\Lambda=\lambda+\mathfrak{n}(\lambda)\subset {}^{\vee}{\mathfrak{g}}$ is a canonical flat.
\item $y^{2}=\exp(2\pi i\lambda)$.
\end{enumerate}
The set of geometric parameters of ${}^{\vee}{G}^{\Gamma}$ is denoted by $X\left({}^{\vee}{G}^{\Gamma}\right)$.
As for Langlands parameters, we make ${}^{\vee}{{G}}$ act on $X\left({}^{\vee}{G}^{\Gamma}\right)$ by conjugation.
Two geometric parameters are called equivalent if they are conjugate
\end{deftn}
Let us give a more explicit description of $X\left({}^{\vee}{G}^{\Gamma}\right)$. For every semisimple $\lambda\in{}^{\vee}\mathfrak{g}$, let
$\mathfrak{n}(\lambda)$ be as in (\ref{eqt:nilpotentpart}) and define
\begin{align}\label{eq:groupslambda}
{}^{\vee}{G}(\lambda)&=\text{centralizer in }{}^{\vee}{G}\text{ of }\exp(2\pi i\lambda)\\
L(\lambda)&=\text{centralizer in }{}^{\vee}{G}\text{ of }\lambda\nonumber\\
N(\lambda)&=\text{connected unipotent subgroup with Lie algebra }\mathfrak{n}(\lambda)\nonumber\\
P(\lambda)&=L(\lambda)N(\lambda).\nonumber
\end{align}
From Proposition 6.5 \cite{ABV} for every $\lambda'\in\Lambda=\lambda + \mathfrak{n}(\lambda)$ we have
$${}^{\vee}{G}(\lambda')={}^{\vee}{G}(\lambda)\quad {L}(\lambda')={L}(\lambda)\quad
{N}(\lambda')={N}(\lambda).$$
Therefore, we can respectively define
${}^{\vee}{G}(\Lambda)={}^{\vee}{G}(\lambda),~{L}(\Lambda)={L}(\lambda),~N(\Lambda)={N}(\lambda)$ and $P(\Lambda)=P(\lambda)$.
Now,
for every ${}^{\vee}{G}$-orbit $\mathcal{O}$ of semi-simple elements in $\mathfrak{g}$
write
$$X\left(\mathcal{O},{}^{\vee}{G}^{\Gamma}\right)=\left\{(y,\Lambda)\in X\left({}^{\vee}{G}^{\Gamma}\right):\Lambda\subset \mathcal{O}\right\}$$
and set
\begin{align*}
\mathcal{F}(\mathcal{O})&=\text{set of canonical flats in }\mathcal{O}.
\end{align*}
Fix $\Lambda\in\mathcal{F}(\mathcal{O})$ and consider the sets
\begin{align*}
\mathcal{C}(\mathcal{O})=\{ge(\Lambda)g^{-1}:g\in {}^{\vee}{G}\}\quad\text{ and }\quad
\mathcal{I}(\mathcal{O})=\{y\in {}^{\vee}{G}^{\Gamma}-{}^{\vee}{G}:y^{2}\in\mathcal{C}(\mathcal{O})\}.
\end{align*}
From Proposition 6.13 \cite{ABV} the set $\mathcal{I}(\mathcal{O})$
decomposes into a finite number of ${}^{\vee}{G}$-orbits. List the orbits as $\mathcal{I}_{1}(\mathcal{O}),\cdots,
\mathcal{I}_{r}(\mathcal{O})$ and for each $1\leq i\leq r$ choose a point
$$y_{i}\in \mathcal{I}_{i}(\mathcal{O})\quad\text{with}\quad y_{i}^{2}=e(\Lambda).$$
Then conjugation by $y_{i}$ defines an involutive automorphism of ${}^{\vee}{G}(\Lambda)$
with fixed point set denoted by $K_{y_i}$.
It is immediate from the definition of $X\left(\mathcal{O},{}^{\vee}{G}^{\Gamma}\right)$ that
\begin{align*}
X\left(\mathcal{O},{}^{\vee}{G}^{\Gamma}\right)=\mathcal{F}(\mathcal{O})\times_{\mathcal{C}(\mathcal{O})}\mathcal{I}(\mathcal{O}).
\end{align*}
Therefore, $X\left(\mathcal{O},{}^{\vee}{G}^{\Gamma}\right)$ is the disjoint union of $r$-closed subvarieties
\begin{align}\label{eq:varieties}
X_{y_i}\left(\mathcal{O},{}^{\vee}{G}^{\Gamma}\right)&=\left\{(y,\mu)\in X\left({}^{\vee}{G}^{\Gamma}\right):y\in {}^{\vee}{G}\cdot y_{i},~\mu\in\mathcal{O}\right\}\\
&=\mathcal{F}(\mathcal{O})\times_{\mathcal{C}(\mathcal{O})}\mathcal{I}_{i}(\mathcal{O})\nonumber
\end{align}
and from Proposition 6.16 \cite{ABV} for each one of this varieties we have
\begin{align}\label{eq:varietiesisomor}
X_{y_i}\left(\mathcal{O},{}^{\vee}{G}^{\Gamma}\right)\cong {}^{\vee}{G}\times_{K_{y_i}}{}^{\vee}{G}(\Lambda)/P(\Lambda).
\end{align}
In particular, the orbits of ${}^{\vee}{G}$ on $X_{y_i}\left(\mathcal{O},{}^{\vee}{G}^{\Gamma}\right)$ are in one-to-one correspondence
with the orbits of $K_{y_i}$ on the partial flag variety
${}^{\vee}{G}(\Lambda)/P(\Lambda)$
and from proposition 7.14 \cite{ABV}
this correspondence preserves their closure relations and the nature of the
singularities of closures. Consequently $X\left(\mathcal{O},{}^{\vee}{G}^{\Gamma}\right)$ has in a natural way the structure of a smooth complex algebraic variety, on which ${}^{\vee}{G}$ acts with a finite number of orbits.\\
Now that the geometric parameters have been introduced, the first question that needs to be answered is whether the Langlands Correspondence (Theorem \ref{theo:4.1}) still holds. That is, whether the ${}^{\vee}{G}$-orbits on $X\left({}^{\vee}{G}^{\Gamma}\right)$
are in bijection with the ones on $P\left({}^{\vee}{G}^{\Gamma}\right)$.
To express this bijection,
a more explicit description of the Langlands parameters of ${}^{\vee}{G}^{\Gamma}$ is needed.
Suppose $\varphi\in P\left({}^{\vee}{G}^{\Gamma}\right)$.
We begin by noticing that the restriction of $\varphi$ to $\mathbb{C}^{\times}$ takes the following form: there exist
$\lambda$, $\lambda'\in {}^{\vee}{\mathfrak{g}}$ semisimple elements with $[\lambda,\lambda']=0$
and exp$(2\pi i(\lambda-\lambda'))=1$
such that
\begin{align}\label{eq:lparameters}
\varphi(z)=z^{\lambda}\bar{z}^{\lambda'}.
\end{align}
Now, write
$$y=\text{exp}(\pi i\lambda)\varphi(j).$$
Since $j$ acts on $\mathbb{C}^{\times}$ by complex multiplication,
we have
$$\text{Ad}(\varphi(j))(\lambda)=\lambda'$$
and because Ad$(\text{exp}(\pi i\lambda))$ fixes $\lambda$, we can write
$$\text{Ad}(y)(\lambda)=\lambda'.$$
Moreover, it's not difficult to prove that
$$y^{2}=\exp(2\pi i\lambda).$$
Thus, to each Langlands parameter $\varphi\in P\left({}^{\vee}{G}^{\Gamma}\right)$ we can attach a couple
\begin{align}\label{eq:descriptionLanglands}
\varphi\mapsto (y,\lambda),
\end{align}
satisfying
\begin{itemize}
\item $y\in{}^{\vee}{G}^{\Gamma}-{}^{\vee}{{G}}$ and $\lambda\in {}^{\vee}{\mathfrak{g}}$ is a semisimple element,
\item $y^{2}=\exp(2\pi i\lambda)$,
\item $[\Lambda,\text{Ad}(y)\lambda]=0$
\end{itemize}
and from Proposition 5.6 \cite{ABV}, the map in (\ref{eq:descriptionLanglands}) defines a bijection.
With this description of Langlands parameters in hand,
we can consider their relation to
geometric parameters and reformulate Langlands classification in a more geometrical setting.\\
For each $\varphi\in P\left({}^{\vee}{G}^{\Gamma}\right)$, let $(y(\varphi),\lambda(\varphi))$ be as in (\ref{eq:descriptionLanglands})
and define $\Lambda(\varphi)$ as in $ii.$ of (\ref{deft:geomP}). Then from Proposition 6.17
\cite{ABV} we have that the map
\begin{align}\label{eq:langlandsgeometric}
p:P\left({}^{\vee}{G}^{\Gamma}\right)&\rightarrow X\left({}^{\vee}{G}^{\Gamma}\right)\\
\varphi &\mapsto (y(\varphi),\Lambda(\varphi))\nonumber
\end{align}
induces a bijection of ${}^{\vee}{G}$-orbits on $P\left({}^{\vee}{G}^{\Gamma}\right)$
onto ${}^{\vee}{G}$-orbits on $X\left({}^{\vee}{G}^{\Gamma}\right)$. Theorem \ref{theo:4.1} gives us a surjection
\begin{align}\label{eq:reciprocity}
\Pi(G/\mathbb{R})\rightarrow \left\{\text{Set~of~conjugacy~classes~of~geometric~parameters~of~} {}^{\vee}{G}^{\Gamma}\right\},
\end{align}
whose fibers are the $L$-packets.
To refine (\ref{eq:reciprocity}) into a bijection,
we need, as for Langlands parameters, to introduce the notion of
complete geometric parameters.
\begin{deftn}[Definition 7.6 \cite{ABV}]\label{deftn:geometricparameter}
Suppose ${}^{\vee}{G}^{\Gamma}$ is a weak $E$-group for $G$ and let
$x\in X\left({}^{\vee}{G}^{\Gamma}\right)$. Set
$${}^{\vee}{G}_{x}=\text{centralizer~in~} {}^{\vee}{G}\text{~of~} x$$
and write
$${}^{\vee}{G}_{x}^{alg}=\text{preimage~of~}{}^{\vee}{G}_{x} \text{~in~the~ algebraic~universal~cover~of~}{}^{\vee}{G}.$$
We define the equivariant fundamental group at $x$
to be the quotient
$$A_{x}={}^{\vee}{{G}}_{x}/({}^{\vee}{{G}}_{x})_{0},$$
where $({}^{\vee}{G}_{x})_0$ denotes the identity component of ${}^{\vee}{G}_{\varphi}$.
Similarly, the universal fundamental group at $x$ is defined as
$$A_{x}^{alg}={}^{\vee}{{G}}_{x}^{alg}/({}^{\vee}{{G}}_{x}^{alg})_{0}.$$
A \textbf{complete geometric parameter} for ${}^{\vee}{G}^{\Gamma}$
is then a pair $(x,\tau)$ with
$x\in X\left({}^{\vee}{G}^{\Gamma}\right)$
and $\tau$ an irreducible representation of $A_{\varphi}^{alg}$.
\end{deftn}
Let $\varphi\in P\left({}^{\vee}{G}^{\Gamma}\right)$ and set $x=p(\varphi)$ for the geometric parameter
attached to $\varphi$ by (\ref{eq:langlandsgeometric}). From Lemma 7.5 \cite{ABV} we have
$$A_{x}=A_{\varphi}\quad\text{ and }\quad A_{x}^{alg}=A_{\varphi}^{alg}.$$
Therefore, (\ref{eq:langlandsgeometric}) provides a bijection of $\Xi\left({}^{\vee}{G}^{\Gamma}\right)$
with the set of conjugacy classes of complete geometric parameters and from
Theorem \ref{theo:4.1}, Local Langlands correspondence can be rephrased as a bijection
$$\Pi^{z}(G/\mathbb{R})\longleftrightarrow\{\text{Set~of~conjugacy~classes~of~complete~geometric~parameters~of~} {}^{\vee}{G}^{\Gamma}\}.$$
From now on, and to simplify notation, this last set will also be denoted $\Xi\left({}^{\vee}{G}^{\Gamma}\right)$
and for each ${}^{\vee}{G}$-orbit $\mathcal{O}$ of semi-simple
elements in $\mathfrak{g}$, we define
$\Xi\left(\mathcal{O},{}^{\vee}G^{\Gamma}\right)$ as the subset of
$\Xi\left({}^{\vee}{G}^{\Gamma}\right)$ consisting of
${}^{\vee}G$-orbits ${}^{\vee}G\cdot (x,\tau)$ with $x\in X\left(\mathcal{O},{}^{\vee}{G}^{\Gamma}\right)$.
In the case of an $E$-group $\left({}^{\vee}G^{\Gamma},\mathcal{S}\right)$ for $G$ with second invariant $z$, we write
$$\Xi^{z}\left(\mathcal{O},{G}/\mathbb{R}\right)=\Xi\left(\mathcal{O},{}^{\vee}{G}^{\Gamma}\right).$$
To end this section suppose ${}^{\vee}{H}^{\Gamma}$ and ${}^{\vee}{G}^{\Gamma}$ are weak $E$-groups, and let
$$\iota_{H,G}:{}^{\vee}{H}^{\Gamma}\rightarrow {}^{\vee}{G}^{\Gamma}$$
be an $L$-homomorphism (i.e. a morphism of algebraic groups, with the property that
the diagram
\begin{align*}
\xymatrix{ {}^{\vee}{H}^{\Gamma} \ar[rr]^{\varphi} \ar[rd] && {}^{\vee}{G}^{\Gamma}\ar[ld] \\ & \Gamma }
\end{align*}
commutes). Then we have the two following results (see Proposition 5.4 and Corollary 6.21 \cite{ABV} for a proof) that relate $L$-parameters and geometric parameters of ${}^{\vee}{H}^{\Gamma}$
with $L$-parameters and geometric parameters of ${}^{\vee}{G}^{\Gamma}$.
\begin{prop}
Suppose ${}^{\vee}{H}^{\Gamma}$ and ${}^{\vee}{G}^{\Gamma}$ are weak $E$-groups, and
$$\iota_{H,G}:{}^{\vee}{H}^{\Gamma}\rightarrow {}^{\vee}{G}^{\Gamma}$$
is an $L$-homomorphism. Then composition with $\iota_{H,G}$ defines
a map
$$P(\iota_{H,G}):P\left({}^{\vee}{H}^{\Gamma}\right)\rightarrow P\left({}^{\vee}{G}^{\Gamma}\right)$$
on Langlands parameters, which descends to a map
$$\Phi(\iota_{H,G}):\Phi\left({}^{\vee}{H}^{\Gamma}\right)\rightarrow \Phi\left({}^{\vee}{G}^{\Gamma}\right)$$
on equivalence classes.
\end{prop}
\begin{prop}\label{prop:mapvarieties}
Suppose ${}^{\vee}{H}^{\Gamma}$ and ${}^{\vee}{G}^{\Gamma}$ are weak $E$-groups, and $\iota_{H,G}:{}^{\vee}{H}^{\Gamma}\rightarrow {}^{\vee}{G}^{\Gamma}$
is an $L$-homomorphism. Then there is
a natural map
$$X(\iota_{H,G}):X\left({}^{\vee}{H}^{\Gamma}\right)\rightarrow X\left({}^{\vee}{G}^{\Gamma}\right)$$
on geometric parameters, compatible with the maps $P(\iota_{H,G})$ and $\Phi(\iota_{H,G})$ of the previous result.
Fix a orbit $\mathcal{O}\subset {}^{\vee}{\mathfrak{h}}$ of semisimple elements, and define
$\iota_{H,G}(\mathcal{O})$ to be the unique orbit containing $dX(\iota_{H,G})(\mathcal{O})$. Then
$X(\iota_{H,G})$
restricts to a
morphism of algebraic varieties
$$X(\mathcal{O},\iota_{H,G}):X\left(\mathcal{O},{}^{\vee}{H}^{\Gamma}\right)\rightarrow X(\iota_{H,G}\left(\mathcal{O}),{}^{\vee}{G}^{\Gamma}\right).$$
If $\iota_{H,G}$ is injective, then $X(\mathcal{O},\iota_{H,G})$ is a closed immersion.
\end{prop}
\begin{comment}
Let's give a classification of the $L$-parameters of ${}^{L}G$:
First we notice that the restriction of $\varphi$ to $\mathbb{C}^{\times}$ takes the following form: there exist
$\Lambda$, $\Lambda'\in \widehat{\mathfrak{g}}$ with,
\begin{itemize}
\item $[\Lambda,\Lambda']=0.$
\item exp$(2\pi i(\Lambda-\Lambda'))=1.$
\end{itemize}
such that,
$$\varphi(z)=z^{\Lambda}\bar{z}^{\Lambda'}\times z.$$
Write,
$$y(\varphi)=\text{exp}(\pi i\lambda(\varphi))\varphi(j).$$
Since $j$ acts on $\mathbb{C}^{\times}$ by complex multiplication,
we have,
$$\text{Ad}(\varphi(j))(\Lambda(\varphi))=\Lambda'$$
and because Ad$(\text{exp}(\pi i\Lambda))$ fixes $\Lambda$ we also have
$$\text{Ad}(y)(\Lambda(\varphi))=\Lambda'.$$
The map,
$$\varphi\mapsto (y(\varphi),\Lambda(\varphi))$$
define a bijection between Langlands parameter of ${}^{L}G$
with the set of pairs $(y,\lambda)$ satisfying,
\begin{itemize}
\item $y\in{}^{L}{G}-\widehat{{G}}$ and $\Lambda\in \widehat{\mathfrak{g}}$ is a semisimple element.
\item $y^{2}=\exp(2\pi i\Lambda)$.
\item $[\Lambda,\text{Ad}(y)\Lambda]=0$.
\end{itemize}
\end{comment}
\section{Micro-packets}
In this section we give a quick review on micro-packets. For a more complete exposition on the subject see Chapters 7, 19 and 22 of \cite{ABV}.\\
For all of this section, let $G^{\Gamma}$ be an extended group for $G$ (Definition \ref{deftn:extendedgroup}),
and let $\left({}^{\vee}{G}^{\Gamma},\mathcal{S}\right)$ be an $E$-group (Definition \ref{deftn:egroup})
for the corresponding inner class of real forms. Suppose $(x,\tau)$ is a complete geometric parameter for ${}^{\vee}{G}^{\Gamma}$ (Definition \ref{deftn:geometricparameter}). Write
$V_{\tau}$ for the space of $\tau$ and
$S_{x}={}^{\vee}{{G}}\cdot x$ for the corresponding ${}^{\vee}{G}$-orbit on $X\left({}^{\vee}{G}^{\Gamma}\right)$.
From Lemma 7.3 \cite{ABV}, by regarding $\tau$ as a representation of ${}^{\vee}{{G}}_{x}^{alg}$
trivial on $({}^{\vee}{{G}}_{x}^{alg})_{0}$, the induced bundle
$$\mathcal{V}_{x,\tau}:={}^{\vee}{{G}}^{alg}\times_{{}^{\vee}{{G}}_{x}^{alg}}V_{\tau}\rightarrow S_{x},$$
carries a ${}^{\vee}{G}^{alg}$-invariant flat connection. Therefore,
$\mathcal{V}_{x,\tau}$ defines an irreducible ${}^{\vee}{G}^{alg}$-equivariant local system on $S_x$.
Moreover, by Lemma 7.3(e) \cite{ABV},
the map
\begin{align}\label{e{appendices}q:bijectionLangGeo}
(x,\tau)\mapsto \xi_{x,\tau}:=(S_{x},\mathcal{V}_{x,\tau})
\end{align}
induces a bijection between $\Xi\left({}^{\vee}G^{\Gamma}\right)$ the set of equivalence classes of complete geometric parameters on $X\left({}^{\vee}G^{\Gamma}\right)$,
and the set of couples $(S,\mathcal{V})$ where $S$ is an orbit of ${}^{\vee}{G}$ on $X\left({}^{\vee}G^{\Gamma}\right)$
and $\mathcal{V}$ an irreducible ${}^{\vee}{G}^{alg}$-equivariant local system on $S$.
Following this bijection, the set of couples $(S,\mathcal{V})$ will also be denoted
$\Xi\left({}^{\vee}G^{\Gamma}\right)$ and
for each ${}^{\vee}{G}$-orbit $\mathcal{O}$ of semi-simple elements in $\mathfrak{g}$, we identify $\Xi\left(\mathcal{O},{}^{\vee}{G}^{\Gamma}\right)$ with the set of couples $(S,\mathcal{V})$ with $S\subset X\left(\mathcal{O},{}^{\vee}{G}^{\Gamma}\right)$.\\
\begin{comment}
\begin{deftn}
Following the notation on \cite{ABV} we define,
\begin{enumerate}[i.]
\item $\mathcal{P}(X,K)$= category of $K$-equivariant perverse sheaves on X.
\item $\mathcal{D}(X,K)$= category of $K$-equivariant regular holonomic sheaves of
$\mathcal{D}_{X}$-modules on X.
\end{enumerate}
Write $K\mathcal{P}(X,K)$ and $K\mathcal{D}(X,K)$ for the respective Grothendieck groups.
\end{deftn}
These two categories are related by the Hilbert correspondence.
\begin{theo}
[Hilbert Correspondence]
There is an equivalence of categories,
\begin{align*}
DR:\mathcal{D}(X,K)\rightarrow\mathcal{P}(X,K)
\end{align*}
this induces an isomorphism of Grothendieck groups,
\begin{align*}
DR:K\mathcal{D}(X,K)\rightarrow K\mathcal{P}(X,K)
\end{align*}
\end{theo}
\end{comment}
Let $\mathcal{O}$ be a ${}^{\vee}{G}$-orbit of semisimple elements in ${}^{\vee}\mathfrak{g}$.
As in Section 2, write
$$\mathcal{P}\left(X\left(\mathcal{O},{}^{\vee}{G}^{\Gamma}\right),{}^{\vee}{{G}}^{alg}\right)\quad\text{and}\quad \mathcal{D}\left(X\left(\mathcal{O},{}^{\vee}{G}^{\Gamma}\right),{}^{\vee}{{G}}^{alg}\right)$$
for the category of ${}^{\vee}{{G}}^{alg}$-equivariant perverse sheaves and
${}^{\vee}{{G}}^{alg}$-equivariant coherent $\mathcal{D}$-modules on $X\left(\mathcal{O},{}^{\vee}{G}^{\Gamma}\right)$. We
define
$\mathcal{P}\left(X\left({}^{\vee}{G}^{\Gamma}\right), {}^{\vee}{{G}}^{alg}\right)$ {and} $\mathcal{D}\left(X\left({}^{\vee}{G}^{\Gamma}\right),
{}^{\vee}{{G}}^{alg}\right)$
to be the direct sum over semisimple orbits $\mathcal{O}\subset {}^{\vee}{\mathfrak{g}}$
of the categories $\mathcal{P}\left(X\left(\mathcal{O},{}^{\vee}{G}^{\Gamma}\right),{}^{\vee}{{G}}^{alg}\right)$ and $\mathcal{D}\left(X\left(\mathcal{O},{}^{\vee}{G}^{\Gamma}\right),{}^{\vee}{{G}}^{alg}\right)$ respectively.
The last necessary
step before the introduction of
micro-packets, is to explain
how the irreducible objects in $\mathcal{P}\left(X\left({}^{\vee}{G}^{\Gamma}\right),{}^{\vee}{{G}}^{alg}\right)$ and $\mathcal{D}\left(X\left({}^{\vee}{G}^{\Gamma}\right),{}^{\vee}{{G}}^{alg}\right)$ are parameterized by the set $\Xi\left({}^{\vee}{G}^{\Gamma}\right)$ of equivalence classes of complete geometric parameters. Fix $\xi=(S,\mathcal{V})\in \Xi\left(\mathcal{O},{}^{\vee}{G}^{\Gamma}\right)$.
Write
$$j:S\rightarrow \overline{S},$$
for the inclusion of $S$ in its closure and
$$i:\overline{S}\rightarrow X\left(\mathcal{O},{}^{\vee}{G}^{\Gamma}\right),$$
for the inclusion of the closure of $S$ in $X(\mathcal{O},{}^{\vee}{G}^{\Gamma})$.
Let $d=d(S)$ be the dimension of $S$. If we regard the local system $\mathcal{V}$ as
a constructible sheaf on $S$, the complex $\mathcal{V}[-d]$, consisting of the single sheaf $\mathcal{V}$
in degree $-d$ defines an ${}^{\vee}{{G}}^{alg}$-equivariant perverse sheaf on $S$.
Applying to it the intermediate extension functor
$j_{!\ast}$, followed by the direct image $i_{\ast}$,
we get an irreducible perverse sheaf on $X(\mathcal{O},{}^{\vee}{G}^{\Gamma})$
\begin{align}\label{eq:irredPerv}
P(\xi)=i_{\ast}j_{!\ast}\mathcal{V}[-d],
\end{align}
the perverse extension of $\xi$, and by applying the de Rham functor (see Theorem \ref{theo:rhcorrespondence}) to it
\begin{align}\label{eq:irredDM}
D(\xi)=DR^{-1}(P(\xi)),
\end{align}
we obtain an irreducible $\mathcal{D}$-module.
That every irreducible ${}^{\vee}{G}^{alg}$-equivariant perverse sheaf on $X({}^{L}G)$ is of this form, follows from Theorem 4.3.1 \cite{BBD}, and it's a consequence of Riemann-Hilbert correspondence (see Theorem \ref{theo:rhcorrespondence}) that the same can be said
in the case of $\mathcal{D}$-modules. The sets
\begin{align*}
\left\{P(\xi):\xi\in \Xi\left({}^{\vee}{G}^{\Gamma}\right)\right\},\quad \left\{D(\xi):\xi\in \Xi\left(({}^{\vee}{G}^{\Gamma}\right)\right\}
\end{align*}
are therefore bases of the Grothendieck group $KX\left({}^{\vee}{G}^{\Gamma}\right)$. We can finally give the
definition of a micro-packet.
\begin{comment}
In other words we have a diagram,
\begin{align*}
\xymatrix{
K\Pi(G/\mathbb{R}) \ar[r]& K\mathcal{D}(X,K)\ar[r] & K\mathcal{P}(X,K)\\
\pi(\xi)\ar@{|->}[r]& D(\xi) \ar@{|->}[r]^{DR}& P(\xi)}
\end{align*}
And using Vogan duality and Beilinson-Bernstein localization theorem these two categories relate
to the set $\Pi(G/\mathbb{R})$ of equivalence classes of irreducible representations of strong real forms of $G$.\\
\end{comment}
\begin{comment}
Suppose $\xi=(S,\mathcal{V})$ is a complete Geometric parameter. Write
$$j:S\rightarrow \overline{S},\quad i:\overline{S}\rightarrow X(\mathbf{G},\mathcal{O}),$$
for the inclusion of $S$ on its closure, and the inclusion of the closure in $X(\mathbf{G},\mathcal{O})$.
Write $d=d(S)$ for the dimension of $S$. If we regard the local system $\mathcal{V}$ as
a constructible sheaf on $S$, the complex $\mathcal{V}[-d]$ consisting of the single sheaf $\mathcal{V}$
in degree $-d$ define an $\widehat{\mathbf{G}}^{alg}$-equivariant perverse sheaf on $S$. Applying to it
the intermediate extension functor $j_{!\ast}$ followed by the direct image $i_{\ast}$,
we get an irreducible perverse sheaf on $X(\mathbf{G},\mathcal{O})$,
$$P(\xi)=i_{\ast}j_{!\ast}\mathcal{V}[-d].$$
$$categories$$
\end{comment}
\begin{deftn}
Let $S$ be a ${}^{\vee}{G}$-orbit in $X\left({}^{\vee}{G}^{\Gamma}\right)$. To every complete geometric parameter
$\xi\in\Xi\left({}^{\vee}{G}^{\Gamma}\right)$ we have attached in (\ref{eq:irredPerv}) a perverse sheaf $P(\xi)$ (and in (\ref{eq:irredDM}) a $\mathcal{D}$-module $D(\xi)$). From Theorem \ref{theo:defcc}, the conormal bundle $T^{\ast}_{S}\left(X({}^{\vee}{G}^{\Gamma})\right)$ has a non-negative integral multiplicity
$\chi_{S}^{mic}(\xi)$ in the characteristic cycle $CC(P(\xi))$ of $P(\xi)$ (or equivalently
in the characteristic cycle $CC(D(\xi))$ of $D(\xi)$).
We define the micro-packet of geometric parameters attached to $S$, to be the set of complete geometric parameters for which this multiplicity is non-zero
\begin{align*}
\Xi\left({}^{\vee}{G}^{\Gamma},{}^{\vee}{G}\right)^{mic}_{S}=\left\{\xi\in\Xi\left({}^{\vee}{G}^{\Gamma}\right):\chi_{S}^{mic}(\xi)\neq 0\right\}.
\end{align*}
\end{deftn}
\begin{deftn}
Suppose
$\left({}^{\vee}{G}^{\Gamma},\mathcal{S}\right)$ is an $E$-group for $G$
with second invariant $z$.
Let $\varphi\in\Phi^{z}({G}/\mathbb{R})$ be an equivalence class of Langlands parameters for ${}^{\vee}{G}^{\Gamma}$ and write $S_{\varphi}$ for the corresponding orbit of ${}^{\vee}{{G}}$ in $X\left({}^{\vee}{G}^{\Gamma}\right)$.
Then we define the micro-packet of geometric parameters attached to $\varphi$ as
\begin{align*}
\Xi^{z}({G}/\mathbb{R})_{\varphi}=\Xi\left(X\left({}^{\vee}{G}^{\Gamma}\right),{}^{\vee}{G}\right)^{mic}_{S_{\varphi}}.
\end{align*}
For any complete parameter $\xi\in \Xi^{z}\left({G}/\mathbb{R}\right)$, let $\pi(\xi)$
be the representation in $\Pi^{z}({G}/\mathbb{R})$
associated to $\xi$ by Theorem \ref{theo:4.1}.
Then the micro-packet of $\varphi$ is defined as
\begin{align*}
\Pi^{z}({G}/\mathbb{R})^{mic}_{\varphi}=\{\pi(\xi'):\xi'\in \Xi^{z}({G}/\mathbb{R})^{mic}_{\varphi}\}.
\end{align*}
Finally, let $\delta$ be a strong real form of $G^{\Gamma}$, then we define the restriction of
$\Pi({G}/\mathbb{R})^{mic}_{\varphi}$ to $\delta$ as
\begin{align*}
\Pi^{z}({G}(\mathbb{R},\delta))^{mic}_{\varphi}=\{\pi\in \Pi^{z}({G}/\mathbb{R})^{mic}_{\varphi}: \pi \text{ is a representation of }G(\mathbb{R},\delta) \}.
\end{align*}
\end{deftn}
We notice that the Langlands packet attached to a Langlands parameter $\varphi$ is always contained in the corresponding micro-packet
\begin{align}\label{eq:contentionoflpackets}
\Pi^{z}({G}/\mathbb{R})_{\varphi}\subset\Pi^{z}({G}/\mathbb{R})^{mic}_{\varphi}.
\end{align}
This is a consequence of $(ii)$ of the following lemma. Point $(i)$ of the lemma
will show to be quite useful later in this section. For a proof, see Lemma 19.14 \cite{ABV}.
\begin{lem}\label{lem:lemorbit}
Let $\eta=(S,\mathcal{V})$ and $\eta'=\left(S',\mathcal{V}'\right)$ be two geometric parameters for
${}^{\vee}{G}^{\Gamma}$.
\begin{enumerate}[i.]
\item If $\eta'\in \Xi\left(X\left({}^{\vee}{G}^{\Gamma}\right),{}^{\vee}{G}\right)_{S}^{mic}$, then $S\subset \overline{S}'$.
\item If $S'=S$, then $\eta'\in \Xi\left(X\left({}^{\vee}{G}^{\Gamma}\right),{}^{\vee}{G}\right)_{S}^{mic}$.
\end{enumerate}
\end{lem}
We are going to be mostly interested in the case of micro-packets attached to
Langlands parameters coming from Arthur parameters.
These types of micro-packets, are going to be called Adams-Barbasch-Vogan packets or simply ABV-packets.
We continue by recalling the definition of an Arthur parameter.\\
An Arthur parameter is a homomorphism
\begin{align*}
\psi:W_{\mathbb{R}}\times\textbf{SL}(2,\mathbb{C})\longrightarrow {}^{\vee}{G}^{\Gamma},
\end{align*}
satisfying
\begin{itemize}
\item The restriction of $\psi$ to $W_{\mathbb{R}}$ is a tempered Langlands parameter (i.e.
the closure of $\varphi(W_{\mathbb{R}})$ in the analytic topology is compact).
\item The restriction of $\psi$ to $\mathbf{SL}(2,\mathbb{C})$ is holomorphic.
\end{itemize}
Two such parameters are called equivalent if they are conjugate by the action of ${}^{\vee}{{G}}$. The set of equivalences classes is written $\Psi\left({}^{\vee}{G}^{\Gamma}\right)$
or $\Psi^{z}({G}/\mathbb{R})$, when we want to specify that
${}^{\vee}{G}^{\Gamma}$ is an $E$-group for $G$ with second invariant $z$.
\begin{comment}
Suppose $\psi$ is an Arthur parameter, we define, ${}^{\vee}{\mathbf{G}}_{\psi}$,
like the centralizer in $\widehat{\mathbf{G}}$ of $\psi(W_{\mathbb{R}}\times\textbf{SL}(2,\mathbb{C}))$
and,
\begin{align*}
A_{\psi}=\widehat{\mathbf{G}}_{\psi}/\widehat{\mathbf{G}}_{0}
\end{align*}
the Arthur component group for $\psi$.
\end{comment}
To every Arthur parameter $\psi$, we can associate a Langlands parameter $\varphi_{\psi}$, by
the following formula (see Section 4 \cite{Arthur89})
\begin{align*}
\varphi_{\psi}&:W_{\mathbb{R}}\longrightarrow {}^{\vee}G^{\Gamma},\\
\varphi_{\psi}(w)&=\psi\left(w,\left(\begin{array}{cc}
|w|^{1/2}& 0\\
0& |w|^{-1/2}
\end{array}\right)\right).
\end{align*}
Now, to $\varphi_{\psi}$ correspond an orbit $S_{\varphi_{\psi}}$ of ${}^{\vee}{G}$ on $X\left({}^{\vee}{G}^{\Gamma}\right)$. We define
\begin{align*}
S_{\psi}=S_{\varphi_{\psi}}.
\end{align*}
\begin{deftn}\label{deftn:abvpackets}
Suppose
$\left({}^{\vee}{G}^{\Gamma},\mathcal{S}\right)$ is an
$E$-group for $G$
with second invariant $z$.
Let $\psi\in\Psi^{z}({G}/\mathbb{R})$ be an Arthur parameter. We define the Adams-Barbasch-Vogan packet $\Pi^{z}({G}/\mathbb{R})_{\psi}^{\mathrm{ABV}}$
of $\psi$, as the micro-packet of the Langlands parameter $\varphi_{\psi}$ attached to $\psi$:
\begin{align*}
\Pi^{z}({G}/\mathbb{R})_{\psi}^{\mathrm{ABV}}=\Pi^{z}({G}/\mathbb{R})_{\varphi_{\psi}}^{mic}.
\end{align*}
\end{deftn}
In other words, $\Pi^{z}({G}/\mathbb{R})_{\psi}^{\mathrm{ABV}}$ is the
set of all irreducible representations with the
property that the corresponding irreducible perverse sheaf contains the conormal bundle $T_{S_{\psi}}^{\ast}\left(X\left({}^{\vee}{G}^{\Gamma}\right)\right)$
in its characteristic cycle.
Micro-packets attached to Arthur parameters satisfy the following important properties.
\begin{theo}\label{theo:abvproperties}
Let $\psi$ be an Arthur parameter for ${}^{\vee}{G}^{\Gamma}$.
\begin{enumerate}[i.]
\item $\Pi^{z}(G/\mathbb{R})_{\psi}^{mic}$ contains the $L$-packet $\Pi^{z}(G/\mathbb{R})_{\varphi_{\psi}}$.
\item $\Pi^{z}(G/\mathbb{R})_{\psi}^{mic}$ is the support of a stable formal virtual character:
$$\eta_{\psi}^{mic}=\sum_{\xi\in\Xi({G}/\mathbb{R})_{\varphi_{\psi}}}e(\xi)(-1)^{d(S_{\xi})-d(S_{\psi})}\chi_{S_{\xi}}^{mic}(P(\xi))\pi(\xi).$$
where for each orbit $S$, $d(S)$ is the dimension of the orbit and $e(\xi)$ the Kottwitz sign attached to the real form of which $\pi(\xi)$ is a representation.
\item $\Pi^{z}(G/\mathbb{R})_{\psi}^{mic}$ satisfies the ordinary endoscopic identities predicted by the theory of endoscopy.
\end{enumerate}
\end{theo}
As explained after (\ref{eq:contentionoflpackets}), point $(i)$ is a consequence of Lemma \ref{lem:lemorbit}$(ii)$. For a proof of the second statement, see Corollary 19.16\cite{ABV} and Theorem 22.7\cite{ABV}. Finally, for a proof and a more precise statement of the last point, see Theorem 26.25 \cite{ABV}.
\subsection{Tempered representations}
In this section we study ABV-packets attached to tempered Langlands parameters. Let's recall that
$\varphi\in P({}^{\vee}{G}^{\Gamma})$ is said to be \textbf{tempered} if the closure of $\varphi(W_{\mathbb{R}})$ in the analytic topology is compact.
\begin{prop}
Let $\varphi$ be a tempered Langlands parameter for ${}^{\vee}{G}^{\Gamma}$, then the corresponding orbit
$S_{\varphi}$ of ${}^{\vee}{G}$ in $X\left({}^{\vee}{G}^{\Gamma}\right)$ is open and dense.
\end{prop}
\begin{proof}
The assertion follows from Proposition 22.9$(b)$ {\cite{ABV}},
applied to an Arthur parameter with trivial $\textbf{SL}(2,\mathbb{C})$ part.
Indeed, suppose $\psi$ is an Arthur parameter with restriction to $W_{\mathbb{R}}$ equal to
$\varphi$ and trivial $\textbf{SL}(2,\mathbb{C})$ part. Let
$(y,\lambda)$ be the couple corresponding to $\varphi$ under Equation (\ref{eq:descriptionLanglands}). Define $P(\lambda)$ as in (\ref{eq:groupslambda}),
and write $K(y)$ for the centralizer of $y$ in ${}^{\vee}G^{\Gamma}$.
Finally, set $$E_{\psi}=d\psi|_{\textbf{SL}(2,\mathbb{C})}
\left(\begin{array}{cc}
0& 1\\
0& 0
\end{array}\right)=0.$$
Then by Proposition 22.9$(b)$ {\cite{ABV}} for each $x\in S_{\varphi}$ the orbit
$(P(\lambda)\cap K(y))\cdot E_{\psi}$ is dense in
$T_{S_{\varphi},x}^{\ast}\left(X\left({}^{\vee}G^{\Gamma}\right)\right)$.
But $E_{\psi}=0$, hence $T_{S_{\varphi},x}^{\ast}\left(X\left({}^{\vee}G^{\Gamma}\right)\right)=0$, that is, the annihilator of $T_x S_{\varphi}$ in $T_x^{\ast}(X({}^{\vee}G^{\Gamma}))$ is
equal to zero.
Therefore, $T_{x}\left(X\left({}^{\vee}G^{\Gamma}\right)\right)=T_x S_{\varphi}$
and the result follows.
\end{proof}
\begin{cor}
Suppose $({}^{\vee}{G}^{\Gamma},\mathcal{S})$
is an $E$-group for $G$
with second invariant $z$.
Let $\varphi$ be a tempered Langlands parameter for ${}^{\vee}G^{\Gamma}$, then
$$\Pi^{z}({G}/\mathbb{R})_{\varphi}^{\mathrm{ABV}}=\Pi^{z}({G}/\mathbb{R})_{\varphi}.$$
where at right, $\Pi^{z}({G}/\mathbb{R})_{\varphi}$ denotes the Langlands packet of $\varphi$.
\end{cor}
\begin{proof}
By Theorem \ref{theo:abvproperties}($i$), the $L$-packet $\Pi^{z}({G}/\mathbb{R})_{\varphi}$ is contained in $\Pi^{z}({G}/\mathbb{R})_{\varphi}^{\text{mic}}$. We only need to show the opposite inclusion.
Let $\pi\in\Pi^{z}({G}/\mathbb{R})_{\varphi}^{\text{ABV}}$ and write $S$
for the orbit of ${}^{\vee}{G}$ in $X({}^{\vee}{G}^{\Gamma})$ corresponding to the Langlands parameter of $\pi$ under the map defined in (\ref{eq:langlandsgeometric}). From Lemma \ref{lem:lemorbit} and the definition of $\Pi^{z}({G}/\mathbb{R})_{\varphi}^{\text{ABV}}$, the orbit $S$ contains $S_{\varphi}$ in its closure. As
$S_{\varphi}$ is open, $S_{\varphi}\cap S\neq \emptyset$ and we have $S_{\varphi}=S$.
By Lemma \ref{lem:lemorbit}($ii$), $\pi\in\Pi^{z}(G/\mathbb{R})_{\varphi}$ and we can conclude the desired inclusion.
\end{proof}
\subsection{Essentially unipotent Arthur parameters}
In this section we give a full description of the ABV-packets attached to essentially unipotent Arthur parameters, such that the image of $\mathbf{SL}(2,\mathbb{C})$ contains a principal unipotent element. Our goal is to give a slight generalization of Theorem 27.18 \cite{ABV} (see Theorem \ref{theo:unipotentparameter} below) that only treats the unipotent case, and prove that
the
ABV-packets corresponding to essentially principal unipotent Arthur parameters
consists of characters, one for each real form of $G$ in our inner class.
This result
will show to be fundamental in the proof,
next section, that the packets defined in \cite{Adams-Johnson} are ABV-packets.\\
Let $\psi$ be an Arthur parameter of ${}^{\vee}G^{\Gamma}$. We say that:\\
$\bullet$ $\psi$ is \textbf{unipotent}, if its restriction to the identity component $\mathbb{C}^{\times}$ of
$W_{\mathbb{R}}$ is trivial.\\
More generally, we say that:\\
$\bullet$ $\psi$ is \textbf{essentially
unipotent}, if the image of its restriction to the identity component $\mathbb{C}^{\times}$ of
$W_{\mathbb{R}}$ is contained in the center $Z({}^{\vee}{G})$ of ${}^{\vee}{G}$.\\
Next, fix a morphism
\begin{align*}
\psi_{1}:\mathbf{SL}(2,\mathbb{C})\longrightarrow {}^{\vee}{G}.
\end{align*}
Then we say that $\psi_1$ is \textbf{principal}, if $\psi(\mathbf{SL}(2,\mathbb{C}))$ contains a principal unipotent element. Finally, the (essentially)
unipotent Arthur parameter $\psi$ is called \textbf{(essentially) principal unipotent Arthur parameter} if
$\psi|_{\mathbf{SL}(2,\mathbb{C})}$ is principal.\\
The next result due to Adams, Barbasch and Vogan (see Theorem 27.18 \cite{ABV} and the remark at the end of page 310 of \cite{ABV})
gives a description of the set $\Pi(G/\mathbb{R})_{\psi}^{\text{ABV}}$ for $\psi$ a principal unipotent Arthur parameter.
\begin{theo}\label{theo:unipotentparameter}
Suppose $\left({}^{\vee}{G}^{\Gamma},\mathcal{S}\right)$
is an $E$-group for $G$
with second invariant $z$.
Fix a principal morphism
\begin{align*}
\psi_{1}:\mathbf{SL}(2,\mathbb{C})\longrightarrow {}^{\vee}{G}.
\end{align*}
\begin{enumerate}[a)]
\item The centralizer $S_{0}$ of $\psi_{1}$ in ${}^{\vee}{G}$ is $Z({}^{\vee}{G})$.
\item Suppose $z=1$ (i.e. $\left({}^{\vee}{G}^{\Gamma},\mathcal{S}\right)$ is an $L$-group). Then the set of equivalence classes of unipotent Arthur parameters attached to $\psi_{1}$ may be identified with
$$H^{1}(\Gamma,Z({}^{\vee}{G}))=\{z\in Z({}^{\vee}{G}):z\theta_{Z}(z)=1\}/\{w\theta_{Z}^{-1}(w):w\in Z({}^{\vee}{G})\}.$$
More generally, if $z\in (1+\theta_{Z})Z({}^{\vee}G)$
then the set of equivalence classes of unipotent Arthur parameters attached to $\psi_{1}$
is a principal homogeneous space for
$H^{1}(\Gamma,Z({}^{\vee}{G})).$
\item The unipotent representations of type $z$ (of some real form $G(\mathbb{R},\delta))$ attached
to $\psi_{1}$ are precisely the projective representations of type z trivial on the identity component $G(\mathbb{R},\delta))_0$.
\item Suppose $z\in (1+\theta_{Z})Z({}^{\vee}G)$.
Let $\delta$ be any strong real form of $G^{\Gamma}$ and write
$$\mathrm{Hom}^{z}_{\mathrm{cont}}({G}^{can}(\mathbb{R},\delta)/{G}^{can}(\mathbb{R},\delta)_{0},\mathbb{C}^{\times}),$$ for the set of characters of type $z$ of
$G^{can}(\mathbb{R},\delta)$ trivial on $G^{can}(\mathbb{R},\delta)_0$.
Then there is a natural surjection
\begin{align}\label{eq:mapunipotent}
H^{1}(\Gamma,Z({}^{\vee}{G}))\rightarrow \mathrm{Hom}^{z}_{\mathrm{cont}}({G}^{can}(\mathbb{R},\delta)/{G}^{can}(\mathbb{R},\delta)_{0},\mathbb{C}^{\times}).
\end{align}
\item
Suppose $\psi$ is a unipotent Arthur parameter attached to $\psi_1$ and $\delta$
is a strong real form of $G^{\Gamma}$. Write $\pi(\psi,\delta)$ for the character of
$G^{can}(\mathbb{R},\delta)$ trivial on $G^{can}(\mathbb{R},\delta)_0$ attached to $\psi$ by composing the bijection of $(b)$ with the surjection of $(c)$. Let $P(\psi,\delta)$, be the perverse
sheaf on $X\left({}^{\vee}G^{\Gamma}\right)$ corresponding to $\pi(\psi,\delta)$
under the Local Langlands Correspondence (see Theorem \ref{theo:4.1}). Then for any perverse sheaf $P$ on $X\left({}^{\vee}G^{\Gamma}\right)$, we have
$$\chi_{S_{\psi}}^{mic}(P)=\left\{\begin{array}{cl}
1& \text{if }P\cong P(\psi,\delta),\quad\text{ for some strong real form }\delta\text{ of }G\\
0& otherwise.
\end{array}\right.
$$
Consequently,
\begin{align}\label{eq:repunipotenttordu}
\Pi^{z}(G/\mathbb{R})_{\psi}^{\mathrm{ABV}}=\{\pi(\psi,\delta)\}_{\delta \text{ strong real form of } G^{\Gamma}}.
\end{align}
\end{enumerate}
\end{theo}
We turn now to the study of ABV-packets attached to
essentially principal unipotent parameters.
We begin
by describing the behaviour of micro-packets under twisting. We record this in
the next result. From this we will be able to give a first generalization of Theorem \ref{theo:unipotentparameter}.
In order to enunciate the result we need first to recall that for each strong real form $\delta$ of
$G^{\Gamma}$, there is a natural morphism
\begin{align}\label{eq:cocylemap}
H^{1}(W_{\mathbb{R}},Z({}^{\vee}{G})) &\rightarrow \mathrm{Hom}_{\mathrm{cont}}({G}(\mathbb{R},\delta),\mathbb{C}^{\times})\\
\mathbf{a}&\mapsto \chi(\mathbf{a},\delta),\nonumber
\end{align}
which is surjective and maps cocyles with compact image to unitary characters of $G(\mathbb{R},\delta)$ (see for example section 2 of \cite{Langlands}).
\begin{prop}\label{prop:torsion}
Suppose $\left({}^{\vee}{G}^{\Gamma},\mathcal{S}\right)$ is an $E$-group for $G$ with second invariant $z$.
Suppose $\mathbf{a}\in H^{1}(W_{\mathbb{R}},Z({}^{\vee}{G}))$ and let the cocycle $\mathrm{a}$ be a representative of $\mathbf{a}$. For each strong real form $\delta$ of
$G^{\Gamma}$ let $\chi(\mathbf{a},\delta)$ be the character of $G(\mathbb{R},\delta)$ attached to $\mathbf{a}$ as in (\ref{eq:cocylemap}).
\begin{enumerate}[i.]
\item If $\varphi$ is a Langlands parameter for ${}^{\vee}G^{\Gamma}$, then the morphism
$$\varphi_{\mathbf{a}}:W_{\mathbb{R}}\rightarrow{}^{\vee}G^{\Gamma},\quad w\mapsto (\mathrm{a}(w),1)\varphi(w)$$
defines a Langlands parameter of ${}^{\vee}G^{\Gamma}$ whose equivalence class depends exclusively on $\mathbf{a}$ and the equivalence class of $\varphi$. Furthermore
\begin{align*}
\Pi^{z}(G/\mathbb{R})^{mic}_{\varphi_{\mathbf{a}}}=\bigsqcup_{\delta \text{ strong real form of } G^{\Gamma}}\Pi^{z}(G(\mathbb{R},\delta))^{mic}_{\varphi_{\mathbf{a}}}
\end{align*}
where
\begin{align*}
\Pi^{z}(G(\mathbb{R},\delta))^{mic}_{\varphi_{\mathbf{a}}}=\{\pi\otimes\chi(\mathbf{a},\delta):\pi\in \Pi^{z}(G(\mathbb{R},\delta))^{mic}_{\varphi}\}.
\end{align*}
\item
If $\psi$ is an Arthur parameter for ${}^{\vee}G^{\Gamma}$, then the morphism
\begin{align}\label{eq:arthurcocycle}
\psi_{\mathbf{a}}:W_{\mathbb{R}}\times\mathbf{SL}(2,\mathbb{C})\rightarrow{}^{\vee}G^{\Gamma},\quad (w, g)\mapsto (a(w), 1)\psi(w,g)
\end{align}
defines an Arthur parameter of ${}^{\vee}G^{\Gamma}$ whose equivalence class depends exclusively on $\mathbf{a}$ and the equivalence class of $\psi$. Furthermore
$$\Pi^{z}(G/\mathbb{R})_{\psi_{\mathbf{a}}}^{\mathrm{ABV}}=\bigsqcup_{\delta \text{ strong real form of } G^{\Gamma}}\Pi^{z}(G(\mathbb{R},\delta))^{mic}_{\psi_{\mathbf{a}}}$$
where
\begin{align*}
\Pi^{z}(G(\mathbb{R},\delta))^{mic}_{\psi_{\mathbf{a}}}=\{\pi\otimes\chi(\mathbf{a},\delta):\pi\in \Pi^{z}(G(\mathbb{R},\delta))^{mic}_{\psi}\}.
\end{align*}
\end{enumerate}
\end{prop}
\begin{proof}
To check that $\varphi_{\mathbf{a}}$ defines a Langlands parameter is straightforward.
Let $\lambda_{\varphi}$ be defined as in (\ref{eq:lparameters}). By the definition of $\varphi_{\mathbf{a}}$
there exist $\lambda_{\mathbf{a}}\in Z({}^{\vee}\mathfrak{g})$ such that $\lambda_{\varphi_{\mathbf{a}}}=
\lambda_{\varphi}+\lambda_{\mathbf{a}}$.
Write
$$\mathcal{O}_{\varphi}={}^{\vee}G\cdot \lambda_{\varphi}\quad
\text{ and }\quad \mathcal{O}_{\varphi_{\mathbf{a}}}={}^{\vee}G\cdot \lambda_{\varphi_{\mathbf{a}}}.$$
Then
\begin{align*}
X\left(\mathcal{O}_{\varphi},{}^{\vee}{G}^{\Gamma}\right)&\rightarrow X\left(\mathcal{O}_{\varphi_{\mathbf{a}}},{}^{\vee}{G}^{\Gamma}\right)\\
(y,\Lambda)&\mapsto (\text{exp}(\pi i \lambda_{\mathbf{a}})y,\Lambda+\lambda_{\mathbf{a}}),
\end{align*}
defines and isomorphism of varieties, that induces a bijection of orbits
\begin{align*}
{}^{\vee}G-\text{orbits on }X(\mathcal{O}_{\varphi},{}^{\vee}{G}^{\Gamma}) &\rightarrow
{}^{\vee}G-\text{orbits on }X(\mathcal{O}_{\varphi_{\mathbf{a}}},{}^{\vee}{G}^{\Gamma})\\
S&\mapsto S',
\end{align*}
and of geometric parameters
\begin{align*}
\Xi^{z}(\mathcal{O}_{\varphi},G/\mathbb{R})&\rightarrow \Xi^{z}(\mathcal{O}_{\varphi_{\mathbf{a}}},G/\mathbb{R})\\
\xi&\mapsto\xi'.
\end{align*}
Furthermore, from the description of irreducibles perverse sheaves given in Equation (\ref{eq:irredPerv}),
we have an isomorphism $KX\left(\mathcal{O}_{\varphi},{}^{\vee}{G}^{\Gamma}\right)\cong KX\left(\mathcal{O}_{\varphi_{\mathbf{a}}},{}^{\vee}{G}^{\Gamma}\right)$ that restrict
for each $\xi\in \Xi^{z}(\mathcal{O}_{\varphi},G/\mathbb{R})$
into an isomorphism $P(\xi)\cong P(\xi')$. Therefore, for each
${}^{\vee}G$-orbit $S$ on $X\left(\mathcal{O}_{\varphi},{}^{\vee}G^{\Gamma}\right)$ we obtain
\begin{align}\label{eq:tensorcycles}
\chi_{S}^{mic}(P(\xi))=\chi_{S'}^{mic}(P(\xi')).
\end{align}
Now, from the properties of $L$-packets described in Section 3 \cite{Langlands}, we deduce
that for each strong real form $\delta\in G^{\Gamma}$, the set of irreducible representations of
$G(\mathbb{R},\delta)$ corresponding to complete geometric parameters for
$X\left(\mathcal{O}_{\varphi_{\mathbf{a}}},{}^{\vee}G^{\Gamma}\right)$, under the Local Langlands Correspondence (see Theorem \ref{theo:4.1}),
is equal to the set of irreducible representations of $G(\mathbb{R},\delta)$
attached to complete geometric parameters for
$X\left(\mathcal{O}_{\varphi},{}^{\vee}G^{\Gamma}\right)$
tensored by $\chi({\mathbf{a}},\delta)$. Point $(i)$ follows then from the definition of micro-packets and equality (\ref{eq:tensorcycles}).
Point $(ii)$ is a direct consequence of point $(i)$ applied to $\varphi_{\psi_{\mathbf{a}}}$ and the definition of ABV-packets.
\end{proof}
The following corollary generalizes Theorem \ref{theo:unipotentparameter}
to the case of essentially principal unipotent Arthur parameters for $E$-groups
with second invariant $z$, satisfying $z\in (1+\theta_{Z})Z({}^{\vee}G)$
(i.e. $E$-groups admitting principal unipotent Arthur parameters).
\begin{comment}
Let $\rho$ be the half sums of positive coroots and $\mathcal{O}\subset\widehat{\mathfrak{g}}$
the orbit of half sums of positive coroots; this corresponds to representations
of $\mathbf{G}(\mathbb{R})$ of infinitesimal character equal to that of the trivial representation.
The conjugacy class,
\begin{align*}
\mathcal{C}(\mathcal{O})=\{g e(\rho)g^{-1}:g\in\widehat{n\mathbf{G}} \}
\end{align*}
consists of the single element,
$$z(\rho)=\chi(2\rho)(-1),\quad\text{ where }\chi(2\rho):\mathbb{C}^{\ast}\rightarrow \mathbf{T}$$
is the sum of positive coroots.
Because $\rho$ is regular and integral, for each canonical flat $\Lambda\in\mathcal{F}(\mathcal{O})$
there is an unique Borel subgroup $B(\lambda)$ which stabilize it. We may therefore
identify the geometric parameters $X(\mathcal{O},\mathbf{G})$ with the set of pairs $(y,B)$ where
$y\in~^{L}(\mathbf{G})-\widehat{\mathbf{G}},~y^{2}=z(\rho)$ and $B$ is a Borel subgroup of $\mathbf{G}$.
Fix a point $(y,B)$ since $\mathbf{G}$ is simply connected (we are in the case of $\mathbf{G}$ adjoint)
the fixed point group $K(y)$ of the involution $\theta_{y}$ is connected. Since $\theta_{y}$ preserve $B$ (the pair $(\theta_{y},B)$ defines the $L$-group
definition 4.14 of ABV), the intersection $K(y)\cap B$ is a Borel subgroup, so it is connected as well.
So it follows that,
$$A_{\varphi}^{loc,alg}=K(y)\cap B/(K(y)\cap B)_{0}.$$
The $L$-packet therefore consist of a single representation, the trivial one.
\end{comment}
\begin{cor}\label{cor:essentialunipotent}
Suppose $\left({}^{\vee}{G}^{\Gamma},\mathcal{S}\right)$ is an $E$-group for $G$. Write $z$ for the second invariant of the $E$-group,
and suppose $z\in (1+\theta_{Z})Z({}^{\vee}G)$.
Let $\psi$ be an essentially unipotent Arthur parameter
for ${}^{\vee}{G}^{\Gamma}$ such that $\psi|_{\mathbf{SL}(2,\mathbb{C})}$ is principal.
Then
there exist a cocycle $\mathbf{a}\in H^{1}(W_{\mathbb{R}},Z({}^{\vee}{G}))$ and a
principal unipotent Arthur parameter $\psi_{u}$ for ${}^{\vee}G^{\Gamma}$,
such that for all
$(w,g)\in W_{\mathbb{R}}\times \mathbf{SL}(2,\mathbb{C})$ we have
$$\psi(w,g)=(\mathrm{a}(w),g)\psi_u(w,g),$$
where $\mathrm{a}$ is a representative of $\mathbf{a}$.
Finally, for each strong real form $\delta$ of $G^{\Gamma}$,
let $\chi(\mathbf{a},\delta)$ be the character
of $G(\delta,\mathbb{R})$ corresponding to
$\mathbf{a}$ under the map in (\ref{eq:cocylemap}), and let
$\pi(\psi_u,\delta)$ be the character
of $G(\delta,\mathbb{R})$ attached to $\psi_u$
in point Theorem \ref{theo:unipotentparameter}$(d)$ .
Define
$$\pi(\psi,\delta)=\chi(\mathbf{a},\delta)\pi(\psi_u,\delta).$$
Then
\begin{align*}
\Pi^{z}(G/\mathbb{R})_{\psi}^{\mathrm{ABV}}=\{\pi(\psi,\delta)\}_{\delta \text{ strong real form of } G^{\Gamma}}.
\end{align*}
\end{cor}
\begin{proof}
Let $\psi$ be an essentially unipotent Arthur parameter
of ${}^{\vee}G^{\Gamma}$ such that $\psi|_{\mathbf{SL}(2,\mathbb{R})}=\psi_1$, is a principal morphism.
Let $\psi_{u}$ be any unipotent Arthur parameter extending $\psi_1$.
It's straightforward to check that
$$(\psi|_{W_{\mathbb{R}}})(\psi_{u}|_{W_{\mathbb{R}}})^{-1},$$
corresponds to a cocycle $a\in Z^{1}(W_{\mathbb{R}},Z({}^{\vee}{G}))$.
The first part of the corollary follows.
For each strong real form $\delta$ of $G^{\Gamma}$ let $\chi(\mathbf{a},\delta)$
and $\pi(\psi_u,\delta)$, be as in the statement of the corollary.
From Theorem \ref{theo:unipotentparameter}$(e)$ we have
$$\Pi^{z}(G/\mathbb{R})_{\psi_u}^{\mathrm{ABV}}=\{\pi(\psi_u,\delta)\}_{\delta \text{ strong real form of } G^{\Gamma}}.$$
Hence from Proposition \ref{prop:torsion}($ii)$ we conclude
\begin{align*}
\Pi(G/\mathbb{R})_{\psi}^{\mathrm{ABV}}
&=\{\pi(\psi_u,\delta)\chi(\mathbf{a},\delta)\}_{\delta \text{ strong real form of } G^{\Gamma}}\\
&=\{\chi(\psi,\delta)\}_{\delta \text{ strong real form of } G^{\Gamma}}
\end{align*}
\end{proof}
The next result fully generalizes Theorem \ref{theo:unipotentparameter} to $E$-groups
admitting essentially principal unipotent Arthur parameters.
The techniques employed in the proof are the same as the one used to show Theorem \ref{theo:unipotentparameter} in \cite{ABV}(see pages 306-310 \cite{ABV}). They are based on results
coming from Chapters 20-21 \cite{ABV}, and still valid
in our framework, and from Chapter 27,
only proved in \cite{ABV} in the case of representations with infinitesimal character $\mathcal{O}$ arising from homomorphisms $\mathbf{SL}(2,\mathbb{C})\rightarrow {}^{\vee}G$.
Since the infinitesimal character of the representations that we consider here
differ from $\mathcal{O}$ by a central element, the results on Chapter 27, easily generalize to our setting
\begin{theo}\label{theo:essentiallyunipotentparameter}
Suppose $\left({}^{\vee}{G}^{\Gamma},\mathcal{S}\right)$ is an $E$-group for $G$. Write $z$ for the second invariant of the $E$-group, and suppose there is $\lambda$, a central element in ${}^{\vee}{\mathfrak{g}}$ with $[\lambda,\theta_{Z}(\lambda)]=0$, such that
$$z\exp(\pi i(\lambda-\theta_{Z}(\lambda)))\in (1+\theta_Z)Z({}^{\vee}G).$$
Fix a principal morphism
\begin{align*}
\psi_{1}:\mathbf{SL}(2,\mathbb{C})\longrightarrow {}^{\vee}{G},
\end{align*}
and write $\mathcal{O}={}^{\vee}G\cdot (\lambda+\lambda_1)$, where
$\lambda_1=d\psi_{1}
\left(\begin{array}{cc}1/2& 0\\
0& 1/2
\end{array}\right)$.
\begin{enumerate}[a)]
\item
The set of equivalence classes of essentially unipotent Arthur parameters attached to $\psi_{1}$,
with corresponding orbit contained in $X\left(\mathcal{O},{}^{\vee}G^{\Gamma}\right)$,
may be identified with
$$H^{1}(\Gamma,Z({}^{\vee}{G}))=\{z\in Z({}^{\vee}{G}):z\theta_{Z}(z)=1\}/\{w\theta_{Z}^{-1}(w):w\in Z({}^{\vee}{G})\}.$$
\item The projective representations (of some real form $G^{can}(\mathbb{R},\delta)$) having infinitesimal character $\mathcal{O}$
attached to $\psi_{1}$ are precisely the projective characters of type $z$ with infinitesimal character $\mathcal{O}$.
\item
Let $\delta$ be any strong real form of $G^{\Gamma}$ and write
$\mathrm{Hom}^{z}_{\mathrm{cont}}({G}^{can}(\mathbb{R},\delta),\mathbb{C}^{\times})$,
for the set of characters of type $z$ of
$G^{can}(\mathbb{R},\delta)$.
Then there is a natural map
\begin{align}\label{eq:mapessentially}
H^{1}(\Gamma,Z({}^{\vee}{G}))\rightarrow \mathrm{Hom}^{z}_{\mathrm{cont}}({G}^{can}(\mathbb{R},\delta),\mathbb{C}^{\times}),
\end{align}
whose image is the set of projective characters of $G^{can}(\mathbb{R},\delta)$ of type $z$,
having infinitesimal character $\mathcal{O}$.
\item
Suppose $\psi$ is an essentially unipotent Arthur parameter attached to $\psi_1$
with corresponding orbit contained in $X\left(\mathcal{O},{}^{\vee}G^{\Gamma}\right)$. Let $\delta$
be a strong real form of $G^{\Gamma}$. Write $\pi(\psi,\delta)$ for the projective character of
type $z$ of $G^{can}(\mathbb{R},\delta)$
attached to $\psi$ by composing the bijection of $a)$ with the map of $c)$. Let $P(\psi,\delta)$, be the perverse
sheaf on $X\left({}^{\vee}G^{\Gamma}\right)$ corresponding to $\pi(\psi,\delta)$
under the Local Langlands Correspondence (see Theorem \ref{theo:4.1}). Then for all perverse sheaf $P$ on $X\left({}^{\vee}G^{\Gamma}\right)$, we have
$$\chi_{S_{\psi}}^{mic}(P)=\left\{\begin{array}{cl}
1& \text{if }P\cong P(\psi,\delta),\quad\text{ for some strong real form }\delta\text{ of }G\\
0& otherwise.
\end{array}\right.
$$
Consequently,
\begin{align}\label{eq:repunipotenttordu}
\Pi^{z}(G/\mathbb{R})_{\psi}^{\mathrm{ABV}}=\{\pi(\psi,\delta)\}_{\delta \text{ strong real form of } G^{\Gamma}}.
\end{align}
\end{enumerate}
\end{theo}
\begin{proof}
For $(a)$ we start by fixing an essentially unipotent Arthur parameter
$\psi_0$ attached to $\psi_1$. Notice that for each essentially unipotent Arthur parameter
$\psi$ attached to $\psi_1$, the equivalence class of the product
$$(\psi|_{W_{\mathbb{R}}})(\psi_{0}|_{W_{\mathbb{R}}})^{-1},$$
defines a cocycle $\mathbf{a}_{\psi}\in H^{1}(\Gamma,Z({}^{\vee}{G}))$. Therefore, each
essentially unipotent Arthur parameter $\psi$ attached to $\psi_1$ can be written as
$$\psi(w,g)=(\mathrm{a}_{\psi}(w),g)\psi_0(w,g),$$
where $\mathrm{a}_{\psi}$ is a representative of $\mathbf{a}$. Point $(a)$ follows.
For $(b)$ we notice that the infinitesimal character corresponding to $\mathcal{O}$ (see Lemma 15.4 \cite{ABV})
is the infinitesimal character of a one dimensional representation,
so the corresponding maximal ideal in $U(\mathfrak{g})$ (see Theorem 21.8 \cite{ABV}) is the annihilator of a one dimensional representation. Point $(b)$, like Theorem \ref{theo:unipotentparameter}(c),
follows from Corollary 27.13 \cite{ABV}. We notice that
Corollary 27.13 \cite{ABV} has been demonstrated in \cite{ABV}, only in the case when the infinitesimal character $\mathcal{O}$ arises from a homomorphism of $\mathbf{SL}(2,\mathbb{C})$ into ${}^{\vee}G$, but is easily generalized to our setting. Indeed, Corollary 27.13 \cite{ABV}
is a consequence of Theorem 27.10 and Theorem 27.12 \cite{ABV},
whose proof are still valid in the case of our infinitesimal character $\mathcal{O}$,
and of Theorem 21.6 and Theorem 21.8 \cite{ABV}, whose proof are valid for any infinitesimal
character.
For (c), we start as in (a) by fixing an essentially unipotent Arthur parameter
$\psi_0$ attached to $\psi_1$. Write $\varphi_{\psi_0}$ for
the corresponding Langlands parameter. Let $\delta$ be a strong real form
of $G^{\Gamma}$, whose associated real form is quasi-split. Then the Langlands packets
$\Pi^{z}(G(\mathbb{R},\delta))_{\varphi_{\psi_0}}$ contains exactly one representation,
a canonical projective character of type $z$ that we denote $\chi(\psi,\delta_0)$.
Next, for each strong real form $\delta$ of $G^{\Gamma}$ define a canonical projective character
$\chi(\psi,\delta)$ of type $z$, as follows; Suppose $T$ is a Cartan subgroup of $G$
with $\text{Ad}(\delta)T=T$, and such that $T(\mathbb{R},\delta)$
is a maximally split Cartan subgroup of $G(\mathbb{R},\delta)$.
Let $g\in G$ be such that $\delta=\text{Ad}(g)\delta$,
then $gTg^{-1}(\mathbb{R},\delta)$ defines a Cartan subgroup of $G(\mathbb{R},\delta)$.
For any $t\in T(\delta',\mathbb{R})$ set
$$\pi(\psi,\delta)(t)=\pi(\psi,\delta_0)(\text{Ad}(g)t).$$
Then the conditions of Lemma 2.5.2 \cite{Adams-Johnson} hold, and $\chi(\psi,\delta)$ extends uniquely to a one-dimensional representation, also denoted $\chi(\psi,\delta)$
of $G(\delta',\mathbb{R})$.
Now, for each essentially unipotent Arthur parameter $\psi$ attached to $\psi_1$,
let
$\mathbf{a}_{\psi}\in H^{1}(\Gamma,Z({}^{\vee}{G}))$ be defined as in $(a)$.
From \ref{theo:unipotentparameter}(d), we know how to attach to each couple
$(\mathbf{a}_{\psi},\delta)$ a character
$\chi(\mathbf{a}_{\psi},\delta)\in \mathrm{Hom}_{\mathrm{cont}}({G}^{can}(\mathbb{R},\delta)/{G}^{can}(\mathbb{R},\delta)_{0},\mathbb{C}^{\times})$.
We can now define the natural map of $(c)$;
for each strong real form $\delta$ of $G^{\Gamma}$ we define (\ref{eq:mapessentially}) by sending
$\mathbf{a}_{\psi}\in H^{1}(\Gamma,Z({}^{\vee}{G}))$ to
$$\mathbf{a}_{\psi}\mapsto \chi(\mathbf{a}_{\psi},\delta)\chi({\psi_0},\delta).$$
The surjectivity of (\ref{eq:mapessentially}) in the set of projective characters of $G^{can}(\mathbb{R},\delta)$ of type $z$,
having infinitesimal character $\mathcal{O}$, is a consequence of the surjectivity of (\ref{eq:mapunipotent}).
Finally, the proof of $(d)$ can be done along the same lines as that of Theorem \ref{theo:unipotentparameter}(e) (see page 309 \cite{ABV}). For each strong real form
of $G$ we just need to replace the characters appearing in \ref{theo:unipotentparameter}(d)
with the ones in the image of the map on (c),
and use point (b) instead of \ref{theo:unipotentparameter}(c).
\end{proof}
\subsection{The Adams-Johnson construction}
In this section we study ABV-packets attached to a particular family of Arthur parameters,
namely those related to representations with cohomology. In \cite{Adams-Johnson}, Adams and Johnson
attached to any Arthur parameter in this family a packet consisting of
representations cohomologically induced from unitary characters. Moreover, they proved that each packet defined in this way is the support of a stable distribution (see Theorem (2.13) \cite{Adams-Johnson}), and that these distributions satisfy the ordinary endoscopic identities predicted by the theory of endoscopy (see Theorem (2.21) \cite{Adams-Johnson}).
The objective of this section is to show that the packets defined by Adams-Johnson are ABV-packets, that is, for the family of Arthur parameters studied in \cite{Adams-Johnson},
the packet associated in (\ref{deftn:abvpackets}) to any
parameter in this family, coincides with the packet defined by Adams and Johnson.
We begin the section by describing the family of Arthur parameters studied in \cite{Adams-Johnson
and the construction of Adams-Johnson packets.
The description of the parameters and of the Adams-Johnson packets that we give here, is inspired by Section 3.4.2.2 \cite{Otaibi} and by Section 5 \cite{Arthur89} (see also Section 5 \cite{AMR} version 1).
The main difference with these two references is that in this article, we use the Galois form of the $L$-group instead of the Weil form. By doing this we are able to describe the Adams-Johnson construction
in the language of extended groups of \cite{ABV}. The only complication by using the Galois form is that to define the packets of Adams-Johnson it will be necessary to work with
an $E$-group of some Levi subgroup of $G$, and consequently to use the canonical cover (see Definition \ref{deftn:canonicalcover}) of this Levi subgroup, and to work with some projective characters of strong real forms of this cover.\\
Suppose
$\left({}^{\vee}{G}^{\Gamma},\mathcal{S}\right)$ is an $L$-group for $G$ (see Definition \ref{deftn:egroup}).
We recall that
$\mathcal{S}$ is a ${}^{\vee}{G}$-conjugacy class of pairs
$\left({}^{\vee}\delta,{}^{d}{B}\right)$, with
${}^{d}{B}$ a Borel subgroup ${}^{d}{B}$ of ${}^{\vee}G$
and
${}^{\vee}\delta$
an element of order two in ${}^{\vee}G^{\Gamma}-{}^{\vee}{G}$ such that conjugation by
${}^{\vee}\delta$ is a distinguished involutive automorphism
$\sigma_{{}^{\vee}\delta}$ of ${}^{\vee}G$
preserving ${}^{d}B$.
Since $\sigma_{{}^{\vee}\delta}$ is distinguished, it preserves an splitting $\left({}^{\vee}G,{}^{d}{B},{}^{d}{T},\{X_{\alpha}\}\right)$ of $^{\vee}G$ that we fix from now on.
We notice that the $L$-group ${}^{\vee}G^{\Gamma}$ can be more explicitly
described as the disjoint union of
${}^{\vee}G$ and the coset ${}^{\vee}G{}^{\vee}\delta$, with multiplication on ${}^{\vee}G^{\Gamma}$ defined by the rules:
$$(g_1{}^{\vee}\delta)(g_2{}^{\vee}\delta)=g_1\sigma_{{}^{\vee}\delta}(g_2) ,\qquad
(g_1{}^{\vee}\delta)(g_2)=g_1\sigma_{{}^{\vee}\delta}(g_2){}^{\vee}\delta$$
and the obvious rules for the other two kinds of products. This explicit description of ${}^{\vee}G^{\Gamma}$, amounts to the
usual description of an $L$-group
as the semi-direct product of ${}^{\vee}G$ with the Galois group.
Now, let $\psi$ be an Arthur parameter for ${}^{\vee}G^{\Gamma}$.
As explained in the case of Langlands parameters, the restriction of
$\psi$ to $\mathbb{C}^{\times}$ takes the following form: there exist
$\lambda$, $\lambda'\in X_{\ast}({}^{d}{T})\otimes\mathbb{C}$ with
$\lambda-\lambda'\in X_{\ast}({}^{d}{T})$
such that
\begin{align}\label{eq:AJparameterequation0}
\psi(z)=z^{\lambda}\bar{z}^{\lambda'}.
\end{align}
We may suppose $\lambda$ is dominant for ${}^{d}{B}$.
Let ${}^{d}{L}$ be the centralizer
of $\psi(\mathbb{C}^{\times})$ in ${}^{\vee}{{G}}$ and write ${}^{d}\mathfrak{l}$ for its lie algebra. We have
$$Z\left({}^{d}{L}\right)_{0}\subset {}^{d}{T}\subset{}^{d}{L}$$
and
$$\lambda,\lambda'\in X_{\ast}\left(Z\left({}^{d}{L}\right)_{0}\right)\otimes\mathbb{C}\quad \text{with}\quad
\lambda-\lambda'\in X_{\ast}\left(Z\left({}^{d}{L}\right)_{0}\right).
$$
Set $\psi(j)=n{}^{\vee}\delta$ where $n\in{}^{\vee}G$. Then
$$\text{Ad}(n{}^{\vee}\delta)(\lambda)=\lambda'\quad \text{and} \quad (n{}^{\vee}\delta)^{2}=\psi(-1)=(-1)^{\lambda+\lambda'}.$$
Set ${}^{d}{B}_{L}={}^{d}{L}\cap {}^{d}{B}$ and write ${}^{\vee}{\rho}_{{}^{d}{L}}$ for the
half-sum of the positive coroots defined by the system of positive roots
$R\left({}^{d}{B}_{{}^{d}{L}},{}^{d}{T}\right)$.
The family of Arthur parameters studied by Adams and Johnson are those which satisfy the following properties:
\begin{enumerate}
\item[AJ1.] The identity component of $Z\left({}^{d}{L}\right)^{\psi(j)}$ is contained in $Z\left({}^{\vee}{G}\right)$.
\item[AJ2.] $\psi(\mathbf{SL}(2,\mathbb{C}))$ contains a principal unipotent element of ${}^{d}{L}$.
\item[AJ3.] $\left<\lambda+{}^{\vee}{\rho}_{{}^{d}{L}},\alpha\right>\neq 0$ for all root
$\alpha\in R\left({}^{\vee}{G},{}^{d}{T}\right)$.
\end{enumerate}
Let $S=\psi(\mathbf{SL}(2,\mathbb{C}))$.
It is immediate from (AJ2), that $S$ is $\mathbf{SL}_2$-principal in ${}^{\vee}{G}$. Write $\mathfrak{s}$
for the lie algebra of $S$, and let $(h,e,f)$ be a $\mathfrak{sl}_{2}$-triplet generating $\mathfrak{s}$. Up to conjugation we can, and will, suppose that
$$h={}^{\vee}{\rho}_{{}^{d}{L}}\in {}^{d}{T}.$$
Since $\psi(j)=n{}^{\vee}\delta$ commutes with $S$, conjugation by $n{}^{\vee}\delta$ fixes $h$, and because $h$ is regular in ${}^{d}{L}$, $n{}^{\vee}\delta$ normalizes ${}^{d}{T}$. It is also
obvious that $n{}^{\vee}\delta$ normalizes ${}^{d}{L}$ and $Z\left({}^{d}{L}\right)$. Now, ${}^{\vee}\delta$ normalizes ${}^{d}{T}$,
so $n$ normalizes ${}^{d}{T}$.
From now on we assume that ${}^{\vee}{G}$ is semi-simple. We make this assumption just to simplify the exposition that follows, the conclusions in
(\ref{eq:AJLparameter}) and (\ref{eq:AJLparameter2})
remain true in the reductive case.
Point (AJ1) is therefore equivalent to\\
$\mathrm{AJ1}^{\prime}$.
$Z({}^{d}{L})_{0}^{\psi(j)}$ is trivial.\\
~\\
As $\mathbb{R}^{\times}$ is the center of $W_{\mathbb{R}}$,
the group $\psi(\mathbb{R}^{\times})$
commutes with $\psi(\mathbb{R}^{\times}\times \mathbf{SL}(2,\mathbb{C}))$. Hence
$\psi(\mathbb{R}^{\times})\subset Z\left({}^{d}{L}\right)^{\psi(j)}$ and
it follows that $\psi(\mathbb{R}_{+}^{\times})\subset Z\left({}^{d}{L}\right)_{0}^{\psi(j)}$
is trivial. Consequently
$$\lambda+\lambda'=0$$
and we can write
$$\psi(z)=z^{\lambda}\bar{z}^{-\lambda}=\left(\frac{z}{|z|}\right)^{2\lambda}.$$
Hence $\psi(j)^{2}=(-1)^{2\lambda}$. In particular Ad$(\psi(j))$
is a linear automorphism of order two of ${}^{\vee}{\mathfrak{g}}$. It is semi-simple
with eigenvalues equal to $\pm 1$. Condition (AJ1$^{\prime}$) is then equivalent to
$$\text{Ad}(\psi(j))|_{\mathfrak{z}({}^{d}\mathfrak{l})}=-\text{Id}|_{\mathfrak{z}({}^{d}\mathfrak{l})}.$$
The following arguments are taken from 3.4.2.2 \cite{Otaibi}.
Let $\mathbf{n}:W\left({}^{\vee}{G},{}^{d}{T}\right)\rtimes \Gamma\rightarrow N\left({}^{\vee}{G},{}^{d}{T}\right)\rtimes \Gamma$ be the section defined in Section 2.1 \cite{Langlands-Shelstad}.
Let $w_0$ be the longest element in $W({}^{\vee}{G},{}^{d}{T})$ and let $w_{{}^{d}{L}}$
be the longest element in $W\left({}^{d}{L},{}^{d}{T}\right)$. Write
\begin{align}\label{eq:n1}
n_{1}{}^{\vee}\delta=\mathbf{n}(w_{0}w_{{}^{d}{L}}{}^{\vee}\delta).
\end{align}
Let $\Delta\left({}^{d}{B}_{{}^{d}{L}},{}^{d}{T}\right)$, be the set of simple roots of the positive root system $R\left({}^{d}{B}_{{}^{d}{L}},{}^{d}{T}\right)$. Since $w_0$ (resp. $w_{{}^{d}{L}}$) sends positive roots in $R\left({}^{d}{B},{}^{d}{T}\right)$ (resp. $R\left({}^{d}{B}_{{}^{d}{L}},{}^{d}{T}\right)$)
to negative roots, and because $\sigma_{{}^{\vee}\delta}$ preserves the splitting $\left({}^{\vee}G,{}^{d}{B},{}^{d}{T},\{{X}_{\alpha}\}\right)$, we can conclude that $n_{1}{}^{\vee}\delta$ preserves $\Sigma\left({}^{d}{B}_{{}^{d}{L}},{}^{d}{T}\right)$.
Moreover, $n_{1}{}^{\vee}\delta$ acts as -Id on $\mathfrak{z}({}^{d}\mathfrak{l})$ and
by $t\mapsto t^{-1}$ in $Z\left({}^{d}{L}\right)$.
In particular, Ad$(n_{1}{}^{\vee}\delta)\lambda=-\lambda=\lambda'$. Thus
$$(n_{1}{}^{\vee}\delta)\psi(z)(n_{1}{}^{\vee}\delta)^{-1}=\psi(\overline{z}),$$
the element $nn_{1}^{-1}$ commutes with $\psi(\mathbb{C}^{\times})$,
and we have $nn_{1}^{-1}\in {}^{d}{L}$.
Furthermore, from Proposition 9.3.5 \cite{Springer}, $w_0w_{{}^{d}{L}}{}^{\vee}\delta$ preserves the splitting
$\left({}^{d}{L},{}^{d}{B}_{{}^{d}{L}},{}^{d}{T},\{{X}_{\alpha}\}_{\alpha\in \Delta\left({}^{d}{B}_{{}^{d}{L}},{}^{d}{T}\right)}\right)$. Hence $n_{1}{}^{\vee}\delta$ commutes with $S$, and we can say the same for $nn_{1}^{-1}$. Now, $S$ is principal in ${}^{\vee}{G}$ so $nn_{1}^{-1}\in Z\left({}^{d}{L}\right)$
and there exists $t\in Z\left({}^{d}{L}\right)$ such that
\begin{align}\label{eq:AJparameterequation01}
tn_1=n.
\end{align}
We compute
\begin{align}\label{eq:ajparameteridentity1}
(-1)^{2\lambda}=\psi(j)^{2}=(n{}^{\vee}\delta)^{2}=&(t n_{1}{}^{\vee}\delta)^{2}=t(n_1{}^{\vee}\delta)t(n_1{}^{\vee}\delta)^{-1}(n_1{}^{\vee}\delta )^{2}\\
=&tt^{-1}(n_1{}^{\vee}\delta)^{2}=(n_1{}^{\vee}\delta)^{2}.\nonumber
\end{align}
Let ${}^{d}{Q}={}^{d}{L}{}^{d}{U}$ be the parabolic subgroup of ${}^{\vee}{G}$ containing
${}^{d}{T}$ and such that the roots of ${}^{d}{T}$ in ${}^{d}{Q}$ are the
$\alpha\in R\left({}^{\vee}{G},{}^{d}T\right)$ satisfying $\left<\lambda,\alpha\right>\geq 0$.
In particular ${}^{d}{L}$ is a Levi factor of ${}^{d}{Q}$.
The section
$\mathbf{n}:W\left({}^{\vee}{G},{}^{d}{T}\right)\rtimes \Gamma\rightarrow N\left({}^{\vee}{G},{}^{d}{T}\right)\rtimes \Gamma$
defined in \cite{Langlands-Shelstad}
has the property that
\begin{align}\label{eq:ajparameteridentity}
(n_{1}{}^{\vee}\delta)^{2}=\mathbf{n}(w_0w_{{}^{d}{L}})=\prod_{\alpha\in R\left({}^{d}{T},{}^{d}{U}\right)}\alpha^{\vee}(-1).
\end{align}
Therefore,
$\prod_{\alpha\in R\left({}^{d}{U},{}^{d}T\right)}\alpha^{\vee}(-1)=(-1)^{2\lambda}$
and from Proposition 1.3.5 \cite{Shelstad81} we can conclude
$$\lambda\in X_{\ast}\left(Z\left({}^{d}{L}\right)\right)+\frac{1}{2}\sum_{\alpha\in R\left({}^{d}{U},{}^{d}T\right)}\alpha^{\vee}.$$
Let
\begin{align}\label{eq:AJLparameter}
\mu=\lambda+{}^{\vee}{\rho}_{{}^{d}{L}},\quad \mu'=\mu=-\lambda+{}^{\vee}{\rho}_{{}^{d}{L}}.
\end{align}
Then
\begin{align}\label{eq:higuest}
\mu,\mu'\in X_{\ast}({}^{d}{T})+\frac{1}{2}\sum_{\alpha\in R\left({}^{d}{B},{}^{d}T\right)}\alpha^{\vee}
\end{align}
and
$$\text{Ad}(\psi(j))\cdot\mu=\mu'.$$
The Langlands parameter corresponding to $\psi$ verifies for every $z\in\mathbb{C}^{\times}$
\begin{align}\label{eq:AJLparameter2}
\varphi_{\psi}(z)=z^{\mu}\bar{z}^{\mu'}.
\end{align}
We point out that from point (AJ3) above,
$\mu$ is regular and from (\ref{eq:higuest}), $\mu$ is the infinitesimal character of a finite-dimensional representation of some real form of $G$ with highest weight (relative to the root system
$R\left({}^{d}{B},{}^{d}T\right)$) equal to
$$\mu-\frac{1}{2}\sum_{\alpha\in R\left({}^{d}{T},{}^{d}{B}\right)}\alpha^{\vee}=\lambda-\frac{1}{2}\sum_{\alpha\in R\left({}^{d}{T},{}^{d}{U}\right)}\alpha^{\vee}.$$
This concludes the description of the Arthur parameters studied in \cite{Adams-Johnson}. To
define the cohomology packets we need first to connect the Levi subgroup ${}^{d}{L}$
of ${}^{\vee}G$ to an extended group for some Levi subgroup $L$ of $G$, and factor $\psi$
through an Arthur parameter $\psi_L$ of some $E$-group for $L$. The second invariant of this $E$-group
will be equal to $(n_1{}^{\vee}\delta)^{2}$, and since this element is not necessarily one, it is not
going to be always possible to factor $\psi$ through the Arthur parameter of
an $L$-group for $L$.
Because of the Local Langlands Correspondence, this will have as a consequence the necessity of using the canonical cover of $L$.
The final step in the construction of the packets is to use Theorem \ref{theo:essentiallyunipotentparameter}
to associate to $\psi_L$ a family of canonical projective characters of strong real forms of $L$ of type $(n_1{}^{\vee}\delta)^{2}$ and apply cohomological induction to them.\\% to obtain a representation of $G$.\\
Let $\delta_{qs}$ be a strong real form of $G^{\Gamma}$,
whose associated real form $\sigma_{\delta_{qs}}$ is quasi-split.
Write $\theta_{\delta_{qs}}$ for the corresponding Cartan involution.
Suppose $T$ is a Cartan subgroup of $G$ stable under $\sigma_{\delta_{qs}}$ and $\theta_{\delta_{qs}}$.
We have a canonical isomorphism between the based root data
$$(X^{\ast}({T}),\Delta^{\vee},X_{\ast}({T}),\Delta)\quad\text{and}
\quad \left(X^{\ast}({}^{d}{T}),{}^{d}\Delta,X_{\ast}({}^{d}{T}),{}^{d}\Delta^{\vee}\right).$$
Using this isomorphism, to the couple $\left({}^{d}{Q},{}^{d}{L}\right)$ we can associate a parabolic group $Q=LU$
of $G$ containing $T$ such that ${}^{\vee}L\cong {}^{d}{L}$.
Now, since conjugation by the element $n_1{}^{\vee}\delta$ defined in (\ref{eq:n1})
preserves the splitting $\left({}^{d}{L},{}^{d}{B}_{{}^{d}{L}},{}^{d}{T},\{{X}_{\alpha}\}_{\alpha\in \Delta\left({}^{d}{B}_{{}^{d}{L}},{}^{d}{T}\right)}\right)$, through the isomorphism
${}^{\vee}L\cong {}^{d}{L}$ it translates into a distinguished involutive automorphism of ${}^{\vee}L$. Write $a_L$ for
the automorphism of the based root datum of ${}^{\vee}L$ (or equivalently
of the based root datum of $L$) induced in this way by $n_1{}^{\vee}\delta$.
We define (see Proposition \ref{prop:classificationextended}($iii$))
\begin{align}\label{eq:extendedgroupL}
L^{\Gamma} \text{ to be the extended group of $L$ with invariants }(a_L,\overline{1}).
\end{align}
Let $a_G$ be the automorphism of the based root datum
of $G$ induced by ${}^{\vee}\delta$ through duality. Since $n_{1}{}^{\vee}\delta$
induces the same automorphism, we have $a_G|_{\Psi_{0}(L)}=a_L$. Consequently,
each real form in the inner class corresponding to $a_L$, is the restriction
to $L$ of a real form in the inner class corresponding to $a_G$,
that is $L^{\Gamma}\subset G^{\Gamma}$.
Next,
write
\begin{align}\label{eq:sinvariant7}
z=(n_1{}^{\vee}\delta)^{2}
\end{align}
and define (see Proposition 4.7(c) \cite{ABV})
\begin{align}\label{eq:e-groupL}
\left({}^{\vee}L^{\Gamma},\mathcal{S}_L\right) \text{ to be the unique } E\text{-group with invariants }(a_L,z).
\end{align}
More precisely,
let ${}^{\vee}{\sigma}_{L}$ be any distinguished automorphism of ${}^{\vee}L$
corresponding to $a_L$. Then the group
${}^{\vee}L^{\Gamma}$ is defined as the union
of ${}^{\vee}{L}$ and the set ${}^{\vee}{L}^{\vee}\delta_{L}$ of formal symbols $l{}^{\vee}\delta_{L}$, with multiplication defined according to the rules:
$$(l_1{}^{\vee}\delta_L)(l_2{}^{\vee}\delta_L)=l_1\sigma_{L}(l_2) z,\qquad (l_1{}^{\vee}\delta_L)(l_2)=l_1\sigma_{L}(l_2){}^{\vee}\delta_L$$
and the obvious rules for the other two kinds of product.
\begin{comment}
Suppose $\delta\in L^{\Gamma}$ is a strong real form of $L$, i.e. $\in$.
and write $\sigma_{\delta}$
for the corresponding real form of $G$.
We notice that the set of real forms
of $L$ that can be extended to $\sigma_{\delta}$ is in correspondence
with
$$W(G(\mathbb{R},\sigma_{\delta}),T(\mathbb{R},\sigma_{\delta}))\setminus W({G},{T})/W({L},{T})$$
\end{comment}
We explain now how to extend the isomorphism ${}^{\vee}{L}\cong{}^{d}{{L}}$
to an embedding
\begin{align}\label{eq:embedding1}
\iota_{{L},{G}}:{}^{\vee}L^{\Gamma}\longrightarrow {}^{\vee}{G}^{\Gamma}.
\end{align}
Since ${}^{\vee}L^{\Gamma}$
is the disjoint union of ${}^{\vee}L$ and the coset ${}^{\vee}L {}^{\vee}\delta_{L}$
we just need to define
$\iota_{{L},{G}}$ on ${}^{\vee}L{}^{\vee}\delta_{L}$.
We do this by sending
each element
$l{}^{\vee}\delta_L\in {}^{\vee}L {}^{\vee}\delta_{L}$ to:
\begin{align}\label{eq:embedding2}
\iota_{{L},{G}}(l{}^{\vee}\delta_L)=ln_1{}^{\vee}\delta.
\end{align}
And because
${}^{\vee}\delta_L^{2}=z=(n_1{}^{\vee}\delta)^{2}$,
is straightforward to verify that
the embedding is well-defined.
Let us go back now to our Arthur parameter $\psi$ satisfying points
(AJ1), (AJ2) and (AJ3), above.
From the properties satisfied by $\psi$ and the definition of ${}^{\vee}{L}^{\Gamma}$
there exists (up to conjugation) a unique essentially unipotent Arthur parameter
\begin{align*}
\psi_{L}:W_{\mathbb{R}}\times SL(2,\mathbb{C})\longrightarrow{}^{\vee}{L}^{\Gamma},
\end{align*}
with restriction to $\mathbf{SL}(2,\mathbb{C})$ equal to the principal morphism,
such that up to conjugation by ${}^{\vee}{{G}}$ we have
\begin{align*}
\psi:W_{\mathbb{R}}\times SL(2,\mathbb{C})\xrightarrow{\psi_{{L}}}{}^{\vee}{L}^{\Gamma} \xrightarrow{\iota_{L,G}}{}^{\vee}{G}^{\Gamma}.
\end{align*}
Indeed, with notation as in (\ref{eq:AJparameterequation0}) and (\ref{eq:AJparameterequation01}),
we just need to define $\psi_L$ by
\begin{align}\label{eq:arthurparameterL}
\psi_{L}(z)=\psi(z)=z^{\lambda}z^{\lambda'}\quad\text{ and }\quad \psi_L(j)=t{}^{\vee}\delta_L.
\end{align}
Next, since $\psi_{{L}}$ is an essentially unipotent Arthur parameter, by
Theorem \ref{theo:essentiallyunipotentparameter}(d) there exists
for each strong real form $\delta$ of $L^{\Gamma}$
a projective character $\chi(\psi_L,\delta)$ of type $z$
of $L^{can}(\delta,\mathbb{R})$.
We notice that since is $\psi_{L}|_{\mathbb{C}^{\times}}$ is bounded, the character
$\chi(\psi_L,\delta)$ is unitary.
Suppose now that $\delta\in L^{\Gamma}$ is a strong real form of $L^{\Gamma}$
so that $\delta^{2}\in Z({}G)$.
Then $\delta$ can be seen as a strong real form of $G^{\Gamma}$.
Write $$O_{\delta}=\{g\delta g^{-1}:g\in N_{G}(L)\}$$ and
make $L$ act on $O_{\delta}$ by conjugation. Denote by $\mathcal{O}_{\delta}$
the set of $L$-conjugacy classes of $O_{\delta}$. Each $L$-orbit in $O_{\delta}$
defines a $L$-conjugacy class of strong real forms in $L^{\Gamma}$, belonging to the same $G$-conjugacy class of strong real forms in $G^{\Gamma}$.
For each $s\in \mathcal{O}_\delta$, let
$\delta_s$
be a representative of the corresponding $L$-conjugacy class,
and as above write $\chi(\psi_L,\delta_s)$
for the unitary character of type $z$ of ${L}^{can}(\mathbb{R},\delta_s)$
corresponding to $\delta_s$ and $\psi_L$. Now, let
\begin{itemize}
\item $K_{\delta_s}^{can}$ be the preimage in $G^{can}$ of $K_{\delta_s}$, the set of
fixed points of the Cartan involution corresponding to $\sigma_{\delta_s}$.
\item $\mathfrak{l}=\text{Lie}({L})$.
\item $i=(1/2)\dim(\mathfrak{k}_{\delta_s}/\mathfrak{l}\cap\mathfrak{k}_{\delta_s})$.
\end{itemize}
and consider the cohomologically induced representation
\begin{align*}
\left(\mathscr{R}_{\mathfrak{l},K_{\delta_s}^{can}\cap L^{can}}^{\mathfrak{g},K_{\delta_s}^{can}}\right)^{i}(\chi(\psi_L,\delta_s)).
\end{align*}
We notice that the $L$-parameter associated to $\left(\mathscr{R}_{\mathfrak{l},K_{\delta_s}^{can}\cap L^{can}}^{\mathfrak{g},K_{\delta_s}^{can}}\right)^{i}(\chi(\psi_L,\delta_s))$
is the $L$-parameter of an $L$-group
obtained by composing the $L$-parameter corresponding to
$\chi(\psi_L,\delta_s)$ with the embedding $\iota_{{L},{G}}:{}^{\vee}L^{\Gamma}\rightarrow {}^{\vee}{G}^{\Gamma}$. Therefore, using the Local Langlands Correspondence (see Theorem \ref{theo:4.1}) we conclude that
$\left(\mathscr{R}_{\mathfrak{l},K_{\delta_s}^{can}\cap L^{can}}^{\mathfrak{g},K_{\delta_s}^{can}}\right)^{i}(\chi(\psi_L,\delta_s))$ is of type 1, that is to say
trivial on $\pi_1(G)^{can}$, and consequently can be seen as a representation of $G(\delta_{s},\mathbb{R})$.
We can know give the definition of the packets defined by Adams and Johnson in \cite{Adams-Johnson}
\begin{deftn}\label{deftn:ajpackets}
Let $\psi$ be an Arthur parameter for ${}^{\vee}G^{\Gamma}$ satisfying points
$\mathrm{(AJ1)}$, $\mathrm{(AJ2)}$ and $\mathrm{(AJ3)}$, above.
The Adams-Johnson packet corresponding to $\psi$ is defined as the set
\begin{align*}
\Pi(G/\mathbb{R})_{\psi}^{\mathrm{AJ}}:=\left\{\left(\mathscr{R}_{\mathfrak{l},K_{\delta_s}^{can}\cap L^{can}}^{\mathfrak{g},K_{\delta_s}^{can}}\right)^{i}(\chi(\psi_L,\delta_s)) :\delta\in L^{\Gamma}-L
\mathrm{~with~} \delta^{2}\in Z(G)\mathrm{~and~}s\in \mathcal{O}_{\delta}\right\}.
\end{align*}
\end{deftn}
The next result enumerates some important properties satisfied
by Adams-Johnson packets.
\begin{theo}\label{ajpproperties}
Let $\psi$ be an Arthur parameter for ${}^{\vee}G^{\Gamma}$ satisfying points
$\mathrm{(AJ1)}$, $\mathrm{(AJ2)}$ and $\mathrm{(AJ3)}$, above.
\begin{enumerate}[i.]
\item $\Pi(G/\mathbb{R})_{\psi}^{\mathrm{AJ}}$ contains the $L$-packet $\Pi(G/\mathbb{R})_{\varphi_{\psi}}$.
\item $\Pi(G/\mathbb{R})_{\psi}^{\mathrm{AJ}}$ is the support of a stable formal virtual character (For a precise description of the stable character see Theorem 2.13 \cite{Adams-Johnson}).
\item $\Pi(G/\mathbb{R})_{\psi}^{\mathrm{AJ}}$ satisfies the ordinary endoscopic identities predicted by the theory of endoscopy (For a more precise statement of this point see Theorem 2.21 \cite{Adams-Johnson}).
\end{enumerate}
\end{theo}
We turn now to the proof that Adams-Johnson packets are ABV-packets. We begin by
noticing that from Theorem \ref{theo:essentiallyunipotentparameter}
we have
$$\Pi(L/\mathbb{R})_{\psi_L}^{\text{ABV}}:=\left\{\chi(\psi_L,\delta):\delta \text{ a strong real form of }L^{\Gamma} \right\}.$$
Thus, we can give a reformulation of Definition \ref{deftn:ajpackets}
in terms of the the ABV-packet attached to $\psi_{L}$ as follows
\begin{align}\label{eq:reformulationAJpackets}
\Pi(G/\mathbb{R})_{\psi}^{\mathrm{AJ}}:=\left\{\left(\mathscr{R}_{\mathfrak{l},K_{\delta_s}^{can}\cap L^{can}}^{\mathfrak{g},K_{\delta_s}^{can}}\right)^{i}(\pi) :\pi\in\Pi^{z}(L(\mathbb{R},\delta_s))_{\psi_L}^{\mathrm{ABV}},~ \delta\in L^{\Gamma}-L
\mathrm{~with~} \delta^{2}\in Z(G)\mathrm{~and~}s\in \mathcal{O}_{\delta}\right\} .
\end{align}
The next two results are going to be needed in what follows. The first is Proposition 1.11 \cite{ABV}.
\begin{prop}\label{prop:quasisplitrepresentation}
Suppose $\sigma$ is a quasisplit real form of $G$. Let $\varphi,~\varphi'\in \Phi({}^{\vee}{G}^{\Gamma})$ be two Langlands parameters, with $S$ and $S'$ as the corresponding ${}^{\vee}{{G}}$-orbits. Then the following conditions are equivalent:
\begin{enumerate}[i.]
\item $S$ is contained in the closure of $S'$.
\item There are irreducible representations $\pi\in \Pi(G(\sigma,\mathbb{R}))_{\varphi}$ and $\pi'\in\Pi(G(\sigma,\mathbb{R}))_{\varphi'}$ with
the property that $\pi'$ is a composition factor of the standard module of which $\pi$ is the unique Langlands quotient.
\end{enumerate}
\end{prop}
We remark that $(ii)$ implies $(i)$ even for $G(\delta,\mathbb{R})$ not quasi-split.
\begin{lem}\label{lem:orbit}
Let $\psi$ be an Arthur parameter for ${}^{\vee}{G}^{\Gamma}$ satisfying points
$\mathrm{(AJ1)}$, $\mathrm{(AJ2)}$ and $\mathrm{(AJ3)}$, above. Let $L^{\Gamma}$ be the extended group defined in (\ref{eq:extendedgroupL}), and let $\left({}^{\vee}{L}^{\Gamma},\mathcal{S}_L\right)$ be the $E$-group defined in (\ref{eq:e-groupL}). Write $\psi_{{L}}$ for the unique (up to conjugation) Arthur parameter of ${}^{\vee}{L}^{\Gamma}$
satisfying $\psi=\iota_{L,G}\circ\psi_{L}$. Let $\iota:X\left({}^{\vee}{L}^{\Gamma}\right)\rightarrow X\left({}^{\vee}{G}^{\Gamma}\right)$
be the closed immersion
induced from the inclusion ${}^{\vee}{L}^{\Gamma}\hookrightarrow {}^{\vee}{G}^{\Gamma}$ (see Proposition \ref{prop:mapvarieties}).
Write $S_{\psi}$ for the ${}^{\vee}{G}$-orbit
in $X\left({}^{\vee}{G}^{\Gamma}\right)$ corresponding to $\psi$ and
$S_{\psi_{L}}$ for the ${}^{\vee}{L}$-orbit
in $X\left({}^{\vee}{G}^{\Gamma}\right)$ corresponding to $\psi_{L}$.
Suppose $S$ is an ${}^{\vee}{G}$-orbit
in $X\left({}^{\vee}{G}^{\Gamma}\right)$ containing $S_{\psi}$ in its closure,
then there exist an orbit $S_{L}$ of
${}^{\vee}{L}$
in $X\left({}^{\vee}{L}^{\Gamma}\right)$ with
$S_{\psi_{L}}\subset \overline{S}_{{L}}$
and such that
$$S={}^{\vee}{G}\cdot \iota(S_{L}).$$
\end{lem}
\begin{proof}
Let $\delta\in G^{\Gamma}$ be a strong real form of $G^{\Gamma}$, whose associated
real form is quasi-split.
Set $\varphi_{\psi}$ to be the Langlands parameter attached to $\psi$.
Suppose $S$ is a ${}^{\vee}{G}$-orbit
in $X\left({}^{\vee}G^{\Gamma}\right)$ containing $S_{\psi}$ in its closure,
and write $\varphi$ for the Langlands parameter corresponding to $S$
under the map defined in (\ref{eq:langlandsgeometric}).
From Proposition \ref{prop:quasisplitrepresentation}
there are irreducible representations $\pi\in \Pi_{\varphi_{\psi}}({G}(\mathbb{R},\delta))$
and $\pi'\in \Pi_{\varphi}({G}(\mathbb{R},\delta))$, with the
property that $\pi'$ is a composition factor of the standard
module $M(\pi)$ of which $\pi$ is the unique quotient.
Let $\Pi({G}(\mathbb{R},\delta))_{\psi}^{\text{AJ}}$ be the restriction to $\delta$ of the Adams-Johnson packet attached to
$\psi.$ By Theorem \ref{ajpproperties}$(i)$,
the $L$-packet $\Pi_{\varphi_{\psi}}({G}(\mathbb{R},\delta))$ is contained in $\Pi_{\psi}^{\text{AJ}}({G}(\mathbb{R},\delta))$.
By Definition \ref{deftn:ajpackets}
there is an element $s\in\mathcal{O}_{\delta}$,
to which we can attach a
strong real form $\delta_s$ of $L^{\Gamma}$, such that $\pi$ is cohomologically induced from the unitary character
$\chi(\psi_L,\delta_s)$ of ${L}^{can}(\mathbb{R},\delta_s)$ corresponding to $\delta_s$ and
$\psi_L$ as above , i.e.
$$\pi=\left(\mathscr{R}_{\mathfrak{l},K_{\delta_s}\cap L}^{\mathfrak{g},K_{\delta_s}}\right)^{i}(\chi(\psi_L,\delta_s)).$$
Now, by dualizing the resolution of Theorem 1 of Section 6 \cite{Johnson}, we find
from the exactness of
$\left(\mathscr{R}_{\mathfrak{l},K_{\delta_s}^{can}\cap L^{can}}^{\mathfrak{g},K_{\delta_s}^{can}}\right)^{i}$
that
$\pi$ possess a resolution by direct sums of standard modules
\begin{align}\label{eq:resolution}
0\leftarrow \pi\leftarrow M(\pi)\leftarrow \cdots \leftarrow M_{0}\leftarrow 0,
\end{align}
cohomologically induced from a resolution by direct sums of standard modules of
$\chi(\psi_L,\delta_s)$
\begin{align}\label{eq:resolutionL}
0\leftarrow \chi(\psi_L,\delta_s)\leftarrow M(\chi(\psi_L,\delta_s))\leftarrow \cdots \leftarrow M_{0}^{L}\leftarrow 0.
\end{align}
Consequently,
every composition factor of $M(\pi)$
is obtained after applying cohomological induction
to a composition factor of $M(\chi(\psi_L,\delta_s))$.
Therefore, there exists $\pi_{L}'$,
the unique quotient of a standard module
appearing in (\ref{eq:resolutionL}), such that
\begin{align}\label{eq:resolutionL2}
\pi'=\left(\mathscr{R}_{\mathfrak{l},K_{\delta_s}^{can}\cap L^{can}}^{\mathfrak{g},K_{\delta_s}^{can}}\right)^{i}(\pi_{L}').
\end{align}
Let $\varphi_{L}$ be the Langlands parameter corresponding to $\pi_{L}'$
and write $S_{{L}}$ for the orbit of ${}^{\vee}{L}$ in $X\left({}^{\vee}{L}^{\Gamma}\right)$ associated to $\varphi_L$
under (\ref{eq:langlandsgeometric}).
Since $\pi_L'$ is a composition factor of $X\left(\chi(\psi_L,\delta_s)\right)$,
the remark after Proposition \ref{prop:quasisplitrepresentation}$(ii)$, implies that
the orbit $S_{\psi_{L}}$ is contained in the closure of ${S_{{L}}}$.
Furthermore, from Equation (\ref{eq:resolutionL2}) and the relation
between the data parameterizing $\pi_L'$ and $\pi$ in \cite{Johnson}
(see page 392 of \cite{Johnson}),
we may verify that the Langlands parameter $\varphi$ factors through
$\varphi_{L}$. Thus by Proposition \ref{prop:mapvarieties}
the orbits $S_L$ and $S$ correspond under the map $\iota:X\left({}^{\vee}{L}^{\Gamma}\right)\rightarrow X\left({}^{\vee}{G}^{\Gamma}\right)$, that is to say
$$S={}^{\vee}{G}\cdot \iota(S_{L}).$$
\end{proof}
\begin{theo}\label{theo:ABV-AJ}
Let $\psi$ be an Arthur parameter for ${}^{\vee}G^{\Gamma}$ satisfying points
$\mathrm{(AJ1)}$, $\mathrm{(AJ2)}$ and $\mathrm{(AJ3)}$, above. Then
$$\Pi(G/\mathbb{R})_{\psi}^{\mathrm{ABV}}=\Pi(G/\mathbb{R})_{\psi}^{\mathrm{AJ}}.$$
\end{theo}
\begin{proof}
Let $L^{\Gamma}$ be the extended group defined in (\ref{eq:extendedgroupL}),
and let $\left({}^{\vee}{L}^{\Gamma},\mathcal{S}_L\right)$ be the $E$-group defined in (\ref{eq:e-groupL}).
Write $\psi_{{L}}$ for the unique (up to conjugation) Arthur parameter of ${}^{\vee}{L}^{\Gamma}$
satisfying $\psi=\iota_{L,G}\circ\psi_{L}$.
Let $\lambda_\psi$ and $\lambda_\psi'$ be defined as in (\ref{eq:AJLparameter}). Then by (\ref{eq:AJLparameter2}), (\ref{eq:arthurparameterL}) and the definition of the embedding ${}^{\vee}{L}^{\Gamma}\hookrightarrow {}^{\vee}{G}^{\Gamma}$, we have
$$\varphi_{\psi_L}(z)=\varphi_{\psi}(z)=z^{\lambda_{\psi}}z^{\lambda'_{\psi}},\quad z\in\mathbb{C}^{\times}.$$
Following the characterization of
Arthur parameters related to Adams-Johnson packets, $\lambda_{\psi}$
is an integral and regular semisimple element of ${}^{\vee}{\mathfrak{g}}$.
Write
$$\mathcal{O}_{\psi}={}^{\vee}{G}\cdot\lambda_{\psi}\quad\text{ and }\quad \mathcal{O}_{\psi_L}={}^{\vee}{L}\cdot\lambda_{\psi}.$$
Let $\xi=\left(\left(y_{\xi},\Lambda_{\xi}\right),\tau_{\xi}\right)$ be a complete geometric parameter for ${}^{\vee}G^{\Gamma}$
such that its corresponding orbit ${S_{\xi}}$ of ${}^{\vee}{G}$ in $X\left({}^{\vee}G^{\Gamma}\right)$,
contains $S_{\psi}$ in its closure. Let $\pi(\xi)$ be the irreducible representation corresponding to $\xi$ under the Local Langlands Correspondence (see Theorem \ref{theo:4.1}) and write
$\delta_{\xi}$ for the strong real form of $G^{\Gamma}$ of which $\pi(\xi)$ is a representation
(i.e. $\pi(\xi)\in \Pi(G(\mathbb{R},\delta_{\xi}))$).
Let $K_\xi$ be the set of fixed points of the Cartan involution corresponding to ${\delta_{\xi}}$.
By Lemma \ref{lem:orbit} there exists an orbit
$S_{L}$ of ${}^{\vee}{L}$ in $X\left({}^{\vee}L^{\Gamma}\right)$ satisfying
\begin{align}\label{eq:orbitsrelation}
S_{\psi_{L}}\subset \overline{S}_{{L}}\quad \text{and}\quad S_{\xi}={}^{\vee}{G}\cdot \iota (S_{L}).
\end{align}
Therefore, the Langlands parameter associated to $\xi$
factors through ${}^{\vee}L^{\Gamma}$ and
$\pi(\xi)$ is cohomologically induced from
an irreducible representation of some real form of $L$. More precisely, there exist a complete geometric parameter $\xi_{L}=((y_{\xi_L},\Lambda_{\xi_L}),\tau_{\xi_L})$ for ${}^{\vee}L^{\Gamma}$ with $S_{\xi_L}=S_{L}$ and such that if we write $\pi_{L}(\xi_{L})$ for
the irreducible representation corresponding to $\xi_{L}$ under Langlands correspondence,
then
\begin{align}\label{eq:LtoG1}
\pi(\xi)=\left(\mathscr{R}_{\mathfrak{l},K_{\xi}^{can}\cap L^{can}}^{\mathfrak{g},K_{{\xi}}^{can}}\right)^{i}(\pi(\xi_{L})).
\end{align}
Now, since $\lambda_{\xi} \in \mathcal{O}_{\psi}$,
point (AJ3) above, and equation (\ref{eq:higuest})
implies that $\lambda_{\xi}$ is a regular and integral element of ${}^{\vee}\mathfrak{g}$.
Similarly, since $\lambda_{\xi_L} \in \mathcal{O}_{\psi_L}$,
we deduce that $\lambda_{\xi_L}$
is a regular and integral element of ${}^{\vee}\mathfrak{l}$.
Consequently, with notation as in
(\ref{eq:groupslambda}),
we obtain that ${}^{\vee}{G}(\Lambda_{\xi})={}^{\vee}{G}$, ${}^{\vee}{L}(\Lambda_{\xi_L})={}^{\vee}{L}$
and that $P(\Lambda_{\xi})$, respectively $P(\Lambda_{\xi_L})$, is a Borel subgroup of ${}^{\vee}{G}$, respectively of ${}^{\vee}{L}$.
Let $X_{y_{\xi}}\left(\mathcal{O}_{\psi},{}^{\vee}G^{\Gamma}\right)$ and $X_{y_{\xi_L}}\left(\mathcal{O}_{\psi_L},{}^{\vee}L^{\Gamma}\right)$ be the smooth subvarieties of $X\left({}^{\vee}G^{\Gamma}\right)$
and $X\left({}^{\vee}L^{\Gamma}\right)$ defined in (\ref{eq:varieties}).
It's clear from the definition of these varieties that
$$S_{\xi}\subset X_{y_{\psi}}\left(\mathcal{O}_{\psi},{}^{\vee}G^{\Gamma}\right)\quad\text{and}\quad S_{\xi_L}\subset X_{y_{\psi_L}}\left(\mathcal{O}_{\psi_L},{}^{\vee}L^{\Gamma}\right).$$
Let ${}^{\vee}K_\xi$ be the set of fixed points of
the involutive automorphism defined by $y_\xi$. Define
${}^{\vee}K_{\xi_L}$ similarly.
We notice that, since $\varphi$ factors trough $\varphi_{L}$, we have
$y_{\xi}=\iota_{L,G}(y_{\xi_L})$ (see Equation \ref{eq:embedding1} and \ref{eq:embedding2}) and thus
${}^{\vee}K_{\xi_L}={}^{\vee}K_{\xi}\cap {}^{\vee}L$.
As explained in (\ref{eq:varietiesisomor})
the varieties $X_{y_{\xi}}\left(\mathcal{O}_{\psi},{}^{\vee}G^{\Gamma}\right)$ and
$X_{y_{\xi_L}}\left(\mathcal{O}_{\psi_L},{}^{\vee}L^{\Gamma}\right)$ satisfy
$$X_{y_{\xi}}\left(\mathcal{O}_{\psi},{}^{\vee}G^{\Gamma}\right)\cong {}^{\vee}{G}\times_{{}^{\vee}K_{\xi}}{}^{\vee}{G}(\Lambda_{\xi})/P(\Lambda_{\xi})\quad\text{and}\quad
X_{y_{\xi_L}}(\mathcal{O}_{\psi_L},{}^{\vee}L^{\Gamma})\cong{}^{\vee}{L}\times_{{}^{\vee}K_{\xi_L}}{}^{\vee}{L}(\Lambda_{\xi_L})/P(\Lambda_{\xi_L}).$$
Thus, denoting the flag varieties of
${}^{\vee}{G}$ and ${}^{\vee}{L}$ by $X_{{}^{\vee}{G}}$ and $X_{{}^{\vee}{L}}$, we can write
$$X_{y_{\xi}}\left(\mathcal{O}_{\psi},{}^{\vee}G^{\Gamma}\right)\cong{}^{\vee}{G}\times_{{}^{\vee}K_\xi}X_{{}^{\vee}{G}} \quad\text{and}\quad
X_{y_{\xi_L}}\left(\mathcal{O}_{\psi_L},{}^{\vee}L^{\Gamma}\right)\cong{}^{\vee}{L}\times_{{}^{\vee}K_{\xi_L}}X_{{}^{\vee}{L}}.$$
Define $\Xi\left(X_{{}^{\vee}{G}}\right)$ to be the set of couples $(S,\mathcal{V})$ with $S$ an orbit of
${}^{\vee}K_\xi$ on $ X_{{}^{\vee}{G}}$ and $\mathcal{V}$ an irreducible ${}^{\vee}K_\xi$-equivariant local system on $S$. Define $\Xi\left(X_{{}^{\vee}{L}}\right)$ similarly.
From Proposition 7.14 \cite{ABV} the map in
Proposition \ref{prop:reduction2}(2) is compatible with the parameterization of irreducibles
by $\Xi_{y_{\xi_L}}\left(\mathcal{O}_{\psi_L},{}^{\vee}L^{\Gamma}\right)$ and $\Xi\left(X_{{}^{\vee}L}\right)$,
respectively by $\Xi_{y_\xi}\left(\mathcal{O}_{\psi},{}^{\vee}G^{\Gamma}\right)$ and $\Xi\left(X_{{}^{\vee}G}\right)$.
In other words we have bijections
\begin{align}\label{eq:bijectioncomparision}
\Xi_{y_{\xi_{L}}}\left(\mathcal{O}_{\psi_L},{}^{\vee}L^{\Gamma}\right)\longrightarrow \Xi(X_{{}^{\vee}L}) \quad\text{ and }\quad
\Xi_{y_\xi}\left(\mathcal{O}_{\psi},{}^{\vee}G^{\Gamma}\right)\longrightarrow \Xi(X_{{}^{\vee}G}).
\end{align}
Let $\xi'$ be the complete geometric parameter of $X_{{}^{\vee}{G}}$
corresponding to $\xi$
under (\ref{eq:bijectioncomparision})
and write
$P(\xi')$
for the irreducible perverse sheaves on $X_{{}^{\vee}{G}}$
defined from $\xi$
as in (\ref{eq:irredPerv}).
Let ${}^{\vee}\pi(\xi')$ be the $({}^{\vee}{\mathfrak{g}},{}^{\vee} K_\xi)$-module corresponding
to $P(\xi')$ under Beilinson-Bernstein localization (see Theorem \ref{theo:bbcorespondance}).
Define $\xi_L'$, $P(\xi_{L}')$ and ${}^{\vee}\pi(\xi_{L}')$ similarly.
Then ${}^{\vee}\pi(\xi')$ and ${}^{\vee}\pi(\xi'_L)$ are Vogan duals (See Theorem 11.9 \cite{ICIV} and Definition 2-28 \cite{Adams}) of $\pi(\xi)$ and
$\pi(\xi_L)$ respectively. Now, Vogan duality
commutes with cohomological induction. Indeed, this is clear from the data parameterizing the representations in Definition 2-9 \cite{Adams},
and is straightforward to translate the data in \cite{Adams} into complete Langlands parameters (Definition \ref{deftn:completeLanglandsparameter}).
Equation (\ref{eq:LtoG1})
implies then that ${}^{\vee}\pi(\xi')$ and ${}^{\vee}\pi(\xi'_L)$ are related trough the equation
\begin{align}
{}^{\vee}\pi(\xi')=\left(\mathscr{R}_{{}^{\vee}{\mathfrak{l}},{}^{\vee}K_{\xi_L}}^{{}^{\vee}{\mathfrak{g}},{}^{\vee}K_{\xi}}\right)^{i}\left({}^{\vee}\pi(\xi'_{L})\right).
\end{align}
Therefore,
from the commutativity of (\ref{eq:cdiagramme1})
we deduce the equality
$$P(\xi')=I_{{}^{\vee}{L}}^{{}^{\vee}{G}}P(\xi'_{L}),$$
where $I_{{}^{\vee}{L}}^{{}^{\vee}{G}}$ denotes the geometric induction functor of Definition \ref{deftn:geominduction}.
On the level of characteristic cycles we have from Proposition \ref{prop:ccLG} that
\begin{align}\label{eq:cycleequation}
CC(P(\xi'))=\sum_{{}^{\vee}K_{\xi_L}-\mathrm{orbits }~S_{{}^{\vee}{L}}~\mathrm{in}~X_{{}^{\vee}{L}}} \chi_{S_{{}^{\vee}{L}}}^{mic}(P_{\xi'_{{L}}})&[\overline{T_{{}^{\vee}K_{\xi}\cdot\iota(S_{{}^{\vee}{L}})}^{\ast}X_{{}^{\vee}{G}}}]\\
&+
\sum_{{}^{\vee}K_{\xi}-\text{orbits}~S'\text{~in~}\partial({}^{\vee}K_{\xi}\cdot \iota(X_{{}^{\vee}{L}}))} \chi_{S'}^{mic}(P(\xi'))[\overline{T_{S'}^{\ast}X_{{}^{\vee}{G}}}].\nonumber
\end{align}
Now,
for each ${}^{\vee}{G}$-orbit $S$ in $X_{y_{\xi}}\left(\mathcal{O}_{\psi},{}^{\vee}G^{\Gamma}\right)$ let
$S'$ be the ${}^{\vee}K_{\xi}$-orbit in $X_{{}^{\vee}{G}}$ corresponding to $S$ under (\ref{eq:bijectioncomparision}).
Then from Proposition \ref{prop:reduction2}$(1)$ and \ref{prop:reduction2}(6) we have
$$S_{\psi}\subset \overline{S}\quad \text{iff}\quad S_{\psi}'\subset \overline{S'}\quad \text{ and that} \quad
\chi_{S}^{\mathrm{mic}}(P(\xi))=\chi_{S'}^{\mathrm{mic}}(P(\xi')).$$
We have similar relations between ${}^{\vee}L^{\Gamma}$-orbits in
$X\left(\mathcal{O}_{\psi_{L}},{}^{\vee}L^{\Gamma}\right)$
and ${}^{\vee}K_{\xi_{L}}$-orbits in $X_{{}^{\vee}{L}}$.
For each orbit $S'$ in the boundary $\partial({}^{\vee}K_{\xi}\cdot \iota(X_{{}^{\vee}{L}}))$
we have $d(S')<d(S_{\psi}')$,
thus from (\ref{eq:cycleequation})
we can conclude
\begin{align}\label{eq:cycleLcycleG}
\chi_{S_{\psi}}^{\mathrm{mic}}(P(\xi))=\chi_{S_{\psi}'}^{\mathrm{mic}}(P(\xi'))\neq 0\quad \text{if and only if}\quad
\chi_{S_{\psi_{L}}}^{\mathrm{mic}}(P(\xi_{L}))=\chi_{S_{\psi_{L}}'}^{\mathrm{mic}}(P(\xi_{L}'))\neq 0.
\end{align}
Hence, from
\ref{lem:lemorbit}($i$)
and
the definition of micro-packets we obtain that
equation (\ref{eq:cycleLcycleG}) translates as:
\begin{align}\label{eq:last}
\text{Let }\xi\in \Xi\left({}^{\vee}G^{\Gamma}\right)\text{, then }\pi(\xi)\in\Pi(G/\mathbb{R})_{\psi}^{\text{ABV}}
\text{ if and only if there exists }\xi_L \in \Xi\left({}^{\vee}L^{\Gamma}\right)\\
\text{ such that } \pi(\xi_{L})\in\Pi^{z}(L/\mathbb{R})_{\psi_{L}}^{\text{ABV}}
\text{ and }\pi(\xi)=\left(\mathscr{R}_{\mathfrak{l},K_{\xi}^{can}\cap L^{can}}^{\mathfrak{g},K_{\xi}^{can}}\right)&^{i}(\pi(\xi_{L})).\nonumber
\end{align}
Finally, the reformulation of the definition of Adams-Johnson packets
expressed in (\ref{eq:reformulationAJpackets})
implies that
Equation (\ref{eq:last}) is equivalent to
$$\Pi(G/\mathbb{R})_{\psi}^{\mathrm{ABV}}=\Pi(G/\mathbb{R})_{\psi}^{\mathrm{AJ}}.$$
\end{proof}
\bibliographystyle{alpha}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 3,485
|
Би́смак Бийо́мбо Су́мба (; род. 28 августа 1992 года, Лубумбаши, Демократическая Республика Конго) — конголезский профессиональный баскетболист, который выступает за команду НБА «Финикс Санз». Был выбран на драфте НБА 2011 года в первом раунде под общим седьмым номером командой «Сакраменто Кингз». Играет на позиции центрового.
Биография
Бисмак родился во втором по величине городе Демократической Республики Конго Лубумбаши. В 16-летнем возрасте на турнире в Йемене его заметил известный испанский тренер Марио Пальма, который сейчас возглавляет сборную Португалии. Последний в то время тренировал иорданскую Сборную и, впечатлившись игрой Бийомбо, дал ему возможность попробовать свои силы в Европе.
Профессиональная карьера
Первый год юный баскетболист провёл во второй команде испанской «Фуэнлабрады» «Ильескасе». По его завершению он был переведён в первую команду, а также начал вызываться в национальную Сборную своей страны.
Дебют Бисмака в чемпионате Испании состоялся 9 января 2010 года против «Ховентута» из Бадалоны. В нём игрок набрал 5 очков и сделал 7 подборов за предоставленные ему 13 минут матча.
На ежегодном матче молодых талантов , в котором Бийомбо выступал за Международную команду против команды США, баскетболист установил первый трипл-дабл (12 очков, 11 подборов и 10 блок-шотов) за все 14 лет проведения матча. После этого многие скауты существенно повысили рейтинг Бисмака среди других потенциальных новичков НБА. В итоге на драфте 2011 года Бисмака Бийомбо под общим седьмым номером выбирает «Сакраменто Кингз», но по условиям договора между «Кингз» и «Шарлотт Бобкэтс», все права на игрока переходят к последним.
После попыток во время локаута в НБА «Фуэнлабрады» в возмещении денежной компенсации для клуба из-за досрочного разрыва контракта, 19 декабря 2011 года Бийомбо всё же подписал контракт с «Шарлотт Бобкэтс».
Статистика
Статистика в НБА
Статистика в других лигах
Примечания
Ссылки
Профиль на сайте Draftexpress.com
Родившиеся в Лубумбаши
Баскетболисты Демократической Республики Конго
Игроки БК «Фуэнлабрада»
Игроки «Шарлотт Бобкэтс»
Игроки «Шарлотт Хорнетс»
Игроки «Торонто Рэпторс»
Игроки «Орландо Мэджик»
Игроки «Финикс Санз»
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 6,883
|
\partial{\partial}
\def\tfrac#1#2{{\textstyle{#1\over #2}}}
\def\tfrac{1}{2}{\tfrac{1}{2}}
\def\times{\times}
\def\otimes{\otimes}
\def\dagger{\dagger}
\def\dagg#1{#1^{\dagger}}
\def\bra#1{{\cal{h}} #1 \mid}
\def\ket#1{\mid #1 {\cal{i}}}
\def\ketbra#1#2{\ket{#1} \bra{#2}}
\def\abs#1{\mid #1 \mid}
\def\norm#1{\| #1 \|}
\def\ensuremath{\mathbbm{1}}{\ensuremath{\mathbbm{1}}}
\def{\mathfrak{Re}} {{\mathfrak{Re}} }
\def{\mathfrak{Im}} {{\mathfrak{Im}} }
\def\overrightarrow{\overrightarrow}
\def\Leftrightarrow{\Leftrightarrow}
\def\Rightarrow{\Rightarrow}
\def\Leftarrow{\Leftarrow}
\def\vect{\nabla}{\overrightarrow{\nabla}}
\def\nabla^2{\nabla^2}
\def\nabla {\bf{\cdot}}{\nabla {\bf{\cdot}}}
\def\nabla \times{\nabla \times}
\def{\alpha^\prime}{{\alpha^\prime}}
\def\mbox{Tr}{\mbox{Tr}}
\def\mbox{Str}{\mbox{Str}}
\def\ensuremath{\Box}{\ensuremath{\Box}}
\def\ensuremath{\mathbb{N}}{\ensuremath{\mathbb{N}}}
\def\ensuremath{\mathbb{Z}}{\ensuremath{\mathbb{Z}}}
\def\ensuremath{\mathbb{Q}}{\ensuremath{\mathbb{Q}}}
\def\ensuremath{\mathbb{R}}{\ensuremath{\mathbb{R}}}
\def\ensuremath{\mathbb{C}}{\ensuremath{\mathbb{C}}}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newenvironment{proof}[1][Proof]{\begin{trivlist}
\item[\hskip \labelsep {\bfseries #1}]}{\end{trivlist}}
\newenvironment{definition}[1][Definition]{\begin{trivlist}
\item[\hskip \labelsep {\bfseries #1}]}{\end{trivlist}}
\newenvironment{example}[1][Example]{\begin{trivlist}
\item[\hskip \labelsep {\bfseries #1}]}{\end{trivlist}}
\newenvironment{remark}[1][Remark]{\begin{trivlist}
\item[\hskip \labelsep {\bfseries #1}]}{\end{trivlist}}
\begin{document}
\preprint{UdeM-GPP-TH-10-189}
\title{Fate of the false monopoles: induced vacuum decay}
\author{Brijesh Kumar$^{1,2}$}
\email{brijesh@phy.iitb.ac.in}
\author{M. B. Paranjape$^2$}
\email{paranj@lps.umontreal.ca}
\author{U. A. Yajnik$^{1,2,3}$}
\email{yajnik@iitb.ac.in}
\affiliation{$^1$ Physics Department, Indian Institute of Technology Bombay, Mumbai, 400076, India}
\affiliation{$^2$Groupe de physique des particules, D\'epartement de
physique, Universit\'e de Montr\'eal, C.P. 6128, Succ. Centre-ville,
Montr\'eal, Qu\'ebec, CANADA, H3C 3J7 }
\affiliation{$^3$ Department of Physics, Ernest Rutherford Physics Building, McGill University, 3600 rue University,
Montr\'eal, Qu\'ebec, CANADA, H3A 2T5}
\begin{abstract}
We study a gauge theory model where there is an intermediate symmetry breaking to a meta-stable vacuum that breaks a simple gauge group to a $U(1)$ factor. Such models admit the existence of meta-stable magnetic monopoles, which we dub false monopoles. We prove the existence of these monopoles in the thin wall approximation. We determine the instantons for the collective coordinate that corresponds to the radius of the monopole wall and we calculate the semi-classical tunneling rate for the decay of these monopoles. The monopole decay consequently triggers the decay of the false vacuum. As the monopole mass is increased, we find an enhanced rate of decay of the false vacuum relative to the celebrated homogeneous tunneling rate due to Coleman \cite{Coleman:1975qj}.
\end{abstract}
\pacs{12.60.Jv,11.27.+d}
\maketitle
\section{Introduction}
Semi-classical solutions with topologically non-trivial boundary conditions in relativistic
field theory \cite{'tHooft:1974qc,Polyakov:1974ek,Polyakov:1976fu} have the interesting property that
they interpolate between two or more alternative translationally invariant vacua of the theory.
For instance the exterior of a monopole or a vortex solution is a phase of broken
symmetry while the interior of the object generically contains a limited region of
unbroken symmetry (for more details and lucid expositions see \cite{Coleman:1975qj} and \cite{Rajaraman:1982is}).
Most of the commonly studied solutions are topologically non-trivial, however non-trivial boundary
conditions are not a guarantee of dynamical stability. In \cite{Preskill:1992ck} for example a large
number of such solutions are constructed in gauge field theories which are generically metastable. The skyrmion is also a classic example of a topologically non-trivial configuration that is unstable without the addition of fourth order Skyrme term \cite{Skyrme:1962vh,Gisiger:1998tv}.
All of the classically stable solutions (allowing for quantum metastabilty), are non-trivial time independent local minima of the effective action of the theory.
The metastability of such solutions can be of significant interest. The implied decay of the
object would be accompanied by the change in phase of the system as a whole. In the context of
cosmology this may imply a change in the cosmic history and determine the abundance of
relic objects. On a more formal footing the question of metastability of vacua has
gained considerable interest in the context of supersymmetric field theories \footnote{See \cite{Terning:2003th}
for discussion of supersymmetric field theories.} where
a non-supersymmetric phase is required on phenomenological grounds but such a phase
is necessarily metastable on theoretical grounds \cite{Dine:1993yw,Intriligator:2006dd}.
In String Cosmology the de Sitter solution obtained is generically meta-stable \cite{Kachru:2003aw}
and its phenomenological viability depends on the tunneling rate being sufficiently slow.
Change in phase due to metastable topological objects is a generalisation of the
following better known mechanism.
When the effective potential of the theory possesses several local minima, all but the
lowest minimum are quantum mechanically unstable. The so called false vacua are then
liable to decay, even in the absence of topological objects, according to a rate given by
a WKB-like formula studied earlier by
\cite{Kobzarev:1974cp} and provided an elegant and lucid footing by Coleman \cite{Coleman:1977py,Coleman:1980aw}.
The cases studied there concerned a transition between two translationally invariant vacua.
The generic scenario of decay consists of spontaneous formation of a small \textit{bubble} of true vacuum,
which can then start growing by semi-classical evolution. In Minkowski space, the formation of
one such bubble is sufficient to convert the phase of the system to the true vacuum.
In the context of an expanding Universe, conversion of the entire Universe to the true vacuum
would require formation of sufficiently large number of such bubbles at an adequate rate.
The existence of topological objects may provide additional sources of metastability.
Phase transitions seeded by topological solutions were studied early in the works of
\cite{Steinhardt:1981ec,Hosotani:1982ii,Yajnik:1986tg,Yajnik:1986wq}. An essential aspect of
these studies is precisely the observations that there exist solutions with non-trivial
boundary conditions which interpolate between two distinct minima of the effective potential.
The importance of this alternative route to decay arises from the fact that it can be
much more rapid than the spontaneous decay of a translationally invariant vacuum. Indeed, for some values of the parameters the decay induced by topological objects may require no tunneling
and therefore would be very prompt in a context where the parameters are changing adiabatically,
as for instance in the early Universe.
Obtaining a general formula characterising this kind of vacuum decay has been rather elusive
although the ideas have been adequately explicated in \cite{Steinhardt:1981ec,Hosotani:1982ii,
Yajnik:1986tg,Yajnik:1986wq}.
More recently, the relevance of the mechanism has been demonstrated in specific examples,
in \cite{Kumar:2008jb} for the mediating sector of a hidden sector scenario of supersymmetry
breaking and in \cite{Kumar:2009pr} in a GUT model with O'Raifeartaigh type direct supersymmetry
breaking. In this paper we explore a model that is amenable to an analytical treatment within
the techniques developed in \cite{Coleman:1978ae}. In doing so we provide a transparent model
in which the generic expectations raised in \cite{Steinhardt:1981ec,Hosotani:1982ii,
Yajnik:1986tg,Yajnik:1986wq} can be realised
and a specific formula can be derived.
We construct an $SU(2)$ gauge model with a triplet scalar field with two possible translationally
invariant vacua, one with $SU(2)$ broken to $U(1)$ and the other with the original gauge symmetry
intact. The former phase permits the existence of monopoles. By appropriate choice of potential
for the triplet it can be arranged that the phase of unbroken symmetry is lower in energy and
represents the true vacuum of the theory. The monopoles interpolate between the true vacuum and
the false vacuum. For a wide range of the parameters, these monopoles are in fact classically stable. In previous work \cite{Steinhardt:1981ec,Hosotani:1982ii} the dissociation of such monopoles was considered, varying the parameters of the theory to critical values where the monopoles were classically unstable due to infinite dilation. This can occur for example in the early Universe where the high temperature phase prefers one
vacuum in which the system starts, but with adiabatic reduction in temperature, a different
phase becomes more favorable. The Universe is then
liable to simply \textit{roll over}, by classical evolution, to the true vacuum.
It was however, overlooked that these monopoles are in fact unstable due to quantum tunneling well before the parameters reach their critical values. We dub such monopoles \textit{false monopoles}.
Working in the thin wall limit for the monopoles \cite{Steinhardt:1981ec}, we show that such monopoles undergo quantum tunneling to larger monopoles, which are then classically unstable by expanding indefinitely, consequently converting all space to the true vacuum, the phase of unbroken $SU(2)$ symmetry. Further, the
formula we derive also recovers the regime of parameter space, within the thin wall monopole limit, where no tunneling is required for the decay but the monopole is simply classically unstable as previously treated \cite{Steinhardt:1981ec,Hosotani:1982ii}.
The rest of the paper is organised as follows. In section \ref{sec:model} we specify the model under
consideration and the monopole ansatz along with the equations of motion. In section \ref{sec:thinwallmonopole} we delineate the conditions under in which there should exist a metastable
monopole solution with a large radius and a thin wall. We find the thin wall monopole solutions and also justify their existence. In section \ref{sec:collco} we use the thin wall approximation which permits a treatment of the solution in terms of a single collective coordinate, the
radius $R$ of the thin wall. We argue that the monopole is unstable to
tunneling to a new configuration of a much larger radius and we determine the existence of the instanton
for this tunneling within the same thin wall approximation. In section \ref{sec:decay} we determine
the Euclidean action for this instanton, the so called bounce $B$ which determines
the tunneling rate for the appearance of the large radius unstable monopole. In section \ref{sec:gutmodel} we relate our findings to a previous study of classical monopole instability in supersymmetric GUT models. In section \ref{sec:conc} we discuss our results and compare our tunneling rate formula with that of the
homogeneous bubble formation case without monopoles. We show that in addition to our tunneling rate being significantly
faster, it also indicates a regime in which the monopoles become unstable,
hence showing that the putative non-trivial vacuum indicated by the effective potential is in fact unstable.
\section{Unstable monopoles in a false vacuum}
\label{sec:model}
Consider an $SU(2)$ gauge theory with a triplet scalar field $\phi$ with the Lagrangian density given by
\begin{equation}
\mathcal{L} = -\frac{1}{4}F_{\mu\nu}^a F^{\mu\nu a} + \frac{1}{2}(D_\mu \phi^a)(D^\mu \phi^a) - V(\phi^a\phi^a)
\end{equation}
where
\begin{equation}
F_{\mu\nu}^a = \partial_\mu A_\nu^a - \partial_\nu A_\mu^a + e\epsilon^{abc}A_\mu^b A_\nu^c,
\end{equation}
and
\begin{equation}
D_\mu \phi^a = \partial_\mu \phi^a + e\epsilon^{abc} A_\mu^b \phi^c.
\end{equation}
The potential we use is a polynomial of order $6$ in $\phi$ and may conveniently be written as
\begin{equation}
V(\phi) = \lambda \phi^2 (\phi^2 - a^2)^2 + \gamma^2 \phi^2 -\epsilon
\label{potential}
\end{equation}
where $\epsilon$ is defined so that the potential vanishes at the meta-stable vacua. The vacuum energy density difference is then equal to $\epsilon$.
Such a potential was numerically analyzed by \cite{Steinhardt:1981mm} as a toy model for the dissociation of monopoles. Here we obtain explicit analytical formulae for the quantum tunneling decay of the monopoles.
The potential has a minimum at $\phi^T\phi=0$ which for $\gamma=0$ is degenerate with the manifold
of vacua at $\phi^T\phi = a^2$. When we set $\gamma \neq 0$, we get a manifold of degenerate metastable
vacua at $\phi^T\phi = \eta^2$ (where the exact value of the VEV, $\eta$, is calculable and satisfies $\eta\approx a$ for small $\gamma$), and the minimum at
$\phi = 0$ becomes the true vacuum.
A plot of the potential for small $\gamma$ as a function of one of the components of $\phi$ is
shown in figure \ref{fig1}. A supersymmetry breaking model \cite{Bajc:2008vk} containing monopoles and
a scalar potential similar to the one given in Eqn. (\ref{potential}) was studied in \cite{Kumar:2009pr}.
\begin{figure}[!htp]
\begin{center}
\includegraphics[width=0.45\textwidth]{ScalarPotential}
\caption{The potential $V(\phi)$ for $\gamma \neq 0$ as a function of one of the components of the field $\phi$,
shifted by an additive constant so that $\phi = \eta$ has vanishing $V$ and the true
vacuum has $V = -\epsilon$.}
\label{fig1}
\end{center}
\end{figure}
The manifold of vacua at $\phi^T\phi=\eta^2$ is topologically an $S^2$ and as spatial infinity is topologically also $S^2$, the appropriate homotopy group of the manifold of the vacua of the
symmetry breaking $SU(2)\rightarrow U(1)$ is $\Pi_2(SU(2)/U(1))$ which is $\mathbf{Z}$. This suggests the existence
of topologically non-trivial solutions of the monopole type which are classically stable. The presence
of the global minimum at $\phi=0$ allows for the possibility that the monopole solution although topologically non-trivial, could be dynamically unstable.
A time independent spherically symmetric ansatz for the monopole can be chosen in the usual way as
\begin{eqnarray}
\phi_a &=& \hat{r}_a \, h(r) \nonumber \\
A_\mu^a &=& \epsilon_{\mu ab}\,\hat{r}_b \,\frac{1 - K(r)}{er} \nonumber \\
A_0 &=& 0
\end{eqnarray}
where $\hat{r}$ is a unit vector in spherical polar coordinates
The energy of the
monopole configuration in terms of the functions $h$ and $K$ is
\begin{align}
E(K,h) = 4\pi \int_0^\infty dr \Big( &\frac{(K')^2}{e^2} + \frac{(1-K^2)^2}{2e^2r^2} + \frac{1}{2}r^2(h')^2 \nonumber \\
&+ K^2h^2 + r^2V(h) \Big)
\label{staticenergy}
\end{align}
where derivatives with respect to $r$ are denoted by primes. The static monopole solution is the minimum
of this functional and the ansatz functions satisfy the equations
\begin{eqnarray}
h'' + \frac{2}{r} h' - \frac{2h}{r^2}K^2 - \frac{\partial V}{\partial h} &=& 0 \label{monopole-equation} \\
K'' - \frac{K}{r^2}(K^2 - 1) - e^2h^2K &=& 0.
\label{eom}
\end{eqnarray}
As $r\rightarrow \infty$ the function $h$ asymptotically approaches $\eta$ and is
zero at $r = 0$ from continuity requirements. On the other hand, $K$ approaches zero at spatial infinity
so that the gauge field decreases as $1/r$, and $K = 1$ at $r = 0$.
\begin{figure}[!htp]
\begin{center}
\includegraphics[width=0.45\textwidth]{ThinWall}
\caption{The monopole profile under the thin wall approximation.}
\label{fig2}
\end{center}
\end{figure}
\section{Thin walled monopoles}
\label{sec:thinwallmonopole}
When the difference between the false and true vacuum energy densities $\epsilon$ is small, the monopole can be
treated as a thin shell, the so called thin wall approximation. Within this approximation, the monopole
can be divided into three regions as shown in figure \ref{fig2}.
There is a region of essentially true vacuum extending from $r = 0$ upto a radius $R$. At $r = R$, there is a thin shell of thickness
$\delta$ in which the field value changes exponentially from the true vacuum to the false vacuum. Outside this shell the monopole is essentially in the false vacuum, and so we have
\begin{eqnarray}
h \approx 0 \,,\, K \approx 1 \qquad \qquad r < R - \frac{\delta}{2} \nonumber \\
h \approx \eta \,,\, K \approx 0 \qquad \qquad r > R + \frac{\delta}{2} \nonumber \\
0 < h < \eta \,,\, 0 < K < 1 \qquad \qquad R - \frac{\delta}{2} \leq r \leq R + \frac{\delta}{2}
\end{eqnarray}
where $\delta$ is a length corresponding to the mass scale of the symmetry breaking.
As we shall see in section \ref{sec:collco}, describing the monopole in this way allows us to study the dynamics in terms of
just one collective coordinate $R$. The energy of the monopole then becomes
a simple polynomial in $R$. Furthermore, due to the spherical symmetry, $R$ is a function of time alone and so the
original field theoretic model in $3 + 1$ dimensions reduces to a one-dimensional problem involving $R(t)$.
We now proceed to elucidate the existence of monopole solutions which have the thin wall behavior described in
the previous subsection. Redefining the couplings appearing in the potential (\ref{potential}) in terms of a mass
scale $\mu$ and expressing $\phi$ in terms of the profile function $h(r)$, we have
\begin{equation}
V = \frac{\tilde{\lambda}}{\mu^2}h^2\big(h^2 - \mu^2\tilde{a}^2\big)^2 + \tilde{\gamma}^2\mu^2h^2-\epsilon
\label{rescaledpotential}
\end{equation}
where a tilde over a variable indicates that it is dimensionless. The vacuum expectation value of $\phi$ or $h$
then becomes $\tilde{\eta}\mu$, where
\begin{equation}
\tilde{\eta} = \sqrt{\frac{2\tilde{a}^2}{3} + \frac{\sqrt{\tilde{a}^4\tilde{\lambda}^2 -
3\tilde{\gamma}^2\tilde{\lambda}}}{3\tilde{\lambda}}}.
\end{equation}
The expression for $V$ can be rearranged as
\begin{equation}
V = \Big( \big(\tilde{\lambda}\tilde{a}^4 + \tilde{\gamma}^2\big)\mu^2 - 2\tilde{\lambda}\tilde{a}^2h^2 \Big)h^2
+ O(h^6).
\end{equation}
The condition that $V$ is approximately quadratic in $h$ is given by
\begin{equation}
\frac{h^2}{\mu^2} << \frac{\tilde{\lambda}\tilde{a}^4 + \tilde{\gamma}^2}{2\tilde{\lambda}\tilde{a}^2}.\label{linear}
\end{equation}
When the above condition is satisfied, $\partial V/ \partial h$ is linear in $h$. The equation of motion for
$h$ given in equation (\ref{eom}) can then be written as
\begin{equation}
h'' + \frac{2}{r}h' - \frac{2h}{r^2} - k^2h = 0
\label{msbe}
\end{equation}
where $k^2 = (\tilde{\lambda}\tilde{a}^4 + \tilde{\gamma}^2)\mu^2$ and $K$
has been set to unity.
Equation (\ref{msbe}) has the
form of the modified spherical Bessel equation whose general form is
\begin{equation}
z^2w'' + 2zw' - [z^2 + l(l+1)]w = 0
\label{msbegen}
\end{equation}
for a function $w(z)$. The primes in the above equation denote derivatives with respect to $z$ and
equation (\ref{msbe}) is obtained from (\ref{msbegen}) with $l = 1$.
The solution of equation (\ref{msbe}) is
\begin{equation}
h(r) = C \Big( \frac{I_{3/2}(kr)}{\sqrt{kr}} \Big) = Ci_1(kr)
\end{equation}
where $I_J$ is the modified Bessel function of the first kind of order $J$, $i_n$ is the modified spherical Bessel
function of the first kind of order $n$, and $C$ is
an arbitrary constant. The function $i_1(kr) \sim e^{kr}/(kr)$ for $kr\gg 1$ and is linear in $kr$ for small $kr\ll 1$.
If we choose $C = e^{-k\xi }$ with arbitrarily large $k\xi $, we see that we can keep Eqn. (\ref{linear}) satisfied and hence stay with the linear equation for $h(r)$ for arbitrarily large $kr$.
The existence of the particular solution with $h(r) = \eta$ at $r=\infty$ can be proven using an argument similar to Coleman's, where he proved in a somewhat different context, the existence of a thin wall instanton, \cite{Coleman:1977py}. We can reinterpret the equation for the monopole profile, Eqn. (\ref{monopole-equation}), as describing the motion of a particle whose position is denoted by $h(r)$ where $r$ is now interpreted as a time coordinate. The particle moves in the presence of friction with a time dependent Stokes coefficient given by the second term in Eqn. (\ref{monopole-equation}) and a time dependent force given by the third term in Eqn. (\ref{monopole-equation}) (setting $K=1$), both of which are singular at $r=0$. The particle also moves in the potential $-V(h)$, obtained by inverting the potential Eqn. (\ref{potential}), as shown in figure \ref{fig4}.
The particle must start at $h = 0$ with a finite velocity and must reach $h = \eta$ as $r\rightarrow\infty$.
We prove the existence of the solution that achieves $h = \eta$ at $r=\infty$ by
proving that initial conditions can be chosen so that the particle can undershoot or overshoot $h = \eta$ for $r\rightarrow\infty$, depending on the choice of the initial velocity. Then by continuity there must exist an appropriate initial condition for which the particle exactly achieves $h=\eta$ at $r\rightarrow\infty$.
\begin{figure}[!htp]
\begin{center}
\includegraphics[width=0.45\textwidth]{InvertedPotential}
\caption{The scalar potential $-V(h)$ which is the Euclidean space equivalent of the potential given
in (\ref{potential}). The potential has zeroes at $h = h_0$ and $h = \eta$.}
\label{fig4}
\end{center}
\end{figure}
In the following, we will assume that $K=1$ is always a good approximation. Indeed, in Eqn, (\ref{monopole-equation}) the term dependent on $K$ is negligible for large $r$ no matter the value of $K$, while for small $r$, $K=1$ is a reasonable approximation. On the other hand, Eqn. (\ref{eom}) for $K$, critically depends on the value of $h(r)\ne 0$, especially for large $kr$. In that sense, the function $h(r)$ does not depend strongly on $K(r)$ whereas, $h(r)$ drives the behaviour of $K(r)$.
\subsection{Overshoot}
The existence of the overshoot can be proven by taking a sufficiently small value of $C$. As explained earlier,
$C$ can be chosen small enough so that Eqn.(\ref{linear}) is valid even for large $kr$, hence the equation remains linear. If $kr$ is large enough, the friction
term $(2/r) h'$ and the term $(2/r^2) h$ in the equation of motion can be neglected in any further evolution and the evolution can be thought of as conservative. Thus with such a choice
of $C$, $h$ increases to $\tilde h<h_0$ at a large value of $kr$ according to the linearised equation ($h_0$ is the zero crossing point of the potential, see Fig. (\ref{fig4})). The motion from then onwards is frictionless. The particle has an energy $E > 0$ at $h = \tilde h$, thus it's energy is still positive when it reaches
$h = \eta$. As a result, it overshoots to $h > \eta$.
\subsubsection{Technical details}
\subsection{Undershoot}
To prove the existence of the undershoot, we start with the full equation for $h(r)$:
\begin{equation}
h'' + \frac{2}{r}h' - \frac{2}{r^2}h - \frac{\partial V}{\partial h} = 0
\end{equation}
which after multiplying both sides by $h'$ can be rewritten as
\begin{eqnarray}
\frac{d}{dr} \Big( \frac{1}{2}(h')^2 -V(h) \Big) &=& \label{energy}
-2h' \Big( \frac{h'}{r} - \frac{h}{r^2} \Big), \\
&=& -2h'\big(\frac{h}{r}\big)'.\label{energy22}
\end{eqnarray}
The quantity on the left hand side of Eqn. (\ref{energy}) can be thought of as the time derivative of the energy $E$. In the linearised regime, it is easy to show that the right hand side is strictly negative for all $r$. It starts with a value of zero at $r=0$ and decreases essentially exponentially for large $kr$. We can chose $C$, which amounts to choosing the initial velocity so that $h$ evolves according to the linearised equation until $kr$ can be taken to be large. However, in contrast to the case of the undershoot, we now require that $E$ becomes negative. This means that the value of $C$ is taken larger than in the case of the overshoot. $E$ is made up of two terms, the kinetic term which is positive semi-definite, and the potential term which becomes negative for $h>h_0$. We impose conditions on the parameters so that $E$ becomes negative and consequently $h>h_0$ within the linearised regime. Now if $kr$ is large enough, as before, the subsequent evolution will be conservative and since the total energy is negative, the subsequent evolution will never be able to overcome the hill at $h=\eta$ and the particle will undershoot.
\subsection{Technical details}
To make the previous arguments more precise and rigorous, we note that when the condition Eqn. (\ref{linear}) is satisfied, the linear regime is valid and $V(h)$ is approximately quadratic in $h$, {\it ie.} $-V(h)\approx\epsilon - (1/2)k^2h^2$ and the equation of motion for $h$ is approximately
\begin{equation}
\frac{d}{dr} \Big( \frac{1}{2}(h')^2 + \epsilon - \frac{1}{2}k^2h^2 \Big)= -2h' \Big( \frac{h'}{r} - \frac{h}{r^2} \Big).
\end{equation}
Using the properties of $i_1(kr)$ we can compute $E$ in the linear regime, we find for large $kr$
\begin{equation}
E\approx \epsilon - \frac{k^2C^2e^{2kr}}{4(kr)^3}\label{energy3}
\end{equation}
which can be evidently taken to be positive or negative by simply choosing the value of $C$. Then in the subsequent evolution, where we can no longer rely on the linear evolution, the right hand side has two competing terms, the friction term, which only reduces the energy and the time dependent force term which tries to increase it. The change in the energy for evolution between $r_0$ and $r_f$ is given by the integral of the right hand side.
\subsubsection{Overshoot}
For the case of the overshoot, we use the expression Eqn. (\ref{energy22}) which gives
\begin{equation}
\Delta E=-2\int_{r_0}^{r_f} \, dr\,h'\big(\frac{h}{r}\big)'.
\end{equation}
Assuming that $h'(r)$ is positive, we will find an estimate for $h'(r)<v$. Then
\begin{eqnarray}
|\Delta E|&<&2v\left|\int_{r_0}^{r_f} \, dr\,\big(\frac{h}{r}\big)'\right|\\
&=&2v\left|\left(\frac{h(r_f)}{r_f}-\frac{h(r_0)}{r_0}\right)\right|\\
&<&2v\left|\left(\frac{\eta}{r_f}-\frac{h(r_0)}{r_0}\right)\right|
\end{eqnarray}
where we replaced $h(r_f)$ with $\eta$ since that is its largest possible value. As long as $v$ is well behaved, as $r_0\to\infty$, $r_f>r_0$ thus the first term vanishes, while the second term can be made small by choosing the value of $C$ to be arbitrarily small. Thus we see that $\Delta E\to 0$ and therefore the change in the energy is arbitrarily small. Thus we necessarily obtain an overshoot since at $r=r_\eta$ such that $h(r_\eta)=\eta$, $V(\eta)=0$, hence the particle has a positive kinetic energy giving an overshoot.
To get the value of $v$, we use Eqn. (\ref{energy})
\begin{eqnarray}
\frac{d}{dr} \Big( \frac{1}{2}(h')^2 -V(h) \Big) &=& -2h' \Big( \frac{h'}{r} - \frac{h}{r^2} \Big), \\
&<&2\frac{hh'}{r^2}<\frac{(h^2)'}{r_0^2}.
\end{eqnarray}
Integrating both sides from $r_0$ to $r_f$ yields
\begin{eqnarray}
(h'(r_f))^2&<&2\left( \frac{1}{r_0^2}\left( h^2(r_f)-h^2(r_0)\right)\right.\\
&+&\left. V(h(r_f))-V(h(r_0))+\frac{1}{2}(h'(r_0))^2\right).
\end{eqnarray}
Thus $v^2$ is given by
\begin{eqnarray}
v^2&=&2\left( \frac{1}{r_0^2}\left( \eta^2-h^2(r_0)\right)\right.\\
&+&\left. \sup\left|V(h(r_f))-V(h(r_0))\right|+\frac{1}{2}(h'(r_0))^2\right)
\end{eqnarray}
which is a bounded function of $r_0$.
\subsubsection{Undershoot}
To prove the undershoot we use the expression Eqn. (\ref{energy}) which gives
\begin{equation}
\Delta E=-2\int_{r_0}^{r_f} \,dr\,\frac{{h'}^2}{r}+2\int_{r_0}^{r_f} \, dr\,\frac{{h'}h}{r^2}.
\end{equation}
Integrating the second term by parts we obtain
\begin{eqnarray}
2\int_{r_0}^{r_f} \, dr\,\frac{{h'}h}{r^2}&=&\int_{r_0}^{r_f} \, dr\,\left(\frac{h^2}{r^2}\right)'+\int_{r_0}^{r_f} \, dr\,\left(\frac{2h^2}{r^3}\right)\\
&<& \left.\left(\frac{h^2}{r^2}\right)\right|_{r_0}^{r_f} -\eta^2\left.\left(\frac{1}{r^2}\right)\right|_{r_0}^{r_f}
\end{eqnarray}
where we obtain the inequality using the fact that we are only interested in the region $h\le \eta$.
We now prove that this contribution to the energy cannot be sufficient push $h$ to $h>\eta$. We take $r_0$ to be the value of $r$ as described after Eqn. (\ref{energy22}), where the energy becomes negative within the linearised regime with $kr_0\gg 1$. We now assume there exists a value $r_f\equiv r_\eta$ for which $h(r_\eta)=\eta$. Then
\begin{eqnarray}
\Delta E &<&-2\int_{r_0}^{r_\eta} \,dr\,\frac{{h'}^2}{r} +\left.\left(\frac{h^2}{r^2}\right)\right|_{r_0}^{r_\eta} -\eta^2\left.\left(\frac{1}{r^2}\right)\right|_{r_0}^{r_\eta}\\
&<&\frac{\eta^2}{r_\eta^2}-\frac{h^2(r_0)}{r_0^2}-\eta^2\left(\frac{1}{r_\eta^2}-\frac{1}{r_0^2}\right)\\
&=&\frac{\eta^2-h^2(r_0)}{r_0^2}
\end{eqnarray}
which is an upper bound to the energy that can be added to the particle. But now it is easy to see that this is insufficient for $kr_0$ large enough. Indeed the energy of the particle at $r=r_0$ is obtained, via the linear regime, by Eqn. (\ref{energy3})
\begin{equation}
E\approx \epsilon - \frac{k^2C^2e^{2kr}}{4(kr)^3}\rightarrow\epsilon -
k\frac{h^2(r_0)}{r_0} .
\end{equation}
This expression is negative. Furthermore, if $kr_0$ is large enough, we will see that $\Delta E$ cannot provide enough energy to increase $E$ to zero, giving a contradiction to the existence of $r_\eta$. To see this, we would require $|E|>\Delta E$ {\it ie.}
\begin{equation}
k\frac{h^2(r_0)}{r_0}-\epsilon >\frac{\eta^2-h^2(r_0)}{r_0^2}.
\end{equation}
The linear approximation assumes $h(r_0)\ll\eta$, hence we get
\begin{equation}
\frac{kh^2(r_0)}{r_0}-\frac{\eta^2}{r_0^2}>\epsilon
\end{equation}
reorganizing the terms, which for small enough $\epsilon$ simply implies
\begin{equation}
h^2(r_0)kr_0> \eta^2.
\end{equation}
Thus we get the the inequality sandwich
\begin{equation}
\frac{\eta^2}{kr_0}<h^2(r_0)<\eta^2.
\end{equation}
Using $h(r_0)\approx Ce^{kr_0}/2kr_0$ we can choose
\begin{equation}
C= \frac{\eta 2kr_0}{e^{kr_0}r_0^{1/4}}
\end{equation} which gives
\begin{equation}
\frac{\eta^2}{kr_0}<\frac{\eta^2}{\sqrt{kr_0}}<\eta^2.
\end{equation}
It is obvious that for large enough $kr_0$ this is easily satisfied. Thus we have established the existence of a choice of $C$ or initial velocity which contradicts the existence of $r_\eta$.
\section{Collective coordinate and the instantons}
\label{sec:collco}
The potential $V(\phi)$ given in (\ref{potential}) can be normalized so that the energy density of the metastable
vacuum is vanishing whereas the energy density of the true vacuum is $-\epsilon$. By making use of the thin-wall
approximation, the expression for the total energy in the static case given in (\ref{staticenergy}) can be expressed as
\begin{eqnarray}
E = 4\pi &\Bigg[& \int_0^{R - \frac{\delta}{2}} dr \,r^2 V(h)
+ \int_{R + \frac{\delta}{2}}^{\infty} dr \, \frac{1}{2e^2r^2} \nonumber \\
&+& \int_{R - \frac{\delta}{2}}^{R + \frac{\delta}{2}} dr \, \Big( \frac{(K')^2}{e^2} + \frac{(1-K^2)^2}{2e^2r^2} \nonumber \\
&+& \frac{1}{2}r^2(h')^2+ K^2h^2 + r^2V(h) \Big) \Bigg].
\label{energy2}
\end{eqnarray}
In the above expression, we have made use of the fact that $V(h)$ is zero for $r > R + \frac{\delta}{2}$, $K = 1$
for $r < R - \frac{\delta}{2}$, $K = 0$ for $r > R + \frac{\delta}{2}$, and that both the derivative terms and the
term $K^2h^2$ are non-zero only when $R - \frac{\delta}{2} < r < R + \frac{\delta}{2}$. Since $\delta$ is small, the
first integral on the right hand side of (\ref{energy2})
gives $-\alpha R^3$ where $\alpha = 4\pi\epsilon/3$ because $V(h) = -\epsilon$ in the domain of integration. The second
integral gives $C/R$ where $C = 2\pi/e^2$. The third integral is due to the energy of the wall and can be written as
$4\pi\sigma R^2$ where $\sigma$ is the surface energy density of the wall given by
\begin{eqnarray}
\sigma &=& \frac{1}{R^2} \int_{R - \frac{\delta}{2}}^{R + \frac{\delta}{2}} dr \, \Big( \frac{(K')^2}{e^2} +
\frac{(1-K^2)^2}{2e^2r^2} \nonumber \\
&+& \frac{1}{2}r^2(h')^2+ K^2h^2 + r^2V(h) \Big).
\label{wallenergy}
\end{eqnarray}
We can thus write the total energy of the monopole as
\begin{equation}
E(R) = -\alpha R^3 + 4\pi\sigma R^2 + \frac{C}{R}.
\label{eofr}
\end{equation}
This function is plotted in figure \ref{fig3}. There is a minimum at $R = R_1$ and this corresponds to the classically stable
monopole solution. This solution has a bubble of true vacuum in its core and the radius $R_1$ of this bubble is obtained
by solving $dE/dR = 0$. However, this monopole configuration can tunnel quantum mechanically through the finite barrier
into a configuration with $R = R_2$ where $E(R_1) = E(R_2)$. Once this occurs, the monopole can continue to lose energy
through an expansion of the core since the barrier which was present at $R_1$ is no longer able to prevent this.
\begin{figure}[!htp]
\begin{center}
\includegraphics[width=0.45\textwidth]{BubbleEnergy}
\caption{The function $E(R)$ plotted versus bubble radius. The classically stable monopole solution has
$R = R_1$. This solution can tunnel quantum mechanically to a configuration with $R = R_2$ and then
expand classically.}
\label{fig3}
\end{center}
\end{figure}
We now proceed to determine the action of the instanton describing the tunneling from $R = R_1$ to $R = R_2$.
In the thin wall approximation, the functions $h$ and $K$ can be written as
\begin{eqnarray}
h &=& h(r - R) \nonumber \\
K &=& K(r - R)
\label{thinwall}
\end{eqnarray}
and the exact forms of the functions $h$ and $K$ will not be required in the ensuing analysis. The only requirement is that
both $h$ and $K$ change exponentially when their argument $(r - R)$ is small. An example of a function with this type of
behaviour is the hyperbolic tangent function. The time derivative of $\phi$ can be written as
\begin{equation}
\dot{\phi}^a = \hat{r}_a \frac{dh}{dR} \dot{R}.
\end{equation}
From (\ref{thinwall}), since $(dh/dR)^2 = (dh/dr)^2$, we have
\begin{equation}
\frac{1}{2} \dot{\phi}^a \dot{\phi}^a = \frac{1}{2} \left(\frac{dh}{dR}\right)^2 \dot{R}^2
= \frac{1}{2} \left(\frac{dh}{dr}\right)^2 \dot{R}^2.
\end{equation}
Similarly,
\begin{equation}
\dot{A}_{\mu}^a = \epsilon_{\mu ab} \hat{r}_b \left(\frac{-1}{er}\right) \frac{dK}{dR} \dot{R}
\end{equation}
and
\begin{equation}
\frac{1}{4} \dot{A}_{\mu}^a \dot{A}_{\mu}^a = \frac{1}{2e^2r^2} \left(\frac{dK}{dr}\right)^2 \dot{R}^2.
\end{equation}
The Lagrangian can then be expressed as
\begin{equation}
L = 2\pi \int_0^{\infty} \Big( r^2 \left(\frac{dh}{dr}\right)^2\dot{R}^2 +
\frac{1}{e^2}\left(\frac{dK}{dr}\right)^2\dot{R}^2 \Big) dr - E(R).
\label{totalLag}
\end{equation}
From (\ref{eom}), for large r, the equation of motion of $h$ can be written as
\begin{equation}
h'' - \frac{\partial V(h)}{\partial h} = 0.
\end{equation}
Multiplying both sides by $h'$ and integrating by parts with respect to $r$, one obtains
\begin{equation}
h' = \sqrt{2V(h)}.
\label{hprime}
\end{equation}
Furthermore, since $dh/dr$ is non-vanishing only in the thin-wall, the value of $r$ in the first integral in
(\ref{totalLag}) can be replaced by $R$ and we have
\begin{eqnarray}
\int_0^{\infty} dr \, r^2 \left(\frac{dh}{dr}\right)^2\dot{R}^2 &=& R^2 \dot{R}^2 \int_0^{\infty} dr
\left(\frac{dh}{dr}\right)\sqrt{2V(h)} \nonumber \\
&=& R^2\dot{R}^2 S_1
\end{eqnarray}
where
\begin{equation}
S_1 = \int_0^{\eta} dh\sqrt{2V(h)}.
\label{s1}
\end{equation}
Defining
\begin{equation}
S_2 = \frac{1}{e^2} \int_0^{\infty} dr \left(\frac{dK}{dr}\right)^2,
\end{equation}
the Lagrangian (\ref{totalLag}) becomes
\begin{equation}
L = 2\pi\dot{R}^2( S_1 R^2 + S_2) - E(R)
\end{equation}
and the action can be written as
\begin{equation}
S = \int_{-\infty}^{\infty} dt \Big( 2\pi\dot{R}^2( S_1 R^2 + S_2) - E(R) \Big).
\end{equation}
In Euclidean space, the expression for the action becomes
\begin{equation}
S_E = \int_{-\infty}^{\infty} d\tau \Big( 2\pi\dot{R}^2( S_1 R^2 + S_2) + E(R) \Big)
\label{euclideanaction}
\end{equation}
where $\tau = it$ is the Euclidean time and $\dot{R}$ is the derivative with respect to $\tau$.
The instanton solution $R(\tau)$ which we are seeking
obeys the boundary conditions $R = R_1$ for $\tau = \pm \infty$, $R = R_2$ for $\tau = 0$, and
$dR/d\tau = 0$ for $\tau = 0$. It can be obtained by solving the equations of motion derived
from (\ref{euclideanaction}). However, the exact form for $R(\tau)$ will not be of interest here
since the decay rate of the monopole is determined ultimately from $S_E$ \cite{Coleman:1977py}.
The calculation of $S_E$ will be the subject of the next section.
\section{Bounce action}
\label{sec:decay}
In this section, we will derive an expression for bounce action $S_E$ for the monopole tunneling
and compare it with the bounce action for the tunneling of the false vacuum to the true vacuum
as discussed in \cite{Coleman:1977py} with no monopoles present.
From (\ref{euclideanaction}), the equation of motion for $R$ can be written
\begin{equation}
(R^2S_1 + S_2)\ddot{R} + S_1R\dot{R}^2 - \frac{1}{4\pi}\frac{\partial E}{\partial R} = 0.
\end{equation}
Multiplying both sides by $\dot{R}$, the equation of motion assumes the form
\begin{equation}
\frac{d}{dt} \left[ \frac{1}{2}(S_2 + R^2S_1)\dot{R}^2 - \frac{E(R)}{4\pi} \right] = 0.
\end{equation}
The term in the square brackets is a constant of motion and can be taken to be zero with loss of
generality. Setting this constant to zero gives
\begin{equation}
E(R) = 2\pi (S_2 + S_1R^2)\dot{R}^2.
\label{com}
\end{equation}
Substituting this in (\ref{euclideanaction}), we have
\begin{equation}
S_E = \int_{-\infty}^{\infty} d\tau \, 4\pi(S_2 + S_1R^2)\dot{R}^2.
\end{equation}
Solving for $\dot{R}$ from (\ref{com}) and using this in the above equation yields
\begin{eqnarray}
S_E &=& \int_{-\infty}^{\infty} d\tau (\frac{dR}{d\tau}) 4\pi (S_2 + S_1R^2) \dot{R} \nonumber \\
&=& \sqrt{32\pi} \int_{R_1}^{R_2} dR \sqrt{(S_2 + S_1R^2)E(R)}.
\end{eqnarray}
Using the expression for $E(R)$ given in (\ref{eofr}) and neglecting $S_2$ in comparison to
$S_1R^2$, the euclidean action of the bounce solution can be written
\begin{equation}
S_E = A \int_{R_1}^{R_2} dR \sqrt{-(\alpha R^5 - 4\pi\sigma R^4 - CR + E_0R^2) }
\end{equation}
where $A = \sqrt{32\pi S_1}$.
In deriving the above expression, the constant $E_0 = E(R_1)$ was subtracted from the expression for
$E(R)$ in (\ref{eofr}) so that the bounce has a finite action. Pulling out a factor of $R$ from the
square root in the integrand, we have
\begin{equation}
S_E = A \int_{R_1}^{R_2} dR \sqrt{R} \sqrt{-J}
\end{equation}
where $J = \alpha R^4 - 4\pi\sigma R^3 - C + E_0R$. The function $J$ has a double root at $R = R_1$,
a positive root at $R = R_2$, and a negative root at $R = R_3$. Since we are working with $\epsilon$
small and $\alpha = 4\pi\epsilon/3$, we can neglect the term containing $\alpha$ while solving $dE/dR = 0$
and obtain $R_1 \approx (4e^2\sigma)^{-1/3}$. To find $R_3$ we also neglect the term containing $\alpha$, and substituting for $E_0$ in terms of the solution for $R_1$, we get a cubic equation for $R_3$, which can be exactly factored, giving $R_3=-2R_1$. Finally, to solve for $R_2$, we solve $J = 0$ neglecting the
constant and linear term in $R$ since $R_2$ is large, obtaining $R_2 \approx 4\pi\sigma/\alpha=3\sigma/\epsilon$.
Factoring $J$, we have
\begin{eqnarray}
S_E &=& A \sqrt{\alpha} \int_{R_1}^{R_2} dR \sqrt{R} \sqrt{-(R-R_1)^2(R-R_2)(R-R_3)} \nonumber \\
&=& A \sqrt{\alpha} \int_{R_1}^{R_2} dR \sqrt{R}(R-R_1) \sqrt{-(R-R_2)(R-R_3)} \nonumber \\
&=& A \sqrt{\alpha} R_2^{7/2} \frac{2}{105} \Big(1 - \frac{R_1}{R_2}\Big)^{5/2} \, I\Big(\frac{R_1}{R_2},\frac{R_3}{R_2}\Big)\label{59}\\
&=&\sqrt{32\pi S_1}\sqrt{4\pi\epsilon/3}R_2^{7/2} \frac{2}{105} \Big(1 - \frac{R_1}{R_2}\Big)^{5/2} \, I\Big(\frac{R_1}{R_2},\frac{R_3}{R_2}\Big).\nonumber
\end{eqnarray}
Here $I$ is a dimensionless function of $R_1/R_2$ and $R_3/R_2$ which is finite everywhere in the domain
$[R_1,R_2]$ and is obtained from the integral defined in Eqn, (\ref{59}) removing the factor of $(1-(R_1/R_2))^{(5/2)}$ and $R_2^{7/2}$ and some numerical factors. It is expressible in terms of elliptic integrals and its explicit expression is not illuminating. As $S_1$ has dimensions of $\mu^3$ and $\epsilon$ has dimensions of $\mu^4$, the expression is dimensionless, as expected. Substituting the value of $R_2$ in $S_E$,
\begin{equation}
S_E = \frac{144\,\pi}{35} \sqrt{2\,S_1} \frac{\sigma^{\frac{7}{2}}}{\epsilon^3} \Big(1 - \frac{R_1}{R_2}\Big)^{5/2}
\, I\Big(\frac{R_1}{R_2},\frac{R_3}{R_2}\Big).
\label{bounce1}
\end{equation}
For small $\epsilon$, the term containing $\tilde{\gamma}$ in the potential (\ref{rescaledpotential}) can be
neglected. Using equation (\ref{s1}) and the fact that $\eta = \tilde{a}\mu$ when $\tilde{\gamma} = 0$,
\begin{eqnarray}
S_1 &=& \frac{\sqrt{2\tilde{\lambda}}}{\mu} \int_0^{\tilde{a}\mu} dh \,\big( h(h^2 - \mu^2\tilde{a}^2) \big) \\
&=& \sqrt{\frac{\tilde{\lambda}}{8}} \tilde{a}^4 \mu^3.
\label{s1value}
\end{eqnarray}
The value of $\sigma$ can be obtained from equation (\ref{wallenergy}) by noting that the terms multiplying $r^2$
are large compared to the terms independent of $r$ and the term multiplying $1/r^2$. Since $\delta$ is small, we
can write $r = R$ and equation (\ref{wallenergy}) becomes
\begin{equation}
\sigma = \int_{R - \frac{\delta}{2}}^{R + \frac{\delta}{2}} dr \, \Big(\frac{1}{2}(h')^2 + V(h) \Big).
\end{equation}
Substituting for $h'$ from equation (\ref{hprime}), $\sigma$ becomes
\begin{eqnarray}
\sigma &=& \int_{R - \frac{\delta}{2}}^{R + \frac{\delta}{2}} dr \, (h')^2 \\
&=& \int_{0}^{\eta} dh \, (h') \\
&=& \int_{0}^{\eta} dh \, \sqrt{2V(h)} \\
&=& S_1.
\label{sigma}
\end{eqnarray}
Using (\ref{sigma}) and (\ref{s1value}) in (\ref{bounce1}) yields
\begin{eqnarray}
S_E &=& \frac{144\,\pi\sqrt{2}}{35} \frac{S_1^4}{\epsilon^3} \Big(1 - \frac{R_1}{R_2}\Big)^{5/2} \, I\Big(\frac{R_1}{R_2},\frac{R_3}{R_2}\Big) \\
&=& \frac{9\sqrt{2}\,\pi}{140} \tilde{\lambda}^2\tilde{a}^{16} \frac{\mu^{12}}{\epsilon^3} \Big(1 - \frac{R_1}{R_2}\Big)^{5/2}
\, I\Big(\frac{R_1}{R_2},\frac{R_3}{R_2}\Big)
\label{finalbounce}
\end{eqnarray}
as the final value of the bounce action. From the values of $R_1$ and $R_2$, we have
\begin{eqnarray}
\frac{R_1}{R_2} &=& \frac{1}{e^{2/3}} \frac{1}{(4\sigma)^{1/3}} \frac{\epsilon}{3\sigma} \\
&=& \frac{1}{(\tilde{\lambda}e)^{2/3}} \big( \frac{16}{27} \big)^{1/3} \frac{\epsilon}{\tilde{a}^{16/3} \mu^4}
\label{r1overr2}
\end{eqnarray}
where the value of $\sigma$ has been expressed in terms of the couplings appearing in the potential using
equations (\ref{sigma}) and (\ref{s1value}). From the expression given in (\ref{finalbounce}), it is evident
that the bounce action $S_E$ is zero when $R_1 = R_2$ as expected. With $\epsilon$ small, $R_1/R_2$ is small,
but it is interesting to note that variations in the couplings can reduce the bounce action. For example, a
reduction in the $U(1)$ gauge coupling $e$ has the effect of increasing the monopole mass and of reducing the bounce action.
We now compare our answer with the well known formula of \cite{Coleman:1977py} relevant to
homogeneous nucleation, i.e. tunneling
of the translation invariant false vacuum to the true vacuum. Denoting this bounce to be $B_0$,
\begin{eqnarray}
B_0 &=& \frac{27\pi^2}{2} \frac{S_1^4}{\epsilon^3} \\
&=& \frac{27\pi^2}{128} \tilde{\lambda}^2 \tilde{a}^{16} \frac{\mu^{12}}{\epsilon^3}.
\end{eqnarray}
Comparing this expression with our bounce $B \equiv S_E$ for the monopole assisted tunneling
given in (\ref{finalbounce}), we see that
\begin{equation}
B = \frac{32\sqrt{2}}{105 \, \pi} B_0 \, \Big(1 - \frac{R_1}{R_2}\Big)^{5/2} \, I\Big(\frac{R_1}{R_2},\frac{R_3}{R_2}\Big).
\end{equation}
We see that unlike the homogeneous case, the bounce can parametrically become indefinitely
small and vanish in the limit $ R_1 \rightarrow R_2 $. The interpretation of this limit
is that the very presence of a monopole in this parameter regime implies the unviability
of a state asymptotically approaching the vacuum deduced by a naive use of the effective potantial.
If the parameters in the effective potential explicitly depend on external variables
such as temperature, it may happen that the limit $ R_1 \rightarrow R_2 $ is reached at a
critical value of this external parameter. In this case, as the external parameter gets tuned to
this critical value, the monopoles will become sites where the true vacuum is nucleated without
any delay and the indefinite growth
of such bubbles will eventually convert the entire system to the true vacuum without
the need for quantum tunneling. Such a phenomenon may be referred to as a \textit{roll-over
transition} \cite{Yajnik:1986wq} characterised by the relevant critical value.
\section{Monopole decay in a supersymmetric SU(5) GUT model}
\label{sec:gutmodel}
The results of this work have direct relevance to a supersymmetric $SU(5)$ model studied in \cite{Bajc:2008vk}
in which supersymmetry symmetry breaking is sought directly through O'Raifeartaigh type breaking. The Higgs sector, which contains two adjoint scalar superfields $\Sigma_1$ and $\Sigma_2$ and the
superpotential, including leading non-renormalizable terms, is of the form
\begin{eqnarray}
W &=& Tr\left[\Sigma_{2}\left(\mu \Sigma_{1} + \lambda \Sigma_{1}^{2} +
\frac{\alpha_{1}}{M}\Sigma_{1}^{3} +
\frac{\alpha_{2}}{M}Tr(\Sigma_{1}^2)\Sigma_{1} \right) \right] \nonumber \\
&=& \sigma_{1}\sigma_{2} \, \left(\mu - \frac{\lambda}{\sqrt{30}}\sigma_{1}
+ (7\alpha_{1} + 30\alpha_{2})\frac{\sigma_{1}^{2}}{30M}\right)
\label{superpot}
\end{eqnarray}
where $\sigma_1$ and $\sigma_2$ are selected components of $\Sigma_1$ and $\Sigma_2$
respectively, relevant to the symmetry breaking. Two mass scales appear in the superpotential,
$\mu$ and $M$, the latter being a larger mass scale whose inverse powers determine
the magnitudes of the coefficients of the non-renormalizable terms. The scalar potential
derived from this superpotential can be written as
\begin{eqnarray}
V &=& \Big( \mu \sigma_{1} - \frac{\lambda \sigma_{1}^2}{\sqrt{30}} +
\frac{7\alpha_{1} \sigma_{1}^3}{30\,M} + \frac{\alpha_{2}\sigma_{1}^3}{M}
\Big)^2 \nonumber \\
&+& \Big( \sigma_2\Big(\mu - \frac{2\lambda\sigma_{1}}{\sqrt{30}}
+ \frac{(7\alpha_{1} + 30\alpha_{2})}{10\,M}\sigma_{1}^2 \Big)\Big)^2.
\label{scalpot}
\end{eqnarray}
In \cite{Kumar:2009pr}, monopole solutions were shown
to exist in this model and the classical instability of the vacuum structure of this theory in the
presence of such monopoles was discussed.
Thin walled monopoles can be obtained in this model under the condition
\begin{equation}
\frac{\sigma_1}{\mu}\ll \frac{\sqrt{30}}{2\lambda}
\end{equation}
which is equivalent to the condition in Eqn. (\ref{linear}), and hence the results of this paper could be applied directly there. In \cite{Kumar:2009pr} the region of parameter space studied did not coincide with this condition, and thus the monopoles were not thin walled. The monopoles were classically unstable when $\epsilon\sim M^4$ was increased beyond a critical value. We can recover this behaviour from Eqn. (\ref{finalbounce}) as $\epsilon$ is increased, however it is important to note that our approximation in this paper becomes invalid for large enough $\epsilon$.
\section{ Discussions and conclusions }
\label{sec:conc}
We have calculated the decay rate for so-called false monopoles in a simple model with a hierarchical structure of symmetry breaking. The toy model that we use has a breaking of $SU(2)$ to $U(1)$ which is the false vacuum, which in principle happens at a higher energy scale, and then a true vacuum which has no symmetry breaking. The symmetry broken false vacuum admits magnetic monopoles. The false vacuum can decay via the usual creation of true vacuum bubbles \cite{Coleman:1977py}, however we find that this decay can be dramatically enhanced in the presence of magnetic monopoles. Even though the false vacuum is classically stable, the magnetic monopoles can be unstable. At the point of instability, the monopoles are said to dissociate. This corresponds to an evolution where the core of the monopole, which contains the true vacuum, dilates indefinitely, \cite{Steinhardt:1981ec,Hosotani:1982ii,Steinhardt:1981mm}. However, before the monopoles become classically unstable, they can be rendered unstable from quantum tunneling. We have computed the corresponding rate and find that as we approach the regime of classical instability, the exponential suppression vanishes. The tunneling amplitude behaves as
\begin{equation}
\frac{\Gamma}{V}\sim \left(\frac{\kappa}{2}\right) \exp\left\{\frac{16}{105}\sqrt{\frac{2 S_1\pi^2\epsilon}{3}}
\mathcal{F}(R_1, R_2, R_3)\right\}
\end{equation}
with
\begin{equation}
\mathcal{F}(R_1, R_2, R_3)= R_2^{7/2} \Big(1 - \frac{R_1}{R_2}\Big)^{5/2} \, I\Big(\frac{R_1}{R_2},\frac{R_3}{R_2}\Big)
\end{equation}
where $\kappa$ contains the determinantal and zero mode factors, and $I$ is defined in Eqn. (\ref{59}). In the limit that $R_1\to R_2$ the tunneling rate is unsuppressed while the homogeneous tunneling rate for the nucleation of true vacuum bubbles as found by Coleman \cite{Coleman:1977py} still remains suppressed. Hence in this limit, the classical false vacuum is classically stable, but subject to quantum instability through the nucleation of true vacuum bubbles, but the rate for such a decay can be quite small. However the existence of magnetic monopole defects render the false vacuum unstable, and in the limit of large monopole mass, the decay rate is unsuppressed.
\maketitle
\section{ACKNOWLEDGEMENTS}
We thank NSERC, Canada for financial support. The visit of BK was made possible by a grant from
CBIE, Canada. The research of UAY is partly supported by a grant from DST, India. The authors
would like to thank R. MacKenzie and P. Ramadevi for useful comments regarding this work.
\bibliographystyle{apsrev}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 6,727
|
\section{Introduction}
Recently there have been much interests in the worldvolume theory of multi M2 branes which is dual to M-theory on $AdS_4\times S^7$ spacetime, initiated by the pioneering works of Bagger and Lambert~\cite{Bagger:2007vi}\cite{Bagger:2007jr} and Gustavsson~\cite{Gustavsson:2007vu} (BLG) based on the, so called, 3-algebra. Especially, the three-dimensional ${\cal N}=6$ superconformal $U(N)\times U(N)$ Chern-Simons gauge theory with level $(k, -k)$ constructed by Aharony, Bergman, Jafferis and Maldacena (ABJM)~\cite{Aharony:2008ug} is conjectured to be dual to M-theory on $AdS_4\times S^7/ {\bf Z}_k $. Some evidences supporting the conjecture include the fact that the classical moduli space is given by $C^4/{\bf Z}_k$ and the model has ${\cal N}=6$ superconformal symmetry.
Since the model is described by a single parameter, the quantized level $k$, one may expect the conformal symmetry persists at the quantum level. Further works have been made on the superconformal index~\cite{Bhattacharya:2008bja}, integrability structures~\cite{Minahan:2008hf}, Wilson loops~\cite{Drukker:2008zx}, non-perturbative monopole instanton~\cite{Hosomichi:2008ip} and relation to the BLG model~\cite{Honma:2008ef}.
One nontrivial test for the ABJM model as the dual field theory of M theory on $AdS_4\times S^7/{\bf Z}_k$ is the study of the membrane scattering amplitude. In the dual gravity description, it would be given by the effective action of the probe M2 brane in the $AdS_4\times S^7/{\bf Z}_k$ background due to the large number of source M2 branes. See \cite{Verlinde:2008di} for the membrane scattering in the context of the BLG model.
For a slowly moving probe membrane, with constant velocity $v^I=\frac{d X^I}{d X^0}$, the effective action in the static gauge can be expressed as the derivative expansions in transverse coordinates:
\[
\Gamma=l_p^{-3}\int d^3\xi \sum_{n=0}^\infty \Big(a_n v^{2n+2} +\cdots\Big)\,,
\]
where $\cdots$ denotes the superpartners of $v^{2n}$ terms. The coefficients $a_n$ depend on the distance $r$ between the source and probe branes. Because of dimensional reasons, their leading terms are expected to be of the form
\[
a_n \sim \Big(\frac{l_p}{r}\Big)^{6n}\,.
\]
In the dual field theory description of M theory on $AdS_4\times S^7$, the computation was done in~\cite{Aharony:1996bh}-\cite{Hyun:1999hf} using the worldvolume theory of multi D2 branes, which is the three-dimensional ${\cal N}=8$ super Yang-Mills theory, and taking the strong coupling limit or decompactification limit of M-circle. It was found that the $v^2$ term and its superpartners are tree-level exact and the $v^4$ term is given by the sum of one-loop correction and the infinite sum of monopole instanton corrections. They match exactly with the results from the supergravity~\cite{Hyun:1998qf}.
The exactness of these results is due to the ${\cal N}=8$ supersymmetry~\cite{Dine:1997nq}\cite{Paban:1998qy}.
One may expect similar behavior in the ABJM model. The model has four complex bifundamental scalar fields $Y^A$ which has mass dimension $\frac{1}{2}$. The vacuum expectation values $b^A\equiv \langle Y^A\rangle$ span the vacuum moduli space, which corresponds to the transverse space of the membranes. Since the only dimensionful parameter at a generic point of moduli space is the vev, $b$, the superconformal invariance would demand the effective action to have the form
\[
\Gamma=\int d^3\xi \sum_{n=0}^\infty\Big[ c_n v^2\Big(\frac{v^2}{b^6}\Big)^{n} +\cdots\Big]\,,
\]
where $c_n$ is a dimensionless constant~\cite{Berenstein:2008dc}. It has been known that the $v^2$ term is one-loop exact with ${\cal N}=4$ supersymmetry~\cite{Seiberg:1996bs}, and tree-level exact with ${\cal N}=8$ supersymmetry~\cite{Paban:1998ea}. This means that we can determine the $v^2$ term exactly by studying one-loop corrections and find out its (non)renormalization.
It is also known that the $v^4$ term is one-loop exact, apart from the possible nonperturbative instanton corrections in the ${\cal N}=8$ supersymmetric field theories~\cite{Dine:1997nq}. Therefore, by studying one-loop corrections, we can determine the $v^4$ term exactly, at least for level $k=1, 2$ where the supersymmetry is believed to be enhanced to ${\cal N}=8$.
In this paper, we compute the one loop corrections to the effective action. We find there is no one-loop correction in the $v^2$ term and thus it is tree-level exact. We also obtain the $v^4$ term in an expected form at one-loop. This should be an exact result if we interpolate level $k$ to 1 or 2.
It would be very nice to see if it is true for general $k$ by using supersymmetry arguments. In any case, these results exactly agree with those from supergravity computations, which supports the correspondence between the ABJM model and M theory on $AdS_4\times S^7/{\bf Z}_k$.
In section 2, we summarize the relevant results from the supergravity. In section 3, we review briefly the ABJM model and then present the basic set-up for our computations including the background configurations and the gauge fixing. In section 4, we present our one-loop calculations. Since the computations are somewhat involved, we mainly focus on the results while the details are deferred to the appendices. In section 5, we give some concluding remarks. In appendix A, we present the gauge fixing for an alternative choice of background configuration. In appendix B, we give details on the one-loop calculations. In appendix C, we present the relevant integrations.
\section{The results from supergravity}
The eleven dimensional metric describing $N$ M2 branes is given by
\begin{equation}
\label{m2metric}
ds^2_{11} = h^{-2/3} (-dt^2+dx_9^2+dx_{10}^2)
+h^{1/3} (dx_1^2 + \cdots +dx_{8}^2) ,
\end{equation}
where $h$ is the transverse eight dimensional harmonic function of $r= ( x_1^2 + \cdots + x_{8}^2 )^{1/2}$,
\begin{equation}
h(r) = 1+ \frac{32\pi^2 Nl_p^6}{r^6} .
\label{m2harmonic}
\end{equation}
The near-horizon limit of (\ref{m2metric}) is given by the $AdS_4\times S^7$ geometry in which the harmonic function becomes
\begin{equation}
h(r)= \frac{32\pi^2 Nl_p^6}{r^6} .
\label{m2harmonic}
\end{equation}
The geometry is maximally supersymmetric with 32 Killing spinors and has the isometry $SO(2,3)\times SO(8)$.
This is the limit where the worldvolume theory of $N$ M2 branes is expected to become the three dimensional ${\cal N}=8$ superconformal field theory.
In addition, one can consider the ${\bf Z}_k$ orbifolding to the transverse space, ${\bf R}^8/{\bf Z}_k={\bf C}^4/{\bf Z}_k$, with $(x^A + i x^{A+4})\sim e^{ i\frac{2\pi}{k}}( x^A + i x^{A+4})$, $A=1,\cdots, 4$.
The effect of ${\bf Z}_k$ orbifolding corresponds to the replacement $N\rightarrow N'=kN$ in the harmonic function $h(r)$~\cite{Aharony:2008ug}. In the near horizon limit, the geometry admits 24 Killing spinors and has the isometry $SO(2,3)\times SO(6)$ for $k\geq 3$. The dual field theory is expected to be an ${\cal N}=6$ superconformal theory.
The Nambu-Goto action for a probe membrane is given by
\begin{equation}
S_2=T_2\Big( -\int d^3\xi \sqrt{ - \det h_{\mu\nu}} + \int H \Big)~,
\end{equation}
where $T_2=1/(4\pi^{2} l_p^{3})$ is the membrane tension and $h_{\mu\nu}$ is the induced metric on the worldvolume. If one uses the static gauge for worldvolume diffeomorphism, the induced metric becomes
\begin{equation}
h_{\mu\nu}=g_{\mu\nu}+\partial_\mu X^I \partial_\nu X^J g_{IJ}~,
\end{equation}
where $X^I$ ($I= 1, \cdots, 8$) are transverse coordinates. We consider the configurations
in which the probe membrane is parallel to the source membranes and is scattered with a constant velocity. Furthermore we restrict ourselves to the case that $X^I$ depends only on time.
After plugging the $AdS_4\times S^7/{\bf Z}_k$ metric into the probe action and expanding it in terms of the velocity, $v^I = \dot{X}^I$, we obtain the effective action of the form
\begin{equation}
S_2=\int d^3\xi \Big( \frac{1}{2} T_2v^2 -V_2 +{\cal O}(v^6)\Big)\,,
\end{equation}
where $V_2$, the interaction potential of $v^4$ order, is given by
\begin{eqnarray}
V_2 = -\frac{1}{8} T_2 h(r) (v^2)^2 = - kN l_p^{3}\frac{ (v^2)^2}{r^6}~. \label{sugrapot}
\end{eqnarray}
To compare with the results from the ABJM model, we introduce complex coordinates in the transverse space as
\begin{eqnarray}
z^A=\frac{1}{2 \sqrt{\pi k l_p^3}}\Big( X^A + i X^{A+4}\Big), \qquad A=1, \cdots, 4~.
\end{eqnarray}
where the orbifolding ${\bf Z}_k$ acts as
$ z^A \rightarrow e^{i\frac{2\pi}{k}}z^A
$.
In these coordinates, the effective action becomes
\begin{equation}
S_2=\int d^3\xi \Big( \frac{k}{2\pi } |v|^2+\frac{N}{4\pi}\frac{|v|^4}{|z|^6} +{\cal O}(|v|^6)\Big)~,
\end{equation}
with $|z|^2\equiv z^A z_{A}^ *$ and $|v|^2\equiv \dot{z}^A \dot{z}_{A}^ *$.
\section{Three-dimensional ${\cal N}=6$ superconformal Chern-Simons theory}
The ABJM model has $OSp(6|4)$ superconformal symmetry for generic $k$. The classical vacuum moduli space is given by ${\bf C}^4/{\bf Z}_k$. These indicates that the model could be the worldvolume theory of $N$ M2-branes, which is dual to $M$ theory on $AdS_4\times S^7/{\bf Z}_k$. It is curious to note that this conjectured duality implies that the model has enhanced supersymmetries to ${\cal N}=8$ for $k=1,2$.
In section 3.1, we give some review on the model, establishing our notation. In section 3.2, we give the basic set-up of the problem, which corresponds to the membrane scattering in the dual gravity descriptions. In section 3.3, we describe the appropriate gauge fixing for the given configuration. In section 3.4, we present the tree level quadratic Lagrangian for gauge/scalar, fermion and ghost fields, respectively.
\subsection{The ABJM Model}
The model contains scalars, fermions and gauge fields. As being a conformal field theory,
the $U(N)\times U(N)$ gauge fields $A_L$, $A_R$ have Chern-Simons action with level $k$ and $-k$, respectively.
The matter fields
consist of four complex scalar fields $Y^A$ and spinor fields $\Psi_A$, which transform as ${\bf 4}$ and $\bar{\bf 4}$ under $SU(4)$ R-symmetry of ${\cal N}=6$ supersymmetry. They have bifundamental representations under the gauge groups.
Our conventions for spinors and their contractions are as follows. The three dimensional worldvolume flat metric and totally antisymmetric $\epsilon$-tensor are taken as $\eta^{\mu\nu}={\rm diag}\,(-,+,+)$ and $\epsilon^{012} =1$. The three dimensional Dirac $\gamma$-matices satisfy $ \gamma^{\mu}\gamma^{\nu} = \eta^{\mu\nu} + \epsilon^{\mu\nu\rho}\gamma_{\rho}\,.$
An explicit realization
may be given by
$ \gamma^{\mu~~ \beta}_{~~\alpha} = (i\sigma^2, \sigma^3,-\sigma^1)\,. $
Indices of three dimensional spinors are raised or lowered by antisymmetric $\epsilon$-matrices, $\epsilon^{12} = \epsilon_{21} = 1$.
We always contract spinor indices from northwest to southeast:
\[ \psi \chi \equiv
\psi^{\alpha}\chi_{\alpha} = \epsilon^{\alpha\beta}\psi_{\beta}\chi_{\alpha} = \chi^{\alpha}\psi_{\alpha}=\chi\psi\,.\]
Similarly, fermion bilinears with $\gamma$-matrices are defined as
\[ \psi\gamma^{\mu}\chi \equiv \psi^{\alpha}\gamma^{\mu~~ \beta}_{~~ \alpha}\chi_{\beta} = -\chi^{\alpha}\gamma^{\mu~~ \beta}_{~~\alpha}\psi_{\beta}=- \chi\gamma^{\mu}\psi\,. \]
The hermitian conjugate is defined as
$ (\psi^{\dagger}\chi)^{\dagger}
= \chi^{\dagger}_{\alpha}\psi^{\alpha} = - \chi^{\dagger}\psi\,.$
Note that $\gamma^{\mu}_{\,\alpha\beta} \equiv \epsilon_{\beta\rho}\gamma^{\mu~~ \rho}_{~~\alpha}=(-{\bf 1},\sigma^1,\sigma^3)$ are real and symmetric.
In order to do one loop computation, we choose the normalization of the scalar and fermion fields so that the classical action has an overall factor of the coupling constant, $\frac{k}{2\pi}$. In these conventions, our starting ABJM action is given by %
\begin{eqnarray} S &=& \frac{k}{2\pi}\int d^3x\bigg[\frac{1}{2} \epsilon^{\mu\nu\rho}{\rm Tr} \bigg\{ A_{L\, \mu}\partial_{\nu}A_{L\, \rho} + \frac{2i}{3}A_{L\, \mu}A_{L\, \nu}A_{L\, \rho} - A_{R\, \mu}\partial_{\nu}A_{R\, \rho}-\frac{2i}{3}A_{R\, \mu}A_{R\, \nu}A_{R\, \rho}\bigg\} \nonumber \\
&&~~~~~ +{\rm Tr}\bigg\{ -(D_{\mu}Y^{\dagger}_{ A})(D^{\mu}Y^{A})+i\Psi^{\dagger\, A}\gamma^{\mu}D_{\mu}\Psi_{A}\bigg\} -V_b-V_f\bigg]\,, \end{eqnarray}
where
\begin{eqnarray}
V_b &=& -\frac{1}{3}{\rm Tr}\bigg[Y^{\dagger}_{A}Y^{A}Y^{\dagger}_{B}Y^{B}Y^{\dagger}_CY^C + Y^{A}Y^{\dagger}_{A}Y^{B}Y^{\dagger}_{B}Y^CY^{\dagger}_C \nonumber\\
&& ~~~~~~+ 4Y^{A}Y^{\dagger}_BY^{C}Y^{\dagger}_{A}Y^{B}Y^{\dagger}_C - 6Y^{A}Y^{\dagger}_{B}Y^{B}Y^{\dagger}_AY^{C}Y^{\dagger}_C \bigg]\,,\\
&& \nonumber \\
V_f &=& i{\rm Tr}\bigg[Y^{\dagger}_{A}Y^{A}\Psi^{\dagger\, B}\Psi_B - Y^{A}Y^{\dagger}_A\Psi_B\Psi^{\dagger\, B}-2Y^{\dagger}_AY^{B}\Psi^{\dagger\, A} \Psi_B + 2 Y^AY^{\dagger}_B\Psi_A\Psi^{\dagger\, B} \nonumber \\
&& ~~~~~~-\epsilon^{ABCD}Y^{\dagger}_{A}\Psi_BY^{\dagger}_C\Psi_D +\epsilon_{ABCD}Y^{A}\Psi^{\dagger\, B}Y^C\Psi^{\dagger\, D}\bigg]\,. \nonumber
\end{eqnarray}
The covariant derivatives are defined by
\begin{eqnarray} D_{\mu}Y^{A} &=& \partial_{\mu}Y^{A} + iA_{L\, \mu}Y^{A} -iY^{A}A_{R\, \mu}\,, \nonumber \\
D_{\mu}\Psi_{A} &=& \partial_{\mu}\Psi_{A} + iA_{L\, \mu}\Psi_{A} -i\Psi_{A}A_{R\, \mu}\,. \nonumber \end{eqnarray}
The model does not have any dimensionful parameter and thus the classical action is conformally invariant. The gauge fields have mass dimension 1, while the scalar and fermion fields have mass dimension $\frac{1}{2}$ and 1, respectively. The only dimensionless parameter is the Chern-Simons level $k$, which plays the role of the coupling constant of the model. As it is quantized, the model is expected to be conformally invariant even at the quantum level.
It also has ${\cal N}=6$ supersymmetry.
Supersymmetry transformations for the ABJM model in our conventions are given by~\cite{Gaiotto:2008cg}
\begin{eqnarray}
\delta\, Y^{A} &=& i{\cal E}^{AB}\Psi_B\,, \nonumber \\
\delta\, Y^{\dagger}_{A} &=& i\Psi^{\dagger\, B}{\cal E}_{AB}\,, \nonumber\\
\delta\, \Psi_{A} &=& \gamma^{\mu}{\cal E}_{AB}D_{\mu}Y^{B} - {\cal E}_{AB}(Y^{C}Y^{\dagger}_{C}Y^{B} - Y^{B}Y^{\dagger}_{C}Y^{C}) + 2 {\cal E}_{CD}Y^{C}Y^{\dagger}_{A}Y^{D}\,, \nonumber \\
\delta\, \Psi^{\dagger\, A} &=& -{\cal E}^{AB}\gamma^{\mu}D_{\mu}Y^{\dagger}_{B} + {\cal E}^{AB}(Y^{\dagger}_{C}Y^{C}Y^{\dagger}_{B} - Y^{\dagger}_{B}Y^{C}Y^{\dagger}_{C}) - 2 {\cal E}^{CD}Y^{\dagger}_{C}Y^{A}Y^{\dagger}_{D}\,, \nonumber \\
\delta\, A_{L
, \mu} &=& {\cal E}^{AB}\gamma_{\mu}\Psi_{A}Y^{\dagger}_{B} + Y^{B}\Psi^{\dagger\, A}\gamma_{\mu}{\cal E}_{AB}\,, \\
\delta\, A_{R\, \mu} &=& Y^{\dagger}_{B}{\cal E}^{AB}\gamma_{\mu}\Psi_{ A}+\Psi^{\dagger\, A}\gamma_{\mu}{\cal E}_{AB}Y^{B}\,, \nonumber
\end{eqnarray}
where ${\cal E}_{AB}$ and ${\cal E}^{AB}$ are supersymmetry variation parameters and should be related as
\[ {\cal E}_{AB} = -\frac{1}{2} \epsilon_{ABCD}{\cal E}^{CD}\,, \qquad {\cal E}^{AB} = -\frac{1}{2} \epsilon^{ABCD}{\cal E}_{CD}\,, \qquad \qquad ({\cal E}^{\alpha}_{AB})^{\dagger} = {\cal E}^{\alpha\, AB}\,. \]
It has been argued that the supersymmetry is enhanced to ${\cal N}=8$ when the level $k$ is 1 or 2, though this is not obvious from the field theory Lagrangian.
\subsection{Set-up}
The appropriate background configurations for a probe membrane scattered by $N$ source membranes in the dual gravity description correspond to $$U(N+1)\times U(N+1)\rightarrow U(N)\times U(N)\times U(1)\times U(1)~.$$
For our purpose, it is enough to consider the case $N=1$.
Let us denote the vacuum expectation values and the quantum fluctuations of scalar fields as $\bar{Y}$ and $\delta Y$, respectively,
\[ Y^{A} = \bar{Y}^{A} + \delta Y^A\,. \]
We make the simplest choice for the vev's as
\begin{equation} \bar{Y}^{A} = \left( \begin{array}{cc} 0 & 0 \\ 0 & b^A \end{array}\right)\,, \qquad \bar{Y}^{\dagger}_{A} = \left( \begin{array}{cc} 0 & 0 \\ 0 & b^{\dagger}_A \end{array}\right)\,.
\label{vev}
\end{equation}
For a different choice of vacua, in which instanton corrections may play some role~\cite{Hosomichi:2008ip}, see appendix A.
To compare with the dual gravity descriptions, we restrict ourselves to the case :
\[ b^A = b_0^A + v^A t~, \]
where $b_0^A$ and $v^A$ are constants. One may note that $b_0$ corresponds to the impact parameter in the membrane scattering and thus satisfies $b_0\cdot v^\dagger\equiv b_0^A v_A^\dagger=0$.
For a generic background $b^A$, the off-diagonal components acquire a mass of order $|b|^2$ and thus, in the low energy, they can be treated as quantum fluctuations and integrated out.
These fluctuations are denoted as
\begin{eqnarray} A_{L\, \mu} &=& \left( \begin{array}{cc} 0 & a_{\mu} \\ a^{\dagger}_{\mu} & 0 \end{array}\right)\,, \qquad~ A_{R\, \mu} = \left( \begin{array}{cc} 0 & \hat{a}_{\mu} \\ \hat{a}^{\dagger}_{\mu} & 0 \end{array}\right)\,, \nonumber \\
\delta Y^A &=& \left( \begin{array}{cc} 0 & y^A \\ \tilde{y}^{A} & 0 \end{array}\right) \,, \qquad
\Psi_{A} = \left( \begin{array}{cc} 0 & \psi_A \\ \tilde{\psi}_{A} & 0 \end{array}\right)\,. \end{eqnarray}
We will integrate out these massive fluctuations and obtain the effective action of diagonal fields.
The resultant effective action will have $U(1)^2\times U(1)^2\times S_2$ gauge symmetry, with the permutation symmetry $S_2$ over the diagonal elements. Those abelian gauge fields can also be integrated out to give the effective action of $b^A$ and their superpartners on the moduli space $(C^4/{\bf Z}_k)^N/S_N$~\cite{Aharony:2008ug}.
\subsection{Gauge fixing}
A convenient gauge fixing in gauge theories with matter fields in the broken phase is the, so-called, $R_\xi$ gauge as the resultant Lagrangian does not have a mixing term between the gauge and scalar fields. The gauge fixing in supersymmetric gauge theories could be even more subtle since the gauge fixing term has to preserve the supersymmetry. One way is to use the supersymmetric $R_\xi$ gauge
in superfield formalism~\cite{Ovrut:1981wa}. Since we are using component fields, we use the $R_\xi$ gauge which may be supplemented with supersymmetric completion.\footnote{One may use an ${\cal N} =2$ superfield formalism~\cite{Benna:2008zy} and use supersymmetric $R_\xi$ gauge.}
Henceforth our gauge fixing functions are given by\footnote{
$U(N)$ indices leftover are contracted and summed. For example, $\delta Y^{\dagger}T^a\bar{Y}= \delta Y^{\dagger \hat{i} i}T^a_{ij}\bar{Y}^{j\hat{i}}$.}
\begin{eqnarray} f_{L}^a &=& -\frac{1}{\sqrt{\xi_L}}\Big(\partial_{\mu}A^{a \mu}_L +i\xi_L \delta Y^{\dagger}T^a\bar{Y} -i\xi_L \bar{Y}^{\dagger}T^a\delta Y \Big)\,,\nonumber \\
f_{R}^a &=&- \frac{1}{\sqrt{\xi_R}}\Big(\partial_{\mu}A^{a \mu}_R +i\xi_R \delta{Y}T^a\bar{Y}^{\dagger} -i\xi_R \bar{Y} T^a\delta Y^{\dagger}\Big) \,, \end{eqnarray}
where $\xi_L$, $\xi_R$ are arbitrary parameters with mass dimension one in {\it three dimensions}.
The gauge fixing Lagrangian is given by
\begin{equation} {\cal L}_{GF} = -\frac{1}{2} f_L^a f_L^a - \frac{1}{2} f_R^a f_R^a\,. \end{equation}
One may note that in this gauge choice we do not need to introduce the Nielsen-Kallosh ghost.
This gauge fixing function is supersymmetric for the configuration (\ref{vev}) with $v=0$ if we take
$ \xi_L =\xi_R = m_0\equiv |b_0|^2$. This is the case since we have chosen the bosonic background, i.e. $\langle \Psi\rangle =0$. The most natural choice for the time dependent background would be to replace $b_0^A$ by $b^A$, wherever applicable. It seems also natural in view of superfield formalism, where $b^A$ might be promoted to a superfield.
Henceforth we use the above gauge fixing function with
\[ \xi_L =\xi_R = m\equiv |b|^2 ~.\]
If we demand the supersymmetry completion among background fields, then, after turning on $v$, we should also include nonvanishing $\langle \Psi\rangle$ as a superpartner.
In this case, in order to have manifest supersymmetry among background fields, the gauge fixng function is needed to have additional terms which are bilinears in fermions, like $\Psi^\dagger\Psi$. These will give rise to terms in the effective action, which depend on $\langle \Psi\rangle$ and thus give the supersymmetric completion.
Since we are only interested in the purely bosonic terms in the effective action, we just use the above gauge function while neglecting those terms involving fermionic background fields. Those terms in the effective action could be obtained by the supersymmetric completion of the purely bosonic terms using the supersymmetry transformation rules for background fields.
\subsection{Quadratic action for quantum fluctuations}
After this choice of gauge fixing terms, one obtains
the quadratic Lagrangian of the bosonic fields, $(a_{\mu}~ y^A)$ and $(\hat{a}_{\mu}~ \tilde{y}^A)$ as
\[{\cal L}_{b,\, quad} = -(a^{\dagger}_{\mu}~~ y^{\dagger}_{A}) {\cal D}^{(\mu A) }_{~~~~\, (\nu\,B)} {a^{\nu}\choose y^B} -(\hat{a}^{\mu\dagger }~~ \tilde{y}^{ A}) \hat{{\cal D}}_{(\mu A) }^{~~~~\, (\nu\,B)} ~ {\hat{a}_{\nu}\choose \tilde{y}^\dagger_B} \]
where
\begin{eqnarray} {\cal D}^{(\mu A) }_{~~~~\, (\nu\,B)} &=& \left(\begin{array}{cc} -\partial^{\mu}\frac{1}{m}\partial_{\nu} + \epsilon^{\mu}_{~\nu\rho}\partial^{\rho} + m\eta^{\mu}_{\nu} & 2i\partial^{\mu}b^\dagger_{B} \\ -2i \partial_{\nu}b^{A}& (-\Box + m^2)\delta^{A}_{B} \end{array}\right)\,,
\end{eqnarray}
and
\begin{eqnarray} \hat{{\cal D}}_{(\mu A) }^{~~~~\, (\nu\,B)}&=& \left(\begin{array}{cc} -\partial_{\mu}\frac{1}{m}\partial^{\nu} - \epsilon_{\mu}^{~\nu\rho}\partial_{\rho} + m\eta_{\mu}^{\nu} & 2i\partial_{\mu}b^{B} \\ -2i\partial^{\nu}b^\dagger_{A} & (-\Box + m^2) \delta_{A}^{~B}\end{array}\right)\,. \end{eqnarray}
Ghost fields are introduced in the standard way as
\begin{eqnarray} C_L = \left( \begin{array}{cc} 0 & c_L\\ \tilde{c}_L & 0 \end{array}\right)\,, \qquad C_R = \left( \begin{array}{cc} 0 & c_R \\ \tilde{c}_R & 0 \end{array}\right)\,. \end{eqnarray}
The quadratic Lagrangian for ghost fields becomes
\begin{equation} {\cal L}_{g, quad} = -c^{\dagger}_{L}{\cal D}^{g}_{L}c_{L} - \tilde{c}^{\dagger}_{L} \tilde{{\cal D}}^{g}_{L}\tilde{c}_{L} -c^{\dagger}_{R}{\cal D}^{g}_{R}c_{R} - \tilde{c}^{\dagger}_{R} \tilde{{\cal D}}^{g}_{R}\tilde{c}_{R}\,, \end{equation}
where
\begin{equation} {\cal D}^{g}_{L} = \tilde{{\cal D}}^{g}_{L} = {\cal D}^{g}_{R} = \tilde{{\cal D}}^{g}_{R} =\frac{1}{\sqrt{m}}(-\Box + m^2)\,. \end{equation}
The quadratic Lagrangian of fermionic fields becomes
\[ {\cal L}_{f, quad} = -\psi^{\dagger\, A}{\cal D}^{~~ B}_{A}\psi_B - \tilde{\psi}^{\dagger\, A}\tilde{{\cal D}}^{~~ B}_{A}\tilde{\psi}_B\,,\]
where
\begin{eqnarray}
({\cal D}^{~~ B}_{A})_{\alpha}^{~ \beta} &=& -i\delta^{~ B}_{A}\slash{\partial}_{\alpha}^{~\beta} + i(m\delta^{~B}_{A}-2b^\dagger_{A}b^{B})\delta_{\alpha}^{~\beta} \,, \nonumber \\
(\tilde{{\cal D}}^{~~ B}_{A})_{\alpha}^{~\beta}&=& -i\delta^{~ B}_{A}\slash{\partial}_{\alpha}^{~\beta} - i(m\delta^{~B}_{A}-2b^\dagger_{A}b^{B})\delta_{\alpha}^{~\beta} \,,\end{eqnarray}
and $\alpha,\beta$ denote spinor indices.
In order to compute one-loop corrections, we need to rescale the gauge and ghost fields such that their kinetic terms are in standard forms. Henceforth, we perform the following time-dependent rescaling,
\[ \frac{1}{\sqrt{m}} a_{\mu} \longrightarrow a_\mu\,, \qquad \frac{1}{m^{1/4}}c_L\longrightarrow c_L\,, \qquad \frac{1}{m^{1/4}}\tilde{c}_L\longrightarrow \tilde{c}_L\,, \]
in which the gauge fields and ghost fields have the same mass dimension as the scalar fields. There are no extra contributions from the path integral measure due to this rescaling since we adopt dimensional regularization.
In addition to the rescaling, we perform the Wick rotation to the Euclidean space.
Time-independent part of the scalar vev's is denoted as
$ m_0 \equiv |b_0|^2\,. $
Then the mass parameter $m$ is given by
\[ m = m_0 + |v|^2\tau^2 \,. \]
One may regard the velocity $v$ has small magnitude and treat it as a perturbation parameter.
The quadratic operators, relevant to the one-loop computation, become of the form
$$Q=Q_0+Q_1$$
where $Q_0=Q_0(m_0)$ and $Q_1=Q_1(v)$. They consist of three parts, each from gauge/scalar fields, fermionic fields and ghost fields. We present those operators below, keeping terms only up to quartic order in $v$.
\underline{\it Gauge/scalar fields}
\begin{eqnarray}
Q^b_0 &=& \left(\begin{array}{cc} -\partial^{\mu}\partial_{\nu} + im_0\epsilon^{\mu}_{~\nu\rho}\partial^{\rho} + m^2_0 \delta^{\mu}_{\, \nu} & 0 \\ 0 & (-\Box + m^2_0)\delta^{A}_{B} \end{array}\right)\,,
\nonumber \\ && \nonumber \\
\qquad Q^b_1 &=& \left(\begin{array}{cc}C^\mu_{\,\nu} & D^{\mu\dagger}_B \\ D_\nu^A & E^A_B\end{array}\right) \,, \end{eqnarray}
where
\begin{eqnarray}
C^\mu_{\,\nu} &=& i|v|^2\tau^2\epsilon^{\mu}_{~ \nu\rho}\partial^{\rho}+ M(\tau)\, \delta^{\mu}_{\,\nu} + \Big(-\frac{|v|^2}{ m_0}+ 4\frac{ |v|^4\tau^2}{m_0^2}\Big) \delta^{\mu}_\tau \delta_{\nu}^\tau-\frac{|v|^2 \tau}{m_0}\Big(\partial^\mu\delta^\tau_\nu-\partial_\nu\delta_\tau^\mu \Big)~, \nonumber \\
D_\nu^A &=& - 2i\sqrt{m}\partial_{\nu}b^A= - 2i\sqrt{m}\, v^A\delta_{\nu}^{\, \tau}~, \\ D^{\mu\dagger}_B &=& 2i\sqrt{m}\partial^{\mu}b^\dagger_B= 2i\sqrt{m}\, v^\dagger_B\delta^{\mu}_{\, \tau}~, \nonumber \\
E^A_B &=& M(\tau)\, \delta^{A}_{B} ~, \nonumber
\end{eqnarray}
and
$$M(\tau)\equiv 2m_0 |v|^2\tau^2 + |v|^4\tau^4\,.$$
One may note that the last two terms in $C^\mu_\nu$ come from the time-dependent rescaling of gauge fields.
There is another part in bosons which may be denoted as $\hat{Q}^b_0 + \hat{Q}^b_1$. This has the same form as $Q^b_0+Q^b_1$ with a replacement $\epsilon^{\mu}_{~ \nu\rho} \rightarrow -\epsilon^{\mu}_{~ \nu\rho}$. It turns out that they give the identical contributions.
\underline{\it Ghost fields}
\begin{eqnarray}
Q^g_0 &=& -\Box + m^2_0\,, \qquad \qquad \qquad ~~~ \Box \equiv \partial^2_{\tau} + \partial^2_{i} \nonumber \\ &&\nonumber\\
Q^g_1(\tau)& =& M(\tau) + \Big(-\frac{ |v|^2}{ 2m_0}+ \frac{5 |v|^4\tau^2}{4m_0^2}\Big)-\frac{ |v|^2 \tau}{m_0}\partial_{\tau}\,. \label{ghostD}
\end{eqnarray}
Note that the last two terms in $ Q^g_1(\tau)$ come from the time-dependent rescaling of the ghost fields.
\underline{\it Fermion fields}
We construct the ``squared'' operator, $Q^f$, in terms of the original operators by
\[
Q^f = {\cal D}\tilde{{\cal D}} = \tilde{{\cal D}}{\cal D}=Q^{f}_{\,0}+ Q^{f}_{\, 1}\,,
\]
where
\begin{eqnarray} Q^{f}_{\,0} &=& (-\Box +m^2_0)\delta^{~ \beta}_{\alpha}\delta^{A}_{B}\,, \nonumber \\ &&\nonumber\\
Q^{f}_{\, 1} &=& M(\tau)\, \delta^{~ \beta}_{\alpha}\delta^{A}_{B} - \slash{\partial}^{~ \beta}_{\alpha}(|b|^2\delta^{A}_{B}-2b^{A}b^\dagger_{B})\,. \end{eqnarray}
\section{One-loop corrections}
The one-loop effective action is given by %
\[ \Gamma_1 = \sum_{fields} (-1)^\epsilon \ln\det (Q_0 + Q_1) \,, \]
where $\epsilon$ is 0 for the bosonic fields and 1 for the fermion and ghost fields.
This can be computed systematically using the Schwinger proper time method\footnote{For a review on the Schwinger formalism, for example, see~\cite{Ball:1988xg}.} as
\[ \Gamma_1 = -\sum (-1)^\epsilon \int^{\infty}_{0} \frac{ds}{s}~{\rm Tr}\, e^{-s(Q_0 + Q_1)}\,, \]
where ${\rm Tr}$ denotes sum over all the indices including coordinates.
The one-loop effective potential in the Euclidean space can be read from the one-loop effective action as
\[ \Gamma_1 = \int d^3x V_1(x) \]
and thus becomes
\[
V_1(x) = V_1^b +V_1^f + V_1^g = - \sum (-1)^\epsilon\int^{\infty}_{0}\frac{ds}{s}~ {\rm tr}\, \langle x | e^{-s(Q_0+Q_1)} |x \rangle \,,
\]
where ${\rm tr}$ denotes the trace over gauge group and Lorentz indicies. Standard Dyson perturbative expansion for small $Q_1$ leads to
\begin{eqnarray} \!\!\!\!\!\!\!V_1 &=& \sum_{n=1}^{\infty} V_{1,n-1}\end{eqnarray}
where
\begin{eqnarray} \!\!\!\!\!\!\!V_{1, n-1} &=& \sum_{fields}(-1)^\epsilon (-1)^{n} \int^{\infty}_{0}\frac{ds_1\cdots ds_n}{(s_1+\cdots + s_n)}\, {\rm tr}\,\langle x| \Big[ e^{-(s_1+\cdots + s_n)Q_0} \prod^{n}_{i=2}Q_1(s_i+\cdots +s_n)\Big] |x \rangle \,, \nonumber \end{eqnarray}
where $Q_1(s)$ is defined by
\[
Q_1(s) \equiv e^{sQ_0}Q_1e^{-sQ_0} \,.
\]
\underline{\it Regularization by dimensional reduction}
The computations typically involve the integration over the momentum $p$ as well as the Schwinger parameters $s_i$. To deal with the divergencies, we adopt dimensional regularization in the momemtum integrals. It is well known that Chern-Simons gauge theories have subtleties in using dimensional regularization~\cite{Martin:1990xv}\cite{Chen:1992ee}. They arise because the theories are sensitive to the dimension they live in, i.e. Chern-Simons term can be defined only in three dimensions. The same kind of subtleties arises in the general supersymmetric theories as well~\cite{Gates:1983nr} because the number of bosonic/fermionic degrees of freedom are sensitive to the spacetime dimension.
In order to avoid this kind of problems, a modified prescription, which is called regularization by dimensional reduction, has been adopted to these theories. The essential point in this modified version is that the usual dimensional regularization rule, with dimensional continuation from three to $n+1$ dimensions, is applied to the momentum integration with divergencies, while all the contractions in tensor and spinor indices
are performed in three dimensions~\cite{Chen:1992ee}\cite{Gates:1983nr}. For example, we have
\begin{eqnarray} \epsilon^{\mu\nu}_{~~ \rho}\epsilon^{\rho}_{~ \lambda\eta} & =& \Big(\delta^{\mu}_{~ \lambda}\delta^{\nu}_{~ \eta} - \delta^{\mu}_{~ \eta}\delta^{\nu}_{~\lambda}\Big)\,, \nonumber \\
\delta^{\mu}_{~ \mu} &=& 3\,,
\end{eqnarray}
in three dimensional Euclidean space. For the spinor indices, we use
\[ \delta^{\alpha}_{~ \alpha} = 2\,. \]
Now we present the results of our computations. We compute only up to $|v|^4$ terms.
At first, we consider the gauge/scalar fields contributions to the one-loop effective potential. In fact it is
the most nontrivial part in computations. Here we present only the results. For details in calculation, see appendix B.
\underline{\it Gauge/scalar fields}
As stated earlier, we have two parts of gauge/scalar fields, each from $Q_0^b+Q_1^b$ and $\hat{Q}_0^b+\hat{Q}_1^b$, which give identical results. Here we present only $Q_0^b+Q_1^b$ part. At the end we should double what we have got.
\begin{eqnarray}
V^b_{1,\, 0} &=& -\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}\, 6\Gamma\Big(-\frac{n+1}{2}\Big)\,,
\nonumber \\ &&\nonumber\\
V^b_{1,\, 1}
&=& \frac{1}{m^2_0}\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}~ \Gamma\Big(\frac{1-n}{2}\Big)\Big[6M(\tau) + |v|^4\tau^4- \frac{|v|^2}{m_0}+4\frac{|v|^4\tau^2}{m^2_0}\Big ]+ {\cal O}(|v|^6)\,,
\nonumber\\&&\nonumber\\
V^b_{1,\, 2}
&=&
-\frac{1}{m^4_0}\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}
\bigg[ \Gamma\Big(\frac{1-n}{2}\Big)\Big(|v|^4\tau^2 +m^2_0|v|^4\tau^4\Big)
\nonumber \\
&& \qquad \qquad \qquad +\, \Gamma\Big(\frac{3-n}{2}\Big)\Big( 4m_0|v|^2 + \frac{|v|^4}{2m^2_0} -|v|^4\tau^2+12 m^2_0|v|^4\tau^4\Big) \nonumber \\
&& \nonumber \\
&& \qquad \qquad \qquad +\Gamma\Big(\frac{5-n}{2}\Big)4|v|^4\tau^2 \bigg] + {\cal O}(|v|^6)\,,
\\&&\nonumber\\
V^b_{1,\, 3} &=& \frac{1}{m^6_0}\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}\, \Gamma\Big(\frac{5-n}{2}\Big)\,\Big( 8m^2_0|v|^4\tau^2 -2 |v|^4 \Big)+ {\cal O}(|v|^6)\,,
\nonumber\\&&\nonumber\\
V^b_{1,\,4} &=& -\frac{1}{m^8_0}\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}\, \Gamma\Big(\frac{7-n}{2}\Big)\, \frac{4}{3}m^2_0|v|^4 + {\cal O}(|v|^6)\,. \nonumber
\end{eqnarray}
\underline{\it Ghost fields}
There are four identical contributions from ghost fields, $c_{L, R}$, $\tilde{c}_{L, R}$. We denote them as
\[
{\rm tr}\, {\bf 1}_g = 4 \,.
\]
All together they become
\begin{eqnarray}
V_{1,0}^g
&=& +\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}\, \Gamma\Big(-\frac{n+1}{2}\Big)~ {\rm tr}\, {\bf 1}_g \nonumber\\&&\nonumber\\
V_{1,1}^g
&=& -\, \frac{1}{m^2_0}\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}} \Gamma\Big(\frac{1-n}{2}\Big) \bigg( M(\tau) - \frac{1}{2} \frac{|v|^2}{m_0}+ \frac{5}{4}\frac{|v|^4\tau^2}{m^2_0}\bigg) ~ {\rm tr}\, {\bf 1}_g + {\cal O}(|v|^6)\nonumber \\&&\nonumber\\
V_{1,2}^g&=& +\, \frac{1}{m^4_0}\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}\Big[\Gamma\Big(\frac{5-n}{2}\Big) \frac{2}{3}|v|^4\tau^2+ \Gamma\Big(\frac{3-n}{2}\Big)\bigg(2m^2_0|v|^4\tau^4 + \frac{1}{8}\frac{|v|^4}{m^2_0}-|v|^4\tau^2\bigg) \nonumber \\
&& \nonumber \\
&& \qquad \qquad \qquad ~ -\,\frac{1}{4} \Gamma\Big(\frac{1-n}{2}\Big)\, |v|^4\tau^2 \Big]~ {\rm tr}\, {\bf 1}_g + {\cal O}(|v|^6)\,.
\end{eqnarray}
\underline{\it Fermionic fields}
The computations in fermionic fields are also straightforward. The results are as follows.
\begin{eqnarray}
V_{1,0}^f
&=& +\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}\, \Gamma\Big(-\frac{n+1}{2}\Big)~ {\rm tr}\, {\bf 1}_f\,,
\nonumber \\&&\nonumber\\
V_{1,1}^f &=&
- \frac{1}{m^2_0}\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}\, \Gamma\Big(\frac{1-n}{2}\Big)\, M(\tau)~ {\rm tr}\, {\bf 1}_f \,, \nonumber \\&&\nonumber\\
V_{1,2}^f&=& + \frac{1}{m^4_0}\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}\, \bigg[ \Gamma\Big(\frac{3-n}{2}\Big) \Big(m_0|v|^2 + 2|v|^4\tau^2 + 2m^2_0|v|^4\tau^4\Big) \nonumber\\ &&\nonumber\\
&& \qquad \qquad \qquad ~ + \Gamma\Big(\frac{5-n}{2}\Big)\frac{2}{3} |v|^4\tau^2 \bigg] {\rm tr}\, {\bf 1}_f + {\cal O}(|v|^6)\,, \nonumber \\&&\nonumber\\
V_{1,3}^f&=& - \frac{1}{m^6_0}\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}\bigg[\Gamma\Big(\frac{5-n}{2} \Big)16m^2_0|v|^4\tau^2\bigg] + {\cal O}(|v|^6) \,,
\nonumber \\&&\nonumber\\
V_{1,4}^f
&=& + \frac{1}{m^8_0}\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}\bigg[\Gamma\Big(\frac{7-n}{2} \Big)\frac{8}{3}m^2_0|v|^4 \bigg]+ {\cal O}(|v|^6) \,,
\end{eqnarray}
where
\[{\rm tr}\, {\bf 1}_f = {\rm tr}\, \delta^{\alpha}_{\, \beta}\delta^{A}_{\, B} =8 \,.\]
\underline{\it The results: summary}
By collecting all the results of one-loop effective potentials, one can easily see that there is a complete cancellation among contributions from the ghosts, fermions and gauge/scalar fields up to order $|v|^2$.
As a result the one-loop effective potential is given by
\begin{eqnarray*}
V_{1} & = & V^{b}_1 + V^g_{1} + V^{f}_{1}= \frac{1}{m^4_0}\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}\Gamma\Big(\frac{3-n}{2}\Big) 14 |v|^4\tau^2 \nonumber \\
&&\qquad \qquad\qquad~~~~- \frac{1}{m^6_0}\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}\bigg[\Gamma\Big(\frac{5-n}{2}\Big) 4 +\Gamma\Big(\frac{3-n}{2}\Big)\frac{1}{2} \bigg]|v|^4 +{\cal O}(|v|^6)\,.
\end{eqnarray*}
Our field theory results correspond to those of a single source brane in the supergravity computations.
The generalization to $N$ source branes is straightforward.
We simply multiply $N$ factor to $V_1$.
Now taking $n=2$, the effective action, including the tree-level part, in field theory side is finally given by
\begin{equation}
\Gamma = \Gamma_{\rm tree} + \Gamma_1 = \int d^3x \bigg(\frac{k}{2\pi}|v|^2 + \frac{7N}{4\pi}\frac{|v|^4\tau^2 }{|b_0|^2}- \frac{5N}{16\pi}\frac{|v|^4}{|b_0|^6} +{\cal O}(|v|^6)\bigg)\,. \end{equation}
Note that the $|v|^4$ term can be eliminated by a suitable shift of $\tau$ with a time reversal symmetry of the effective action.
The effective action up to $v^4$ terms obtained from $D=3$ ${\cal N}=8$ super Yang-Mills theory is coincident with the probe action on $AdS_4\times S^7$ in the {\it static gauge}. In general, this doesn't have to be the case.
We find complete agreement between the results from our field theory computations and those from the dual supergravity, if we choose the following gauge for the worldvolume diffeomorphism in the supergravity:
\begin{equation} X^0 = \frac{1}{\sqrt{7}|b_0|^2}\ln ( |b_0|^2 \xi^0)\,, \qquad X^{9} = (\sqrt{7}|b_0|^2 \xi^0)^{-1}\, \xi^1 \,, \qquad X^{10} =\xi^2\,. \end{equation}
In order to see this, note that the transverse coordinates $z^A$ are identified with $b^A$ and worldvolume coordinates in supergravity, $(\xi^0, \xi^1,\xi^2)$, are identified with field theory coordinates, $(t,x^1,x^2)$. Since
the velocities in supergravity and field theory are defined as
\begin{equation} v^{A}_{\rm sugra} \equiv \frac{d z^{A}}{d X^{0}} = \sqrt{7}|b_0|^2\xi^0 \frac{d z^{A}}{d \xi^0} \,, \qquad v^{A}_{\rm field} \equiv \frac{d b^{A}}{d t}\,, \end{equation}
one can see that
\[ v^{A}_{\rm sugra} = \sqrt{7} |b_0|^2t~ v^A_{\rm field}\,, \]
and
\[ S_2 = \int d^3x\Big( \frac{k}{2\pi}|v_{\rm field}|^2 + \frac{7N}{4\pi}\frac{|v_{\rm field}|^4t^2 }{|b_0|^2}\Big)\,. \]
This shows that supergravity results are in complete agreement with field theory ones.
\section{Conclusion}
In this paper we took a first step toward the understanding of the quantum correction in the ABJM model, which would give a nontrivial test for the $AdS/CFT$ correspondence. We used the $R_\xi$ gauge, which preserves the supersymmetry if the vev is time-independent, to perform one-loop computations. We found complete agreement in membrane scattering dynamics between the results from the ABJM model and those from the dual supergravity on $AdS_4\times S^7/{\bf Z}^k$ in a specific gauge for worldvolume diffeomorphism.
As a result we find that there is no correction in the $v^2$ term. As stated earlier, ${\cal N}=4$ supersymmetry in three dimensions guarantees that the $v^2$ term is one-loop exact. Our result, supplemented with the supersymmetry, shows that there is non-renormalization in the $v^2$ term, i.e. tree-level exact. It would be very nice to show that it is indeed the case by using supersymmetry arguments for ${\cal N}=6$. One may note that this also reflects the conformal symmetry at the quantum level.
We also find that the $v^4$ term appears at one-loop, which agrees with the supergravity computations in the special choice of gauge for worldvolume diffeomorphism. There is a
non-renormalization theorem, at least for ${\cal N}=8$ supersymmetry, which states that the $v^4$ term appears only at one-loop with possible non-perturbative instanton corrections. Since there is no monopole-instanton for our configurations, we expect our result is exact, at least for $k=1$ and 2. If we start with the background configuration shown in the appendix A, we need to include the instanton corrections to reproduce the results from supergravity.
It would be very interesting to reexamine the problem using the superfield formalism with the supersymmetric $R_\xi$ gauge.
It would be also very interesting to see whether there is a, perturbative, non-renormalization theorem for the ABJM model with generic $k$. This might be determined by studying the supersymmetry completion. Another way to see this is to study two-loop corrections to the effective action.
{\bf Acknowledgments}
We would like to thank the theory group at UBC, KEK and KIAS for hospitality. J.H.B, S.H. and S.H.Y were supported by
the Korea Research Foundation Grant funded
by Korea Government(MOEHRD, Basic Reasearch Promotion Fund)
(KRF-2005-070-C00030).
W.J was supported by
the Science Research Center Program of the
Korea Science and Engineering Foundation through the Center for
Quantum Spacetime \textbf{(CQUeST)} of Sogang University with grant
number R11-2005-021.
\section*{Appendix}
\subsection*{A. Supersymmetry and Gauge Fixing}
In this appendix, we consider different background configurations and the corresponding gauge fixing.
Consider the following vacuum expectation values with real $d^A$ and $b^A$
\begin{equation} \bar{Y}^{A} = \left( \begin{array}{cc} d^A & 0 \\ 0 & b^A \end{array}\right)\,, \qquad \bar{Y}^{\dagger}_{A} = \left( \begin{array}{cc} d_{A} & 0 \\ 0 & b_A \end{array}\right)\,. \end{equation}
The bosonic part in the quadratic Lagrangian is given by
\begin{eqnarray} {\cal L}_b = - (y^{\dagger}~ \tilde{y})\left(\begin{array}{cc} {\cal M} & {\cal N} \\ {\cal Q} & {\cal P} \end{array}\right) {y\choose \tilde{y}^{\dagger}}\,,
\end{eqnarray}
where
\begin{eqnarray} {\cal M}^{A}_{~ B} &=& \Big[-\Box + (b^2+d^2)^2 -4(b\cdot d)^2\Big]~ \delta^{A}_{\, B} - (b^2+d^2)(b^Ab_B+d^Ad_B) + 2(b\cdot d)(b^Ad_B +d^Ab_B) \,,
\nonumber \\
{\cal N}^{AB} &=& (b^2+d^2) (b^Ad^B +d^Ab^B) -2 (b\cdot d)(b^Ab^B+d^Ad^B)\,, \nonumber \\
{\cal P}_{A}^{~\, B} &=& {\cal M}_{A}^{~\, B}\,, \qquad {\cal Q}_{AB} = {\cal N}_{AB}\,. \end{eqnarray}
The fermionic part is
\begin{eqnarray} - (\psi^{\dagger},\, \tilde{\psi})\, {\cal D}_f \, { \psi \choose \tilde{\psi}^{\dagger}} \equiv - (\psi^{\dagger\, A},\, \tilde{\psi}_C)\left(\begin{array}{cc} F^{~ B}_{A} & -2i\epsilon_{ADPQ}b^Pd^Q \\ 2i\epsilon^{CBPQ}b_Pd_Q & - F^{C}_{~ D} \end{array}\right) { \psi_{B} \choose \tilde{\psi}^{\dagger\, D}} \,, \end{eqnarray}
where
\[ F^{~ B}_{A}= \Big[-i\slash{\partial} + i(b^2-d^2)\Big]~\delta^{~ B}_{A} -2i(b_Ab^{B}-d_{A}d^{B})\,. \]
For constant $b$ and $d$, one gets
\[ Q_f \equiv {\cal D}_f{\cal D}^{\dagger}_f ={\cal D}_f^{\dagger}{\cal D}_f = \Big[-\Box+(b^2+d^2)^2 -4(b\cdot d)^2\Big]{\bf 1}\,. \]
To get a covariant gauge fixing term which respects the supersymmetry, we introduce
\[ A_{\pm\, \mu} \equiv \frac{1}{2} \Big(A_{L\,\mu} \pm A_{R\, \mu}\Big) \,, \]
and take the $R_\xi$ gauge for these gauge fields. The gauge fixing terms are given by
\begin{eqnarray} {\cal L}_{GF} &=& -\frac{1}{2\xi_{+}}{\rm Tr}\Big(\partial_{\mu}A^{\mu}_{+} + \frac{i}{\sqrt{2}}\xi_{+} [\bar{Y}, \delta Y^{\dagger}] + \frac{i}{\sqrt{2}} \xi_{+}[\bar{Y}^{\dagger}, \delta Y] \Big)^2 \nonumber \\
&& -\frac{1}{2\xi_{-}}{\rm Tr}\Big(\partial_{\mu}A^{\mu}_{-} + \frac{i}{\sqrt{2}}\xi_{-} \{\bar{Y}, \delta Y^{\dagger}\} - \frac{i}{\sqrt{2}} \xi_{-}\{\bar{Y}^{\dagger}, \delta Y\} \Big)^2
\,. \nonumber \end{eqnarray}
One can show that they are supersymmetric if $\xi_\pm$ are given by
\[ \xi_{+} = b^2+d^2 + 2\, b\cdot d\,, \qquad \xi_{-} = b^2+d^2 - 2\, b\cdot d\,. \]
Note that $d=0$ case reduces to the same gauge fixing Lagrangian given in the main text.
\subsection*{B. Some details in one-loop computations}
Our normalization conventions for the plane wave basis in the computation of the one-loop effective action are
\[
\langle x | p \rangle = \frac{1}{(2\pi)^{n+1}}\, e^{ip\cdot x}\,, \]
with the completeness relations
\[
\int d^{n+1}x~ |x \rangle \langle x| = {\bf 1}\,, \qquad \int d^{n+1}p~ |p \rangle \langle p| = {\bf 1}\,. \]
Here we present the calculational details of the bosonic part contributions. First of all, it is convenient to introduce
\[ P^{\mu}_{~ \nu} = \delta^{\mu}_{\, \nu} - \frac{p^{\mu}p_{\nu}}{p^2}\,, \qquad R^{\mu}_{~ \nu} = \epsilon^{\mu}_{~ \nu\rho}\frac{p^{\rho}}{p}\,, \]
which give
\begin{equation} e^{aP} = 1 - P + e^{a}P\,, \qquad e^{aR} = 1 -P + \cos a\, P + \sin a\, R\,. \nonumber \end{equation}
They have nice properties such as
\begin{equation} P^2 = P\,, \qquad R^2 = -P\,, \qquad PR = RP = R\,, \qquad {\rm tr}\, P = 2 \,, \qquad {\rm tr}\, R=0\,. \nonumber \end{equation}
Let us define
\begin{eqnarray} e^{-sQ^{b}_{0}(p)} \equiv \left( \begin{array}{cc} e^{-sA(p)} & 0 \\ 0 & e^{-sB(p)} \end{array}\right)\,, \end{eqnarray}
where the quadratic operator $A$ and $B$ are given by
%
\begin{eqnarray} A^{\mu}_{\, \nu} &=& p^{\mu}p_{\nu} - m_0 \epsilon^{\mu}_{~ \nu\rho}p^{\rho} + m^2_0\delta^{\mu}_{~ \nu} = (p^2+m^2_0)\delta^{\mu}_{~ \nu} -p^2P^{\mu}_{~ \nu}-m_0pR^{\mu}_{~ \nu}\,, \nonumber\\
B &=& (p^2 +m^2_0)\delta^{A}_{B}\,. \nonumber \end{eqnarray}
Then we obtain
\[ e^{-sA} = e^{-s(p^2+m^2_0)}(1-P) + e^{-sm^2_0}\Big( P\, \cos(smp) + R\, \sin(smp)\Big)\,. \]
The above relations facilitate the various calculations involving products of $e^{-sQ^b_0}Q^b_1$'s.
The computation of the zeroth order, in $Q_1^b$ insertion, is straightforward and goes as follows:
%
\begin{eqnarray}
V^b_{1,\, 0} &=& -\int^{\infty}_{0}\frac{ds}{s}\, {\rm tr}\, \langle x| e^{-sQ^b_0} |x \rangle = -\int^{\infty}_{0}\frac{ds}{s}\int \frac{d^{n+1}p}{(2\pi)^{n+1}}\, {\rm tr}\, \langle p| e^{-sQ^b_0} |p \rangle \nonumber\\
&=& -\int^{\infty}_{0}\frac{ds}{s}\, e^{-sm^2_0} \int \frac{d^{n+1}p}{(2\pi)^{n+1}}\, \Big[ 5e^{-sp^2} + 2\cos m_0ps\Big] \nonumber\\
&=&-\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}\int^{\infty}_{0} ds \, e^{-s}\, \bigg[ \frac{5}{s^{(n+3)/2}} - \frac{4}{s^{n+2}}\frac{\Gamma(n+1)}{\Gamma\big(\frac{n+1}{2}\big)}\sin \frac{n\pi}{2}\bigg]
\nonumber \\
&=&
-\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}\, 6\Gamma\Big(-\frac{n+1}{2}\Big)\,,
\end{eqnarray}
where ${\rm tr}$ denotes sum over gauge, Lorentz and $SU(4)_R$ indices and we have used
\[ \frac{\Gamma(n+1)\Gamma(-n-1)}{\Gamma\big(\frac{n+1}{2}\big)} = -\frac{\Gamma\big(-\frac{n+1}{2}\big)}{4\sin\frac{n}{2}\pi }\,. \]
In what follows, we integrate over Schwinger parameters $s_i$ first, and then calculate momentum integrals.
The first order part is also straightforward and is given by
\begin{eqnarray}
V^b_{1,\, 1} &=& \int^{\infty}_{0}\frac{ds_1ds_2}{s_1+s_2}\, {\rm tr} \Big\langle x \Big| e^{-(s_1+s_2)A}C + e^{-(s_1+s_2)B}E \Big|x \Big\rangle \nonumber\\
&=& \int \frac{d^{n+1}p}{(2\pi)^{n+1}}\, \int^{\infty}_{0}\frac{ds_1ds_2}{s_1+s_2}\, e^{-(s_1+s_2)m^2_0} \bigg\{\Big[\frac{1}{n+1}\, e^{-(s_1+s_2)p^2} + \frac{n}{n+1}\cos m_0p(s_1+s_2) \Big] \nonumber \\ && \nonumber \\
&&\qquad \qquad \qquad ~~~ \times \Big(3M(\tau) - \frac{|v|^2}{m_0}+4\frac{|v|^4\tau^2}{m^2_0}\Big) + \Big[ p \sin m_0p(s_1+s_2)\Big]2|v|^2\tau^2 + 4M(\tau) \bigg\} \nonumber \\
&&\nonumber \\
&=& \frac{1}{m^2_0}\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}~ \Gamma\Big(\frac{1-n}{2}\Big)\Big[6M(\tau) + |v|^4\tau^4- \frac{|v|^2}{m_0}+4\frac{|v|^4\tau^2}{m^2_0}\Big ]\,,
\end{eqnarray}
where we have used
\[ p_\mu p_\nu = \frac{1}{n+1} p^2 \delta_{\mu\nu}
\]
in the momentum integral.
The calculations of integrals are quite involved starting from the second order computations. We present all those integrals in appendix C.
The second order in the perturbation consists of three parts,
\begin{equation}
V^b_{1,\, 2} = [CC]+ [EE]+[DD]\,,
\end{equation}
where each represents the contribution from the gauge-gauge, the scalar-scalar and the gauge-scalar fields.
Using those integral formulae given in appendix C, we find the contributions from the second order part as follows:
\begin{eqnarray}
\big[CC\big] &\equiv& -\int^{\infty}_{0}\frac{ds_1ds_2ds_3}{s_1+s_2+s_3}\, {\rm tr}\, \Big\langle x \Big| e^{-(s_1+s_3)A}Ce^{-s_2A}C \Big|x \Big\rangle \nonumber \\
&=&
-\frac{1}{m^4_0}\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}
\bigg[ \Gamma\Big(\frac{1-n}{2}\Big)\Big(|v|^4\tau^2 +m^2_0|v|^4\tau^4\Big)
+ \Gamma\Big(\frac{3-n}{2}\Big)\Big( \frac{|v|^4}{2m^2_0} -5|v|^4\tau^2+ 4|v|^4\tau^4\Big)\nonumber \\
&& \qquad \qquad \qquad \, \, \, +\Gamma\Big(\frac{5-n}{2}\Big)\frac{4}{3}|v|^4\tau^2 \bigg] + {\cal O}(|v|^6)\,, \nonumber \\
\big[EE\big] &\equiv& -\int^{\infty}_{0}\frac{ds_1ds_2ds_3}{s_1+s_2+s_3}\, {\rm tr}\, \Big\langle x \Big| e^{-(s_1+s_3)B}Ee^{-s_2B}E \Big|x \Big\rangle \\
&=& -\frac{1}{m^4_0}\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}\, \bigg[ \Gamma\Big(\frac{3-n}{2}\Big) 8m^2_0|v|^4\tau^4 + \Gamma\Big(\frac{5-n}{2}\Big)\frac{8}{3} |v|^4\tau^2 \bigg]+ {\cal O}(|v|^6)\,, \nonumber \\ && \nonumber \\ %
\big[DD\big] &\equiv& -\int^{\infty}_{0}\frac{ds_1ds_2ds_3}{s_1+s_2+s_3}\, {\rm tr}\, \Big\langle x \Big| e^{-(s_1+s_3)A}D^{\dagger}e^{-s_2B}D + e^{-(s_1+s_3)B}De^{-s_2A}D^{\dagger} \Big|x \Big\rangle \nonumber \\
&=& -\frac{1}{m^4_0}\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}\, \Gamma\Big(\frac{3-n}{2}\Big) 4(m_0|v|^2 + |v|^4\tau^2) + {\cal O}(|v|^6)\,. \nonumber \\ && \nonumber %
\end{eqnarray}
Similarly, the cubic order can be found to be
\begin{equation}
V^b_{1,\, 3} = [CDD] + [EDD]\,, \end{equation}
where
\begin{eqnarray}
\big[CDD\big] &\equiv& \int^{\infty}_{0}\frac{ds_1ds_2ds_3ds_4}{s_1+s_2+s_3+s_4}\, {\rm tr}\, \Big\langle x \Big| e^{-(s_1+s_4)A}C e^{-s_2A}D^{\dagger} e^{-s_3B}D
+ e^{-(s_1+s_4)A}D^{\dagger} e^{-s_2B}D e^{-s_3A}C\nonumber \\
&&\qquad \qquad \qquad \qquad \qquad ~~~ + e^{-(s_1+s_4)B}De^{-s_2A} Ce^{-s_3A}D^{\dagger} \Big| x \Big\rangle \nonumber \\
&=& \frac{1}{m^6_0}\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}\, \Gamma\Big(\frac{5-n}{2}\Big)\,\Big( 4m^2_0|v|^4\tau^2 -2 |v|^4 \Big)+ {\cal O}(|v|^6)\,, \nonumber \\ && \nonumber \\
\big[EDD\big] &\equiv& \int^{\infty}_{0}\frac{ds_1ds_2ds_3ds_4}{s_1+s_2+s_3+s_4}\, {\rm tr}\, \Big\langle x \Big| e^{-(s_1+s_4)B}E e^{-s_2B}D e^{-s_3A}D^{\dagger} + e^{-(s_1+s_4)B} De^{-s_2A}D^{\dagger} e^{-s_3B} E \nonumber \\
&& \qquad \qquad \qquad \qquad \qquad ~~~+ e^{-(s_1+s_4)A}D^{\dagger} e^{-s_2B}E e^{-s_3B}D \Big| x \Big\rangle \nonumber \\
&=& \frac{1}{m^6_0}\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}\, \Gamma\Big(\frac{5-n}{2}\Big)\, 4m^2_0|v|^4\tau^2 + {\cal O}(|v|^6)\,. \nonumber \end{eqnarray}
Finally, the fourth order one-loop effective potential is given by
\begin{eqnarray}
V^b_{1,\,4} &=& -\int^{\infty}_{0}\frac{\prod_{i=1}^5ds_i}{\sum_{i=1}^5s_i}\, \Big\langle x \Big| e^{-(s_1+s_5)A}D^{\dagger}e^{-s_2B}De^{-s_3A}D^{\dagger}e^{-s_4B}D + (A\leftrightarrow B\,, \, D\leftrightarrow D^{\dagger})\Big|x\Big\rangle \nonumber \\
&=&-\frac{1}{m^8_0}\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}\, \Gamma\Big(\frac{7-n}{2}\Big)\, \frac{4}{3}m^2_0|v|^4 + {\cal O}(|v|^6)\,. \end{eqnarray}
\subsection*{C. Useful Integrals}
In this appendix we collect all the nontrivial integral formulae used. Note that, after the $s_i$ integrations, we are left with the momentum integrals of the form:
\[
\int \frac{d^{n+1}p}{(2\pi)^{n+1}}\, \frac{p^m}{(p^2+m^2)^r} = \frac{1}{m^{2r-m}}\Big[\frac{m^2}{4\pi}\Big]^{\frac{n+1}{2}}\frac{\Gamma\big(\frac{n+m+1}{2}\big)\Gamma\big(r-\frac{n+m+1}{2}\big)}{\Gamma\big(\frac{n+1}{2}\big)\Gamma(r)}\,.
\]
\noindent Now we present various integral formulae for $s_i$ parameters.
\noindent \underline{\it Symmetric case}
\[ \int^{\infty}_0 \frac{ds_1\cdots ds_n}{(s_1+\cdots + s_n)^m}\, f(s_1+\cdots + s_n) = \frac{1}{\Gamma(n)} \int^{\infty}_{0} ds\, s^{n-m-1}f(s)\,. \]
\underline{\it Triple integrals over $s_i$ parameters}
\[ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \int^{\infty}_{0} \frac{ds_1ds_2ds_3}{s_1+s_2+s_3}\, e^{-(s_1+s_2+s_3)m^2} \Big[\cos mp(s_1+s_3)\,\cos mps_2 \Big] = \frac{1}{2}\frac{1}{(p^2+m^2)^2}
\,. \]
\[ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \int^{\infty}_{0} \frac{ds_1ds_2ds_3}{s_1+s_2+s_3}\, e^{-(s_1+s_2+s_3)m^2} \Big[\sin mp(s_1+s_3)\,\sin mps_2 \Big] = \frac{1}{2}\frac{p^2}{m^2}\frac{1}{(p^2+m^2)^2}
\,. \]
\[\!\!\!\!\! \int^{\infty}_{0} \frac{ds_1ds_2ds_3}{s_1+s_2+s_3}\, e^{-(s_1+s_2+s_3)m^2} \Big[ e^{-s_2p^2}\cos mp(s_1+s_3) + e^{-(s_1+s_3)p^2}\cos mps_2 \Big] = \frac{1}{(p^2+m^2)^2}
\,. \]
\[ \int^{\infty}_{0} \frac{ds_1ds_2ds_3}{s_1+s_2+s_3}\, e^{-(s_1+s_2+s_3)m^2} \Big[ 2\cos mp(s_1+s_2+s_3) + \frac{p}{m} \sin mp(s_1+s_2+s_3) \Big] = \frac{1}{(p^2+m^2)^2}
\,.\]
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \int^{\infty}_{0} \frac{ds_1ds_2ds_3}{s_1+s_2+s_3}\, e^{-(s_1+s_2+s_3)m^2} \bigg[ \Big\{ e^{-(s_1+s_3)p^2}\Big(2\cos mqs_2 + \frac{q}{m}\sin mqs_2\Big) \nonumber \\
&& \qquad \qquad \qquad\qquad \qquad +\, e^{-s_2q^2}\Big(2\cos mp(s_1+s_3) + \frac{p}{m} \sin mp(s_1+s_3)\Big) \Big\} + \Big\{ p \leftrightarrow q \Big\} \bigg] \nonumber \\
&& \qquad \qquad \qquad ~~~~~ =\, \frac{2}{(p^2+m^2)(q^2+m^2)} + \frac{1}{m^2}\Big[\frac{1}{p^2+m^2} + \frac{1}{q^2+m^2}\Big]
\,. \nonumber
\end{eqnarray}
\[ \int^{\infty}_{0} \frac{ds_1ds_2ds_3}{s_1+s_2+s_3}\, e^{-(s_1+s_2+s_3)m^2} \Big[ e^{-(s_1+s_3)p^2 -s_2q^2} + e^{-(s_1+s_3)q^2 -s_2p^2} \Big] = \frac{1}{(p^2+m^2)(q^2+m^2)}
\,. \]
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \int^{\infty}_{0} \frac{ds_1ds_2ds_3}{s_1+s_2+s_3}\, e^{-(s_1+s_2+s_3)m^2} \bigg[ \Big\{ \Big(2\cos mp(s_1+s_3) + \frac{p}{ m}\sin mp(s_1+s_3)\Big) \nonumber \\
&& \qquad \qquad \qquad \qquad \qquad \qquad \times \, \Big(2\cos mqs_2 + \frac{q}{m} \sin mqs_2 \Big) \Big\} + \Big\{ p \leftrightarrow q \Big\} \bigg] \nonumber \\
&& \qquad \qquad \qquad ~~~ =\, \frac{p^2q^2}{m^4}\frac{1}{(p^2+m^2)(q^2+m^2)} + \frac{2}{m^2}\Big[\frac{1}{p^2+m^2} + \frac{1}{q^2+m^2}\Big]
\,. \nonumber
\end{eqnarray}
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \int^{\infty}_{0} \frac{ds_1ds_2ds_3}{s_1+s_2+s_3}\, e^{-(s_1+s_2+s_3)m^2} \bigg[ \Big\{ \Big(2\sin mp(s_1+s_3) - \frac{p}{m}\cos mp(s_1+s_3)\Big) \nonumber \\
&& \qquad \qquad \qquad \qquad \qquad \qquad \times \, \Big(2\sin mqs_2 - \frac{q}{m} \cos mqs_2 \Big) \Big\} + \Big\{ p \leftrightarrow q \Big\} \bigg] \nonumber \\
&& \qquad \qquad \qquad ~~~ =\, \frac{pq}{m^2}\frac{1}{(p^2+m^2)(q^2+m^2)}
\,. \nonumber
\end{eqnarray}
\underline{\it Quadruple integrals over $s_i$ parameters}
\begin{eqnarray} &&\int^{\infty}_{0}\frac{\prod_{i=1}^{4}ds_i}{\sum_{i=1}^{4}s_i}e^{-m^2\sum_{i=1}^{4}s_i}\bigg\{ e^{-(s_2+s_3)p^2}\cos mp(s_1+s_4) + e^{-(s_1+s_3+s_4)p^2}\cos mps_2 \nonumber \\
&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ~~~ +\, e^{-(s_1+s_2+s_4)p^2} \cos mps_3 \bigg\} = \frac{1}{(p^2+m^2)^3}\,. \nonumber
\end{eqnarray}
\begin{eqnarray} &&\int^{\infty}_{0}\frac{\prod_{i=1}^{4}ds_i}{\sum_{i=1}^{4}s_i}e^{-m^2\sum_{i=1}^{4}s_i}\bigg\{ e^{-s_2 p^2}\Big[\cos mp(s_1+s_3+s_4) + \frac{p}{2m}\sin mp(s_1+s_3+s_4)\Big] \nonumber \\
&& \qquad \qquad \qquad ~~~~~ +\, e^{-s_3p^2}\Big[\cos mp(s_1+s_2+s_4) + \frac{p}{2m}\sin mp(s_1+s_2+s_4)\Big] \nonumber \\
&& \qquad \qquad \qquad ~~~~~ +\, e^{-(s_1+s_4)p^2}\Big[ \cos mp(s_2+s_3) + \frac{p}{2m}\sin mp(s_2+s_3)\Big] \bigg\} = \frac{1}{(p^2+m^2)^3}\,. \nonumber
\end{eqnarray}
\underline{\it Quintic integrals over $s_i$ parameters}
\begin{eqnarray} &&\int^{\infty}_{0}\frac{\prod_{i=1}^{5}ds_i}{\sum_{i=1}^{5}s_i}e^{-m^2_0\sum_{i=1}^{5}s_i}\bigg\{ \Big[ e^{-(s_1+s_5)p^2}\cos mps_3 + e^{-s_3p^2}\cos mp(s_1+s_5)\Big]e^{-(s_2+s_4)p^2} \nonumber \\
&& \qquad \qquad \qquad \qquad \qquad +\, \Big[ e^{-s_2p^2}\cos mps_4 + e^{-s_4p^2}\cos mps_2 \Big]e^{-(s_1+s_3+s_5)p^2} \bigg\} = \frac{1}{(p^2+m^2)^4}\,. \nonumber
\end{eqnarray}
\begin{eqnarray} &&\int^{\infty}_{0}\frac{\prod_{i=1}^{5}ds_i}{\sum_{i=1}^{5}s_i}e^{-m^2_0\sum_{i=1}^{5}s_i}\bigg\{ \Big[ \cos mps_3 \cdot \cos mp(s_1+s_5)\Big]e^{-(s_2+s_4)p^2} \nonumber \\
&& \qquad \qquad \qquad \qquad \qquad +\, \Big[ \cos mps_2 \cdot \cos mps_4 \Big]e^{-(s_1+s_3+s_5)p^2} \bigg\} = \frac{1}{2} \frac{1}{(p^2+m^2)^4}\,. \nonumber
\end{eqnarray}
\thebibliography{999}
\bibitem{Bagger:2007vi}
J.~Bagger and N.~Lambert,
``Comments On Multiple M2-branes,''
JHEP {\bf 0802}, 105 (2008)
[arXiv:0712.3738 [hep-th]].
\bibitem{Bagger:2007jr}
J.~Bagger and N.~Lambert,
``Gauge Symmetry and Supersymmetry of Multiple M2-Branes,''
Phys.\ Rev.\ D {\bf 77}, 065008 (2008)
[arXiv:0711.0955 [hep-th]].
\bibitem{Gustavsson:2007vu}
A.~Gustavsson,
``Algebraic structures on parallel M2-branes,''
[arXiv:0709.1260 [hep-th]].
\bibitem{Aharony:2008ug}
O.~Aharony, O.~Bergman, D.~L.~Jafferis and J.~Maldacena,
``N=6 superconformal Chern-Simons-matter theories, M2-branes and their
gravity duals,''
[arXiv:0806.1218 [hep-th]].
\bibitem{Bhattacharya:2008bja}
J.~Bhattacharya and S.~Minwalla,
``Superconformal Indices for ${\cal N}=6$ Chern Simons Theories,''
[arXiv:0806.3251 [hep-th]];
F.~A.~Dolan,
``On Superconformal Characters and Partition Functions in Three Dimensions,''
arXiv:0811.2740 [hep-th];
J.~Choi, S.~Lee and J.~Song,
``Superconformal Indices for Orbifold Chern-Simons Theories,''
[arXiv:0811.2855 [hep-th]].
\bibitem{Minahan:2008hf}
J.~A.~Minahan and K.~Zarembo,
``The Bethe ansatz for superconformal Chern-Simons,''
arXiv:0806.3951 [hep-th];
G.~Grignani, T.~Harmark and M.~Orselli,
``The SU(2) x SU(2) sector in the string dual of N=6 superconformal
Chern-Simons theory,''
arXiv:0806.4959 [hep-th];
G.~Grignani, T.~Harmark, M.~Orselli and G.~W.~Semenoff,
``Finite size Giant Magnons in the string dual of N=6 superconformal
Chern-Simons theory,''
JHEP {\bf 0812}, 008 (2008)
[arXiv:0807.0205 [hep-th]];
N.~Gromov and P.~Vieira,
``The AdS4/CFT3 algebraic curve,''
arXiv:0807.0437 [hep-th];
N.~Gromov and P.~Vieira,
``The all loop AdS4/CFT3 Bethe ansatz,''
arXiv:0807.0777 [hep-th];
D.~Astolfi, V.~G.~M.~Puletti, G.~Grignani, T.~Harmark and M.~Orselli,
``Finite-size corrections in the SU(2) x SU(2) sector of type IIA string
theory on $AdS_4 \times CP^3$,''
arXiv:0807.1527 [hep-th];
C.~Ahn and R.~I.~Nepomechie,
``N=6 super Chern-Simons theory S-matrix and all-loop Bethe ansatz
equations,''
[arXiv:0807.1924 [hep-th]];
D.~Bak and S.~J.~Rey,
``Integrable Spin Chain in Superconformal Chern-Simons Theory,''
[arXiv:0807.2063 [hep-th]];
B.~H.~Lee, K.~L.~Panigrahi and C.~Park,
JHEP {\bf 0811}, 066 (2008)
[arXiv:0807.2559 [hep-th]].
C.~Ahn, P.~Bozhilov and R.~C.~Rashkov,
``Neumann-Rosochatius integrable system for strings on $AdS_4 \times CP^3$,''
arXiv:0807.3134 [hep-th];
T.~McLoughlin and R.~Roiban,
``Spinning strings at one-loop in $AdS_4 \times P^3$,''
arXiv:0807.3965 [hep-th];
C.~Krishnan,
``AdS4/CFT3 at One Loop,''
JHEP {\bf 0809}, 092 (2008)
[arXiv:0807.4561 [hep-th]];
N.~Gromov and V.~Mikhaylov,
arXiv:0807.4897 [hep-th];
D.~Bak, D.~Gang and S.~J.~Rey,
``Integrable Spin Chain of Superconformal U(M)xU(N) Chern-Simons Theory,''
[arXiv:0808.0170 [hep-th]];
C.~Ahn and R.~I.~Nepomechie,
``An alternative S-matrix for N=6 Chern-Simons theory ?,''
[arXiv:0810.1915 [hep-th]];
C.~Ahn and P.~Bozhilov,
``Finite-size Effect of the Dyonic Giant Magnons in N=6 super Chern-Simons
Theory,''
[arXiv:0810.2079 [hep-th]];
C.~Ahn and P.~Bozhilov,
``M2-brane Perspective on N=6 Super Chern-Simons Theory at Level k,''
[arXiv:0810.2171 [hep-th]].
\bibitem{Drukker:2008zx}
N.~Drukker, J.~Plefka and D.~Young,
``Wilson loops in 3-dimensional N=6 supersymmetric Chern-Simons Theory and
their string theory duals,''
JHEP {\bf 0811} (2008) 019
[arXiv:0809.2787 [hep-th]].
~
B.~Chen and J.~B.~Wu,
``Supersymmetric Wilson Loops in N=6 Super Chern-Simons-matter theory,''
[arXiv:0809.2863 [hep-th]].
~
S.~J.~Rey, T.~Suyama and S.~Yamaguchi,
``Wilson Loops in Superconformal Chern-Simons Theory and Fundamental Strings
in Anti-de Sitter Supergravity Dual,''
arXiv:0809.3786 [hep-th].
\bibitem{Hosomichi:2008ip}
K.~Hosomichi, K.~M.~Lee, S.~Lee, S.~Lee, J.~Park and P.~Yi,
``A Nonperturbative Test of M2-Brane Theory,''
arXiv:0809.1771 [hep-th].
\bibitem{Honma:2008ef}
Y.~Honma, S.~Iso, Y.~Sumitomo, H.~Umetsu and S.~Zhang,
``Generalized Conformal Symmetry and Recovery of SO(8) in Multiple M2 and D2
Branes,''
arXiv:0807.3825 [hep-th].
~
Y.~Honma, S.~Iso, Y.~Sumitomo and S.~Zhang,
``Scaling limit of N=6 superconformal Chern-Simons theories and Lorentzian
Bagger-Lambert theories,''
Phys.\ Rev.\ D {\bf 78} (2008) 105011
[arXiv:0806.3498 [hep-th]].
~
E.~Antonyan and A.~A.~Tseytlin,
``On 3d N=8 Lorentzian BLG theory as a scaling limit of 3d superconformal N=6
ABJM theory,''
arXiv:0811.1540 [hep-th].
J.~Bagger and N.~Lambert,
``Three-Algebras and N=6 Chern-Simons Gauge Theories,''
arXiv:0807.0163 [hep-th].
S.~Cherkis and C.~Saemann,
Phys.\ Rev.\ D {\bf 78}, 066019 (2008)
[arXiv:0807.0808 [hep-th]].
\bibitem{Verlinde:2008di}
H.~Verlinde,
``D2 or M2? A Note on Membrane Scattering,''
arXiv:0807.2121 [hep-th].
\bibitem{Aharony:1996bh}
O.~Aharony and M.~Berkooz,
Nucl.\ Phys.\ B {\bf 491}, 184 (1997)
[arXiv:hep-th/9611215].
\bibitem{Lifschytz:1996rw}
G.~Lifschytz and S.~D.~Mathur,
``Supersymmetry and membrane interactions in M(atrix) theory,''
Nucl.\ Phys.\ B {\bf 507}, 621 (1997)
[arXiv:hep-th/9612087].
\bibitem{Berenstein:1997vm}
D.~Berenstein and R.~Corrado,
``M(atrix)-theory in various dimensions,''
Phys.\ Lett.\ B {\bf 406} (1997) 37
[arXiv:hep-th/9702108].
\bibitem{Polchinski:1997pz}
J.~Polchinski and P.~Pouliot,
``Membrane scattering with M-momentum transfer,''
Phys.\ Rev.\ D {\bf 56}, 6601 (1997)
[arXiv:hep-th/9704029].
\bibitem{Becker:1997xw}
K.~Becker, M.~Becker, J.~Polchinski and A.~A.~Tseytlin,
``Higher order graviton scattering in M(atrix) theory,''
Phys.\ Rev.\ D {\bf 56} (1997) 3174
[arXiv:hep-th/9706072].
\bibitem{Chepelev:1997fk}
I.~Chepelev and A.~A.~Tseytlin,
Nucl.\ Phys.\ B {\bf 515} (1998) 73
[arXiv:hep-th/9709087].
\bibitem{Kabat:1997im}
D.~N.~Kabat and W.~Taylor,
``Spherical membranes in matrix theory,''
Adv.\ Theor.\ Math.\ Phys.\ {\bf 2}, 181 (1998)
[arXiv:hep-th/9711078].
\bibitem{Paban:1998ea}
S.~Paban, S.~Sethi and M.~Stern,
``Constraints from extended supersymmetry in quantum mechanics,''
Nucl.\ Phys.\ B {\bf 534}, 137 (1998)
[arXiv:hep-th/9805018].
\bibitem{Paban:1998qy}
S.~Paban, S.~Sethi and M.~Stern,
``Supersymmetry and higher derivative terms in the effective action of
Yang-Mills theories,''
JHEP {\bf 9806}, 012 (1998)
[arXiv:hep-th/9806028].
\bibitem{Paban:1998mp}
S.~Paban, S.~Sethi and M.~Stern,
``Summing up instantons in three-dimensional Yang-Mills theories,''
Adv.\ Theor.\ Math.\ Phys.\ {\bf 3}, 343 (1999)
[arXiv:hep-th/9808119].
\bibitem{Hyun:1998qf}
S.~Hyun, Y.~Kiem and H.~Shin,
``Effective action for membrane dynamics in DLCQ M theory on a two-torus,''
Phys.\ Rev.\ D {\bf 59} (1999) 021901
[arXiv:hep-th/9808183].
\bibitem{Hyun:1999hf}
S.~Hyun, Y.~Kiem and H.~Shin,
``Supersymmetric completion of supersymmetric quantum mechanics,''
Nucl.\ Phys.\ B {\bf 558} (1999) 349
[arXiv:hep-th/9903022].
\bibitem{Dine:1997nq}
M.~Dine and N.~Seiberg,
``Comments on higher derivative operators in some SUSY field theories,''
Phys.\ Lett.\ B {\bf 409}, 239 (1997)
[arXiv:hep-th/9705057].
\bibitem{Berenstein:2008dc}
D.~Berenstein and D.~Trancanelli,
``Three-dimensional N=6 SCFT's and their membrane dynamics,''
arXiv:0808.2503 [hep-th].
\bibitem{Seiberg:1996bs}
N.~Seiberg,
``IR dynamics on branes and space-time geometry,''
Phys.\ Lett.\ B {\bf 384}, 81 (1996)
[arXiv:hep-th/9606017];
N.~Seiberg and E.~Witten,
``Gauge dynamics and compactification to three dimensions,''
arXiv:hep-th/9607163;
D.~E.~Diaconescu and R.~Entin,
``A non-renormalization theorem for the d = 1, N = 8 vector multiplet,''
Phys.\ Rev.\ D {\bf 56}, 8045 (1997)
[arXiv:hep-th/9706059].
\bibitem{Gaiotto:2008cg}
D.~Gaiotto, S.~Giombi and X.~Yin,
``Spin Chains in N=6 Superconformal Chern-Simons-Matter Theory,''
arXiv:0806.4589 [hep-th].
~
K.~Hosomichi, K.~M.~Lee, S.~Lee, S.~Lee and J.~Park,
``N=5,6 Superconformal Chern-Simons Theories and M2-branes on Orbifolds,''
arXiv:0806.4977 [hep-th].
~
S.~Terashima,
``On M5-branes in N=6 Membrane Action,''
JHEP {\bf 0808}, 080 (2008)
[arXiv:0807.0197 [hep-th]].
~
M.~A.~Bandres, A.~E.~Lipstein and J.~H.~Schwarz,
``Studies of the ABJM Theory in a Formulation with Manifest SU(4)
R-Symmetry,''
arXiv:0807.0880 [hep-th].
\bibitem{Ovrut:1981wa}
B.~A.~Ovrut and J.~Wess,
``Supersymmetric R(Xi) Gauge And Radiative Symmetry Breaking,''
Phys.\ Rev.\ D {\bf 25}, 409 (1982).
\bibitem{Benna:2008zy}
M.~Benna, I.~Klebanov, T.~Klose and M.~Smedback,
``Superconformal Chern-Simons Theories and $AdS_4/CFT_3$ Correspondence,''
arXiv:0806.1519 [hep-th].
\bibitem{Ball:1988xg}
R.~D.~Ball,
``Chiral Gauge Theory,''
Phys.\ Rept.\ {\bf 182}, 1 (1989).
\bibitem{Martin:1990xv}
C.~P.~Martin,
``DIMENSIONAL REGULARIZATION OF CHERN-SIMONS FIELD THEORY,''
Phys.\ Lett.\ B {\bf 241} (1990) 513.
\bibitem{Chen:1992ee}
W.~Chen, G.~W.~Semenoff and Y.~S.~Wu,
``Two loop analysis of nonAbelian Chern-Simons theory,''
Phys.\ Rev.\ D {\bf 46}, 5521 (1992)
[arXiv:hep-th/9209005].
\bibitem{Gates:1983nr}
S.~J.~Gates, M.~T.~Grisaru, M.~Rocek and W.~Siegel,
Front.\ Phys.\ {\bf 58}, 1 (1983)
[arXiv:hep-th/0108200].
\end{document}
N=4 setup--------------------------------------------------------------------
\bibitem{Gaiotto:2008sd}
D.~Gaiotto and E.~Witten,
``Janus Configurations, Chern-Simons Couplings, And The Theta-Angle in N=4
Super Yang-Mills Theory,''
arXiv:0804.2907 [hep-th].
\bibitem{Hosomichi:2008jd}
K.~Hosomichi, K.~M.~Lee, S.~Lee, S.~Lee and J.~Park,
``N=4 Superconformal Chern-Simons Theories with Hyper and Twisted Hyper
Multiplets,''
JHEP {\bf 0807} (2008) 091
[arXiv:0805.3662 [hep-th]].
-----------------------------------------------------------------------------
ABJM Susy transformation --------------------------------------------
-------------------------------------------------------------------------
Mass deformation----------------------------------------------
\bibitem{Gomis:2008cv}
J.~Gomis, A.~J.~Salim and F.~Passerini,
``Matrix Theory of Type IIB Plane Wave from Membranes,''
JHEP {\bf 0808} (2008) 002
[arXiv:0804.2186 [hep-th]].
\bibitem{Hosomichi:2008qk}
K.~Hosomichi, K.~M.~Lee and S.~Lee,
``Mass-Deformed Bagger-Lambert Theory and its BPS Objects,''
arXiv:0804.2519 [hep-th].
\bibitem{Bergshoeff:2008ix}
E.~A.~Bergshoeff, M.~de Roo, O.~Hohm and D.~Roest,
``Multiple Membranes from Gauged Supergravity,''
arXiv:0806.2584 [hep-th].
\bibitem{Gomis:2008vc}
J.~Gomis, D.~Rodriguez-Gomez, M.~Van Raamsdonk and H.~Verlinde,
``A Massive Study of M2-brane Proposals,''
arXiv:0807.1074 [hep-th].
----------------------------------------------------------------
Lorentzian BL --------------------------------------------------------------
\bibitem{Gomis:2008uv}
J.~Gomis, G.~Milanesi and J.~G.~Russo,
``Bagger-Lambert Theory for General Lie Algebras,''
JHEP {\bf 0806} (2008) 075
[arXiv:0805.1012 [hep-th]].
\bibitem{Benvenuti:2008bt}
S.~Benvenuti, D.~Rodriguez-Gomez, E.~Tonni and H.~Verlinde,
``N=8 superconformal gauge theories and M2 branes,''
arXiv:0805.1087 [hep-th].
\bibitem{Ho:2008ei}
P.~M.~Ho, Y.~Imamura and Y.~Matsuo,
``M2 to D2 revisited,''
JHEP {\bf 0807} (2008) 003
[arXiv:0805.1202 [hep-th]].
---------------------------------------------------------------------------
Spin chain -------------------------------------------------------
----------------------------------------------------------------------------
BL BPS states---------------------------------------------------------------
\bibitem{Jeon:2008zj}
I.~Jeon, J.~Kim, N.~Kim, B.~H.~Lee and J.~H.~Park,
``M-brane bound states and the supersymmetry of BPS solutions in the
Bagger-Lambert theory,''
arXiv:0809.0856 [hep-th].
\bibitem{Jeon:2008bx}
I.~Jeon, J.~Kim, N.~Kim, S.~W.~Kim and J.~H.~Park,
``Classification of the BPS states in Bagger-Lambert Theory,''
JHEP {\bf 0807}, 056 (2008)
[arXiv:0805.3236 [hep-th]].
--------------------------------------------------------------------------
Ghost Free (D2) action----------------------------------------------
\bibitem{Bandres:2008kj}
M.~A.~Bandres, A.~E.~Lipstein and J.~H.~Schwarz,
``Ghost-Free Superconformal Action for Multiple M2-Branes,''
JHEP {\bf 0807}, 117 (2008)
[arXiv:0806.0054 [hep-th]].
\bibitem{Gomis:2008be}
J.~Gomis, D.~Rodriguez-Gomez, M.~Van Raamsdonk and H.~Verlinde,
``Supersymmetric Yang-Mills Theory From Lorentzian Three-Algebras,''
arXiv:0806.0738 [hep-th].
-----------------------------------------------------------------------
BL on orbifold--------------------------------------------------
\bibitem{Distler:2008mk}
J.~Distler, S.~Mukhi, C.~Papageorgakis and M.~Van Raamsdonk,
``M2-branes on M-folds,''
JHEP {\bf 0805}, 038 (2008)
[arXiv:0804.1256 [hep-th]].
\bibitem{Lambert:2008et}
N.~Lambert and D.~Tong,
``Membranes on an Orbifold,''
Phys.\ Rev.\ Lett.\ {\bf 101} (2008) 041602
[arXiv:0804.1114 [hep-th]].
-----------------------------------------------------------------------
BL as $U(2)\times U(2)$------------------------------------------------
\bibitem{VanRaamsdonk:2008ft}
M.~Van Raamsdonk,
``Comments on the Bagger-Lambert theory and multiple M2-branes,''
JHEP {\bf 0805}, 105 (2008)
[arXiv:0803.3803 [hep-th]].
-------------------------------------------------------------------------
Muhki M2 D2------------------------------------------------------------
\bibitem{Mukhi:2008ux}
S.~Mukhi and C.~Papageorgakis,
``M2 to D2,''
JHEP {\bf 0805} (2008) 085
[arXiv:0803.3218 [hep-th]].
\bibitem{Ezhuthachan:2008ch}
B.~Ezhuthachan, S.~Mukhi and C.~Papageorgakis,
``D2 to D2,''
JHEP {\bf 0807} (2008) 041
[arXiv:0806.1639 [hep-th]].
----------------------------------------------------------------------
\bibitem{Bandres:2008vf}
M.~A.~Bandres, A.~E.~Lipstein and J.~H.~Schwarz,
``N = 8 Superconformal Chern--Simons Theories,''
JHEP {\bf 0805} (2008) 025
[arXiv:0803.3242 [hep-th]].
\bibitem{Gaiotto:2007qi}
D.~Gaiotto and X.~Yin,
``Notes on superconformal Chern-Simons-matter theories,''
JHEP {\bf 0708} (2007) 056
[arXiv:0704.3740 [hep-th]].
\bibitem{Schnabl:2008wj}
M.~Schnabl and Y.~Tachikawa,
``Classification of N=6 superconformal theories of ABJM type,''
arXiv:0807.1102 [hep-th].
\bibitem{Honma:2008jd}
Y.~Honma, S.~Iso, Y.~Sumitomo and S.~Zhang,
``Scaling limit of N=6 superconformal Chern-Simons theories and Lorentzian
Bagger-Lambert theories,''
arXiv:0806.3498 [hep-th].
\bibitem{Nishioka:2008gz}
T.~Nishioka and T.~Takayanagi,
``On Type IIA Penrose Limit and N=6 Chern-Simons Theories,''
JHEP {\bf 0808} (2008) 001
[arXiv:0806.3391 [hep-th]].
\bibitem{Ahn:2008ya}
C.~Ahn,
``Holographic Supergravity Dual to Three Dimensional N=2 Gauge Theory,''
JHEP {\bf 0808} (2008) 083
[arXiv:0806.1420 [hep-th]].
3-algebra classification:---------------------------------------
\bibitem{Papadopoulos:2008sk}
G.~Papadopoulos,
``M2-branes, 3-Lie Algebras and Plucker relations,''
JHEP {\bf 0805} (2008) 054
[arXiv:0804.2662 [hep-th]].
\bibitem{Gauntlett:2008uf}
J.~P.~Gauntlett and J.~B.~Gutowski,
``Constraining Maximally Supersymmetric Membrane Actions,''
arXiv:0804.3078 [hep-th].
-----------------------------------------------------------------
In the following, we present only the contribution from $Q^b_0 + Q^b_1$, because the other part gives the identical contribution.
That is, the one-loop effective potential of the bosonic part is given by
\[ V^{Total}_{1} = V_1 + \hat{V}_1 =2V_1\,. \]
For Fermions
Component form of ${\cal D}_0$ is
\begin{eqnarray}
{\cal D}_0 = \left(\begin{array}{cccc} -\partial^2_{\tau} +m^2_0 + \omega^2_g\tau^2 & -\partial_{\tau}\partial_1 + im_0 \partial_2 & -\partial_\tau\partial_2-im_0\partial_1 & 0 \\
-\partial_1\partial_\tau -im_0\partial_2 & -\partial^2_1 + m^2_0 + \omega^2_g\tau^2 & -\partial_1\partial_2 + im_0\partial_\tau & 0 \\
-\partial_2\partial_\tau + im_0\partial_1 & -\partial_2\partial_1 -im_0\partial_{\tau} & -\partial^2_2 +m^2_0 +\omega^2_2\tau^2 & 0 \\
0 & 0& 0& (-\Box + m^2_0)\delta^{A}_{B} \end{array}\right)\,. \end{eqnarray}
\[ Q_b + Q_{b\, 1} = ({\cal D}_0 + M)\,, \qquad (\hat{{\cal D}}_0+\hat{M}) + \cdots
,.\]
In $n+1$ dimensions,
\[ \delta^{\mu}_{~ \mu} = n+1\,, \qquad \qquad
\gamma^{\mu}\gamma_{\mu} = n+1\,. \]
&=& -\int^{\infty}_{0}\frac{ds_1ds_2ds_3}{s_1+s_2+s_3}\, e^{-(s_1+s_2+s_3)m^2_0} \int d^{n+1}x'\int \frac{d^{n+1}p}{(2\pi)^{n+1}}\frac{d^{n+1}p'}{(2\pi)^{n+1}} \nonumber \\
&& \qquad \qquad \qquad \qquad \qquad \Big[ \tau^2(\tau + \tau')^2\, e^{i(p-p')\cdot x'}\, e^{-(s_1+s_3)p^2-s_2p'^2}\Big] 16m^2_0|v|^4~ \nonumber \\
-\int^{\infty}_{0}\frac{ds_1ds_2ds_3}{s_1+s_2+s_3}\, e^{-(s_1+s_2+s_3)m^2_0}\bigg\{ \nonumber \\
&& \int \frac{d^{n+1}p}{(2\pi)^{n+1}} \Big[ \frac{2n}{n+1}e^{-(s_1+s_2+s_3)p^2} + \frac{1}{n+1}\Big\{e^{-s_2p^2}\cos m_0p(s_1+s_3) \nonumber \\
&& \qquad \qquad \qquad \qquad \qquad ~~~~~~~~~~~~ +\, e^{-(s_1+s_3)p^2}\cos m_0ps_2 \Big\} \Big]\, 4(m_0|v|^2 +|v|^4\tau^2) \bigg\} \nonumber \\
\begin{eqnarray}
V^b_{1,\, 2} &=&
-\int^{\infty}_{0}\frac{ds_1ds_2ds_3}{s_1+s_2+s_3}\, e^{-(s_1+s_2+s_3)m^2_0}\bigg\{ \nonumber \\
&& +\int \frac{d^{n+1}p}{(2\pi)^{n+1}} \Big[ \frac{2n}{n+1}e^{-(s_1+s_2+s_3)p^2} + \frac{1}{n+1}\Big\{e^{-s_2p^2}\cos m_0p(s_1+s_3) \nonumber \\
&& \qquad \qquad \qquad \qquad \qquad ~~~~~~~~~~~~ +\, e^{-(s_1+s_3)p^2}\cos m_0ps_2 \Big\} \Big]\, 4(m_0|v|^2 +|v|^4\tau^2) \nonumber \\
&& +\, \int d^{n+1}x'\int \frac{d^{n+1}p}{(2\pi)^{n+1}}\frac{d^{n+1}p'}{(2\pi)^{n+1}}\Big[ \tau^2(\tau + \tau')^2\, e^{i(p-p')\cdot x'}\, e^{-(s_1+s_3)p^2-s_2p'^2}\Big] 16m^2_0|v|^4~ \nonumber \\
&& +\,\int \frac{d^{n+1}p}{(2\pi)^{n+1}} \Big[ e^{-(s_1+s_2+s_3)p^2} +2 \Big(1-\frac{p^2}{4m^2_0}\Big)\cos (mp \sum_{i=1}^{3} s_i) + \frac{2p}{m_0}\sin (mp\sum_{i=1}^{3} s_i) \Big] 4m^2_0|v|^4\tau^4\nonumber \\
&& +\, \int d^{n+1}x'\int\frac{d^{n+1}p}{(2\pi)^{n+1}}\frac{d^{n+1}p'}{(2\pi)^{n+1}}\Big[\frac{1}{2} m^2_0|v|^4\tau^2\tau'^2\, I(p,p')~ e^{ix'\cdot (p'-p)} \Big]\,, \nonumber \\
&& +\,\int \frac{d^{n+1}p}{(2\pi)^{n+1}} \Big[e^{-(s_1+s_3)p^2}\cos m_0ps_2 + e^{-s_2p^2}\cos m_0p(s_1+s_3)\Big]\Big(-\frac{p^2}{m^2_0}\frac{n}{n+1}\Big)|v|^4\tau^2 \nonumber \\ && +\,\int \frac{d^{n+1}p}{(2\pi)^{n+1}} \Big[ \frac{1}{(n+1)^2}\, e^{-(s_1+s_2+s_3)p^2} + \frac{n}{(n+1)^2}\Big(e^{-(s_1+s_3)p^2}\cos m_0ps_2 + e^{-s_2p^2}\cos m_0p(s_1+s_3)\Big) \nonumber \\
&& ~~~~~~ + \frac{n^2}{(n+1)^2}\cos m_0p(s_1+s_3)\cos m_0ps_2\Big]\frac{|v|^4}{m^2_0} \nonumber \\
&& +\,\int \frac{d^{n+1}p}{(2\pi)^{n+1}}\Big[\frac{1}{n+1} e^{-(s_1+s_2+s_3)p^2}+ \frac{n}{n+1}\Big(\cos m_0p(s_1+s_2+s_3) \nonumber \\
&& \qquad \qquad ~~~ + \frac{p}{2m_0}\sin m_0p(s_1+s_2+s_3)\Big)\Big]\Big(-4\frac{|v|^4\tau^2}{m_0}\Big) ~~~ \bigg\}\end{eqnarray}
where
\begin{eqnarray}
I(p,p') &\equiv& 1 + \Big(\frac{p\cdot p'}{pp'}\Big)^2 + \bigg[5-2\frac{p\cdot p'}{m^2_0} +\Big(\frac{p\cdot p'}{pp'}\Big)^2\bigg]\frac{1}{(p^2+m^2_0)(p'^2+m^2_0)} \nonumber \\
&& +\frac{1}{m^2_0}\bigg[3-\Big(\frac{p\cdot p'}{pp'}\Big)^2\bigg] \Big(\frac{1}{p^2+m^2_0} + \frac{1}{p'^2+m^2_0}\Big)
\end{eqnarray}
Using the formulae given in the appendix, one can obtain
\begin{eqnarray} V^b_{1,\, 2}
&=&
-\frac{1}{m^4_0}\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}
\bigg[ \Gamma\Big(\frac{1-n}{2}\Big)\Big(|v|^4\tau^2 +m^2_0|v|^4\tau^4\Big)
\nonumber \\
&& \qquad \qquad \qquad +\, \Gamma\Big(\frac{3-n}{2}\Big)\Big( 4m_0|v|^2 + \frac{|v|^4}{2m^2_0} -|v|^4\tau^2+12 m^2_0|v|^4\tau^4\Big) \nonumber \\
&& \nonumber \\
&& \qquad \qquad \qquad +\Gamma\Big(\frac{5-n}{2}\Big)4|v|^4\tau^2 \bigg] + {\cal O}(|v|^6)\,,
\end{eqnarray}
\begin{eqnarray}
V^b_{1,\, 3} &=& \int^{\infty}_{0}\frac{ds_1ds_2ds_3ds_4}{s_1+s_2+s_3+s_4}\, e^{-(s_1+s_2+s_3+s_4)m^2_0} \nonumber \\
&& \qquad \qquad \int \frac{d^{n+1}p}{(2\pi)^{n+1}} \bigg[\Big(F_1(s_i,p) + F_2(s_i,p)\Big) 8m^2_0|v|^4\tau^2 - F_3(s_i,p)4|v|^4 \bigg]+ {\cal O}(|v|^6)\,, \end{eqnarray}
where $F_1$ and $F_2$ are given by
\begin{eqnarray}
F_1(s_i,p) &=& \frac{3}{n+1}\, e^{-(s_1+s_2+s_3+s_4)p^2} + \frac{n}{n+1}\Big[ e^{-(s_2+s_3)m^2_0}\cos m_0p(s_1+s_4) \nonumber \\
&& \qquad \qquad \qquad \qquad \qquad ~~~~~ + e^{-(s_1+s_3+s_4)p^2}\cos m_0ps_2 + e^{-(s_1+s_2+s_4)p^2}\cos m_0ps_3 \Big]\,. \nonumber \\
&& \nonumber \\
F_2(s_i,p) &=& \frac{3}{n+1}\, e^{-(s_1+s_2+s_3+s_4)p^2} \nonumber \\
&& +\, \frac{n}{n+1}\bigg[ e^{-s_2p^2}\Big(\cos m_0p(s_1+s_3+s_4) + \frac{p}{2m_0}\sin m_0p(s_1+s_3+s_4)\Big) \nonumber \\
&& \qquad \qquad +\, e^{-s_3p^2} \Big( \cos m_0p(s_1+s_2+s_4)
+ \frac{p}{2m_0}\sin m_0p (s_1+s_2+s_4) \Big) \nonumber \\
&& \qquad \qquad +\, e^{-(s_1+s_4)p^2}\Big(\cos m_0p(s_2+s_3) + \frac{p}{2m_0}\sin m_0p(s_2+s_3)\Big)\bigg]\,, \nonumber \\
F_3(s_i,p)&=& 3\Big(\frac{1}{n+1}\Big)^2 e^{-(s_1+s_2+s_3+s_4)p^2} + \frac{2n}{(n+1)^2}\Big(e^{-(s_2+s_3)p^2}\cos m_0p(s_1+s_4) \nonumber \\ && ~~~ + e^{-(s_1+s_3+s_4)p^2}\cos m_0ps_2 + e^{-(s_1+s_2+s_)p^2}\cos m_0ps_3\Big) \nonumber \\
&& ~~~ + \Big(\frac{n}{n+1}\Big)^2\Big(e^{-s_3p^2}\cos m_0p(s_1+s_4)\cos m_0ps_2 + e^{-(s_1+s_4)p^2}\cos m_0ps_2\cos m_0ps_3 \nonumber \\
&& ~~~ + e^{-s_2p^2}\cos m_0ps_3\cos m_0p(s_1+s_4)\Big)\,.
\end{eqnarray}
Using the integrals in the appendix, one can see that
\begin{equation}
V^b_{1,\, 3} = \frac{1}{m^6_0}\Big[\frac{m^2_0}{4\pi}\Big]^{\frac{n+1}{2}}\, \Gamma\Big(\frac{5-n}{2}\Big)\,\Big( 8m^2_0|v|^4\tau^2 -2 |v|^4 \Big)+ {\cal O}(|v|^6)\,.
\end{equation}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 9,785
|
Q: How to uninstall kerio vpn client? I want to delete this problem(uninstall kerio vpn client)
It appears every time I want to update/install an app
A: Kerio VPN client is not a application from the official Ubuntu mirrors. This implies, that you have installed this package by your self. From the Kerio VPN client installation guide, the Kerio package must be installed via dpkg -i and there are not official repos.
To remove the package, search for the installed package via dpkg
$ dpkg -l | grep -i kerio
The output will be something like:
ii kerio-control-vpnclient-###-linux 1.2.3 amd64
Then remove this package with:
$ sudo apt-get --purge --yes remove kerio-control-vpnclient-xxx
by replacing the package name with the output of dpkg -l.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 4,271
|
layout: "openstack"
page_title: "OpenStack: openstack_lb_pool_v1"
sidebar_current: "docs-openstack-resource-lb-pool-v1"
description: |-
Manages a V1 load balancer pool resource within OpenStack.
---
# openstack\_lb\_pool_v1
Manages a V1 load balancer pool resource within OpenStack.
## Example Usage
```
resource "openstack_lb_pool_v1" "pool_1" {
name = "tf_test_lb_pool"
protocol = "HTTP"
subnet_id = "12345"
lb_method = "ROUND_ROBIN"
monitor_ids = ["67890"]
}
```
## Complete Load Balancing Stack Example
```
resource "openstack_networking_network_v2" "network_1" {
name = "network_1"
admin_state_up = "true"
}
resource "openstack_networking_subnet_v2" "subnet_1" {
network_id = "${openstack_networking_network_v2.network_1.id}"
cidr = "192.168.199.0/24"
ip_version = 4
}
resource "openstack_compute_secgroup_v2" "secgroup_1" {
name = "secgroup_1"
description = "Rules for secgroup_1"
rule {
from_port = -1
to_port = -1
ip_protocol = "icmp"
cidr = "0.0.0.0/0"
}
rule {
from_port = 80
to_port = 80
ip_protocol = "tcp"
cidr = "0.0.0.0/0"
}
}
resource "openstack_compute_instance_v2" "instance_1" {
name = "instance_1"
security_groups = ["default", "${openstack_compute_secgroup_v2.secgroup_1.name}"]
network {
uuid = "${openstack_networking_network_v2.network_1.id}"
}
}
resource "openstack_compute_instance_v2" "instance_2" {
name = "instance_2"
security_groups = ["default", "${openstack_compute_secgroup_v2.secgroup_1.name}"]
network {
uuid = "${openstack_networking_network_v2.network_1.id}"
}
}
resource "openstack_lb_monitor_v1" "monitor_1" {
type = "TCP"
delay = 30
timeout = 5
max_retries = 3
admin_state_up = "true"
}
resource "openstack_lb_pool_v1" "pool_1" {
name = "pool_1"
protocol = "TCP"
subnet_id = "${openstack_networking_subnet_v2.subnet_1.id}"
lb_method = "ROUND_ROBIN"
monitor_ids = ["${openstack_lb_monitor_v1.monitor_1.id}"]
}
resource "openstack_lb_member_v1" "member_1" {
pool_id = "${openstack_lb_pool_v1.pool_1.id}"
address = "${openstack_compute_instance_v2.instance_1.access_ip_v4}"
port = 80
}
resource "openstack_lb_member_v1" "member_2" {
pool_id = "${openstack_lb_pool_v1.pool_1.id}"
address = "${openstack_compute_instance_v2.instance_2.access_ip_v4}"
port = 80
}
resource "openstack_lb_vip_v1" "vip_1" {
name = "vip_1"
subnet_id = "${openstack_networking_subnet_v2.subnet_1.id}"
protocol = "TCP"
port = 80
pool_id = "${openstack_lb_pool_v1.pool_1.id}"
}
```
## Argument Reference
The following arguments are supported:
* `region` - (Required) The region in which to obtain the V2 Networking client.
A Networking client is needed to create an LB pool. If omitted, the
`OS_REGION_NAME` environment variable is used. Changing this creates a new
LB pool.
* `name` - (Required) The name of the pool. Changing this updates the name of
the existing pool.
* `protocol` - (Required) The protocol used by the pool members, you can use
either 'TCP, 'HTTP', or 'HTTPS'. Changing this creates a new pool.
* `subnet_id` - (Required) The network on which the members of the pool will be
located. Only members that are on this network can be added to the pool.
Changing this creates a new pool.
* `lb_method` - (Required) The algorithm used to distribute load between the
members of the pool. The current specification supports 'ROUND_ROBIN' and
'LEAST_CONNECTIONS' as valid values for this attribute.
* `tenant_id` - (Optional) The owner of the pool. Required if admin wants to
create a pool member for another tenant. Changing this creates a new pool.
* `monitor_ids` - (Optional) A list of IDs of monitors to associate with the
pool.
* `member` - (Optional) An existing node to add to the pool. Changing this
updates the members of the pool. The member object structure is documented
below. Please note that the `member` block is deprecated in favor of the
`openstack_lb_member_v1` resource.
The `member` block supports:
* `address` - (Required) The IP address of the member. Changing this creates a
new member.
* `port` - (Required) An integer representing the port on which the member is
hosted. Changing this creates a new member.
* `admin_state_up` - (Required) The administrative state of the member.
Acceptable values are 'true' and 'false'. Changing this value updates the
state of the existing member.
* `tenant_id` - (Optional) The owner of the member. Required if admin wants to
create a pool member for another tenant. Changing this creates a new member.
## Attributes Reference
The following attributes are exported:
* `region` - See Argument Reference above.
* `name` - See Argument Reference above.
* `protocol` - See Argument Reference above.
* `subnet_id` - See Argument Reference above.
* `lb_method` - See Argument Reference above.
* `tenant_id` - See Argument Reference above.
* `monitor_id` - See Argument Reference above.
* `member` - See Argument Reference above.
## Notes
The `member` block is deprecated in favor of the `openstack_lb_member_v1` resource.
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 9,097
|
Arab Spring: Ten years on, has Tunisia's uprising delivered on its promises?
Tunisians had high hopes when they toppled the Ben Ali regime, but some say the corruption now is worse than before the revolution
A picture of Mohamed Bouazizi, a street vendor who set himself alight 10 years ago on 17 December 2010, is displayed on the post office building in Sidi Bouzid, Tunisia on 8 December 2020 (Reuters)
Massinissa Benlakehal
Published date: 17 December 2020 10:28 UTC | Last update: 2 years 1 month ago
On 17 December 2010, Mohamed Bouazizi, a 26-year-old street vendor from the central Tunisian town of Sidi Bouzid, set himself on fire in protest against the corruption of his local police force, which regularly humiliated him and confiscated his wares.
The act of defiance, which cost Bouazizi his life, became the catalyst for an uprising that took down President Zine El Abidine Ben Ali after 23 years of repressive rule, and spread across North Africa and the Middle East in what would become known as the "Arab Spring".
From uprisings to political crisis: How Tunisia's democracy came to be at risk
As Tunisia marks a decade since the historic popular movement, the anniversary is bittersweet for many Tunisians who say little has changed. Despite elections, successive governments and a slew of political promises, many denounce the country's political elite.
Wahid Benchahla, 37, is an artist who created the "Tunisian Association of Young Artists" with a group of friends.
"Judging what we have experienced during the last ten years, it is obvious that the country's political class has significantly proved it does not really care about the population's interests and isn't qualified to do the job," he tells Middle East Eye. "It is very disappointing and frustrating."
Like many of his peers in the post-revolution era, corruption - which was already widespread under Ben Ali - is a phenomenon he faces regularly, whether in his daily life or in administrative procedures.
"Things that are usually very simple to have or do, like obtaining a document at the municipality, feel like climbing a mountain because of the bribery phenomenon," he says.
Benchahla is even categorical on this point: "Corruption is worse than before the revolution."
Amid high unemployment, mounting debt, unfulfilled reforms and the looming threat of the pandemic, some Tunisians tell MEE of their disillusionment.
At an impasse
A decade later, many in the country see the events that started on 17 December 2010 as an unfinished revolution, with the country's growing democracy inching along over the years but now stuck at an impasse, Nadia Mesghouni, a Tunisian geopolitical analyst, tells MEE.
"After ten years, we still can't see our way out considering how things have been managed and dealt with by the different governments we have seen, and it doesn't look like the situation is going to improve soon," she says.
The Tunisian Forum for Economic and Social Rights (FTDES) has recorded 6,500 protests in the country since the beginning of 2020, motivated by economic, social and environmental demands.
The civil society movement, which was created in 2011, also estimates that 12,500 Tunisians crossed the Mediterranean Sea into Italy while another 10,000 were intercepted attempting to cross, since the beginning of the year, seeking better opportunities in Europe despite the risks.
People queue as they wait to get cooking gas outside a main outlet in Sidi Bouzid, Tunisia, on 8 December 2020 (Reuters)
During the last few months alone, the country has seen several movements of social unrest erupting across different regions, mainly triggered by unemployment, a struggling economy, rampant inflation and growing public debt.
Public anger comes amid a significant wave of undocumented migration from Tunisia to Italy, explained in large part by the worsening economic, social and political crisis in the country.
"We face the impatience of people who have demands, while the government has no programme to deal with a broken down and unproductive economy, with a social and political climate that is dominated by partisan interests and fraudulent consensus," FTDES said in a statement.
"It is evident in the broadening of the number of people claiming their rights and the diversity of their forms of protest."
FTDES explains that the ongoing crisis affects every category within Tunisian society, whether fresh graduates, doctorate holders, public health agents, young doctors, curators, engineers, journalists, judges, sailors or farmers.
Crippling corruption
Amid Tunisia's transition to democracy, corruption has been a destabilising force in the North African country, infecting all levels of its security, economy and political system.
"I've seen and experienced many situations where bribery was a must, whether directly or indirectly," Benchahla tells MEE. One example he cites is the day he wanted to install an internet line at his home in downtown Tunis.
"I went to the public phone company to purchase an internet line, which the company says it would take them two weeks to come and install," he recalls. "One of the technicians told me that if I gave him some cash, he would come and install it the next day."
Between condolences and 'good riddance,' Tunisians reflect on Ben Ali's death
The latest major case, revealed by a local news channel in early November, is that of the import of waste from Italy by a Tunisian company. The still-unidentified company imported 282 containers of waste in Sousse, south of Tunis, including 212 containers of household waste, without authorisation, according to official statements by the Ministry of Environment.
The ministry launched an administrative investigation, as it is believed government officials within the department were involved.
In October, a Swiss consultancy firm published the results of a survey it conducted targeting law firms, banks, industrialists, transcontinental companies, decision-makers and governmental organisations in Tunisia.
Stratege Consulting's survey shows that 59 percent of those polled say the government doesn't make enough efforts to reduce corruption.
"Only one in ten Tunisians considers the state's fight against corruption to be promising, which is an insufficient percentage," the firm's executive director, Sami Jalouli, told the press.
In July 2018, the Tunisian parliament approved a law aimed at combating illicit enrichment as part of the country's efforts to fight against corruption.
A year prior, the government had launched a significant operation arresting twenty businessmen on suspicion of corruption, confiscating their property and freezing their bank accounts.
This was the first time such a major operation was conducted by the authorities to demonstrate their good faith.
In the first year after Ben Ali's ouster, the transitional government set up an anti-corruption committee, but it lacked the resources to effectively achieve its aims.
A Tunisian employee of the prime ministry removes a portrait of former Tunisian President Zine El Abidine Ben Ali on 17 January 2011 (AFP)
Today, more than ever, dealing with such a phenomenon requires youth-led activism and broad and sustained public support, the International Centre for Transitional Justice legal officer Mohamed A Zouari, has argued.
Tunisia was ranked 74 out of 198 countries in the 2019 Corruption Perceptions Index compiled by Transparency International. The organisation attributed the country's trouble with corruption to the weakness of state institutions and laxity of authorities to end this phenomenon.
Eight governments in 10 years
Since 2011, Tunisia has seen eight successive governments, including two interim ones, with an average lifespan of one year each, due to fragmented politics and a lack of a clear majority. The government of Prime Minister Youssef Chahed was the only exception, lasting for three-and-a-half years between August 2016 and January 2020.
While the North African country may have been the birthplace of a regional movement for change, few reforms have succeeded in getting it on track.
'Thinking of harqa': Why young Tunisians are still leaving, eight years after Ben Ali
Government instability has contributed to weaken the chain of command within the ministries necessary to implement reforms, says Michael Ayari, a senior analyst at the International Crisis Group for Tunisia and Algeria. International partners, he explains, have complained of the frequent change of interlocutors and today they would seem to prefer stability.
The country's economy continues to hold up thanks to foreign donations and loans obtained since 2011 – but most economic prospects present grim forecasts for the near and distant future of the country's economy, according to Ayari.
Unlike its neighbours, Tunisia doesn't hold abundant natural resources. One-third of its principal mineral resource, phosphate, is exported but fails to provide a significant enough source of income.
Prime Minister Hichem Mechichi told Parliament in September that his government's priorities would be "to restore the normal pattern of oil, gas and phosphate production" to pre-November levels when a protest movement decided to block production over unemployment and the general social and economic situation.
Risk of shocks
Since 2017, the government had failed to resolve its dispute with protesters in southern Tunisia's El-Kamour, until earlier this month, when it finally signed an agreement with the group's representatives to put an end to the blockade.
Meanwhile, the clock is ticking for Tunis. Between 2020 and 2021, Tunisia will be required to start reimbursing 123 external loans dating from 2012 to 2016, according to a report released last March by the country's accounting office authority.
Between 2021 and 2025, the accounting authority said reimbursements would reach around $1bn a year, and that Tunisia could be paying back these loans well into 2055.
Meanwhile, "public debt, the majority external (70 percent), increased by 95 percent between 2010 and 2019, placing Tunisia at risk of serious shocks and reducing liquidity available to the private sector", the African Development Bank (ADB) warned.
Unemployment is also a point of concern. According to the institution's data, unemployment among people ages 15 to 24 is 34.3 percent. The poverty rate, which dropped from 20.5 percent in 2010 to 15.2 percent in 2016, later significantly increased due to increased living costs, the ADB reported.
Continuing inequalities, the multilateral development finance institution says, are destabilising the social climate and impeding investment and growth.
Second wave of Covid-19
As Tunisia's political consensus cracks, IMF austerity may hit the rocks
The global coronavirus pandemic has only made things worse for the recent government that took office on 2 September.
The country's doctors, interns and residents in medicine launched a large protest movement to alert people to their working conditions amid a second wave of the Covid-19 pandemic that has further strained hospitals struggling to cope with the rise in cases amid a lack of adequate equipment.
The economic effects of the pandemic will put Tunisia in even more dire straits vis-a-vis its debt, the International Monetary Fund (IMF) warned.
"Tunisia's debt burden will increase significantly as the country grapples with the Covid-19 shock, reflecting the steep fall in growth and the deterioration of the primary fiscal balance as a result of lower revenues and the crisis-response measures," the international institution wrote, adding that public and external debt were expected to reach 89 and 110 percent of the GDP in 2020 respectively.
"Once the Covid-19 crisis abates," the IMF added, "the Tunisian government will take measures to improve Tunisia's fiscal accounts, including through policies to further reduce energy subsidies in a socially conscious way and to contain the civil service wage bill that remains among the highest in the world."
Tunisian demonstrators shout slogans on 20 January 2011 in Sidi Bouzid (AFP)
As uncertainty persists ahead of the celebration of the 10th anniversary, many observers expect an intensification of protest movements ahead of 14 January, the anniversary of Ben Ali's ouster.
Videographer Malek Abdelrahman was full of hope when he participated in and documented the protests that started in December 2010. Today, he says frustration and disappointment permeate Tunisian society.
"You see it and smell it when you take a walk in the neighbourhoods and markets, as life gets more expensive and living costs increase every year," he told MEE. "As seen at this time of the year for the last ten years, we'll have a lot of protests across the country as 14 January approaches.
Egypt Turmoil
Burying our loved ones does not mean burying the truth': A father's account of the Rabaa massacre
Ahmed Abd el Aziz
How Rabaa and its symbol changed Turkish-Egyptian relations
Inside Egypt
Why Egypt's civilian opposition to blame for seven decades of military rule
Abdel-Fattah Mady
Tunisia Politics
Beji Caid Essebsi: Skillful, polarising, freely elected
Coups, fake footage and proxy parties: What is the UAE doing in Tunisia?
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 1,453
|
\section{Introduction}
Hyperproperties are system properties that relate multiple execution traces of a system \cite{ClarksonS08} and commonly arise, e.g., in information-flow policies \cite{McCullough88}, the verification of code optimizations \cite{BarrettFGHPZ05}, and robustness of software \cite{ChaudhuriGL12}.
Consequently, many methods for the automated verification of hyperproperties have been developed \cite{ShemerGSV19,UnnoTK21,FarzanV19,SousaD16}.
Almost all previous approaches verify a class of hyperproperties called $k$-safety, i.e., properties that stipulate the absence of a bad interaction between any $k$ traces in the system.
For example, we can express a simple form of non-interference as a $2$-safety property by stating that any \emph{two} traces that agree on the low-security inputs should produce the same observable output.
The vast landscape of hyperproperties does, however, stretch far beyond $k$-safety.
The overarching limitation of $k$-safety (or, more generally, of hypersafety \cite{ClarksonS08}) is an implicit \emph{universal} quantification over all executions.
By contrast, many properties of interest, ranging from applications in information-flow control to robust cleanness, require a combination of universal and existential quantification.
For example, consider the reactive program in \Cref{fig:gniex}, where $\star_{\mathbb{N}}$ denotes a nondeterministic choice of a natural number.
We assume that $h$, $l$, and $o$ are a high-security input, a low-security input, and a low-security output, respectively.
This program violates the simple $2$-safety non-interference property given above as the non-determinism influences the output.
Nevertheless, the program is ``secure'' in the sense that an attacker that observes low-security inputs and outputs cannot deduce information about the high-security input.
To capture this formally, we use a relaxed notion of non-interference, in the literature often referred to as generalized non-interference (GNI) \cite{McCullough88}.
We can, informally, express GNI in a temporal logic as follows:
\begin{align*}
\forall \pi. \forall \pi'. \exists \pi''\mathpunct{.} \LTLglobally \big( o_\pi = o_{\pi''} \land l_\pi = l_{\pi''} \land h_{\pi'} = h_{\pi''} \big)
\end{align*}
This property requires that for any two traces $\pi, \pi'$, there exists some trace $\pi''$ that, globally, agrees with the low-security inputs and outputs on $\pi$ but the high-security inputs on $\pi'$.
Phrased differently, any observation on the low-security input-output behavior is compatible with every possible high-security input.
The program in \Cref{fig:gniex} satisfies GNI.
Crucially, GNI is no longer a hypersafety property (and, in particular, no $k$-safety property for any $k$) as it requires a combination of universal and \emph{existential} quantification.
\setlength{\columnsep}{0pt}%
\setlength{\intextsep}{8pt}%
\begin{wrapfigure}{r}{0.25\textwidth}
\vspace{-5mm}
\small
\begin{algorithmic}
\RepeatFive
\State \textbf{readInput}($h, l$)
\IfOne{$h > l$}
\State $o \leftarrow l + \star_{\mathbb{N}}$
\ElseThree
\State $x \leftarrow \star_{\mathbb{N}}$
\IfOne{$x \geq l$}
\State $o \leftarrow x$
\ElseOne
\State $o \leftarrow l$
\end{algorithmic}
\vspace{-1mm}
\caption{}
\label{fig:gniex}
\end{wrapfigure}
\subsection{Verification Beyond $k$-Safety}
Instead, GNI falls in the general class of $\forall^*\exists^*$ properties.
Concretely, a $\forall^k\exists^l$ safety property (using $k$ universal and $l$ existential quantifiers) stipulates that for any $k$ traces, there exist $l$ traces such that the resulting $k+l$ traces do not interact badly.
$k$-safety properties are the \emph{special case} where $l = 0$.
We study the verification of such properties in infinite-state systems arising, e.g., in software.
In contrast to $k$-safety, where a broad range of methods has been developed \cite{ShemerGSV19,FarzanV19,SousaD16,Benton04,UnnoTK21}, no method for the automated verification of \emph{temporal} $\forall^*\exists^*$ properties in infinite-state systems exists (we discuss related approaches in \Cref{sec:relatedWork}).
Our novel verification method is based on a game-based reading of existential quantification \emph{combined} with the search for a program reduction.
The game-based reading of existential quantification instantiates existential trace quantification with an explicit strategy and constitutes the first practicable method for the verification of $\forall^*\exists^*$-properties in finite-state systems \cite{CoenenFST19}.
Program reductions are a well-established technique to align executions of independent program fragments (such as the individual program copies in a self-composition) to obtain proofs with easier invariants \cite{Lipton75,ShemerGSV19,FarzanV19}.
So far, both techniques are limited to their respective domain, i.e., the game-based approach has only been applied to finite-state systems and synchronous specifications, and reductions have (mostly) been used for the verification of $k$-safety.
We combine both techniques yielding an effective (and first) verification technique for hyperproperties beyond $k$-safety in infinite-state systems arising in software.
Notably, our search for reduction and strategy-based instantiation of existential quantification is \emph{mutually dependent}, i.e., a particular strategy might depend on a particular reduction and vice versa.
\subsection{Contributions and Structure}
The starting point of our work is a new temporal logic called \emph{Observation-based HyperLTL} (OHyperLTL for short).
Our logic extends the existing hyperlogic HyperLTL \cite{ClarksonFKMRS14} with capabilities to reason about asynchronous properties (i.e., properties where the individual traces are traversed at different speeds), and to specify properties using assertions from arbitrary background theories (to reason about the infinite domains encountered in software) (\Cref{sec:ohyperltl}).
To automatically verify $\forall^k\exists^l$ OHyperLTL properties, we combine program reductions with a strategy-based instantiation of existential quantification; both in the context of a fixed predicate abstraction.
To facilitate this combination, we first present a game-based approach that automates the search for a reduction.
Concretely, we construct an abstract game where a winning strategy for the verifier directly corresponds to a reduction with accompanying proof.
As a side product, our game-based interpretation simplifies the search for a reduction in a given predicate abstraction as, e.g., studied by Shemer et al.~\cite{ShemerGSV19} (\Cref{sec:ksafety}).
Our strategic (game-based) view on reductions allows us to combine them with a game-based instantiation of existential quantification.
Here, we view the existentially quantified traces as being constructed by a strategy that, iteratively, reacts to the universally quantified traces.
As we phrase both the search for a reduction and the search for existentially quantified traces as a game, we can frame the search for both as a combined abstract game.
We prove the soundness of our approach, i.e., a winning strategy for the verifier constitutes both a strategy for the existentially quantified traces and accompanying (mutually dependent) reduction.
Despite its finite nature, constructing the abstract game is expensive as it involves many SMT queries.
We propose an inner refinement loop that determines the winner of the game (without constructing it explicitly) by computing iterative approximations (\Cref{sec:beyondksafety}).
We have implemented our verification approach in a prototype tool called \texttt{HyPA} (short for \textbf{Hy}perproperty Verification with \textbf{P}redicate \textbf{A}bstraction)
and evaluate \texttt{HyPA} on $k$-safety properties (that can already be handled by existing methods) and on $\forall^*\exists^*$ benchmarks that cannot be handled by any existing tool (\Cref{sec:eval}).
\paragraph{Contributions.}
In short, our contributions include the following:
\begin{itemize}[leftmargin=*]
\item We propose a temporal hyperlogic that can specify asynchronous hyperproperties in infinite-state systems;
\item We propose a game-based interpretation of a reduction (improving and simplifying previous methods for $k$-safety \cite{ShemerGSV19});
\item We combine a strategy-based instantiation of existentially quantified traces with the search for a reduction. This yields a flexible (and first) method for the verification of temporal $\forall^*\exists^*$ properties.
We propose an iterative method to solve the abstract game that avoids an expensive explicit construction;
\item We provide and evaluate a prototype implementation of our method.
\end{itemize}
\section{Overview: Reductions and Quantification as a Game}
\label{sec:overview}
Our verification approach hinges on the observation that we can express both a reduction and existential trace quantification as a game.
In this section, we provide an overview of our game-based interpretations.
We begin by outlining our game-based reading of a reduction (illustrating this in the simpler case of $k$-safety) in \Cref{sec:overview_ksafety} and then extend this to include a game-based interpretation of existential quantification in \Cref{sec:overview_beyond}.
\subsection{Reductions as a Game}
\label{sec:overview_ksafety}
Consider the two programs in \Cref{fig:univ} and the specification that both programs produce the same output (on initially identical values for $x$).
We can formalize this in our logic OHyperLTL (formally defined in \Cref{sec:ohyperltl}) as follows:
\begin{align*}
\forall^{\texttt{P1}} \pi_1 :(\mathit{pc} = 2). \; \forall^{\texttt{P2}} \pi_2 : (\mathit{pc} = 2). \; (x_{\pi_1} = x_{\pi_2}) \rightarrow \LTLglobally (x_{\pi_1} = x_{\pi_2})
\end{align*}
The property states that for all traces $\pi_1$ in \texttt{P1} and $\pi_2$ in \texttt{P2} the LTL specification $(x_{\pi_1} = x_{\pi_2}) \rightarrow \LTLglobally (x_{\pi_1} = x_{\pi_2})$ holds (where $x_\pi$ refers to the value of $x$ on trace $\pi$).
Additionally, the observation formula $\mathit{pc} = 2$ marks the positions at which the LTL property is evaluated:
We only observe a trace at steps where $\mathit{pc} = 2$ (i.e., where the program counter is at the output position).
\begin{figure}[t]
\begin{subfigure}{0.25\textwidth}
\scalebox{0.9}{
\begin{minipage}{\textwidth}
\begin{algorithmic}[1]
\RepeatThree
\State \textbf{print$(x)$}
\State $y \leftarrow 2 x$
\WhileTwo{$y > 0$}
\State $y \leftarrow y - 1$
\State $x \leftarrow 2x$
\end{algorithmic}
\end{minipage}
}
\subcaption{Program \texttt{P1}}\label{fig:p1}
\end{subfigure}%
\begin{subfigure}{0.25\textwidth}
\scalebox{0.9}{
\begin{minipage}{\textwidth}
\begin{algorithmic}[1]
\RepeatThree
\State \textbf{print$(x)$}
\State $y \leftarrow x$
\WhileTwo{$y > 0$}
\State $y \leftarrow y - 1$
\State $x \leftarrow 4x$
\end{algorithmic}
\end{minipage}
}
\subcaption{Program \texttt{P2}}\label{fig:p2}
\end{subfigure}%
\begin{subfigure}{0.5\textwidth}
\centering
\scalebox{0.8}{
\small
\begin{tikzpicture}[scale=0.9]
\node[draw, rectangle, thick,align=center, rounded corners=3pt,label={[label distance=-0.5mm]west:{\scriptsize$\{1,2\}$}}] at (2.5,2) (n0) {$(2,2)$\\ $x_1 = x_2$};
\node[draw, rectangle, thick,align=center, rounded corners=3pt,label={[label distance=-0.5mm]west:{\scriptsize$\{1,2\}$}}] at (2.5,0) (n1) {$(4,4)$\\ $x_1 = x_2$\\$y_1 = 2 y_2$};
\node[draw, rectangle, thick,align=center, rounded corners=3pt,label={[label distance=-0.5mm]north:{\scriptsize$\{1,2\}$}}] at (5,0) (n2) {$(5,5)$\\ $x_1 = x_2$\\$y_1 = 2 y_2$};
\node[draw, rectangle, thick,align=center, rounded corners=3pt,label={[label distance=-0.5mm]north:{\scriptsize$\{1,2\}$}}] at (7.5,0) (n3) {$(6,6)$\\ $x_1 = x_2$\\$y_1 = 2 y_2 + 1$};
\node[draw, rectangle, thick,align=center, rounded corners=3pt,label={[label distance=-0.5mm]east:{\scriptsize$\{1,2\}$}}] at (5,2) (n4) {$(3,3)$\\ $x_1 = x_2$};
\node[draw, rectangle, thick,align=center, rounded corners=3pt,label={[label distance=-0.5mm]west:{\scriptsize$\{1\}$}}] at (2.5,0-2) (n5) {$(6,4)$\\ $x_1 = 2x_2$\\$y_1 = 2 y_2$};
\node[draw, rectangle, thick,align=center, rounded corners=3pt,label={[label distance=-0.5mm]north:{\scriptsize$\{1\}$}}] at (5,-2) (n6) {$(5,4)$\\ $x_1 = 2x_2$\\$y_1 = 2 y_2 + 1$};
\node[draw, rectangle, thick,align=center, rounded corners=3pt,label={[label distance=-0.5mm]east:{\scriptsize$\{1\}$}}] at (7.5,-2) (n7) {$(4,4)$\\ $x_1 = 2x_2$\\$y_1 = 2 y_2 + 1$};
\draw[->, thick] (n1) -- (n2);
\draw[->, thick] (n2) -- (n3);
\draw[->, thick] (n3) -- (n7);
\draw[->, thick] (n7) -- (n6);
\draw[->, thick] (n6) -- (n5);
\draw[->, thick] (n5) --(n1);
\draw[->, thick] (n0) -- (n4);
\draw[->, thick] (n4) -- (n1);
\draw[->, thick] (n1) -- (n0);
\draw[->, thick] (n0) + (-0.8, -0.8) -- (n0);
\end{tikzpicture}
}
\subcaption{Winning strategy for the verifier.} \label{fig:winningStrat}
\end{subfigure}
\caption{Two output-equivalent programs \texttt{P1} and \texttt{P2} in \Cref{fig:p1,fig:p2}. In \Cref{fig:winningStrat} a possible winning strategy for the verifier is given.
Each abstract state contains the value of the program counter of both copies (given as the pair at the top) and the predicates that hold in that state.
For sake of readability we omit the trace variables and write, e.g., $x_1$ for $x_{\pi_1}$.
We mark the initial state with an incoming arrow.
The outer label at each state gives the scheduling $M \subseteq \{1, 2\}$ chosen by the strategy in that state.\vspace{-0.3cm}}\label{fig:univ}
\end{figure}
The verification of our property involves reasoning about two copies of our system (in this case, one of \texttt{P1} and one of \texttt{P2}) on \emph{disjoint} state spaces.
Consequently, we can interleave the statements of both programs (between two observation points) without affecting the behavior of the individual copies.
We refer to each interleaving of both copies as a \emph{reduction}.
As is well known, the choice of a reduction drastically influences the complexity of the needed invariants \cite{FarzanV19,ShemerGSV19,Lipton75}.
Given an initial abstraction of the system \cite{GrafS97,ShemerGSV19}, we aim to discover a suitable reduction \emph{automatically}.
Our first observation is that we can phrase the search for a reduction as a game as follows:
In each step, the verifier decides on a \emph{scheduling} (i.e., a non-empty subset $M \subseteq \{1, 2\}$) that indicates which of the copies should take a step (i.e., $i \in M$ iff copy $i$ should make a program step).
Afterward, the refuter can choose an abstract successor state compatible with that scheduling, after which the process repeats.
This naturally defines a finite-state two-player safety game that we can solve efficiently\footnote{The LTL specification is translated to a symbolic safety automaton that moves alongside the game. For sake of readability, we omitted the automaton from the following discussion.}.
If the verifier wins, a winning strategy directly corresponds to a reduction and accompanying inductive invariant for the safety property within the given abstraction.
For our example, we give (parts of) a possible winning strategy in \Cref{fig:winningStrat}.
In each abstract state, the strategy chooses a scheduling (written next to the state), and all abstract states compatible with that scheduling are listed as successors.
Note that whenever the program counter is $(2,2)$ (i.e., both programs are at their output position), it holds that $x_1 = x_2$ (as required).
The example strategy schedules in lock-step for the most part (by choosing $M = \{1, 2\}$), but lets \texttt{P1} take the inner loop \emph{twice}, thereby maintaining the linear invariants $x_1 = x_2$ and $y_1 = 2y_2$.
In particular, the resulting reduction is property-based \cite{ShemerGSV19} as the scheduling is based on the current (abstract) state.
Note that the program cannot be verified with only linear invariants in a sequential or parallel (lock-step) reduction.
\subsection{Beyond $k$-Safety: Quantifiers as a Game}
\label{sec:overview_beyond}
We build upon this game-based interpretation of a reduction to move beyond $k$-safety.
As a second example, consider the two programs \texttt{Q1} and \texttt{Q2} in \Cref{fig:exist} (where $\star_\tau$ denotes a nondeterministic choice of type $\tau \in \{\mathbb{N}, \mathbb{B}\}$).
We wish to check that \texttt{Q1} refines \texttt{Q2}, i.e., all output behavior of \texttt{Q1} is also possible in \texttt{Q2}.
We can express this in our logic as follows:
\begin{align*}
\forall^{\texttt{Q1}} \pi_1 : (\mathit{pc} = 2). \; \exists^{\texttt{Q2}} \pi_2 : (\mathit{pc} = 2). \; \LTLglobally (a_{\pi_1} = a_{\pi_2})
\end{align*}%
The property states that for every trace $\pi_1$ in \texttt{Q1} there \emph{exists} a trace $\pi_2$ in \texttt{Q2} that outputs the same value.
The quantifiers range over infinite traces of variable assignments (with infinite domains), making a direct verification of the quantifier alternation challenging.
In contrast to alternation-free formulas, we cannot reduce the verification to verification on a self composition \cite{BartheDR11,FinkbeinerRS15}.
Instead, we adopt, yet another, game-based interpretation by viewing the existentially quantified execution as being resolved by a \emph{strategy} (called the witness strategy) \cite{CoenenFST19}.
That is, instead of trying to find a witness traces $\pi_2$ in \texttt{Q2} when given the \emph{entire} trace $\pi_1$, we interpret the $\forall\exists$ property as a game between verifier and refuter.
The refuter moves through the state space of \texttt{Q1} (thereby producing a trace $\pi_1$), and the verifier reacts to each move by choosing a successor for its state in the state space of \texttt{Q2} (thereby producing a trace $\pi_2$).
If the verifier can assure that the resulting traces$\pi_1, \pi_2$ satisfy $ \LTLglobally (a_{\pi_1} = a_{\pi_2})$, the $\forall\exists$ property holds.
However, this game-based interpretation fails in many instances.
There might exist a witness trace $\pi_2$, but the trace cannot be produced by a witness strategy as it requires knowledge of \emph{future} moves of the refuter.
Let us discuss this on the example programs in \Cref{fig:exist}.
A simple (informal) solution to construct a witness trace $\pi_2$ (when given the entire $\pi_1$) would be to guarantee that in \texttt{Q2}:4 (meaning location 4 of \texttt{Q2}) and line \texttt{Q1}:6 the value of $x$ in both programs agrees (i.e., $x_1 = x_2$ holds) and then simply resolve the nondeterminism at \texttt{Q2}:6 with $0$.
However, to follow this idea, the witness strategy for the verifier, when at \texttt{Q2}:3, would need to know the future value of $x_1$ when \texttt{Q1} is at location \texttt{Q1}:6.
\begin{figure}[t]
\begin{subfigure}{0.25\textwidth}
\scalebox{0.9}{
\begin{minipage}{\textwidth}
\begin{algorithmic}[1]
\RepeatFive
\State \textbf{print($a$)}
\State $x \leftarrow \star_\mathbb{N}$
\WhileOne{$\star_\mathbb{B}$}
\State $x \leftarrow x + 1$
\State $y \leftarrow x$
\WhileTwo{$y > 0$}
\State $a \leftarrow a + x$
\State $y \leftarrow y - 1$
\end{algorithmic}
\end{minipage}
}
\subcaption{Program \texttt{Q1}}\label{fig:p1'}
\end{subfigure}%
\begin{subfigure}{0.25\textwidth}
\scalebox{0.9}{
\begin{minipage}{\textwidth}
\begin{algorithmic}[1]
\RepeatFour
\State \textbf{print($a$)}
\State $x \leftarrow \star_\mathbb{N}$
\State $y \leftarrow x$
\WhileTwo{$y > 0$}
\State $a \leftarrow a + x + \star_\mathbb{N}$
\State $y \leftarrow y - 1$
\end{algorithmic}
\end{minipage}
}
\subcaption{Program \texttt{Q2}}\label{fig:p2'}
\end{subfigure}%
\begin{subfigure}{0.5\textwidth}
\scalebox{0.85}{
\small
\centering
\begin{tikzpicture}[scale=0.9]
\node[draw, rectangle, thick,align=center, rounded corners=3pt, fill=white,label=-45:{\scriptsize$ \{1\}$}] at (0,0) (n0) {$\boldsymbol{\alpha_1:}(5,3)$\\ $a_1 = a_2$};
\node[draw, rectangle, thick,align=center, rounded corners=3pt, fill=white,label=south:{\scriptsize$\{1\}$}] at (2.5,0) (n1) {$\boldsymbol{\alpha_2:}(3,3)$\\ $a_1 = a_2$};
\node[draw, rectangle, thick,align=center, rounded corners=3pt, fill=white,label=south:{\scriptsize$\{1, 2\}$}] at (5,0) (n2) {$\boldsymbol{\alpha_3:}(2,2)$\\ $a_1 = a_2$};
\node[draw, rectangle, thick,align=center, rounded corners=3pt, fill=white,label={[xshift=-0.4cm]south:{\scriptsize$\{1\}$}}] at (0,-2.5) (n3) {$\boldsymbol{\alpha_4:}(4,3)$\\ $a_1 = a_2$};
\node[draw, rectangle, thick,align=center, rounded corners=3pt, fill=white,label=north:{\scriptsize$\{1, 2\}$}] at (2.5,-2.5) (n4) {$\boldsymbol{\alpha_5:}(7,5)$\\ $a_1 = a_2$\\$x_1 = x_2$\\$y_1 = y_2$};
\node[draw, rectangle, thick,align=center, rounded corners=3pt, fill=white,label={[align=center]north:{\scriptsize$\{1,2\}$}}] at (5,-2.5) (n5) {$\boldsymbol{\alpha_6:}(9,7)$\\ $a_1 = a_2$\\$x_1 = x_2$\\$y_1 = y_2$};
\node[draw, rectangle, thick,align=center, rounded corners=3pt, fill=white,label={[xshift=0.7cm,yshift=-0.8mm]north:{\scriptsize$\{2\}, \{\alpha_8\}$}}] at (0,-5) (n6) {$\boldsymbol{\alpha_7:}(6,3)$\\ $a_1 = a_2$};
\node[draw, rectangle, thick,align=center, rounded corners=3pt, fill=white,label={[xshift=-0.4cm]north:{\scriptsize$\{1,2\}$}}] at (2.5, -5) (n7) {$\boldsymbol{\alpha_8:}(6,4)$\\ $a_1 = a_2$\\$x_1 = x_2$};
\node[draw, rectangle, thick,align=center, rounded corners=3pt, fill=white,label={[xshift=0.75cm,yshift=-0.8mm]north:{\scriptsize$\{1,2\}$, $\{\alpha_6\}$}}] at (5,-5) (n8) {$\boldsymbol{\alpha_9:}(8,6)$\\ $a_1 = a_2$\\$x_1 = x_2$\\$y_1 = y_2$};
\draw[->, thick] (n2) + (1.3, 0) -- (n2);
\draw[->, thick] (n2) -- (n1);
\draw[->, thick] (n1) -- (n3);
\draw[->, thick,transform canvas={xshift=0.9ex}] (n0) -- (n3);
\draw[->, thick,transform canvas={xshift=-0.9ex}] (n3) -- (n0);
\draw[->, thick] (n3) -- (n6);
\draw[->, thick] (n6) -- (n7);
\draw[->, thick] (n7) -- (n4);
\draw[->, thick] (n4) -- (n2);
\draw[->, thick] (n8) -- (n5);
\draw[->, thick] (n4) -- (n8);
\draw[->, thick] (n5) -- (n4);
\end{tikzpicture}
}
\subcaption{Winning strategy for the verifier.} \label{fig:winningStratEx}
\end{subfigure}
\caption{Two programs \texttt{Q1} and \texttt{Q2} in \Cref{fig:p1',fig:p2'}. In \Cref{fig:winningStratEx} a possible winning strategy for the verifier is depicted, showing that the $\forall\exists$-property holds.
The outer label gives the scheduling $M \subseteq \{1, 2\}$ and, if applicable, the restriction chosen by the strategy. \vspace{-0.3cm}}\label{fig:exist}
\end{figure}
Our insight in this paper is that we can turn the strategy-based interpretation of the witness trace $\pi_2$ into a useful verification method by \emph{combining} it with a program reduction.
As we express both searches strategically, we can phrase the combined search as a combined game.
In particular, both the reduction and the witness strategy are controlled by the verifier and can thus \emph{collaborate}.
In the resulting game, the verifier chooses a scheduling (as in \Cref{sec:overview_ksafety}) and, additionally, whenever the existentially quantifier copy is scheduled, the verifier also decides on the successor state of that copy.
We depict a possible winning strategy in \Cref{fig:winningStratEx}.
This strategy formalizes the interplay of reduction and witness strategy.
Initially, the verifier only schedules $\{1\}$ until \texttt{Q1} has reached program location \texttt{P1}:6 (at which point the value of $x$ is fixed).
Only then does the verifier schedule $\{2\}$, at which point the witness strategy can decide on a successor state for \texttt{Q2}.
In our case, the strategy chooses a value for $x$ such that $x_1 = x_2$ holds.
As we work in an abstraction of the actual system, we formalize this by restricting the abstract successor states.
In particular, in state $\alpha_7$ the verifier schedules $\{2\}$ and simultaneously restricts the successors to $\{\alpha_8\}$ (i.e., the abstract state where $x_1 = x_2$ holds), even though abstract state $[(6, 4), a_1 = a_2, x_1 \neq x_2]$ is also a valid successors under scheduling $\{2\}$.
We formalize when a restriction is valid in \Cref{sec:beyondksafety}.
The resulting strategy is winning and therefore denotes both a reduction \emph{and} witness strategy for the existentially quantified copies.
Importantly, both reduction and witness strategy are mutually dependent.
Our tool \texttt{HyPA} is able to verify both properties (in \Cref{fig:univ} and \Cref{fig:exist}) in a matter of a few seconds (cf.~\Cref{sec:eval}).
\section{Preliminaries}
\label{sec:prelim}
We begin by introducing basic preliminaries, including our basic model of computation and background on (finite-state) safety games.
\paragraph{Symbolic Transition Systems.}
We assume some fixed underlying first-order theory.
A \emph{symbolic transition systems} (STS) is a tuple $\mathcal{T} = (X, \mathit{init}, \mathit{step})$ where $X$ is a finite set of variables (possibly sorted), $\mathit{init}$ is a formula over $X$ describing all initial states, and $\mathit{step}$ is a formula over $X \uplus X'$ (where $X' \coloneqq \{x' \mid x \in X\}$ is the set of primed variables) describing the transitions of the system.
A concrete state $\mu$ in $\mathcal{T}$ is an assignment to the variables in $X$.
We write $\mu'$ for the assignment over $X'$ given by $\mu'(x') \coloneqq \mu(x)$.
A trace in $\mathcal{T}$ is an infinite sequence of assignment $\mu_0\mu_1\cdots$ such that $\mu_0 \models \mathit{init}$ and for ever $i \in \mathbb{N}$, $\mu_i \uplus \mu_{i+1}' \models \mathit{step}$.
We write $\traces{\mathcal{T}}$ for the set of all traces in $\mathcal{T}$.
We can naturally interpret programs as STS by making the program counter explicit.
\paragraph{Formula Transformations.}
For the remainder of this paper, we fix the set of system variables $X$.
We also fix a finite set of trace variables $\mathcal{V} = \{\pi_1, \ldots, \pi_k\}$.
For a trace variable $\pi \in \mathcal{V}$ we define $X_\pi \coloneqq \{x_\pi \mid x \in X\}$ and write $\vec{X}$ for $X_{\pi_1} \cup \cdots \cup X_{\pi_k}$.
For a formula $\theta$ over $X$, we define $\theta_{\langle \pi \rangle}$ as the formula over $X_\pi$ obtained by replacing every variable $x$ with $x_\pi$.
Similarly, we define $k$ fresh disjoint copies $\vec{X}' = X'_{\pi_1} \cup \cdots \cup X'_{\pi_k}$ (where $X'_\pi \coloneqq \{x'_\pi \mid x \in X\}$).
For a formula $\theta$ over $\vec{X}$ we define $\theta^{\langle'\rangle}$ as the formula over $\vec{X}'$ obtained by replacing every variable $x_\pi$ with $x_\pi'$.
\newcommand{\texttt{SAFE}}{\texttt{SAFE}}
\newcommand{\texttt{REACH}}{\texttt{REACH}}
\paragraph{Safety Games.}
A \emph{safety game} is a tuple $\mathcal{G} = (S_\texttt{SAFE}, S_\texttt{REACH}, S_0, T, B)$ where $S = S_\texttt{SAFE} \uplus s_\texttt{REACH}$ is a set of game states, $S_0 \subseteq S$ a set of initial state, $T \subseteq S \times S$ a transition relation, and $B \subseteq S$ a set of bad states.
We assume that for every $s \in S$ there exists at least one $s'$ with $(s, s') \in T$.
States in $S_\texttt{SAFE}$ are controlled by player $\texttt{SAFE}$ and those in $S_\texttt{REACH}$ by player $\texttt{REACH}$.
A play is an infinite sequence of states $s_0s_1\cdots$ such that $s_0 \in S_0$, and $(s_i, s_{i+1}) \in T$ for every $i \in \mathbb{N}$.
A positional strategy $\sigma$ for player $p \in \{\texttt{SAFE}, \texttt{REACH}\}$ is a function $\sigma : S_p \to S$ such that $(s, \sigma(s))$ for every $s \in S_p$.
A play $s_0s_1\cdots$ is compatible with strategy $\sigma$ for player $p$ if $s_{i+1} = \sigma(s_i)$ for every $s_i \in S_p$.
The safety player wins $\mathcal{G}$ if there is a strategy $\sigma$ for $\texttt{SAFE}$ such that all $\sigma$-compatible plays never visit a state in $B$.
In particular, $\texttt{SAFE}$ needs to win from \emph{all} initial states.
\section{A Temporal Logic for Asynchronous Hyperproperties}
\label{sec:ohyperltl}
In this section, we present OHyperLTL (short for observation-based HyperLTL).
Our logic builds upon HyperLTL \cite{ClarksonFKMRS14} which itself extends linear time temporal-logic (LTL) with explicit trace quantification.
In OHyperLTL, we include predicates from the background theory (to reason about infinite variable domains) and explicit observations (to express asynchronous properties).
Formulas in OHyperLTL are given by the following grammar:\footnote{For the examples in \Cref{sec:overview}, we additionally annotated quantifiers with an STS if we want to reason about different STSs within the same formula.
In the following, we assume that all quantifiers range over traces in the same STS to simplify notation.}
\begin{align*}
\varphi &\coloneqq \forall \pi : \xi\mathpunct{.} \varphi \mid \exists \pi : \xi\mathpunct{.} \varphi \mid \phi \\
\phi &\coloneqq \theta \mid \neg \phi \mid \phi_1 \land \phi_2 \mid \LTLnext \phi \mid\phi_1 \LTLuntil \phi_2
\end{align*}%
Here $\pi \in \mathcal{V}$ is a trace variable, $\theta$ is a formula over $\vec{X}$, and $\xi$ is a formula over $X$ (called the observation formula).
For ease of notation, we assume that all variables in $\mathcal{V}$ occur in the quantifier prefix \emph{exactly} once.
We also use the standard Boolean connectives $\wedge$, $\rightarrow$, $\leftrightarrow$ and constants $\top, \bot$, as well as the derived LTL operators eventually $\LTLeventually \phi \coloneqq \top \LTLuntil \phi $, and globally $\LTLglobally \phi \coloneqq \neg \LTLeventually \neg \phi$.
\newcommand{\filterdef}[2]{\mathit{valid}(#1,#2)}
\newcommand{\mathbb{T}}{\mathbb{T}}
\paragraph{Semantics.}
A trace $t$ is an infinite sequence $\mu_0\mu_1\cdots$ of assignments to $X$.
For $i \in \mathbb{N}$, we write $t(i)$ to denote the $i$th value in $t$.
A trace assignment $\Pi$ is a partial mapping of trace variables in $\mathcal{V}$ to traces.
Given a trace assignment $\Pi$ and $i \in \mathbb{N}$, we define $\Pi(i)$ to be the assignment to $\vec{X}$ given by $\Pi(i)(x_\pi) \coloneqq \Pi(\pi)(i)(x)$, i..e, the value of $x_\pi$ is the value of $x$ on the trace assigned to $\pi$.
For the LTL body of an OHyperLTL formula we define:
\begin{align*}
\Pi, i &\models \theta &\text{ iff } \quad &\Pi(i) \models \theta \\
\Pi, i &\models \neg \phi &\text{ iff } \quad & \Pi, i \not\models \phi \\
\Pi, i &\models \phi_1 \land \phi_2 &\text{ iff } \quad &\Pi, i \models \phi_1 \text{ and } \Pi, i \models \phi_2\\
\Pi, i&\models \LTLnext \phi &\text{ iff } \quad & \Pi , i + 1 \models \phi \\
\Pi, i&\models \phi_1 \LTLuntil \phi_2 &\text{ iff } \quad & \exists j \geq i\mathpunct{.} \Pi, j\models \phi_2 \text{ and } \forall i \leq k < j\mathpunct{.} \Pi, k \models \phi_1
\end{align*}
The distinctive feature of OHyperLTL over HyperLTL are the explicit observations.
Given an observation formula $\xi$ and trace $t$, we say that $\xi$ is a \emph{valid observation on} $t$ (written $\filterdef{t}{\xi}$) if there are infinitely many $i \in \mathbb{N}$ such that $t(i) \models \xi$.
If $\filterdef{t}{\xi}$ holds, we write $\filter{t}{\xi}$ for the trace obtained by projecting on those positions $i$ where $t(i) \models \xi$, i.e., $\filter{t}{\xi}(i) \coloneqq t(j)$ where $j$ is the $i$th index that satisfies $\xi$.
Given a set of traces $\mathbb{T}$ we resolve trace quantification as follows:
\begin{align*}
\Pi &\models_{\mathbb{T}} \phi &\text{ iff } \quad &\Pi, 0 \models \phi \\
\Pi &\models_{\mathbb{T}} \forall \pi : \xi\mathpunct{.} \varphi &\text{ iff } \quad &\forall t \in \{t \in \mathbb{T} \mid \filterdef{t}{\xi}\}\mathpunct{.} \Pi[\pi \mapsto \filter{t}{\xi}] \models_{\mathbb{T}} \varphi \\
\Pi &\models_{\mathbb{T}} \exists \pi : \xi\mathpunct{.} \varphi &\text{ iff } \quad &\exists t \in \{t \in \mathbb{T} \mid \filterdef{t}{\xi}\}\mathpunct{.} \Pi[\pi \mapsto \filter{t}{\xi}] \models_{\mathbb{T}} \varphi
\end{align*}
The semantics mostly agrees with that of HyperLTL\cite{ClarksonFKMRS14} but projects each trace to the positions where the observation holds.
Given an STS $\mathcal{T}$ and HyperLTL formula $\varphi$, we write $\mathcal{T} \models \varphi$ if $\emptyset \models_{\mathit{Traces}(\mathcal{T})} \varphi$ where $\emptyset$ is the empty assignment.
\paragraph{The Power of Observations.}
The explicit observations in OHyperLTL facilitate the specification of asynchronous hyperproperties, i.e., properties where traces are traversed at different speeds.
For the example in \Cref{sec:overview_ksafety}, the explicit observations allow us to compare the output of both programs even though the actual step at which the output occurs (in a synchronous semantics) differs between both programs (as \texttt{P1} takes the inner loop twice as often as \texttt{P2}).
As the observations are part of the specification, we can model a broad spectrum of properties ranging, e.g., from timing-insensitive properties (by placing observations only at output locations) to timing-sensitive specifications \cite{GeYCH18} (by placing observations at closer intervals).
Functional (opposed to temporal) $k$-safety properties specified by pre-and postcondition \cite{Benton04,ShemerGSV19,UnnoTK21} can easily be encoded as $\forall^k$-OHyperLTL properties by placing observations at the start and end of each program.
By setting $\xi = \top$, i.e., observing \emph{every} step, we can express synchronous properties; OHyperLTL thus subsumes HyperLTL.
\paragraph{Finite-State Model Checking.}
Many mechanisms used to express asynchronous hyperproperties render finite-state model checking undecidable \cite{GutsfeldMO21,BozzelliPS21,BaumeisterCBFS21}.
In contrast, the simple mechanism used in OHyperLTL maintains decidable finite-state model checking.
\begin{restatable}{theorem}{finiteStateR}
Assume a STS $\mathcal{T}$ with finite variable domains and decidable background theory and an OHyperLTL formula $\varphi$. It is decidable if $\mathcal{T} \models \varphi$.
\label{theo:finiteState}
\end{restatable}
\begin{proofSketch}
Under the assumptions, we can view $\mathcal{T}$ as an explicit (instead of symbolic) finite-state transition system.
Given an observation formula $\xi$ we can effectively compute an explicit finite-state system $\mathcal{T}'$ such that $\traces{\mathcal{T}'} = \{ \filter{t}{\xi} \mid t \in \traces{\mathcal{T}} \land \filterdef{t}{\xi}\}$.
This reduces OHyperLTL model checking on $\mathcal{T}$ to (standard) HyperLTL model checking on $\mathcal{T}'$ which is decidable \cite{FinkbeinerRS15}.
\qed
\end{proofSketch}
Note that for infinite-state (symbolic) systems, we cannot effectively compute $\mathcal{T}'$ as in the proof of \Cref{theo:finiteState}.
In fact, there may not even exist a system $\mathcal{T}'$ with the desired property that is expressible in the same background theory.
The finite-state result in \Cref{theo:finiteState} is of little relevance for the present paper.
Nevertheless, it indicates that our logic is well suited for verification of infinite-state (software) systems as the (inevitable) undecidability stems from the infinite domains in software programs and not already from the logic itself.
\paragraph{Safety.}
In this paper we assume that the hyperproperty is temporally safe \cite{BeutnerCFHK22}, i.e., the temporal body of any OHyperLTL formula denotes a \emph{safety property}.
Note that, as we support quantifier alternation, we can still express hyperliveness properties \cite{ClarksonS08,CoenenFST19}.
For example, GNI is both temporally safe and hyperliveness.
We model the body of a formula by a symbolic safety automaton \cite{DAntoniV17}, which is a tuple $\mathcal{A} = (Q, q_0, \delta, B)$ where $Q$ is a finite set of states, $q_0 \in Q$ the initial state, $B \subseteq Q$ a set of bad-states, and $\delta$ a finite set of automaton edges of the form $(q, \theta, q')$ where $q, q' \in Q$ are states and $\theta$ is a formula over $\vec{X}$.
Given a trace $t$ over assignments to $\vec{X}$, a run of $\mathcal{A}$ on $t$ is an infinite sequence of states $q_0q_1\cdots$ (starting in $q_0$) such that for every $i$ there exists an edge $(q_i, \theta_i, q_{i+1}) \in \delta$ such that $t(i) \models \theta_i$.
A word is accepted by $\mathcal{A}$ if it has \emph{no} run that visits a state in $B$.
The automaton is \emph{deterministic} if for every $q \in Q$ and for every assignments $\mu$ to $\vec{X}$ there exists exactly one edge $(q, \theta, q') \in \delta$ with $\mu \models \theta$.
\section{Reductions as Games}
\label{sec:ksafety}
After having formally defined our temporal logic, we turn our attention to the automatic verification of OHyperLTL formulas on STSs.
In this section, we begin by formalizing our game-based interpretation of a reduction.
To illustrate this, we consider $\forall^k$ OHyperLTL formulas, which, as the body of the formula is a safety property, always denote $k$-safety properties.
\subsection{Game Construction}
\label{sec:reduction1}
Our search for a reduction is based in the scope of a fixed predicate abstraction \cite{GrafS97,JhalaPR18}.
The idea of predicate abstraction is to abstract a system by only keeping track of the truth value of a few selected predicates that (ideally) identify properties that are relevant to prove the property in question.
Let $\mathcal{T} = (X, \mathit{init}, \mathit{step})$ be a STS and let $\varphi = \forall \pi_1 : \xi_1 \ldots \forall \pi_k : \xi_k\mathpunct{.} \phi$ be the ($k$-safety) OHyperLTL we wish to verify.
Let $\mathcal{A}_\phi = (Q_\phi, q_{\phi, 0}, \delta_\phi, B_\phi)$ be a deterministic safety automaton for $\phi$.
A \emph{relational} predicate $p$ is a formula over $\vec{X}$ that identifies a property of the combined state space of $k$ system copies.
Let $\mathcal{P} = \{p_1, \ldots, p_n\}$ be a finite set of relational predicates.
We say a formula over $\vec{X}$ is \emph{expressible in $\mathcal{P}$}, if it is equivalent to a boolean combination of the predicates in $\mathcal{P}$.
We assume that all edge formulas in the automaton $\mathcal{A}_\phi$, and formulas $\mathit{init}_{\langle \pi_i \rangle}$ and $(\xi_{i})_{\langle \pi_i \rangle}$ for $\pi_i \in \mathcal{V}$ are expressible in $\mathcal{P}$.
Note that we can always add missing predicates to $\mathcal{P}$.
Given the set of predicates $\mathcal{P}$ the state-space of the abstraction w.r.t.~$\mathcal{P}$ is given by $\mathbb{B}^n$ where for each abstract state $\hat{s} \in \mathbb{B}^n$, the $i$th position $\hat{s}_i \in \mathbb{B}$ tracks whether or not predicate $p_i$ holds.
To simplify notation, we write $\mathit{ite}(b, \theta, \theta')$ to be formula $\theta$ if $b = \top$, and $\theta'$ otherwise.
For each abstract state $\hat{s} \in \mathbb{B}^n$ we define $\predSat{\hat{s}} \coloneqq \bigwedge_{i = 1}^n \mathit{ite}(\hat{s}_i , p_i, \neg p_i)$, i.e., $\predSat{\hat{s}}$ is a formula over $\vec{X}$ that capture all concrete states that are abstracted to $\hat{s}$.
To incorporate reductions in our abstraction, we parametrize the abstraction transition relation by a scheduling.
Let $M \subseteq \{\pi_1, \ldots, \pi_k\}$ be a \emph{scheduling}. We lift the $\mathit{step}$ formula from our STS by defining
\begin{align*}
\textstyle\mathit{step}_M \coloneqq \bigwedge_{\pi_i \in M} \mathit{step}_{\langle \pi_i \rangle} \land \bigwedge_{\pi_i \not\in M, x \in X} x_{\pi_i}' = x_{\pi_i}.
\end{align*}
That is all copies in $M$ take a step while all other copies remain unchanged.
Given two abstract states $\hat{s}_1, \hat{s}_2$ we say that $\hat{s}_2$ is a \emph{$M$-successor} of $\hat{s}_1$, written $\hat{s}_1 \xrightarrow{M} \hat{s}_2$, if $\predSat{\hat{s}_1} \land \predSat{\hat{s}_2}^{\langle'\rangle} \land \mathit{step}_M$ is satisfiable, i.e., we can transition from $\hat{s}_1$ to $\hat{s}_2$ by only progressing the copies in $M$.
For an abstract state $\hat{s}$ we define $\mathit{obs}(\hat{s}) \in \mathbb{B}^k$ as the boolean vector that indicates which copy (of $\pi_1, \ldots, \pi_k$) is currently at an observation point, i.e., $\mathit{obs}(\hat{s})_i = \top$ iff $\predSat{\hat{s}} \land (\xi_{i})_{\langle \pi_i \rangle}$ is satisfiable.
Note that as $(\xi_{i})_{\langle \pi_i \rangle}$ is, by assumption, expressible in $\mathcal{P}$, either all or none of the concrete states in $\predSat{\hat{s}}$ satisfy $(\xi_{i})_{\langle \pi_i \rangle}$.
Building on the parametrized abstract transition relation we can construct a (finite-state) safety game where winning strategies for the verifier correspond to valid reductions with accompanying proofs.
The nodes in our game have two forms:
Either they are of the form $(\hat{s}, q, b)$ where $\hat{s} \in \mathbb{B}^n$ is an abstract state, $q \in Q_\phi$ a state of the safety automaton, and $b \in \mathbb{B}^k$ a boolean vector indicating which copy has moved since the last automaton step.
Or of the form $(\hat{s}, q, b, M)$ where $\hat{s}$, $q$ and $b$ are as before and $\emptyset \neq M \subseteq \{\pi_1, \ldots, \pi_k\}$ is a scheduling.
The initial states are all states $(\hat{s},q_{\phi, 0}, \top^k)$ where $\predSat{\hat{s}} \land \bigwedge_{i=1}^k \mathit{init}_{\langle \pi_i \rangle}$ is satisfiable (recall that $\mathit{init}_{\langle \pi_i \rangle}$ is expressible in $\mathcal{P}$).
We mark a state $(\hat{s}, q, b)$ or $(\hat{s}, q, b, M)$ as losing iff $q \in B_\phi$.
For automaton state $q \in Q_\phi$ and abstract state $\hat{s}$ we define $\delta_\phi(q, \hat{s})$ as the \emph{unique} state $q'$ such that there is an edge $(q, \theta, q') \in \delta_\phi$ such that $\predSat{\hat{s} }\land \theta$ is satisfiable.
Uniqueness follows from the assumption that $\mathcal{A}_\phi$ is deterministic and all edge formulas are expressible in $\mathcal{P}$.
The transition relation of our game are now given by the following rules:
\noindent
\begin{minipage}{0.31\textwidth}
\def\hskip .1in{\hskip .1in}
\def0pt{0pt}
\scalebox{0.85}{
\begin{minipage}{\textwidth}
\begin{prooftree}
\AxiomC{$\forall \pi_i \in M\mathpunct{.} \neg b_i \lor \neg \mathit{obs}(\hat{s})_i$}
\RightLabel{\footnotesize(A)}
\UnaryInfC{$(\hat{s}, q, b) \rightsquigarrow (\hat{s}, q, b, M)$}
\end{prooftree}
\end{minipage}
}
\end{minipage}%
\begin{minipage}{0.34\textwidth}
\def\hskip .1in{\hskip .1in}
\def0pt{0pt}
\scalebox{0.85}{
\begin{minipage}{\textwidth}
\begin{prooftree}
\AxiomC{$\mathit{obs}(\hat{s}) = \top^k$}
\AxiomC{$q' = \delta_\phi(q, \hat{s})$}
\RightLabel{\footnotesize(B)}
\BinaryInfC{$(\hat{s}, q, \top^k) \rightsquigarrow (\hat{s}, q', \bot^k)$}
\end{prooftree}
\end{minipage}
}
\end{minipage}%
\begin{minipage}{0.33\textwidth}
\def\hskip .1in{\hskip .1in}
\def0pt{0pt}
\scalebox{0.85}{
\begin{minipage}{\textwidth}
\begin{prooftree}
\vspace{2mm}
\AxiomC{$\hat{s} \xrightarrow{M} \hat{s}'$}
\AxiomC{$b' = b[i \mapsto \top]_{\pi_i \in M}$}
\RightLabel{\footnotesize(C)}
\BinaryInfC{$(\hat{s}, q, b, M) \rightsquigarrow (\hat{s}', q, b')$}
\end{prooftree}
\end{minipage}
}
\end{minipage}
\vspace{1mm}
\noindent
In rule (A) we select any scheduling that schedules only copies that have not reached an observation point or have not moved since the last automaton step.
In particular, we cannot schedule any copy that has moved and already reached an observation point.
In rule (B) all copies reached an observation point and have moved since the last update (i.e., $b = \top^k$) so we progress the automaton and reset $b$.
Lastly, in rule (C) we select a $M$-successor of $\hat{s}$ and update $b$ for all copies that take part in the step.
In our game, player $\texttt{SAFE}$ takes the role of the verifier and player $\texttt{REACH}$ that of the refuter.
It is the safety player's responsibility to select a scheduling in each step, so we assign nodes of the form $(\hat{s}, q, b)$ to $\texttt{SAFE}$.
Nodes of the form $(\hat{s}, q, b, M)$ are controlled by $\texttt{REACH}$ who can choose an abstract $M$-successor.
Let $\gameE{\mathcal{T}}{\varphi}{\mathcal{P}}$ be the resulting (finite-state) safety game.
A winning strategy for the safety-player in $\gameE{\mathcal{T}}{\varphi}{\mathcal{P}}$ picks, in each abstract state, a valid scheduling between two observation points and avoids a visit to a losing state; all within the abstraction given by $\mathcal{P}$.
We can thus show:
\begin{restatable}{theorem}{soundnessksafety}
If player $\texttt{SAFE}$ wins $\gameE{\mathcal{T}}{\varphi}{\mathcal{P}}$, then $\mathcal{T} \models \varphi$.
\end{restatable}
\begin{proofSketch}
Assume $\sigma$ is a winning strategy for $\texttt{SAFE}$ in $\gameE{\mathcal{T}}{\varphi}{\mathcal{P}}$.
Let $t_1, \ldots, t_k \in \traces{\mathcal{T}}$ be arbitrary.
We, iteratively, construct stuttered versions $t'_1, \ldots, t'_k$ of $t_1, \ldots, t_k$ by using the schelduing selected by $\sigma$ on abstracted prefixes.
By construction of $\gameE{\mathcal{T}}{\varphi}{\mathcal{P}}$ the stuttered traces $t'_1, \ldots, t'_k$ align at observation points.
In particular, we have $[\pi_1 \mapsto \filter{t_1}{\xi_1}, \ldots, \pi_k \mapsto \filter{t_k}{\xi_k}] \models \phi$ iff $[\pi_1 \mapsto \filter{t'_1}{\xi_1}, \ldots, \pi_k \mapsto \filter{t'_k}{\xi_k}] \models \phi$.
Moreover, the sequence of abstract states in $\gameE{\mathcal{T}}{\varphi}{\mathcal{P}}$ forms an abstraction of $t'_1, \ldots, t'_k$ and shows that $\mathcal{A}_\phi$ cannot reach a bad state when reading $\filter{t'_1}{\xi_1}, \ldots, \filter{t'_k}{\xi_k}$ (as $\sigma$ is winning).
This already shows that $[\pi_1 \mapsto \filter{t'_1}{\xi_1}, \ldots, \pi_k \mapsto \filter{t'_k}{\xi_k}] \models \phi$ and thus $[\pi_1 \mapsto \filter{t_1}{\xi_1}, \ldots, \pi_k \mapsto \filter{t_k}{\xi_k}] \models \phi$.
As this holds for all traces $t_1, \ldots, t_k \in \traces{\mathcal{T}}$ we get $\mathcal{T} \models \varphi$ as required.
\qed
\end{proofSketch}
\paragraph{Game Construction and Complexity.}
If the background theory is decidable, $\gameE{\mathcal{T}}{\varphi}{\mathcal{P}}$ can be constructed effectively using at most $2^{|\mathcal{P}|+1} \cdot 2^k$ queries to an SMT solver.
Checking if $\texttt{SAFE}$ wins $\gameE{\mathcal{T}}{\varphi}{\mathcal{P}}$ can be done with a simple fixpoint computation of the attractor in linear time.
Our game-based method to find a reduction in a given abstraction is closely related to the notation of a \emph{property-directed self-composition} \cite{ShemerGSV19}.
The previously only known algorithm for finding such a reduction is based on an optimized enumeration \cite{ShemerGSV19}, which, in the worst case, requires $\mathcal{O}(2^{|\mathcal{P}|+1} \cdot 2^k)$ many enumerations.
Our worst-case complexity thus matches the bounds inferred by \cite{ShemerGSV19}, but avoids the explicit enumeration of reductions (and the concomitant repeated construction of the abstract state-space) and is, as we believe, conceptually simpler to comprehend.
Moreover, our game-based technique is the key stepping stone for extending our method beyond $k$-safety in \Cref{sec:beyondksafety}.
\section{Verification Beyond $k$-Safety}
\label{sec:beyondksafety}
Building on the strategic interpretation of a reduction, we extend our verification beyond $\forall^*$ properties to support $\forall^*\exists^*$ properties.
We accomplish this by \emph{combining} the game-based reading of a reduction (as discussed in the previous section) with a strategic reading of existential quantification.
For the remainder of this section, fix a STS $\mathcal{T} = (X, \mathit{init}, \mathit{step})$ and let $\varphi = \forall \pi_1 : {\xi_1} \ldots \forall \pi_l : {\xi_l}. \exists \pi_{l+1} : {\xi_{l+1}} \ldots \exists \pi_k : {\xi_{k}}\mathpunct{.} \phi$ be the OHyperLTL we wish to check, i.e., we universally quantify over $l$ traces followed by an existential quantification over $k-l$ traces.
We assume that for every existential quantification $\exists \pi_i : \xi_i$ occurring in $\varphi$, $\filterdef{t}{\xi_i}$ holds for every $t \in \traces{\mathcal{T}}$ (we discuss this later in \Cref{rem:assumption}).
\subsection{Reductions and Witness Strategy as a Game}
The idea of a game-based verification of $\forall^*\exists^*$ properties is to consider a $\forall^*\exists^*$-property as a game between a verifier and a refuter \cite{CoenenFST19}.
The refuter controls the $l$ universally quantified traces by moving thorough $l$ copies of the system (thereby producing traces $\pi_1, \ldots, \pi_l$) and the verifier reacts by, incrementally, moving through $k-l$ copies of the system (thereby producing traces $\pi_{l+1}, \ldots, \pi_k$).
If the verifier can ensure that the resulting traces satisfy $\phi$, $\mathcal{T} \models \varphi$ holds.
We combine this game-based reading of existential quantification with our game-based interpretation of a reduction, by, additionally, letting the verifier control the scheduling of the system.
When played on the \emph{concrete} state-space of $\mathcal{T}$ the game proceeds in three stages as follows:
\textbf{(1)} The verifier selects a valid scheduling $M \subseteq \{\pi_1, \ldots, \pi_k\}$;
\textbf{(2)} The refuter selects successor states for all universally quantified copies by fixing an assignment to $X_{\pi_1}', \ldots, X_{\pi_l}'$ (only moving copies scheduled by $M$);
\textbf{(3)} The verifier reacts by choosing successor states for the existentially quantified copies by fixing an assignment to $X_{\pi_{l+1}}', \ldots, X_{\pi_k}'$ (again, only moving copies scheduled by $M$).
Afterward, the process repeats.
As we work within a fixed abstraction of $\mathcal{T}$, the verifier can, however, not choose concrete successor states directly but only work in the precision captured by the abstraction.
Following the general scheme of abstract games, we, therefore, underapproximate the moves available to the verifier \cite{AlfaroGJ04}.
Formally, we abstract the three-stage game outlined before (which was played at the level of concrete states), to a simpler abstract game (consisting of only two stages).
In the first stage, the verifier selects both a scheduling $M$ and a \emph{restriction} on the set of abstract successor states, i.e., a set of abstract states $A$.
In the second stage, the refuter cannot choose any abstract successor state (any $M$-successor in the terminology from \Cref{sec:ksafety}), but only successors contained in the restriction $A$.
To guarantee the soundness of this approach, we ensure that the verifier can only pick restrictions that are \emph{valid}, i.e., restrictions that underapproximate the possibilities of the verifier on the level of concrete states.
\paragraph{Game Construction.}
We modify our game from \Cref{sec:ksafety} as follows.
States are either of the form $(\hat{s}, q, b)$ (as in \Cref{sec:ksafety}) or of the form $(\hat{s}, q, b, M, A)$ where $\hat{s}$, $q$, $b$ and $M$ are as in \Cref{sec:ksafety} and $A \subseteq \mathbb{B}^n$ is a subset of abstract states (the restriction).
To reflect the restriction we modify transition rules (A) and (B) (rule (B) stays as it is):
\noindent
\begin{minipage}{0.55\textwidth}
\def\hskip .1in{\hskip .2in}
\def0pt{0pt}
\scalebox{1}{
\begin{minipage}{\textwidth}
\begin{prooftree}
\AxiomC{$\forall \pi_i \in M\mathpunct{.} \neg b_i \lor \neg \mathit{obs}(\hat{s})_i$}
\AxiomC{$\validRes{\hat{s}}{M}{A}$}
\RightLabel{\footnotesize(A)}
\BinaryInfC{$(\hat{s}, q, b) \rightsquigarrow (\hat{s}, q, b, M, A)$}
\end{prooftree}
\end{minipage}
}
\end{minipage}%
\begin{minipage}{0.45\textwidth}
\def\hskip .1in{\hskip .1in}
\def0pt{0pt}
\scalebox{1}{
\begin{minipage}{\textwidth}
\begin{prooftree}
\AxiomC{$\hat{s}' \in A$}
\AxiomC{$b' = b[i \mapsto \top]_{i \in M}$}
\RightLabel{\footnotesize(C)}
\BinaryInfC{$(\hat{s}, q, b, M, A) \rightsquigarrow (\hat{s}', q, b')$}
\end{prooftree}
\end{minipage}
}
\end{minipage}
\vspace{1mm}
\noindent
In rule (1), the safety player (who, again, takes the role of the verifier) selects both a scheduling $M$ and a restriction $A$ such that $\validRes{\hat{s}}{M}{A}$ holds (which we define later).
The reachable player (who takes the role of the refuter) can, in rule (3), not select any successor but only those allowed by $A$.
\paragraph{Valid Restriction.}
The above game construction depends on the definition of $\validRes{\hat{s}}{M}{A}$ which should hold whenever $A$ is a valid restriction for the safety player from state $\hat{s}$ under scheduling $M$.
Intuitively, $A$ is a valid restriction if it underapproximates the possibilities of a witness strategy that can pick concrete successor states for all existentially quantified traces.
That is, for every concrete state in $\hat{s}$, a strategy (on the level of concrete states) can guarantee a move to a concrete state that is abstracted to an abstract state within $A$.
Formally we define $\validRes{\hat{s}}{M}{A}$ as follows:
\vspace{-5mm}
\begin{align*}
&\forall \{X_{\pi_i}\}_{i=1}^{k}. \forall \{X_{\pi_i}'\}_{i=1}^l. \; \predSat{\hat{s}} \land \bigwedge_{i=1}^l \mathit{ite}\Big(\pi_i \in M, \mathit{step}_{\langle \pi_i \rangle}, \bigwedge_{x \in X} x_{\pi_i}' = x_{\pi_i}\Big)\\[-0.3cm]
&\quad\Rightarrow \exists \{X_{\pi_i}'\}_{i=l+1}^k. \bigwedge_{i=l+1}^k \mathit{ite}\Big(\pi_i \in M, \mathit{step}_{\langle \pi_i \rangle}, \bigwedge_{x \in X} x_{\pi_i}' = x_{\pi_i}\Big) \land \bigvee\limits_{\hat{s}' \in A} \predSat{\hat{s}'}^{\langle'\rangle}
\end{align*}
The formula expresses that for all concrete states in $\predSat{\hat{s}}$ (assignments to $\{X_{\pi_i}\}_{i=1}^{k}$), and for all concrete successor states for the universally quantified copies (assignments to $\{X_{\pi_i}'\}_{i=1}^l$), there exists a successor state for the existentially quantified copies ($\{X_{\pi_i}'\}_{i=l+1}^k$) such that one of the abstract states in $A$ is reached.
\begin{example}
With this definition at hand, we can validate the restrictions chosen by the strategy in \Cref{fig:winningStratEx}:
For example, in state $\alpha_7$ the strategy schedules $M = \{2\}$ and restricts the successor states to $\{\alpha_8\}$ even though abstract state $\big[(6, 4), a_1 = a_2, x_1 \neq x_2\big]$ is also a $\{2\}$-successor of $\alpha_7$.
If we spell out $\validRes{\alpha_7}{\{2\}}{\{\alpha_8\}}$ we get\\[-2mm]
%
\scalebox{0.9}{\parbox{\linewidth}{
\begin{align*}
&\forall X_1 \!\cup\! X_2 \!\cup\! X_1'\mathpunct{.} \; \underbrace{a_1 = a_2}_{\predSat{\alpha_7}} \land \Big(\!\bigwedge_{z \in X} \! z'_1 = z_1\Big) \Rightarrow \exists X_2'\mathpunct{.} \; \underbrace{a_2' = a_2 \land y_2' = y_2}_{\mathit{step_{\langle 2 \rangle}}} \land \underbrace{\big(a_1 = a_2 \land x_1 = x_2\big)}_{\predSat{\alpha_8}}
\end{align*}
}}\\[-2mm]
%
where $X = \{a, x, y\}$.
Here we assume that $\mathit{step} \coloneqq \big(a' = a \land y' = y\big)$ is the update performed on instruction $x \leftarrow \star_\mathbb{N}$ from \texttt{Q2}:3 to \texttt{Q2}:4.
The above formula is valid.
\end{example}
\paragraph{Correctness.}
Call the resulting game $\gameEF{\mathcal{T}}{\varphi}{\mathcal{P}}$.
The game combines the search for a reduction with that of a witness strategy (both within the precision captured by $\mathcal{P}$).\footnote{In particular, $\gameEF{\mathcal{T}}{\varphi}{\mathcal{P}}$ (strictly) generalizes the construction of $\gameE{\mathcal{T}}{\varphi}{\mathcal{P}}$ from \Cref{sec:ksafety}:
If $k = l$ (i.e, the property is a $\forall^*$-property) the unique minimal valid restriction from $\hat{s}, M$ is $A = \{\hat{s}' \mid \hat{s} \xrightarrow{M} \hat{s}'\}$, i.e., the set of all $M$-successors of $\hat{s}$.
The safety player can thus not be more restrictive than allowing \emph{all} $M$-successors.}
We can show:
\begin{restatable}{theorem}{soundessBeyond}\label{theo:soundessKSafety}
If player $\texttt{SAFE}$ wins $\gameEF{\mathcal{T}}{\varphi}{\mathcal{P}}$, then $\mathcal{T} \models \varphi$.
\label{theo:soundnessBeyond}
\end{restatable}
\begin{proofSketch}
Let $\sigma$ be a winning strategy for $\texttt{SAFE}$ in $\gameEF{\mathcal{T}}{\varphi}{\mathcal{P}}$.
Let $t_1, \ldots, t_l \in \traces{\mathcal{T}}$ be arbitrary.
We use $\sigma$ to incrementally construct witness traces $t_{l+1}, \ldots, t_k$ by simulating prefixes of $t_1, \ldots, t_l$ in the game and querying $\sigma$.
In every abstract state $\hat{s}$ where $\sigma$ selects a scheduling $M$ and a restriction $A$, $\validRes{\hat{s}}{M}{A}$ holds.
We can plug the current \emph{concrete} state into the universal quantification of $\validRes{\hat{s}}{M}{A}$ and get a (concrete) witness for the existential quantification that is a valid successor in $\mathcal{T}$.
We use these concrete assignments to incrementally construct the traces $t_{l+1}, \ldots, t_k$.
\qed
\end{proofSketch}
\begin{remark}\label{rem:assumption}
Recall that we assume that for every existential quantification $\exists \pi_i : \xi_i$ occurring in $\varphi$ and all $t \in \traces{\mathcal{T}}$, $\filterdef{t}{\xi_i}$ holds.
This is important to ensure that the $\exists$-player cannot avoid observation points forever.
We could drop this assumption by strengthening the winning condition in $\gameEF{\mathcal{T}}{\varphi}{\mathcal{P}}$ and explicitly state that, in order to win, $\texttt{SAFE}$ needs to visit observations points on existentially quantified traces infinitely many times.
\end{remark}
\paragraph{Clairvoyance vs.~Abstraction.}
The cooperation between reduction (the ability of the verifier to select schedulings) and witness strategy (the ability to select restrictions on the successor) can be seen as limited form of prophecy \cite{AbadiL91,BeutnerF22}.
By first scheduling the universal copies, the witness strategy can peek at future moves before committing on a successor state, as we e.g., saw in \Cref{fig:exist}.
The ``theoretically optimal'' reduction is thus a sequential one that first schedules only the universal traces (until an observation point is reached) and thereby provide maximal information for the witness strategy.
However, in the context of a fixed abstraction, this reduction is not always optimal.
For example, in \Cref{fig:exist} the strategy schedules the loop in lockstep which is crucial to generate a proof with simple (linear) invariants.
In particular, \Cref{fig:exist}, does not admit a witness strategy in the lock-step reduction and does not admit a proof with linear invariants in a sequential reduction.
Our verification framework, therefore, strikes a delicate balance between clairvoyance needed by the witness strategy and precision captured in the abstraction, further emphasising why the searches for reduction and witness strategy need to be mutually dependent.
\subsection{Constructing and Solving $\gameEF{\mathcal{T}}{\varphi}{\mathcal{P}}$}\label{sec:solvingInner}
\begin{wrapfigure}{R}{0.57\textwidth}
\vspace{-0.6cm}
\scalebox{1}{
\begin{minipage}{0.55\textwidth}
\begin{algorithm}[H]
\caption{Iterative solver for $\gameEF{\mathcal{T}}{\varphi}{\mathcal{P}}$.
}\label{alg:cap}
\begin{algorithmic}[1]
\State \textbf{Input: } $\mathcal{T}, \varphi, \mathcal{P}$
\State $\tilde{\mathcal{G}}\coloneqq \mathit{constructApproximateGame}(\mathcal{T}, \varphi, \mathcal{P})$
\RepeatOne
\MatchTwo{$\mathit{Solve}(\tilde{\mathcal{G}})$}
\State \textbf{case }$\texttt{REACH}(\sigma)$\textbf{:} \textbf{return} $\texttt{REACH}$
\CaseTwo{$\texttt{SAFE}(\sigma)$}
\ForAllOne{$(\hat{s}, M, A) \in \mathit{Restrictions}(\sigma)$}
\IfTwo{$\neg \validRes{\hat{s}}{M}{A}$}
\ForAllOne{$A' \subseteq A$ }
\State $\tilde{\mathcal{G}}\coloneqq \mathit{Remove}(\tilde{\mathcal{G}}, (\hat{s}, M, A'))$
\State \textbf{goto } 4
\State \textbf{return} $\texttt{SAFE}$
\end{algorithmic}
\end{algorithm}
\end{minipage}
}
\vspace{-0.2cm}
\end{wrapfigure}
Constructing the game graph of $\gameEF{\mathcal{T}}{\varphi}{\mathcal{P}}$ requires the identification of all valid restrictions (of which there are exponentially many in the number of abstract states and thus double exponentially many in the number of predicates) each of which requires to solve a quantified SMT query.
We propose a more effective algorithm that solves $\gameEF{\mathcal{T}}{\varphi}{\mathcal{P}}$ without constructing it explicitly.
Instead, we iteratively refine an abstraction $\tilde{\mathcal{G}}$ of $\gameEF{\mathcal{T}}{\varphi}{\mathcal{P}}$.
Our method hinges on the following easy observation:
\begin{lemma}\label{lem:closed}
For any $\hat{s}$ and $M$, $\{A \mid \validRes{\hat{s}}{M}{A}\}$ is upwards closed (w.r.t.~$\subseteq$).
\end{lemma}
Our initial abstraction consists of all possible restrictions (even those that might be invalid), i.e., we allow all restrictions of the form $(\hat{s}, M, A)$ where $A \subseteq \{\hat{s}' \mid \hat{s} \xrightarrow{M} \hat{s}'\}$.\footnote{Note that $\{\hat{s}' \mid \hat{s} \xrightarrow{M} \hat{s}'\}$ is always a valid restriction. Importantly, we can compute $ \{\hat{s}' \mid \hat{s} \xrightarrow{M} \hat{s}'\}$ locally by iterating over abstract states opposed to \emph{sets} of abstract states. }
This overapproximates the power of the safety player, i.e., a winning strategy for $\texttt{SAFE}$ in $ \tilde{\mathcal{G}}$ may not be valid in $\gameEF{\mathcal{T}}{\varphi}{\mathcal{P}}$.
To remedy this, we propose the following inner refinement loop:
If we find a winning strategy $\sigma$ for $\texttt{SAFE}$ in $\tilde{\mathcal{G}}$ we check if all restrictions chosen by $\sigma$ are valid, i.e., we check validity only on a \emph{need-to-know-now} basis.
If all are valid, $\sigma$ is also winning for $\gameEF{\mathcal{T}}{\varphi}{\mathcal{P}}$ and we can apply \Cref{theo:soundnessBeyond}.
If we find an invalid restriction $(\hat{s}, M, A)$ used by $\sigma$ we refine $\tilde{\mathcal{G}}$ by removing not only the restriction $(\hat{s}, M, A)$ but \emph{all} $(\hat{s}, M, A')$ with $A' \subseteq A$ (which is justified by \Cref{lem:closed}).
The algorithm is sketched in \Cref{alg:cap}.
The subroutine $\mathit{Restrictions}(\sigma)$ returns all restrictions used by $\sigma$, i.e., all tuples $(\hat{s}, M, A)$ such that $\sigma$ uses an edge $(\hat{s}, q, b) \rightsquigarrow (\hat{s}, q, b, M, A)$ for some $q, b$.
$\mathit{Remove}\linebreak[1](\tilde{\mathcal{G}},\linebreak[1] (\hat{s}, M, A'))$ removes from $\tilde{\mathcal{G}}$ all edges of the form $(\hat{s}, q, b) \rightsquigarrow (\hat{s}, q, b, M, A')$ for some $q, b$ and $\mathit{Solve}$ solves a finite-state safety game.
To improve the algorithm further, in line 4 we always compute a maximal safety strategy, i.e., a strategy that selects maximal restrictions (w.r.t.~$\subseteq$) and therefore allows us to eliminate many invalid restrictions from $\tilde{\mathcal{G}}$ simultaneously.
For safety games there always exists such a maximal winning strategy (see e.g.~\cite{BernetJW02}).
Note that while $\tilde{\mathcal{G}}$ is large, solving this finite-state game can be done very efficiently.
The running time of solving $\gameEF{\mathcal{T}}{\varphi}{\mathcal{P}}$ is dominated by the SMT queries of which our refinement loop, in practice, requires very few.
\section{Implementation and Evaluation}
\label{sec:eval}
\begin{wraptable}{R}{0.5\textwidth}
\vspace{-3mm}
\captionsetup{width=\linewidth-4mm}
\caption{Evaluation of \texttt{HyPA} on $k$-safety. We give the size of the abstract game-space (Size), the time taken to compute the abstraction ($t_\mathit{abs}$), and the overall time taken by \texttt{HyPA} ($t$).
Times are given in seconds.
} \label{tab:ksafetyeval}
\vspace{-1mm}
\centering
\small
\def1.3{1.3}
\setlength\tabcolsep{1.7mm}
\setlength\dashlinedash{1pt}
\setlength\dashlinegap{2pt}
\setlength\arrayrulewidth{0.7pt}
\begin{tabular}{lccc}
\toprule[1pt]
\textbf{Instance} & \textbf{Size} & $\boldsymbol{t}_\mathit{abs}$ & $\boldsymbol{t}$ \\
%
\cmidrule[1pt](r{1mm}){1-1}
\cmidrule[1pt](l{1mm}r{1mm}){2-2}
\cmidrule[1pt](l{1mm}r{1mm}){3-3}
\cmidrule[1pt](l{1mm}r{0mm}){4-4}
%
DoubleSquareNI & 819 & 92.3 & 92.8 \\
%
\hdashline
%
HalfSquareNI & 1166 & 85.9 & 86.5 \\
%
\hdashline
%
SquaresSum & 286 & 29.8 & 29.9 \\
%
\hdashline
%
ArrayInsert & 213 & 28.2 & 28.2 \\
%
\hdashline
%
Exp1x3 & 112 & 4.5 & 4.5 \\
%
\hdashline
%
Fig3& 268 & 11.9 & 12.0 \\
%
\hdashline
%
DoubleSquareNIff & 121 & 9.8 & 9.9 \\
%
\hdashline
%
\Cref{fig:univ} & 333 & 23.7 & 23.8 \\
%
\hdashline
%
ColIitem-Symm & 494 & 24.0 & 24.1 \\
%
\hdashline
%
Counter-Det & 216 & 10.2 & 10.3 \\
%
\hdashline
%
MultEquiv & 757 & 18.9 & 19.0 \\
\bottomrule[1pt]
\end{tabular}
\vspace{-3mm}
\end{wraptable}
When combining \Cref{theo:soundnessBeyond} and our iterative solver from \Cref{sec:solvingInner} we obtain an effective algorithm to verify $\forall^*\exists^*$ properties within a given abstraction.
We have implemented a prototype of our method in a tool we call \texttt{HyPA}.
We use \texttt{Z3} \cite{MouraB08} to discharge SMT queries.
The input of our tool is provided as an arbitrary STS in the SMTLIB format \cite{barrett2010smt} making it \emph{language independent}.
In our programs we make the program counter explicit, allowing us to track predicates locally \cite{HenzingerJMS02}.
\paragraph{Evaluation for $k$-Safety.}
As a special case of $\forall^*\exists^*$ properties, \texttt{HyPA} is also applicable to $k$-safety verification.
We collected an exemplifying suite of programs and $k$-safety properties from the literature \cite{ShemerGSV19,FarzanV19,UnnoTK21,SousaD16,UnnoTK21} and manually translated them into STS (this can be automated easily).
The results are given in \Cref{tab:ksafetyeval}.
As done by Shemer et al.~\cite{ShemerGSV19}, we already provide a set of predicates that is sufficient for \emph{some} reduction (but not necessarily the lockstep or sequential one), the search for which is then automated by \texttt{HyPA}.
Our results show the game-based search for a reduction can verify interesting $k$-safety properties from the literature.
We also note, that currently the vast majority of time is spent on the construction of the abstract system.
If we would move to fixed language the computation time of the initial abstraction could be reduced by using existing (heavily optimized) abstraction tools \cite{HenzingerJMS02,ChakiCGJV04}.
\begin{table}[!t]
\caption{Evaluation of \texttt{HyPA} on $\forall^*\exists^*$ verification instances.
We give the size and construction time of the initial abstraction (Size and $t_\mathit{abs}$).
For both the na\"ive (direct) and lazy (\Cref{alg:cap}) solver we give the time to construct (and solve) the game ($t_\mathit{solve}$) and the overall time ($t = t_\mathit{abs} + t_\mathit{solve}$).
For the lazy solver we, additionally, give the number of refinement iterations (\#Ref).
Times are given in seconds.
TO indicates a timeout after 5 minutes.} \label{tab:beyondeval}
\vspace{2mm}
\centering
\small
\def1.3{1.3}
\setlength\tabcolsep{1.7mm}
\setlength\dashlinedash{1pt}
\setlength\dashlinegap{2pt}
\setlength\arrayrulewidth{0.7pt}
\begin{tabular}{lcc@{\hspace{7mm}}cc@{\hspace{7mm}}ccc}
\toprule[1pt]
&&& \multicolumn{2}{@{}c@{\hspace{7mm}}}{\textbf{Na\"ive}} &\multicolumn{3}{c}{\textbf{Lazy}} \\
%
\cmidrule[1pt](l{-1mm}r{6mm}){4-5}
\cmidrule[1pt](l{-1mm}r{0mm}){6-8}
%
\textbf{Instance} & \textbf{Size} & $\boldsymbol{t}_\mathit{abs}$ & $\boldsymbol{t}_\mathit{solve}$ & $\boldsymbol{t}$ & \textbf{\#Ref} & $\boldsymbol{t}_\mathit{solve}$ & $\boldsymbol{t}$ \\
%
\cmidrule[1pt](r{1mm}){1-1}
\cmidrule[1pt](l{0.5mm}r{1mm}){2-2}
\cmidrule[1pt](l{0.5mm}r{6mm}){3-3}
\cmidrule[1pt](l{-1mm}r{1mm}){4-4}
\cmidrule[1pt](l{0.5mm}r{6mm}){5-5}
\cmidrule[1pt](l{-1mm}r{1mm}){6-6}
\cmidrule[1pt](l{0.5mm}r{1mm}){7-7}
\cmidrule[1pt](l{0.5mm}r{0mm}){8-8}
%
NonDetAdd & 4568 & 3.5 & TO & TO & 4 & 1.0 & 4.6 \\
%
\hdashline
%
CounterSum & 479 & 5.3 & 9.1 & 14.4 & 17 & 0.9 & 6.3 \\
%
\hdashline
%
AsynchGNI & 437 & 6.1 & 6.9 & 13.0 & 1 & 0.1 & 6.2 \\
%
\hdashline
%
CompilerOpt1 & 354 & 2.4 & 2.3 & 4.7 & 2 & 0.2 & 2.6 \\
%
\hdashline
%
CompilerOpt2 & 338 & 2.8 & 2.4 & 5.2 & 2 & 0.2 & 3.0 \\
%
\hdashline
%
Refine & 1357 & 6.1 & TO & TO & 4 & 0.7 & 6.8 \\
%
\hdashline
%
Refine2 & 1476 & 5.6 & TO & TO & 5 & 0.6 & 6.3 \\
%
\hdashline
%
Smaller& 327 & 2.3 & 4.0 & 6.3 & 11 & 0.4 & 2.8 \\
%
\hdashline
%
CounterDiff & 959 & 8.5 & 18.3 & 26.8 & 19 & 1.1 & 9.7 \\
%
\hdashline
%
\Cref{fig:exist} & 3180 & 11.1 & TO & TO & 22 & 2.9 & 14.1 \\
%
\hdashline
%
P1$_{\text{simple}}$ & 83 & 2.0 & 1.4 & 3.4 & 1 & 0.1 & 2.0 \\
%
\hdashline
%
P1$_{\text{GNI}}$ & 34793 & 17.0 & TO & TO & 72 & 95.7 & 112.7 \\
%
\hdashline
%
P2$_{\text{GNI}}$ & 15753 & 10.2 & TO & TO & 7 & 5.1 & 15.4 \\
%
\hdashline
%
P3$_{\text{GNI}}$ & 1429 & 6.6 & 20.9 & 27.5 & 7 & 0.6 & 7.2 \\
%
\hdashline
%
P4$_{\text{GNI}}$ & 7505 & 16.5 & TO & TO & 72 & 13.2 & 29.7 \\
\bottomrule[1pt]
\end{tabular}
\end{table}
\paragraph{Evaluation Beyond $k$-Safety.}
The main novelty of \texttt{HyPA} lies in its ability to, for the first time, verify temporal properties beyond $k$-safety.
As none of the existing tools can verify such properties, we compiled a collection of very small example programs and $\forall^*\exists^*$ properties.
Additionally, we modified the boolean programs from \cite{BeutnerF21} (where they checked GNI on boolean programs) by including data from infinite domains.
The properties we checked range from refinement properties for compiler optimizations, over general refinement of nondeterministic programs, to generalized non-interference.
Verification often requires a non-trivial combination of reduction and witness strategy (as the reduction must, e.g., compensate for branches of different length).
As before we provide a set of predicates and let \texttt{HyPA} automatically search for a witness strategy with accompanying reduction.
We list the results in \Cref{tab:beyondeval}.
To highlight the effectiveness of our inner refinement loop, we apply both the direct (na\"ive) construction of $\gameEF{\mathcal{T}}{\varphi}{\mathcal{P}}$ and the iterative (lazy) solver in \Cref{alg:cap}.
Our lazy solver in \Cref{alg:cap} clearly outperforms an explicit construction and is oftentimes the only method to solve the game in reasonable time.
In particular, we require very few refinement iterations and therefore also few expensive SMT queries.
Unsurprisingly, the problem of verifying properties beyond $k$-safety becomes much more challenging (compared to $k$-safety verification) as it involves the \emph{synthesis} of a witness function which is already 2\texttt{EXPTIME}-hard for finite-state systems \cite{PnueliR89,CoenenFST19}.
We emphasize, that no other existing tool can verify any of the benchmarks.
\section{Related Work}
\label{sec:relatedWork}
\paragraph{Asynchronous Hyperproperties.}
Recently, many logics for the formal specification of asynchronous hyperproperties have been developed \cite{BaumeisterCBFS21,GutsfeldMO21,BozzelliPS21,BeutnerF21}.
Our logic OHyperLTL is closely related to stuttering HyperLTL (HyperLTL$_S$) \cite{BozzelliPS21}.
In HyperLTL$_S$ each temporal operator is endowed with a set of temporal formulas $\Gamma$ and steps where the truth values of all formulas in $\Gamma$ remain unchanged are ignored during the operator's evaluation.
As for most mechanism used to design asynchronous hyperlogics \cite{BaumeisterCBFS21,GutsfeldMO21,BozzelliPS21}, finite-state model checking of HyperLTL$_S$ is undecidable.
By contrast, in OHyperLTL we always observe the trace at a fixed location, which is key for ensuring decidable finite-state model checking (cf.~\Cref{theo:finiteState}).
\paragraph{$k$-Safety Verification.}
The literature on $k$-safety verification is rich.
Many approaches verify $k$-safety by using a form of self-composition \cite{BartheDR11,eilers2019modular,churchill2019semantic,FinkbeinerRS15} and often employ reductions to obtain compositions that are easier to verify.
Our game-based interpretation of a reduction (\Cref{sec:ksafety}) is related to Shemer et al.~\cite{ShemerGSV19} who study $k$-safety verification within a given predicate abstraction using an enumeration-based solver (see \Cref{sec:ksafety} for a discussion).
Farzan and Vandikas \cite{FarzanV19} present a counterexample-guided refinement loop that simultaneously searches for a reduction and a proof.
Sousa and Dillig \cite{SousaD16} facilitate reductions at the source-code level in program logic.
\paragraph{$\forall^*\exists^*$-Verification.}
Barthe et al.~\cite{BartheCK13} describe an asymmetric product of the system such that only a subset of the behavior of the second system is preserved, thereby allowing the verification of $\forall^*\exists^*$ properties.
Constructing an asymmetric product and verifying its correctness (i.e., showing that the product preserves all behavior of the first, universally quantified, system) is challenging.
Unno et al.~\cite{UnnoTK21} present a constraint-based approach to verify functional (opposed to temporal) $\forall^*\exists^*$ properties in infinite-state systems using an extension of constraint Horn clauses called pfwCHC.
The underlying verification approach is orthogonal to ours: pfwCHC allows for a clean separation of the actual verification and verification conditions whereas our approach combines both.
For example, our method can prove the existence of a witness strategy without ever formulating precise constraints on the strategy (which seems challenging).
Coenen et al.~\cite{CoenenFST19} introduce the game-based reading of existential quantification to verify temporal $\forall^*\exists^*$ in a synchronous and finite-state setting.
By contrast, our work constitutes the first verification method for temporal $\forall^*\exists^*$ properties in \emph{infinite-state} systems.
The key to our method is a careful combination of reduction and witness strategy (which is not possible in a synchronous setting).
For finite-state systems (where the abstraction is precise) and synchronous specifications (where we observe every step), our method subsumes the one in \cite{CoenenFST19}.
Beutner and Finkbeiner \cite{BeutnerF22} use prophecy variables to ensure that the game-based reading of existential quantification is complete in a finite-state setting.
Automatically constructing prophecies for infinite-state systems is interesting future work.
Pommellet and Touili \cite{PommelletT18} study the verification of HyperLTL in infinite-state systems arising from pushdown systems.
By contrast, we study verification in infinite-state systems that arise from the infinite variables domains used in software.
\paragraph{Game Solving.}
Our game-based interpretations are naturally related to infinite-state game solving \cite{FarzanK18,BaierCFFJS21,WalkerR14,BeyeneCPR14}.
State-of-the-art solvers for infinite-state games unroll the game \cite{FarzanK18}, use necessary subgoals to inductively split a game into subgames \cite{BaierCFFJS21}, encode the game as a constraint system \cite{BeyeneCPR14}, and iteratively refine the controllable predecessor operator \cite{WalkerR14}.
We tried to encode our verification approach directly as an infinite-state linear-arithmetic game.
However, existing solvers (which, importantly, work \emph{without} a user-provided set of predicates) could not solve the resulting game \cite{FarzanK18,BaierCFFJS21}.
Our method for encoding the witness-strategy for general $\forall^*\exists^*$ properties using \emph{restrictions} corresponds to hyper-must edges in general abstract games \cite{AlfaroGJ04,AlfaroR07}.
Our inner refinement loop for solving a game with hyper-must edges without explicitly identifying all edges (\Cref{alg:cap}) is thus also applicable in general abstract games.
\section{Conclusion}
In this work, we have presented the first verification method for temporal hyperproperties beyond $k$-safety in infinite-state systems arising in software.
Our method is based on a game-based interpretation of reductions and existential quantification and allows for mutual dependence of both.
A direct next step would be to integrate our approach in a counter-example guided refinement loop that automatically refines the current abstraction.
Moreover, in future work, it is interesting to study if, and to what extent, the numerous other methods developed for $k$-safety verification of infinite-state systems (apart from reductions) are applicable to the vast landscape of hyperproperties that lies beyond $k$-safety.
\subsubsection{Acknowledgments}
This work was partially supported by the DFG in project 389792660 (Center for Perspicuous Systems, TRR 248).
R.~Beutner carried out this work as a member of the Saarbrücken Graduate School of Computer Science.
\bibliographystyle{splncs04}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 2,760
|
Man arrested and banned from Madison Square Garden after allegedly assaulting two men after hockey game
Around 10:50 p.m. Thursday, following the New York Rangers and Tampa Bay Lightning game, police said two men got into a verbal dispute that quickly became physical.
One of the men — James Anastasio — allegedly punched the other in the face, causing him to fall to the ground and lose consciousness, police said.
The suspect then attempted to flee, police said, but was chased by a man they described as a good Samaritan, whom Anastasio also allegedly punched in the face, police said. Both victims refused medical attention and Anastasio was immediately arrested, they said.
The incident was partially captured in a cell phone video, which was posted online.
Anastasio, 29, was arraigned around midnight Friday and charged with two counts of assault in the third degree, two counts of attempted assault in the third degree, and two counts of harassment in the second degree, the Manhattan District Attorney's Office told CNN.
He was granted supervised release and is slated to appear in court June 29, the DA's office added.
Jason Goldman, Anastasio's defense attorney, said his client "repeatedly tried to defuse and deescalate the situation, one where he was immensely outnumbered and ultimately threatened by a group of highly intoxicated individuals."
"We believe that Garden surveillance will confirm the events immediately prior to the short, viral clip captured on cellphone camera," Goldman said. "While he is anxious for the rest of his story to be revealed, he also trusts and respects the Court process. We are confident that James will be vindicated once all the facts come to light."
MSG bans alleged assailant for life
MSG called the incident "an abhorrent assault" and praised the good Samaritan's actions as brave intervention, according to a statement. The venue also banned Anastasio for life.
"All guests — no matter what team they support — should feel safe and respected in The Garden. This has and always will be our policy," MSG said in the statement.
Robert Joseph Kaplan captured the incident on camera. Kaplan told CNN that while he's been to plenty of games in the past, he's never seen anything like this.
Prior to the alleged assault, Kaplan said he saw Anastasio and another man angrily insulting Tampa Bay Lightning fans as they were descending the venue's escalators. He described the insults as "intense" and claimed there was a lack of security by the escalators.
An MSG spokesperson told CNN the alleged assailant was taken into custody within 30 seconds by NYPD security detail and MSG security. There were 183 uniformed MSG security officers, 14 supervisors, 18 NYPD officers and two NYPD sergeants on paid detail inside the venue Thursday night, the spokesperson said.
Amid the alleged insults, Kaplan said an empty beer cup was thrown in the direction of Anastasio and the man he was with. He said the two waited for the victim and confronted him, asking if he was the person who had thrown the cup. It's at this point Kaplan said he took out his phone to record the scene because he sensed something bad was about to happen.
"I thought that was pretty disgusting," Kaplan told CNN. "Something like this shouldn't happen."
Tags: Man arrested and banned from Madison Square Garden after allegedly assaulting two men after hockey game - CNN us
Previous Iran is closer than ever to a nuclear weapon as Biden runs out of options
Next 'Inundated with American travelers': End of testing rule opens summer travel floodgates
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 9,471
|
package com.haulmont.cuba.desktop.gui.components;
import com.haulmont.cuba.core.entity.FileDescriptor;
import com.haulmont.cuba.core.global.Messages;
import com.haulmont.cuba.desktop.App;
import com.haulmont.cuba.desktop.TopLevelFrame;
import com.haulmont.cuba.gui.components.AbstractAction;
import com.haulmont.cuba.gui.components.Action.Status;
import com.haulmont.cuba.gui.components.Component;
import com.haulmont.cuba.gui.components.Frame;
import com.haulmont.cuba.gui.executors.BackgroundWorker;
import com.haulmont.cuba.gui.export.ExportDataProvider;
import com.haulmont.cuba.gui.export.ExportDisplay;
import com.haulmont.cuba.gui.export.ExportFormat;
import com.haulmont.cuba.gui.export.FileDataProvider;
import org.apache.commons.io.FilenameUtils;
import org.apache.commons.io.IOUtils;
import org.apache.commons.lang3.StringUtils;
import org.springframework.context.annotation.Scope;
import javax.inject.Inject;
import javax.swing.*;
import java.awt.*;
import java.io.File;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
/**
* Allows to show exported data in external desktop app or download it.
*/
@org.springframework.stereotype.Component(ExportDisplay.NAME)
@Scope("prototype")
public class DesktopExportDisplay implements ExportDisplay {
private static final String RESERVED_SYMBOLS = "\\/:*?\"<>|";
@Inject
protected BackgroundWorker backgroundWorker;
@Inject
protected Messages messages;
protected Frame frame;
public DesktopExportDisplay() {
}
/**
* Show/Download resource at client side
*
* @param dataProvider {@link ExportDataProvider}
* @param resourceName ResourceName for client side
* @param format {@link ExportFormat}
* @see com.haulmont.cuba.gui.export.FileDataProvider
* @see com.haulmont.cuba.gui.export.ByteArrayDataProvider
*/
@Override
public void show(final ExportDataProvider dataProvider, String resourceName, ExportFormat format) {
backgroundWorker.checkUIAccess();
String fileName = resourceName;
if (format != null) {
if (StringUtils.isEmpty(getFileExt(fileName)))
fileName += "." + format.getFileExt();
}
String dialogMessage = messages.getMessage(getClass(), "export.saveFile");
String correctName = StringUtils.replaceChars(fileName, RESERVED_SYMBOLS, "_");
dialogMessage = String.format(dialogMessage, correctName);
final String finalFileName = correctName;
String fileCaption = messages.getMessage(getClass(), "export.fileCaption");
getFrame().getWindowManager().showOptionDialog(fileCaption, dialogMessage, Frame.MessageType.CONFIRMATION,
new com.haulmont.cuba.gui.components.Action[]{
new AbstractAction("action.openFile", Status.PRIMARY) {
@Override
public void actionPerform(Component component) {
openFileAction(finalFileName, dataProvider);
}
},
new AbstractAction("action.saveFile") {
@Override
public void actionPerform(Component component) {
saveFileAction(finalFileName, getFrame(), dataProvider);
}
},
new AbstractAction("actions.Cancel") {
@Override
public void actionPerform(Component component) {
// do nothing
}
}
});
}
private void saveFileAction(String fileName, JFrame frame, ExportDataProvider dataProvider) {
JFileChooser fileChooser = new JFileChooser();
fileChooser.setSelectedFile(new File(fileName));
if (fileChooser.showSaveDialog(frame) == JFileChooser.APPROVE_OPTION) {
File selectedFile = fileChooser.getSelectedFile();
boolean success = saveFile(dataProvider, selectedFile);
TopLevelFrame mainFrame = App.getInstance().getMainFrame();
if (success) {
mainFrame.showNotification(messages.getMessage(DesktopExportDisplay.class, "export.saveSuccess"), Frame.NotificationType.TRAY);
} else {
mainFrame.showNotification(messages.getMessage(DesktopExportDisplay.class, "export.saveError"), Frame.NotificationType.ERROR);
}
}
}
private void openFileAction(String finalFileName, ExportDataProvider dataProvider) {
File destFile = null;
try {
destFile = File.createTempFile("get_" + FilenameUtils.getBaseName(finalFileName), "." + getFileExt(finalFileName));
} catch (IOException e) {
String message = messages.getMessage(DesktopExportDisplay.class, "export.tempFileError");
getFrame().getWindowManager().showNotification(message, Frame.NotificationType.WARNING);
}
if (destFile != null) {
if (Desktop.isDesktopSupported() && saveFile(dataProvider, destFile)) {
try {
Desktop.getDesktop().open(destFile);
} catch (IOException ex) {
String message = messages.getMessage(DesktopExportDisplay.class, "export.openError");
getFrame().getWindowManager().showNotification(message,
Frame.NotificationType.WARNING);
}
}
}
}
/**
* Show/Download resource at client side
*
* @param dataProvider {@link ExportDataProvider}
* @param resourceName ResourceName for client side
* @see com.haulmont.cuba.gui.export.FileDataProvider
* @see com.haulmont.cuba.gui.export.ByteArrayDataProvider
*/
@Override
public void show(ExportDataProvider dataProvider, String resourceName) {
String extension = getFileExt(resourceName);
ExportFormat format = ExportFormat.getByExtension(extension);
show(dataProvider, resourceName, format);
}
/**
* Show/Download file at client side
*
* @param fileDescriptor File descriptor
* @param format {@link ExportFormat}
*/
@Override
public void show(FileDescriptor fileDescriptor, ExportFormat format) {
show(new FileDataProvider(fileDescriptor), fileDescriptor.getName(), format);
}
@Override
public void show(FileDescriptor fileDescriptor) {
ExportFormat format = ExportFormat.getByExtension(fileDescriptor.getExtension());
show(fileDescriptor, format);
}
@Override
public void setFrame(Frame frame) {
this.frame = frame;
}
@Override
public boolean isShowNewWindow() {
return true;
}
@Override
public void setShowNewWindow(boolean showNewWindow) {
// ignored
}
protected boolean saveFile(ExportDataProvider dataProvider, File destinationFile) {
try {
if (!destinationFile.exists()) {
boolean crateResult = destinationFile.createNewFile();
if (!crateResult)
throw new IOException("Couldn't create file");
}
InputStream fileInput = null;
try {
fileInput = dataProvider.provide();
FileOutputStream outputStream = new FileOutputStream(destinationFile);
IOUtils.copy(fileInput, outputStream);
IOUtils.closeQuietly(fileInput);
IOUtils.closeQuietly(outputStream);
} finally {
if (fileInput != null) {
IOUtils.closeQuietly(fileInput);
}
}
} catch (IOException e) {
String message = messages.getMessage(DesktopExportDisplay.class, "export.saveError");
getFrame().getWindowManager().showNotification(message, com.haulmont.cuba.gui.components.Frame.NotificationType.WARNING);
return false;
}
return true;
}
protected String getFileExt(String fileName) {
int i = fileName.lastIndexOf('.');
if (i > -1)
return StringUtils.substring(fileName, i + 1, i + 20);
else
return "";
}
protected TopLevelFrame getFrame() {
if (frame != null) {
return DesktopComponentsHelper.getTopLevelFrame(frame);
} else {
return App.getInstance().getMainFrame();
}
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 8,343
|
package org.apache.causeway.viewer.wicket.ui.components.tree.themes;
import org.apache.wicket.behavior.Behavior;
import org.apache.causeway.viewer.wicket.ui.components.tree.themes.bootstrap.WktBootstrapTreeTheme;
public class TreeThemeProviderDefault implements TreeThemeProvider {
// other options as provided by wicket-extensions: WindowsTheme, HumanTheme
private final Behavior bootstrapTheme = new WktBootstrapTreeTheme();
@Override
public Behavior treeThemeFor(Object model) {
return bootstrapTheme;
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 3,160
|
Q: How to determine UTM zone from UTM co-ordinates I am trying to convert the UTM coordinates ( 552123.07 4182282.11 ) into Latitude and Longitude so That i can plot it on a Map provider like google or leaflet.
Can anybody please help me in determining how to convert UTM coordinates to Latitude/Longitude?
Your help will be highly appreciated!
I have the following code snippet which does this but it needs a zone number for every UTM coordinate.
public static void UTMToLatLongDSP(double X, double Y, out double latitude, out double longitude)
{
double[] XY = new double[2];
XY[0] = X;
XY[1] = Y;
double[] Z = new double[1];
Z[0] = 1;
string utmStr = "+proj=utm +zone=30 +ellps=WGS84 +datum=WGS84 +units=m +no_defs ";
}
ProjectionInfo projIn = ProjectionInfo.FromProj4String(utmStr);
ProjectionInfo projOut = KnownCoordinateSystems.Geographic.World.WGS1984;
Reproject.ReprojectPoints(XY, Z, projIn, projOut, 0, 1);
longitude = XY[0];
latitude = XY[1];
}
A: Converting UTM coordinates to latitude/longitude without a zone number is not possible. This is due to the fact that UTM coordinates are measured from a reference point of origin of the given zone number.
But, given a latitude/longitude, you can determine the UTM coordinates (northing, easting, zone number)
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 2,036
|
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>W29744_text</title>
<link rel="stylesheet" type="text/css" href="style.css" />
</head>
<body>
<div style="margin-left: auto; margin-right: auto; width: 800px; overflow: hidden;">
<div style="float: left;">
<a href="page48.html">«</a>
</div>
<div style="float: right;">
</div>
</div>
<hr/>
<div style="position: absolute; margin-left: 54px; margin-top: 54px;">
<p class="styleSans1.0000<enum PANGO_WEIGHT_NORMAL of type PangoWeight><enum PANGO_STYLE_NORMAL of type PangoStyle>"></p>
</div>
<div style="position: absolute; margin-left: 274px; margin-top: 274px;">
<p class="styleSans642.0000<enum PANGO_WEIGHT_NORMAL of type PangoWeight><enum PANGO_STYLE_NORMAL of type PangoStyle>">API : . State 5792 41 -1 5 11T aninfi; <br/>WELLBORE SCHEMATIC FIELD: Cottonwood <br/>' 19 5B" Cement Job Lead: 635 SKS g 12.: we - 257 BBLS Tall: :16 ans 914.2 PPG - 19 am <br/>Floab Held <br/>Bumped plug 9 1300 psi 137 bbls of cement to surface <br/>9 518" 361! J-55 LTC 9 2.163' MD 13 '/z " Diem. Hole <br/>TD —2,210' <br/>OASIS PETROLEUM NA LLC <br/>State 5792 41-15 11T Status: Surface Drilled and Cemented T157N R92W Sec 15 Mountail. North Dakota 514' FSL & 292' FWL Sec. 15 Updated: 2/1/2015 MJD <br/> </p>
</div>
<div style="position: absolute; margin-left: 164px; margin-top: 0px;">
<p class="styleSans5.0000<enum PANGO_WEIGHT_NORMAL of type PangoWeight><enum PANGO_STYLE_NORMAL of type PangoStyle>"></p>
</div>
</body>
</html>
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 35
|
@implementation KBFileLabel
- (void)setFile:(KBFile *)file {
[self.nameLabel setText:file.name font:[self.class fontForStyle:self.style] color:KBAppearance.currentAppearance.textColor alignment:NSLeftTextAlignment lineBreakMode:NSLineBreakByTruncatingTail];
self.imageView.image = file.icon;
[self setNeedsLayout];
}
@end
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 3,635
|
well, i seem to be entering the dreaded super uncomfortable stage of the third trimester - my back is aching and twingey and i'm up to four bathroom visits a night, each with a lengthly period of sleeplessness directly after. last night i went to bed at 10pm and got up three times between then and 3:30am when, upon returning to the coziness of bed, i found myself unable to sleep until around 6am. boo. i can't imagine 8 more weeks of this getting progressively worse. the only, very minor, plus side to the sore back is that sitting for too long in any one position is unbearable, forcing me to get up and do various tasks all day long. the nesting thing is in full force so it's sort of working out well, lots getting accomplished but with a fair amount of discomfort and exhaustion.
i managed to finish two much needed cushion covers, using scraps leftover from the quilt i'm working on. i had a rough idea of how i wanted them to look and just randomly started sewing patchwork strips and then sewing them together. i used a lovely, soft grey/blue corduroy on the back and more of the hand-dyed grey/blue cotton for the binding. i also managed to whip up a new dog bed for baxter, which has been on my to-do list for a very, very long time but haven't been able to get any decent photos of it yet.
I remember with my first daughter someone saying to me that the discomfort you feel in the late stages of pregnancy is a way of getting you ready for the lack of sleep once baby arrives! Personally I would have thought it kinder to let us enter a period of absolute perfect sleep, unlimited energy and still able to do up our pants!
Oh poor you - I remember that stage! Wheat packs were my friend, yoga was great - really great to stretch out those sore bits, and I also got into reflexology which was fantastic and was a huge relief to feeling all squashed up!
Yay to ticking things off your to-do list.
Your cushion looks fantastic, really beautiful.
oh my goodness, again, i think i could have written the first paragraph myself!
it is so much like what i am experiencing at this very moment in my pregnancy! so, know this: you're totally not alone!
Thanks Leslie. I hope you're able to get some sleep soon! The cushion is just lovely!
Well, in spite of ll your aches and pains, you've made a beautiful cushion. I feel restful just looking at it.
The cushion is lovely - very clever.
Regarding the interrupted sleeps - it may not necessarily get worse or continue for the rest of the pregnancy. The baby will probably shift position and take pressure off your bladder and you'll get better sleep.
I had a similar nights sleep last night - but it wasn't from something as wonderful as pregnancy. My jetlagged husband kept tossing and turning and woke me up as many times as you described. I think it will be early to bed after vegetating in front of Desperate Housewives with some hand sewing tonight.
I hope you get a better night's sleep tonight.
oh i can't stand getting out of bed to wee. not that i wee in bed... anyways i hope things become more comfortable.
not long to go, hope you manage to catch up on some sleep tonight!
the cushion is lovely, beautiful colours and gorgeous design!
Even though you may be experiencing sleep deprevation you are still making the most gorgeous pieces :) Such a beautiful arrangement of colours on that cushion. Pure genius!
Late pregnancy is just plain hard work. But it does make you friends with sleep deprivation.Been my best friend for almost five years now.
I've never heard of magnolia square market. Sounds fantastic!
Girl, I feel your pain. I have no bladder control at all anymore and my back is constantly aching! And you're farther along than I am, so the light at the end of your tunnel is closer. And don't even get me started on the swollen feet...it's summertime here and pregnancy + humidity do not mix!
Third trimester. Hmmm yup. Don't miss that discomfort AT ALL!!!!! It's all worth it in the long run but I found pregnancy a hard slog!
You're being so super productive, the cushion looks gorgeous!
Lovely colours, I'm loving grey/blue at the moment. HOpe your feeling less tired today. Maybe ice-cream and a midday movie may help.
Oh Leslie - I love this cushion cover. I am not sure what to say about being pregnant the last 8 weeks. Um, yes, it does get a bit uncomfortable depending on which way the baby decides to be. But sometime they turn and you feel better. Maybe she's on something. Just put those feet up when you can!
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 789
|
package org.magcruise.citywalk.model.json.app.task;
import java.util.ArrayList;
import java.util.List;
import org.magcruise.citywalk.model.json.app.TaskJson;
public class SelectionTask extends TaskContent {
private List<String> selections = new ArrayList<>();
private List<Integer> answerIndexes = new ArrayList<>();
public SelectionTask() {
}
@Override
public TaskJson toTaskJson(String id) {
TaskJson j = super.toTaskJson(id);
j.setSelections(selections);
j.setAnswerIndexes(answerIndexes);
return j;
}
public SelectionTask(String label, List<String> selections, List<Integer> answerIndexes,
double point, boolean checkIn, int activeArea) {
super(checkIn, point, label, activeArea);
this.selections.addAll(selections);
setAnswerIndexes(answerIndexes);
}
public List<String> getSelections() {
return selections;
}
public void setSelections(List<String> selections) {
this.selections.addAll(selections);
}
public List<Integer> getAnswerIndexes() {
return answerIndexes;
}
public void setAnswerIndexes(List<Integer> answerIndexes) {
this.answerIndexes = answerIndexes;
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 7,846
|
{"url":"https:\/\/ask.wingware.com\/question\/602\/update-check\/","text":"# update check\n\nHi,\n\nI am using a local proxy to connect to the internet. Using the same proxy configuration Wing IDE 6 update works fine, but Wing IDE 7 fails, as listed below.\n\nAs far as I can see this is a problem with Wing IDE 7 as proxy config is not all that difficult to understand, but I am always open to the possibility that I just don't know enough to make something work :-)\n\nDoes anyone have any ideas how to debug this more or options to try and make the getaddrinfo work?\n\nCheers Mark\n\nwing - 2019\/04\/29 11:42:17 - Starting update check...\nwing - 2019\/04\/29 11:42:41 - Fill is true but expand is false in pack\nwing - 2019\/04\/29 11:42:41 - Runtime failure details:\nwing - 2019\/04\/29 11:42:41 - Exception: \"<class 'urllib2.URLError'>\"\nwing - 2019\/04\/29 11:42:41 - Value = '<urlopen error [Errno 11004] getaddrinfo failed>'\nwing - 2019\/04\/29 11:42:41 - Traceback:\nwing - 2019\/04\/29 11:42:41 - File \"C:\\Program Files\\Wing Pro 7.0\\bootstrap\\wing.py\", line 242, in <module>\nwing - 2019\/04\/29 11:42:41 - startwing.Main(argv, importer, err_list)\nwing - 2019\/04\/29 11:42:41 - File \"C:\\src\\ide\/bin\/ide-2.7\/src\\startwing.pyo\", line 957, in Main\nwing - 2019\/04\/29 11:42:41 - File \"C:\\src\\ide\/bin\/ide-2.7\/src\\startwing.pyo\", line 685, in Run\nwing - 2019\/04\/29 11:42:41 - File \"C:\\src\\ide\/bin\/ide-2.7\/src\\startwing.pyo\", line 680, in start\nwing - 2019\/04\/29 11:42:41 - File \"C:\\src\\ide\/bin\/ide-2.7\/src\\main.pyo\", line 168, in main\nwing - 2019\/04\/29 11:42:41 - File \"C:\\src\\ide\/bin\/ide-2.7\/src\\wingide\\wingapp.pyo\", line 816, in Run\nwing - 2019\/04\/29 11:42:41 - File \"C:\\src\\ide\/bin\/ide-2.7\/src\\guimgr\\guimanager.pyo\", line 1204, in RunGUI\nwing - 2019\/04\/29 11:42:41 - File \"C:\\src\\ide\/bin\/ide-2.7\/src\\guiutils\\qt_utils.pyo\", line 1513, in main\nwing - 2019\/04\/29 11:42:41 - File \"C:\\src\\ide\/bin\/ide-2.7\/src\\wingide\\updatemgr.pyo\", line 243, in __CB_Step\nwing - 2019\/04\/29 11:42:41 - File \"C:\\src\\ide\/bin\/ide-2.7\/src\\wingide\\updatemgr.pyo\", line 223, in __CB_Step\nwing - 2019\/04\/29 11:42:41 - File \"C:\\src\\ide\/bin\/ide-2.7\/src\\wingide\\updatemgr.pyo\", line 306, in __Iteration\nwing - 2019\/04\/29 11:42:41 - File \"C:\\src\\ide\/bin\/ide-2.7\/src\\wingutils\\urlutils.pyo\", line 51, in safe_urlopen\nwing - 2019\/04\/29 11:42:41 - File \"C:\\Program Files\\Wing Pro 7.0\\bin\\__os__\\win32\\runtime-python2.7\\lib\\urllib2.py\", line 154, in urlopen\nwing - 2019\/04\/29 11:42:41 - return opener.open(url, data, timeout)\nwing - 2019\/04\/29 11:42:41 - File \"C:\\Program Files\\Wing Pro 7.0\\bin\\__os__\\win32\\runtime-python2.7\\lib\\urllib2.py\", line ...\nedit retag close merge delete\n\nSort by \u00bb oldest newest most voted\n\nI think I see the problem. Wing 7 switched to always using https and the proxy setup is only being done for http. We'll try to fix this in the next update.\n\nmore\n\nThere is a tickbox to de\/select HTTPS when connecting to wingware.com in Wing 7 and I have tried selected and not but the connection fails in both cases. If Wing 7 ignores this tickbox and always connects using HTTPS then it would be a good idea to remove this tickbox too.\n\n( 2019-05-01 02:06:12 -0500 )edit\n\nIt looks like it should be working to substitute in http for https when that option is unchecked, but perhaps there is some other problem. I didn't go into the Python code yet but it's possible that I was wrong above and that installing proxy info for http and covers https. If so, it may be the second answer in https:\/\/stackoverflow.com\/questions\/4... is relevant -- namely, that on Windows it's ignoring our proxy setup and using different settings from IE or the OS. If so, it may be that defining http_proxy=http:\/\/userid:pswd@proxyurl.com:port in the system env and restarting Wing will solve it.\n\n( 2019-05-01 08:49:09 -0500 )edit","date":"2022-07-05 09:31:57","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.20173992216587067, \"perplexity\": 8914.91738978412}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-27\/segments\/1656104542759.82\/warc\/CC-MAIN-20220705083545-20220705113545-00314.warc.gz\"}"}
| null | null |
\section{Introduction}
\label{sect:intro}
Sensitive searches with the Atacama Large Millimeter/submillimeter Array (ALMA) are steadily adding to the number of disks with dust levels typical of debris disks but hosting detected CO gas, now comprising 15 systems \citep{Zuckerman1995,Moor2011,Dent2014,Moor2015,Marino2016, Lieman-Sifry2016,Greaves2016,Matra2017b,Marino2017a,Moor2017}. All but two of these are relatively young ($\sim$10-100 Myr-old), which would naively suggest that the presence of CO is related to the protoplanetary era of disk evolution. However, the recent detections of CO around the 440 Myr-old star Fomalhaut \citep{Matra2017b} and the 1-2 Gyr-old star $\eta$ Corvi \citep{Marino2017a}, as well as the tentative detection around the 125-130 Myr-old eclipsing binary HD23642 \citep{Pericaud2017}, show that at least in some of the systems the CO cannot be primordial. Instead, the wide range of system ages advocates for CO replenishment through gas release from exocometary ices
This exocomet scenario has now been confirmed as the origin of the observed CO gas in four systems, including relatively young stars $\beta$ Pictoris \citep{Dent2014,Matra2017a} and HD181327 \citep{Marino2016}, as well as Fomalhaut and $\eta$ Corvi. While the exact CO release mechanism itself remains unknown, the similarly clumpy CO and dust distributions in the disk around $\beta$ Pictoris suggest that collisions play an important role \citep{Dent2014}. A collisional cascade is expected to result in efficient gas release as long as the smallest grains blown out of the system by radiation pressure from the central star cannot retain their ice \citep{Matra2017b}. In this case, assuming steady state in both the collisional cascade and in the production/photodestruction of gas allows us to extract the CO ice fraction in exocomets from simple observables \citep{Zuckerman2012,Matra2015,Matra2017b}. This is a particularly powerful tool to probe the bulk volatile composition of exocomets, allowing us to place Solar System comets in the larger context of planetary systems around stars with different properties.
In the debris disk systems where an exocometary origin of CO has been confirmed (two A- and one F-type stars), measured CO mass fractions of exocomets are consistent with Solar System comets \citep{Marino2016,Matra2017a,Matra2017b}. This consistency suggests similar comet formation conditions in the original protoplanetary disks and the Solar Nebula. The next step in investigating this similarity requires to both expand this sample of exocometary CO abundance measurements and probe relative ice abundances through detection of other molecules. As we discuss in this paper (\S\ref{sect:disc}), detection of other molecules can uniquely shed light on both the exocometary outgassing mechanism (where outgassing refers to any gas release process from solid bodies of any size) and the chemical history of cometary volatiles.
Here we present the first molecular survey of the exocometary gas disk around $\beta$ Pictoris. In \S\ref{sect:obs}, we describe new SMA and archival ALMA observations of CO, CN, HCN, HCO$^+$, N$_2$H$^+$, H$_2$CO, H$_2$S, CH$_3$OH, SiO and DCN. The continuum and CO J=2-1 detections with the SMA as well as SMA and ALMA upper limits for all other molecules are presented in \S\ref{sect:res}. We interpret the observations with a steady state exocometary release model in \S\ref{sect:anal}. In particular, we focus on CN and CO as the longest-lived observable gas-phase species in the disk, and describe the connection between the constraint on the observed CN/CO flux ratio upper limit and an upper limit in the HCN/(CO+CO$_2$) ratio of outgassing rates. We compare this ratio with measurements from Solar System comets and discuss the implications of this comparison on the release mechanism and/or on the HCN/(CO+CO$_2$) ice abundance in the exocomets in \S\ref{sect:disc}. Finally, we use our exocometary release model to make predictions for molecular detection in future surveys from far-IR to radio wavelengths, and conclude with a summary in \S\ref{sect:concl}.
\section{Observations}
\label{sect:obs}
\begin{deluxetable*}{lcc}
\tablewidth{0pt}
\tablecaption{SMA Observational Parameters}
\tablehead{
\colhead{Parameter} & \colhead{2014 Nov 11} & \colhead{2015 Dec 15}}
\startdata
No. Antennas & 7 & 8 \\
$\tau_{\rm 225~GHz}$ & 0.07 & 0.04 \\
Pointing center &
\multicolumn{2}{c}{
R.A. $05^{h}47^{m}17\fs0877$, Dec. $-51^{h}03^{m}59\fs44$
(J2000) } \\
Min/Max baseline & 6 to 67 meters & 6 to 67 meters \\
Gain Calibrators & J0522-364 (2.7 Jy) & J0522-364 (4.9 Jy) \\
& J0538-440 (1.7 Jy) & J0538-440 (1.5 Jy) \\
Passband Calibrator & 3C84, Uranus & 3C84, Uranus \\
Flux Calibrator & Callisto & Uranus \\
Spectral Lines & ~ & ~ \\
~~transition, frequency & HCN J=3--2, 265.88643 GHz
& CO J=2-1, 230.538 GHz \\
~~ & HCO$^+$ J=3--2, 267.55763 GHz
& CN N= 2-1, 226.87478 GHz \\
~~ & N$_2$H$^+$ J=3--2, 279.51176 GHz
& (J=5/2-3/2 component) \\
~~ & H$_2$CO J=$4_{14}-3_{13}$, 281.52693 GHz
& \\
Primary Beam FWHM & $43''$ & $55''$ \\
Synthesized Beam FWHM\tablenotemark{a}
& $7\farcs7 \times 3\farcs3$, p.a. $12^{\circ}$
& $8\farcs7 \times 3\farcs7$, p.a. $-10^{\circ}$ \\
rms noise (line images) & 0.09 Jy beam$^{-1}$ @ 265.9 GHz & 0.06 Jy beam$^{-1}$ @ 230.5 GHz \\
& (for 0.92 km/s channels) & (for 1.06 km/s channels)\\
rms noise (continuum image) & 1.4 mJy beam$^{-1}$ & 0.57 mJy beam$^{-1}$ \\
\enddata
\tablenotetext{a}{for natural weighting plus a
$3''$ Gaussian taper in the east-west direction}
\label{tab:obs_sma}
\end{deluxetable*}
\subsection{SMA}
\label{sect:obssma}
We observed the $\beta$~Pictoris debris disk with the Submillimeter Array
(SMA) on Mauna Kea, Hawaii in two tracks in a compact antenna configuration
to search for line emission from a suite of molecular species.
Table~\ref{tab:obs_sma} provides a summary of the observational parameters.
The first track was obtained on 2014 November 11 using 7 antennas.
The ASIC correlator was configured with 128 channels (0.8125 MHz spacing)
in each of 48 chunks that together spanned $\pm(4-8)$~GHz from an LO frequency
of 273.732 GHz. This setup was aimed at simultaneous coverage of spectral
lines of HCN and HCO$^+$ (LSB) and N$_2$H$^+$ and H$_2$CO (USB).
The second track was obtained on 2015 December 15 with 8 antennas.
The ASIC correlator was again configured with 128 channels (0.8125 MHz
spacing) in each of 48 chunks, spanning $\pm(4-8)$~GHz from an LO frequency
of 221.654 GHz. In addition, the partly deployed SWARM correlator
(operating at $8/11$ speed) provided an additional 2 chunks of 16384 channels
each ($0.8125/8$~MHz spacing) that spanned $\pm(8-12)$~GHz from the LO
frequency with a gap centered at $10$~GHz of width $1.15$~GHz. This setup
was aimed at simultaneous coverage of spectral lines of CO and CN (USB).
The southern location of $\beta$~Pictoris (Dec. $-51\degr$) makes it a
challenging target for the SMA, since it is available only at high airmass,
rising above an elevation of 15 degrees for only about 3 hours.
However, for both of these tracks the weather conditions were very good,
with stable atmospheric phase and low opacity at 225~GHz as measured by
the nearby tipping radiometer. The basic observing sequence consisted of a
loop of 2 minutes each on the quasars J0522-364 and J0538-440 and 8 minutes
on $\beta$~Pictoris. Passband calibration was obtained with observations
of a combination of available strong sources. The absolute flux scale for
the two tracks were set using observations of Callisto and Uranus, with
an estimated accuracy of 20\%. All of the calibration was performed
using the {\tt MIR\footnote{\url{https://www.cfa.harvard.edu/~cqi/mircook.html}}} software package. Continuum and spectral line imaging
and clean deconvolution were done in the {\tt miriad} package.
The 2014 dataset presented a systematic astrometric offset of $\sim 2.4\arcsec$ (a fraction of the $8\farcs7\times3\farcs7$ beam size) with respect to location of the star. We corrected this offset using the 274 GHz continuum image from the same 2014 dataset, by measuring the centroid of continuum emission and spatially shifting the image to ensure that this centroid lies at the expected stellar location.
\subsection{ALMA}
\label{sect:obsalma}
We extracted ALMA Band 6 observations of $\beta$ Pictoris from the archive (project number 2012.1.00142.S) in order to search for line emission from several additional molecular species. In summary, observations were performed between 2013 and 2015 with the 12-m array in a compact and extended configuration, as well as with the Atacama Compact Array (ACA), with total on-source times of 28, 114, and 50 mins, respectively. We direct the reader to \citet{Matra2017a} for a detailed description of the calibration and CO imaging procedures. Two of the spectral windows were set in frequency division mode and allowed us to test for the presence of spectral line emission from H$_2$S (2$_{2,0}$-2$_{1,1}$, 216.710 GHz), CH$_3$OH (5$_{1,4}$-4$_{2,2}$, 216.946 GHz) SiO (J=5-4, 217.105 GHz) and DCN (J=3-2, 217.239 GHz), as well as CO (J=2-1, 230.538 GHz). We here repeated the imaging step for all molecules in the same way as previously done for the CO transition \citep{Matra2017a}; this resulted in 4 line datasets with a native channel spacing of 488.281 kHz (corresponding to a resolution of 1.35 km/s at the rest frequency of the SiO line). To allow direct comparison with CO (\S\ref{sect:filtering}), we applied a taper to the visibilities in order to produce datacubes with exactly the same synthesized beam size ($0\farcs30\times0\farcs26$, or 13.4$\times$10.7 au at the distance of $\beta$ Pictoris) and position angle (p.a., -83$\fdg$9) as the CO dataset.
\section{Results}
\label{sect:res}
\subsection{SMA continuum and CO J=2-1 detections}
\begin{figure*}
\hspace{15mm}
\includegraphics*[scale=0.49]{biplot.pdf}
\vspace{-6mm}
\caption{\textit{Left}: SMA 1.35 mm continuum image of the $\beta$ Pictoris disk. Contours represent [-2,2,4,6,8] times the rms noise level of 0.57 mJy beam$^{-1}$. \textit{Right}: Contours show the moment 0 (spectrally integrated) map of the CO J=2-1 line, at [-2,2,4,6] times the rms noise level of 0.18 Jy km s$^{-1}$ beam$^{-1}$. The colour scale (moment 1 map) represents the velocity field of the disk, i.e. the centroid velocity of the line (with respect to the star) at each spatial location. In both panels, the ellipse in the bottom left represents the synthesized beam.}
\label{fig:biplot}
\end{figure*}
Continuum emission from dust within the $\beta$ Pictoris disk is detected in both SMA datasets at 222 and 274 GHz (1.35 and 1.09 mm). Figure~\ref{fig:biplot} (left) shows the SMA image with the highest signal-to-noise ratio (1.35 mm), presenting a double-peaked morphology consistent with the 1.09 mm dataset, as well as previous SMA 236 GHz \citep[1.27 mm,][]{Wilner2011} and ALMA 339 GHz \citep[0.89 mm,][]{Dent2014} datasets, interpreted as a belt viewed almost exactly edge-on.
Flux density measurements from the SMA observations at 222 and 274 GHz are $21.0\pm3.0$~mJy and $25.9\pm6.4$~mJy, respectively
(obtained by fitting a Gaussian to the visibilities at baselines shorter than 30~k$\lambda$). These values are consistent with
the ALMA measurement at 339 GHz ($60\pm6$~mJy), given the large uncertainties and steep spectral index of the emission \citep[$\alpha\sim$2.81,][]{Ricci2015}. The previously reported SMA 236 GHz continuum measurement remains anomalously low ($13\pm1.4$~mJy), perhaps resulting from calibration challenges at very low elevations.
CO J=2-1 emission is clearly detected in the SMA channel maps, centered around a velocity of $\sim 1.4$ km/s with respect to the rest frequency of 230.538 GHz in the Local Standard of Rest (LSR) reference frame. We created a moment 0 image (contours in Figure~\ref{fig:biplot}, right) by spectrally integrating the datacube within $\pm 7$ km/s of the stellar velocity (the latter being consistent with the centroid velocity reported above). We used this image to create a spatial mask large enough to include all of the emission from the disk. Then, we applied this mask to the original datacube as a spatial filter; in other words, we spatially integrated emission within this mask. This produced the 1D spectrum shown in Figure~\ref{fig:allmolspec} (top left, lilac line).
The CO line profile is spectrally resolved and appears asymmetric; the enhancement at blueshifted radial velocities is consistent with previous ALMA observations showing that the majority of CO emission originates in a clump on the south-west (SW) side of the edge-on disk, where this SW side is rotating towards Earth \citep{Dent2014,Matra2017a}. This is confirmed by the moment 1 map (color map in Figure~\ref{fig:biplot} right), which shows the line-of-sight velocity centroid of CO emission at spatial locations where the disk is detected (at the 4$\sigma$ level). This centroid was obtained by fitting Gaussians to spectra at all spatial locations. Within the measurement uncertainties, the CO J=2-1 line shape and integrated line flux are consistent with the previous ALMA measurements (see ALMA CO spectrum as the lilac line in Figure~\ref{fig:allmolspec}, top right, as well as CO line fluxes in Table \ref{tab:mols}).
\subsection{Upper limits from molecular survey: spectro-spatial filtering using CO as a template}
\label{sect:filtering}
\begin{figure*}
\centering
\includegraphics*[scale=0.5]{allmols1d_allobs.pdf}
\vspace{-4mm}
\caption{Spectra of the lines targeted in our SMA survey (left) and in archival ALMA data (right). For the CO J=2-1 line (top row), we show the effect of our spectro-spatial filtering technique (see \S\ref{sect:filtering}). Lilac represents the spectrum of the disk spatially integrated over a mask covering all significant emission from the disk. The red line is the result of our technique, which additionally shifts spectra to the zero velocity at each spatial location, leading to a significant S/N boost at the zero velocity. For all other species, we only display spectra after filtering in the same way as done for the CO. Note the different flux scale of the two columns, and of the CO J=2-1 line observed by ALMA with respect to all other lines. Line fluxes and upper limits are reported in Table \ref{tab:mols}.}
\label{fig:allmolspec}
\end{figure*}
No significant signal is detected in the SMA datacubes created around the spectral location corresponding to the CN N=2-1, J=$\frac{5}{2}$-$\frac{3}{2}$ line (composed of the 3 blended F=$\frac{7}{2}$-$\frac{5}{2}$, $\frac{5}{2}$-$\frac{3}{2}$ and $\frac{3}{2}$-$\frac{1}{2}$ hyperfine transitions), HCN J=3-2, HCO$^{+}$ J=3-2 , N$_2$H$^{+}$ J=3-2 and H$_2$CO J=4$_{1,4}$-$3_{1,3}$ lines. Similarly, no significant emission is observed in the ALMA datacubes around the H$_2$S 2$_{2,0}$-2$_{1,1}$, CH$_3$OH 5$_{1,4}$-4$_{2,2}$, SiO J=5-4 and DCN J=3-2 lines.
To optimize detectability, we implement a spectro-spatial filtering technique similar to that introduced in \citet{Matra2015} and subsequently applied to achieve CO detections in the HD181327, $\eta$~Corvi and Fomalhaut debris disks \citep{Marino2016, Marino2017a, Matra2017b}. The method involves choosing a prior on the spectro-spatial distribution of the gas emission, and using this prior to select pixels/channel combinations where emission is expected. This maximizes the signal-to-noise ratio by effectively removing noise-dominated pixels/channels.
We demonstrate this technique by applying it first to the CO J=2-1 line already detected without need for filtering by both the SMA and ALMA. In practice, the process involves two steps. First, we apply spectral filtering by shifting all 1D spectra at each spatial location in the datacube by the negative of the centroid CO velocity displayed in the moment 1 map (Figure~\ref{fig:biplot}, right). This causes the spectra to align such that the bulk of the CO emission now lies at the stellar velocity at all spatial locations. Second, we apply spatial filtering by spatially integrating over the previously created mask covering the region where CO emission is detected. This produces the CO spectra shown in
Figure~\ref{fig:allmolspec} (top, red line) for the SMA (left) and ALMA (right) observations.
This technique significantly improves the signal-to-noise of the detection with respect to the CO spectra directly extracted from the spectral image cube (lilac lines). The improvement is clearer for the ALMA dataset since the line is better resolved both spectrally and spatially, increasing the number of spatial locations that are spectrally independent of one another. In other words, if the emission is better resolved spatially, then the spectral lines at each spatial location will be narrower, and the spectral shifting will produce a better alignment and consequential signal--noise ratio boost.
For simplicity, we here assume emission from all molecules to have similar spectro-spatial distributions of emission as observed for the CO molecule (though see caveats in \S\ref{sect:pred}). This allows us to use the observed CO emission as a template for the spectro-spatial filtering technique. Figure~\ref{fig:allmolspec} shows the resulting filtered spectra for all observed molecules. There are no significant detections of any molecular lines other than CO. We measure the filtered integrated flux upper limits by multiplying the filtered spectrum's root mean square (RMS, measured away from the line frequency) by the number of independent spectral channels covering the line width across which CO is detected. The $3\sigma$ limits derived are listed in Table \ref{tab:mols}. As this assumes that all molecules have the same spectro-spatial distribution as CO, we also calculate upper limits in a more agnostic way by spatially integrating the unfiltered data cubes over regions where the mm continuum is detected, and spectrally integrating between $\pm$5 km/s of the stellar velocity (values in parentheses in Table 2). Whether these are more or less stringent than the CO-filtered limits depends on both the on-sky area over which the continuum versus CO are detected, and on the spectral width of the CO line before versus after spectral filtering, which differs for the SMA and ALMA datasets (see top row in Fig. \ref{fig:allmolspec}).
\begin{deluxetable}{cccccc}
\tabletypesize{\scriptsize}
\tablecaption{Observations of molecular line emission around $\beta$ Pictoris at millimeter wavelengths \label{tab:mols}}
\tablewidth{0pt}
\tablehead{
\colhead{Species} & \colhead{Transition} & \colhead{Freq.} & \colhead{Line Flux} & \colhead{Instr.} &
\colhead{Ref.} \\
\colhead{ } & \colhead{ } & \colhead{(GHz)} & \colhead{(Jy km/s)} & \colhead{ } &
\colhead{ }
}
\startdata
H$_2$S & 2$_{2,0}$-2$_{1,1}$ & 216.710 & $\leqslant$0.10 (0.18) & ALMA & 1 \\
CH$_3$OH & 5$_{1,4}$-4$_{2,2}$ & 216.946 & $\leqslant$0.11 (0.13) & ALMA & 1 \\
SiO & J=5-4 & 217.105 & $\leqslant$0.11 (0.16) & ALMA & 1 \\
DCN & J=3-2 & 217.239 & $\leqslant$0.13 (0.16) & ALMA & 1 \\
CO & J=2-1 & 230.538 & 4.5$\pm$0.5 & ALMA & 2 \\
& & & 3.9$\pm$0.8 & SMA & 1 \\
& J=3-2 & 345.796 & 5.8$\pm$0.6 & ALMA & 2, 3 \\
CN & N=2-1 & 226.875 & $\leqslant$1.0 (1.2) & SMA & 1 \\
HCN & J=3-2 & 265.886 & $\leqslant$2.0 (1.4) & SMA & 1 \\
HCO$^{+}$ & J=3-2 & 267.558 & $\leqslant$1.5 (1.1) & SMA & 1 \\
N$_2$H$^{+}$ & J=3-2 & 279.512 & $\leqslant$1.7 (1.2) & SMA & 1 \\
H$_2$CO & J=4$_{1,4}$-$3_{1,3}$ & 281.527 & $\leqslant$1.6 (1.2) & SMA & 1 \\
\enddata
\tablecomments{For CN, the upper limits refer to the J=$\frac{5}{2}$-$\frac{3}{2}$ transition. Line flux upper limits reported are at the 3$\sigma$ level, after spectro-spatial filtering. Values in parentheses are upper limits obtained from spatial integration over the region where the continuum is detected at a level $>2\sigma$ in each respective dataset.} References: (1) This work; (2) \citet{Matra2017a}; (3) \citet{Dent2014}
\end{deluxetable}
\section{Analysis}
\label{sect:anal}
We investigate how to connect the line flux upper limits measured in \S\ref{sect:res} to exocometary ice compositions relative to CO, the only molecular species detected so far in exocometary gas disks. In particular, we aim to understand the significance of these upper limits in relation to Solar System cometary compositions. We put a special emphasis on the CN/CO ratio upper limit and its consequences for HCN and CO in $\beta$ Pictoris' exocometary ices, for reasons which will become apparent in \S\ref{sect:lifetimes}.
\subsection{Expanding the steady state cometary release model}
\label{sect:expandmodel}
In order to understand the factors that affect remote detection of molecular species released in the gas phase from exocometary ices, we
use the steady state exocometary gas release model of \citet{Matra2015,Matra2017b} and expand it to include molecular species other than CO. Within the framework of this model, the mass production rate $\dot{M_{\rm i}^+}$ of molecular species i will equal its destruction rate $\dot{M_{\rm i}^-}$,
\begin{equation}
\dot{M_{\rm i}^+}=m_{\rm i}\dot{N_{\rm i}^+}=\dot{M_{\rm i}^-}=\frac{M_{\rm i}}{\tau_{\rm i}^-},
\end{equation}
where $m_{\rm i}$ is its molecular mass in kg, $\dot{N_{\rm i}^+}$ its production rate in number of molecules per second, $M_{\rm i}$ its observed gas mass, and $\tau_{\rm i}^-$ its destruction timescale in the gas phase, which we here assume to be dominated by photodissociation, neglecting any other gas phase chemical reactions ($\tau_{\rm i}^-=\tau_{\rm i, phd}^-$).
Then, the gas mass of species i we would expect to observe relative to CO is
\begin{equation}
\frac{M_{\rm i}}{M_{\rm CO}}=\frac{\tau_{\rm i, phd}^-m_{\rm i}}{\tau_{\rm CO, phd}^-m_{\rm CO}}\frac{\dot{N_{\rm i}^+}}{\dot{N_{\rm CO}^+}}.
\end{equation}
If CO and species i are released in the gas at the same rate \textit{per molecule} (an assumption that we will revisit in \S\ref{sect:depl}) we can directly link an observed gas mass ratio to an ice abundance ratio in the exocomets. Furthermore, this relation shows the crucial impact of the photodissociation timescale on the observable mass of exocometary gas species (see \S\ref{sect:lifetimes}).
The next step is to link the molecular gas mass to line fluxes for given transitions as observed in this work. Following e.g. Eq. 2 in \citet{Matra2015}, and assuming optically thin transitions (see \S\ref{sect:pred}),
\begin{equation}
\label{eq:releaseratetoflux}
\frac{F_{\nu_{\rm u\rightarrow l, i}}}{F_{\nu_{\rm u\rightarrow l, CO}}}=\frac{ A_{\rm u\rightarrow l, i}}{A_{\rm u\rightarrow l, CO}}\frac{x_{\rm u, i}}{x_{\rm u, CO}}\frac{\tau_{\rm i, phd}^-}{\tau_{\rm CO, phd}^-}\frac{\dot{N_{\rm i}^+}}{\dot{N_{\rm CO}^+}},
\end{equation}
where, for both CO and species i, $F_{\nu_{\rm u\rightarrow l}}$ are integrated line fluxes in Jy km/s, $A$ and $x_{\rm u}$ are Einstein A coefficients and upper level fractional populations of the observed transition.
We note that the steady state relation in Eq. \ref{eq:releaseratetoflux} remains the same for daughter molecular species such as CN and OH, as long as their production is dominated by a single parent species (such as HCN and H$_2$O).
\subsection{The effect of photodissociation: lifetime of Solar System cometary molecules}
\label{sect:lifetimes}
\begin{figure*}
\vspace{-0mm}
\hspace{-2mm}
\includegraphics*[scale=0.65]{phdcombo_8.pdf}
\vspace{-5mm}
\caption{\textit{Left:} Unshielded photodissociation timescales of molecular gas species observed in Solar System comets. Species are listed in order of their timescale, from longest lived to shortest lived, and are therefore in the same vertical order as the corresponding lines (though note that the C$_2$H$_2$ line overlaps with the CH$_3$OH line). The transparency of solid lines is set to indicate their log-scale abundance in Solar System cometary ice with respect to H$_2$O, darker for more abundant species. The purple line is for N$_2$, by far the least abundant of all listed species from measurements in comet 67P \citep{Rubin2015}. Orange dashed lines represent daughter species (CN, OH) or species that we observed but are not detected in Solar System comets (SiO). \textit{Right:} Example of our derivation of photodissociation timescales for CO; the top panel shows cross sections at relevant wavelengths obtained from the \citet{Heays2017} database. The bottom panel shows the mean intensity felt by a molecule at 85 AU in the $\beta$ Pic disk (assuming the gas to be optically thin to UV photons). We show contributions from both the ISRF and the star, highlighting the difference between the observed stellar flux of $\beta$ Pictoris and that expected from a model of the same spectral type as the star.
}
\label{fig:timescales}
\end{figure*}
In Figure~\ref{fig:timescales} (left), we present unshielded photodissociation timescales for molecules observed around $\beta$ Pic as well as other species observed to be abundant in Solar System comets. We do not include HCO$^+$ and N$_2$H$^+$ as these ions, if present at all, would likely recombine through free electrons faster than they photodissociate; taking the example of HCO$^+$, using an electron density of 10$^2$-10$^3$ cm$^{-3}$ \citep{Matra2017a}, temperatures between 20 and 100 K and reaction rates from the KIDA \footnote{\url{http://kida.obs.u-bordeaux1.fr}} database \citep{Wakelam2012}, its recombination timescale is 0.2 to 5 days, much shorter than the $3.7\times10^3$ yr photodissociation timescale at 85 AU from the star.
Photodissociation timescales for all molecules were calculated using photodissociation cross sections from \citet[][]{Heays2017}\footnote{\url{http://home.strw.leidenuniv.nl/~ewine/photo/}} and references therein, neglecting isotope-selective photodissociation for rarer isotopologues. The radiation field includes both the observed spectrum of the $\beta$ Pic star and the interstellar radiation field (ISRF). The stellar spectrum (A. Brandeker, priv. comm.) was obtained by superimposing observed spectra from HST/STIS \citep{Roberge2000} and FUSE \citep{Bouret2002,Roberge2006} on the photospheric model spectrum of \citet{Fernandez2006}. For the ISRF we adopted the formulation of \citet{Draine1978} with the long wavelength expansion of \citet{vanDishoeck2006}.
We note that the $\beta$ Pictoris star is surprisingly active for an A star, with X-ray observations presenting evidence for thermal emission from a cool corona \citep{Hempel2005, Gunther2012}, as well as UV chromospheric emission \citep{Deleuil2001, Bouret2002}. The latter effectively causes a considerable UV flux `excess' compared to a typical main sequence A-star; this acts to shorten the survival lifetime of molecules against photodissociation, particularly for molecules whose photodissociation bands lie at shorter UV wavelengths (such as CO, see Figure~\ref{fig:timescales}, top right). This leads to a considerable change in the unshielded CO photodissociation timescale for the $\beta$ Pictoris gas disk; Figure~\ref{fig:timescales} (bottom right) shows that photodissociation at the clump radial location ($\sim$85 AU) has a significant contribution by the strong UV field of $\beta$ Pictoris as well as by the ISRF \citep[where the latter would instead dominate for a typical A5V stellar model, ][]{Kamp2000}.
We therefore revise the unshielded CO photodissociation lifetime from 120 yrs \citep{Kamp2000, Dent2014} down to 42 yrs.
We note that dust shielding is negligible for optically thin dust in debris disks \citep[e.g.][]{Kamp2000}, and H$_2$, if at all present, should have a low enough column density not to provide shielding either \citep{Matra2017a}. CO is however abundant enough that self-shielding cannot be neglected \citep{Matra2017a}. When this is taken into account, the CO lifetime increases from 42 to $\sim$105 yrs.
This is accounted for in all calculations involving CO.
We neglect self-shielding and shielding by CO for all other molecules due to their unknown spatial distributions and shielding functions. Neglecting self-shielding should be a good approximation given the much lower predicted abundance of molecules other than CO.
Overall, Figure~\ref{fig:timescales} (left) shows that the harsh UV radiation field around $\beta$ Pic causes photodestruction of the vast majority of volatile species to take place in timescales of days to months.
Therefore, photodissociation alone can explain most of our upper limits, as we confirm more quantitatively in \S\ref{sect:pred}, where we make predictions for future surveys (see Figure~\ref{fig:molpreds}, right).
N$_2$, CO and CN are by far the longest lasting molecules with unshielded photodissociation timescales of 96, 42 and 16 years, at least an order of magnitude longer than for any other molecular species.
This means that the only observable molecule with a survival timescale similar to that of the detected CO is CN, which we therefore focus on for the rest of this work. In Solar System comets CN gas is produced mainly by photodissociation of HCN gas \citep{Fray2005}; we will here assume that the same applies for exocomets around $\beta$ Pictoris.
One potential caveat to this assumption is that although for simplicity we have neglected gas-phase chemical production/destruction pathways other than photodissociation, CN could be destroyed by reaction with atomic oxygen \ion{O}{1}, which has been detected in the disk around $\beta$ Pictoris \citep[e.g.][]{Brandeker2016, Kral2016}. Unfortunately, the detection being unresolved spectrally and spatially combined with excitation and optical depth effects make number density estimates at the locations where CN is expected to lie very uncertain. Nonetheless, combining a number density value of $2\times10^2$ cm$^{-3}$ at 85 AU from the best-fit dynamical evolution model of \citet{Kral2016} with a reaction rate of $5\times10^{11}$ cm$^{3}$ s$^{-1}$ from the KIDA database yields a reaction timescale of $\sim$3 years. This is shorter than the CN photodissociation timescale (16 years), indicating that reaction with oxygen could introduce an uncertainty of a factor of a few to our calculations, increasing the inferred HCN/CO outgassing ratio. Given the uncertainty on the oxygen abundance, however, we opt to neglect this destruction pathway of CN in this work.
\subsection{The effect of molecular excitation including fluorescence}
\label{sect:exc}
The only unknowns remaining to link the observed $F_{\rm CN\ N=2-1,\ J=5/2-3/2}$ / $F_{\rm CO\ J=2-1}$ flux ratio to the $\dot{N_{\rm HCN}^+}$/$\dot{N_{\rm CO}^+}$ ratio of the HCN and CO outgassing rates are the fractional populations of the upper levels $x_{\rm N=2,\ J=5/2,\ CN}$ and $x_{\rm J=2,\ CO}$ of the observed line transitions. \citet{Matra2015} showed that, at least for CO, the assumption of local thermodynamic equilibrium (LTE) is unlikely to apply in the low gas-density environments present in debris disks. This prediction was confirmed by CO line ratio observations in $\beta$ Pictoris \citep{Matra2017a}, and observations of atomic gas species \citep{Kral2016,Kral2017}.
To determine the CO and CN level populations we therefore carry out a full non-local thermodynamic equilibrium (NLTE) calculation which includes solving the equations of statistical equilibrium for both molecules. NLTE level populations depend on the local radiation field within the disk, and on the local density of the dominant collider species (here taken to be electrons) $n_{e^-}$ and the local kinetic temperature $T_k$. We direct the reader to \citet{Matra2015} for details.
\begin{figure}
\hspace{-6.5mm}
\includegraphics*[scale=0.31]{CNoverCOuplevfracpops_vs_ncoll_Tk_bPic.pdf}
\vspace{-8mm}
\caption{Fractional populations $x$ of the upper level of the observed transitions for CO (dashed lines) and CN (solid lines), assuming the radiation field felt by molecules at 85 AU from the star. Fainter lines show the case where fluorescence is not taken into account. In general, the populations depend on the density of the dominant collider species \citep[here electrons, x axis,][]{Matra2015}, and the kinetic temperature of the gas (colours). Increasingly darker colours indicate increasingly lower temperatures.
\label{fig:CNvsCOfracpops}
\end{figure}
Thin dashed and solid lines in Figure~\ref{fig:CNvsCOfracpops} show the resulting dependence of the CO and CN upper level fractional populations $x$ on the electron density and temperature in the gas disk, when applying our previous NLTE model \citep{Matra2015}. Upper level populations vary between two limiting regimes, where excitation of the molecules is dominated by either radiation (far left, low $n_{\rm e^-}$) or collisions (i.e. LTE, far right, high $n_{\rm e^-}$).
As expected, the calculated excitation of the considered CO and CN levels drops by several orders of magnitude in the radiation-dominated regime. This is because the model so far included only low-energy rotational transitions at far-IR/mm wavelengths, which can only be excited by the faint CMB and disk continuum emission.
In Appendix \ref{app:a}, we expand this NLTE excitation model to include vibrational and electronic as well as rotational levels for both CO and CN. This allows us to account for the effect of fluorescence, which in low-density regimes such as Solar System comets and exocometary gas disks, where radiation can dominate excitation, becomes an important excitation mechanism. Fluorescence works through absorption of stellar UV (IR) photons from low rotational levels in the ground electronic and vibrational state to excited electronic (vibrational) levels. This is followed by rapid relaxation to the ground electronic and vibrational state, but to higher rotational levels than the molecule started from. As we can see in Fig. \ref{fig:rotpops} (e.g. green versus red lines),
this effectively leads to enhanced population of rotational levels above ground, where this can be orders of magnitude higher compared to when fluorescence is not considered. We note that fluorescence matters only in the low-density, radiation-dominated regime (left in Fig. \ref{fig:CNvsCOfracpops}, thick versus thin lines), whereas populations in the LTE, collision-dominated regime (right in Fig. \ref{fig:CNvsCOfracpops}) remain unchanged. The extent to which fluorescence impacts molecular excitation depends on the stellar flux received by the molecule and its spectral shape from UV to IR wavelengths.
In general, we find that taking fluorescence into account is crucial when using NLTE excitation models to derive molecular gas masses from observed fluxes. This impacts previous CO mass measurements where this effect was not taken into account; while the exact impact depends on the UV and IR flux received by a molecule which is different for each observed system, the overall effect of the enhanced excitation will be to lower the upper limits in the range of NLTE-derived masses. The total CO mass around $\beta$ Pic remains largely unchanged as it is tightly constrained observationally through the J=3-2/J=2-1 CO line ratio. For completeness, we report an updated value of $3.6^{+0.9}_{-0.6}\times10^{-5}$ M$_{\oplus}$, calculated in the same way as \citet{Matra2017a}. Including fluorescence will also lower the electron densities derived in Fig. 11 of that work by a factor of a few, bringing them closer to the model predictions of \citet{Kral2016}, and flatten their radial distribution in the inner regions of the CO disk.
Taking fluorescence at the clump location into account, we find that the ratio in fractional upper level populations between CN and CO varies between 0.49 and 0.71, with the lowest value for electron densities of 10$^2$-10$^3$ cm$^{-3}$ and temperatures above 50 K. We note that this electron density - temperature dependence of the ratio between CN and CO fractional upper level populations is much weaker than for each individual upper level population, due to the similarity in the structure of rotational energy levels between CO and CN.
\subsection{Deriving the HCN/(CO+CO$_2$) ratio of outgassing rates}
\label{sect:hcnovco}
Plugging CN/CO ratios of upper level populations, photodissociation timescales and observed fluxes into Eq. \ref{eq:releaseratetoflux} allows us to derive an upper limit on the HCN/CO ratio of outgassing rates, which will carry the same dependence on electron density and temperature as the ratio between the CN and CO fractional populations mentioned above (coloured lines in Fig. \ref{fig:CNvsCOfracpops}). The lowest CN/CO ratio in upper level fractional populations sets the most conservative upper limit of 2.5\% on the HCN/CO ratio of outgassing rates
An important point to make is that CO gas is also produced via rapid photodissociation of CO$_2$ gas. This implies that if CO$_2$ is being released from cometary ice together with CO, which we deem likely (see discussion in \S\ref{sect:depl}), the CO gas observed has contributions from both the CO and CO$_2$ ice reservoirs. From now on, we therefore consider our upper limit of 2.5\% to trace the HCN/(CO+CO$_2$) rather than the HCN/CO ratio of outgassing rates.
With this limit, in \S\ref{sect:disc} we draw the comparison between $\beta$ Pictoris and Solar System comets, and discuss the origin of a potential abundance discrepancy.
\section{Discussion}
\label{sect:disc}
\subsection{HCN/(CO+CO$_2$) outgassing rates in $\beta$ Pictoris exocomets vs. Solar System comets}
\label{sect:depl}
\begin{figure}
\hspace{-6.5mm}
\includegraphics*[scale=0.37]{HCNoverCOiceratio_bPic_vs_comets.pdf}
\vspace{-8mm}
\caption{Comparison between our measured upper limit on the HCN/(CO+CO$_2$) ratio of outgassing rates in $\beta$ Pictoris (red) with Solar System comets.
For Solar System comets, symbols represent the HCN/(CO+CO$_2$) outgassing ratios (orange, where available) or HCN/CO outgassing ratios (blue) reported by \citet{LeRoy2015} and references therein. We use filled circles where a single value is reported, and vertical bars where a range of values is reported. Upward and downward pointing triangles represent lower and upper limits, respectively. Green points are HCN/CO measurements for comets observed at large distances from the Sun \citep[Hale-Bopp at 4.6-6.8 AU, Chiron at 8.5-11 AU, and 29P at 5.8 AU,][]{Womack2017}, where different thermal outgassing rates are expected for CO and HCN.}
\label{fig:HCNoverCOoutgasrat}
\end{figure}
Figure~\ref{fig:HCNoverCOoutgasrat} shows our conservative upper limit of 2.5\% on the ratio of the HCN/(CO+CO$_2$) outgassing rate in $\beta$ Pic (red) compared to HCN/(CO+CO$_2$) and HCN/CO values measured in Solar System comets from \citet{LeRoy2015} and references therein (orange and blue symbols, respectively). We find that the upper limit we set in $\beta$ Pic is at the low end of the values derived for comets as they approach the Earth in our own Solar System.
Although our current upper limit per se does not yet indicate a significant depletion compared to Solar System values, we here consider how a low value (and, if confirmed in future, a true depletion) could be linked to either the outgassing mechanism itself, or to the intrinsic HCN/(CO+CO$_2$) abundance in the ice.
The spatial distribution of CO provides an important clue about the gas release mechanism, as CO emission is peaked at a clump located $\sim$85 AU from the star \citep{Dent2014,Matra2017a} that coincides with a dust enhancement most clearly observed in mid-IR imaging \citep{Telesco2005}. This dust enhancement should be associated with higher collisional rates compared to the rest of the belt, and it strongly suggests both the CO and dust enhancements originate from collisions, directly or indirectly.
The measured outgassing rates of molecular species (integrated over all grain sizes) should reflect the ice composition \textit{as long as both ice species are released as gas before they reach the bottom size of the collisional cascade} \citep{Matra2017b}.
Whether this is the case depends on the details of the release process. Gas may be released during collisions through at least three mechanisms:
(1) direct collisional release of species that should have already sublimated at the temperature characteristic of the belt, but were trapped beneath a layer of refractories or within other ice matrices prior to the collision;
(2) thermal sublimation following collisional heating due to e.g. hypervelocity impacts \citep[e.g.][]{Czechowski2007}, or (3) UV photodesorption of resurfaced ice, which was previously buried below a layer of refractories \citep{Grigorieva2007}.
While CO is considered a supervolatile and is most likely fully lost by the time grains reach the blow-out size \citep{Matra2017b}, HCN is not and will not sublimate even at the temperature of the smallest grains in the $\beta$ Pic belt (which dominate the emission and lead to a directly measurable value of $\sim$86 K from the spectral energy distribution). Such temperature dependence on the relative outgassing rates is observed in Solar System comets, where sublimation drives the outgassing and causes the HCN/CO ratio of outgassing rates to decrease with increasing distance from the Sun \citep{Womack2017}. For comparison, we note that $\beta$ Pictoris is much more luminous than our own Sun \citep[$L_{\star}$=8.7 L$_{\odot}$,][]{KennedyWyatt2014}, and that we are observing exocometary CO at distances of $\sim$50-220 of AU, where blackbody temperatures are in the range $\sim$30-70 K. The same temperatures are attained in the Solar System at a distances of 19-39 AU from the Sun, i.e. from the orbit of Uranus out to the Kuiper belt. This means that $\beta$ Pic's exocomets are much colder than observed active comets in our Solar System, even compared to the distantly active ones mentioned above (observed at $\sim$5-11 AU from the Sun). This implies that we would expect CO but not HCN to sublimate from $\beta$ Pic's exocomets. The question is then whether thermal sublimation is the dominant gas release mechanism in the $\beta$ Pic disk.
Given the strong UV field from the central A-type star, UV photodesorption of ice resurfaced after collisions is a promising mechanism to release ice species less volatile than CO to the gas phase. The calculations of \citet[][]{Grigorieva2007} show that the timescale for complete UV photodesorption of a pure H$_2$O ice grain is shorter than the collision timescale for grains below $\sim$20 $\mu$m in size. Assuming HCN has a similar photodesorption yield as water, and neglecting any trapping within a refractory layer, this implies that all of the HCN ice will be released from grains before reaching the smallest size in the collisional cascade. A similar argument applies to CO$_2$, which has a volatility in between that of CO and HCN \citep{Matra2017b}. UV photodesorption is therefore most likely at play, and if dominant that would imply that HCN and CO$_2$ should be released in the gas phase together with CO, at rates reflecting their ice abundances.
Even if HCN does not photodesorb and survives in the ice-phase on the smallest grains in the cascade, it can be thermally sublimated as these unbound grains collide on their way out of the system \citep{Czechowski2007}. This is because stellar radiation pressure accelerates them to sufficiently high velocities for collisions to provide sufficient heating for sublimation. Application of this model to the $\beta$ Pic disk shows that this process can release ice (of unspecified composition) in the gas phase \citep{Czechowski2007}, with gas production rates of the same order as those estimated for CO in the belt. Then, similarly to UV photodesorption, and as long as all ice species are fully sublimated in each impact, high velocity collisions of unbound grains would also produce gas release rates reflecting ice abundances.
In summary, a low HCN/(CO+CO$_2$) ratio of outgassing rates could be attributed to the gas release mechanism only if this is driven by low velocity collisions, as this induces release of trapped supervolatiles alone. However, we would expect outgassing of all molecules independent of their sublimation temperatures if UV photodesorption and/or high velocity collisions with unbound grains are the dominant outgassing mechanisms.
Detailed modeling results \citep{Grigorieva2007,Czechowski2007} show that these processes are most likely at play for small grains in the $\beta$ Pic belt. Then, unless most CO is being released through low velocity \textit{non-catastrophic} collisions between large bodies at the top of the cascade \citep[which are not taken into account by our model, see][]{Matra2017b}, we would expect ratios of outgassing rates to reflect cometary ice abundances. These can then be directly compared with relative abundances from Solar System comets, measured as they come close enough to the Sun for their ratios of outgassing rates to reflect ice compositions.
In the scenario where outgassing in the $\beta$ Pic belt is dominated by UV photodesorption or high velocity collisions of small grains, our upper limit on the HCN/(CO+CO$_2$) ratio of outgassing rates for $\beta$ Pic's exocomets suggests an HCN/(CO+CO$_2$) ice abundance ratio that is at the low end of the range observed in Solar System comets, although still not exceptionally low.
Future, deeper observations (Sect. \S\ref{sect:pred}) are required to conclusively determine whether the abundance of $\beta$ Pic's exocomets is truly abnormal compared to Solar System comets.
Determining the abundance of HCN in exocomets is especially interesting because of its connections to the origins of life chemistry \citep[e.g.][]{Powner2009}. The large scatter in HCN abundances observed amongst Solar System comets \citep[][and references therein]{Mumma2011}, together with observation of cyanides in protoplanetary disks \citep{Oberg2015, Guzman2017, Huang2017}, indicate that cometary cyanides may not be directly inherited from the ISM, but undergo active chemical processing during the epoch of planet formation. Whether the endpoint of this chemistry is different for different planetary systems is yet unclear and can only be resolved by future, deeper observations of exocometary cyanides.
\subsection{Predictions for molecular surveys of gas-bearing debris disks}
\label{sect:pred}
\begin{figure*}
\hspace{-3mm}
\includegraphics*[scale=0.4]{molpredictionsplusresults.pdf}
\vspace{-8mm}
\caption{\textit{Left:} Predicted flux ratios compared to the detected CO J=2-1 line in the $\beta$ Pic disk, for Solar System cometary species observable at far-IR to radio wavelengths. For each species, vertical bars represent the expected flux ratios for the range of abundances with respect to CO observed in Solar System comets. Dashed vertical bars assume optically thin emission, whereas solid vertical bars take into account the optical thickness of the line (where this is only significant for OH). LTE excitation is assumed for all species, with a temperature of 52 K. The transition with the best `effective S/N' (see main text) is selected for each species. Green horizontal bars show the 3$\sigma$ threshold achievable with available facilities (SOFIA, ALMA, and the VLA) for 1h of integration and typical weather. Lighter green bars assume unresolved emission (best case scenario), whereas darker ones consider the loss in sensitivity expected for resolving the disk if it has the same on-sky extent as the CO disk, for the lowest spatial resolution of the instrument at the frequency in question (worst case scenario). Labels on the top x axis show the instrument and frequency best suited for detection of each species. \textit{Right:} The same predictions are now compared to the upper limits (red triangles) achieved in this work for molecular transitions observed with the SMA, ALMA and \textit{Herschel}.
}
\label{fig:molpreds}
\end{figure*}
To guide future observations, we
use Eq. \ref{eq:releaseratetoflux} to calculate expected line flux ratios around $\beta$ Pictoris for all common cometary species taking into account the range of abundances observed in the Solar System. We make two main assumptions. First, we assume all species are co-located with CO, leading to similar spatial distributions and excitation conditions. Since the bulk of the CO emission originates from the SW clump at 85 AU from the star, we use the stellar radiation field at 85 AU as representative of the bulk of the gas disk. Second, we calculate the excitation of all species using the LTE approximation \citep[apart from CO itself, where the level populations are very well constrained by the existing J=3-2/J=2-1 line ratio,][]{Matra2017a}. Figure~\ref{fig:CNvsCOfracpops} shows that the LTE assumption does not affect the calculated populations of CN and CO by more than a factor of a few for low-energy levels, but we cannot exclude that this may have a significant effect for other molecules and higher transitions. We vary the gas kinetic temperature between 52 K \citep[as predicted by thermodynamical models at the clump location,][]{Kral2016} and 210 K \citep[as the upper limit estimated through the resolved CO scale height at the clump location,][]{Matra2017a}.
We focus on molecular transitions observable with currently available facilities at far-IR to radio wavelengths, namely the Stratospheric Observatory for Infrared Astronomy (SOFIA), ALMA and the Karl G. Jansky Very Large Array (VLA). We select the best observable transition for each temperature and molecule to be that with the highest `effective S/N'. We construct this by dividing a transition's flux ratio with respect to CO (as calculated above) by the sensitivity of each telescope facility considered, at the frequency of the given transition. In addition, we consider the telescope's resolution by dividing this effective S/N by $\sqrt{N_{\rm beams}}$, where $N_{\rm beams}$ is the number of resolution elements covering the gas disk. As this acts to penalize facilities that over-resolve the gas disk, we choose a resolution closest to the CO disk's on-sky diameter \citep[$\sim16.5\arcsec$][]{Matra2017a} for the ALMA and VLA interferometers. This is a conservative approach as the spectral stacking technique described in \S\ref{sect:filtering} \citep[][Loomis et al. in press]{Matra2015} should significantly alleviate the sensitivity loss caused by over-resolving the disk.
Having selected the best transition for each molecule and temperature, we show its range of fluxes expected for Solar System cometary abundances in Figure~\ref{fig:molpreds}, left (vertical bars). We only show predictions for a single temperature of 52 K, as we find that the maximum effective S/N achievable for most species is largely independent of temperature (since e.g. for higher temperatures, higher transitions will emit more strongly and can then be selected for observations).
These predictions can then be compared with each telescope's 3$\sigma$ detection thresholds (green bars), where these also take into account that the disk will be resolved (darker as opposed to lighter bars for unresolved emission). We find CN (as expected), but also HCN and OH to be promising molecules for detection around $\beta$ Pic. This is largely due to their transitions having Einstein A coefficients orders of magnitude larger than CO, compensating for their lower gas masses caused by their shorter photodissociation timescales.
Since the predicted line fluxes can, in a few cases, be up to orders of magnitude higher than CO, we need to consider that these transitions may be optically thick, particularly since CO itself has modest optical depth at the clump location ($\tau_{\rm CO\ J=2-1}\sim$0.12). To estimate optical depth, we first take the CO column density at the clump location as determined through J=3-2/J=2-1 line ratio observations \citep{Matra2017a}. Then, we scale this by the abundance ratio of molecule $i$ with respect to CO in the gas, to find the column density of molecule $i$.
This can be combined with LTE fractional populations to solve for the optical depth through its definition \citep[Eq. 3,][]{Matra2017a}, scaling the velocity width of the line to the Doppler-broadened width of the molecule in question. We find that the predictions, when accounting for optical depth effects (solid as opposed to dotted vertical bars in
Figure~\ref{fig:molpreds}), change significantly only for the OH $^2\Pi_{3/2}\ \rm{J}=5/2^+-3/2^-$ line at 2.51 THz.
Overall, if cometary volatiles are being released at rates reflecting their ice composition, and if their composition reflects that of Solar System comets, we expect CN and HCN to be readily detectable with ALMA around $\beta$ Pictoris. Given that CN photodissociation takes place through photons of a similar wavelength to those causing CO photodissociation, we expect the CN/CO flux prediction to hold for exocometary gas disks around different stars as well.
An interesting result of our calculation is that OH detection should be possible with SOFIA. This encouraged us to check the \textit{Herschel} Science Archive, where we found unpublished PACS observations of $\beta$ Pictoris covering the OH line. We extracted the point-source-corrected spectrum from the central 7$\arcsec$ spaxel of the PACS Level 3 data product\footnote{\textit{Herschel} Observation ID: 1342198170}, and scaled it to the total flux of the central 3x3 spaxels. Then, we carried out continuum subtraction through a second order polynomial, leading to a measured RMS noise level of 0.24 Jy. Given the coarse spectral resolution of the data (295.2 km/s at 119.3 $\mu$m), we assume the line is unresolved to set a 3$\sigma$ upper limit on the integrated line flux of 2.1$\times$10$^2$ Jy km/s. We find that this is already at the very low end of our predicted range. However, given the significant optical thickness of the line, our prediction is very sensitive to the choice of OH temperature or spatial distribution, which may be oversimplified here.
We find detection of rotational lines from the H$_2$$^{18}$O water isotopologue, hydrogen sulfide (H$_2$S) and ammonia (NH$_3$) to be very unlikely with current instrument capabilities, although ammonia may be detectable in future with the sensitivity improvement brought by the Next Generation Very Large Array\footnote{\url{http://library.nrao.edu/public/memos/ngvla/NGVLA_17.pdf}}. On the other hand, we can already probe Solar System cometary levels for molecules such as the main isotopologue of water (H$_2$O), heavy water (HDO), formaldehyde (H$_2$CO) and methanol (CH$_3$OH), which could be detected if their respective ices are at the high abundance end of Solar System composition compared to CO. We note that for these species, like OH, the short photodissociation timescale would cause an even more pronounced clump compared to CO observations (an aspect which we have neglected here for simplicity). This would mean that molecular emission would have a higher optical depth than predicted here. At the same time, the more compact emission would be less over-resolved compared to CO, leading to higher spatially integrated S/N, and lower upper limits as derived here through our spectro-spatial filtering technique.
Finally, we note that an increased optical depth \textit{to UV photons} could also prolong the survival lifetime of these molecules, leading to more favourable conditions for detection in future surveys.
\section{Conclusions}
\label{sect:concl}
We have presented the first molecular survey of the exocometary gas disk around $\beta$ Pictoris with the SMA, and combined it with archival ALMA data to obtain coverage of 10 different molecular species, namely CO, CN, HCN, HCO$^+$, N$_2$H$^+$, H$_2$CO, H$_2$S, CH$_3$OH, SiO and DCN. We reported continuum and CO J=2-1 detections and used upper limits on all other species as the basis to expand an exocometary release model beyond CO. We obtain the following conclusions:
\begin{enumerate}
\item The (optically thin) line fluxes for a given molecule depend on its exocometary outgassing rate, its photodissociation lifetime (neglecting other gas-phase chemical reactions), its intrinsic line strength, and the fractional population of the upper level of the transition (Eq. \ref{eq:releaseratetoflux}). Knowledge of this upper level population requires NLTE excitation modelling which we here expand to include the effect of fluorescence. We find that the latter produces orders of magnitude higher excitation in the low gas density, NLTE regime, and needs to be taken into account for robust derivation of gas masses from observed line fluxes.
\item CO is the longest-lived cometary molecule against photodissociation that is observable at millimeter wavelengths. This long lifetime explains, in part, why it is the only one that has been detected so far. Other observable molecules have survival lifetimes several orders of magnitude below that of CO, except for CN. This makes CN the most promising molecule for detection in debris disks after CO.
\item CN is a daughter molecule, which we assumed to be produced mostly through photodissociation of HCN gas released from the ice phase. CO gas on the other hand, is both a parent and daughter species produced by CO$_2$ photodissociation. Therefore, we used our upper limit on the CN/CO flux ratio to derive an upper limit on the HCN/(CO+CO$_2$) ratio of outgassing rates of 2.5\%. This is still consistent with, although below most of the values measured in Solar System comets as observed near Earth.
If deeper CN and/or HCN observations push this HCN/(CO+CO$_2$) outgassing rate upper limit down to a significant depletion compared to Solar System comets, we show that this may be caused by either the outgassing mechanism or a true depletion in the ice abundance.
\item Outgassing ratios of molecular species reflect their ice compositions as long as ices cannot survive on grains down to the blow-out size in the collisional cascade and be removed from the system. This is likely the case in $\beta$ Pictoris, as 1) collisions will allow ice layers to resurface, where they become subject to rapid UV photodesorption \citep{Grigorieva2007}, and 2) the smallest unbound grains will undergo hypervelocity collisions on the way out of the system leading to vaporization of any ice that may have survived on these grains \citep{Czechowski2007}. In general, these processes will allow release of molecules less volatile than CO that would otherwise survive in the ice phase on the smallest grains and be dynamically removed by radiation pressure from the central star.
\item Then, unless non-catastrophic collisions at the top of the cascade are the dominant gas release mechanism, a low HCN/(CO+CO$_2$) outgassing rate is attributed to a low HCN/(CO+CO$_2$) abundance in the exocometary ice. If confirmed by deeper observations, this would indicate that active cyanide chemistry during planet formation produces a wide variety of cometary compositions across planetary systems.
\item We present predictions for future detection of cometary species around $\beta$ Pictoris, assuming Solar System abundances and LTE excitation. We find that ALMA should readily detect CN and HCN line emission, even for the most HCN-poor Solar System abundances. Other species such as H$_2$O, HDO, H$_2$CO, CH$_3$OH and DCN may be detectable, depending on their specific composition, spatial distribution and excitation conditions.
\item Finally, we report \textit{Herschel} archival upper limits on the OH $^2\Pi_{3/2}\ \rm{J}=5/2^+-3/2^-$ line at 2.51 THz that are already at the very low end of the range predicted assuming LTE and Solar System cometary H$_2$O abundances. The line is optically thick and hence the flux prediction is sensitive to the assumed on-sky distribution and temperature of OH gas, which are likely oversimplified here. Further modelling taking these effects into account is needed to evaluate whether this upper limit indicates an underabundance of water in $\beta$ Pic's exocomets, and whether SOFIA may be able to detect OH from photodestruction of water in exocometary gas disks.
\end{enumerate}
\acknowledgments
The authors are grateful to A. Brandeker and W.-F. Thi for providing the spectrum of the star and the complete rate matrix of CO including vibrational and electronic levels. The authors would also like to thank Quentin Kral and Gianni Cataldi for helpful notes on the submitted manuscript. LM acknowledges support from the Smithsonian Institution as a Submillimeter Array (SMA) Fellow. KI\"O acknowledges support from the Simons Collaboration on Origins of Life (SCOL).
The Submillimeter Array is a joint project between the Smithsonian
Astrophysical Observatory and the Academia Sinica Institute of Astronomy and
Astrophysics and is funded by the Smithsonian Institution and the Academia
Sinica.
This paper makes use of ALMA data ADS/JAO.ALMA\#2012.1.00142.S. ALMA is a partnership of ESO (representing its member states), NSF (USA) and NINS (Japan), together with NRC (Canada), NSC and ASIAA (Taiwan), and KASI (Republic of Korea), in cooperation with the Republic of Chile. The Joint ALMA Observatory is operated by ESO, AUI/NRAO and NAOJ. \textit{Herschel} is an ESA space observatory with science instruments provided by European-led Principal Investigator consortia and with important participation from NASA.
\facility{SMA, ALMA, \textit{Herschel}}.
\software{
MIR, miriad \citep{Sault1995},
CASA \citep{McMullin2007},
Matplotlib \citep{Hunter2007}
}
\bibliographystyle{apj}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 2,910
|
Le diffusionnisme est, en anthropologie et sociologie, une appréhension des cultures humaines par leur distribution dans l'aire, leur historicité et les dynamiques géographiques associées. Le diffusionnisme va s'institutionnaliser en tant que courant de pensée à la fin du et au début du notamment en Allemagne, en Angleterre et aux États-Unis. Première critique de l'évolutionnisme, le diffusionnisme est considéré comme la deuxième grande théorie anthropologique après celle-ci.
Théorie
Selon cette théorie, l'évolution de l'humanité ne se fait pas selon une logique évolutionniste monolinéaire mais plutôt plurilinéaire : au niveau géographique, les cultures développent leurs propres jeux de connaissances, puis avec le jeu des flux migratoires humains, les connaissances se rencontrent et s'enrichissent à la hauteur de l'échelle géographique de la diffusion. Le rôle du diffusionnisme est alors d'étudier les transmissions inter-culturelles entre foyers de diffusion pour comprendre les phénomènes d'hybridation des connaissances qui composent l'évolutionnisme.
Cette approche rompt avec les problématiques évolutionnistes. Il n'est en effet plus possible de se réclamer de la conception du progrès héritée de l'évolutionnisme d'alors qui voit dans les transformations culturelles et sociales des inventions parallèles et convergentes résultant d'une unique loi d'évolution des sociétés humaines (évolution parallèle et indépendante chez différents peuples).
Cependant, le diffusionnisme n'exclut pas forcément une conception évolutionniste de l'histoire. Il ne fait que remplacer le parallélisme et la loi des évolutions convergentes par une théorie des diffusions des traits culturels. Ainsi, si le diffusionnisme put servir de vecteur critique de tout évolutionnisme postulant une hiérarchie des sociétés (aux États-Unis notamment), il s'en accommoda ailleurs assez facilement.
Méthode
Le diffusionnisme insiste donc sur la théorisation des contacts interculturels. Cela donna lieu à un grand nombre d'études comparatives et cartographiques, ayant le plus souvent pour but d'établir la séquence de filiation d'un fait culturel et de circonscrire le « foyer culturel » dans lequel aurait émergé l'élément en question. La constitution de cartes géographiques porteuses d'un savoir anthropologique va tenir une place importante dans les méthodes du diffusionnisme.
Histoire et développement du diffusionnisme
Origines
Précurseur
C'est chez des penseurs évolutionnistes que l'on voit apparaître, au milieu du , les premières idées diffusionnistes. En Allemagne, Adolf Bastian, bien que soutenant une unité psychique de l'humanité et des lois de développement universelles, propose les processus de diffusion (associés aux stimuli du milieu) comme l'un des facteurs secondaires expliquant l'évolution des sociétés. En Angleterre, Edward Tylor introduit largement le principe de diffusion dans ses ouvrages. Il applique tout d'abord ces idées à l'étude des techniques préhistoriques avant de les généraliser à une anthropologie générale et aux éléments non matériels de la culture (mythologie, théorie de la maladie,...).
L'Anthropogéographie
Principales écoles d'anthropologie diffusionnistes
Les idées diffusionnistes sont reprises par trois écoles :
l'école historico-culturelle germano-autrichienne dite de Vienne, par Fritz Graebner (théorie des )
l'école américaine, notamment dans les premiers travaux de Franz Boas et d'Alfred Louis Kroeber (plus portée sur les données empiriques)
l'école britannique et l', principalement par Grafton Elliot Smith et William James Perry
Dans son ouvrage L'origine des systèmes familiaux, Emmanuel Todd a utilisé des concepts diffusionnistes (conservation des zones périphériques ...) pour expliquer la répartition géographique des systèmes familiaux en Eurasie.
Critiques
Le diffusionnisme sera l'objet d'importantes critiques, d'une part parce qu'il ne peut rendre compte de l'émergence de phénomènes culturels semblables chez des populations n'ayant jamais pu être l'objet d'un contact (par exemple, appartenant à des époques fort éloignées), mais plus encore, parce qu'il néglige le contexte et la signification des éléments culturels qui auraient été l'objet d'une diffusion, s'en tenant exclusivement à la similitude de leur forme.
Il est également reproché aux diffusionnistes leur dogmatisme, leur réductionnisme géographique et leur schématisation excessive de faits sociaux.
Articles connexes
Appropriation culturelle
Enrichissement culturel
Migration humaine
Liste des courants de l'anthropologie
Evolutionnisme (anthropologie)
Friedrich Ratzel
Hyperdiffusionisme
Notes et références
Anthropologie sociale et culturelle
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 1,090
|
May 28, 2014 By Marcellino D'Ambrosio, Ph.D.Sunday Scripture Reflections
Photography © by Andy Coan
The celebration of the Ascension used to leave me a bit flat. It was clear what Good Friday did for me. And Easter Sunday's benefits were indisputable. But as for the Ascension, what's in it for me?
Christianity is about a kind of love we call agape or charity. It is love that looks away from itself to another and gives itself away for another. The Divine Word did not become man or endure the cross because something was in it for Him.
Charity shares in the beloved's joys and sorrows (Jn 14:28). The first thing to remember about the Ascension is that it is about sharing in Jesus' joy. It is about celebrating his return to the heavenly glory to which he refused to cling (Phil 2:6-11). It is about rejoicing that his crown of thorns has been replaced with the kingly crown, that the mocking crowd at Calvary has been replaced with myriads of adoring angels. The Ascension is about Jesus's triumph and glorification. If we get our attention off ourselves and allow the Holy Spirit's love of the Son to animate our souls, we'll experience greater joy than when we see our child hit a home run or graduate from college.
But the Ascension is not just about charity. It is also a feast of hope. Yes, there is something in it for us. He goes to prepare a place for us (Jn 14:2). We will also one day wear crowns made of gold instead of thorns.
For us to endure until that blessed moment, we need divine power. That's another reason we ought to rejoice in his Ascension. He takes his place at God's right hand so that he can pour out the promise of the Father, the Holy Spirit, upon his disciples (Eph 4:10).
As he ascends, he tells the disciples to wait for this power. But notice that he does not tell them to wait passively for the rapture. He does not instruct them to pour over Bible prophecies, debating about how and when he will return. In fact in Acts 1:11, after the Lord ascends out of their sight, the angels ask why the disciples just stand there, staring into space.
The waiting is not to be a squandering of precious time. It is waiting for a purpose, nine days of prayer (the first novena!) leading to empowerment. Why empowerment? Because they have challenging work to do. "Go, therefore, and make disciples of all nations." (Matt 28: 16-20).
We used to think that evangelization was something that happened in mission countries far away, carried out by priests and religious. But the Second Vatican Council told us that our own neighborhoods are mission territory, and that every single Catholic is called to be an evangelist. Pope St. John Paul II proclaimed this as the "New Evangelization" because the place is new — right next door — and the missionaries are new since they include all us all.
I'm really not sure that St. Francis of Assisi ever said, "Preach the Gospel always; when necessary, use words." But if he did, note this: Francis often thought it very necessary to use words. His words could be heard in marketplaces, on street corners, in Churches, wherever there were people. Of course, preaching without an authentic witness of life is certainly counterproductive. But forget about the idea that just the witness of our lives is enough. It is not. You may not called to preach on street corners, but Vatican II and subsequent popes, echoing 1 Pet 3:15, say that we all must be ready to articulate what Jesus has done for us, what he means to us, and why he is the answer to the world's problems.
Feel inadequate to the task? You're in good company. Benedict XVI's first public statement was an admission of his inadequacy. Do as he did — pray for the power of the Holy Spirit to move in and through you, and take the time to keep learning more about your faith so that you can share it with ever greater confidence.
Editor's Note: Reflection on the Mass readings for the Solemnity of the Ascension of the Lord (Year ABC) — Acts 1:1-11; Psalms 47:2-3, 6-7, 8-9; Ephesians 1:17-23; Matthew 28:16-20.
If you liked this article, please share it with your friends and family using the Recommend and Social Media buttons below and via email. We value your comments and encourage you to leave your thoughts below. Thank you! – The Editors
Dr. Marcellino D'Ambrosio writes from Texas. For his resources on parenting and family life or information on his pilgrimages to Rome and the Holy Land, visit www.crossroadsinitiative.com or call 1.800.803.0118.
From a colorful and varied background as a professor of theology, a father of five, business owner, and professional performer Marcellino D'Ambrosio (aka "Dr. Italy") crafts talks, blog posts, books, and videos that are always fascinating, practical, and easy to understand. He is a TV and radio personality, New York Times best-selling author, speaker, and pilgrimage host who has been leading people on a journey of discovery for over thirty years. For complete bio and video, visit the Dr. Italy page.
Connect with Dr. Italy:
Sunday Scripture ReflectionsAscension, Call to Holiness, Catholic Church, Easter, Evangelization, featured, God, Gospel, Holy Spirit, Jesus Christ, Paraclete, Scripture
Encountering the Word -- Wisdom 18:14-15a
Advice on Happiness from the Smartest Man in the World
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 2,025
|
{"url":"https:\/\/fr.maplesoft.com\/support\/help\/maplesim\/view.aspx?path=geometry\/MajorAxis","text":"MajorAxis - Maple Help\n\ngeometry\n\n MajorAxis\n find the length of the major axis of a given ellipse\n\n Calling Sequence center(e)\n\nParameters\n\n e - ellipse\n\nDescription\n\n \u2022 The routine returns the length of the major axis of the given ellipse e.\n \u2022 The line segment through the foci and across the ellipse is called the major axis.\n \u2022 The command with(geometry,MajorAxis) allows the use of the abbreviated form of this command.\n\nExamples\n\n > $\\mathrm{with}\\left(\\mathrm{geometry}\\right):$\n > $\\mathrm{ellipse}\\left(\\mathrm{e1},2{x}^{2}+{y}^{2}-4x+4y=0,\\left[x,y\\right]\\right):$\n > $\\mathrm{MajorAxis}\\left(\\mathrm{e1}\\right)$\n ${2}{}\\sqrt{{6}}$ (1)","date":"2023-02-07 12:05:43","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 4, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9788268208503723, \"perplexity\": 736.7558762459036}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2023-06\/segments\/1674764500456.61\/warc\/CC-MAIN-20230207102930-20230207132930-00091.warc.gz\"}"}
| null | null |
Guitar Amplifiers – How Many Watts Are Enough?
October 28, 2014 October 28, 2014 gordon
If you're an active guitar player, you've certainly noticed the continuing trend towards smaller, lower powered amplifiers. This has occurred for a number of reasons: Popularity of boutique clones of early amplifier designs, older players downsizing to focus on playing at home rather than performing, and the advent of very good affordable sound reinforcement systems.
Historically, the first guitar amplifiers were pretty small because the guitar was not necessarily the lead instrument of the band, and in many cases the band itself was not amplified. Plus the technology of the time – tubes – dictated smaller low power systems.
As rock music and the electric guitar became more popular, both the volume levels and the size of the venues increased. But true high quality "house" sound systems like today did not exist. So the guitar amplifier was not for "tone", but was also the primary vehicle for what the audience heard. Amplifiers also did double or triple duty: Check out the input panel of many early amplifiers and you'll see input jacks for multiple guitars and microphones. Adjusted for today's dollars, pro gear back in the day was vastly more expensive than now, and sharing was a practical necessity.
So like the Space Race, the power race was on, and by the end of the 50's the larger guitar amplifiers were pushing 50 watts or more. Fender's introduction of the blackface amps in the early 60's addressed the need for louder, cleaner sound. The blackface amps differed from their tweed predecessors in a number of ways, but features such as fixed bias design, higher plate voltages, and solid-state rectification had more to do with volume and headroom than tone. The largest amps topped out at 100 watts, which is really the practical limit of four 6L6 or EL-34 power tubes. This is pretty much true today for tube amps, and anything more than that gets very heavy and hot (bass players had it tough then). Today a bass player can get a 500-watt Class D solid-state amplifier that's the size of a phone book.
But back then if you were going to play an arena, you needed stacks of amplifiers because they were doing a lot of the heavy lifting. Plus it's kind a macho thing and looks really cool too. Today, you can play an arena with a 15-watt amplifier, and some performers do. While there is a certain visceral sensation to the sound of a 4×12, the need for a row of Marshall stacks is essentially visual. And unless you're Yngwie Malmsteen, many of those cabs on stage aren't even on.
So how much power do you actually need? Unless you require extremely high levels of clean volume without the assist of a PA, 50 watts is the most you'll ever need. How much volume an amplifier produces is a function of its design: Fixed versus cathode bias, amount of negative feedback, plate voltage, rectifier type, etc. It's hard to generalize, but a 15-watt amplifier with no negative feedback and a solid-state rectifier can be very loud and clean. My main amp head has (4) 6V6 tubes, solid state rectifier and a 20/40-watt switch. The only time it's on 40 watts is when the band is playing outside.
If you are playing clubs and typically put the guitars through the PA, 15-30 watts will likely do it. While early amplifier designs were guided by power output, choosing an amplifier today is more about how you want it to sound, rather than how loud it will go. If you play mostly at home or jam with friends, 10-30 watts is where a lot of the amplifier market is targeted these days. Finding the right amount clean headroom – which is important if you use pedals – is an important selection criteria. If you regularly jam with a drummer, 30 watts is probably a better choice than 10. Five-watt amplifiers can be fun, if all you want is loose, old school grind. But with a humbucker-equipped guitar, there will be little in the way of decent clean volume.
Many modern amp designs have the ability to vary total amplifier output. Some do this by actually dropping out power tubes (4-to-2 for example) while others vary the amount of voltage to the power tubes or phase inverter. These features cut volume as well as headroom, allowing the ability to clip the power tubes at reasonable volumes. The Traynor Ironhorse amplifier has a fixed/cathode bias switch that changes the output of the amp from 37 to 17 watts, respectively. This not only affects total volume and headroom, but also the feel (I like the softer nature of cathode bias).
If you have a large amp that you don't want to part with, there are of course power attenuators, which are available as an add-on accessory. These work by absorbing some of the energy that would normally go to your speakers. In effect, you can crank the amplifier but the attenuator "soaks up" some of the energy (volume). Attenuators work by placing some type of resistance/inductance network in the signal path to the speakers. Without getting technical, even the best ones mess with the feel of the amp, and how the guitar interacts with the amplifier. It's hard to explain but it's a disconnected feeling. They sound good on YouTube, but so does everything. My suggestion is to buy a smaller amplifier.
The trend towards lower stage volumes, and the affordability of good sound reinforcement and monitoring systems has been a boon to amateur and pro players alike. Using a guitar amp as the sole amplification source is very rare, and your band will actually sound better if you turn down and let the PA and the monitors do their job. And your band mates will appreciate it. Which brings us to the guitar player's favorite lament of "I can't hear myself." Which is a topic we'll address shortly.
Amplifiers and Accessories, General Items Blackface, Dr. Z, Fender, Marshall, Mesa Boogie, rectifier, Rivera, Traynor, Tweed, UpFront Guitars
Two neat EL34 guitar amplifiers from Traynor
July 5, 2014 gordon
If you crave the sound of the classic big British amplifiers, then you should check out amplifiers using the EL34 type power tube. The EL34 power tube was originally released by Mullard-Phillips in the 1950's and in most common amplifier applications produces about 25 watts per tube. This is the power tube associated with the big British amps of the 60's and 70's produced by Marshall, Orange, Hiwatt and others.
The EL34 power tube has great harmonic detail, and a broad upper midrange that really fattens up the unwound strings of a guitar without making it sound muddy. Single note lines tend to have a lot of "meat" to them making amplifiers using these types of tubes sound both bright and thick at the same time. Traynor has two affordable amplifiers using the EL34 power tube, each with it's own unique features.
Traynor Ironhorse – This Ironhorse is a "lunchbox" style of amplifier that uses (2) El34 power tubes and is switchable for either cathode or fixed bias. The fixed bias mode produces 40 watts, and is the tightest/brightest setting with the most available clean headroom. The cathode bias setting produces fewer than 20 watts and while still providing a good amount of clean headroom, has a little softer attack and a bit more "feel" to it. With a gain, master volume control and a three-way tone stack control it's easy to dial in some natural crunch, or keep it clean and get your jollies with pedals. The Ironhorse is honest and tasty El-34 tone for great classic rock crunch in an affordable and lightweight package. It's on the mark for all sorts of live situations, and with the money you save on this amp you can go get yourself a nice set of new or NOS power tubes. It comes with a nice soft gig bag too.
Traynor YBA Mod 1 – In Traynor-speak, YBA means "Yorkville Bass Amplifier, which is what the 40-watt YBA was originally designed as. But the YBA found favor with guitar players too, and the Mod 1 version adds the ability to place the bright and normal channels in series or parallel, and adds an attenuator. In series the volume controls act like a master/gain control, while in parallel they function like normal/bright channels that you can blend just like a Tweed Fender. The attenuator can vary the ouptut power from .5 watts all the way up to 40 watts. At low wattage settings you can get all sorts of massive rock clang at very reasonable levels, while 20 watts is good for even the loudest of stage volumes. It's hard to imagine even needing the 40 watt setting. Like the Ironhorse, the YBA has the complex crunch and texture that EL-34 tubes do best, and the added attenuation feature makes it good for practice and recording. Not quite as portable as the Ironhorse and pretty darn loud, but the attenuator can essentially size this amp from a large club down to a bedroom.
If you are looking to change up your sound, think about your amplifier too, and not just your guitar. With the EL84 dominating the small amplifier market, many players have never sampled the brawnier more harmonically rich tones of the EL34. So for less than a high quality solid body, you can get a taste of the tone that powered the British Invasion.
Amplifiers and Accessories, General Items, Guitars and Accessories Dr. Z, EL34, Fender, Marshall, Traynor
Pedal Amps – What's a Pedal Amp?
September 22, 2013 February 1, 2014 gordon
Nearly every guitar player today uses some type of effect pedal, either for practice, recording or playing live. Safe to say there is hardly anyone who does not own some type of effect pedal, making them both a great market for manufacturers, but also a real consideration when deciding what type of amplifier to use.
One of the frequent questions asked these days on the gear pages is what's a good "Pedal Amp?" So what is a Pedal Amp? I would define a Pedal Amp as an amplifier that does not add extreme tonal coloration, and is able to handle high signal inputs without adding additional coloration or distortions. To some that does not sound like a particularly good amplifier, as for many old-school players the amplifier is an essential part of the sound equation. But for players that increasingly use various types of effects and digital modeling, the amplifier becomes more of an "amplification system" and less of a tone source.
Going back a few decades, the early amplifiers were instrumental to the developing sound of rock music. The happy accident of distortion, and then the use of lots of distortion as the essential rock guitar sound was not what the Founding Fathers intended. But as recording techniques, sound systems, and musical styles evolved, the concept of a pure unaffected guitar tone became increasingly rare. From the early days of cranking up small wattage amplifiers to get grindy tone, practically everyone today — well maybe not Neil Young — is using some type of effect to generate anything from mild to insane distortion. And while there are zillions of multi-channel amps out there, for flexibility's sake pedals just allow much more room to mix and match tone.
So what makes a good Pedal Amp? In a word: Headroom. From a design standpoint, early amplifiers were notoriously short of headroom, both in the preamp and power sections. This of course gave them their warm creamy tone, but pump a high gain pedal into a Fender Tweed and the net result will be mushy distortion with very loose undefined low end. Practically speaking the pedal is creating distortion, and the higher input signal from the pedal is also distorting the preamp of the guitar amplifier. Distortion-on-distortion is not always desirable or musical.
Generally speaking, low powered cathode biased amplifiers (Tweeds, small Vox's, lots of other low power EL-84 amplifiers) are not super candidates for pedals that have the capability of generating fairly high input levels. Even the relatively brawny 45 watt Fender Bassman won't handle a lot of input signal without getting floppy. Hot input signals can come from distortion pedals or frequency modulation pedals (chorus, flangers) that tend to increase the signal level. Now boost pedals are made specifically to increase the signal, often for the purpose of overdriving the front end of an amplifier. But a boost pedal it typically only increasing the signal, and not adding its own distortion or other artifacts.
The boutique amp craze, with its plethora low power Tweed and Vox inspired designs (Dr. Z, Matchless, Badd Cat, Victoria etc) created some awesome sounding amplifiers that well-heeled baby boomers were craving. However they were not necessarily great at handling pedals, and even at 18 watts a Maz 18 is still damn loud. And this inspired the attenuator craze….and now everybody just buys pedals.
Fixed biased amplifiers — like Fender Blackface or similar designs — by virtue of their circuit topology have higher headroom and tolerate pedals better. Fender of course was trying to make louder and cleaner amplifiers to fill the larger venues that rock bands were playing. For that reason amplifiers that follow the higher powered Fender Blackface 6L6 tube lineage tend to be pretty good pedal amplifiers.
Once amplifier designers discovered master volume techniques and cascading gain (preamp distortion) techniques, amplifier designs became "stiffer" cleaner and louder. The general elimination of tube rectifiers in favor of diode rectifiers also increased headroom, and made the amplifiers sag less, and play cleaner under heavy loads. Distortion was now a design goal, not a by-product of marginal design or power handling capability. But to some, all these improvements — including dreaded solid state — took away some of the "organic" nature of the early amplifier sound.
Fast forward to the boutique amp craze and builders were putting all this "marginal" stuff back into amplifiers: Cathode bias, low power, and tube rectifiers. And at the opposite end of the spectrum some players are now using a totally digital preamp source — like an Axe Effect Fractal or Eleven Rack — and a powered full-range speaker system from JBL, QSC, or EV.
So back to the original topic: Good Pedal Amps tend to be more modern or higher powered designs that can tolerate strong signal inputs, and if they use a tube power amp section, have a solid state rectifier. If you are playing live or play at high volumes and want to use gain pedals, it's advisable to avoid lower powered designs in the mold of a Tweed or Vox. Nothing against these amps — Robert Cray through a Matchless is a great sound — but it's not a pedal sound. There are always exceptions to the rules of course, and some of the boutique designs using the EF-86 preamp tube (Dr Z. Z-28 for example) have quite of bit of clean headroom despite modest power outputs. It's always dangerous to generalize.
Speaking of which, what about the Mesa Dual and Triple-Rec designs. Don't they also violate the low power/tube rectifier rule? Yes, sort of. Up around 100 watts, tube rectifiers are pretty marginal AC-to-DC converters for creating the high voltage DC that power tubes need. By wiring two rectifiers in parallel, each rectifier is only carrying half the current, and therefore can share the load and maintain headroom (Fender did this on some 50's amps for the same reason). A Triple-Rec adds one more rectifier for handling even higher powered designs. Mesa could have just used solid state diodes, but a Mesa Diode Roadking lacks marketing pizazz. Most Mesa — and modern metal — amplifiers are characterized by very "clean" clean channels and the distortion is produced by various combinations of tube and solid state wizardry.
Finally, here's a personal example with the two amps I like to play the most: A Fender Reissue Bassman, and a Dr. Z Remedy head plugged into a Mojo Pine 4×10 cabinet with Jensen P10R/Eminence Blueframe speakers. Both are using virtually identical speaker arrangements, speakers, and cabinet materials. But the Bassman is an early cathode bias design with a tube rectifier, and the Remedy is a solid state rectifier design using four 6V6 tubes (not a "reissue" design but billed as having Marshall Plexi-style tone). They both have the same power output, about 40 watts. I love the Bassman tone, but in gigging situations using pedals for various levels of gain and effects, the amp loses definition, attack, and can get sloppy. Even on half power, the Remedy has better attack, low end firmness, and is overall tighter. Crunchy gain is crunchy gain. On full power the Remedy is really too clean for most situations except outdoor gigs. But both are 40 watt amps.
My general rule of thumb for any amplifier selection is to find the best clean tone that makes you happy and then go pedal shopping. If your favorite tone is clean to slightly crunchy, you may never need pedals and a smaller lowered powered "old school" amplifier may be the ticket. But if you are like most players, make sure the amplifier of your dreams has sufficiently stout headroom to serve as a suitable platform for whatever pedals you decide to use down the road.
For more information on platform and pedal options at Upfront Guitars: www.upfrontguitars.com
Amplifiers and Accessories, Guitars and Accessories Blackface, Dr. Z, effect pedals, guitar amplifiers, Marshall, Mesa Boogie, Tweed, UpFront Guitars, ValveTrain
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 4,808
|
In the year 2002, > 20% of all CABG operations were performed using beating-heart techniques without the use of CPB. A variety of beating-heart techniques for bypass surgery have been described, including minimally invasive direct coronary artery bypass with a small anterior thoracotomy incision, OPCAB using a conventional sternotomy incision, and port-access and robotically assisted CABG.> These techniques have been popularized based on the assumption of technically equivalent results coupled with improved short-term outcomes.
Because at least some portion of postoperative AF has been thought to be due to intraoperative atrial ischemia, there was some reason to believe that the beating-heart approach to CABG (ie, without arrest of the heart) would result in less postoperative AF. Several authors have reported a reduced incidence of postoperative AF in patients undergoing beating-heart CABG. But other authors have reported no difference. In the current evidence review, two randomized, controlled trials and one large-scale concurrent cohort study were identified. Ascione et al reported a significantly lower rate of postoperative AF in the OPCAB group than in the on-pump CABG group. In contrast, though, Van Dijk et al reported no difference in AF frequency for the two surgical approaches in a randomized, prospective study of similar size. The large-scale concurrent cohort study reported by Hernandez et al found a small but statistically significant difference in the frequency of postoperative AF between on-pump CABG patients and OPCAB patients favoring the OPCAB group.
In conventional CABG operations with CPB and an arrested heart, the CABGs are created after cardioplegic arrest. As an alternative, intermittent aortic cross-clamping can be used to arrest the heart for short periods of time to allow for a quiet operative field during the creation of distal anastomoses. Some percentage of postoperative atrial arrhythmias are thought to be due to inadequate myocardial preservation during cardioplegic arrest. Several investigators have studied the effects of various methods of myocardial protection on the frequency of postoperative atrial arrhythmias, but no convincing benefit for any one particular protection strategy has been identified in retrospective reports, either in terms of postoperative AF or other conduction ab-normalities. In four of the five identified randomized, controlled trials related to cardioplegia or myocardial protection strategy, Butler et al identified no difference between cold potassium cardioplegia and intermittent aortic cross-clamping, and Fontan et al found no benefit to any particular cardioplegic technique. Hynninen et al found no benefit to insulin-enhanced cardioplegia, and Wand-schneider et al found no benefit to blood cardioplegia vs crystalloid cardioplegia. In a small study, Pehkonen et al found a lower incidence of postoperative AF in the group that received cold crystalloid cardioplegia.
The (3-adrenergic receptor antagonists have been shown in most randomized, controlled studies to reduce the incidence of postoperative AF in cardiac surgery patients. It has been postulated that increased sympathetic tone may be one mechanism that is responsible for AF. There has been interest in the use of TEA as an adjunct to conventional anesthesia because it has been shown to decrease both heart rate and BP variability, suggesting effective sympathetic blockade. Nonrandomized studies of TEA have produced conflicting results. Two randomized, controlled studies have also reported conflicting results. Jideus et al reported no difference in the rate of postoperative AF, but Scott et al reported a significant reduction in the rate of postoperative AF with the use of TEA.
In most CABG operations, the pericardium is usually opened longitudinally in its anterior aspect. This opening provides unobstructed access to the underlying heart and proximal great vessels. The pericardium is usually left open, although some surgeons choose to close a portion of the pericardium. A second, or auxiliary, incision in the posterior pericardium has been used to facilitate the drainage of blood into the chest cavity where it can be evacuated with chest tubes. This technique has been shown in nonrandomized trials to reduce the incidence of both postoperative pericardial effusion and postoperative supraventricular tachycardia. In contrast, Asimakopoulos et al found no association between the use of posterior pericardiotomy and the incidence of postoperative AF. In the single randomized, controlled trial identified in the evidence review, Kuralay et al reported a significant reduction in postoperative AF for patients who underwent posterior pericardiotomy. Although this study reports a benefit, it is important to keep in mind that there are multiple other reasons why this effect could be a surrogate for other (and unidentified) intraoperative factors.
Metabolic substrate enhancement with glucose or other energy sources during periods of myocardial ischemia and reperfusion has been proposed as one strategy to limit myocardial necrosis. As early as 1965, Sodi-Pollaris et al noted that GIK solutions administered to patients experiencing acute myocardial infarction limited the subsequent ECG changes. In addition, animal models of myocardial ischemia/ reperfusion have indicated that GIK solution infusion limited tissue necrosis, resulted in less myocardial acidosis, reduced the frequency of ventricular arrhythmias, and resulted in fewer wall motion abnormalities. Two randomized, controlled trials of GIK solution infusion in patients undergoing heart surgery were identified in the evidence review. Wistbacka et al reported that in patients undergoing CABG surgery there was no difference in the rate of postoperative AF associated with the use of GIK solutions and that the use of GIK solutions was associated with serious adverse effects such as hypoglycemia. In contrast, Lazar et al found in unstable patients with angina who were undergoing CABG that GIK infusion enhanced myocardial performance and also resulted in less postoperative AF.
CPB has long been associated with a variety of deleterious systemic inflammatory effects mediated by a generalized systemic inflammatory response. Heparin-coated CPB circuits have been developed to reduce the systemic inflammatory response associated with CPB, as measured by less complement activation, less leukocyte activation, a reduction in the release of cytokines, and the need for less systemic anticoagulation therapy. Proponents have postulated that the use of heparin-coated CPB circuits would result in less postoperative bleeding and fewer thromboembolic complications, but the reported results have been mixed. Two randomized, controlled trials of the use of heparin-coated circuits were identified. The data in these studies that addressed the impact on bleeding and neurologic injury were mixed, but both of these reports documented at least some evidence for a reduction in the rate of postoperative AF with the use of a heparin-coated circuit. Ovrum et al reported significantly less postoperative AF in the heparin-coated circuit group (a reduction of approximately 50%). Sven-marker et al studied different types of heparin-coated circuits and found that one type was associated with less postoperative AF. Both groups of investigators questioned whether their observation of less postoperative AF in the group of patients in whom heparin-coated circuit had been used was truly due to the type of circuit or to other" unmeasured factors.
It is important to keep in mind that our recommendations are based on a relatively small number of studies. In the case of posterior pericardiotomy and the use of mild hypothermia" our recommendations are based on only single randomized controlled studies. We understand that" when making clinical decisions for the individual patient" the reader must place our recommendations in the proper context.
AF remains a significant complication following cardiac surgery. This arrhythmia is associated with an increased hospital length of stay" increased costs" and an increased risk for thromboembolic complications. On the basis of the findings of this and the other reports in this series" we suggest that a multifactorial approach" involving appropriate prophylaxis and treatment for this arrhythmia" will best serve our cardiac surgery patients. Below are the recommendations for the management of intraoperative interventions. A summary of these clinical recommendations and grades of evidence is presented in Table 2.
frequency of postoperative AF (strength of recommendation" A; evidence grade" fair; net benefit" substantial).
(strength of recommendation" B; evidence grade" fair; net benefit" intermediate).
3. OPCAB cannot be recommended to decrease postoperative AF because of conflicting results reported from randomized, controlled trials or large-scale concurrent cohort studies (strength of recommendation" I; evidence grade" fair; net benefit" conflicting).
(strength of recommendation" I; evidence grade" good; net benefit" none).
(strength of recommendation" I; evidence grade" fair; net benefit" conflicting).
7. We recommend the use of heparin-coated circuits to reduce the rate of postoperative AF (strength of recommendation" B; evidence grade" fair; net benefit" intermediate).
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 4,428
|
Kerala Assembly pays tributes to Mukherjee, C.F. Thomas
The Assembly on Monday paid glowing tributes to former President Pranab Mukherjee and senior Kerala Congress leader and MLA C.F. Thomas who passed away last year. The Assembly, in an obituary reference, remembered Mukherjee, who passed away on August 31, 2020, as a President who upheld the prestige of the nation on the international stage. Speaker P. Sreeramakrishnan said Mukherjee left his mark as a statesman and an able administrator. A soft-spoken personality, Mukherjee, as the 13th President, was able to take decisions rooted in Constitutional values, which were also not limited by considerations of party politics, Mr. Sreeramakrishnan said.Chief Minister Pinarayi Vijayan remembered Mukherjee as a politician and President who upheld secular and democratic values. He stood for the protection and strengthening of Constitutional values. A direct heir to the Nehruvian political legacy, he adopted anti-imperialist and pro-socialist stances at critical junctures. Mukherjee was a rare politician who combined in himself a remarkable depth of knowledge and the value of experience, Vijayan said. Mukherjee exemplified Machiavelli's precept that "It is not titles that honour men, but men that honour titles," Leader of Opposition Ramesh Chennithala said.
Mr. Thomas, who passed away on September 27 last year, was remembered as a politician who personified clean politics. An MLA who represented Changanassery since 1980, Mr. Thomas was Minister for Registration, Rural Development and Khadi in the A.K. Antony and Oommen Chandy ministries, Mr. Sreeramakrishnan said in his obituary reference. Mr. Thomas voiced the issues and plight of high range farmers in the House, the Chief Minister said. He was strongly opposed to liberalised trade pacts that were inimical to the interests of farmers, Mr. Vijayan said. Mr. Chennithala remembered Mr. Thomas as a mature politician who was not interested in power politics.
The Assembly adjourned for the day after taking up the obituary references.
Posted in IndiaTagged assembly, Kerala, pays, Tributes
Bring on the night: Music inspired by the moon over the years
Samantha blown away by Rakul Preet Look
Japan's PM Shinzo Abe resigns: here's how his tenure was and what happens next
Thrissur registers 168 new cases
Schools to open after 315-day COVID break
UP: In Phase 1, govt to build five industrial parks along expressways
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 1,277
|
\section{ Introduction}
In 1993, Camassa and Holm derived the well-known Camassa-Holm (CH) equation \cite{CH}
\begin{eqnarray}
m_t+2m u_x+m_xu=0, \quad m=u-u_{xx}+k,
\label{CH}
\end{eqnarray}
(with $k$ being an arbitrary constant) with the aid of an asymptotic approximation to the Hamiltonian of the Green-Naghdi equations.
Since the work of Camassa and Holm \cite{CH}, more diverse studies on this equation have remarkably been developed \cite{CH2}-\cite{CGI}.
The most interesting feature of the CH equation (\ref{CH}) is that it admits peakon solutions in the case $k=0$.
The stability and interaction of peakons were discussed in several references \cite{CS1}-\cite{JR}.
In addition to the CH equation, other similar integrable models with peakon solutions were found \cite{DP1,DP2}.
Recently, there are two integrable peakon equations found with
cubic nonlinearity. They are the following cubic equation \cite{OR,Fo,Fu,Q1}
\begin{eqnarray}
m_t+\frac{1}{2}\left[ m(u^2-u^2_x)\right]_x=0, \quad m=u-u_{xx},\label{mCH}
\end{eqnarray}
and the Novikov's equation \cite{NV1,HW1}
\begin{eqnarray}
m_t=u^2m_x+3uu_xm, \quad m=u-u_{xx}.\label{cCHN}
\end{eqnarray}
There is also much attention in studying integrable multi-component peakon equations.
For example, in \cite{HI}, the authors proposed a multi-component generalization of the CH equation, and
in \cite{QSY}, multi-component extensions of the cubic nonlinear equation (\ref{mCH}) were studied.
In this paper, we propose the following multi-component system
\begin{eqnarray}
\left\{\begin{split}
m_{j,t}=&(m_jH)_x+m_jH+\frac{1}{(N+1)^2}\sum_{i=1}^{N}[m_{i}(u_j-u_{j,x})(v_{i}+v_{i,x})+m_j(u_i-u_{i,x})(v_{i}+v_{i,x})],
\\
n_{j,t}=&(n_jH)_x-n_jH-\frac{1}{(N+1)^2}\sum_{i=1}^{N}[n_{i}(u_i-u_{i,x})(v_{j}+v_{j,x})+n_j(u_i-u_{i,x})(v_{i}+v_{i,x})],
\\
m_{j}=&u_{j}-u_{j,xx}, \quad n_{j}=v_{j}-v_{j,xx}, \quad 1\leq j\leq N,
\end{split}\right. \label{meq}
\end{eqnarray}
where $H$ is an arbitrary smooth function of $u_j$, $v_j$, $1\leq j\leq N$, and their derivatives.
For $N=1$, this system is reduced to the standard CH equation (\ref{CH}) as $v_{1}=2$, $H=-u_1$ and to the cubic nonlinear CH equation (\ref{mCH}) as $u_1=v_{1}$, $H=-\frac{1}{2}(u_{1}^{2}-u_{1,x}^{2})$.
Therefore, it is a kind of multi-component combination of the CH equation (\ref{CH}) and the cubic nonlinear CH equation (\ref{mCH}).
The system (\ref{meq}) contains an arbitrary function $H$, thus it is actually a large class of multi-component equations.
We remark that, very recently, Li, Liu and Popowicz proposed a four-component peakon equation which also contains an arbitrary function \cite{LLP}.
They derived the Lax pair and infinite conservation laws for their four-component equation,
and presented a bi-Hamiltonian structure for their equation in the case that the arbitrary function is taken to be zero.
In this paper, we show that the multi-component system (\ref{meq}) admits Lax representation and infinitely many conservation laws.
Due to the presence of the arbitrary function $H$, we do not expect the system (\ref{meq}) is bi-Hamiltonian integrable system in general.
However, we demonstrate that for some special choices of $H$ we may find the corresponding bi-Hamiltonian structures.
As examples, we derive the peakon solutions of this system in the case $N=2$.
In particular, we obtain a new integrable model which admits stationary peakon solutions.
The whole paper is organized as follows. In section 2, the Lax
pair and conservation laws of equation (\ref{meq}) are presented.
In section 3, the Hamiltonian structures and peakon solutions of equation (\ref{meq}) in the case $N=2$ are discussed.
Some conclusions and open problems are addressed in section 4.
\section{Lax pair and conservation laws}
We first introduce the $N$-component vector potentials $\vec{u}$, $\vec{v}$ and $\vec{m}$, $\vec{n}$ as
\begin{eqnarray}
\begin{split}
\vec{u}=&(u_1,u_2,\cdots,u_N),
\quad
\vec{v}=(v_1,v_2,\cdots,v_N),
\\
\vec{m}=&\vec{u}-\vec{u}_{xx}, \quad \vec{n}=\vec{v}-\vec{v}_{xx}.
\end{split}
\label{um}
\end{eqnarray}
Using this notation, equation (\ref{meq}) is expressed in the following vector form
\begin{eqnarray}
\left\{\begin{split}
\vec{m}_{t}=&(\vec{m}H)_x+\vec{m}H+\frac{1}{(N+1)^2}[\vec{m}(\vec{v}+\vec{v}_x)^T(\vec{u}-\vec{u}_x)+(\vec{u}-\vec{u}_x)(\vec{v}+\vec{v}_x)^T\vec{m}],
\\
\vec{n}_{t}=&(\vec{n}H)_x-\vec{n}H-\frac{1}{(N+1)^2}[\vec{n}(\vec{u}-\vec{u}_x)^T(\vec{v}+\vec{v}_x)+(\vec{v}+\vec{v}_x)(\vec{u}-\vec{u}_x)^T\vec{n}],
\\
\vec{m}=&\vec{u}-\vec{u}_{xx}, \quad \vec{n}=\vec{v}-\vec{v}_{xx},
\end{split}\right.
\label{veq}
\end{eqnarray}
where the symbol $T$ denotes the transpose of a vector.
Let us introduce a pair of $(N+1)\times(N+1)$ matrix spectral problems
\begin{eqnarray}
\phi_x=U\phi,\quad \phi_t=V\phi,
\label{LP}
\end{eqnarray}
with
\begin{eqnarray}
\begin{split}
\phi&=(\phi_1,\phi_{21},\cdots,\phi_{2N})^T,\\
U&=\frac{1}{N+1}\left( \begin{array}{cc} -N & \lambda \vec{m}\\
\lambda \vec{n}^T & I_N \\ \end{array} \right),
\\
V&=\frac{1}{N+1}\left( \begin{array}{cc} -N\lambda^{-2}+\frac{1}{N+1}(\vec{u}-\vec{u}_x)(\vec{v}+\vec{v}_x)^T &
\lambda^{-1}(\vec{u}-\vec{u}_x)+\lambda \vec{m}H
\\ \lambda^{-1}(\vec{v}+\vec{v}_x)^T+\lambda \vec{n}^TH & \lambda^{-2}I_N-\frac{1}{N+1}(\vec{v}+\vec{v}_x)^T(\vec{u}-\vec{u}_x)\\ \end{array} \right),
\label{UV}
\end{split}
\end{eqnarray}
where $\lambda$ is a spectral parameter, $I_N$ is the $N\times N$ identity matrix,
$\vec{u}$, $\vec{v}$, $\vec{m}$ and $\vec{n}$ are the vector potentials shown in (\ref{um}).
\begin{proposition}
(\ref{LP}) provides the Lax pair for the multi-component system (\ref{meq}).
\end{proposition}
{\bf Proof} \quad
It is easy to check that the compatibility condition of (\ref{LP}) generates
\begin{eqnarray}
U_t-V_x+[U,V]=0.\label{cc}
\end{eqnarray}
From (\ref{UV}), we have
\begin{eqnarray}
\begin{split}
U_t&=\frac{1}{N+1}\left( \begin{array}{cc} 0 & \lambda \vec{m}_t\\
\lambda \vec{n}_t^T & 0 \\ \end{array} \right),
\\
V_x&=\frac{1}{N+1}\left( \begin{array}{cc} \frac{1}{N+1}[\vec{m}(\vec{v}+\vec{v}_x)^T-(\vec{u}-\vec{u}_x)\vec{n}^T] &
\lambda^{-1}(\vec{u}_x-\vec{u}_{xx})+\lambda (\vec{m}H)_x
\\ \lambda^{-1}(\vec{v}_x+\vec{v}_{xx})^T+\lambda (\vec{n}^TH)_x & \frac{1}{N+1}[\vec{n}^T(\vec{u}-\vec{u}_x)-(\vec{v}+\vec{v}_x)^T\vec{m}]\\ \end{array} \right),
\label{UVd}
\end{split}
\end{eqnarray}
and
\begin{eqnarray}
\begin{split}
[U,V]=UV-VU=\frac{1}{(N+1)^2}\left( \begin{array}{cc} \Gamma_{11} &
\Gamma_{12}
\\ \Gamma_{21} & \Gamma_{22}\\ \end{array} \right),
\label{UVc}
\end{split}
\end{eqnarray}
where
\begin{eqnarray*}
\begin{split}
\Gamma_{11}&= \vec{m}(\vec{v}+\vec{v}_x)^T-(\vec{u}-\vec{u}_x)\vec{n}^T,
\\
\Gamma_{12}&=(N+1)[\lambda^{-1}(\vec{u}_x-\vec{u}_{xx})-\lambda\vec{m}H]-
\frac{\lambda}{N+1}[\vec{m}(\vec{v}+\vec{v}_x)^T(\vec{u}-\vec{u}_x)+(\vec{u}-\vec{u}_x)(\vec{v}+\vec{v}_x)^T\vec{m}],
\\
\Gamma_{21}&=(N+1)[\lambda^{-1}(\vec{v}_x+\vec{v}_{xx})^T+\lambda\vec{n}^TH]+
\frac{\lambda}{N+1}[\vec{n}^T(\vec{u}-\vec{u}_x)(\vec{v}+\vec{v}_x)^T+(\vec{v}+\vec{v}_x)^T(\vec{u}-\vec{u}_x)\vec{n}^T],
\\
\Gamma_{22}&= \vec{n}^T(\vec{u}-\vec{u}_x)-(\vec{v}+\vec{v}_x)^T\vec{m}.
\label{gma}
\end{split}
\end{eqnarray*}
We remark that (\ref{UVc}) is written in the form of block matrix. As shown above, the element $\Gamma_{11}$ is a scalar function, the element $\Gamma_{12}$ is a $N$-component row vector function, the element $\Gamma_{21}$ is a $N$-component column vector function, and the element $\Gamma_{22}$ is a $N \times N$ matrix function.
Substituting the expressions (\ref{UVd}) and (\ref{UVc}) into (\ref{cc}), we find that (\ref{cc}) gives rise to
\begin{eqnarray}
\left\{\begin{split}
\vec{m}_{t}=&(\vec{m}H)_x+\vec{m}H+\frac{1}{(N+1)^2}[\vec{m}(\vec{v}+\vec{v}_x)^T(\vec{u}-\vec{u}_x)+(\vec{u}-\vec{u}_x)(\vec{v}+\vec{v}_x)^T\vec{m}],
\\
\vec{n}^T_{t}=&(\vec{n}^TH)_x-\vec{n}^TH-\frac{1}{(N+1)^2}[\vec{n}^T(\vec{u}-\vec{u}_x)(\vec{v}+\vec{v}_x)^T+(\vec{v}+\vec{v}_x)^T(\vec{u}-\vec{u}_x)\vec{n}^T],
\\
\vec{m}=&\vec{u}-\vec{u}_{xx}, \quad \vec{n}^T=\vec{v}^T-\vec{v}_{xx}^T,
\end{split}\right.
\label{veq2}
\end{eqnarray}
which is nothing but the vector equation (\ref{veq}). Hence, (\ref{LP}) exactly gives the Lax pair of multi-component equation (\ref{meq}).
Now let us construct the conservation laws of equation (\ref{meq}).
We write the spacial part of the spectral problems (\ref{LP}) as
\begin{eqnarray}
\left\{\begin{array}{l}
\phi_{1,x}=\frac{1}{N+1}(-N\phi_{1}+\lambda\sum_{i=1}^Nm_i\phi_{2i}),
\\
\phi_{2j,x}=\frac{1}{N+1}\left(\lambda n_j\phi_{1}+\phi_{2j}\right),
\end{array}\right.
1\leq j\leq N.
\label{slp}
\end{eqnarray}
Let $\Omega_j=\frac{\phi_{2j}}{\phi_{1}}$, $1\leq j\leq N$, we obtain the following system of Riccati equations
\begin{eqnarray}
\Omega_{j,x}=\frac{1}{N+1}\left[\lambda n_j+(N+1)\Omega_{j}-\lambda\Omega_{j}\sum_{i=1}^Nm_i\Omega_{i}\right], ~~1\leq j\leq N.
\label{ric}
\end{eqnarray}
Making use of the relation $(\ln\phi_1)_{xt}=(\ln\phi_1)_{tx}$ and (\ref{LP}), we arrive at the conservation law
\begin{eqnarray}
\left(\sum_{i=1}^Nm_i\Omega_{i}\right)_t=
\left(\lambda^{-2}\sum_{i=1}^N(u_i-u_{i,x})\Omega_{i}+\frac{1}{N+1}\lambda^{-1}\sum_{i=1}^N(u_i-u_{i,x})(v_{i}+v_{i,x})+H\sum_{i=1}^Nm_i\Omega_{i}\right)_x.
\label{CL}
\end{eqnarray}
Equation (\ref{CL}) means that $\sum_{i=1}^Nm_i\Omega_{i}$ is a generating function of the conserved densities. To derive the explicit forms of conserved densities, we expand $\Omega_j$ into the negative power series of $\lambda$ as
\begin{equation}
\Omega_j=\sum_{k=0}^{\infty}\omega_{jk}\lambda^{-k}, \quad 1\leq j\leq N.\label{oe1}
\end{equation}
Substituting (\ref{oe1}) into the Riccati system (\ref{ric}) and equating the coefficients of powers of $\lambda$, we obtain
\begin{eqnarray}
\begin{split}
\omega_{j0}&=n_j\left(\sum_{i=1}^Nm_in_i\right)^{-\frac{1}{2}},
\\
\omega_{j1}&=(N+1)\left[\omega_{j0}-\omega_{j0,x}-
\frac{1}{2}n_j\left(\sum_{i=1}^Nm_i(\omega_{i0}-\omega_{i0,x})\right)\left(\sum_{i=1}^Nm_in_i\right)^{-1}\right]\left(\sum_{i=1}^Nm_in_i\right)^{-\frac{1}{2}},
\end{split}
\label{wj1}
\end{eqnarray}
and the recursion relations for $\omega_{j(k+1)}$, $k\geq 1$,
\begin{eqnarray}
\begin{split}
\omega_{j(k+1)}&=(N+1)\left[\omega_{jk}-\omega_{jk,x}-\frac{1}{2}n_j\left(\sum_{i=1}^Nm_i(\omega_{ik}-\omega_{ik,x})\right)
\left(\sum_{i=1}^Nm_in_i\right)^{-1}\right]\left(\sum_{i=1}^Nm_in_i\right)^{-\frac{1}{2}}.
\end{split}
\label{wj}
\end{eqnarray}
Inserting (\ref{oe1}), (\ref{wj1}) and (\ref{wj}) into (\ref{CL}), we finally obtain the following infinitely many conserved densities $\rho_j$
and the associated fluxes $F_j$:
\begin{eqnarray}
\begin{split}
\rho_{0}&=\sum_{i=1}^Nm_i\omega_{i0}=\left(\sum_{i=1}^Nm_in_i\right)^{\frac{1}{2}}, ~~F_0=H\sum_{i=1}^Nm_i\omega_{i0}=H\left(\sum_{i=1}^Nm_in_i\right)^{\frac{1}{2}},
\\
\rho_{1}&=\sum_{i=1}^Nm_i\omega_{i1}, ~~ F_1=\frac{1}{N+1}\sum_{i=1}^N(u_i-u_{i,x})(v_{i}+v_{i,x})+H\sum_{i=1}^Nm_i\omega_{i1},
\\
\rho_{2}&=\sum_{i=1}^Nm_i\omega_{i2}, ~~ F_2=\sum_{i=1}^N(u_i-u_{i,x})\omega_{i0}+H\sum_{i=1}^Nm_i\omega_{i2},
\\
\rho_{j}&=\sum_{i=1}^Nm_i\omega_{ij}, ~~F_{j}=\sum_{i=1}^N(u_i-u_{i,x})\omega_{i(j-2)}+H\sum_{i=1}^Nm_i\omega_{ij},\quad j\geq 3,
\end{split}
\label{rjj}
\end{eqnarray}
where $\omega_{ij}$, $1\leq i\leq N$, $j\geq 0$ is given by (\ref{wj1}) and (\ref{wj}).
\vspace*{0.2cm}
{\bf Remark 1.} The $2N$-component system (\ref{meq}) containing an arbitrary function $H$ does possess Lax representation and infinitely many conservation laws. Such a system is interesting since different choices of $H$ lead to different peakon equations (see examples in the following section). Let us look back why an arbitrary smooth function may be involved in system (\ref{meq}). System (\ref{meq}) is produced by the compatibility condition (\ref{cc}) of the spectral problems (\ref{LP}) where such an arbitrary function is included in $V$ part (see (\ref{UV})). The appearance of this arbitrary function can be explained as that the Lax equation is an over determined system by choosing the appropriate $V$ to match $U$.
\section{Examples for $N=2$}
For $N=1$, equation (\ref{meq}) becomes
\begin{eqnarray}
\left\{\begin{split}
m_{1,t}=&(m_1H)_x+m_1H+\frac{1}{2}m_{1}(u_1-u_{1,x})(v_{1}+v_{1,x}),
\\
n_{1,t}=&(n_1H)_x-n_1H-\frac{1}{2}n_{1}(u_1-u_{1,x})(v_{1}+v_{1,x}),
\\
m_{1}=&u_{1}-u_{1,xx}, \quad n_{1}=v_{1}-v_{1,xx},
\end{split}\right. \label{teq0}
\end{eqnarray}
where $H$ is an arbitrary smooth function of $u_1$, $v_1$, and their derivatives.
This system is exactly the synthetical two-component peakon model we proposed in \cite{XQZ}, where one may see the details of the Lax pair, bi-Hamiltonian structures and peakon solutions of this model.
Let us study the case of $N=2$. In this case, equation (\ref{meq}) is cast into the following four-component model
\begin{eqnarray}
\left\{
\begin{split}
m_{1,t}=&(m_1H)_x+m_1H
\\&+\frac{1}{9}\{m_{1}[2(u_{1}-u_{1,x})(v_{1}+v_{1,x})+(u_{2}-u_{2,x})(v_{2}+v_{2,x})]+m_{2}(u_{1}-u_{1,x})(v_{2}+v_{2,x})\},
\\
m_{2,t}=&(m_2H)_x+m_2H
\\&+\frac{1}{9}\{m_{1}(u_{2}-u_{2,x})(v_{1}+v_{1,x})+m_{2}[(u_{1}-u_{1,x})(v_{1}+v_{1,x})+2(u_{2}-u_{2,x})(v_{2}+v_{2,x})]\},
\\
n_{1,t}=&(n_1H)_x-n_1H
\\&-\frac{1}{9}\{n_{1}[2(u_{1}-u_{1,x})(v_{1}+v_{1,x})+(u_{2}-u_{2,x})(v_{2}+v_{2,x})]+n_{2}(u_{2}-u_{2,x})(v_{1}+v_{1,x})\},
\\
n_{2,t}=&(n_2H)_x-n_2H
\\&-\frac{1}{9}\{n_{1}(u_{1}-u_{1,x})(v_{2}+v_{2,x})+n_{2}[(u_{1}-u_{1,x})(v_{1}+v_{1,x})+2(u_{2}-u_{2,x})(v_{2}+v_{2,x})]\},
\\ m_{1}=&u_{1}-u_{1,xx},\quad m_{2}=u_{2}-u_{2,xx}, \quad n_{1}=v_{1}-v_{1,xx},\quad n_{2}=v_{2}-v_{2,xx},
\end{split}
\right.
\label{teq1}
\end{eqnarray}
where $H$ is an arbitrary smooth function of $u_1$, $u_2$, $v_1$, $v_2$, and their derivatives. This system admits the following $3\times 3$ Lax pair
\begin{eqnarray}
U=\frac{1}{3}\left( \begin{array}{ccc} -2 & \lambda m_1 & \lambda m_2\\
\lambda n_1 & 1 & 0
\\ \lambda n_2 & 0 & 1 \\ \end{array} \right),
\quad
V=\frac{1}{3}\left( \begin{array}{ccc} V_{11} & V_{12} & V_{13} \\
V_{21} & V_{22} & V_{23} \\ V_{31} & V_{32} & V_{33}\\ \end{array} \right),
\label{UV1}
\end{eqnarray}
where
\begin{eqnarray}
\begin{split}
V_{11}&=-2\lambda^{-2}+\frac{1}{3}[(u_{1}-u_{1,x})(v_{1}+v_{1,x})+(u_{2}-u_{2,x})(v_{2}+v_{2,x})],
\\
V_{12}&=\lambda^{-1}(u_1-u_{1,x})+\lambda m_1H, ~~V_{13}=\lambda^{-1}(u_2-u_{2,x})+\lambda m_2H,
\\
V_{21}&=\lambda^{-1}(v_1+v_{1,x})+\lambda n_1H, ~~V_{22}=\lambda^{-2}-\frac{1}{3}(u_{1}-u_{1,x})(v_{1}+v_{1,x}),
\\
V_{23}&=-\frac{1}{3}(u_{2}-u_{2,x})(v_{1}+v_{1,x}), ~~V_{31}=\lambda^{-1}(v_2+v_{2,x})+\lambda n_2H,
\\
V_{32}&=-\frac{1}{3}(u_{1}-u_{1,x})(v_{2}+v_{2,x}), ~~V_{33}=\lambda^{-2}-\frac{1}{3}(u_{2}-u_{2,x})(v_{2}+v_{2,x}).
\label{v}
\end{split}
\end{eqnarray}
Due to the appearance of arbitrary function $H$, we do not expect that (\ref{teq1}) is bi-Hamiltonian in general.
But we find that it is possible to figure out the bi-Hamiltonian structures for some special cases, which we will show in the following examples.
\subsection*{Example 1.~~A new integrable model with stationary peakon solutions}
Taking $H=0$, equation (\ref{teq1}) becomes the following system
\begin{eqnarray}
\left\{\begin{array}{l}
m_{1,t}=\frac{1}{9}\{m_{1}[2(u_{1}-u_{1,x})(v_{1}+v_{1,x})+(u_{2}-u_{2,x})(v_{2}+v_{2,x})]+m_{2}(u_{1}-u_{1,x})(v_{2}+v_{2,x})\},
\\
m_{2,t}=\frac{1}{9}\{m_{1}(u_{2}-u_{2,x})(v_{1}+v_{1,x})+m_{2}[(u_{1}-u_{1,x})(v_{1}+v_{1,x})+2(u_{2}-u_{2,x})(v_{2}+v_{2,x})]\},
\\
n_{1,t}=-\frac{1}{9}\{n_{1}[2(u_{1}-u_{1,x})(v_{1}+v_{1,x})+(u_{2}-u_{2,x})(v_{2}+v_{2,x})]+n_{2}(u_{2}-u_{2,x})(v_{1}+v_{1,x})\},
\\
n_{2,t}=-\frac{1}{9}\{n_{1}(u_{1}-u_{1,x})(v_{2}+v_{2,x})+n_{2}[(u_{1}-u_{1,x})(v_{1}+v_{1,x})+2(u_{2}-u_{2,x})(v_{2}+v_{2,x})]\},
\\ m_{1}=u_{1}-u_{1,xx},\quad m_{2}=u_{2}-u_{2,xx},
\quad n_{1}=v_{1}-v_{1,xx},\quad n_{2}=v_{2}-v_{2,xx}.
\end{array}\right. \label{teq2}
\end{eqnarray}
Let us introduce a Hamiltonian pair
\begin{eqnarray}
\begin{split}
J&=\left( \begin{array}
{cccc} 0 & 0 & \partial+1 & 0 \\ 0 & 0 & 0 & \partial+1 \\ \partial-1 & 0 & 0 &0 \\ 0 & \partial-1 & 0 &0 \\
\end{array} \right),
~~K=\frac{1}{9}\left( \begin{array}
{cccc} K_{11} & K_{12} & K_{13} & K_{14} \\
K_{21} & K_{22} & K_{23} & K_{24} \\
K_{31} & K_{32} & K_{33} & K_{34} \\
K_{41} & K_{42} & K_{43} & K_{44} \\
\end{array} \right),
\label{JK1}
\end{split}
\end{eqnarray}
where
\begin{eqnarray}
\begin{split}
K_{11}=& -2m_1\partial^{-1}m_1, ~~K_{12}=-m_2\partial^{-1}m_1-m_1\partial^{-1}m_2,
\\ K_{13}=& 2m_1\partial^{-1}n_1+m_2\partial^{-1}n_2, ~~K_{14}=m_1\partial^{-1}n_2,
\\ K_{21}=&-K_{12}^{\ast}=-m_1\partial^{-1}m_2-m_2\partial^{-1}m_1, ~~K_{22}=-2m_2\partial^{-1}m_2,
\\ K_{23}=& m_{2}\partial^{-1} n_{1}, ~~K_{24}=m_1\partial^{-1}n_1+2m_2\partial^{-1}n_2,
\\ K_{31}=&-K_{13}^{\ast}=2n_1\partial^{-1}m_1+n_2\partial^{-1}m_2, ~~K_{32}=-K_{23}^{\ast}=n_1\partial^{-1} m_2,
\\ K_{33}=&-2 n_{1}\partial^{-1} n_{1}, ~~K_{34}=-n_{1}\partial^{-1} n_{2}-n_{2}\partial^{-1} n_{1},
\\ K_{41}=&-K_{14}^{\ast}=n_2\partial^{-1} m_1, ~~K_{42}=-K_{24}^{\ast}=n_1\partial^{-1}m_1+2n_2\partial^{-1}m_2,
\\ K_{43}=&-K_{34}^{\ast}=-n_2\partial^{-1}n_1-n_1\partial^{-1}n_2, ~~K_{44}=-2n_2\partial^{-1}n_2.
\end{split}
\label{K1}
\end{eqnarray}
By direct but tedious calculations, we arrive at
\begin{proposition}
Equation (\ref{teq2}) can be rewritten in the following bi-Hamiltonian form
\begin{eqnarray}
\left(m_{1,t}, ~m_{2,t}, ~n_{1,t}, ~n_{2,t}\right)^{T}
=J \left(\frac{\delta H_2}{\delta m_{1}},~\frac{\delta H_2}{\delta m_{2}},~\frac{\delta H_2}{\delta n_{1}},~\frac{\delta H_2}{\delta n_{2}}\right)^{T}
=K \left(\frac{\delta H_1}{\delta m_{1}},~\frac{\delta H_1}{\delta m_{2}},~\frac{\delta H_1}{\delta n_{1}},~\frac{\delta H_1}{\delta n_{2}}\right)^{T},
\label{BH}
\end{eqnarray}
where $J$ and $K$ are given by (\ref{JK1}), and
\begin{eqnarray}
\begin{split}
H_1&=\int_{-\infty}^{+\infty}[(u_{1,x}-u_{1})n_{1}+(u_{2,x}-u_{2})n_{2}]dx,
\\ H_2&=\frac{1}{9}\int_{-\infty}^{+\infty}[(u_{1}-u_{1,x})^2(v_{1}+v_{1,x})n_1+(u_{1}-u_{1,x})(u_{2}-u_{2,x})(v_{2}+v_{2,x})n_1
\\&~~~~~~~~~~~~ +(u_{1}-u_{1,x})(u_{2}-u_{2,x})(v_{1}+v_{1,x})n_2+(u_{2}-u_{2,x})^2(v_{2}+v_{2,x})n_2]dx.
\end{split}
\label{H}
\end{eqnarray}
\end{proposition}
Suppose an $N$-peakon solution of (\ref{teq2}) is in the form
\begin{eqnarray}
\begin{split}
u_{1}&=\sum_{j=1}^N p_j(t)e^{-\mid x-q_j(t)\mid}, ~~u_{2}=\sum_{j=1}^N r_j(t)e^{-\mid x-q_j(t)\mid},
\\
v_{1}&=\sum_{j=1}^N s_j(t)e^{-\mid x-q_j(t)\mid}, ~~v_{2}=\sum_{j=1}^N w_j(t)e^{-\mid x-q_j(t)\mid}.
\end{split}
\label{NP2}
\end{eqnarray}
Then, in the distribution sense, one can get
\begin{eqnarray}
\begin{split}
u_{1,x}&=-\sum_{j=1}^N p_jsgn(x-q_j)e^{-\mid x-q_j\mid}, \quad m_{1}=2\sum_{j=1}^N p_j\delta(x-q_j),
\\
u_{2,x}&=-\sum_{j=1}^N r_jsgn(x-q_j)e^{-\mid x-q_j\mid}, \quad m_{2}=2\sum_{j=1}^N r_j\delta(x-q_j),
\\
v_{1,x}&=-\sum_{j=1}^N s_jsgn(x-q_j)e^{-\mid x-q_j\mid}, \quad n_{1}=2\sum_{j=1}^N s_j\delta(x-q_j),
\\
v_{2,x}&=-\sum_{j=1}^N w_jsgn(x-q_j)e^{-\mid x-q_j\mid}, \quad n_{2}=2\sum_{j=1}^N w_j\delta(x-q_j).
\end{split}
\label{Npd}
\end{eqnarray}
Substituting (\ref{NP2}) and (\ref{Npd}) into (\ref{teq2}) and integrating against test functions with compact support, we arrive at the $N$-peakon dynamical system as follows:
\begin{eqnarray}
\begin{split}
q_{j,t}=&0,\\
p_{j,t}=&-\frac{1}{9}\{\frac{2}{3}p_j(p_js_j+r_jw_j)
\\&-\sum_{i,k=1}^N \left(p_j(2p_is_k+r_iw_k)+r_jp_iw_k\right)\left(sgn(q_j-q_i)-sgn(q_j-q_k)\right)e^{ -\mid q_j-q_i\mid-\mid q_j-q_k\mid}
\\&+\sum_{i,k=1}^N \left(p_j(2p_is_k+r_iw_k)+r_jp_iw_k\right)\left(sgn(q_j-q_i)sgn(q_j-q_k)-1\right)e^{ -\mid q_j-q_i\mid-\mid q_j-q_k\mid}\},\\
r_{j,t}=&-\frac{1}{9}\{\frac{2}{3}r_j(r_jw_j+p_js_j)
\\&-\sum_{i,k=1}^N \left(r_j(2r_iw_k+p_is_k)+p_jr_is_k\right)\left(sgn(q_j-q_i)-sgn(q_j-q_k)\right)e^{ -\mid q_j-q_i\mid-\mid q_j-q_k\mid}
\\&+\sum_{i,k=1}^N \left(r_j(2r_iw_k+p_is_k)+p_jr_is_k\right)\left(sgn(q_j-q_i)sgn(q_j-q_k)-1\right)e^{ -\mid q_j-q_i\mid-\mid q_j-q_k\mid}\},\\
s_{j,t}=&\frac{1}{9}\{\frac{2}{3}s_j(p_js_j+r_jw_j)
\\&-\sum_{i,k=1}^N \left(s_j(2p_is_k+r_iw_k)+w_jr_is_k\right)\left(sgn(q_j-q_i)-sgn(q_j-q_k)\right)e^{ -\mid q_j-q_i\mid-\mid q_j-q_k\mid}
\\&+\sum_{i,k=1}^N \left(s_j(2p_is_k+r_iw_k)+w_jr_is_k\right)\left(sgn(q_j-q_i)sgn(q_j-q_k)-1\right)e^{ -\mid q_j-q_i\mid-\mid q_j-q_k\mid}\},\\
w_{j,t}=&\frac{1}{9}\{\frac{2}{3}w_j(p_js_j+r_jw_j)
\\&-\sum_{i,k=1}^N \left(w_j(2r_iw_k+p_is_k)+s_jp_iw_k\right)\left(sgn(q_j-q_i)-sgn(q_j-q_k)\right)e^{ -\mid q_j-q_i\mid-\mid q_j-q_k\mid}
\\&+\sum_{i,k=1}^N \left(w_j(2r_iw_k+p_is_k)+s_jp_iw_k\right)\left(sgn(q_j-q_i)sgn(q_j-q_k)-1\right)e^{ -\mid q_j-q_i\mid-\mid q_j-q_k\mid}\}.
\end{split}
\label{dNcp2}
\end{eqnarray}
The formula $q_{j,t}=0$ in (\ref{dNcp2}) implies that the peakon position is stationary and the solution is in the form of separation of variables.
Especially, for $N=1$, (\ref{dNcp2}) becomes
\begin{eqnarray}
\left\{\begin{array}{l}
q_{1,t}=0,
\\
p_{1,t}=\frac{4}{27}p_1(p_1s_1+r_1w_1),
\\
r_{1,t}=\frac{4}{27}r_1(p_1s_1+r_1w_1),
\\
s_{1,t}=-\frac{4}{27}s_1(p_1s_1+r_1w_1),
\\
w_{1,t}=-\frac{4}{27}w_1(p_1s_1+r_1w_1),
\end{array}\right. \label{N1}
\end{eqnarray}
which has the solution
\begin{eqnarray}
q_{1}=C_1,\quad p_{1}=A_4e^{\frac{4}{27}(A_2+A_3)t},\quad r_{1}=\frac{1}{A_1}p_1,\quad
s_{1}=\frac{A_2}{A_4}e^{-\frac{4}{27}(A_2+A_3)t},
\quad
w_{1}=\frac{A_3}{r_1},
\label{N1s}
\end{eqnarray}
where $C_1$ and $A_1$, $\cdots$, $A_4$ are integration constants. Thus, the stationary single-peakon solution becomes
\begin{eqnarray}
\begin{array}{l}
u_{1}=A_4e^{\frac{4}{27}(A_2+A_3)t}e^{-\mid x-C_1\mid},\quad u_{2}=\frac{u_1}{A_1},
\\
v_{1}=\frac{A_2}{A_4}e^{-\frac{4}{27}(A_2+A_3)t}e^{-\mid x-C_1\mid},\quad v_{2}=\frac{A_1A_3}{A_2}v_1.
\end{array}
\label{ocpnp2}
\end{eqnarray}
See Figure \ref{F1} for the stationary single-peakon of the potentials $u_{1}(x,t)$ and $v_{1}(x,t)$ with $C_1=0$, $A_2=A_4=1$ and $A_3=2$.
\begin{figure}
\begin{minipage}[t]{0.5\linewidth}
\centering
\includegraphics[width=2.2in]{F1.eps}
\caption{\small{ The stationary single-peakon solution of the potentials $u_{1}(x,t)$ and $v_{1}(x,t)$ given by (\ref{ocpnp2}) with $C_1=0$, $A_2=A_4=1$ and $A_3=2$. Solid line: $u_{1}(x,t)$; Dashed line: $v_{1}(x,t)$; Black: $t=1$; Blue: $t=2$. }}
\label{F1}
\end{minipage}
\hspace{2.0ex}
\begin{minipage}[t]{0.5\linewidth}
\centering
\includegraphics[width=2.2in]{F2.eps}
\caption{\small{ The single-peakon solution of the potentials $u_{1}(x,t)$ and $v_{1}(x,t)$ given by (\ref{N2s}) with $A_4=0$, $A_2=A_5=1$ and $A_3=2$. Solid line: $u_{1}(x,t)$; Dashed line: $v_{1}(x,t)$; Black: $t=-2$; Blue: $t=2$. }}
\label{F2}
\end{minipage}
\end{figure}
\subsection*{Example 2.~~A new integrable four-component system with peakon solutions}
Choosing
$$H=-\frac{1}{9}[(u_{1}-u_{1,x})(v_{1}+v_{1,x})+(u_{2}-u_{2,x})(v_{2}+v_{2,x})],$$
equation (\ref{teq1}) is cast into
\begin{eqnarray}
\left\{\begin{array}{l}
m_{1,t}=(m_{1}H)_{x}+\frac{1}{9}m_1(u_{1}-u_{1,x})(v_{1}+v_{1,x})+\frac{1}{9}m_{2}(u_{1}-u_{1,x})(v_{2}+v_{2,x}),
\\
m_{2,t}=(m_{2}H)_{x}+\frac{1}{9}m_1(u_{2}-u_{2,x})(v_{1}+v_{1,x})+\frac{1}{9}m_{2}(u_{2}-u_{2,x})(v_{2}+v_{2,x}),
\\
n_{1,t}=(n_{1}H)_{x}-\frac{1}{9}n_1(u_{1}-u_{1,x})(v_{1}+v_{1,x})-\frac{1}{9}n_{2}(u_{2}-u_{2,x})(v_{1}+v_{1,x}),
\\
n_{2,t}=(n_{2}H)_{x}-\frac{1}{9}n_1(u_{1}-u_{1,x})(v_{2}+v_{2,x})-\frac{1}{9}n_{2}(u_{2}-u_{2,x})(v_{2}+v_{2,x}),
\\ m_{1}=u_{1}-u_{1,xx},\quad m_{2}=u_{2}-u_{2,xx}, \quad n_{1}=v_{1}-v_{1,xx},\quad n_{2}=v_{2}-v_{2,xx}.
\end{array}\right. \label{teq3}
\end{eqnarray}
Let us set
\begin{eqnarray}
\begin{split}
K=\frac{1}{9}\left( \begin{array}
{cccc} K_{11} & K_{12} & K_{13} & K_{14} \\
K_{21} & K_{22} & K_{23} & K_{24} \\
K_{31} & K_{32} & K_{33} & K_{34} \\
K_{41} & K_{42} & K_{43} & K_{44} \\
\end{array} \right),
\label{JK2}
\end{split}
\end{eqnarray}
where
\begin{eqnarray}
\begin{split}
K_{11}=& \partial m_1\partial^{-1}m_1\partial-m_1\partial^{-1}m_1, ~~K_{12}=\partial m_1\partial^{-1}m_2\partial-m_2\partial^{-1}m_1,
\\ K_{13}=& \partial m_1\partial^{-1}n_1\partial+m_1\partial^{-1}n_1+m_2\partial^{-1}n_2, ~~K_{14}=\partial m_1\partial^{-1}n_2\partial,
\\ K_{21}=&-K_{12}^{\ast}=\partial m_2\partial^{-1}m_1\partial-m_1\partial^{-1}m_2, ~~K_{22}=\partial m_2\partial^{-1}m_2\partial-m_2\partial^{-1}m_2,
\\ K_{23}=& \partial m_{2}\partial^{-1} n_{1}\partial, ~~K_{24}=\partial m_2\partial^{-1}n_2\partial+m_1\partial^{-1}n_1+m_2\partial^{-1}n_2,
\\ K_{31}=&-K_{13}^{\ast}=\partial n_1\partial^{-1}m_1\partial+n_1\partial^{-1}m_1+n_2\partial^{-1}m_2, ~~K_{32}=-K_{23}^{\ast}=\partial n_1\partial^{-1} m_2\partial,
\\ K_{33}=&\partial n_{1}\partial^{-1} n_{1}\partial-n_{1}\partial^{-1} n_{1}, ~~K_{34}=\partial n_{1}\partial^{-1} n_{2}\partial-n_{2}\partial^{-1} n_{1},
\\ K_{41}=&-K_{14}^{\ast}=\partial n_2\partial^{-1} m_1\partial, ~~K_{42}=-K_{24}^{\ast}=\partial n_2\partial^{-1}m_2\partial+n_1\partial^{-1}m_1+n_2\partial^{-1}m_2,
\\ K_{43}=&-K_{34}^{\ast}=\partial n_2\partial^{-1}n_1\partial-n_1\partial^{-1}n_2, ~~K_{44}=\partial n_2\partial^{-1}n_2\partial-n_2\partial^{-1}n_2.
\end{split}
\label{K2}
\end{eqnarray}
Direct calculations yield that
\begin{proposition}
Equation (\ref{teq3}) can be rewritten in the following Hamiltonian form
\begin{eqnarray}
\left(m_{1,t}, ~m_{2,t}, ~n_{1,t}, ~n_{2,t}\right)^{T}
=K \left(\frac{\delta H_1}{\delta m_{1}},~\frac{\delta H_1}{\delta m_{2}},~\frac{\delta H_1}{\delta n_{1}},~\frac{\delta H_1}{\delta n_{2}}\right)^{T},
\label{BH2}
\end{eqnarray}
where $K$ are given by (\ref{JK2}), and
\begin{eqnarray}
\begin{split}
H_1&=\int_{-\infty}^{+\infty}[(u_{1,x}-u_{1})n_{1}+(u_{2,x}-u_{2})n_{2}]dx.
\end{split}
\label{H2}
\end{eqnarray}
\end{proposition}
We believe that the equation (\ref{teq3}) could be cast into a bi-Hamiltonian system. But we didn't
find another Hamiltonian operator yet that is compatible with the Hamiltonian operator (\ref{JK2}).
Suppose $N$-peakon solution of (\ref{teq3}) is expressed also in the form of (\ref{NP2}). Then,
we obtain the $N$-peakon dynamical system of (\ref{teq3}):
\begin{eqnarray}
\begin{split}
q_{j,t}=&\frac{1}{9}\{-\frac{1}{3}(p_js_j+r_jw_j)
+\sum_{i,k=1}^N \left(p_is_k+r_iw_k\right)\left(sgn(q_j-q_i)-sgn(q_j-q_k)\right)e^{ -\mid q_j-q_i\mid-\mid q_j-q_k\mid}
\\&+\sum_{i,k=1}^N \left(p_is_k+r_iw_k\right)\left(1-sgn(q_j-q_i)sgn(q_j-q_k)\right)e^{ -\mid q_j-q_i\mid-\mid q_j-q_k\mid}\},\\
p_{j,t}=&\frac{1}{9}\{-\frac{1}{3}p_j(p_js_j+r_jw_j)
+\sum_{i,k=1}^N \left(p_jp_is_k+r_jp_iw_k\right)\left(sgn(q_j-q_i)-sgn(q_j-q_k)\right)e^{ -\mid q_j-q_i\mid-\mid q_j-q_k\mid}
\\&+\sum_{i,k=1}^N \left(p_jp_is_k+r_jp_iw_k\right)\left(1-sgn(q_j-q_i)sgn(q_j-q_k)\right)e^{ -\mid q_j-q_i\mid-\mid q_j-q_k\mid}\},\\
r_{j,t}=&\frac{1}{9}\{-\frac{1}{3}r_j(p_js_j+r_jw_j)
+\sum_{i,k=1}^N \left(r_jr_iw_k+p_jr_is_k\right)\left(sgn(q_j-q_i)-sgn(q_j-q_k)\right)e^{ -\mid q_j-q_i\mid-\mid q_j-q_k\mid}
\\&+\sum_{i,k=1}^N \left(r_jr_iw_k+p_jr_is_k\right)\left(1-sgn(q_j-q_i)sgn(q_j-q_k)\right)e^{ -\mid q_j-q_i\mid-\mid q_j-q_k\mid}\},\\
s_{j,t}=&\frac{1}{9}\{\frac{1}{3}s_j(p_js_j+r_jw_j)
-\sum_{i,k=1}^N \left(w_jr_is_k+s_jp_is_k\right)\left(sgn(q_j-q_i)-sgn(q_j-q_k)\right)e^{ -\mid q_j-q_i\mid-\mid q_j-q_k\mid}
\\&+\sum_{i,k=1}^N \left(s_jp_is_k+w_jr_is_k\right)\left(sgn(q_j-q_i)sgn(q_j-q_k)-1\right)e^{ -\mid q_j-q_i\mid-\mid q_j-q_k\mid}\},\\
w_{j,t}=&\frac{1}{9}\{\frac{1}{3}w_j(p_js_j+r_jw_j)
-\sum_{i,k=1}^N \left(s_jp_iw_k+w_jr_iw_k\right)\left(sgn(q_j-q_i)-sgn(q_j-q_k)\right)e^{ -\mid q_j-q_i\mid-\mid q_j-q_k\mid}
\\&+\sum_{i,k=1}^N \left(w_jr_iw_k+s_jp_iw_k\right)\left(sgn(q_j-q_i)sgn(q_j-q_k)-1\right)e^{ -\mid q_j-q_i\mid-\mid q_j-q_k\mid}\}.
\end{split}
\label{dNcp3}
\end{eqnarray}
For $N=1$, (\ref{dNcp3}) becomes
\begin{eqnarray}
\left\{\begin{array}{l}
q_{1,t}=\frac{2}{27}(p_1s_1+r_1w_1),
\\
p_{1,t}=\frac{2}{27}p_1(p_1s_1+r_1w_1),
\\
r_{1,t}=\frac{2}{27}r_1(p_1s_1+r_1w_1),
\\
s_{1,t}=-\frac{2}{27}s_1(p_1s_1+r_1w_1),
\\
w_{1,t}=-\frac{2}{27}w_1(p_1s_1+r_1w_1).
\end{array}\right. \label{N2}
\end{eqnarray}
We may solve this equation as
\begin{eqnarray}
q_{1}=\frac{2}{27}(A_2+A_3)t+A_4,~p_{1}=A_5e^{\frac{2}{27}(A_2+A_3)t},~r_{1}=\frac{1}{A_1}p_1,~
s_{1}=\frac{A_2}{A_5}e^{-\frac{2}{27}(A_2+A_3)t},~w_{1}=\frac{A_3}{r_1},
\label{N2s}
\end{eqnarray}
where $A_1$, $\cdots$, $A_5$ are integration constants.
See Figure \ref{F2} for the single-peakon of the potentials $u_{1}(x,t)$ and $v_{1}(x,t)$ with $A_4=0$, $A_2=A_5=1$ and $A_3=2$.
\section {Conclusions and discussions}
In our paper, we propose a multi-component generalization of the Camassa-Holm equation, and provide its Lax representation and infinitely many conservation laws.
This system contains an arbitrary smooth function $H$, thus it is actually a large class of multi-component peakon equations.
Due to the presence of the arbitrary function, we do not expect that the system is bi-Hamiltonian in the general case.
But we show it is possible to find the bi-Hamiltonian structures for the special choices of $H$.
In particular, we study the peakon solutions of this system in the case $N=2$, and obtain a new integrable system which admits stationary peakon solutions.
As mentioned above, Li, Liu and Popowicz proposed a four-component system which also contains an arbitrary function \cite{LLP}.
In contrast with the usual soliton equations, the peakon equations with arbitrary functions seem to be unusual.
We believe that there are much investigations deserved to do for both our generalized peakon system and Li-Liu-Popowicz's system.
The following topics seem to be interesting:
\\ (1) Is there a gauge transformation that can be applied to the Lax pair to remove the arbitrary function $H$?
\\ (2) Can the inverse scattering transforms be applied to solve the systems in general?
\\ (3) Do there exist infinitely many commuting symmetries for the systems?
\section*{ACKNOWLEDGMENTS}
This work was partially supported by the National Natural Science Foundation of China (Grant Nos. 11301229, 11271168, and 11171295), the Natural Science Foundation of the Jiangsu Province (Grant No. BK20130224), the Natural Science Foundation of the Jiangsu Higher Education Institutions of China (Grant No. 13KJB110009), and the China state administration of foreign experts affairs system under the affiliation of China University of Mining and Technology.
\vspace{1cm}
\small{
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 36
|
Q: How to add table to table? I want to add table 'mm' to table 'fmenu.pages' but it doesnt work at all. Error: attempt to index a nil value (field 'main2'). Its about last line. Code:
local fmenu = {
selected_button = 0,
menu = {
font = 1,
},
pages = {
["main"] = {
name = "name",
id = 1,
btns = {
{name = "name I", id = 1}
}
}
}
}
local mm = {
["main2"] = {
name = "name2",
id = 2,
btns = {
{name = "name I", id = 1},
{name = "name II", id = 2}
}
}
}
table.insert(fmenu.pages, mm)
print(fmenu.pages["main2"].name)
A: How about using table.merge from lua-stdlib?
local table = require"std.table"
local fmenu = {
selected_button = 0,
menu = {
font = 1,
},
pages = {
["main"] = {
name = "name",
id = 1,
btns = {
{name = "name I", id = 1}
}
}
}
}
local mm = {
["main2"] = {
name = "name2",
id = 2,
btns = {
{name = "name I", id = 1},
{name = "name II", id = 2}
}
}
}
table.merge(fmenu.pages, mm)
print(fmenu.pages["main2"].name)
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 3,668
|
A fasori evangélikus templom Budapest VII. kerületében a Városligeti fasor és a Bajza utca sarkán áll. A főváros legdíszesebb protestáns temploma, a Deák téri evangélikus templom után a második legnagyobb evangélikus templom Budapesten. Egy épülettömböt alkot a fasori evangélikus gimnázium épületével.
Leírása
A templom neogótikus stílusú, kéthajós épület. Tornya a főhajó mellett áll, 55 méter magas, és három harang található benne, melyek közül a legnagyobb 710 kilogrammos, a középső 270, a legkisebb 150 kilogramm súlyú. A nagyharang 1999-ben készült a második világháborúba elvitt 550 kilogrammos nagyharang helyett. Kapuja fölött Róth Miksa mozaikja látható: "Krisztus és a gyermekek", valamint a szintén általa alkotott öt méter átmérőjű rózsaablak.
A két hajót monumentális oszlop választja el, a falakat növényi motívumokkal mintázott pillérek és vékony oszlopok díszítik. Az oltárképet Benczúr Gyula festette 1913-ban, témája: "A napkeleti bölcsek hódolása a kisded Jézus előtt". Az oltár mögött három táblából álló, 21 négyzetméteres festett üvegablak Krisztust és a négy evangélistát ábrázolja. A kétmanuálos orgonát a pécsi Angster cég készítette 1906-ban, 1989-ben 21 regiszteresre bővítették.
Története
A pesti evangélikusok a 19. század végén új gimnázium építését határozták el, helyszínül a kertes villákkal övezett Városligeti fasort választották. 1903-ban vásárolták meg a telket. Ugyanebben az évben pályázatot írtak ki, ekkor már egy gimnázium és egy templom épületegyüttesének megtervezésére. Nyolc pályamű érkezett be, de a bíráló bizottság egyiket sem ítélte megvalósítandónak. Végül Pecz Samut kérték fel a végleges tervek elkészítésére, amik alapján 1903 nyarán el is kezdődött az építkezés. A gimnázium 1904 decemberére készült el, a templom pedig 1905-re. 1905. október 8-án szentelték fel.
A fasori templom volt az első budapesti evangélikus templom, melyben csak magyar nyelven folytak az istentiszteletek.
Budapest ostromakor megsemmisültek a Róth Miksa által festett üvegablakok és megsérült a toronyóra.
1973-74-ben teljes belső, 1996-ban pedig külső felújításra került sor.
Az üvegablakok pótlását Kiss Miklós üvegművész végezte 1997-ben, 1999-ben és 2002-ben.
Külső hivatkozások
A gyülekezet honlapja
A templom története
A templom leírása az evangélikus egyház honlapján
Budapest keresztény templomai, kolostorai
Magyarország evangélikus templomai
Budapest 20. századi építményei
Magyarország neogótikus vallási építményei
Budapest VII. kerülete
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 4,856
|
Labeatis Catenae una formació geològica del tipus catena del quadrangle Tharsis de Mart, situat amb coordenades planetocèntriques a 20.5 ° latitud N i 268.34 ° longitud E. Té un diàmetre de 220.6 km i va rebre el nom d'una característica clàssica d'albedo. El nom va ser fet oficial per la UAI l'any 1988. El terme "Catena" fa referència a una cadena de cràters.
Referències
Enllaços externs
Labeatis
Labeatis
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 6,713
|
{"url":"https:\/\/iescoders.com\/category\/seismology\/general-seismology\/","text":"## Solving nth order differential equation for a given boundary\u00a0condition\n\nWe are going to solve the\u00a0differential equation with the boundary conditions\n\n$\\psi_{xx} +[100 - \\beta ]\\psi = 0$\n\n$\\implies \\psi_{xx} = [\\beta - 100] \\psi$\n\nLet\u2019s take the simpler boundary condition\n\n$\\psi(x=\\pm 1) =0$.\n\nNow, the first thing which we wanna do is to convert this equation in the first order form. After that, we can simply use the 4th order Runge-Kutta to obtain higher order solution.\n\nTo transform our 2nd order differential equation into first order, we define two variables. To transform nth order equations we need n such variables. This is the standard process to convert any differential equation into the system of first order differential equations.\n\nLets take\n\n$y_1 = \\psi$ (1)\n\n$\\implies y_1' = \\psi_x$\n\n$\\implies y_1' = y_2$\n\n$y_2= \\psi_x$ (2) [derivative of $\\psi$]\n\n$\\implies y_2' = \\psi_{xx}$\n\n$\\implies y_2' = [\\beta - 100] \\psi$\n\nSo, we have transformed the 2nd order differential equation into two first order differential equations.\n\nAnd our boundary condition will become $y_1(\\pm 1) =0$\n\nNow for the differential equations of the type\n\n$\\psi_{xx} = [\\beta - 100] \\psi$\n\n$\\implies \\psi_{xx} = a \\psi$\n\nhave the solution $\\psi = e^{\\pm \\sqrt{a}x}$ (for positive $a$). Now this solution doesn\u2019t satisfy our condition $\\psi(x=\\pm 1) =0$. Exponential functions are monotonic and hence cannot be 0 at two points. So, this tells us that $a$ is not positive, i.e., $\\beta \\ngtr 100$.\n\nNow, if $\\beta < 100$, then our eqn becomes\n\n$\\psi_{xx} = - a \\psi$\n\nThe solution to this is of the form of cosines and sines, which has potential to satisfy the boundary condition. So, we have $\\beta < 100$.\n\nDefining Function in MATLAB:\n\nfunction [ rhs ] = yfunc( x,ic,dummy,beta )\ny1=ic(1);\ny2=ic(2);\n\nrhs = [y2; (beta -100)*y1];\nend\n\nCalculating Solutions:\n\nclear; close all; clc\nset(0,'defaultlinelinewidth',2);\nset(0,'defaultaxeslinewidth',1.5);\nset(0,'defaultpatchlinewidth',2);\nset(0, 'DefaultAxesFontSize', 14)\n\ntol=10^(-4); % define a tolerance level to be achieved\n% by the shooting algorithm\n\nxspan=[-1 1]; %boundary conditions % define the span of the computational domain\n\nA=1; % define the initial slope at x=-1\nic=[0 A]; % initial conditions: x1(-1)=0, x1?(-1)=A\n\nn0=100; % define the parameter n0\ndbeta=1;\nbetasol=[];\nbeta_start=n0; % beginning value of beta\nfor jj=1:5 % begin mode loop\nbeta=beta_start; % initial value of eigenvalue beta\ndbeta=n0\/100; % default step size in beta\nfor j=1:1000 % begin convergence loop for beta\n[t,y]=ode45('yfunc',xspan, ic,[],beta); % solve ODEs\nif y(end,1)*((-1)^(jj+1)) > 0 %this checks if the beta needs to be higher or lower for convergence\nbeta = beta-dbeta;\nelse\nbeta = beta + dbeta\/2; %uses bisection to converge to the solution\ndbeta=dbeta\/2;\nend\nif abs(y(end,1)) < tol % check for convergence\nbetasol=[betasol beta]; % write out eigenvalue\nbreak % get out of convergence loop\nend\nend\nbeta_start=beta-0.1; % after finding eigenvalue, pick\n% new starting value for next mode\nnorm=trapz(t,y(:,1).*y(:,1)); % calculate the normalization\nplot(t,y(:,1)\/sqrt(norm)); hold on\nend\nbetasol\nlegend(sprintf('\\\\beta_1 = %.4f',betasol(1)),sprintf('\\\\beta_2 = %.4f',betasol(2)),sprintf('\\\\beta_3 = %.4f',betasol(3)),...\nsprintf('\\\\beta_4 = %.4f',betasol(4)),sprintf('\\\\beta_5 = %.4f',betasol(5)));\nxlabel('x'); ylabel('\\psi (x)','FontWeight','bold')\nsaveas(gcf,'DiffEqnSol','jpg')\n\n## Introduction to Genetics Algorithm (GA) (Part\u00a02)\n\nTo find a basic introduction of GA, the first part can be found here.\n\nIII. Examples using Genetics Algorithm\n\nIn these examples, we will use Matlab and its function ga to apply GA for the optimization problem. For the manual of using this function, you can find it at\u00a0https:\/\/www.mathworks.com\/help\/gads\/ga.html or type in Matlab:\n\nhelp ga\n\nIII.1. Optimization of a polynomial function\n\nProblem: find the integer x \u00a0value that minimizes function:\n\ny = 0.2x2 + 50\/x\n\nFirst, we define the function:\n\nfunction y = simple_fitness(x)\ny = 0.2*x.^2+(50.\/x);\n\nAnd use Matlab to finish the job:\n\nclear; close all; clc\n\nx=linspace(0,40,100);\ny=simple_fitness(x);\nplot(x,y);\n\nrng(1,'twister') % for reproducibility\nlb=[1];\n[x,fval] = ga(@simple_fitness,1,[],[],[],[],lb\n\nxlabel('Size of cup');\nylabel('Objective Function Value');\ntitle(sprintf('Minimum of the function is %.2f at x = %.2f',min(y), x))\ngrid on\n\nAnd here is the result:\n\nIII.2. Find the best location of the earthquakes\n\nProblem: we have a set of 5 seismic stations with coordinates: [-2 3 0; 1 3 0; -2 -1 0; 0 -3 0; 2 -2 0]\n\nThe travel time of seismic ray from EQ to stations is calculated from:\n\nTo simplify the problem, we choose v = 6 km\n\nclear; close all; clc\nstnx=[-2 1 -2 0 2];\nstny=[3 3 -1 -3 -2];\nstnz=[0 0 0 0 0];\nEQ=[2 2 -2];\n\nto=sqrt((EQ(1)-stnx).^2+(EQ(2)-stny).^2+(EQ(3)-stnz).^2)\/6\nto=to+0.05*randn(size(to))\nrms = @(mp) sum(sqrt((1.\/(length(to)-4)).*((sqrt((mp(1)-stnx).^2+(mp(2)-stny).^2+(mp(3)-stnz).^2)\/6)-to).^2));\n% rms([2 2 -2])\nub=[10 10 0]; %upper bound\nlb=[0 0 -10]; %lower bound\noptions = gaoptimset('Display','iter','PlotFcn',@gaplotbestf,'Generations',150)\n% options = gaoptimset('Display','iter','PlotFcn',@gaplotbestf,'TolCon',1e-3);\n\nmp = ga(rms,3,[],[],[],[],lb,ub,[],options)\n\nfigure; hold on; grid on;\nscatter3(stnx, stny, stnz, 'b', 'filled', '^');\nscatter3(mp(1), mp(2), mp(3), 'r', '*');\nscatter3(EQ(1), EQ(2), EQ(3), 'g', '*');\nview(3); xlabel('X axis'); ylabel('Y axis'); zlabel('Z axis');\nlegend('Stations','EQ location from GA','Real EQ location');\n\n\n\nUtpal Kumar\n\nIESAS\n\n## Simple 1D velocity model inversion from P arrival\u00a0time\n\nRefer to Chapter 5, Introduction to Seismology, Shearer 2009.\n\nProblem:\n\nFrom the P-wave travel time data below (note that the reduction velocity of 8km\/s), inverse for the 1D velocity model using T(X) curve fitting (fit the T(X) curve with lines, then invert for the ray parameter p and delay time\u00a0\u03c4(p), then solve for the velocity and depth).\n\nSolution:\n\nWe construct the T(X) plot from the data: y = T \u00a0\u2013 X\/8 => T = y + X\/8 and then divide the T(X) curve into fitting lines (Fig. 2).\n\nThe ray parameter and delay is calculated for each line by using:\n\np = dT\/dX\n\n\u03c4 = T \u2013 pX\n\nThese parameters p and \u03c4 are slope and intercept of a line, respectively.\n\nThe\u00a0\u03c4(p) can be written as:\n\nas in the matrix form of:\n\nor, in short,\n\n$\\tau = Gh$\n\nWe can invert for the thickness of each layer using least square inversion:\n\n$h = (G^{_{T}}G)^{-1}G^{T}\\tau$\n\nIf we assume ray slowness u(i) = p(i) for each line, we can solve the h and v for the 1D velocity model.\n\nHere is the result of the inversion:\n\nThe sharp increase in Vp from ~6.6 km\/s to 7.7 km\/s at ~ 30km depth suggest for the Moho discontinuity.\n\nThe codes for the problem is written in python and can be downloaded: source code\u03c4 inversion code, data points.\n\nNguyen Cong Nghia, IESAS\n\n## Ray tracing through a 1-D velocity\u00a0model\n\nRefer to Chapter 4 of Shearer, Introduction to Seismology.\n\nFor a ray piercing through Earth, the ray parameter\u00a0(or horizontal slowness) p is defined by several expressions:\n\nwhere u = 1\/v is the slowness,\u00a0\u03b8 is the ray incidence angle, T is the travel time, X is the horizontal range and utp is the slowness at the ray turning point.\n\nThe vertical slowness is defined as:\n\nand integral expressions for the surface-to-surface travel time are:\n\nWith these equations we can calculate the travel time (T) and horizontal distance (X) for a given ray with ray parameter p and velocity model v(z).\n\nApply these to a problem (Exercise 4.8 Shearer):\n\n(COMPUTER) Consider MARMOD, a velocity-versus-depth model, which is typical of much of the oceanic crust (Table 4.1). Linear velocity gradients are assumed to exist at intermediate depths in the model; for example, the P velocity at 3.75 km is 6.9 km\/s. Write a computer program to trace rays through this model and produce a P-wave T(X) curve, using 100 values of the ray parameter p equally spaced between 0.1236 and 0.2217 s\/km. You will find it helpful to use subroutine LAYERXT (provided in Fortran in Appendix D and in the supplemental web material as a Matlab script), which gives dx and dt as a function of p for layers with linear velocity gradients. Your program will involve an outer loop over ray parameter and an inner loop over depth in the model. For each ray, set x and t to zero and then, starting with the surface layer and proceeding downward, sum the contributions, dx and dt, from LAYERXT for each layer until the ray turns. This will give x and t for the ray from the surface to the turning point. Multiply by two to obtain the total surface-to-surface values of X(p) and T(p). Now produce plots of: (a) T(X) plotted with a reduction velocity of 8 km\/s, (b) X(p), and (c) \u03c4(p). On each plot, label the prograde and retrograde branches. Where might one anticipate that the largest amplitudes will occur?\n\nUsing the LAYERXT subroutine (Appendix D\u00a0of Shearer), the FORTRAN and MATLAB can be downloaded from here. In this example, we will use python to calculate and plot the result.\n\nThe codes can be downloaded here: main program, python LAYERXT subroutine, velocity model.\n\nYou need to install numpy, pandas and matplotlib package to run this. Normally it can be installed in the\u00a0terminal by (presume python has been installed):\n\npip install numpy pandas matplotlib\n\nNguyen Cong Nghia \u2013 IESAS\n\n## Modeling a wave on a string using finite\u00a0differences\n\nBased on problem 3.7 chapter 3 of Introduction to Seismology (Shearer)\n\n(COMPUTER) In the case of plane-wave propagation in the x direction within a uniform medium, the homogeneous momentum equation (3.9) for shear waves can be expressed as\n\n,where u is the displacement. Write a computer program that uses finite differences to solve this equation for a bar 100 km in length, assuming \u03b2 = 4 km\/s. Use dx = 1 km for the length spacing and dt = 0.1 s for the time spacing. Assume a source-time function \u00a0at u (50 km) of the form:\n\nApply a stress-free boundary condition at u(0 km) and a fixed boundary condition at u (100 km). \u00a0Approximate the second derivatives using the finite difference scheme:\n\nPlot u(x) at 4 s intervals from 1 to 33 s. Verify that the pulses travel at velocities of 4 km\/s. What happens to the reflected pulse at each endpoint? What happens when the pulses cross?\n\n!**********************************************************************\n!* *\n!* 1-D Plane wave propagation with fixed and free boundaries *\n!* *\n!**********************************************************************\n\n!----initialzie t, dx, dt, tlen, beta, and u1, u2, u3 arrays----c\n\nimplicit none\ninteger:: i, counter, nx, nx1\ninteger:: it, ntmax\nreal(8):: t, dt, dt2, dx, dx2, beta, beta2, tlen, tmax, rhs\nreal*8, allocatable:: u1(:), u2(:), u3(:)\ncharacter(10):: data\ncharacter(4):: nf\n!----parameters ----c\n! dt must be less than (dx\/beta) for numerical stability\n\nnx = 100 ! number of grid\ndt = 0.10d0 ! time interval\ndx = 1.0d0 ! grid interval\nbeta = 4.0d0 ! wave velocity\ntlen = 5.0d0 ! time length of wave source\nntmax = 1000 ! maximum time step (finish at it = ntmax, or)\ntmax = 33.0d0 ! maximum calculation time (finish at t< tmax)\n\n!---- allocate dimension variables ----c\n\nnx1 = nx + 1\nallocate (u1(nx1), u2(nx1), u3(nx1))\n\n!---- initialize t, counter, u1, u2, u3 ----c\n\nt = 0.0d0\ncounter = 1\n\ndo i = 1, nx1\nu1(i) = 0.0d0\nu2(i) = 0.0d0\nu3(i) = 0.0d0\nend do\n\n!---- calculate c^2, dt^2, dx^2 ----c\nbeta2 = beta**2\ndt2 = dt**2\ndx2 = dx**2\n\n! ============= Time marching loop ===========\n\ndo it = 1, ntmax\n\nt = t + dt\n\n!---- calculate u3(i) ----c\n\ndo i= 2, nx\nrhs=beta2*2*(u2(i+1)-2.0d0*u2(i)+u2(i-1))\/dx2\nu3(i)=dt2*rhs+2.0d0*u2(i)-u1(i)\nend do\n\n!---- free boundary (du\/dx=0) ----c\nu3(1)=u3(2)\n!---- fixed boundary (u=0) ----c\nu3(nx1)=0.0d0\n!---- source time function c----c\n\nif (t.le.tlen) then\nu3(51) = sin(3.1415927*t\/tlen)**2\nend if\n\n!---- change new and old variables ----c\n\ndo i=1,nx1\nu1(i) = u2(i)\nu2(i) = u3(i)\nend do\n!---- make data file ----c\nwrite(nf,'(i3.3)') counter\ndata = \"data\"\/\/nf\n\nopen(7,file=data,status='replace')\ndo i=1, nx1\nwrite(7,*) i, u2(i)\nend do\n!---- output u2 at desired intervals, stop when t is big enough\n\nwrite(6,'(a5,1x,i3,1x,f6.3)') 'loop', counter, t\n\nif (t.gt.tmax) exit\n\ncounter = counter + 1\n\nend do\n\nstop\nend\n\nThe following is used to plot the results:\n\n#!\/usr\/bin\/bash\ngnuplot << eof\nset term gif animate\nset output \"animate.gif\"\nn=9 #n frames\nset xrange [0:102]\nset yrange [-2:2]\ni=0\nlist = system('ls data*')\ndo for [file in list] {\nplot file w lines lt 1 lw 1.5 t sprintf(\"t=%i sec\",i\/10)\ni = i + 1\n}\nset output\neof\n\nNguyen Cong Nghia \u2013 IESAS\n\n## Calling SAC(Seismic Analysis Code) (in\u00a0Perl)\n\nFor seismologists, using a SAC for sac data manipulation is essential (though there are few alternatives). Here, we see how can we call SAC from a perl script:\n\n#!\/usr\/bin\/perl\nopen(SAC, \"| sac \") or die \"Error opening sac\\n\";\nprint SAC qq[\necho on\n*fg seismogram\u00a0\u00a0 \u00a0#sample seismic signal in SAC's memory\nfg sine 2 npts 2000 delta 0.01\n*fg impulse npts 100 delta 0.01\nbandpass bessel corner 0.1 0.3 npole 4\nppk\nfft\nplotsp am\nsave sine_fft.pdf\n];\nprint SAC \"quit\\n\";\nclose(SAC);\n\nExample script to call SAC functions in perl\n\n## Understanding Seismograms\n\nSeismograms are basic information about earthquakes, chemical and nuclear explosions, mining induced earthquakes, rock bursts and other events generating seismic waves.\n\nSeismograms reflect the combined influence of the seismic source, the propagation paths, the frequency response of the recording instrument and the ambient noise at the recording site.\n\nu(t) = s(t)*g(t)*i(t)*n(t). \u00a0 \u00a0 \u00a0 \u00a0 \u201c*\u201d means convolution.\n\nwhere, u is the seismic record, s is the source effect, g is the propagation effect, i is the instrument effect and n is the noise.\n\nSo, by understanding seismogram we can understand the seismic source, earth structure, and the noise in the medium.\n\nThis is our short attempt to understand the seismograms.\n\nUnderstanding Seismograms 1\n\nFor understanding the seismogram, there are two main things we should notice- record duration and the dispersion. Due to different nature of the propagation velocity of the seismic waves and the different propagation paths taken by them to the station, travel time differences between the main group usually grows with distance. Since the body wave groups do not disperses, so their individual duration remains more or less constant, only the time difference between them changes with distance.\n\nFigure 1: Event-Station Distribution map for the event in Mindanao, Phillipines Island (2005-02-05, Mw-7). The source is 540 km deep.\n\nFigure 2\n\nFigure 3\n\nFigure 4\n\nThe epicentral distance for QIZ, BJT and DGAR is ~ 19, 35, 52 degrees respectively. For the station BJT (Figure 3), the epicentral distance is 35.2 degrees and for the station DGAR (Figure 4), the epicentral distance is 52.4 degrees. The P-S arrival times for the station BJT is nearly 300 sec and for the station DGAR, the P-S arrival time is nearly 400 sec. For the station QIZ, it is only around 200 sec. So, with the increase in epicentral distance, the difference in individual phase arrival increases. Also for the station DGAR, the P arrival is later than the BJT.\n\nThe time difference between main body wave onsets is:\n\n Distance (degrees) less than Time-difference in body wave onsets (P-S) in sec 10 180 60 (Fig 1,2,3) 960 100 1800 180 2700\n\nFigure 5 : Travel-Time curve for depth 540 km\n\nFor the deep events, we do not see the first arrival as P wave (Figure 6) for short distances (D\n\nFigure 6\n\nFigure 6, is the seismogram at 2 degrees epicentral distance from the deep source. Here, we can notice that the first arrival is not P but the core reflected phase PcP. It is also cevident from the travel time curve(Figure 5).\n\nFigure 7: Event-Station Distribution Map of Near Coast of Peru (2001-06-23, Mw-8.3). The source is at 2 km depth.\n\nWhen the source is near the surface then we observe the surface waves as well in the seismogram.The surface waves from earthquakes at intermediate (> 70 km) or great depth (> 300 km) may have amplitudes smaller than those of body waves or may not even be detected on seismic records.\n\nIn contrast to the body waves, the velocity of surface waves is frequency dependent and hence it is dispersed. The duration of Love and Rayleigh waves increases with distance. As for station NNA (Figure 8), the duration of the first group of Rayleigh wave is about 230 seconds and for station HKT (Figure 9), the duration is around 600 seconds.\n\nWe can also notice that for shallower earthquake event, P-wave is the first arrival for stations less than 10 degrees of the source (Figure 10).\n\nFigure 8\n\nIn Figure 8, we can notice that for surface waves, the largers periods arrive first than the lowers periods. This is the general case for normal layering. The larger periods sample the higher depths and since the velocity for higher depths is larger so it arrives before the shorter period waves.\n\nFigure 9\n\nIn figure 9, for the surface waves, we can observe the wave groups or the beating effect (analogous to acoustics). This is because of the interference of the two or more harmonic waves with closer frequencies or periods.\n\nFigure 10: Travel time curve for depth 2.2km\n\nFor certain distance ranges, the travel time curves for different types of seismic phases are close to each other or can even overlap (Figure 5,11).\n\n(a)\n\n(b)\n\nFigure 11: S, SKS and ScS are very close to each other.\n\nThe amplitude of P-wave depends not only on the epicentral distance but on the site effect (underground geology and crustal heterogeneity), azimuth etc.\n\nThe most obvious indication on a seismogram that a large earthquake has a deep focus is the small amplitude of the surface waves with respect to the body-wave amplitudes and the P and S waveforms often have impulsive onsets.\n\nA more precise determination of the depth h of a seismic source, however, requires either the availability of a seismic network with at least one station being very near to the source, e.g., at an epicentral distance D < h (because only in the near range the travel time t(D, h) of the direct P wave varies strongly with source depth h), or the identification of seismic depthphases on the seismic record.\n\nAt distant seismograph stations, the depth phases pP or sP follow the direct P wave by a time interval that changes only slowly with distance but rapidly with depth.","date":"2018-09-20 18:54:58","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 21, \"codecogs_latex\": 2, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6606974601745605, \"perplexity\": 4193.505270166515}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-39\/segments\/1537267156554.12\/warc\/CC-MAIN-20180920175529-20180920195929-00405.warc.gz\"}"}
| null | null |
Q: Retrieving command line argument of process at driver level Hello I am writing a minifilter driver for intercepting all the irp packets from a certain process say a.exe .
So , in the driver code it can be done by applying a check on the command line arguments that started the process.
Does anyone know how can i retrieve the command line argument ??
Thanks in advance .
A: There's no supported way to do this from within kernel-mode. In fact, trying to access user-mode process information from the kernel is a pain in general. I would suggest firing up a request to a user-mode service, which can then find that information and pass it back down to your kernel component.
However, there an undocumented method to do it. If you can get a handle to an EPROCESS struct for the target process, you can get at a pointer to the PEB (process environment block) struct within it, which then has a pointer to an RTL_USER_PROCESS_PARAMETERS structure, which has a member called CommandLine.
Example:
UNICODE_STRING* commandLine = epProcess->Peb->ProcessParameters->CommandLine;
The downside to this is that EPROCESS is almost entirely opaque and PEB is semi-opaque too, meaning that it may change in future versions of Windows. I certainly wouldn't advocate trying this in production code.
A: Try using the NtQueryInformationProcess or ZwQueryInformationProcess function with the PROCESSINFOCLASS parameter as ProcessBasicInformation. The output parameter, ProcessInformation, will be a struct of type PROCESS_BASIC_INFORMATION. As Polynomial mentioned, this struct has a pointer to the process's PEB struct, which contains the information you are looking for in its ProcessParameters field.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 4,604
|
22 Apr April 22, 2019
Posted at 23:38h in Weekly eBulletin by Gabriela.Islas 0 Comments
CMS' 2020 ACA Marketplace Rule Expects 70,000 Fewer Customers
CMS has released the final payment and benefit rules for the ACA Marketplaces that included a number of changes geared to move people away from the exchange plans and to attempt to lower drug prices. The final rule will allow insurers to exclude manufacturer coupons from counting towards a patient's annual limitation on cost sharing if a medically appropriate generic drug is available. Additionally, the rule makes a technical change to how ACA subsidies are calculated, which will raise insurance costs for some customers. CMS is also finalizing its proposal to reduce the exchange fee that insurers pay by half a percentage point. That should reduce premiums slightly since insurers typically add that fee into their rates. Overall, CMS expects the changes in the final rule will result in $980 million less in federal financial assistance in 2020 and 70,000 fewer Marketplace customers. Click here for the CMS final rule, and here for the fact sheet.
According to the Congressional Budget Office, the number of unisured under 65 climbed from 27.5 million in 2016 to 28.9 million in 2018, click here.
Most Medicaid money goes to Medicaid heath plans. In 2018, states and the federal government paid health insurance companies $307 billion to run Medicaid programs, according to the latest numbers from Health Management Associates, click here.
Fear of Medicare for All on Wall Street is pummeling health care stocks, at least for now. Click here for details. Click here for the WSJ report.
CMS Will Boost SNF Payments by 2.5%, Hospice Payments by 2.7%
CMS is proposing to raise Medicare payments for skilled nursing facilities by 2.5 percent, about $887 million, in fiscal year 2020. CMS also estimated that a 2019 policy change to value-based purchasing would reduce aggregate payments to facilities by $213.6 million over 2020. The agency also proposed expanding data collection for quality reporting programs to all skilled nursing facility residents, regardless of their payer. Click here for details. CMS in a separate rule proposed a 2.7 percent increase in hospice payments next year, or about $540 million. Click here for the CMS summary.
CMS Offers Payment Increases for IRFs and Inpatient Psychiatric Facilities
CMS also released a couple of proposed rules for 2020 payments for Inpatient Rehabilitation Facilities and Inpatient Psychiatric Facilities, offering both an increase in payments for 2020. For the approximately 1,600 psychiatric facilities, CMS is proposing to increase payments by an estimated 1.7 percent, or $75 million. This pays for services in psychiatric hospitals and excludes psychiatric units of an acute care hospital or critical access hospital. IRFs payments will see a proposed increase of 2.3 percent (or $195 million) for FY 2020, relative to payments in FY 2019. CMS is proposing various quality reporting requirement changes for both types of facilities as well. Click here for the IRF fact sheet, and here for the IRF rule. Click here for the psychiatric facilities fact sheet, here for the rule.
Federal Surprise Billing Legislation Would Supersede State Laws: CRS
That is according to the Congressional Research Service, the non-partisan public policy research arm of Congress. The group suggested that lawmakers should consider state laws already in place when drafting federal legislation, as well as the potential for litigation over any set payment rates. Balance billing, when an out-of-network provider bills the patient for charges that exceed her health insurance plan's payment, is currently governed by a patchwork of federal and state laws that provide varying degrees of protection to the patients and guidance on the underlying payment question. Over the past few months several congressional committees have investigated the practice and are considering legislation to clamp down. To read the full CRS study, click here.
Senate Leader McConnell Will Move To Raise Tobacco Age to 21 when Congress Returns
Senate Majority Leader Mitch McConnell (R-KY) announced plans at a press conference to introduce legislation to raise the nationwide minimum age to buy tobacco products from 18 to 21, stating that it is his top priority this spring. Even though tobacco has long been a main export from his home state of Kentucky, McConnell said he wants to change the law to discourage vaping and teenage nicotine addiction. The legislation will be similar to the current system, where retailers have the responsibility to verify the age of anyone buying tobacco products and will follow eleven states who have already enacted laws to raise the purchasing age of tobacco products to 21. The bill will also include an exemption for members of the military. Click here for more from McConnell's website.
Energy and Commerce Chairman Rep. Frank Pallone (D-NJ) and Rep. Donna Shalala (D-FL) introduced a bill last week that would allow FDA to collect user fees for e-cigarettes, raise the minimum age to purchase tobacco to 21, prohibit online sales of tobacco products, among other policies, click here.
Primary Care Accounts for Two-Percent of Medicare Spending
A new study released by the RAND Corp. found that care provided by primary care practitioners accounts for a small fraction of total spending among Medicare beneficiaries. As low as only about 2% of total Medicare spending is dedicated to primary care under a narrow definition of the term. The researchers looked at both narrow and broad definitions of the term and find that when using the broad definition the percentage went up to nearly five-percent of the annual care for Medicare beneficiaries. RAND researchers contend that the estimates are important because health system orientation toward primary care has been associated with higher quality, better outcomes and lower costs. To read the study, click here.
Washington State Likely To Be First To Create an Employee-Paid LTC Program
Washington State is poised to become the first state to establish an employee-paid program creating an insurance benefit to help offset the costs of long-term care, a step advocates say will help an aging population that is likely not prepared for the increasing costs needed for daily assistance. The measure creates a benefit for those who pay into the program, with a lifetime maximum of $36,500 per person, indexed to inflation, paid for by an employee payroll premium. It has cleared both the House and the Senate. Click here for more.
Bills Introduced that Would Exempt APMs from Stark Law Restrictions
Reps. Raul Ruiz (D-CA), Larry Bucshon (R-IN), Ron Kind (D-WI) and Kenny Marchant (R-TX) have reintroduced the Medicare Care Coordination Improvement Act of 2019 in the House shortly after Sen. Rob Portman (R-OH) and Michael Bennet (D-CO) introduced their version in the Senate. The bills would give CMS the authority to exempt alternative pay models and those under development from some Stark, or physician self-referral, law provisions by providing APMs the same waivers that accountable care organizations were granted by the ACA. Additionally, the legislation would give providers developing a new demo that CMS determines "constitutes significant progress toward establishing (an APM) model" a three-year waiver. To view the bill, click here.
Major Opioid Takedown in Appalachia Joint Task Force, Includes Dozens of Medical Professionals
The Appalachian Regional Opioid Strike Force, a coordinated effort between HHS, the Justice Department, and multiple law enforcement partners, announced last week a major takedown in the region that involved 60 charged defendants across 11 federal districts, including 31 doctors, seven pharmacists, eight nurse practitioners and seven other medical professionals, for their allegedly illegally prescribing and distributing opioids and other narcotics as well as for health care fraud schemes. The charges involve over 350,000 prescriptions and over 32 million pills distributed in West Virginia, Ohio, Kentucky, Alabama, and Tennessee to more than 24,000 patients in the region over the past 2 years. Click here for more from the Justice Department including sate-by-state figures.
This Flu Season One of the Longest
This flu season has been officially going for 21 weeks, according to reports collected through last week and released Friday by the Centers for Disease Control and Prevention. That makes it among the longest seen since the government started tracking flu season duration more than 20 years ago. Some experts likened the unusual double waves to having two different flu seasons compressed, back-to-back, into one. The previous longest recent flu season was 20 weeks, which occurred in 2014-2015. The current season began the week of Thanksgiving, a typical start time. But in mid-February, a nastier strain started causing more illnesses and driving up hospitalizations. Still, this flu season is not nearly as bad as last winter's 19-week season, the deadliest in at least four decades. An estimated 80,000 Americans died of flu and its complications last season. The CDC is estimating that flu-related deaths this season in the range of 35,000 to 55,000. Click here for more from the CDC.
A Light at the End of the Tunnel for Electronic Health Records
For technology more associated with physician dissatisfaction and burnout, there may be a change in the tide. Studies show that when organizations meet Stage 7 of Meaningful Use of the HIMSS Analytics EMR Adoption Model physicians are more satisfied and have more of their clinical decisions based off of feedback from the EHR system. While varying somewhat by vendor, in 2017, 94% of U.S. hospitals used their EHR data in ways that inform clinical practice, up from 87% in 2015 and equal to 2016 usage. Conversely, patient engagement in accessing their health records electronically has been slow to catch on. However the data shows that only about 8% of patients even activate their electronic access to their records. For more information on Hospitals' Use of Electronic Health Records Data, click here, and for how patients are utilizing electronic access to their records, click here.
CMS' Hospital Readmissions Reduction Program Not Having Desired Impact; Rural ED Visits Jump
A new study by researchers at University of California has found that no significant differences were seen after implementation of hospital penalty payments, suggesting that any benefits of the HRRP are modest at best. The study examined U.S. and Canada hospitalization length of stay and 30-day readmission rates of patients with heart failure from 2005 through 2015 and found that a little over 18-percent of the Canadian patients were readmitted and almost 20-percent of those in the U.S. were readmitted. Canada does not have an equivalent readmission program to the U.S. Clinically, the readmitted patients from both countries bore similarities in that they had more comorbidities and hospitalizations in the prior 6 months versus patients who were not readmitted. To view the study, click here.
Visits to the emergency department in rural areas jumped by more than 50% since 2005 according to researchers at University of New Mexico, click here.
Global Measles Crisis Reaching Critical Numbers
Data from the World Health Organization suggests that within the first few months of 2019, 112,163 measles cases have been reported. This number is up by almost 300 percent from the same period last year. Last week, New York City declared a public health emergency due to the rapid spread of measles cases. This disease is very infectious, and in those who are non-immune, it is estimated to infect 9 out 10 people exposed to it. Anti-vaxxers are also being blamed for the surge in cases, especially impacting children who cannot be vaccinated until 12 months of age. To read the study, click here.
$350 Million in HEAL Grants Awarded to Kentucky, Massachusetts, New York and Ohio
Awarded in as part of the HEAL [Helping to End Addiction Long-term] initiative, the NIH gave Kentucky, Massachusetts, New York and Ohio $350M in an effort to support addiction prevention and treatment in primary care settings, mental health services and the criminal justice system. The grants were given with a goal to reduce opioid related heaths by 40% over the next 3 years. "We believe this effort will show truly dramatic reductions in overdose deaths are possible," HHS Secretary Alex Azar said. To view the press release from the NIH click here. Click here for the Wall Street Journal article.
The Patient-Centered Outcomes Research Institute last week announced $44.6 million to fund 12 studies looking into opioid research, how to treat pain in cancer patients, and ways to enhance prenatal care, click here.
FDA Pulls Surgical Mesh for Transvaginal Repair Off the Market
The Food and Drug Administration has ordered the manufacturers of surgical mesh products for the transvaginal repair of pelvic organ prolapse to stop selling and distributing their products in the U.S. immediately. This comes as the FDA has determined that the manufacturers, Boston Scientific and Coloplast, have not demonstrated a reasonable assurance of safety and effectiveness for the devices and have 10 days to submit their plan to withdraw these products from the market. Click here for more from the FDA.
Syphilis Becoming a Problem in Rural America
Public health officials say rural counties across the Midwest and West are showing a rapid increase in syphilis infections. While syphilis is still concentrated in major cities, it has continued spread into places like Missouri, Iowa, Kansas and Oklahoma which creates a new set of challenges as compared with urban areas, rural populations tend to have less access to public health resources, less experience with syphilis and less willingness to address. In Missouri alone, the number of syphilis patients has more than quadrupled since 2012, going from 425 to 1,896 cases last year. Click here for more on the issue.
Pregnant Women Less Likely to Seek Treatment for Depression
That's according to new research published in Psychiatric Services in Advance, based on a survey of 12,360 women through the National Survey on Drug Use and Health between 2011 and 2016. Researchers looked at pregnant and nonpregnant women between 18 and 44, all of whom had an episode of major depression. The study found that 51 percent of women and 43 percent of nonpregnant women did not seek help during a depressive episode. The study also found in the month before the survey was administered, 23 percent of pregnant women used alcohol and 17 percent used marijuana. To read the full study, click here.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 7,496
|
Canadian Forces Station Cobourg (also CFS Cobourg) was a military logistics base located in Cobourg, Ontario.
The facility was created due to the expansion of Canadian military capability brought about by the Korean War and Cold War. Logistics and supply facilities for the Army were being expanded across Canada, and Cobourg was chosen as a site for a new supply depot. The choice of Cobourg for a location was partly due to its proximity to major rail lines. Land was purchased in 1951 and construction proceeded on several buildings over the next two years. In 1954, the headquarters of No. 26 Ordnance Depot of the Royal Canadian Ordnance Corps was relocated from Ottawa to Cobourg. The base existed mainly to provide supplies to other military facilities, but also included the Canadian Army's only respirator assembly plant and a detachment of No. 22 Works Company of the Royal Canadian Engineers.
When the three arms of the Canadian military were integrated to create the Canadian Forces, No. 26 COD became No. 26 Canadian Forces Supply Depot and the base was renamed Canadian Forces Base Cobourg (CFB Cobourg) in 1966. However, as Cobourg didn't house two or more major units, it didn't qualify as a "base" under Canadian Forces' criteria and was renamed Canadian Forces Station Cobourg a short time later.
In 1969, the Canadian Forces supply system was reorganized and CFS Cobourg was identified as surplus. The supply duties of the base were assumed by depots at CFB Downsview in the Toronto area and at the Longue Point site of CFB Montreal; the station was closed and decommissioned on August 31, 1970.
The depot was sold to the government of Ontario, which converted it to an industrial park, and later sold it to the Town of Cobourg.
Sources
Cobourg: Early Days and Modern Times. Cobourg, ON: Cobourg Book Committee, 1981. .
Ozorak, Paul. Abandoned Military Installations of Canada: Volume I: Ontario. 1991. .
Cobourg
Cobourg
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 4,084
|
Q: Do some calculations of the values from two table and store it in one of the table First Table Name: table1
| email | value |
----------------------------
| abc@gmail.com |0.12 |
| dsv@gmail.com |0.23 |
| rthgmail.com | 0.45 |
| hfg@gmail.com |0.56 |
| yyt@gmail.com | 0.78 |
| hjg@gmail.com | 0.35 |
Second Table Name: table2
| email | result |
----------------------------
| abc@gmail.com |0.3 |
| dsv@gmail.com |0.6 |
| rthgmail.com | 0.7 |
| hfg@gmail.com |0.8 |
| yyt@gmail.com | 0.1 |
| hjg@gmail.com | 0.3 |
Now, I want to perform this mathematical operation
value(table1) = result(table2)+value(table1) /2
for email="abc@gmail.com"
I want to use the UPDATE query.
Thank you!!!
A: You can join tables in update the same way you do in a select:
UPDATE table1
JOIN table2 ON table1.email=table2.email
SET table1.value = (table2.result + table1.value) / 2
WHERE table1.email = "abc@gmail.com"
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 6,094
|
/*
* Project:Easy Web Framework
* Description:
* EasyFK stands for Easy Web Framework.It's an open source product for E-Business / E-Commerce.It
* was launched by a chinese Hezhiping(QQ:110476592) in 2015.The goal of EasyFK is to provide a
* foundation and starting point for reliable, secure , simple-to-use ,cost-effective ,scalable
* and suitable-for-Chinese E-Business / E-Commerce solutions. With EasyFK, you can get started
* right away without the huge deployment and maintenance costs of E-Business / E-Commerce systems.
* Of course, you can customize it or use it as a framework to implement your most challenging business needs.
* EasyFk is licensed under the Apache License Version 2.0. You may obtain a copy of the License at
* http://www.apache.org/licenses/LICENSE-2.0
* Author:hezhiping Email:110476592@qq.com
*/
package cn.gorun8.easyfk.base.util.collections;
import java.util.Map;
import cn.gorun8.easyfk.base.util.UtilObject;
/** this class can go away when easyfk switches to java 1.6, replaced by
* AbstractMap.SimpleImmutableEntry
*/
public class ReadOnlyMapEntry<K, V> implements Map.Entry<K, V> {
protected final K key;
protected final V value;
public ReadOnlyMapEntry(K key, V value) {
this.key = key;
this.value = value;
}
public K getKey() {
return key;
}
public V getValue() {
return value;
}
public V setValue(V value) {
throw new UnsupportedOperationException();
}
@Override
public boolean equals(Object o) {
if (!(o instanceof Map.Entry<?, ?>)) return false;
if (this == o) return true;
Map.Entry<?, ?> other = (Map.Entry<?, ?>) o;
return UtilObject.equalsHelper(getKey(), other.getKey()) && UtilObject.equalsHelper(getValue(), other.getValue());
}
@Override
public int hashCode() {
return UtilObject.doHashCode(getKey()) ^ UtilObject.doHashCode(getValue());
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 8,033
|
package models;
import org.codehaus.jackson.JsonNode;
import play.libs.F.Promise;
import play.libs.WS;
import play.libs.WS.Response;
import com.google.common.base.Preconditions;
public class UserEmail {
private static final String USER_EMAIL_URL = "https://www.googleapis.com/oauth2/v1/userinfo";
/**
* return the user email address
*/
public String getUserEmailAddress(String accessToken) {
Preconditions.checkNotNull(accessToken, "accessToken cannot be null");
Promise<Response> promiseResponse = WS.url(USER_EMAIL_URL).setQueryParameter("access_token", accessToken).get();
JsonNode jsonResponse = promiseResponse.get().asJson();
String email = jsonResponse.findPath("email").asText();
if (email == null || email.isEmpty()) {
throw new RuntimeException("email cannot be null");
}
return email;
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 3,383
|
Az afrikai pálmacibet (Nandinia binotata) az emlősök (Mammalia) osztályának ragadozók (Carnivora) rendjébe, ezen belül a pálmacibetfélék (Nandiniidae) családjába tartozó faj.
Korábban a cibetmacskafélék (Viverridae) családjába sorolták be; azonban az áthelyezés miatt manapság családjának és nemének az egyetlen fajává vált.
Előfordulása
Az afrikai pálmacibet Afrikában, a Szaharától délre eső térségekben fordul elő. Guineától és Dél-Szudántól kezdve, Angoláig és Zimbabwe keleti részéig sokfelé megtalálható. Akár 2500 méteres tengerszint feletti magasságban is fellelhető.
Alfajai
Nandinia binotata arborea Heller, 1913
Nandinia binotata binotata Gray, 1830
Nandinia binotata gerrardi Thomas, 1893
Nandinia binotata intensa Cabrera & Ruxton, 1926
Megjelenése
Viszonylag kistestű, erős felépítésű faj, kis, kerek fülekkel és erőteljes lábakkal, rajta hajlott, visszahúzható karmokkal. Hosszú farka van, melyet a fákon való közlekedés során egyensúlyozásra használ. A kifejlett állat körülbelül 1,70-2,10 kilogramm tömegű.
Életmódja
Magányosan élő, éjszakai állat, amely az erdőket választotta élőhelyül. A folyók mentén és a szavannák ligeterdeiben is fellelhető. Mindenevő, de étrendjének nagy részét gyümölcsök alkotják; étrendjét kiegészítheti rágcsálókkal, denevérekkel, rovarokkal, madarakkal, tojásokkal, gyíkokkal, békákkal és dögökkel. Főleg hajnalkor és napnyugtakor mozog. Általában magányos, de ha bőséges a táplálékkínálat, akkor akár 15 állat is összegyűlhet. A hasán kettő és mindegyik lábán a harmadik és negyedik ujjak között, egy-egy szagmirigy van; valószínűleg ezekkel jelöli meg a területét.
Körülbelül 4 évig él.
Szaporodása
Ez az emlősfaj évente kétszer is szaporodhat. Először májusban, aztán ha a körülmények továbbra is megfelelőek októberben is világra hoz egy almot. A vemhesség 2-3 hónapig tart. Az alomnagyság általában 4 kölyökből áll. Az elválasztás 3 hónap után következik be. A nőstény emlői a tej mellett narancssárga levet is termel, mely megszínezi az anyaállat és a kölykök bundáját; a kutatók szerint ez visszataszító a párosodókedvű hímek számára.
Az afrikai pálmacibet és az ember
Állatkertekben nagyon ritka fajnak számít. A szabad természetben az élőhelyének elvesztése és a vadászat veszélyezteti. Továbbá a háziszárnyasok védelmére, valamint a hagyományos orvoslásban játszott szerepe miatt is irtják. Mindezek ellenére a Természetvédelmi Világszövetség (IUCN) nem fenyegetett fajnak minősíti.
Képek
Jegyzetek
Források
Ronald M. Nowak: Walker's Mammals of the World. Johns Hopkins University Press, 1999
Wilson, D. E., and D. M. Reeder: Mammal Species of the World. Johns Hopkins University Press, 2005.
Mammal Species of the World. Don E. Wilson & DeeAnn M. Reeder (szerkesztők). 2005. Mammal Species of the World. A Taxonomic and Geographic Reference (3. kiadás)
További információk
Ragadozók
Emlősfajok
A nyugattrópusi-afrikai erdőség emlősei
A nyugatközéptrópusi-afrikai erdőzóna emlősei
A kelettrópusi-afrikai szavanna és sztyeppe emlősei
A déltrópusi-afrikai szavanna és sztyeppe emlősei
Monotipikus emlőscsaládok
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 6,053
|
{"url":"https:\/\/math.stackexchange.com\/questions\/1835506\/sum-of-series-using-mean-value-theorem","text":"sum of series using mean value theorem\n\nLet $f(x)$ be a function which is differentiable on $[0,1]$ with $f(0)=0$ and $f(1)=1$. Show that for every $n\\in \\Bbb N$ there exists numbers $x_1,x_2,\\ldots,x_n\\in [0,1]$ such as $$\\sum_{k = 1}^n \\frac{1}{f' (x_k)} = n$$ I think the mean value theorem should be applied. So there exists $x_1$ in $[0,1]$ such that $f ' (x_1) = \\frac{f(1) - f(0)}{1-0} =1$ and there exists $x_2$ in $[0,x_1]$ such that $f ' (x_2) = \\frac{f(x1) - f(0)}{x1-0} = \\frac{f(x1)}{x1}$, so on and so forth for $x_3 ,x_4, \\ldots,x_n$ and we have the sum $$1+\\frac{x_1}{f(x1)} + \\frac{x_2}{f(x2)} +\\cdots+\\frac{x_{n-1}}{f(x_{n-1})}$$ and from here I have no idea what to do . I was wondering if anyone could be so kind to help ?\n\n\u2022 I edited your question to make the math more readable. If I made any mistakes, let me know, or try to fix it yourself. For future reference, check this link for more info on how to get math to render more beautifully. \u2013\u00a0Arthur Jun 22 '16 at 8:53\n\u2022 \"differentiable\" (has a linear approximation) is not the same as \"derivable\" (can be proven). \u2013\u00a0user21820 Jun 22 '16 at 8:55\n\u2022 Should we assume that each $x_i$ is distinct? If not, the found $x_1$ for $n=1$ gives a correct answer for any $x_i$ for any $n$. \u2013\u00a0Maarten Hilferink Jun 22 '16 at 9:00\n\u2022 yes each x i is distinct. Thank you \u2013\u00a0alana Jun 22 '16 at 9:01\n\u2022 Duplicate of math.stackexchange.com\/questions\/1834917\/\u2026? \u2013\u00a0Maarten Hilferink Jun 22 '16 at 9:03\n\nBy the intermediate value property of continuous functions, there are $0=x_0,x_1,\\ldots,x_{n-1},x_n=1\\in[0,1]$ such that: $$f(x_i) = \\frac{i}{n}$$ for every $i\\in[0,n]$, and we may assume WLOG $x_0<x_1<\\ldots<x_n$.\nMoreover, there is some $\\xi_i$ in the interval $(x_{i-1},x_i)$ such that: $$\\frac{f(x_{i})-f(x_{i-1})}{x_{i}-x_{i-1}}=f'(\\xi_i)$$ that is: $$\\frac{1}{f'(\\xi_i)} = n(x_i-x_{i-1}).$$ By summing those identities over $i\\in[1,n]$ we get:\n$$\\sum_{i=1}^{n}\\frac{1}{f'(\\xi_i)}= n(x_n-x_0) = \\color{red}{n}$$","date":"2019-10-14 05:36:01","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8795492649078369, \"perplexity\": 118.47627755659019}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.3, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-43\/segments\/1570986649232.14\/warc\/CC-MAIN-20191014052140-20191014075140-00361.warc.gz\"}"}
| null | null |
Q: requested an insecure XMLHttpRequest endpoint. This request has been blocked; the content must be served over HTTPS I am making an ajax call from Sharepoint site to access data from a public facing website.
$.ajax({
type: "POST",
contentType: "application/json; charset=utf-8",
url: 'http://license.mypublicfacingwebsite.com/my.svc/productValidation/https%E2%82%ACMYSPSITE.sharepoint.com,123AppName,15',
success: function (data) {
$($.parseJSON(data.d)).each(function (index, value) {
});
},
error: function (result) {
alert(result);
}
});
I get the error of
"requested an insecure XMLHttpRequest endpoint
http://license.mypublicfacingwebsite.com/my.svc/productValidation/https%E2%82%ACMYSPSITE.sharepoint.com,123AppName,15".This
request has been blocked; the content must be served over HTTPS.
What should I do in order to make this WCF call work.?
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 5,382
|
package org.rabix.executor.engine;
import java.util.ArrayList;
import java.util.List;
import java.util.concurrent.Executors;
import java.util.concurrent.ScheduledExecutorService;
import java.util.concurrent.TimeUnit;
import org.rabix.bindings.model.Job;
import org.rabix.common.engine.control.EngineControlFreeMessage;
import org.rabix.common.engine.control.EngineControlMessage;
import org.rabix.common.engine.control.EngineControlStopMessage;
import org.rabix.executor.service.ExecutorService;
import org.rabix.executor.service.FileService;
import org.rabix.transport.backend.Backend;
import org.rabix.transport.backend.HeartbeatInfo;
import org.rabix.transport.mechanism.TransportPlugin;
import org.rabix.transport.mechanism.TransportPlugin.ErrorCallback;
import org.rabix.transport.mechanism.TransportPlugin.ReceiveCallback;
import org.rabix.transport.mechanism.TransportPluginException;
import org.rabix.transport.mechanism.TransportQueue;
import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
public abstract class EngineStub<Q extends TransportQueue, B extends Backend, T extends TransportPlugin<Q>> {
protected final Logger logger = LoggerFactory.getLogger(getClass());
protected B backend;
protected T transportPlugin;
protected ScheduledExecutorService scheduledHeartbeatService = Executors.newSingleThreadScheduledExecutor();
protected Q sendToBackendQueue;
protected Q sendToBackendControlQueue;
protected Q receiveFromBackendQueue;
protected Q receiveFromBackendHeartbeatQueue;
protected FileService fileService;
protected ExecutorService executorService;
public void start() {
transportPlugin.startReceiver(sendToBackendQueue, Job.class, new ReceiveCallback<Job>() {
@Override
public void handleReceive(Job job) throws TransportPluginException {
executorService.start(job, job.getRootId());
}
}, new ErrorCallback() {
@Override
public void handleError(Exception error) {
logger.error("Failed to receive message.", error);
}
});
transportPlugin.startReceiver(sendToBackendControlQueue, EngineControlMessage.class, new ReceiveCallback<EngineControlMessage>() {
@Override
public void handleReceive(EngineControlMessage controlMessage) throws TransportPluginException {
switch (controlMessage.getType()) {
case STOP:
List<String> ids = new ArrayList<>();
ids.add(((EngineControlStopMessage)controlMessage).getId());
executorService.stop(ids, controlMessage.getRootId());
break;
case FREE:
executorService.free(controlMessage.getRootId(), ((EngineControlFreeMessage)controlMessage).getConfig());
break;
default:
break;
}
}
}, new ErrorCallback() {
@Override
public void handleError(Exception error) {
logger.error("Failed to execute control message.", error);
}
});
scheduledHeartbeatService.scheduleAtFixedRate(new Runnable() {
@Override
public void run() {
transportPlugin.send(receiveFromBackendHeartbeatQueue, new HeartbeatInfo(backend.getId(), System.currentTimeMillis()));
}
}, 0, 1, TimeUnit.SECONDS);
}
public void stop() {
scheduledHeartbeatService.shutdown();
}
public void send(Job job) {
transportPlugin.send(receiveFromBackendQueue, job);
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 7,781
|
Home › Guillain-Barre Syndrome (Flu GBS) › Do Flu Vaccines Cause GBS? The Government's Plan to Add GBS to the Table
Do Flu Vaccines Cause GBS? The Government's Plan to Add GBS to the Table
By leahdurantlaw on October 7, 2015 • ( 0 )
GBS vaccine attorney Leah Durant discusses the government's proposal to add GBS to the Vaccine Injury Table and the potential link between flu shots and GBS
The federal government has recently opened a period for public comment to determine whether Guillain Barre Syndrome (GBS) should be added to the Vaccine Injury Table. While it is currently possible for patients who contract GBS after being vaccinated to receive compensation through the Vaccine Injury Compensation Program (VICP), adding GBS to the Vaccine Injury Table could streamline the process significantly.
About Guillain Barre Syndrome (GBS)
Guillain Barre Syndrome is an acute form of paralysis that results from dysfunction of the peripheral nervous system. Approximately 3,000 to 4,000 new cases of GBS are reported every year. While most people who contract GBS are able to fully recover, some patients suffer from permanent disabilities. In some cases, GBS can even be fatal. Sadly, GBS is more prevalent among the elderly, and senior citizens who contract GBS often face less-favorable prognoses.
Chronic inflammatory demyelinating polyneuropathy (CIDP) is a severe form of GBS with symptoms that include weakness, impaired motor function, paralysis, and symmetrical loss of sensory response.
Why Add GBS to the Vaccine Injury Table?
Although it has not yet been scientifically proven, there is evidence to suggest that in some rare cases, GBS can be caused by flu shots. In its proposal to add GBS to the Vaccine Injury Table, the U.S. Health Resources and Services Administration (HRSA) points specifically to research concerning the H1N1 flu vaccine of 2009. The scientists who conducted the study concluded that they could neither "accept [nor] reject a causal relationship between seasonal influenza vaccine and GBS." So, while uncertainty remains, the possibility of a link between the flu vaccine and such a serious complication certainly warrants consideration.
While there are currently multiple types of flu vaccines on the Vaccine Injury Table, the table does not provide coverage for GBS resulting from the administration of seasonal flu vaccines. By adding GBS to the Vaccine Injury Table, all patients who contract GBS within a certain period after being administered the seasonal flu vaccine would be eligible for compensation under the Vaccine Injury Compensation Program. The current proposal to add GBS to the Vaccine Injury Table would put this time period at anywhere between three and 42 days after administration.
What to do if You Have Contracted GBS
If you have recently received a flu shot and are now suffering from GBS or CIPD, you may be entitled to financial compensation. To find out more, you should speak with an experienced vaccine injury lawyer as soon as possible. While you are not legally required to have a lawyer in order to pursue a vaccine claim, one of the many benefits of the program is that hiring a lawyer comes at no financial cost to you. By hiring an attorney, you can make sure that you have the best possible chance at obtaining maximum compensation for your injuries.
To Speak with an Experienced Attorney About Your Vaccine Injury Claim, Contact the Law Offices of Leah Durant Today
At the Law Offices of Leah V. Durant, we strongly support the government's proposal to add GBS to the Vaccine Injury Table. We are passionate about helping victims of vaccine injuries, and if you have contracted GBS or CIPD after receiving a flu shot, we want to help you claim the compensation you deserve. To find out more, please call (202) 800-1711 or contact us online today.
‹ How to File a Lawsuit for Your Injuries
What Vaccines are Available for Hepatitis? What are the Risks? ›
Categories: Guillain-Barre Syndrome (Flu GBS), Health Resources and Services Administration, National Vaccine Injury Compensation Program, vaccine injury
Tags: flu vaccine, Guillain-Barré Syndrome, National Vaccine Injury Compensation Program, vaccine attorney
Tweets by @LeahDurantLaw
CDC "You Call the Shots" Infographic Explains How to Avoid SIRVA
VICP Eligibility: Does Your Vaccine Shoulder Injury Qualify for Compensation?
HPV Vaccination Rates Among Teens are Increasing
It's the Peak of Flu Season. Is it Too Late to Get Vaccinated?
Brachial Neuritis
Chronic Inflammatory Demyelinating Polyneuropathy (CIDP)
Guillain-Barre Syndrome (Flu GBS)
Health Resources and Services Administration
Human Papilloma Virus
National Vaccine Injury Compensation Program
Shoulder Injury Related to Vaccine Administration or SIRVA
U.S. Court of Federal Claims
Vaccine Adverse Event Reporting System
vaccine injury
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 9,000
|
Results of The Fact Music Awards 2020
The awards ceremony The Fact Music Awards 2020 announced the winners of a series of awards and the results are not unpredictable.
On 12/12, The Fact Music Awards 2020 (TMA 2020) officially took place. However, like other events, the ceremony must also be held online, with no audience due to the epidemic. Although it is only the second year, TMA brings together many famous artists such as BTS, TWICE, ITZY, MAMAMOO, Kang Daniel, IZ * ONE, Super Junior, TXT, Stray Kids, SEVENTEEN, …
TMA 2020 award ceremony only awarded 1 Daesang. It's no surprise that BTS is the owner of this award thanks to a series of outstanding achievements such as setting a sales record with the album MAP OF THE SOUL: 7, being at #1 on the Billboard Hot 100 chart with 2 songs Dynamite and Life Goes On nominated at the prestigious Grammy Awards. The 7 boys also won 3 other titles: Artist of the Year, Global Icon, and Listener's Choice.
However, like the 2019 ceremony, TMA 2020 confuses viewers with the Artists Of The Year award. At other awards ceremonies, this title is in the Daesang system, but at TMA the Artist of the Year is equivalent to the Bonsang Award and is awarded to 11 singers and groups.
In addition, instead of calling the award for new artists Rookie of the Year like many other awards ceremonies, TMA 2020 named this title as Next Leader. Weeekly, CRAVITY, and ENHYPEN are the 3 winning groups. ITZY and TXT – 2 representatives who were previously named the Next Leaders gave the awards to their juniors.
Notably, this is the first award for rookie ENHYPEN from Big Hit's survival show even though they debuted for 12 days. Meanwhile, TREASURE – a new boy group from YG did not attend and was also empty-handed at this year's ceremony despite their great achievements. Similarly, their senior group, BLACKPINK, did not win in any category.
Summary of awards at TMA 2020
Bonsang Artist of the Year (Year's Artist): MAMAMOO, Hwasa, Kang Daniel, BTS, GOT7, TWICE, NU'EST, IZ*ONE, MONSTA X, SEVENTEEN, Super Junior
TMA Popularity Award: Super Junior
Next Leader: Weeekly, CRAVITY, ENHYPEN
Global Hottest: Stray Kids, (G)I-DLE, The Boyz, ATEEZ
Best Performer: ITZY, TXT, Jessi
Worldwide Icon: SEVENTEEN, BTS
Fan N Star: Super Junior, Hwang Chi Yeol
Fan N Star Trot Popularity: Lim Young Woong
Listener's Choice: BTS
Daesang: BTS
Source:kenh14
Related Items:BTS, ITZY, mamamoo, The Fact Music Awards 2020, TMA 2020, TWICE
HOT: TWICE will appear at a special event of the famous magazine – TIME
What happened to GOT7 got fans thinking about TWICE: The real reason why TWICE only focus on group activities instead of promoting solo even after 6 years of debut
Artists were treated disrespectfully at MAMA 2020
Fans were surprised to see someone like Soojin ((G) I-DLE) among dancers of aespa
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 381
|
Q: example of chinese remainder theorem The chinese remainder theorem says that if I consider $\mathbb{Z}/{6\mathbb{Z}}$ I can write it as $\mathbb{Z}/{2\mathbb{Z}}\times\mathbb{Z}/{3\mathbb{Z}}$ since $(2,3)=1$. If $\alpha$ goes a system of representatives mod $6$ does $\alpha$ also go a system mod $3$ ?
A: It means that, when $\alpha$ goes through a system of representatives of congruence classes modulo $6$, the pairs $(\alpha\bmod 2,\alpha\bmod 3)$ go through the set of all pairs of (representatives of) congruence classes mod $2$ and mod $3$.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 6,888
|
\section{Introduction}
Throughout this paper, let $R$ denote a commutative Noetherian
ring (with identity) and $I$ an ideal of $R$. For an $R$-module $L$,
the $i^{th}$ local cohomology module of $L$ with respect to $I$ is
defined as$$H^i_I(L) = \underset{n\geq1} {\varinjlim}\,\,
\text{Ext}^i_R(R/I^n, L).$$ We refer the reader to \cite{Gr1} or \cite{BS} for more details about local cohomology.
For any finitely generated $R$-module $M$, the notion $f_I(M)$, the {\it finiteness dimension} of $M$ relative to $I$, is defined
to be the least integer $i$ such that $H^i_I(M)$ is not finitely generated, if there exist such $i$'s and $\infty$ otherwise, i.e.
$$f_I(M):=\inf\{i\in \Bbb{N}_0|\,H^i_I(M)\,\,{\rm is}\,\,{\rm not}\,\,{\rm finitely}\,\,{\rm generated} \}.$$
Our objective in this paper is to investigate the finiteness dimension $f_I(R)$, when $R$ is a local Cohen-Macaulay ring. More precisely, as
a main result we shall show that:
\begin{thm} Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring and $I$ a non-nilpotent ideal of $R$. Then
$f_I(R)={\rm max}\{1,\height I \}.$
\end{thm}
One of our tools for proving Theorem 1.1 is the following, which will play a key role in the proof of that theorem.
\begin{prop}
Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring and let $X$ and $Y$ be non-empty subsets of $\Ass_R(R)$ such that $\Ass_R(R)=X\cup Y$ and $X\cap Y=\emptyset$. Then $R/(I+J)$ is an equidimensional local ring of dimension $\dim R-1$, where
$I=\cap_{\p\in X}\p$ and $J=\cap_{\p \in Y}\p.$
\end{prop}
Recall that a Noetherian ring $R$, of finite Krull dimension $d$, is called {\it equidimensional} if $\dim R/\p=d$ for every minimal
prime ideal $\p$ of $R$. As an another main result, we shall show that:
\begin{thm}
Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring and let $I$ be a non-nilpotent ideal of $R$ such that
$\mAss_R(R/I)\subseteq \Ass_R(R)$. Then $R/ (I+\cup_{n\geq 1}(0:_RI^n))$ is an equidimensional local ring of dimension $\dim R-1$.
\end{thm}
In \cite{HaS}, Hartshorne and Speiser, proved that if $(R, \m, k)$ is a regular local ring, contains a field of characteristic $p>0$, and $H^i_I(R)$
is supported only at the maximal ideal, then $\Hom_R(k, H^i_I(R)$ is a finitely generated $R$-module and, moreover, $H^i_I(R)$ is injective.
Also, Huneke and Sharp in \cite{HS} made a remarkable breakthrough. They generalized Hartshorne-Speiser's result by proving that if $R$ is
any regular ring containing a field of characteristic $p>0$, then ${\rm inj}\dim H_I^i(R)\leq \dim\Supp H_I^i(R),$ where ${\rm inj}\dim H_I^i(R)$ denotes
the injective dimension of $H_I^i(R)$ and $\dim\Supp H_I^i(R)$ stands for the dimension of the support of $H_I^i(R)$ in $\Spec R$. Finally, in
\cite{Ly1}, Lyubeznik generalized the above-mentioned result of Hartshorne-Speiser by proving that if $R$ is any regular ring containing a field of characteristic
zero and $Y\subseteq \Spec R$ is a locally closed subscheme, then ${\rm inj}\dim H_Y^i(R)\leq \dim\Supp H_Y^i(R)$.
As a final main result, we able to obtain a lower bound for the injective dimension of the local cohomology module $H^{\height I}_I(R)$, in the case
$(R, \frak m)$ is a complete equidimensional local ring. More precisely, we show that:
\begin{thm}
Let $(R, \m)$ be a complete local equidimensional ring and $I$ an ideal of $R$. Then
${\rm inj}\dim H_I^{\height I}(R)\geq \dim R-\height I.$ In particular, if $R$ is a regular local ring containing a field, then
${\rm inj}\dim H_I^{\height I}(R)=\dim R-\height I$.
\end{thm}
Finally, we will end the paper with an example, which shows that Theorem 1.4 does not hold in general. \\
For each $R$-module $L$, we denote by
${\rm Ass h}_R(L)$ (resp. ${\rm mAss}_RL$) the set $\{\frak p\in \Ass
_R(L):\, \dim R/\frak p= \dim L\}$ (resp. the set of minimal primes of
$\Ass_RL$). Also, the set of all zerodivisors on $L$ is denoted by $Z_R(L)$. For any ideal $\frak{b}$ of
$R$, {\it the radical of} $\frak{b}$, denoted by $\Rad(\frak{b})$,
is defined to be the set $\{x\in R \,: \, x^n \in \frak{b}$ for some
$n \in \mathbb{N}\}$ and we denote $\{{\frak p} \in {\Spec} (R) :
{\frak p} \supseteq {\frak b}\}$ by $V({\frak b})$. Finally, for any
ideal $\frak{b}$ of $R$, {\it the cohomological dimension} of an
$R$-module $M$, with respect to $\frak{b}$ is defined as
$${\rm cd}(\frak{b}, M):=\sup \{i\in \mathbb{Z} : H^{i}_{\frak{b}}(M)\neq 0 \}.$$
For any unexplained notation and terminology we refer the reader
to \cite{BS} and \cite{Mat}.
\section{The Results}
The following lemmas will be quite useful in the proof of the main results. Following $D:=\Hom_R(\bullet, E_R(R/\m))$ (resp. $\omega_R$) denotes the Matlis duality functor (resp. the canonical module for $R$) (see \cite[3.3]{BH}.
\begin{lem}
\label{2.1}
Let $(R,\mathfrak{m})$ be a local Noetherian ring and $M$ a finitely
generated $R$-module. Let $\frak p$ be a prime ideal of $R$ such
that $\dim R/{\frak p}=1$ and let $t\geq1$ be an integer. Then
$H^{t}_{\frak m}(M)$ is ${\frak p}$-cofinite if and only if $(H^{t-1}_{\frak p}(M))_{\frak p}=0$.
\end{lem}
\proof See \cite[Lemma 2.1]{BAB}. \qed\\
\begin{lem}
Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring of dimension $d$. Then the $R$-module $H^d_{\m}(R)$ is indecomposable.
\end{lem}
\proof Without loss of generality, we may assume that $R$ is a complete Cohen-Macaulay local ring. Now, we suppose the contrary and we look for a contradiction. Let $H^d_{\m}(R)=A\oplus B$, where $A$ and $B$ are two non-zero Artinian $R$-modules. Then we have $\omega_R\cong D(A)\oplus D(B)$, where $\omega_R$ denotes the canonical module of $R$. So as the $R$-module $\omega_R$ is indecomposable, it follows that $D(A)=0$ or $D(B)=0$. Hence $A=0$ or $B=0$, which is a contradiction. \qed\\
The following result will be useful in the proof of the main results in this section.
\begin{thm}
Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring and let $X$ and $Y$ be non-empty subsets of $\Ass_R(R)$ such that $\Ass_R(R)=X\cup Y$ and $X\cap Y=\emptyset$. Set $$I:=\cap_{\p\in X}\p\,\,\,\,\,{\rm and}\,\,\,\,\,J:=\cap_{\q \in Y}\q.$$ Then $R/(I+J)$ is an equidimensional local ring of dimension $\dim R-1$.
\end{thm}
\proof It follows from the hypothesis $X\cap Y=\emptyset$ that $\height(I+J)\geq 1$. Now, we show that $\height(I+J)= 1$. To do this, suppose the contrary is true. Then
there exists a minimal prime ideal $\frak p$ over $I+J$ such that $\height \frak p :=n> 1$.
Since $\Ass_R(R)=X\cup Y$ it follows that $I\cap J={\rm nil}(R)$, and so $I\cap J$ is a nilpotent ideal of $R$. Therefore
$$H^{n-1}_{I\cap J}(R)=0 = H^n_{I\cap J}(R).$$
Now, in view of the Mayer-Vietoris sequence (see e.g., \cite[Theorem 3.2.3]{BS}) we obtain the isomorphism $$H^n_{I+ J}(R) \cong H^n_I(R) \oplus H^n_J(R).$$
Therefore
$$H^n_{\frak pR_{\frak p}}(R_{\frak p})= H^n_{(I+J)R_{\frak p}}(R_{\frak p})\cong H^n_{IR_{\frak p}}(R_{\frak p}) \oplus H^n_{JR_{\frak p}}(R_{\frak p}).$$
Now, using Lemma 2.2, we deduce that
$$H^n_{{\frak p}R_{\frak p}}(R_{\frak p})\cong H^n_{IR_{\frak p}}(R_{\frak p})\,\,\,\,\,{\rm or}\,\,\,\,\, H^n_{{\frak p}R_{\frak p}}(R_{\frak p})\cong H^n_{JR_{\frak p}}(R_{\frak p}). $$
Consequently, in view of \cite[Proposition 5.1]{Me}, $H^n_{{\frak p}R_{\frak p}}(R_{\frak p})$ is an $IR_{\frak p}$ or
$JR_{\frak p}$-cofinite $R_{\frak p}$-module. Next, as $\height \frak p > 1$, it is easy to see that there exists a prime ideal $\frak q \in V(I)$ or
$\frak q \in V(J)$ such that $\frak q \subseteq \frak p$ and $\height \frak p /\frak q= 1$. Now, using \cite[Proposition 4.1]{Me}, one easily sees that the
$R_{\frak p}$-module $H^n_{{\frak p}R_{\frak p}}(R_{\frak p})$ is ${\frak q}R_{\frak p}$-cofinite. Therefore, it follows from Lemma 2.1 that $H^{n-1}_{{\frak q}R_{\frak q}}(R_{\frak q})=0$. On the other hand, as $R$ is catenary, it follows that
$\height \frak p /\frak q= \height \p - \height \frak q$, and so $$\height \frak q= \height \frak p -1= n-1.$$ Hence in view of Grothendieck's non-vanishing theorem we have $H^{n-1}_{{\frak q}R_{\frak q}}(R_{\frak q})\neq 0$, which is a contradiction. Therefore $\height \frak p = 1$, and so
$\height(I+J)= 1$. Now, as $R$ is Cohen-Macaulay, it follows easily that $R/(I+J)$ is an equidimensional ring of dimension $\dim R-1$, as required. \qed\\
\begin{cor} Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring and let $x_1,\dots, x_t$ be an $R$-regular sequence. Let $X$ and $Y$
be non-empty subsets of $\Ass_R(R/(x_1,\dots, x_t))$ such that $\Ass_R(R/(x_1,\dots, x_t))=X\cup Y$ and $X\cap Y=\emptyset$. Set
$$I:=\cap_{\p\in X}\p\,\,\,\,\,{\rm and}\,\,\,\,\,J:=\cap_{\q \in Y}\q.$$Then $R/(I+J)$ is an equidimensional local ring of dimension $\dim R-t-1$.
\end{cor}
\proof
Since $R/(x_1,\dots, x_t)$ is a Cohen-Macaulay local ring, the assertion follows easily from Theorem 2.3.\qed\\
\begin{lem}
Let $R$ be a Noetherian ring and $I$ an ideal of $R$ such that ${\rm cd}(I, R)=n>0$. Then the $R$-module $H^n_I(R)$ is not finitely generated.
\end{lem}
\proof Since by the definition we have $H^n_I(R)\neq 0$, it follows that $\Supp H^n_I(R)\neq \emptyset$. Let $\p\in \Supp H^n_I(R)$. Then it is easy to see that ${\rm cd}(IR_{\p},R_{\p})=n>0$. So replacing of the ring $R$ with the local ring $(R_{\p},\p R_{\p})$, we may assume that $(R,\m)$ is a Noetherian local ring and $I$ is an ideal of $R$ such that ${\rm cd}(I,R)=n>0$. Then using \cite[Exercise 6.1.8]{BS} and Grothendieck's vanishing theorem we have: $$H^n_I(R)/\m H^n_I(R)\cong H^n_I(R)\otimes_R R/\m\cong H^n_I(R/\m)=0.$$Therefore, $H^n_I(R)=\m H^n_I(R)$ and hence using Nakayama's lemma we can deduce that the $R$-module $H^n_I(R)$ is not finitely generated.\qed\\
We are now in a position to state and prove the first main result of this paper, which investigates the finiteness
dimension $f_I(R)$ over a Cohen-Macaulay local ring.
\begin{thm} Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring and $I$ a non-nilpotent ideal of $R$. Then
$$f_I(R)={\rm max}\{1,\height I \}.$$
\end{thm}
\proof To prove there are two cases to consider:\\
{\bf Case 1.} Suppose that $\height I=0$. Put
\begin{center}
$X:=\Ass_R(R)\cap V(I)$ and $Y:=\Ass_R(R)\backslash V(I)$.
\end{center} Let
$J:= \cap_{\p\in X}\p$ and $K:=\cap_{\q\in Y}\q$. Since $I$ is not nilpotent it follows that $Y\neq \emptyset$. Also, as $\height I=0$, it follows that $X \neq \emptyset$. Moreover, it is easy to see that $\Ass_R(R)=X\cup Y$. Hence, in view of the proof of Theorem 2.3, we have $\height (J+K)=1$. Therefore, there exists a minimal prime ideal $\frak p$ over $J+K$ such that $\height \frak p=1$. Since $K\subseteq \p$, there exists an ideal $\q \in Y$ such that $\q \subseteq \p$. As $I\subseteq J \subseteq \p$, it follows that $I+\q \subseteq \p$. Moreover, as $I \not\subseteq \q$ it follows that $\height(I+\q)>0$. Therefore, $\height(I+\q)=\height \p=1$. Thus, $\p$ is a minimal prime ideal over $I+\q$ and so $IR_{\p}+\q R_{\p}$ is a $\frak p R_{\frak p}$-primary ideal. Hence, by Grothendieck's non-vanishing theorem we have $H^1_{IR_{\p}}(R_{\p}/\q R_{\p})\neq 0$. Consequently, it follows from Grothendieck's vanishing theorem that ${\rm cd}(IR_{\p}, R_{\p}/\q R_{\p})=1$. Now, as $\Supp(R_{\p}/\q R_{\p})\subseteq \Spec R_{\p}$, it follows from \cite[Theorem 2.2]{DNT} that
$${\rm cd}(IR_{\p}, R_{\p})\geq {\rm cd}(IR_{\p}, R_{\p}/\q R_{\p})=1.$$ By using Grothendieck's vanishing theorem we can deduce that
${\rm cd}(IR_{\p}, R_{\p})=1$ and so by Lemma 2.5 the $R_{\p}$-module $H^1_{IR_{\p}}(R_{\p})\cong (H^1_I(R))_{\p}$ is not finitely generated. In particular, the $R$-module $H^1_I(R)$ is not finitely generated. Now, as the $R$-module $H^0_I(R)$ is finitely generated, it follows that $$f_I(R)=1={\rm max}\{1,0\}={\rm max}\{1,\height I\},$$ as required.
{\bf Case 2.} Now suppose that $\height I=n\geq 1$. Then we have $\grade I=n$ and so in view of \cite[Theorem 6.2.7]{BS}, $f_I(R)\geq n$. Moreover, by the definition there exists a minimal prime ideal $\frak q$ over $I$ such that $\height \q=n$. Hence, in view of Grothendieck's vanishing and non-vanishing theorems we have $${\rm cd}(IR_{\q},R_{\q})={\rm cd}({\q}R_{\q},R_{\q})=n.$$ Thus, by Lemma 2.5, the $R_{\q}$-module $H^n_{IR_{\q}}(R_{\q})\cong (H^n_I(R))_{\q}$ is not finitely generated. In particular, the $R$-module $H^n_I(R)$ is not finitely generated. Hence in view of the definition we have $$f_I(R)=n={\rm max}\{1,n\}={\rm max}\{1,\height I\},$$ and this completes the proof.\qed\\
The next theorem is the second main result of this paper.
\begin{thm}
Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring and let $I$ be a non-nilpotent ideal of $R$ such that
$\mAss_R(R/I)\subseteq \Ass_R(R)$. Then $R/(I+\Gamma_I(R))$ is an equidimensional local ring of dimension $\dim R-1$.
\end{thm}
\proof
Since $I$ is not nilpotent, it is clear that $\Gamma_I(R)\subseteq Z_R(R)$ and so it follows from \cite[Theorem 17.4]{Mat} that $\dim R/\Gamma_I(R)=\dim R$. Moreover, as $$\Ass_R(R/\Gamma_I(R))=\Ass_R(R) \setminus V(I),$$ it follows that $I$ contains an $R/\Gamma_I(R)$-regular element $x$, and so
$$\dim R/(xR+\Gamma_I(R))=\dim R/\Gamma_I(R)-1.$$
Hence $\dim R/(I+\Gamma_I(R))\leq d-1$.
Next, in view of the Artin-Rees lemma there exists a positive integer $s$ such that $I^s\cap \Gamma_I(R)=0$ and so
$$H^{n-1}_{I^s\cap \Gamma_I(R)}(R)=0 = H^n_{I^s\cap \Gamma_I(R)}(R).$$
Hence, the Mayer-Vietoris sequence (see e.g., \cite[Theorem 3.2.3]{BS}) yields the isomorphism
$$H^n_{I+ \Gamma_I(R)}(R)=H^n_{I^s+ \Gamma_I(R)}(R) \cong H^n_{I^s}(R) \oplus H^n_{\Gamma_I(R)}(R) \cong H^n_{I}(R) \oplus H^n_{\Gamma_I(R)}(R).$$
Now, suppose that $\p$ is a minimal prime ideal over $I+\Gamma_I(R)$ such that $\height \p=n> 1$.
Then, as $\p$ is minimal over $I+\Gamma_I(R)$ we get the following isomorphism
$$H^n_{\p R_{\p}}(R_{\p})\cong H^n_{IR_{\p}}(R_{\p}) \oplus H^n_{\Gamma_{IR_{\p}}}(R_{\p}).$$
Now, using Lemma 2.2, we deduce that
$$H^n_{\p R_{\p}}(R_{\p})\cong H^n_{IR_{\p}}(R_{\p})\,\,\,\,\,{\rm or}\,\,\,\,\,H^n_{\p R_{\p}}(R_{\p})\cong H^n_{\Gamma_{IR_{\p}}(R_{\p})}(R_{\p}).$$
Assume that $H^n_{{\p}R_{\p}}(R_{\p})\cong H^n_{IR_{\p}}(R_{\p})$. Then, in view of \cite[Proposition 5.1]{Me}, $H^n_{{\frak p}R_{\frak p}}(R_{\frak p})$ is an $IR_{\frak p}$-cofinite $R_{\frak p}$-module. Next, as $\height \frak p > 1$ and $\mAss_R(R/I)\subseteq \Ass_R(R)$, it is easy to see that there exists a prime ideal $\frak q \in V(I)$ such that $\frak q \subseteq \frak p$ and $\height \frak p /\frak q= 1$. Now, using \cite[Proposition 4.1]{Me}, it follows easily that the $R_{\frak p}$-module $H^n_{{\frak p}R_{\frak p}}(R_{\frak p})$ is ${\frak q}R_{\frak p}$-cofinite. Therefore, it follows from Lemma 2.1 that $H^{n-1}_{{\frak q}R_{\frak q}}(R_{\frak q})=0$. On the other hand, as $R$ is catenary, it follows that
$\height \frak p /\frak q= \height \frak p - \height \frak q$, and so $$\height \frak q= \height \frak p -1= n-1.$$ Hence in view of Grothendieck's non-vanishing theorem we have $H^{n-1}_{{\frak q}R_{\frak q}}(R_{\frak q})\neq 0$, which is a contradiction.
Now, assume that $H^n_{\p R_{\p}}(R_{\p})\cong H^n_{\Gamma_{IR_{\p}}(R_{\p})}(R_{\p})$. Then, again using the fact that
$$\Ass_R(R/\Gamma_I(R))=\Ass_R(R)\backslash V(I)\subseteq \Ass_R(R),$$
and repeating the above argument we derive a contradiction. Therefore $\height \frak p = 1$, and so
$\height(I+\Gamma_I(R))= 1$.
Now, as $R$ is Cohen-Macaulay, it follows easily that $R/(I+\Gamma_I(R))$ is an equidimensional local ring of dimension $\dim R-1$, as required. \qed\\
\begin{cor}
Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring and let $A$ be a non-empty proper subset of $\Ass_R(R)$.Then $R/(I+\Gamma_I(R))$
is an equidimensional local ring of dimension $\dim R-1$, where $I=\cap_{\p\in A}\p$.
\end{cor}
\proof The assertion follows easily from Theorem 2.7.\qed\\
\begin{prop}
Let $(R,\m)$ be a Cohen-Macaulay local ring and let $\Ass_R(R)=\{\p_1,\dots,\p_n\}$, $n\geq 2$. Let $A_j=\Ass_R(R)\backslash \{\p_j\}$ and $I_j=\cap_{\p\in A_j}\p$, for all $1\leq j \leq n$. Then $0=\cap_{j=1}^n\Gamma_{I_j}(R)$ is the unique reduced primary decomposition of the zero ideal $0$ in $R$, $\Gamma_{I_j}(R)$ is a $\frak p_j$-primary ideal of $R$ and $R/(I_j+\Gamma_{I_j}(R))$ is an equidimensional local ring of dimension $\dim R-1$.
\end{prop}
\proof
As $$\Ass_R(R/\Gamma_{I_j}(R))=\Ass_R(R)\backslash V(I_j)=\{\p_j\},$$ it follows that $\Gamma_{I_j}(R)$ is a $\p_j$-primary ideal of $R$. Now,
we show that $\cap_{j=1}^n\Gamma_{I_j}(R)=0$. To this end, we assume that $\cap_{j=1}^n\Gamma_{I_j}(R)\neq 0$ and derive a contradiction.
Let $a\in \cap_{j=1}^n\Gamma_{I_j}(R)$ be such that $a\neq 0$. Then $(0:_Ra)\subseteq Z_R(R)$, and so there exists $\p_j\in \Ass_R(R)$ such that
$(0:_Ra) \subseteq \p_j$. Next, as $a\in\Gamma_{I_j}(R)$ it follows that there exists a positive integer $k$ such that
$I_j^k\subseteq (0:_Ra)\subseteq \p_j$, and so $I_j\subseteq \p_j$. Therefore, there exists $\p_i\in A_j$ such that $\p_i\subsetneqq \p_j$,
which is a contradiction, (note that $\Ass_R(R)= \mAss_R(R)$). Now, using \cite[Theorem 6.8]{Mat} we see that $\p_j$-primary component
$\Gamma_{I_j}(R)$ of the zero ideal $0$ of $R$ is uniquely determined. That is,
$0=\cap_{j=1}^n\Gamma_{I_j}(R)$ is the unique reduced primary decomposition of the zero ideal $0$ in $R$. Moreover, it follows from Corollary 2.8 that the ring $R/(I_j+\Gamma_{I_j}(R))$ is equidimensional local of dimension $\dim R-1$. \qed\\
The following lemma is needed in the proof of Theorem 2.11.
\begin{lem}
Let $(R, \m)$ be a local ring and $M$ an arbitrary $R$-module. Let $x$ be an element of $\m$ such that
$x\not\in \bigcup_{\p \in \Ass_R (M)\backslash V(\m)} \p$. Then $\Gamma_{Rx}(M)=\Gamma_{\m}(M)$.
\end{lem}
\proof As $x\in \m$, it is enough to show that $\Gamma_{Rx}(M)\subseteq \Gamma_{\m}(M)$. To do this, let $w\in \Gamma_{Rx}(M)$.
Then $x\in \Rad(0:_Rw)$. Since $\mAss_RR/(0:_Rw)\subseteq \Ass_R(M)$, it follows from the assumption
$x\not\in \bigcup_{\p \in \Ass_R (M)\backslash V(\m)} \p$
that $\Rad(0:_Rw)=\m$, and so there exists $n\in \mathbb{N}$ such that $\m^nw=0$. Thus $w\in \Gamma_{\m}(M)$, as required. \qed\\
The following theorem is in preparation for the third main result of this paper, which gives us a lower bound of injective dimension of $ H_I^{\height I}(R)$.
Here $D_I(R)$
denotes the ideal transform of $R$ with respect to $I$ (see \cite[2.2.1]{BS}).
\begin{thm}
Let $(R, \m)$ be a complete local equidimensional ring of dimension $d$ and $I$ an ideal of $R$ such that $\height I=t$. Then
$H_{\m}^{d-t}(H_I^t(R))\neq0$.
In particular, $${\rm inj}\dim H_I^t(R)\geq d-t.$$
\end{thm}
\proof As $R$ is catenary, it follows from \cite[ Lemma 2, P. 250]{Mat} that
$$\height J+ \dim R/J=\dim R,$$
for every ideal $J$ of $R$. In particular, we have $\dim R/I=d-t$. We now use induction on $d-t$. When $d=t$, the ring $R/I$ is Artinian and so $\Rad(I)=\m$.
Hence $H_I^t(R)=H_{\m}^t(R)$ and so as $H^0_{\m}(H_I^t(R))= H_{\m}^t(R)$, the assertion follows from Grothendieck's non-vanishing theorem (see \cite[Theorem 6.1.4]{BS}) in this case.
Assume, inductively, that $d-t>0$ and that the result has been proved for the ideals $J$ with $\dim R/J=0,1, \dots, d-t-1$. Since the sets $\Ass_R(H^t_I(R))$ and
$\Ass_R (H^{t+1}_I(R))$ are countable, it follows from \cite[Lemma 3.2]{MV} that
$$\m\nsubseteq (\bigcup_{\p \in \Ass_R(H^t_I(R))\backslash V(\m)} \p)\, \bigcup\, (\bigcup_{\p \in \Ass_R(H^{t+1}_I(R))\backslash V(\m)} \p)\,\bigcup\, (\bigcup_{\p \in {\rm Assh}_R(R/I)} \p).$$
Whence, there exists $x\in \frak m$ such that
$$x\not\in (\bigcup_{\p \in \Ass_R(H^t_I(R))\backslash V(\m)} \p)\, \bigcup\, (\bigcup_{\p \in \Ass_R(H^{t+1}_I(R))\backslash V(\m)} \p)\,\bigcup\, (\bigcup_{\p \in {\rm Assh}_R(R/I)} \p).$$
Then it follows easily from $x\not\in \bigcup_{\p \in {\rm Assh}_R(R/I)} \p$ that $$\dim R/(I+Rx)=d-t-1,$$
and in view of Lemma 2.10 we have
\begin{center}
$\Gamma_{Rx}(H_I^t(R))=\Gamma_{\m}(H_I^t(R))$ \,\,\,\,\, and \,\,\,\,\, $\Gamma_{Rx}(H_I^{t+1}(R))=\Gamma_{\m}(H_I^{t+1}(R))$.
\end{center}
Moreover, there is an exact sequence
$$0\longrightarrow H^{1}_{Rx}(H^{t}_{I}(R))\longrightarrow H^{t+1}_{I+Rx}(R) \longrightarrow H^{0}_{Rx}(H^{t+1}_{I}(R))\longrightarrow 0, \hspace{13mm}(\dag)$$
(see \cite[Corollary 3.5]{Sc}).
Now, if $\dim R/I=1$ then in view of \cite[Theorem 2.6]{BN1} the $R$-module $H^{t}_{I}(R)=H^{d-1}_{I}(R)$ is $I$-cofinite. Next, it is easy
to see that $\dim\Supp H_I^{d-1}(R)=1$, note that $\dim R/I=1$. Hence, it follows from \cite[Theorem 2.9]{Ma} that $H^1_{\m}(H_I^{d-1}(R))\neq 0$,
and so the result has been proved in this case. Therefore, we assume that $\dim R/I\geq2$. Then $$\dim R/(I+Rx)=d-t-1\geq1,$$ and so in view of Grothendieck's vanishing theorem $$H_{\m}^{d-t-1}(\Gamma_{\m}(H_I^{t+1}(R)))=0.$$
Hence by using the exact sequence $(\dag)$ we obtain the following exact sequence
$$H_{\m}^{d-t-1}(H^{1}_{Rx}(H^{t}_{I}(R)))\longrightarrow H_{\m}^{d-t-1}(H^{t+1}_{I+Rx}(R) )\longrightarrow 0.$$
Thus by the inductive hypothesis $H_{\m}^{d-t-1}(H^{1}_{Rx}(H^{t}_{I}(R)))\neq0$.
On the other hand, since $d-t>0$, it yields that
$$H_{\m}^{d-t}(H^{t}_{I}(R))\cong H_{\m}^{d-t}(H^{t}_{I}(R)/\Gamma_{\m}(H^{t}_{I}(R))).$$
Now, let $T:=H^{t}_{I}(R)/\Gamma_{\m}(H^{t}_{I}(R))$. It is thus sufficient for us to show that $H^{d-t}_{\m}(T)\neq0$.
To do this, in view of \cite[Remark 2.2.17]{BS}, there is the exact sequence
$$0 \longrightarrow T \longrightarrow D_{Rx}(T) \longrightarrow H^{1}_{Rx}(T)\longrightarrow 0. \hspace{13mm}(\dag\dag)$$
Also, in view of \cite[Theorem 2.2.16]{BS}, we have $ D_{Rx}(T)\cong T_x$, and so $$ D_{Rx}(T)\stackrel{x}\longrightarrow D_{Rx}(T),$$
is an $R$-isomorphism. Therefore, for all $i\geq0$, $$ H_{\m}^i(D_{Rx}(T))\stackrel{x}\longrightarrow H_{\m}^i(D_{Rx}(T)),$$
is an $R$-isomorphism, and hence $ H_{\m}^i(D_{Rx}(T))=0$, for all $i\geq0$. Consequently, it follows from the exact sequence $(\dag\dag)$ that
$$H^{d-t}_{\m}(T)\cong H^{d-t-1}_{\m}(H^1_{Rx}(T)).$$
As $H^{d-t-1}_{\m}(H^1_{Rx}(T))\neq0$, this completes the inductive step. \qed\\
\begin{cor} Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring of dimension $d$ and $I$ an ideal of $R$ such that $\height I=t$. Then
$H_{\m}^{d-t}(H_I^t(R))\neq0$. In particular, ${\rm inj}\dim H_I^t(R)\geq d-t$.
\end{cor}
\proof Let $\hat{R}$ denote the completion of $R$ with respect to the $\frak m$-adic topology. Then, as $(\hat{R},\mathfrak{m}\hat{R})$ is a complete local equidimensional ring of dimension $d$, the assertion follows from Theorem 2.11, the faithfully flatness of the homomorphism $R\longrightarrow \hat{R}$ and the fact that $$\height I=\grade I= {\rm grade}\, I \hat{R}=\height I\hat{R}.$$ \qed
\begin{lem}
Let $(R, \m)$ be a regular local ring containing a field and $I$ an ideal of $R$. Then, for any integer $n$ with $H^n_I(R)\neq0$,
$${\rm inj}\dim H_I^n(R)\leq \dim\Supp H_I^n(R).$$
\end{lem}
\proof The result follows from \cite{HS} and \cite{Ly1}. \qed\\
\begin{cor}
Let $(R, \m)$ be a regular local ring containing a field and $I$ an ideal of $R$ such that $\height I=t$. Then
$${\rm inj}\dim H_I^t(R)=\dim R-t.$$
\end{cor}
\proof In view of Corollary 2.12 and Lemma 2.13, it is enough to show that $$\dim\Supp H_I^t(R)=\dim R-t.$$
To this end, as $\Supp H_I^t(R)\subseteq V(I)$ and $\dim R/I=\dim R-t$, we have $$\dim\Supp H_I^t(R)\leq\dim R-t.$$
On the other hand, since $\height I=t$ there exists a minimal prime $\p$ over $I$ such that $\height \p=t$. Now, in view of
\cite[Theorems 4.3.2 and 6.1.4]{BS} we deduce that
$$(H_I^t(R))_{\p}\cong H_{IR_{\p}}^t(R_{\p})\cong H_{\p R_{\p}}^t(R_{\p})\neq0.$$
Thus $\p \in \Supp H_I^t(R)$, and so as $\dim R/\p=\dim R-t$, it follows that $$\dim\Supp H_I^t(R)\geq\dim R-t.$$
This completes the proof. \qed\\
We end the paper with the following example, which shows that Corollary 2.14 does not hold in general.
\begin{exam}
Let $(R, \m)$ be a regular local ring containing a field with $\dim R=d\geq 3$, $\p$ a prime ideal of $R$ such that $\dim R/\p=1$ and
$x\in \m\backslash \p$. Then ${\rm inj}\dim H^{\dim R-1}_{Rx \cap \frak p }(R)=0,$ and $\dim\Supp H^{\dim R-1}_{Rx \cap \p }(R)=1.$
\end{exam}
\proof Since $\Rad(\p+Rx)=\m$, it follows from the Mayer-Vietoris sequence (see e.g., \cite[Theorem 3.2.3]{BS}) that
$$0\longrightarrow H^{d-1}_{\p}(R)\longrightarrow H^{d-1}_{x\p}(R) \longrightarrow H^d_{\m}(R)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\dag\dag\dag)$$
is an exact sequence. Since, in view of the proof of Corollary 2.14, $\dim\Supp H^{d-1}_{\p}(R)=1$ and $H^d_{\m}(R)$ is Artinian, it follows that $\dim\Supp H^{d-1}_{x\p}(R)=1$.
On the other hand, the exact sequence
$$0\longrightarrow R \stackrel{x} \longrightarrow R \longrightarrow R/xR \longrightarrow 0$$
induces the exact sequence
$$ H^{d-2}_{x\p}(R/xR) \longrightarrow H^{d-1}_{x\p}(R) \stackrel{x} \longrightarrow H^{d-1}_{x\p}(R) \longrightarrow H^{d-1}_{x\p}(R/xR).$$
Since $\Gamma_{ x\p}(R/xR)=R/xR$ and $d\geq3$, it follows that
$$H^{d-2}_{ x\p}(R/xR)=0=H^{d-1}_{ x\p}(R/xR).$$
Therefore, the $R$-homomorphism $$H^{d-1}_{x\p}(R)\stackrel{x} \longrightarrow H^{d-1}_{x\p}(R)$$
is an isomorphism, and so $(H^{d-1}_{x\p}(R))_x\cong H^{d-1}_{x\p}(R)$.
On the other hand, from the exact sequence $ (\dag\dag\dag)$, we have $$(H^{d-1}_{x\p}(R))_x\cong (H^{d-1}_{\p}(R))_x.$$
Moreover, the exact sequence $$0 \longrightarrow H^{d-1}_{\p }(R) \longrightarrow E_R(R/\p) \longrightarrow E_R(R/\m), $$
implies that $$(H^{d-1}_{\p}(R))_x \cong (E_R(R/\p))_x \cong E_R(R/\p).$$ Therefore $H^{d-1}_{x\p}(R) \cong E_R(R/\p)$, and so
${\rm inj}\dim H^{d-1}_{x\p}(R)=0$, as required. \qed\\
\begin{center}
{\bf Acknowledgments}
\end{center}
The authors are deeply grateful to the referee for his/her careful reading and helpful suggestions on
the paper. We also would like to thank Professor Hossein Zakeri for his reading of the first draft and valuable discussions.
Finally, the authors would like to thank from the Institute for Research in Fundamental Sciences (IPM) for the financial support.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 6,125
|
Джованни Конти (Giovanni Conti) — католический церковный деятель XII века. На консистории 1150 года был провозглашен кардиналом-священником с титулом церкви Санти-Джованни-э-Паоло. Участвовал в выборах папы 1153 (Анастасий IV), 1154 (Адриан IV), 1159 (Александр III) и 1181 (Луций III) годов. В качестве папского легата посылался на переговоры с королём Амори I. По возвращении с востока, в апреле 1178 года стал римским викарием.
Примечания
Литература
Brixius, Johannes Matthias. Die Mitglieder des Kardinalkollegiums von 1130-1181. Berlin : R. Trenkel, 1912, p. 55, no. 13
col. 1046
"Essai de liste générale des cardinaux. Les cardinaux du XIIè siècle". Annuaire Pontifical Catholique 1928. Paris : Maison de la Bonne Presse, 1928, p. 138
Кардиналы Италии
Кардиналы Евгения III
Кардиналы-протопресвитеры
Церковные деятели XII века
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 5,688
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.