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# Apparent solubility of Ag2C2O4 in a buffer solution

We take the following quantities and mix them together:

• $$\pu{20 mL}$$ solution of $$\ce{AgNO3}$$ $$\pu{0.001 mol//dm3}$$
• $$\pu{20 mL}$$ solution of $$\ce{H2C2O4}$$ $$\pu{0.200 mol//dm3}$$
• $$\pu{40 mL}$$ buffer solution $$\mathrm{pH} = 5.00$$

What is the apparent solubility of the insoluble salt formed, knowing: $$K_\mathrm{a1}=6.50\cdot 10^{-2}$$, $$K_\mathrm{a2}=6.10\cdot 10^{-5}$$ (for $$\ce{H2C2O4}$$) and $$K_\mathrm{s}=1.10\cdot 10^{-12}$$ (for $$\ce{Ag2C2O4}$$)?

The initial concentrations for $$\ce{Ag+}$$ and $$\ce{H2C2O4}$$ are: \begin{align} [\ce{Ag+}]&=\pu{2.50E-4 mol//dm3}\\ [\ce{H2C2O4}]&=\pu{5.00E-2 mol//dm3} \text{(excess)} \end{align}

We're dealing with the following equilibria: \begin{align} \ce{Ag2C2O4 &<=> 2Ag+ + C2O4^{2-}}\\ \ce{H2C2O4 &<=> HC2O4- + H+}\\ \ce{HC2O4- &<=> C2O4^{2-} + H+} \end{align}

The pH of the solution is above $$\mathrm{p}K_\mathrm{a2}$$, therefore $$\ce{H2C2O4}$$ is found mostly as $$\ce{C2O4^{2-}}$$. By using the Henderson–Hasselbalch equation, we find that $$[\ce{C2O4^{2-}}]=\pu{4.30E-2 mol//dm3}$$.

Since we started with an excess of acid, we can consider that the variation of concentration for that particular species (after precipitation completes) is negligible. By using $$K_\mathrm{s}$$, we find that:

$$2S=[\ce{Ag+}]=\pu{5.06E-6 mol//dm3}$$

Is my approach correct? I'm confused because of their approach, where they invoke the following equation:

$$S=\frac{1}{2}\ce{[Ag+]}=[\ce{H2C2O4}]+[\ce{HC2O4-}]+[\ce{C2O4^{2-}}]$$

Which doesn't really make sense since we started with an excess of $$\ce{H2C2O4}$$.

In my opinion, this equation would make sense if we made a solution only with the buffer and an arbitrary quantity of insoluble salt.

• Your work looks good to me. (I just followed your logic and didn't check the math.) I also don't understand the equation $$S=\frac{1}{2}\ce{[Ag+]}=[\ce{H2C2O4}]+[\ce{HC2O4-}]+[\ce{C2O4^{2-}}]$$ – MaxW May 1 '16 at 15:45
• @MaxW Thanks for checking out! If we consider a certain case, the equation can be used to work out the apparent solubility: in a flask containing a buffer solution of arbitrary $pH$, we add an arbitrary quantity of $\ce{Ag2C2O4}$. We get the equilibria as seen in the post and that relation represents a simple conservation law. The equation becomes: $$\frac{1}{2}[\ce{Ag+}]=\frac{1}{2}\sqrt{\frac{K_{s}}{[\ce{C2O4^{2-}}]}}=\frac{[\ce{C2O4^{2-}}][\ce{H+}]^2}{K_{a1}K_{a2}}+\frac{[\ce{C2O4^{2-}}][\ce{H+}]}{K_{a2}}+[\ce{C2O4^{2-}}]$$ Which is solvable and gives us the apparent solubility. :D – L3ul May 1 '16 at 19:44
• Ah.. yes since there are two silver ions, the molar solubility of the salt would be half the silver ion concentration. So $$S(\ce{Ag2C2O4})=2.53\cdot 10^{-6} \frac{\mathrm{moles}}{\mathrm{dm^3}}$$ – MaxW May 1 '16 at 19:52
• Oups, my solution forgot to take that into account. Corrected. And as a matter of fact, considering their approach which applies to the case I presented, we get: $$S=6.84\cdot10^{-5} \frac{\mathrm{moles}}{\mathrm{dm^3}}$$ Close, but not quite. – L3ul May 1 '16 at 20:01


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How to draw knots with Latex? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T06:29:27Z http://mathoverflow.net/feeds/question/21424 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21424/how-to-draw-knots-with-latex How to draw knots with Latex? Hailong Dao 2010-04-15T06:33:05Z 2013-04-22T19:47:11Z
I am writing an exam for my students, and the topic is intro knots theory. I have no idea how to put knots into the file, but I know many MO users who can draw amazing diagrams in their papers.

Can someone please provide some hints on what can be used, preferably with some example codes? I do not need complicated diagrams, just some simple knots and links with few crossings. Thanks in advance.
 http://mathoverflow.net/questions/21424/how-to-draw-knots-with-latex/21428#21428 Answer by Sam Nead for How to draw knots with Latex? Sam Nead 2010-04-15T07:41:50Z 2011-02-22T13:42:56Z
EDIT - Following the suggestions of several people I have, somewhat painfully, switched from what is described below to using just pdflatex and pinlabel. I still use xfig for figures, exporting them to both eps and pdf to make pinlabel happy.

Old answer -

I draw knots in a different program (xfig), save them as eps files, and then include them via

\begin{figure}[htbp] $$\begin{array}{c} \epsfig{file=Figures/foo.eps, width = 10 cm} \\ \epsfig{file=Figures/bar.eps, width = 10 cm} \end{array}$$ \caption{Top: A picture of a foo. Bottom: a bar.} \label{Fig:FooBar} \end{figure}

Then latex, dvips -o, ps2pdf produces the pdf file. This technique allows one to use psfrag to add labels inside of figures.
 http://mathoverflow.net/questions/21424/how-to-draw-knots-with-latex/21430#21430 Answer by Niel de Beaudrap for How to draw knots with Latex? Niel de Beaudrap 2010-04-15T07:56:49Z 2011-02-22T19:54:41Z

I would seriously consider the photographic technology suggested by Kevin Buzzard in the comments, if you happen to have a decent digital camera handy (and if you trust your drawing skills). Or, as Peter Shor suggests in the comments, you can use a scanner if you have one handy.† Multiple 2½-dimension diagrams, without tons of regularity, are not fun to draw on a computer --- unless perhaps you have access to touch-screens.

If you want special effects or other benefits of computer-aided graphics badly enough, you can do it directly in LaTeX with TiKZ. I recommend their fairly comprehensive and reasonably readable manual. For sufficiently complicated diagrams, I find that drawing diagrams with TiKZ is relatively painless, and more robust to changes (e.g. in how you want lines to curve, or the positions of nodes if you have them) than ones drawn in xfig.
 The over/under relationships can be achieved by drawing the "over" curve-segments twice: once with a thicker white line (or a line matching the background color), and once with a thinner line corresponding to the color of the knot. If you use the PGF extensions (technology which accompanies the TiKZ package), you can quite probably do colorful things such as gradients along the knot curve, if this serves to emphasize features of your knot diagrams.

Good luck with your exam!

  †Added after Peter's comment — my workplace at the time of my original post has the odd feature that it is more likely to have convenient cameras than convenient scanners.
 http://mathoverflow.net/questions/21424/how-to-draw-knots-with-latex/21431#21431 Answer by Sammy Black for How to draw knots with Latex? Sammy Black 2010-04-15T07:58:00Z 2010-04-15T07:58:00Z
Aaron Lauda has a nice description using the package XY-pic here. There are commands that generate pieces of knots (such as crossings in various orientations), although I prefer just using the \crv "curve" command to make splines.
 http://mathoverflow.net/questions/21424/how-to-draw-knots-with-latex/21433#21433 Answer by David Lehavi for How to draw knots with Latex? David Lehavi 2010-04-15T08:48:26Z 2010-04-15T08:48:26Z
The knot atlas gang have a mathematica package for this - you paste the pictures it produces.
 http://mathoverflow.net/questions/21424/how-to-draw-knots-with-latex/21442#21442 Answer by Andrew Stacey for How to draw knots with Latex? Andrew Stacey 2010-04-15T10:41:54Z 2010-04-15T10:41:54Z
Here's a, perhaps silly, suggestion that perhaps isn't quite what you're looking for:

Draw them properly in 3D using something like Blender, and then render it to a 2D picture.

The second animation on this page is of a variant on the borromean rings.
 http://mathoverflow.net/questions/21424/how-to-draw-knots-with-latex/21443#21443 Answer by David Jordan for How to draw knots with Latex? David Jordan 2010-04-15T11:15:30Z 2010-04-15T11:15:30Z
I would discourage the method of drawing them by hand and importing them via scanner etc. as they won't be scalable. Many journals would be unhappy with such an image, and in any case it would look bad and make your pdf filesize huge. Instead, I am a big fan of the free program Inkscape, which you can get for unix/linux and also mac with the X server, and probably Windows, if you're into that kind of thing. It produces pdf files, in which one can zoom in on the curves and they look nice. Also they print quickly, whereas if you include an image file in a pdf it will take a long time to print on a laser printer (which prefers postscript).

Inkscape is similar to xy-pic, only it's very easy to use. You can start by drawing the knots free hand using the pen tool. Then you click a tool which highlights all the data points of a line segment, and you can start deleting them (alternatively, you can use the command "Path->Simplify"). It also has two extra compatibilities with latex: First, you can get a script to actually run snippets of latex in inkscape to make fonts and such for labels. Alternatively, you can just compile the font you want into a pdf and import it. It will import in a scalable way. Then when you're done you export to pdf, and you can include it back into a new pdf if you are using pdflatex to compile your .tex document. If you use ordinary latex, then you need to save it to .eps; I prefer the former.

Since they only have much fancier screenshots on their homepage, here is a paper with diagrams done in Inkscape.

http://arxiv.org/pdf/0908.3013
 http://mathoverflow.net/questions/21424/how-to-draw-knots-with-latex/21456#21456 Answer by Paul for How to draw knots with Latex? Paul 2010-04-15T13:11:22Z 2010-04-15T13:11:22Z
Knotinfo has .png files of all knots of 12 crossings or less.
 http://www.indiana.edu/~knotinfo
 http://mathoverflow.net/questions/21424/how-to-draw-knots-with-latex/23172#23172 Answer by Pádraig Ó Conbhuí for How to draw knots with Latex? Pádraig Ó Conbhuí 2010-05-01T11:47:22Z 2010-05-01T12:21:43Z
I didn't really like Simon's link. The tutorial there seems to be based solely on drawing the curves by specifying coordinates. I find that a more, say, tabular or rigidly based approach is much easier to work with.

There's a good xy-pic tutorial for creating knots at http://lf.starlogik.de/res/XYPic-Knot-Intro.pdf

There's a great xymatrix tutorial at http://www.ctan.org/tex-archive/macros/generic/diagrams/xypic/xy/doc/xyguide.pdf

There are only 10 pages to read in each manual and you'll have mastered them both. You could get away without reading the xymatrix tutorial, but I'd reccommend reading it because it goes more in-depth into some of the parameter modifiers that aren't actually explained in the knot manual.
 http://mathoverflow.net/questions/21424/how-to-draw-knots-with-latex/30863#30863 Answer by Nikos Apostolakis for How to draw knots with Latex? Nikos Apostolakis 2010-07-07T06:46:43Z 2010-07-07T06:46:43Z
Have a look at the knot program by Kodama:

http://www.math.kobe-u.ac.jp/~kodama/knot.html

You can draw and manipulate knot diagrams using the mouse and there is an option to export the figures to pstricks. It also comes with all the prime knots of up to 10 crossings.
 http://mathoverflow.net/questions/21424/how-to-draw-knots-with-latex/30883#30883 Answer by Kevin O'Bryant for How to draw knots with Latex? Kevin O'Bryant 2010-07-07T12:58:02Z 2010-07-07T12:58:02Z
Wolfram|Alpha will draw knots with Alexander-Briggs input. Mathematica will do it with almost any sort of input, including Dowker notation and Conway notation, and most any named knot. The command to look up in the help files is "KnotData".

From Wolfram|Alpha, I'm sure there's some way to grab the picture into a file. Mathematica itself will export to JPG, PNG, PDF, EPS, GIF, etc.
 http://mathoverflow.net/questions/21424/how-to-draw-knots-with-latex/30887#30887 Answer by Henry Segerman for How to draw knots with Latex? Henry Segerman 2010-07-07T13:24:06Z 2010-07-07T13:24:06Z
I gave a talk on the subject of drawing knots using computers (without much detail, but I cover both 2D and 3D methods), the slides might be of interest.
 http://mathoverflow.net/questions/21424/how-to-draw-knots-with-latex/56278#56278 Answer by Andrew Stacey for How to draw knots with Latex? Andrew Stacey 2011-02-22T13:01:37Z 2013-04-22T19:32:14Z
Update 2013-04-22: I've now uploaded a TikZ library to CTAN expressly for the purpose of drawing knots. It is already in MikTeX (according to CTAN), TeXLive might have to wait until TL2013 has been released. It is released as part of the spath3 library. Here's the link to CTAN. The hobby package also contains some useful code for drawing knots. For drawing braids, there's the braids package. And here's a short blog post I wrote explaining the variety of methods.

This is a follow-up to Niels' answer, namely to use TikZ. I've recently had occasion to draw a few knots and links using TikZ and have quite a workable system. I'll not post code here (ask by email or on http://tex.stackexchange.com if you want). Using it, I can produce SVGs (as evidenced by some nLab pages: http://ncatlab.org/nlab/show/Borromean+link, http://ncatlab.org/nlab/show/Whitehead+link, and some others) as well as PDFs, PNGs, and so forth. As an example, here's the trefoil knot:

 http://mathoverflow.net/questions/21424/how-to-draw-knots-with-latex/58170#58170 Answer by Vafa Khalighi for How to draw knots with Latex? Vafa Khalighi 2011-03-11T15:51:34Z 2011-03-11T15:51:34Z
You can draw knots with metapost:

\documentclass{article} \usepackage{mpgraphics} \begin{document} \begin{mpdisplay} for i=0 step 1 until 360: pair P; P = 2cm* (sind(2*i), cosd(3*i)); fill fullcircle scaled 4mm shifted P withcolor white; draw fullcircle scaled 4mm shifted P; endfor; picture p; p:=nullpicture; for i=-180 step 1 until 180: pair P; P = 2cm* (sind(2*i), cosd(3*i)); addto p contour fullcircle scaled 4mm shifted P withcolor white; addto p doublepath fullcircle scaled 4mm shifted P withpen pencircle scaled .5bp; endfor; clip p to (.5cm,2.5cm) -- (.5cm,1.5cm)-- (-.5cm,1.5cm) -- (-.5cm,2.5cm)--cycle; draw p; \end{mpdisplay} \end{document}

or use pst-knot package (using postscript).

format: plain


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122 views

We are given a graph $G$ along with a matching $M$ and a vertex cover $C$ in it such that $|M|=|C|$. Consider the following statements:

1. $M$ is a maximum matching in $G$.
2. $C$ is a minimum vertex cover in $G$.
3. $G$ is a bipartite graph.

Which of the following is $\text{TRUE}$?

1. Only statement $(1)$ is correct
2. Only statement $(2)$ is correct
3. Only statement $(3)$ is correct
4. Only statements $(1)$ and $(2)$ are correct
5. All the three statements $(1), (2),$ and $(3)$ are correct

reshown

@Deepak Poonia sir can you pls provide answer to this

It is ALREADY Covered in GO Classes Graph Theory Course.

Learn the WHOLE Concept, with Proof HERE:

Theorem 1:

The cardinality of ANY Matching is less than or equal to the cardinality of ANY Vertex Cover.

Proof: Watch This.

Corollary of Theorem 1:

Size of Maximum Matching is less than or equal to the size of Minimum Vertex Cover.

Proof: Watch the above lecture.

So, the MAXIMUM matching you can create, will have size less than or equal to the size of MINIMUM Vertex Cover.

So, when can a Matching $M$ & Vertex Cover $C$ be Same??

That is possible only when $M$ is a maximum matching & $C$ is a minimum vertex cover.

So, Statement (1),(2) are correct.

Konig’s Theorem:

For a bipartite graph, the Matching number (i.e., size of a maximum matching) is equal to the vertex cover number (i.e., size of a minimum vertex cover).

BUT this is only ONE WAY. i.e. If Matching number is same as Covering number, it doesn’t imply that the graph is Bipartite.

For counter example, Take a “complete graph K4 with one edge removed” graph, it has Matching Number 2, Vertex Covering Number 2, But it is Not bipartite.

So, Statement 3 is False.

EVERY Concept used in this question is covered, With Proof, in this lecture.

WATCH $\color{red}{\text{The COMPLETE Graph Theory Course, with ALL the Proofs, Questions, Variations}}$ etc, HERE

This playlist is ALL you need for ANY exam, and proper knowledge of Graph Theory.


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Next: Customizing the Configuration Files Up: Bacula Main Reference Previous: Critical Items to Implement   Contents   Index

Subsections

# A Brief Tutorial

This chapter will guide you through running Bacula. To do so, we assume you have installed Bacula, possibly in a single file as shown in the previous chapter, in which case, you can run Bacula as non-root for these tests. However, we assume that you have not changed the .conf files. If you have modified the .conf files, please go back and uninstall Bacula, then reinstall it, but do not make any changes. The examples in this chapter use the default configuration files, and will write the volumes to disk in your /tmp directory, in addition, the data backed up will be the source directory where you built Bacula. As a consequence, you can run all the Bacula daemons for these tests as non-root. Please note, in production, your File daemon(s) must run as root. See the Security chapter for more information on this subject.

The general flow of running Bacula is:

1. cd install-directory
2. Start the Database (if using MySQL or PostgreSQL)
3. Start the Daemons with ./bacula start
4. Start the Console program to interact with the Director
5. Run a job
6. When the Volume fills, unmount the Volume, if it is a tape, label a new one, and continue running. In this chapter, we will write only to disk files so you won't need to worry about tapes for the moment.
7. Test recovering some files from the Volume just written to ensure the backup is good and that you know how to recover. Better test before disaster strikes

Each of these steps is described in more detail below.

# Before Running Bacula

Before running Bacula for the first time in production, we recommend that you run the test command in the btape program as described in the Utility Program Chapterbtape of this manual. This will help ensure that Bacula functions correctly with your tape drive. If you have a modern HP, Sony, or Quantum DDS or DLT tape drive running on Linux or Solaris, you can probably skip this test as Bacula is well tested with these drives and systems. For all other cases, you are strongly encouraged to run the test before continuing. btape also has a fill command that attempts to duplicate what Bacula does when filling a tape and writing on the next tape. You should consider trying this command as well, but be forewarned, it can take hours (about four hours on my drive) to fill a large capacity tape.

# Starting the Database

If you are using MySQL or PostgreSQL as the Bacula database, you should start it before you attempt to run a job to avoid getting error messages from Bacula when it starts. The scripts startmysql and stopmysql are what I (Kern) use to start and stop my local MySQL. Note, if you are using SQLite, you will not want to use startmysql or stopmysql. If you are running this in production, you will probably want to find some way to automatically start MySQL or PostgreSQL after each system reboot.

If you are using SQLite (i.e. you specified the --with-sqlite=xxx option on the ./configure command, you need do nothing. SQLite is automatically started by Bacula.

# Starting the Daemons

Assuming you have built from source or have installed the rpms, to start the three daemons, from your installation directory, simply enter:

./bacula start

The bacula script starts the Storage daemon, the File daemon, and the Director daemon, which all normally run as daemons in the background. If you are using the autostart feature of Bacula, your daemons will either be automatically started on reboot, or you can control them individually with the files bacula-dir, bacula-fd, and bacula-sd, which are usually located in /etc/init.d, though the actual location is system dependent. Some distributions may do this differently.

Note, on Windows, currently only the File daemon is ported, and it must be started differently. Please see the Windows Version of BaculaWin32Chapter Chapter of this manual.

The rpm packages configure the daemons to run as user=root and group=bacula. The rpm installation also creates the group bacula if it does not exist on the system. Any users that you add to the group bacula will have access to files created by the daemons. To disable or alter this behavior edit the daemon startup scripts:

• /etc/bacula/bacula
• /etc/init.d/bacula-dir
• /etc/init.d/bacula-sd
• /etc/init.d/bacula-fd

and then restart as noted above.

The installation chapterInstallChapter of this manual explains how you can install scripts that will automatically restart the daemons when the system starts.

# Using the Director to Query and Start Jobs

To communicate with the director and to query the state of Bacula or run jobs, from the top level directory, simply enter:

./bconsole

Alternatively to running the command line console, if you have Qt4 installed and used the --enable-bat on the configure command, you may use the Bacula Administration Tool (bat):

./bat

Which has a graphical interface, and many more features than bconsole.

Two other possibilities are to run the GNOME console bgnome-console or the wxWidgets program bwx-console.

For simplicity, here we will describe only the ./bconsole program. Most of what is described here applies equally well to ./bat, ./bgnome-console, and to bwx-console.

The ./bconsole runs the Bacula Console program, which connects to the Director daemon. Since Bacula is a network program, you can run the Console program anywhere on your network. Most frequently, however, one runs it on the same machine as the Director. Normally, the Console program will print something similar to the following:

[kern@polymatou bin]$./bconsole Connecting to Director lpmatou:9101 1000 OK: HeadMan Version: 2.1.8 (14 May 2007) *  the asterisk is the console command prompt. Type help to see a list of available commands: *help Command Description ======= =========== add add media to a pool autodisplay autodisplay [on|off] -- console messages automount automount [on|off] -- after label cancel cancel [<jobid=nnn> | <job=name>] -- cancel a job create create DB Pool from resource delete delete [pool=<pool-name> | media volume=<volume-name>] disable disable <job=name> -- disable a job enable enable <job=name> -- enable a job estimate performs FileSet estimate, listing gives full listing exit exit = quit gui gui [on|off] -- non-interactive gui mode help print this command list list [pools | jobs | jobtotals | media <pool=pool-name> | files <jobid=nn>]; from catalog label label a tape llist full or long list like list command memory print current memory usage messages messages mount mount <storage-name> prune prune expired records from catalog purge purge records from catalog python python control commands quit quit query query catalog restore restore files relabel relabel a tape release release <storage-name> reload reload conf file run run <job-name> status status [[slots] storage | dir | client]=<name> setdebug sets debug level setip sets new client address -- if authorized show show (resource records) [jobs | pools | ... | all] sqlquery use SQL to query catalog time print current time trace turn on/off trace to file unmount unmount <storage-name> umount umount <storage-name> for old-time Unix guys update update Volume, Pool or slots use use catalog xxx var does variable expansion version print Director version wait wait until no jobs are running [<jobname=name> | <jobid=nnn> | <ujobid=complete_name>] *  Details of the console program's commands are explained in the Console Chapter_ConsoleChapter of this manual. # Running a Job At this point, we assume you have done the following: • Configured Bacula with ./configure --your-options • Built Bacula using make • Installed Bacula using make install • Have created your database with, for example, ./create_sqlite_database • Have created the Bacula database tables with, ./make_bacula_tables • Have possibly edited your bacula-dir.conf file to personalize it a bit. BE CAREFUL! if you change the Director's name or password, you will need to make similar modifications in the other .conf files. For the moment it is probably better to make no changes. • You have started Bacula with ./bacula start • You have invoked the Console program with ./bconsole Furthermore, we assume for the moment you are using the default configuration files. At this point, enter the following command: show filesets  and you should get something similar to: FileSet: name=Full Set O M N I /home/kern/bacula/regress/build N E /proc E /tmp E /.journal E /.fsck N FileSet: name=Catalog O M N I /home/kern/bacula/regress/working/bacula.sql N  This is a pre-defined FileSet that will backup the Bacula source directory. The actual directory names printed should correspond to your system configuration. For testing purposes, we have chosen a directory of moderate size (about 40 Megabytes) and complexity without being too big. The FileSet Catalog is used for backing up Bacula's catalog and is not of interest to us for the moment. The I entries are the files or directories that will be included in the backup and the E are those that will be excluded, and the O entries are the options specified for the FileSet. You can change what is backed up by editing bacula-dir.conf and changing the File = line in the FileSet resource. Now is the time to run your first backup job. We are going to backup your Bacula source directory to a File Volume in your /tmp directory just to show you how easy it is. Now enter: status dir  and you should get the following output: rufus-dir Version: 1.30 (28 April 2003) Daemon started 28-Apr-2003 14:03, 0 Jobs run. Console connected at 28-Apr-2003 14:03 No jobs are running. Level Type Scheduled Name ================================================================= Incremental Backup 29-Apr-2003 01:05 Client1 Full Backup 29-Apr-2003 01:10 BackupCatalog ====  where the times and the Director's name will be different according to your setup. This shows that an Incremental job is scheduled to run for the Job Client1 at 1:05am and that at 1:10, a BackupCatalog is scheduled to run. Note, you should probably change the name Client1 to be the name of your machine, if not, when you add additional clients, it will be very confusing. For my real machine, I use Rufus rather than Client1 as in this example. Now enter: status client  and you should get something like: The defined Client resources are: 1: rufus-fd Item 1 selected automatically. Connecting to Client rufus-fd at rufus:8102 rufus-fd Version: 1.30 (28 April 2003) Daemon started 28-Apr-2003 14:03, 0 Jobs run. Director connected at: 28-Apr-2003 14:14 No jobs running. ====  In this case, the client is named rufus-fd your name will be different, but the line beginning with rufus-fd Version ... is printed by your File daemon, so we are now sure it is up and running. Finally do the same for your Storage daemon with: status storage  and you should get: The defined Storage resources are: 1: File Item 1 selected automatically. Connecting to Storage daemon File at rufus:8103 rufus-sd Version: 1.30 (28 April 2003) Daemon started 28-Apr-2003 14:03, 0 Jobs run. Device /tmp is not open. No jobs running. ====  You will notice that the default Storage daemon device is named File and that it will use device /tmp, which is not currently open. Now, let's actually run a job with: run  you should get the following output: Using default Catalog name=MyCatalog DB=bacula A job name must be specified. The defined Job resources are: 1: Client1 2: BackupCatalog 3: RestoreFiles Select Job resource (1-3):  Here, Bacula has listed the three different Jobs that you can run, and you should choose number 1 and type enter, at which point you will get: Run Backup job JobName: Client1 FileSet: Full Set Level: Incremental Client: rufus-fd Storage: File Pool: Default When: 2003-04-28 14:18:57 OK to run? (yes/mod/no):  At this point, take some time to look carefully at what is printed and understand it. It is asking you if it is OK to run a job named Client1 with FileSet Full Set (we listed above) as an Incremental job on your Client (your client name will be different), and to use Storage File and Pool Default, and finally, it wants to run it now (the current time should be displayed by your console). Here we have the choice to run (yes), to modify one or more of the above parameters (mod), or to not run the job (no). Please enter yes, at which point you should immediately get the command prompt (an asterisk). If you wait a few seconds, then enter the command messages you will get back something like: 28-Apr-2003 14:22 rufus-dir: Last FULL backup time not found. Doing FULL backup. 28-Apr-2003 14:22 rufus-dir: Start Backup JobId 1, Job=Client1.2003-04-28_14.22.33 28-Apr-2003 14:22 rufus-sd: Job Client1.2003-04-28_14.22.33 waiting. Cannot find any appendable volumes. Please use the "label" command to create a new Volume for: Storage: FileStorage Media type: File Pool: Default  The first message, indicates that no previous Full backup was done, so Bacula is upgrading our Incremental job to a Full backup (this is normal). The second message indicates that the job started with JobId 1., and the third message tells us that Bacula cannot find any Volumes in the Pool for writing the output. This is normal because we have not yet created (labeled) any Volumes. Bacula indicates to you all the details of the volume it needs. At this point, the job is BLOCKED waiting for a Volume. You can check this if you want by doing a status dir. In order to continue, we must create a Volume that Bacula can write on. We do so with: label  and Bacula will print: The defined Storage resources are: 1: File Item 1 selected automatically. Enter new Volume name:  at which point, you should enter some name beginning with a letter and containing only letters and numbers (period, hyphen, and underscore) are also permitted. For example, enter TestVolume001, and you should get back: Defined Pools: 1: Default Item 1 selected automatically. Connecting to Storage daemon File at rufus:8103 ... Sending label command for Volume "TestVolume001" Slot 0 ... 3000 OK label. Volume=TestVolume001 Device=/tmp Catalog record for Volume "TestVolume002", Slot 0 successfully created. Requesting mount FileStorage ... 3001 OK mount. Device=/tmp  Finally, enter messages and you should get something like: 28-Apr-2003 14:30 rufus-sd: Wrote label to prelabeled Volume "TestVolume001" on device /tmp 28-Apr-2003 14:30 rufus-dir: Bacula 1.30 (28Apr03): 28-Apr-2003 14:30 JobId: 1 Job: Client1.2003-04-28_14.22.33 FileSet: Full Set Backup Level: Full Client: rufus-fd Start time: 28-Apr-2003 14:22 End time: 28-Apr-2003 14:30 Files Written: 1,444 Bytes Written: 38,988,877 Rate: 81.2 KB/s Software Compression: None Volume names(s): TestVolume001 Volume Session Id: 1 Volume Session Time: 1051531381 Last Volume Bytes: 39,072,359 FD termination status: OK SD termination status: OK Termination: Backup OK 28-Apr-2003 14:30 rufus-dir: Begin pruning Jobs. 28-Apr-2003 14:30 rufus-dir: No Jobs found to prune. 28-Apr-2003 14:30 rufus-dir: Begin pruning Files. 28-Apr-2003 14:30 rufus-dir: No Files found to prune. 28-Apr-2003 14:30 rufus-dir: End auto prune.  If you don't see the output immediately, you can keep entering messages until the job terminates, or you can enter, autodisplay on and your messages will automatically be displayed as soon as they are ready. If you do an ls -l of your /tmp directory, you will see that you have the following item: -rw-r----- 1 kern kern 39072153 Apr 28 14:30 TestVolume001  This is the file Volume that you just wrote and it contains all the data of the job just run. If you run additional jobs, they will be appended to this Volume unless you specify otherwise. You might ask yourself if you have to label all the Volumes that Bacula is going to use. The answer for disk Volumes, like the one we used, is no. It is possible to have Bacula automatically label volumes. For tape Volumes, you will most likely have to label each of the Volumes you want to use. If you would like to stop here, you can simply enter quit in the Console program, and you can stop Bacula with ./bacula stop. To clean up, simply delete the file /tmp/TestVolume001, and you should also re-initialize your database using: ./drop_bacula_tables ./make_bacula_tables  Please note that this will erase all information about the previous jobs that have run, and that you might want to do it now while testing but that normally you will not want to re-initialize your database. If you would like to try restoring the files that you just backed up, read the following section. # Restoring Your Files If you have run the default configuration and the save of the Bacula source code as demonstrated above, you can restore the backed up files in the Console program by entering: restore all  where you will get: First you select one or more JobIds that contain files to be restored. You will be presented several methods of specifying the JobIds. Then you will be allowed to select which files from those JobIds are to be restored. To select the JobIds, you have the following choices: 1: List last 20 Jobs run 2: List Jobs where a given File is saved 3: Enter list of comma separated JobIds to select 4: Enter SQL list command 5: Select the most recent backup for a client 6: Select backup for a client before a specified time 7: Enter a list of files to restore 8: Enter a list of files to restore before a specified time 9: Find the JobIds of the most recent backup for a client 10: Find the JobIds for a backup for a client before a specified time 11: Enter a list of directories to restore for found JobIds 12: Cancel Select item: (1-12):  As you can see, there are a number of options, but for the current demonstration, please enter 5 to do a restore of the last backup you did, and you will get the following output: Defined Clients: 1: rufus-fd Item 1 selected automatically. The defined FileSet resources are: 1: 1 Full Set 2003-04-28 14:22:33 Item 1 selected automatically. +-------+-------+----------+---------------------+---------------+ | JobId | Level | JobFiles | StartTime | VolumeName | +-------+-------+----------+---------------------+---------------+ | 1 | F | 1444 | 2003-04-28 14:22:33 | TestVolume002 | +-------+-------+----------+---------------------+---------------+ You have selected the following JobId: 1 Building directory tree for JobId 1 ... 1 Job inserted into the tree and marked for extraction. The defined Storage resources are: 1: File Item 1 selected automatically. You are now entering file selection mode where you add and remove files to be restored. All files are initially added. Enter "done" to leave this mode. cwd is: /$

where I have truncated the listing on the right side to make it more readable. As you can see by starting at the top of the listing, Bacula knows what client you have, and since there was only one, it selected it automatically, likewise for the FileSet. Then Bacula produced a listing containing all the jobs that form the current backup, in this case, there is only one, and the Storage daemon was also automatically chosen. Bacula then took all the files that were in Job number 1 and entered them into a directory tree (a sort of in memory representation of your filesystem). At this point, you can use the cd and ls ro dir commands to walk up and down the directory tree and view what files will be restored. For example, if I enter cd /home/kern/bacula/bacula-1.30 and then enter dir I will get a listing of all the files in the Bacula source directory. On your system, the path will be somewhat different. For more information on this, please refer to the Restore Command ChapterRestoreChapter of this manual for more details.

To exit this mode, simply enter:

done

and you will get the following output:

Bootstrap records written to
/home/kern/bacula/testbin/working/restore.bsr
The restore job will require the following Volumes:

TestVolume001
1444 files selected to restore.
Run Restore job
JobName:         RestoreFiles
Bootstrap:      /home/kern/bacula/testbin/working/restore.bsr
Where:          /tmp/bacula-restores
Replace:        always
FileSet:        Full Set
Backup Client:  rufus-fd
Restore Client: rufus-fd
Storage:        File
JobId:          *None*
When:           2005-04-28 14:53:54
OK to run? (yes/mod/no):

If you answer yes your files will be restored to /tmp/bacula-restores. If you want to restore the files to their original locations, you must use the mod option and explicitly set Where: to nothing (or to /). We recommend you go ahead and answer yes and after a brief moment, enter messages, at which point you should get a listing of all the files that were restored as well as a summary of the job that looks similar to this:

28-Apr-2005 14:56 rufus-dir: Bacula 2.1.8 (08May07): 08-May-2007 14:56:06
Build OS:               i686-pc-linux-gnu suse 10.2
JobId:                  2
Job:                    RestoreFiles.2007-05-08_14.56.06
Restore Client:         rufus-fd
Start time:             08-May-2007 14:56
End time:               08-May-2007 14:56
Files Restored:         1,444
Bytes Restored:         38,816,381
Rate:                   9704.1 KB/s
FD Errors:              0
FD termination status:  OK
SD termination status:  OK
Termination:            Restore OK
08-May-2007 14:56 rufus-dir: Begin pruning Jobs.
08-May-2007 14:56 rufus-dir: No Jobs found to prune.
08-May-2007 14:56 rufus-dir: Begin pruning Files.
08-May-2007 14:56 rufus-dir: No Files found to prune.
08-May-2007 14:56 rufus-dir: End auto prune.

After exiting the Console program, you can examine the files in /tmp/bacula-restores, which will contain a small directory tree with all the files. Be sure to clean up at the end with:

rm -rf /tmp/bacula-restore

# Quitting the Console Program

Simply enter the command quit.

If you have gotten the example shown above to work on your system, you may be ready to add a second Client (File daemon). That is you have a second machine that you would like backed up. The only part you need installed on the other machine is the binary bacula-fd (or bacula-fd.exe for Windows) and its configuration file bacula-fd.conf. You can start with the same bacula-fd.conf file that you are currently using and make one minor modification to it to create the conf file for your second client. Change the File daemon name from whatever was configured, rufus-fd in the example above, but your system will have a different name. The best is to change it to the name of your second machine. For example:

...
#
# "Global" File daemon configuration specifications
#
FileDaemon {                          # this is me
Name = rufus-fd
FDport = 9102                  # where we listen for the director
WorkingDirectory = /home/kern/bacula/working
Pid Directory = /var/run
}
...

would become:

...
#
# "Global" File daemon configuration specifications
#
FileDaemon {                          # this is me
Name = matou-fd
FDport = 9102                  # where we listen for the director
WorkingDirectory = /home/kern/bacula/working
Pid Directory = /var/run
}
...

where I show just a portion of the file and have changed rufus-fd to matou-fd. The names you use are your choice. For the moment, I recommend you change nothing else. Later, you will want to change the password.

Now you should install that change on your second machine. Then you need to make some additions to your Director's configuration file to define the new File daemon or Client. Starting from our original example which should be installed on your system, you should add the following lines (essentially copies of the existing data but with the names changed) to your Director's configuration file bacula-dir.conf.

#
# Define the main nightly save backup job
#   By default, this job will back up to disk in /tmp
Job {
Name = "Matou"
Type = Backup
Client = matou-fd
FileSet = "Full Set"
Schedule = "WeeklyCycle"
Storage = File
Messages = Standard
Pool = Default
Write Bootstrap = "/home/kern/bacula/working/matou.bsr"
}
# Client (File Services) to backup
Client {
Name = matou-fd
FDPort = 9102
Catalog = MyCatalog
File Retention = 30d                # 30 days
Job Retention = 180d                # six months
AutoPrune = yes                     # Prune expired Jobs/Files
}

Then make sure that the Address parameter in the Storage resource is set to the fully qualified domain name and not to something like "localhost". The address specified is sent to the File daemon (client) and it must be a fully qualified domain name. If you pass something like "localhost" it will not resolve correctly and will result in a time out when the File daemon fails to connect to the Storage daemon.

That is all that is necessary. I copied the existing resource to create a second Job (Matou) to backup the second client (matou-fd). It has the name Matou, the Client is named matou-fd, and the bootstrap file name is changed, but everything else is the same. This means that Matou will be backed up on the same schedule using the same set of tapes. You may want to change that later, but for now, let's keep it simple.

The second change was to add a new Client resource that defines matou-fd and has the correct address matou, but in real life, you may need a fully qualified domain name or an IP address. I also kept the password the same (shown as xxxxx for the example).

At this point, if you stop Bacula and restart it, and start the Client on the other machine, everything will be ready, and the prompts that you saw above will now include the second machine.

To make this a real production installation, you will possibly want to use different Pool, or a different schedule. It is up to you to customize. In any case, you should change the password in both the Director's file and the Client's file for additional security.

For some important tips on changing names and passwords, and a diagram of what names and passwords must match, please see Authorization ErrorsAuthorizationErrors in the FAQ chapter of this manual.

# When The Tape Fills

If you have scheduled your job, typically nightly, there will come a time when the tape fills up and Bacula cannot continue. In this case, Bacula will send you a message similar to the following:

rufus-sd: block.c:337 === Write error errno=28: ERR=No space left
on device

This indicates that Bacula got a write error because the tape is full. Bacula will then search the Pool specified for your Job looking for an appendable volume. In the best of all cases, you will have properly set your Retention Periods and you will have all your tapes marked to be Recycled, and Bacula will automatically recycle the tapes in your pool requesting and overwriting old Volumes. For more information on recycling, please see the Recycling chapterRecyclingChapter of this manual. If you find that your Volumes were not properly recycled (usually because of a configuration error), please see the Manually Recycling Volumesmanualrecycling section of the Recycling chapter.

If like me, you have a very large set of Volumes and you label them with the date the Volume was first writing, or you have not set up your Retention periods, Bacula will not find a tape in the pool, and it will send you a message similar to the following:

rufus-sd: Job kernsave.2002-09-19.10:50:48 waiting. Cannot find any
appendable volumes.
Please use the "label"  command to create a new Volume for:
Storage:      SDT-10000
Media type:   DDS-4
Pool:         Default

Until you create a new Volume, this message will be repeated an hour later, then two hours later, and so on doubling the interval each time up to a maximum interval of one day.

The obvious question at this point is: What do I do now?

The answer is simple: first, using the Console program, close the tape drive using the unmount command. If you only have a single drive, it will be automatically selected, otherwise, make sure you release the one specified on the message (in this case STD-10000).

Next, you remove the tape from the drive and insert a new blank tape. Note, on some older tape drives, you may need to write an end of file mark (mt -f /dev/nst0 weof) to prevent the drive from running away when Bacula attempts to read the label.

Finally, you use the label command in the Console to write a label to the new Volume. The label command will contact the Storage daemon to write the software label, if it is successful, it will add the new Volume to the Pool, then issue a mount command to the Storage daemon. See the previous sections of this chapter for more details on labeling tapes.

The result is that Bacula will continue the previous Job writing the backup to the new Volume.

If you have a Pool of volumes and Bacula is cycling through them, instead of the above message "Cannot find any appendable volumes.", Bacula may ask you to mount a specific volume. In that case, you should attempt to do just that. If you do not have the volume any more (for any of a number of reasons), you can simply mount another volume from the same Pool, providing it is appendable, and Bacula will use it. You can use the list volumes command in the console program to determine which volumes are appendable and which are not.

If like me, you have your Volume retention periods set correctly, but you have no more free Volumes, you can relabel and reuse a Volume as follows:

• Do a list volumes in the Console and select the oldest Volume for relabeling.
• If you have setup your Retention periods correctly, the Volume should have VolStatus Purged.
• If the VolStatus is not set to Purged, you will need to purge the database of Jobs that are written on that Volume. Do so by using the command purge jobs volume in the Console. If you have multiple Pools, you will be prompted for the Pool then enter the VolumeName (or MediaId) when requested.
• Then simply use the relabel command to relabel the Volume.

To manually relabel the Volume use the following additional steps:

• To delete the Volume from the catalog use the delete volume command in the Console and select the VolumeName (or MediaId) to be deleted.

• Use the unmount command in the Console to unmount the old tape.
• Physically relabel the old Volume that you deleted so that it can be reused.
• Insert the old Volume in the tape drive.
• From a command line do: mt -f /dev/st0 rewind and mt -f /dev/st0 weof, where you need to use the proper tape drive name for your system in place of /dev/st0.
• Use the label command in the Console to write a new Bacula label on your tape.
• Use the mount command in the Console if it is not automatically done, so that Bacula starts using your newly labeled tape.

# Other Useful Console Commands

status dir
Print a status of all running jobs and jobs scheduled in the next 24 hours.

status
The console program will prompt you to select a daemon type, then will request the daemon's status.

status jobid=nn
Print a status of JobId nn if it is running. The Storage daemon is contacted and requested to print a current status of the job as well.

list pools
List the pools defined in the Catalog (normally only Default is used).

list media
Lists all the media defined in the Catalog.

list jobs
Lists all jobs in the Catalog that have run.

list jobid=nn
Lists JobId nn from the Catalog.

list jobtotals
Lists totals for all jobs in the Catalog.

list files jobid=nn
List the files that were saved for JobId nn.

list jobmedia
List the media information for each Job run.

messages
Prints any messages that have been directed to the console.

unmount storage=storage-name
Unmounts the drive associated with the storage device with the name storage-name if the drive is not currently being used. This command is used if you wish Bacula to free the drive so that you can use it to label a tape.

mount storage=storage-name
Causes the drive associated with the storage device to be mounted again. When Bacula reaches the end of a volume and requests you to mount a new volume, you must issue this command after you have placed the new volume in the drive. In effect, it is the signal needed by Bacula to know to start reading or writing the new volume.

quit
Exit or quit the console program.

Most of the commands given above, with the exception of list, will prompt you for the necessary arguments if you simply enter the command name.

# Debug Daemon Output

If you want debug output from the daemons as they are running, start the daemons from the install directory as follows:

./bacula start -d100

This can be particularly helpful if your daemons do not start correctly, because direct daemon output to the console is normally directed to the NULL device, but with the debug level greater than zero, the output will be sent to the starting terminal.

To stop the three daemons, enter the following from the install directory:

./bacula stop

The execution of bacula stop may complain about pids not found. This is OK, especially if one of the daemons has died, which is very rare.

To do a full system save, each File daemon must be running as root so that it will have permission to access all the files. None of the other daemons require root privileges. However, the Storage daemon must be able to open the tape drives. On many systems, only root can access the tape drives. Either run the Storage daemon as root, or change the permissions on the tape devices to permit non-root access. MySQL and PostgreSQL can be installed and run with any userid; root privilege is not necessary.

# Patience When Starting Daemons or Mounting Blank Tapes

When you start the Bacula daemons, the Storage daemon attempts to open all defined storage devices and verify the currently mounted Volume (if configured). Until all the storage devices are verified, the Storage daemon will not accept connections from the Console program. If a tape was previously used, it will be rewound, and on some devices this can take several minutes. As a consequence, you may need to have a bit of patience when first contacting the Storage daemon after starting the daemons. If you can see your tape drive, once the lights stop flashing, the drive will be ready to be used.

The same considerations apply if you have just mounted a blank tape in a drive such as an HP DLT. It can take a minute or two before the drive properly recognizes that the tape is blank. If you attempt to mount the tape with the Console program during this recognition period, it is quite possible that you will hang your SCSI driver (at least on my Red Hat Linux system). As a consequence, you are again urged to have patience when inserting blank tapes. Let the device settle down before attempting to access it.

# Difficulties Connecting from the FD to the SD

If you are having difficulties getting one or more of your File daemons to connect to the Storage daemon, it is most likely because you have not used a fully qualified domain name on the Address directive in the Director's Storage resource. That is the resolver on the File daemon's machine (not on the Director's) must be able to resolve the name you supply into an IP address. An example of an address that is guaranteed not to work: localhost. An example that may work: megalon. An example that is more likely to work: magalon.mydomain.com. On Win32 if you don't have a good resolver (often true on older Win98 systems), you might try using an IP address in place of a name.

If your address is correct, then make sure that no other program is using the port 9103 on the Storage daemon's machine. The Bacula port numbers are authorized by IANA, and should not be used by other programs, but apparently some HP printers do use these port numbers. A netstat -a on the Storage daemon's machine can determine who is using the 9103 port (used for FD to SD communications in Bacula).

# Daemon Command Line Options

Each of the three daemons (Director, File, Storage) accepts a small set of options on the command line. In general, each of the daemons as well as the Console program accepts the following options:

-c file
Define the file to use as a configuration file. The default is the daemon name followed by .conf i.e. bacula-dir.conf for the Director, bacula-fd.conf for the File daemon, and bacula-sd for the Storage daemon.

-d nn
Set the debug level to nn. Higher levels of debug cause more information to be displayed on STDOUT concerning what the daemon is doing.

-f
Run the daemon in the foreground. This option is needed to run the daemon under the debugger.

-g <group>
Run the daemon under this group. This must be a group name, not a GID.

-s
Do not trap signals. This option is needed to run the daemon under the debugger.

-t
Read the configuration file and print any error messages, then immediately exit. Useful for syntax testing of new configuration files.

-u <user>
Run the daemon as this user. This must be a user name, not a UID.

-v
Be more verbose or more complete in printing error and informational messages. Recommended.

-?
Print the version and list of options.

# Creating a Pool

Creating the Pool is automatically done when Bacula starts, so if you understand Pools, you can skip to the next section.

When you run a job, one of the things that Bacula must know is what Volumes to use to backup the FileSet. Instead of specifying a Volume (tape) directly, you specify which Pool of Volumes you want Bacula to consult when it wants a tape for writing backups. Bacula will select the first available Volume from the Pool that is appropriate for the Storage device you have specified for the Job being run. When a volume has filled up with data, Bacula will change its VolStatus from Append to Full, and then Bacula will use the next volume and so on. If no appendable Volume exists in the Pool, the Director will attempt to recycle an old Volume, if there are still no appendable Volumes available, Bacula will send a message requesting the operator to create an appropriate Volume.

Bacula keeps track of the Pool name, the volumes contained in the Pool, and a number of attributes of each of those Volumes.

When Bacula starts, it ensures that all Pool resource definitions have been recorded in the catalog. You can verify this by entering:

list pools

to the console program, which should print something like the following:

*list pools
Using default Catalog name=MySQL DB=bacula
+--------+---------+---------+---------+----------+-------------+
| PoolId | Name    | NumVols | MaxVols | PoolType | LabelFormat |
+--------+---------+---------+---------+----------+-------------+
| 1      | Default | 3       | 0       | Backup   | *           |
| 2      | File    | 12      | 12      | Backup   | File        |
+--------+---------+---------+---------+----------+-------------+
*

If you attempt to create the same Pool name a second time, Bacula will print:

Error: Pool Default already exists.
Once created, you may use the {\bf update} command to
modify many of the values in the Pool record.

Bacula requires that each Volume contains a software label. There are several strategies for labeling volumes. The one I use is to label them as they are needed by Bacula using the console program. That is when Bacula needs a new Volume, and it does not find one in the catalog, it will send me an email message requesting that I add Volumes to the Pool. I then use the label command in the Console program to label a new Volume and to define it in the Pool database, after which Bacula will begin writing on the new Volume. Alternatively, I can use the Console relabel command to relabel a Volume that is no longer used providing it has VolStatus Purged.

Another strategy is to label a set of volumes at the start, then use them as Bacula requests them. This is most often done if you are cycling through a set of tapes, for example using an autochanger. For more details on recycling, please see the Automatic Volume RecyclingRecyclingChapter chapter of this manual.

If you run a Bacula job, and you have no labeled tapes in the Pool, Bacula will inform you, and you can create them "on-the-fly" so to speak. In my case, I label my tapes with the date, for example: DLT-18April02. See below for the details of using the label command.

# Labeling Volumes with the Console Program

Labeling volumes is normally done by using the console program.

1. ./bconsole
2. label

If Bacula complains that you cannot label the tape because it is already labeled, simply unmount the tape using the unmount command in the console, then physically mount a blank tape and re-issue the label command.

Since the physical storage media is different for each device, the label command will provide you with a list of the defined Storage resources such as the following:

The defined Storage resources are:
1: File
2: 8mmDrive
3: DLTDrive
4: SDT-10000
Select Storage resource (1-4):

At this point, you should have a blank tape in the drive corresponding to the Storage resource that you select.

It will then ask you for the Volume name.

Enter new Volume name:

If Bacula complains:

Media record for Volume xxxx already exists.

It means that the volume name xxxx that you entered already exists in the Media database. You can list all the defined Media (Volumes) with the list media command. Note, the LastWritten column has been truncated for proper printing.

+---------------+---------+--------+----------------+-----/~/-+------------+-----+
| VolumeName    | MediaTyp| VolStat| VolBytes       | LastWri | VolReten   | Recy|
+---------------+---------+--------+----------------+---------+------------+-----+
| DLTVol0002    | DLT8000 | Purged | 56,128,042,217 | 2001-10 | 31,536,000 |   0 |
| DLT-07Oct2001 | DLT8000 | Full   | 56,172,030,586 | 2001-11 | 31,536,000 |   0 |
| DLT-08Nov2001 | DLT8000 | Full   | 55,691,684,216 | 2001-12 | 31,536,000 |   0 |
| DLT-01Dec2001 | DLT8000 | Full   | 55,162,215,866 | 2001-12 | 31,536,000 |   0 |
| DLT-28Dec2001 | DLT8000 | Full   | 57,888,007,042 | 2002-01 | 31,536,000 |   0 |
| DLT-20Jan2002 | DLT8000 | Full   | 57,003,507,308 | 2002-02 | 31,536,000 |   0 |
| DLT-16Feb2002 | DLT8000 | Full   | 55,772,630,824 | 2002-03 | 31,536,000 |   0 |
| DLT-12Mar2002 | DLT8000 | Full   | 50,666,320,453 | 1970-01 | 31,536,000 |   0 |
| DLT-27Mar2002 | DLT8000 | Full   | 57,592,952,309 | 2002-04 | 31,536,000 |   0 |
| DLT-15Apr2002 | DLT8000 | Full   | 57,190,864,185 | 2002-05 | 31,536,000 |   0 |
| DLT-04May2002 | DLT8000 | Full   | 60,486,677,724 | 2002-05 | 31,536,000 |   0 |
| DLT-26May02   | DLT8000 | Append |  1,336,699,620 | 2002-05 | 31,536,000 |   1 |
+---------------+---------+--------+----------------+-----/~/-+------------+-----+

Once Bacula has verified that the volume does not already exist, it will prompt you for the name of the Pool in which the Volume (tape) is to be created. If there is only one Pool (Default), it will be automatically selected.

If the tape is successfully labeled, a Volume record will also be created in the Pool. That is the Volume name and all its other attributes will appear when you list the Pool. In addition, that Volume will be available for backup if the MediaType matches what is requested by the Storage daemon.

When you labeled the tape, you answered very few questions about it - principally the Volume name, and perhaps the Slot. However, a Volume record in the catalog database (internally known as a Media record) contains quite a few attributes. Most of these attributes will be filled in from the default values that were defined in the Pool (i.e. the Pool holds most of the default attributes used when creating a Volume).

It is also possible to add media to the pool without physically labeling the Volumes. This can be done with the add command. For more information, please see the Console Chapter_ConsoleChapter of this manual.

Next: Customizing the Configuration Files Up: Bacula Main Reference Previous: Critical Items to Implement   Contents   Index
Eric 2011-06-08


text:
## Bus Assignment

#### 문제

The 2024 UCPC summer contest is finally over! The day after the contest, participants will join an excursion to Kyungbok palace to blow off the steam of the contest. Everyone is looking forward to it, and the organizing committee arranged everything for the trip.

However one small problem remains; the transportation. The committee plans to rent two buses to take participants to the palace and back - a Tayo bus, and a Tobot bus. And each participant has a different preference on which bus to ride. If the i$i$th participant takes the Tayo bus, (s)he will get a satisfaction value of A[i]$A[i]$. Taking the Tobot bus instead will give him(her) a satisfaction value of B[i]$B[i]$. (Each participant can take only one bus.) The total satisfaction between all participants is calculated by the sum of individual satisfaction values.

Okay, this sounds like an easy problem - everyone can take the bus they prefer(each bus is big enough for all participants)! However, the problem is complicated by the fact people's satisfaction decreases when two friends take different buses. If i$i$th participant and j$j$th participant take different buses, the total satisfaction will decrease by H[i,j]$H[i,j]$! However, if either i$i$ or j$j$ doesn't go to the excursion at all, there will be no satisfaction decreases.

The organizing committee wants to find a plan that maximizes the overall satisfaction. To do this, they can assign each participant to one of the buses, or exclude him(her) from the excursion at all.

Write a program that calculates the maximum satisfaction.

#### 입력

The input consists of multiple test cases. The first line of the input will contain the number of test cases T$T$.

Each test case begins with a line containing N$N$ (2 \le N \le 200$2 \le N \le 200$), the number of participants. Each of the next N$N$ lines will contain two integers, A[i]$A[i]$ and B[i]$B[i]$. The next N$N$ lines will give you the friendship information between the participants. Each line will contain N$N$ integers, and the j$j$th number of i$i$th line will be H[i,j]$H[i,j]$.

You can assume the following:

• All preferences (A[i]$A[i]$, B[i]$B[i]$, H[i,j]$H[i,j]$) are nonnegative integers less than or equal to 1000.
• H[i,j] = H[j,i]$H[i,j] = H[j,i]$ for all i$i$ and j$j$.
• H[i,j] = 0$H[i,j] = 0$ when i=j$i=j$.

#### 출력

For each test case, print a single line with the maximum satisfaction achievable.

#### 예제 입력

2
2
1 1
1 1
0 0
0 0
3
1 1
5 2
4 7
0 9 2
9 0 1
2 1 0

#### 예제 출력

2
11

#### 노트

The optimal solution for the second test case:

• Participant 1 doesn't go to the excursion.
• Participant 2 takes the Tayo bus.
• Participant 3 takes the Tobot bus.

The individual satisfaction from participant 2 and 3 are 5 and 7, respectively. Also, since the two participants are taking different buses, the total satisfaction is decreased by H[2,3] = 1$H[2,3] = 1$.

Description 작성: JongMan


text:
Home > PC > Chapter 4 > Lesson 4.1.3 > Problem4-44

4-44.

Sketch the graph of $f ( x ) = \left\{ \begin{array} { c l } { - x ^ { 2 } + 4 } & { \text { for } x < 0 } \\ { x ^ { 2 } - 1 } & { \text { for } x \geq 0 } \end{array} \right.$. Is $f(x)$ continuous? Explain.

1. Find the height at $0$ for both parts of the piece-wise function.
2. If they are the same, it is continuous.
3. Plot the points found above and continue with the appropriate graph on either side.


text:
0
Research Papers

A New Incremental Harmonic Balance Method With Two Time Scales for Quasi-Periodic Motions of an Axially Moving Beam With Internal Resonance Under Single-Tone External Excitation

[+] Author and Article Information
J. L. Huang

Department of Applied Mechanics and Engineering,
Sun Yat-sen University,
Guangzhou, 510275, China
e-mail: huangjl@mail.sysu.edu.cn

W. D. Zhu

Department of Mechanical Engineering,
University of Maryland, Baltimore County,
1000 Hilltop Circle,
Baltimore, MD 21250
e-mail: wzhu@umbc.edu

1Corresponding author.

Contributed by the Technical Committee on Vibration and Sound of ASME for publication in the JOURNAL OF VIBRATION AND ACOUSTICS. Manuscript received June 18, 2016; final manuscript received October 19, 2016; published online February 15, 2017. Assoc. Editor: Izhak Bucher.

J. Vib. Acoust 139(2), 021010 (Feb 15, 2017) (15 pages) Paper No: VIB-16-1307; doi: 10.1115/1.4035135 History: Received June 18, 2016; Revised October 19, 2016

Abstract

Quasi-periodic motion is an oscillation of a dynamic system characterized by $m$ frequencies that are incommensurable with one another. In this work, a new incremental harmonic balance (IHB) method with only two time scales, where one is one of the $m$ frequencies, referred to as a fundamental frequency, and the other is an interval distance of two adjacent frequencies, is proposed for quasi-periodic motions of an axially moving beam with three-to-one internal resonance under single-tone external excitation. It is found that the interval frequency of every two adjacent frequencies, located around the fundamental frequency or one of its integer multiples, is fixed due to nonlinear coupling among resonant vibration modes. Consequently, only two time scales are used in the IHB method to obtain all incommensurable frequencies of quasi-periodic motions of the axially moving beam. The present IHB method can accurately trace from periodic responses of the beam to its quasi-periodic motions. For periodic responses of the axially moving beam, the single fundamental frequency is used in the IHB method to obtain solutions. For quasi-periodic motions of the beam, the present IHB method with two time scales is used, along with an amplitude increment approach that includes a large number of harmonics, to determine all the frequency components. All the frequency components and their corresponding amplitudes, obtained from the present IHB method, are in excellent agreement with those from numerical integration using the fourth-order Runge–Kutta method.

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References

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Ling, F. H. , 1991, “ Quasi-Periodic Solutions Calculated With the Simple Shooting Technique,” J. Sound Vib., 144(2), pp. 293–304.
Kim, Y. B. , and Noah, S. T. , 1996, “ Quasi-Periodic Response and Stability Analysis for a Non-Linear Jeffcott Rotor,” J. Sound Vib., 190(2), pp. 239–253.
Guennoun, K. , Houssni, M. , and Belhaq, M. , 2002, “ Quasi-Periodic Solutions and Stability for a Weakly Damped Nonlinear Quasi-Periodic Mathieu Equation,” Nonlinear Dyn., 27(3), pp. 211–236.
Pusenjak, R. R. , and Oblak, M. M. , 2004, “ Incremental Harmonic Balance Method With Multiple Time Variables for Dynamical Systems With Cubic Non-Linearities,” Int. J. Numer. Methods Eng., 59(2), pp. 255–292.
Zhou, B. , Thouverez, F. , and Lenoir, D. , 2015, “ A Variable-Coefficient Harmonic Balance Method for the Prediction of Quasi-Periodic Response in Nonlinear Systems,” Mech. Syst. Signal Process., 64–65, pp. 233–244.
Avramov, K. , Shulzhenko, M. , Borysiuk, O. , and Pierre, C. , 2015, “ Influence of Periodic Excitation on Self-Sustained Vibrations of One Disk Rotors in Arbitrary Length Journals Bearings,” Int. J. Non-Linear Mech., 77, pp. 274–280.
Ni, Q. , Wang, Y. K. , Tang, M. , Luo, Y. Y. , Yan, H. , and Wang, L. , 2015, “ Nonlinear Impacting Oscillations of a Fluid-Conveying Pipe Subjected to Distributed Motion Constraints,” Nonlinear Dyn., 81(1-2), pp. 893–906.
Oberst, S. , and Lai, J. C. S. , 2015, “ Nonlinear Transient and Chaotic Interactions in Disc Brake Squeal,” J. Sound Vib., 342(28), pp. 272–289.
Nayfeh, A. H. , and Mook, D. T. , 1979, Nonlinear Oscillations, Wiley, New York.
Belhaq, M. , and Houssni, M. , 1999, “ Quasi-Periodic Oscillations, Chaos and Suppression of Chaos in a Nonlinear Oscillator Driven by Parametric and External Excitations,” Nonlinear Dyn., 18(1), pp. 1–24.
Summers, J. L. , 1995, “ Variable-Coefficient Harmonic Balance for Periodically Forced Nonlinear Oscillators,” Nonlinear Dyn., 7(1), pp. 11–35.
Schilder, F. , Vogt, W. , Schreiber, S. , and Osinga, H. M. , 2006, “ Fourier Methods for Quasi-Periodic Oscillations,” Int. J. Numer. Methods Eng., 67(5), pp. 629–671.
Wang, L. , 2009, “ A Further Study on the Non-Linear Dynamics of Simply Supported Pipes Conveying Pulsating Fluid,” Int. J. Non-Linear Mech., 44(1), pp. 115–121.
Balachandran, B. , and Nayfeh, A. H. , 1991, “ Observations of Modal Interactions in Resonantly Forced Beam-Mass Structures,” Nonlinear Dyn., 2(2), pp. 77–117.
Raghothama, A. , and Narayanan, S. , 1999, “ Bifurcation and Chaos in Geared Rotor Bearing System by Incremental Harmonic Balance Method,” J. Sound Vib., 226(3), pp. 469–492.
Yang, X. D. , and Chen, L. Q. , 2005, “ Bifurcation and Chaos of an Axially Accelerating Viscoelastic Beam,” Chaos Solitons Fractals, 23(1), pp. 249–258.
Ding, H. , Yan, Q. Y. , and Chen, L. Q. , 2013, “ Chaotic Dynamics in the Forced Nonlinear Vibration of an Axially Accelerating Viscoelastic Beam,” Acta Phys. Sin., 62(20), p. 200502.
Chen, L. H. , Zhang, W. , and Yang, F. H. , 2010, “ Nonlinear Dynamics of Higher-Dimensional System for an Axially Accelerating Viscoelastic Beam With In-Plane and Out-of-Plane Vibrations,” J. Sound Vib., 329(25), pp. 5321–5345.
Lu, L. F. , Wang, Y. F. , Liu, X. R. , and Liu, Y. X. , 2010, “ Delay-Induced Dynamics of an Axially Moving String With Direct Time-Delayed Velocity Feedback,” J. Sound Vib., 329(26), pp. 5434–5451.
Ghayesh, M. H. , 2012, “ Stability and Bifurcations of an Axially Moving Beam With an Intermediate Spring Support,” Nonlinear Dyn., 69(1), pp. 193–210.
Huang, J. L. , Su, R. K. L. , Li, W. H. , and Chen, S. H. , 2011, “ Stability and Bifurcation of an Axially Moving Beam Tuned to Three-to-One Internal Resonances,” J. Sound Vib., 330(3), pp. 471–485.
Farokhi, H. , Ghayesh, M. H. , and Hussain, S. , 2015, “ Three-Dimensional Nonlinear Global Dynamics of Axially Moving Viscoelastic Beams,” ASME J. Vib. Acoust., 138(1), p. 011007.
Lau, S. L. , and Cheung, Y. K. , 1981, “ Amplitude Incremental Variational Principle for Nonlinear Structural Vibrations,” ASME J. Appl. Mech., 48(4), pp. 959–964.
Xu, G. Y. , and Zhu, W. D. , 2010, “ Nonlinear and Time-Varying Dynamics of High-Dimensional Models of a Translating Beam With a Stationary Load Subsystem,” ASME J. Vib. Acoust., 132(6), p. 061012.
Wang, X. F. , and Zhu, W. D. , 2017, “ Dynamic Analysis of an Automotive Belt-Drive System With a Noncircular Sprocket by a Modified Incremental Harmonic Balance Method,” ASME J. Vib. Acoust., 139(1), p. 011009.
Cheung, Y. K. , Chen, S. H. , and Lau, S. L. , 1990, “ Application of the Incremental Harmonic Balance Method to Cubic Non-Linearity Systems,” J. Sound Vib., 140(2), pp. 273–286.
Nayfeh, A. H. , and Balachandran, B. , 1995, Applied Nonlinear Dynamics: Analytical, Computational, and Experimental Methods, Wiley, New York.
Ma, Y. T. , 1997, “ An Accurate Error Analysis Model for Fast Fourier Transform,” IEEE Trans. Signal Processing, 45(6), pp. 1641–1645.
Adireddy, S. , and Prabhu, K. M. M. , 1999, “ A Note on an Accurate Error Analysis Model for Fast Fourier Transform,” IEEE Trans. Signal Process., 47(6), pp. 1743–1744.

Figures

Fig. 1

Schematic of an axially moving beam

Fig. 2

Example spectrum of quasi-periodic motion described by Eq. (24) with nc0=4, ns0=3, and nf=9

Fig. 3

Scenarios depicting Floquet multipliers moving outside the unit circle in the complex plane for different local bifurcations: (a) Hopf bifurcation, and (b) saddle-node bifurcation

Fig. 4

Frequency response curves of periodic solutions when ω̃20≈3ω̃10 and the excitation frequency Ω is close to ω̃10 with f1=0.0055 and μ11=μ22=0.04: (a) Ω/ω̃10∼A110, and (b) Ω/ω̃10∼A220. Solid lines represent stable periodic solutions obtained from the IHB method; dashed and dash-dotted lines represent unstable periodic solutions with a real Floquet multiplier moving outside the unit circle along the +1 direction and a complex conjugate pair of Floquet multipliers moving outside the unit circle, respectively; and denotes numerical integration (NI), denotes a saddle-node bifurcation point, and denotes a Hopf bifurcation point.

Fig. 5

Fourier spectra of the periodic motion of (a) q1 and (b) q2 with Ω/ω̃10=1.23779 from numerical integration (NI) and the IHB method

Fig. 6

Time histories of the quasi-periodic motion with Ω/ω̃10=1.23334 from numerical integration (NI) and the present IHB method with nc0=5 and nf=9: ((a) and (b)) time histories of q1 and q2, respectively; enlarged views of two zones highlighted in (a) and (b) are shown in (c) and (d), respectively

Fig. 7

Phase plane portraits of the quasi-periodic motion and the unstable periodic solution of (a) q1 and (b) q2 with Ω/ω̃10=1.23334 from numerical integration (NI), the IHB method, and the present IHB method with nc0=5 and nf=9

Fig. 8

Fourier spectra of the quasi-periodic motion of (a) q1 and (b) q2 with Ω/ω̃10=1.23334 from numerical integration (NI) and the present IHB method with nc0=5 and nf=9

Fig. 9

Poincaré sections of the quasi-periodic motion of (a) q1 and (b) q2 with Ω/ω̃10=1.23334 from numerical integration (NI) and the present IHB method with different nc0 and nf

Fig. 10

Frequency response curves of periodic motions with f2=0.0055 and μ11=μ22=0.04 when ω̃20≈3ω̃10 and the excitation frequency Ω is near ω̃20 : (a) Ω/ω̃20∼A110, and (b) Ω/ω̃20∼A220. The legends are the same as those in Fig. 4.

Fig. 11

Fourier spectra of the periodic motion of (a) q1 and (b) q2 with Ω/ω̃20=1.17967 from numerical integration (NI) and the IHB method

Fig. 12

Time histories of the quasi-periodic motion with Ω/ω̃20=1.16936 from numerical integration (NI) and the present IHB method with nc0=5 and nf=9: ((a) and (b)) time histories of q1 and q2, respectively; enlarged views of two zones highlighted in (a) and (b) are shown in (c) and (d), respectively

Fig. 13

Phase plane portraits of the quasi-periodic motion of (a) q1 and (b) q2 with Ω/ω̃20=1.16936 from numerical integration (NI) and the present IHB method with nc0=5 and nf=9

Fig. 14

Fourier spectra of the quasi-periodic motion of (a) q1 and (b) q2 with Ω/ω̃20=1.16936 from numerical integration and the present IHB method with nc0=5 and nf=9

Fig. 15

Poincaré sections of the quasi-periodic motion of (a) q1 and (b) q2 with Ω/ω̃20=1.16936 from numerical integration (NI) and the present IHB method with nc0=5 and nf=9

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text:
# zbMATH — the first resource for mathematics

Asymptotic behaviour of a difference equation with complex-valued coefficients. (English) Zbl 1122.39006
Motivated by his previous results concerning the differential equation $$z'= f(t, z)$$, the author investigates the first order difference equation $$\Delta z_n = f(n, z_n)$$, where $$z_n \in \Omega$$, $$\Omega$$ being a simply connected region in $$\mathbb C$$ containing $$z=0$$. It is supposed that the function $$f(n, z)$$ is for every $$n\in \mathbb N$$ closed, in a certain sense, to a function $$h(z)$$ satisfying $$h'(0)\neq 0$$ and $$h(z)=0$$ iff $$z=0$$. Using the function $$h$$, a certain Lyapunov like function $$V$$ and certain subsets in $$\Omega$$ are defined and then used to study asymptotic properties of solutions of $$\Delta z_n = f(n, z_n)$$.
##### MSC:
 39A22 Growth, boundedness, comparison of solutions to difference equations
Full Text:


text:
# Analyticity for Multi-Regge Limits of the Bern-Dixon-Smirnov Amplitudes - High Energy Physics - Theory

Analyticity for Multi-Regge Limits of the Bern-Dixon-Smirnov Amplitudes - High Energy Physics - Theory - Descarga este documento en PDF. Documentación en PDF para descargar gratis. Disponible también para leer online.

Abstract: As a consequence of the AdS-CFT correspondence, planar ${\cal N} =4$ superYang-Mills SUN theory is expected to exhibit stringy behavior and multi-Reggeasymptotic. In this paper we extend our recent investigation to consider issuesof analyticity, a central feature of Regge asymptotics. We contrast flat-spaceopen string theory in the planar limit with the ${\cal N}=4$ super Yang-Millstheory, as represented by the Bern, Dixon and Smirnov \cite{Bern:2005iz} BDSconjecture for n-gluon scattering, believed to be exact for $n=4,5$ andmodified only by a function of cross-ratios for $n\geq 6$. It is emphasizedthat multi-Regge factorization should be applied to trajectories with definitesignature. A variety of analyticity and factorization constraints realized inflat space string theory are not satisfied by the BDS conjecture, at least whenthe exponential factors are truncate in the infra-red regulator below$O\epsilon$.

Autor: Richard C. Brower, Horatiu Nastase, Howard J. Schnitzer, Chung-I Tan

Fuente: https://arxiv.org/


text:
# Solved – Compute quantile function from a mixture of Normal distribution

I have this mixture of normal distribution:

$$X sim frac{1}{2}mathcal{N}(mu_{x_1}=10,,sigma_{x_1}^{2}=1)+frac{1}{2}mathcal{N}(mu_{x_2}=13,,sigma_{x_2}^{2}=1)$$
How can i compute the quantile function of this distributionin `R`?
I have to prove this function:
`qmist2n=function(y,mu1=10,sigma1=1,pi=0.5,mu2=13,sigma2=1){ (1 - pi) * qnorm(y, mu1, sigma1) + pi * qnorm(y, mu2, sigma2) }#qmist2n`

I'm not sure if this code is correct. Thanks for the help in advance!!

Contents

The question has already been discussed here.

Possibly after it's been dealt with in 2011, the following canned solution has become available in `R`:

``library(KScorrect) qmixnorm(.95, c(0,2,5,10), c(1,1,1,1), c(.5,.25,.2,.05)) ``

Here, the first entry is the quantile, followed by the vector of means, standard deviations and weights (see https://rdrr.io/cran/KScorrect/man/dmixnorm.html).

``> qmixnorm(.95, c(0,2,5,10), c(1,1,1,1), c(.5,.25,.2,.05)) [1] 7.745521 ``


text:
## iGreen one year ago Solved

1. iGreen

What are the expected number of flights on time?

2. acxbox22

use proportions

3. mathstudent55

Since 84% of the flights in the random sample are on time, we expect the same percentage to be on time in general. 84% means 84 out of 100.

4. iGreen

Whoa..major lag

5. mathstudent55

$$\dfrac{84}{100} = \dfrac{x}{203}$$

6. Astrophysics

7. iGreen

I came up with .84 * 203 or 170.52

8. Astrophysics

That sounds right

9. mathstudent55

You can think of the proportion as follows: 84 flights out of 100 are on time just like x flights out of 203 are on time.

10. Astrophysics

170.52

11. iGreen

I was trying to say that before anyone else commented..for some reason I can only make one comment and it glitches and I have to reload..

12. mathstudent55

Correct. I'd round it off to 171 because I don't know what 0.52 of a flight is.

13. iGreen

Okay, now it asks: "What is the standard deviation?"

14. iGreen

Any idea on how to find that? @mathstudent55

15. iGreen

16. Astrophysics

Standard deviation is the average difference between any point of data and the mean, been a long time since I did statistics but I'm sure we can figure it out. $\sigma = \sqrt{\frac{ 1 }{ N }\sum_{i=1}^{N}(x_i-\mu)^2}$

17. Astrophysics

But for this the easiest way I see is we can just use $\sigma = \sqrt{npq}$ where $\mu = \sqrt{np}$ that should be easy enough :P

18. iGreen

Ahhh

19. iGreen

I got 5.22

20. Astrophysics

That sounds good!

21. iGreen

$$\sf \sqrt{260 \times 0.84 \times 0.16}$$

22. iGreen

Thanks!

23. Astrophysics

Np


text:
What is the domain and range of the graph f(x)=1/x?

Jun 6, 2018

Both the domain and range are: all real numbers except for zero.

Explanation:

Domain is all the possible x-values that can be plugged in and range is all the possible y-values that can be outputs.

$f \left(x\right) = \frac{1}{x}$ can have any number as an input except for zero.

If we plug in zero for $x$, then we would be dividing by zero which is impossible.

Thus the domain is all real numbers except for zero.

The range is easier to see on the graph:

graph{1/x [-10, 10, -5, 5]}

Since the function goes up forever and down forever vertically, we can say that the range too is all real numbers except for zero.


text:
# Non-commutativity of a limiting sum

1. Jul 16, 2010

### Unit

I read this in the appendix of Hardy's pure mathematics textbook and an explanation was not given:

$$\lim_{x \to 1} \sum_1^\infty \left( x^{n-1}-x^n \right) = \lim_{x \to 1} \left( 1 - x + (x - x^2) + \ldots \right) = \lim_{x \to 1} 1 = 1$$

$$\sum_1^\infty \lim_{x \to 1} \left( x^{n-1}-x^n \right) = \sum_1^\infty \left( 1 - 1 \right) = 0 + 0 + 0 + \ldots = 0$$

Are both answers correct? I thought that associativity only applied to a finite series of operations, which is why I think the first equation is wrong.

2. Jul 16, 2010

### disregardthat

What's going on when you are taking the limit in the first equation? It seems to me that it also should be 0.

Note however that the infinite sums only make sense when |x| <= 1.

Remember that $$\lim_{x \to a } \sum^\infty_{n=1} f_n(x)$$ and $$\sum^\infty_{n=1} \lim_{x \to a } f_n(x)$$ are essentially two different things, and are only equal under certain conditions.

Last edited: Jul 16, 2010
3. Jul 16, 2010

### Unit

$$\lim_{x \to 1} \left( 1 - x + (x - x^2) + \ldots \right) = \lim_{x \to 1} \left( 1 + (-x + x) + (-x^2 + x^2) + \ldots \right) = \lim_{x \to 1} \left( 1 + 0 + 0 + \ldots \right) = 1$$

This is what I meant by my associativity remark.
In the limits, x approaches 1, meaning both $\lim_{x \to 1^+}$ and $\lim_{x \to 1^-}$. Surely at least the "from below" limit is meets the |x| <= 1 condition?

4. Jul 16, 2010

### disregardthat

Note that $$\left( 1 - x + (x - x^2) + \ldots \right)$$ and $$\left( 1 + (-x + x) + (-x^2 + x^2) + \ldots \right)$$ are very different sums! They just happen to converge to the same value under certain conditions. That does not mean they necessarily behave equally. It must be clear what the consecutive terms are in the sum for it to make sense.

Yes, but my point was that in the first case you are taking the limit of a sum which requires that |x| <= 1 in the first place.

Last edited: Jul 16, 2010
5. Jul 16, 2010

### Office_Shredder

Staff Emeritus
To avoid confusion: assume the limit has been removed (it doesn't exist) and instead we are just evaluating the series at x=1 (which does exist)

For infinite sums, associativity is sketchy. In particular, you can only do infinitely many re-arrangements if your sum converges absolutely, otherwise you might be changing the value.

For this particular example, the formal definition of an infinite series is being obscured to confuse you. Formally, an infinite series is a sequence of partial sums

$$S_1,S_2,S_3,...$$ where $$S_i$$ is the sum of the first $$i$$ terms. The series converges to whatever number the partial sums converge to

Writing $$\sum_1^\infty \left( x^{n-1}-x^n \right$$ means that $$S_1=1-x$$ and $$S_2=(1-x)+(x-x^2)$$ and $$S_3=(1-x)+(x-x^2)+(x^2-x^3)$$. In general, $$S_n=1-x^n$$ and this converges to 0 as x goes to 1

The series being used to get a value of 1 (which is not properly written out) is
$$S_1=1$$ and $$S_2=1 + (-x + x)$$ and $$S_3= 1 + (-x + x) + (-x^2+x^2)$$ which of course has $$S_k=1$$ for all values of k.

To summarize, these are two entirely different infinite series. What makes them look similar is the fact that they can be disguised as being the same by looking at a third divergent series
$$S_1=1$$ and $$S_2=1-x$$ and $$S_3=1-x+x$$ and $$S_4=1-x+x-x^2$$ etc. Basically what has happened here is that there is a natural divergent series, and two different subsequences converging to different values were picked out of it. Because we're not very good intuitively at noticing this slight of hand, we assume that the two subsequences should converge to the same value, and are surprised when it doesn't

As a final aside, you should always be careful about whether you can really move a limit from outside of an infinite series to inside of it.

6. Jul 16, 2010

### Unit

Wow, Office_Shredder -- that was exceptionally well put. I understand completely! Thanks a lot!


text:
# If $S$ is subnormal in $G$, $S$ is simple and nonabelian and $S \subseteq H \subseteq G$, then $S \subseteq \textrm{Soc}(H)$.

I suspect I have missed some easier way to show this claim, and there might be a mistake in my approach. I know this is a very lengthy proof, but I spent a lot of time on trying to solve this problem and I have tried to make my solution clear and easy to read, so I would really appreciate if someone checked my approach. Still, a different solution will be appreciated as well.

This is a part of exercise 2A.7. in Isaacs' finite group theory, which is:

If $$S$$ is subnormal in a finite group $$G$$ and $$S$$ is simple and nonabelian, then $$S^G$$ is a minimal normal subgroup of $$G$$.

The hint points out that we should first show that if a subgroup $$H$$ of $$G$$ satisfies $$S \subseteq H$$, then $$S \subseteq \textrm{Soc}(H)$$, and that we should should do this by induction on the order of $$G$$. I can do the rest of this exercise, but I've had some trouble showing this claim mentioned in the hint is true.

My attempt at proving this claim:

1, Suppose $$G = S$$. Then $$S$$ normal in $$G$$, and thus it's a minimal normal subgroup, and so it is contained in the socle of $$G$$, so the claim is true in this case.

2, By induction, suppose the claim is true in any $$\tilde{G}$$ of order less than $$k$$ satisfying the assumptions, and suppose $$G$$ is of order $$k$$ and satisfies the assumptions. Let $$H$$ be a subgroup such that $$S \subseteq H < G$$. Then $$S$$ is subnormal in $$H$$ by Lemma 2.3., and since the order of $$H$$ is lower than the order of $$G$$, we see that $$S$$ is contained in the socle of $$H$$ by the induction hypothesis.

3, It remains to show that $$S \subseteq \textrm{Soc}(G)$$. If $$S$$ is normal in $$G$$, then this is clear, so we can assume it is not.

4, First we will show that $$S$$ is contained in a normal subgroup $$M$$ of $$G$$, and this $$M$$ is a direct product of non-abelian simple groups.

Since $$S$$ is subnormal in $$G$$, we can find a subgroup $$S_1$$ such that $$S$$ is subnormal in $$S_1$$ and $$S_1$$ is normal in $$G$$, and we can assume that $$S_1$$ is strictly smaller than $$G$$. By the induction hypothesis, we see that $$S$$ is contained in the socle of $$S_1$$.

By an previous exercise, we know that the socle of a finite group can be expressed as a direct product of simple groups, so we have $$S \subseteq A_1 \times ... \times A_n$$. If we consider the projection $$\pi_i : A_1 \times ... A_n \rightarrow A_i$$, then we see that $$\pi_i (S)$$ is a subgroup of $$A_i$$, and because it's an homomorphic image of $$S$$, it is either $$S$$ or $$1$$, because $$S$$ is simple. It follows that if $$A_i$$ is an abelian simple group, then $$\pi_i(S)$$ must be $$1$$, as otherwise $$A_i$$ would contain the nonabelian subgroup $$S$$. Thus, we can conclude that $$S$$ is contained in some $$A_{l_1} \times ... \times A_{l_m} \subseteq A_1 \times ... \times A_n = \textrm{Soc}(S_1)$$, where $$A_{l_j}$$ are non-abelian simple groups.

Notice that the socle of $$S_1$$ is normal in $$G$$, because it is characteristic in $$S_1$$ and $$S_1$$ is normal in $$G$$. Consider the subgroup $$M \subseteq \textrm{Soc}(S_1)$$ that is generated by all non-abelian simple subgroups that are normal in $$\textrm{Soc}(S_1)$$. This subgroup is characteristic in $$\textrm{Soc}(S_1)$$ by definition, it contains $$A_{l_1} \times ... \times A_{l_m}$$, and it is also expressible as a direct product of simple non-abelian groups. It follows that $$S \subseteq A_{l_1} \times ... \times A_{l_m} \subseteq M_1 \times ... M_r = M$$ (in fact, $$A_{l_1} \times ... \times A_{l_m}= M$$ because by the same argument as previously, the projection of $$M$$ to $$A_i$$ is $$1$$ for all $$A_i$$ abelian, but that isn't necessary to show), where $$M_i$$ are simple, non-abelian, normal subgroups of $$\textrm{Soc}(S_1)$$.

5, We will now show that $$M \subseteq \textrm{Soc}(G)$$, which also shows that $$S\subseteq \textrm{Soc}(G)$$.

5.1, Suppose that some $$M_i$$ aren't contained in $$\textrm{Soc}(S_1)$$. Then each such $$M_i$$ satisfies $$M_i \cap \textrm{Soc}(G) = 1$$, because $$M_i \cap \textrm{Soc}(G)$$ is normal in $$M_i$$ (because $$\textrm{Soc}(G)$$ is normal in $$G$$), so the intersection must be $$1$$ or $$M_i$$, as $$M_i$$ is simple. Take the subgroup $$L=M_{k_1} \times ... \times M_{k_t} \subseteq M$$ of all such $$M_i$$. Then $$L \cap \textrm{Soc}(G) = 1$$, because the this intersection is a normal subgroup of $$L$$, its intersection with each $$M_{k_i}$$ is trivial, and thus all the projections of $$L \cap \textrm{Soc}(G)$$ into $$M_{k_i}$$ are contained in the center of $$M_{k_i}$$, which is trivial. For more detail, see

Why is a normal subgroup of $$G_1\times G_2$$ with trivial intersections with $$G_1$$ and $$G_2$$ is abelian?

5.2 We have shown that $$L = M_{k_1} \times .. \times M_{k_t} \subseteq M$$ has a trivial intersection with $$\textrm{Soc}(G)$$. We will now show that $$L$$ is normal, which is a contradiction - it contains a minimal normal subgroup that has a trivial intersection with $$\textrm{Soc}(G)$$, which is a contradiction.

Consider an element $$g \in G$$. The subgroup $$M$$ is normal in $$G$$, and so conjugation of $$M$$ by $$g$$ can be expressed as automorphism $$\varphi : M \rightarrow M$$. We will show that $$\varphi$$ permutes the subgroups $$M_i$$. An automorphism preserves normality, so $$\varphi(M_i)$$ is a normal, simple subgroup of $$M$$, and all $$M_i$$ are normal and simple in $$M$$ too, so either $$\varphi(M_i) \cap M_j =1$$, or $$\varphi(M_i) = M_j$$. If $$\varphi(M_i) = M_j$$ for some $$j$$, we are done, so we will suppose $$\varphi(M_i) \cap M_j =1$$ for all $$j$$ and derive a contradiction. The subgroup $$\varphi(M_i)$$ is a normal subgroup of $$M$$, it has a trivial intersection with all $$M_i$$, and so it is contained in the direct product of the centers of $$M_i$$ - for more detail, again see

Why is a normal subgroup of $$G_1\times G_2$$ with trivial intersections with $$G_1$$ and $$G_2$$ is abelian?

i.e. $$\varphi(M_i) \subseteq Z(M_1) \times ... \times Z(M_n)$$. But $$M_i$$ are all non-abelian and simple, so $$\varphi(M_i)$$ is the trivial subgroup, which is a contradiction. Thus, we see that $$\varphi(M_i) = M_j$$ for some $$j$$.

A nice, simple example to give some intuition behind this claim - for a non-abelian simple group $$Q$$, if we take the diagonal injection $$q \mapsto (q,q)$$ of $$Q$$ into $$Q \times Q$$, then this subgroup is not normal in $$Q \times Q$$.

5.3, Now since $$\textrm{Soc}(G)$$ is normal, we see that if $$M_i \subseteq \textrm{Soc}(G)$$, then $$\varphi(M_i) =(M_i)^g \subseteq \textrm{Soc}(G)$$. It follows that if we consider the group $$L$$ described in 5.1, $$L=M_{k_1} \times ... \times M_{k_t}$$ where $$M_{k_i}$$ are all such $$M_i$$ in $$M=M_1 \times ... \times M_n$$ that are not contained in $$\textrm{Soc}(G)$$, then by the previous sentence we see that $$\varphi(M_{k_i}) =(M_{k_i})^g =M_{k_j} \subseteq L$$, so $$\varphi(L) = L$$, and thus $$L$$ is normal in $$G$$.

This shows that $$L$$ is a non-trivial, normal subgroup of $$G$$, that has a trivial intersection with $$\textrm{Soc}(G)$$, which is a contradiction. Thus, there cannot be any $$M_i$$ in $$M$$ that is not contained in $$\textrm{Soc}(G)$$, and so $$M \subseteq \textrm{Soc}(G)$$. Since $$S \subseteq M$$, it follows that $$S \subseteq \textrm{Soc}(G)$$, which finishes the proof.

• In 4, after your consideration of projections $\pi_i$, I wonder how to derive that "if $A_i$ is abelian simple, then $\pi_i(S)=1$", as $\pi_i(S)=A_i$ seems not to be able to get $A_i\supseteq S$ Apr 7 '19 at 16:10
• @WembleyInter I haven't been doing any math for the last half year or so now, and I also haven't seen this question for around that time, so I'm sorry if this won't make sense: $\pi_i(S)$ is isomorphic to either $1$ or $S$, and it's contained in $A_i$. If it were isomorphic to $S$, then we have that $A_i$ contains $\pi_i(S) \simeq S$, but $S$ is non-abelian, which is a contradiction, since $A_i$ is abelian. Apr 7 '19 at 19:32
• You're right. I was a little bit fool. Thanks! Apr 9 '19 at 9:57

I think it might be easier to prove the whole thing by induction on $$|G|$$. We prove that for any subnormal nonabelian simple subgroup $$S$$ of $$G$$, $$S^G$$ is a minimal normal subgroup of $$G$$ and is the direct product of the distinct conjugates of $$S$$ in $$G$$.

So suppose $$S \lhd \lhd H \lhd G$$, and let $$K = S^H$$. Then by induction $$K = S_1 \times \cdots \times S_r$$ is minimal normal in $$H$$, where the $$S_i$$ are the distinct conjugates of $$S$$ in $$H$$.

Now let $$K = K_1,K_2,\ldots,K_t$$ be the distinct conjugates of $$K$$ in $$G$$. These are all minimal normal in $$H$$, disjoint, and have trivial centres, so they generate their direct product, and hence $$S^G = K^G = K_1 \times \cdots \times K_t$$.

The critical fact that you need is that the only normal subgroups of a direct product of nonabelian simple groups $$T_i$$ are the direct products of some of the direct factors. To prove that, consider the projection of a normal subgroup $$N$$ onto one of the factors $$T_i$$. This is either trivial or the whole $$T_i$$, and in the second case, we get $$[N,T_i] =T_i \le N$$.

So we see now that $$S^G = K_1 \times \cdots \times K_t$$ is the direct product of the distinct conjugates of $$S$$ in $$G$$ and, since the only normal subgroups of $$S^G$$ are direct products of some of these conjugates, it follows that $$S ^G$$ is a minimal normal subgroup of $$G$$. This also proves the claim in the hint that $$S \le {\rm Soc}(G)$$.

Answers to questions: For any two groups $$G$$ and $$H$$, we have $$Z(G \times H) = Z(G) \times Z(H)$$. Nonabelian simple groups have trivial centres, and hence so do direct products of nonabelian simple groups. So $$Z(K_i) = 1$$.

Yes, you are correct in saying that the fact that $$\langle K^G \rangle = K_1 \times \cdots \times K_t$$ requires $$Z(K_i) = 1$$, and your proof of this is also correct. This is not true in general for abelian simple groups.

• Thank you for the answer. I see the approach is quite similar - you use the property of a direct product of simple non-abelian groups you described. One thing I don't understand - in the third paragraph you mention that these groups $K_i$ have trivial centres. I don't see whether that's somehow obvious - I see it as a consequence of the property in the following paragraph. Second, we can take the direct product of $K_i$ in $H$, just from the fact that they are all minimal normal in $H$ and again using the property in the following paragraph. Oct 14 '18 at 4:26
• Lastly, in what way is the fact that these groups are centerless used? Does $Z(K_i) =1$, $K_i$ normal in $H$ and $K_i, K_j$ pairwise disjoint imply that $K_1 \cap \langle K_2,...,K_n \rangle =1$, and thus we can take the direct product of these groups? I can see this is true using induction and the fact that normal subgroups of $K_2 \times ... \times K_n$ that have trivial intersection pairwise with all $K_j$, $j>2$ are contained in the direct product $Z(K_2) \times ... \times Z(K_n)$, but I might be missing a simpler way to show this? Oct 14 '18 at 4:38
• I have edited the answer and tried to answer your questions. Oct 14 '18 at 11:45


text:
Conference

### The Impact of Galaxy Formation on the Diffuse Background Radiation

Abstract:

The far infrared background is a sink for the hidden aspects of galaxy formation. At optical wavelengths, ellipticals and spheroids are old, even at $z \sim 1.$ Neither the luminous formation phase nor their early evolution is seen in the visible. We infer that ellipticals and, more generally, most spheroids must have formed in dust-shrouded starbursts. In this article, we show how separate tracking of disk and spheroid star formation enables us to infer that disks dominate near the peak in t...

Publication status:
Published

### Authors

Devriendt, J More by this author
Issue:
204
Pages:
423-435
Publication date:
2000-10-23
ISSN:
0074-1809
URN:
uuid:00060eaa-8257-4fb8-b0d8-0f279ca6f4a6
Source identifiers:
6064
Local pid:
pubs:6064
ISBN:
1-58381-062-5
Keywords: