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I would like to customize Ubuntu live CD by installing some additional packages. I have followed   but it doesn't seems to work. Can anyone provide clear instructions?   Note: I do not prefer Remastersys, manual way will be appreciated.  Customization  Packages that I want to install:   Thunderbird    Samba    SSH       Changes that I need:   Remove Games menu from the Application menu Firefox shortcut on Desktop Radiance as the default Theme Different default Ubuntu Wallpaper   Configuration file changes   I want the panel to be placed at the bottom I want to paste my Samba configuration file instead of default Samba configuration I have few Firefox shortcuts and folders I would like to show that in Desktop Also it will be nice if you say me how to change the icon sets   Recent Updates   I have customized Ubuntu 10.10 with Firefox shortcuts and few folders on desktops. Everything went smooth. But the installer gets crashes after choosing the timezone. How do i fix this issue? Also setting wallpaper affects the login screen. The wallpaper which i set is displayed on the login screen also. I just want the default one for the login screen. Divisibility Proof with Induction - Stuck on Induction Step I'm working on a problem that's given me the run around for about a weekend. The statement:  For all $m$ greater than or equal to $2$ and for all $n$ greater than or equal to $0$, $m - 1$ divides $m^n - 1$.  Using induction over $n$, the base step comes easy since $m^n - 1$ is $0$ when $n = 0$.  My induction hypothesis is letting $k \geq 0$ and assuming that $m - 1$ divides $m^k - 1$. In order to show that $m - 1$ divides $m^{k+1} - 1$, I obviously need to use the induction hypothesis. However, no matter where I try to use the fact that $m^k - 1 = (m-1)a$ for some $a$ in the integers, the expression $m^{k+1} - 1$ always becomes more difficult to get to $m^{k+1} - 1$ being equal to $(m-1)b$ for some $b$ in the integers.   In other words, I can't figure out any actually helpful way to apply the induction hypothesis with the goal of proving the next step.  Any help would be appreciated! Clone Method not working I am trying to Clone a lead when certain condition is arrived using before trigger but i am not able to see cloned record.   for( Lead ld : Trigger.new ) {                          if(ld.Integration_Source__c == 'Hubspot' &amp;&amp; Trigger.isUpdate){         Lead LeadBeforeUpd = Trigger.oldMap.get( ld.Id );         if(ld.Program__c != LeadBeforeUpd.Program__c ){            Lead ClonedLead = new Lead();            ClonedLead = LeadBeforeUpd.clone(false, false, false, false);         }     }  } How do you sort du output by size? How do the bonus perks from the Advanced Warfare Center work? Modular exponentiation by repeated squaring Copy files from a list to a folder Connect iPhone to Mac Without Charging I am developing iOS apps and I hate, that always when I connect my iPhone to my Mac to install and test an app, the iPhone gets charged. Is there any possibility to connect my iPhone to my Mac without charging the phone?    My Environment: -  iPhone 4S (No Jailbreak) -  MacBook Air3,2 (Late 2010) LibVLC.NET in GTK# NULL values inside NOT IN clause Is the following scheme for generating $p_n=(1/3)^n$ stable or not. $p_n=(5/6)p_{n-1}-(1/6)p_{n-2}$. I have a problem with the endfloat package and unicode characters.  When I compile the following, everything works as expected:  \documentclass{article} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage{graphicx}  \begin{document}  \begin{figure} \caption{Ö} \end{figure}  \end{document}   However, when I also load the endfloat package (\usepackage{endfloat}), I get the following error message:  ! Package inputenc Error: Unicode char \u8:ÖÖ not set up for use with LaTeX.   This could be a bug, but my suspicion is that I'm making a mistake somewhere.  I'm hardly the first person to use special characters with the endfloat package.  However, the above example is so simple, I don't see where I could have made a mistake.  Any advice is appreciated.  Edit: The file is of course encoded using utf-8. Give an example of a relation R on $A^2$ which is reflexive, symmetric, and not transitive I am just looking for some clarification on this exercise:  Let $A = \{a,b,c,d\}$. Give an example of a relation $R$ on $A^2$ which is reflexive, symmetric, and not transitive.  I understand that if I was only dealing with $A$ and and not $A^2$ I could give an answer similar to:  $R = \{(a,a),(a,b),(a,d),(b,a),(b,b),(c,c),(d,a),(d,d)\}$  Here $R$ is reflexive, symmetric, and not transitive.  What is confusing me is the fact that we have $A^2$. I am not sure what I am missing here. I am looking through my book to see if I can figure this out, but if someone could help shed some light on this it would be appreciated.  Thanks,  Tony If I have bought X tickets, do I have X hand baggage allowances? If I have bought 3 tickets for the same flight, can I bring three times the cabin allowance allowed per ticket?  The way I see it I am doing the airline a favor  Related: There are two directories shown by 'ls'.  Normally directories anywhere are blue on black background.  But the first one is blue on green and impossible to read.  Why is this?  How to make it blue on black, or at least something light on something dark?  This is on Ubuntu 12.04, using bash in Gnome Terminal.  In Konsole, the blue is slightly darker, and possible to read, though could be way better. What would you call a person who doesn't want to learn anything new? Prove the identity $\binom{2n+1}{0} + \binom{2n+1}{1} + \cdots + \binom{2n+1}{n} = 4^n$ I've worked out a proof, but I was wondering about alternate, possibly more elegant ways to prove the statement. This is my (hopefully correct) proof:  Starting from the identity $2^m = \sum_{k=0}^m \binom{m}{k}$ (easily derived from the binomial theorem), with $m = 2n$:  $2^{2n} = 4^n = \binom{2n}{0} + \binom{2n}{1} + \cdots + \binom{2n}{2n-1} + \binom{2n}{2n}$  Applying the property $\binom{m}{k} = \binom{m}{m-k}$ to the second half of the list of summands in RHS above:  $4^n = \binom{2n}{0} + \binom{2n}{1} + \cdots + \binom{2n}{n-1} + \binom{2n}{n} + \underbrace{\binom{2n}{n-1} +\cdots \binom{2n}{1} + \binom{2n}{0}}_{\binom{m}{k} = \binom{m}{m-k} \text{ has been applied}}$  Rearranging the above sum by alternately taking terms from the front and end of the summand list in RHS above (and introducing the term $\binom{2n}{-1} = 0$ at the beginning just to make explicit the pattern being developed):  $4^n = (\binom{2n}{-1} + \binom{2n}{0}) + (\binom{2n}{0} + \binom{2n}{1}) + \cdots + (\binom{2n}{n-1} + \binom{2n}{n})$  Finally, using the property $\binom{m}{k} + \binom{m}{k-1} = \binom{m+1}{k}$ on the paired summands, we get the desired result:  $4^n = \binom{2n+1}{0} + \binom{2n+1}{1} + \cdots + \binom{2n+1}{n}$ Does Python have a ternary conditional operator? If Python does not have a ternary conditional operator, is it possible to simulate one using other language constructs? Elegant way of obtaining the envelope of oscillating function Comments must be at least 15 characters long, which is supposed to make comments like &quot;...&quot; impossible to submit. This works fine, however, there is a way around this, by using comment replies, even &quot;fake&quot; ones e.g. &quot;@somelongname ...&quot;. Can the system strip the comment reply part (e.g. &quot;@somelongname&quot;) before performing the minimum characters check?