Every single mirror is missing the Quantal dists.  NOTICE:  Before you comment, I do realize 12.10 is in End of Life status!  But to totally restrict further usage?  Unacceptable!
Recently I have installed an older version of Ubuntu on my old machine. Whenever I try to install any software, I get an error saying it couldn't be found:  $ sudo apt-get install vlc Reading package lists... Done                Building dependency tree        Reading state information... Done     E: Couldn't find package vlc
Copy/duplicate database without using mysqldump 15 year mortgage vs 30 year paid off in 15 I've found similar questions, but they all seem to focus on a preexisting 30 year mortgage and whether or not it makes sense to refinance to a 15.  My question is a bit different.  I have not yet bought a house but I am trying to decide which mortgage to go with.  As a result, I don't have to consider refinance fees and a change in interest rate.   The situation is that I can afford the payment on a 15 year mortgage and would prefer not to pay off a house for the next 30 years.  So I was going to go with a 15.  However, in looking at some mortgage calculators (which it is possible I am looking at incorrectly) it seems that there is no difference in getting a 30 year mortgage and just paying it off in 15 years.  The calculators seem to say that--assuming the same interest rate--you will pay the exact same interest over 15 years.  As a result, my thinking is, I can get the 30 and pay it like a 15.  If there really is no drawback then this gives me the ability to pay the normal 30 year payment if needed due to financial hardship (loss of employment etc.)  So my question is, am I missing something?  If not, why would anyone ever get a 15 year instead of just paying off a 30 year in 15 years?  In summary: Does it really cost the same amount of money overall to pay off a 15 year mortgage vs paying off a 30 year mortgage in 15 years?  PS: Please don't suggest taking the 30 year and investing the extra money.  I know this is an option and the one many would suggest.  However, for the purposes of this question please only consider 15 year vs 30 year paid off in 15. How to edit a file in vi editor (arrow keys prints letters)? I am open a file in vi:  vi index.php   Here the problem is: when I press the down arrow and it prints 'B'. Why it prints 'B'? Stupidly, I had been using a condition like this as part of a script:  if [ $(ls FOO* 2&gt; /dev/null) ] # if files named "FOO*" were downloaded  then     echo "Files found"      # ... process and email results else     echo "Not found"     # ... email warning that no files were found (against expectations) fi   That works for zero and one files named FOO*, but fails if there are more than one. From logs I found several different error messages stemming from this:  [: FOO_20131107_082920: unary operator expected [: FOO_20131108_070203: binary operator expected [: too many arguments   My question is: what is the correct way to check, in a Bash if condition, whether one or more files whose name begins with FOO exist?  GNU bash, version 4.2.25(1)-release (x86_64-pc-linux-gnu) Finding the convergence interval of $\sum_{n=0}^\infty\frac{n!x^n}{n^n}$. I want to find the convergence interval of the infinite series $\sum\limits_{n=0}^\infty \dfrac{n!x^n}{n^n}$. I will use the ratio test: if I call $u_n = \dfrac{n!x^n}{n^n}$, the ratio test says that, if the following is true for some values of $x$, the series will be convergent for these values of $x$: $$\lim_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right|&lt;1$$ So, I will first calculate the value of $\left|\dfrac{u_{n+1}}{u_n}\right|$: $$\left|\dfrac{\dfrac{(n+1)!x^{n+1}}{(n+1)^{n+1}}}{\dfrac{n!x^n}{n^n}}\right|=\dfrac{(n+1)!|x|^{n+1}}{(n+1)^{n+1}}\times\dfrac{n^n}{n!|x|^n}=\frac{(n+1)n^n|x|}{(n+1)^{n+1}}=|x|\left(\frac{n}{n+1}\right)^n$$ So, $\lim\limits_{n\to+\infty}\left|\dfrac{u_{n+1}}{u_n}\right|$ becomes: $$\lim_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right|=\lim_{n\to+\infty}|x|\left(\frac{n}{n+1}\right)^n=|x|\lim_{n\to+\infty}\left(\frac{n}{n+1}\right)^n$$ Now I must evaluate the value of $\lim\limits_{n\to+\infty}\left(\dfrac{n}{n+1}\right)^n$. For this, let $y = \left(\dfrac{n}{n+1}\right)^n$; so, instead of calculating $\lim\limits_{n\to+\infty}y$, I will first calculate $\lim\limits_{n\to+\infty}\ln y$: $$\lim_{n\to+\infty}\ln y=\lim_{n\to+\infty}\ln \left(\dfrac{n}{n+1}\right)^n=\lim_{n\to+\infty}n\ln\left(\frac{n}{n+1}\right) =\lim_{n\to+\infty}\frac{\ln\left(\frac{n}{n+1}\right)}{\frac{1}{n}}$$ Applying L'Hôpital's rule: $$\lim_{n\to+\infty}\frac{\ln\left(\frac{n}{n+1}\right)}{\frac{1}{n}} =\lim_{n\to+\infty}\frac{\frac{1}{n(n+1)}}{-\frac{1}{n^2}} =\lim_{n\to+\infty}\left(-\frac{n}{n+1}\right)=-1$$ Now, since we know that $\lim\limits_{n\to+\infty}\ln y = -1$, we have that: $$\lim_{n\to+\infty}y=\lim_{n\to+\infty}e^{\ln y} = e^{-1} = \frac{1}{e}$$. Substituting this back into the expression $\lim\limits_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right| = |x|\lim\limits_{n\to+\infty}\left(\frac{n}{n+1}\right)^n$, we have that the limit of $\left|\dfrac{u_{n+1}}{u_n}\right|$ as $n\to+\infty$ is: $$\lim_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right|=\frac{|x|}{e}$$ Therefore, the series will certainly be convergent for the values of $x$ for which $\dfrac{|x|}{e}&lt;1$, that is, $|x|&lt;e$. So, I know that the series is convergent for $-e &lt; x &lt; e$, but I have to test whether the series is convergent at $x = e$ or $x = -e$. That is, I have to test whether $\sum\limits_{n=0}^{\infty} \dfrac{n!e^n}{n^n}$ and $\sum\limits_{n=0}^{\infty} \dfrac{(-1)^nn!e^n}{n^n}$ are convergent. Since these limits don't approach zero, I know they are both divergent, but I'm not sure how to find the limits, because of the factorial function. Also, I can't use integral test here, because of the factorial. Probably I should use comparison test, but I haven't found any divergent series to which to compare it. Any hints? Thank you in advance.  Edit: Using the suggestion by Ragib Zaman in the answer below, since the Taylor polynomial $P_n(x)$ of $e^x$ at $a=0$ is $$e^x = 1 + x + \dfrac{x^2}{2!} + \cdots + \dfrac{x^n}{n!}+\cdots,$$ if we substitute $n$ for $x$ we see that $e^n&gt;\dfrac{n^n}{n!}$; therefore, $\dfrac{n!e^n}{n^n} &gt; 1$, and, thus, we show that $\sum\limits_{n=0}^{\infty} \dfrac{n!e^n}{n^n}$ is divergent, because its term doesn't approach zero. $\sum\limits_{n=0}^{\infty} \dfrac{(-1)^nn!e^n}{n^n}$ is also divergent, because, although the absolute value of the ratio between two successive terms, $|e|\left(\frac{n}{n+1}\right)^n$, approaches 1 as $n\to\infty$, it approaches 1 from values bigger than 1; therefore, the absolute value of a term is always greater than the absolute value of the previous term. Is there anything to do during an 8 hour after-midnight layover in Dubai? We are planning a trip with an 8 hour layover in Dubai spanning from 1am-9am. I am thinking about going to see the city, but is it worth it to do it given the very late night hours? Will at least the outside attractions accessible?   Will Burj Khalifa be visible in the middle of the night?  Will at least some of the malls be open?  Will there be anything to do at night?  I have seen other questions about layovers in Dubai (), but none of them is asking about a night-only stay. Storable Action - System.LimitException: Too many DML statements: 1 @AuraEnabled(cacheable=true) public static String  deleteContacts(List&lt;String&gt; contactIds) {     List&lt;Contact&gt; returnList = new List&lt;Contact&gt; ();     String query = ' SELECT Id,Name, LastName,Department,MobilePhone, Email FROM Contact WHERE id IN : contactIds ';     for (Contact thisContact:  Database.Query(query)) {          returnList.add(thisContact);     }     try {       delete returnList;       return'deleted successfully';     }     catch(Exception ex){        return 'Problem occoured';     }  }  JS Method deleteSelected(){     deleteContacts({         contactIds :this.selectedIds     })      .then(result =&gt; {         this.dispatchEvent(             new ShowToastEvent({                 title: 'Success',                 message: result,                 variant: 'success',             }),         );         return refreshApex(this.getContactList, { searchKeyWord: '$searchKeyWord' });     })     .catch(error =&gt; {         this.error = error;     }); I know that I can make certain blocks only placeable on a specified block, but what is the command to give an item that can only BREAK certain blocks? How do we see? Where do the photons disappear? Uncaught TypeError: 'set' on proxy: trap returned falsish for property Name Am not doing anything fancy, just trying to update a field on Contact in JS of LWC but getting this exception.  Uncaught TypeError: 'set' on proxy: trap returned falsish for property 'Name' throws at mydomain/auraFW/javascript/mhontaYdOya4Y_lBu7v9yg/aura_prod.js:2:27687  HTML Code:   &lt;template&gt;      &lt;template if:true={wiredContact}&gt;          {wiredContact.Name}          &lt;lightning-input value={wiredContact.Name} onchange={updateName}&gt;&lt;/lightning-input&gt;     &lt;/template&gt;  &lt;/template&gt;   JS:   import { LightningElement ,wire,track,api } from 'lwc'; import myContact from "@salesforce/apex/ContactController.fetchContact";  export default class Myrefreshapextest extends LightningElement {    @track wiredContact;   @wire (myContact)         fetchedContact({error, data}){             if(data){                 console.log(JSON.stringify(data));                 this.wiredContact = data;             }else if (error){                 console.log(error);             }     }      updateName (event){         console.log(JSON.stringify(event.detail.value));         console.log(JSON.stringify(this.wiredContact));         this.wiredContact.Name = event.detail.value;     }  }   Apex:  public class ContactController {      @AuraEnabled(cacheable=true)     public static Contact fetchContact(){         return [SELECT Id,Name FROM COntact LIMIT  1];     } }   On top of my head, am not doing anything wrong, anyone has idea what's wrong with my code?  When I print console.log(JSON.stringify(this.wiredContact)); I get old values so am pretty sure it exists.  I tried with @track and @api, but same response. Can anyone shed some light? Knowing $|z|$ and $|w|$ (and possibly $z\overline{w}$), what are $|z+w|^2$, $|zw|^2$, $|z-w|^2$, $|z/w|^2$? Question 1: Let $z$ and $w$ be complex numbers satisfying $|z| = 4$ and $|w| = 2$. Then enter in the numbers $$ |z+w|^2, |zw|^2, |z-w|^2, \left| \dfrac{z}{w} \right|^2 $$ below, in the order listed above. If any of these cannot be uniquely determined from the information given, enter in a question mark.  I plugged in $4$ and $2$ to each equation and got $36$, $64$, $4$, $4$, for question one. It was wrong.  Question 2: Let $z$ and $w$ be complex numbers satisfying $|z| = 5, |w| = 2,$ and $z\overline{w} = 6+8i.$ Then enter in the numbers $$ |z+w|^2, |zw|^2, |z-w|^2, \left| \dfrac{z}{w} \right|^2 $$ below, in the order listed above. If any of these cannot be uniquely determined from the information given, enter in a question mark.  I don't know how to even start on question 2 because of the $z\overline{w}$. How do you prove that $e=\sum_{n=0}^{\infty}\frac{1}{n!}$? Here I am assuming $e:=\lim_{n\to\infty}(1+\frac{1}{n})^n$. Do you have any good PDF file or booklet available online on this? I do not like how my analysis text handles this... I'm getting the following error when trying to update a taxonomy field value via REST api:  {   "error": {     "code": "-1, Microsoft.SharePoint.Client.ClientServiceException",     "message": {       "lang": "en-US",       "value": "Cannot deserialize data for type Microsoft.SharePoint.Taxonomy.TaxonomyFieldValueCollection."     }   } }   The request:  https://server/_api/web/GetFileByServerRelativeUrl('/Shared Documents/file.txt')/ListItemAllFields  {   "__metadata": {     "type": "SP.Data.Shared_x0020_DocumentsItem"   },   "Title": "some title",   "Project": {     "results": [       {         "Label": "Term1",         "TermGuid": "897bdf7f-08d6-44b9-a71a-ddc4cf131664",         "WssId": 5       }     ]   } } I was wondering if there is a simpler way to find the last to digits of a power such as $3^{45}$. I reduced it modulo 100 to get the answer, which is 43. But I was curious if there was a simpler, or more eloquent way to get this result.  Here's my method:  Using the fact that $3^4=81\equiv -19 \pmod{100}$ and $19^5=2,476,099\equiv -1 \pmod{100}$  $$ 3^{45}\equiv 3(3^4)^{11}\equiv 3(-19)^{11}\equiv 3(-19)(19^5)^2\equiv 3(-19)(-1)^2 \equiv -57 \equiv 43 \pmod {100} $$  I just felt like this was sort of a messy way to find out that the last two digits are 43. Could the granting lag for each badge be added to the FAQ? How to show $(a^b)^c=a^{bc}$ for arbitrary cardinal numbers? One of the basic (and frequently used) properties of cardinal exponentiation is that $(a^b)^c=a^{bc}$.  What is the proof of this fact?  As Arturo pointed out in his comment, in computer science this is called . How to create UEFI-only bootable USB live media? $$ z = 1+2i \ (complex \ number) \\ z^n = a_n + b_ni \ \ \ (a_n, b_n \in \mathbb{Z}, n \in \mathbb{N}^*) $$ Prove that $ b_{n+2} - 2b_{n+1} + 5b_n = 0$  How can I solve this? Thank you!  EDIT: Or please tell me your ideas. How to prove by induction that $\frac{a^n+b^n}{2}\geq\left(\frac{a+b}{2}\right)^n$? I'm about to prove that for any $a,b&gt;0$ and $n\in\mathbb{N},$ the inequality: $\frac{a^n+b^n}{2}\geq\left(\frac{a+b}{2}\right)^n$ holds.   By induction I get: $$\left(\frac{a+b}{2}\right)\cdot\frac{a^n+b^n}{2}$$ $$=\frac{1}{2}\cdot\frac{a^{n+1}+b^{n+1}+ab^n+ba^n}{2}$$ $$=\frac{1}{2}\left(\frac{a^{n+1}+b^{n+1}}{2}+\frac{ab^n+ba^n}{2}\right).$$  Now I have to prove that $$\frac{1}{2}\left(\frac{a^{n+1}+b^{n+1}}{2}+\frac{ab^n+ba^n}{2}\right)$$ $$\leq\frac{a^{n+1}+b^{n+1}}{2}$$ $$\Leftrightarrow a^{n+1}+b^{n+1}\geq ab^n +ba^n.$$ But I don't know how to prove that one, is it possible to be proved by induction ? Thank you. dist$(x,A)=0$ if and only if $x\in \overline{A}$ (closure of $A$) Let $(X,d)$ be  a metric space and non-empty $A\subset X$. The distance from $x$ to $A$ is defined as $$\mbox{dist}(x,A)=\inf\{d(x,a):a\in A\}$$  I want to show that dist$(x,A)=0$ if and only if $x$ is in the closure of $A$.     (=>): dist$(x,A)=0$ implies $\inf \{d(x,a):a\in A\}=0$. Suppose $x\not\in \overline{A}$, then $x\in X\setminus \overline{A}$, which is an open set. Thus $\exists\epsilon &gt;0$ such that punctured open ball $B(x;\epsilon)\setminus\{x\}= \{y\in X:0&lt;d(x,y)&lt;\epsilon\} \subset X\setminus\overline{A}$. But then dist$(x,A)=\inf\{d(x,a):a\in A\}\ge \inf\{d(x,y):y\in B(x,\epsilon)\setminus\{x\}\}&gt;0$. This is a contradiction! Hence, $x\in \overline{A}$.      (&lt;=): Suppose $x\in \overline{A}$. If $x\in A^\circ$ (the interior of $A$) then $\inf\{d(x,a):a\in A\}=d(x,x)=0$.   If $x\not\in A^\circ$ then $\exists $ a sequence $(a_k)_k\subset A$ such that $a_k$ converges to $x$. Then, $\forall \epsilon &gt; 0, \exists N\in\mathbb{N}$ such that $k\ge N\implies d(a_k,x)&lt;\epsilon$.      So that dist$(x,A)=\inf\{d(x,a_k)\}=\lim\limits_{k\to\infty} d(x,a_n) = 0$.   Please let me know if my proof is rigorous / correct. I'm not completely sure, think it's better to check.