I am unsuccessfully trying to implement biblatex in my project, so that I can cite other content. I get an error that looks loke this:  Undefined control sequence. &lt;argument&gt; \mkbibdateapalongextra                                    {year}{month}{day}\iffieldundef {endyear}{... l.15     \printbibliography[heading=bibempty]                                               The control sequence at the end of the top line of your error message was never \def'ed. If you have misspelled it (e.g., `\hobx'), type `I' and the correct spelling (e.g., `I\hbox'). Otherwise just continue, and I'll forget about whatever was undefined.   Can you help me to solve the problem? A minimum code example would look like this:  \documentclass{article} \usepackage[utf8]{inputenc} \usepackage{hyperref} \usepackage[style=apa,citestyle=numeric,sorting=nyt,sortcites=true,autopunct=true,hyperref=true,abbreviate=false,backref=true,backend=biber]{biblatex} \addbibresource{bibliography.bib} % BibTeX bibliography file \defbibheading{bibempty}{}  \begin{document}     My text with a cite\cite{1_nielsen_2011}.      \printbibliography[heading=bibempty] \end{document}   And the bibliography.bib file has this content:  @misc{1_nielsen_2011, author={Nielsen, Oskar Herrik}, title={Oldmoney.dk | Din genvej til gamle penge}, url={http://www.oldmoney.dk/}, urldate={2015-9-11}, journal={Oldmoney.dk}, year={2011} }
I'm having trouble getting basic APA bibliographies to work. The same thing works with nejm style, but changing the style to apa gives errors:  ! Undefined control sequence. &lt;argument&gt; \mkbibdateapalongextra  {year}{month}{day}\iffieldundef {endyear}{...  Package biblatex Warning: Macro 'name:apa:last-first' undefined. (biblatex)                Using \newbibmacro.  Package biblatex Warning: Macro 'name:apa:first-last' undefined. (biblatex)                Using \newbibmacro.   I'm using texlive 2012, as you can see I'm using biber, yes, I reran biber, and also tried delting the .bbl file and rerunning tools. My source markup is:  \documentclass[12pt]{article}  \usepackage[hidelinks]{hyperref} \usepackage[backend=biber,style=apa]{biblatex}  \addbibresource{BibResource.bib}  \begin{document}  test\parencite{ormrod2004human}  \clearpage  \printbibliography  \end{document}
I have an Asus R500VD laptop which came with the preinstalled Windows 8. As actually this computer will be used by my grandmother (and she's only using win 7) I'd like to downgrade it.  Thanks to other thread here, I was able to successfully boot computer from Windows 7 USB stick, but unfortunately during install I got a message:  A required CD/DVD drive device driver is missing. (...)   and I can't get pass it... This is what I've done:   I've downloaded all drivers from Asus Website (), unzip and copied to visible location for windows 7 installation - I've picked each one of them, none has helped. As I belive it's some kind of problem with SATA/RAID/AHCI drivers I went to Intel site and downloaded various "Intel Rapid Storage Technology" drivers - a lot of them were recognized by win7, but none of them helped.   Does anyone have an idea what can I do next? Any help would be really appreciated. This or That person I'm using  to go through JavaScript, and it's returning many suggestions to replace == (two equals signs) with === (three equals signs) when doing things like comparing idSele_UNVEHtype.value.length == 0 inside of an if statement.  Is there a performance benefit to replacing == with ===?   Any performance improvement would be welcomed as many comparison operators exist.  If no type conversion takes place, would there be a performance gain over ==? prove every connected manifold is path connected manifold .       my thought:      connected space : Let $ X$  be a topological space. A separation of $ X $ is a pair $U, V$ of disjoint    nonempty open subsets of $ X $ whose union is $X$. The space $ X $ is said to be connected    if there does not exist a separation of  $X$ .      Components: Given  $X$ , define an equivalence relation on $X$ by setting x~y if there    is a connected subspace of $X$  containing both $ x$  and $ y$ . The equivalence classes are    called the components (or the "connected components") of $ X$.       Path Component: I define another equivalence relation on the space $ X$  by defining x ~ у    if there is a path in $ X$  from $ x$  to $ y$ . The equivalence classes are called the path components of $ X$ .       Theorem  : The path components of $ X$  are path connected disjoint subspaces    of $X$  whose union is $X$ , such that each nonempty path connected subspace of $X$  intersects only one of them.       thank you so much Auto-sort column Google Spreadsheet on incoming Google Form response How can I get the Google Spreadsheet to automatically order the incoming form responses alphabetically (2nd column)?   I know how to do this manually or by going in and editing a cell, but cannot get it to do this as an onevent for incoming response. I'm trying to work with  in Java.   I want to implement arithmetic functions.  For this, I will first require a way to normalize the functions.  I know I can't add 1/6 and 1/2 until I have a common denominator. I will have to add 1/6 and 3/6.  A naive approach would have me add 2/12 and 6/12 and then reduce.  How can I achieve a common denominator with the least performance penalty?  What algorithm is best for this?    Version 8 (thanks to ):     Improvements include:         the equals() method is now consistent with the compareTo() method      final class Fraction extends Number {     private int numerator;     private int denominator;      public Fraction(int numerator, int denominator) {         if(denominator == 0) {             throw new IllegalArgumentException("denominator is zero");         }         if(denominator &lt; 0) {             numerator *= -1;             denominator *= -1;         }         this.numerator = numerator;         this.denominator = denominator;     }      public Fraction(int numerator) {         this.numerator = numerator;         this.denominator = 1;     }      public int getNumerator() {         return this.numerator;     }      public int getDenominator() {         return this.denominator;     }      public byte byteValue() {         return (byte) this.doubleValue();     }      public double doubleValue() {         return ((double) numerator)/((double) denominator);     }      public float floatValue() {         return (float) this.doubleValue();     }      public int intValue() {         return (int) this.doubleValue();     }      public long longValue() {         return (long) this.doubleValue();     }      public short shortValue() {         return (short) this.doubleValue();     }      public boolean equals(Fraction frac) {         return this.compareTo(frac) == 0;     }      public int compareTo(Fraction frac) {         long t = this.getNumerator() * frac.getDenominator();         long f = frac.getNumerator() * this.getDenominator();         int result = 0;         if(t&gt;f) {             result = 1;         }         else if(f&gt;t) {             result = -1;         }         return result;     } }     I have removed all previous versions.  My thanks to: QGIS - export a project with dependencies Block quote with big quotation marks Login to Stack Exchange account doesn't use HTTPS/SSL? How can I get the list of available databases on a SQL Server instance? I'm planning to make a list of them in a combo box in VB.NET. How do i prove invertibility of this linear transformation Let $T$ be a linear transformation from a vector space $V$  over reals into $V$ such that $T-T^2=I$. Show that $T$ is invertible I was just wondering since i have the 12.04 lts ubuntu edition if it's possible to upgrade it to 13.04 or even 14.04 after it gets released without losing any of my files and installed apps?  That's actually the main reason i chose the LTS version because i didn't know if i had to re-install all my apps every time a new ubuntu version gets released. Reading the reviews of the  on Google Play,  said that since Chrome supports HTML5, he had managed to persuade the BBC iPlayer website to play him streams meant for iPad/iPhone devices by spoofing an IOS user agent string on his Nexus 7.  How might I go about getting the same thing to work myself?  I have looked at  and  but both seem to concentrate on the Dolphin browser. Also, please mention if a solution will require me to root my Nexus-7.  Incidentally, I also tried using  with Firefox, but although Firefox supports HTML5 and the iPlayer website shows pages without either install flash or use external app messages, it still won't play the streams (it just sits swirling its progress wheel), hence looking for a solution for Chrome. Automatic code reindent button "The queen of England's crown" or "The queen's of England crown"? Possible Duplicate:          What is the correct way of these two sentences?   The queen of England's crown The queen's of England crown   Strictly linguistically, sentence 2 should be the correct one, since the crown belongs to the queen and not to England but it sounds really awkward to pronounce. Correcting for multiple kruskal wallis tests I am running a kruskal wallis test a number of times (37) and wanted to know if I have to adjust the p-values of the kruskal tests before I look any further into the post hocs? Additional Drivers for Ubuntu 17.10 The Additional Drivers settings interface is no longer available in Ubuntu 17.10  How can I install proprietary drivers in Ubuntu 17.10?  I need drivers for Wifi &amp; Bluetooth (BCM43142) and Intel Graphics.  The earlier Additional Drivers program showed a package named jockey but it no longer seem to exist in repositories. System of linear equations having a real solution has also a rational solution. I saw this question     Let $A ∈ M_{m\times n}(\mathbb{Q})$ and $b ∈ \mathbb{Q}^m$. Suppose   that the system of linear equations $Ax = b$ has a solution in   $\mathbb{R}^n$. Does it necessarily have a solution in $\mathbb{Q}^n$?   and I thought I'd give an interesting, possibly wrong, approach to solving it. I'm not sure if such things can be done, if not maybe you can help me refine.    I considered the form of the equality as  $$ A^{(1)}x_1+\cdots+A^{(n)}x_n=b, $$  where $A^{(i)}$ is a column vector of $A$. I then noticed that for $x_i\in\mathbb{R\setminus Q}=\mathbb{T}$ then, and this is where I think I'm doing something forbidden, each $x$ has the represenation  $$ x_1  =  k_{11}\tau_1+\cdots+k_{1p}\tau_p \\  x_2= k_{21}\tau_1+\cdots+k_{2p}\tau_p \\ \vdots \hspace{4cm} \vdots \\ x_n=k_{n1}\tau_1+\cdots+k_{np}\tau_p,$$  where $\tau_i$ is a distinct irrational number, $k_{ij}\in\mathbb{R}$, and $p$ is the number of such distinct irrational numbers. I wound this out, but there may be a discrepancy with $p$ and $m$. I feel this method can lead me to the answer, but I'm not sure where to go from here.  EDIT$^1$:  I end up getting something like this, I believe, after substitution:  $$ A(k^{(1)}\tau_1+\cdots+k^{(p)}\tau_p)=b$$  Here, $k^{(i)}$ is the vector  $$\begin{pmatrix} k_{1i} \\ \vdots \\ k_{ni} \end{pmatrix}.$$  EDIT$^2$:  I think there is no discrepancy with $p$ and $m$ because $A\in M_{m\times n}(\mathbb{Q})$, $K\in M_{n\times p}(\mathbb{R})$, and $\tau\in M_{p\times 1}(\mathbb{T})$, so  $$ (m\times n)\cdot (n\times p)\cdot (p\times 1) = m\times 1. $$ MATLAB/OCTAVE Extracting elements from different sized vectors in a cell array I have an elliptic curve $E$ over $\mathbb{F}_{11}$ defined by $y^2=x^3+4x$ with the point at infinity $\mathcal{O}$  I have a divisor of $E$, defined by $$D=\left[(0,0)\right]+\left[(2,4)\right]+\left[(4,5)\right]+\left[(6,3)\right]-4\left[\mathcal{O}\right]$$  I know that $\text{deg}(D)=1+1+1+1-4=0$ and it is clear that $\text{sum}(D)=\infty$  Therefore $D$ is the divisor of a function - we want to find this function  I am given \begin{align}\text{div}(y-2x)&amp;=\left[(0,0)\right]+2\left[(2,4)\right]-3\left[\mathcal{O}\right]\\ \text{div}(x-2)&amp;=\left[(2,4)\right]+\left[(2,-4)\right]-2\left[\mathcal{O}\right] \end{align}  I am then told that we can express $D$ as follows:  $$D = \left[(2,4)\right]+\text{div}\left(\frac{y-2x}{x-2}\right)+\left[(4,5)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]$$  However, when I expand this back out, I get the following:  \begin{align}D &amp;= \left[(2,4)\right]+\text{div}\left(\frac{y-2x}{x-2}\right)+\left[(4,5)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]\\ &amp;= \left[(2,4)\right]+\text{div}(y-2x)-\text{div}(x-2)+\left[\left(4,5\right)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]\\ &amp;= \left[(2,4)\right]+\left(\left[(0,0)\right]+2\left[(2,4)\right]-3\left[\mathcal{O}\right]\right)-\left(\left[(2,4)\right]+\left[(2,-4)\right]-2\left[\mathcal{O}\right]\right)+\left[\left(4,5\right)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]\\ &amp;= \left[(0,0)\right]+2\left[(2,4)\right]+\left[(4,5)\right]+\left[(6,3)\right]+2\left[\mathcal{O}\right]-\left[(2,-4)\right] \end{align}  which clearly isn't equivalent to the original $D$ unless $$\left[(2,4)\right]+2\left[\mathcal{O}\right]-\left[(2,-4)\right]=-4\left[\mathcal{O}\right]$$  Can anyone either explain what I've done wrong or show me how I can obtain the above equation please    This question comes in three parts:   Step 1 - This one Step 2 -  Step 3 -