To typeset a collection of equations with alignment, I am using align* from amsmath.  Some of these equations have right-hand sides that have multiple lines that should be indented.  So far I am using a nested aligned environment and \MoveEqLeft from the mathtools package:  \documentclass{article} \usepackage{amsmath} \usepackage{mathtools}  \begin{document} \begin{align*}   x+y &amp;= z+w \\     a &amp;= \begin{aligned}[t]            \MoveEqLeft[0.5]            b+c+d \\            &amp;+e+f \\            &amp;+g+h+i+j          \end{aligned} \end{align*} \end{document}   The problem is that the aligned environment seems to introduce extra horizontal space to the left (and possibly on the right as well).   How can I eliminate the extra space from aligned?   I am open to other solutions that might avoid aligned, but I do not want to need to manually insert spaces to indent the lines.
Near the beginning of the definition of the amsmath environments aligned and gathered (and also mathtools' multlined) there's an explicit thin space \,. This often  has to be cancelled  with  \!, as has been observed  various  times:         I'd like to know why this explicit \, is there &ndash; what would go wrong without it?    From amsmath.dtx:  \newcommand{\start@aligned}[2]{%     \RIfM@\else         \nonmatherr@{\begin{\@currenvir}}%     \fi     \savecolumn@ % Assumption: called inside a group %    \end{macrocode} %    The \cs{null} here is to keep the \cs{,} glue from causing the %    invocation of the clause in \tex/'s built-in tag placement %    algorithm that can cause an equation to be shifted all the way over %    to the margin. %    \begin{macrocode}     \null\,%
Is it true that the order of $ab$ is always equal to the order of $ba$? How do I prove that if $a$, $b$ are elements of group, then $o(ab) = o(ba)$?   For some reason I end up doing the proof for abelian(ness?), i.e., I assume that the order of $ab$ is $2$ and do the steps that lead me to conclude that $ab=ba$, so the orders must be the same. Is that the right way to do it? I have trouble comparing 2 double in Excel VBA  suppose that I have the following code  Dim a as double Dim b as double a = 0.15 b = 0.01   After a few manipulations on b, b is now equal to 0.6  however the imprecision related to the double data type gives me headache because  if a = b then  //this will never trigger end if   Do you know how I can remove the trailing imprecision on the double type? I think it's well established that for the basic premise in the Back to the Future Trilogy to work, we have a system of alternate timelines.  If this is the case though, then why would Marty in Back to the Future (the original) begin to fade?  Logically, there would simply be no Marty in the alternate timeline created when Marty travels back to 1955 for the first time; why should this affect the original Marty from the original timeline? Let's say I fire a bus through space at (almost) the speed of light in vacuum. If I'm inside the bus (sitting on the back seat) and I run up the aisle of the bus toward the front, does that mean I'm traveling faster than the speed of light? (Relative to Earth that I just took off from.) Tensor product in dual-space I am a bid confused regarding the notation for tensor products when going into dual-space  If $\left| \Psi \rangle \right. = \left| A \rangle \right. \left| B \rangle \right.$ is $ \left. \langle \Psi \right| = \left. \langle A \right| \left. \langle B \right|$ or $\left. \langle B  \right| \left. \langle A \right|$?  My guess is the last choice, since operators should act on   $\left| 0 \rangle \right. = \left( \begin{matrix}   1 \\   0  \end{matrix} \right) $  $\left| 1 \rangle \right. = \left( \begin{matrix}   0 \\   1  \end{matrix} \right) $  so  $\left| 01 \rangle \right. = \left| 0 \rangle \right. \left| 1 \rangle \right. = \left( \begin{matrix}   0 \\   1 \\   0 \\   0  \end{matrix} \right) $  But then, for me, it should be  $\left( \begin{matrix}   0 &amp; 1 &amp; 0 &amp; 0  \end{matrix} \right) = \left \langle 10 \right|$ But this is not correct if i calculate the tensor product, since   $\left \langle 1 \right| \left \langle 0 \right| = \left( \begin{matrix}   0 &amp; 0 &amp; 1 &amp; 0  \end{matrix} \right)$  For instance, shoulden't it be  $\left| 01 \rangle \right. \left \langle 10 \right| = \left( \begin{matrix}   0 &amp; 0 &amp; 0 &amp; 0\\   0 &amp; 1 &amp; 0 &amp; 0\\   0 &amp; 0 &amp; 0 &amp; 0\\   0 &amp; 0 &amp; 0 &amp; 0\\  \end{matrix} \right) $ ?  I hope I have made my confusion clear Cases environment compresses fractions? what should I do to stop the cases environment compressing the first line of the following? \begin{equation} \alpha_{mn} = \begin{cases} \frac{16\mu^2}{mn\pi^2\left[\pi^2(m^2/a^2 + n^2/b^2) - \mu^2\right]} &amp;\mbox{for } m \mbox{ and } n \mbox{ odd;}  \\ 0 &amp;\mbox{for } m \mbox{ or } n \mbox{ even;} \end{cases} \\[1ex] \end{equation}  I'm not sure to do tex on this site, but if anyone tries it they see the first case is compressed because it's a fraction. Hello [Comma?] John, Does "until [date]" mean "before that date"? I have a dictionary of values read from two fields in a database: a string field and a numeric field. The string field is unique, so that is the key of the dictionary.  I can sort on the keys, but how can I sort based on the values?  Note: I have read Stack Overflow question here  and probably could change my code to have a list of dictionaries, but since I do not really need a list of dictionaries I wanted to know if there is a simpler solution to sort either in ascending or descending order. Is a transit visa required to pass through Bangkok Airport? When is it appropriate to use the original pronunciation of a foreign word versus the English pronunciation? When reading to an audience, or speaking in conversation, when is it appropriate to use the original pronunciation of a foreign word versus the English pronunciation (assuming you know the appropriate pronunciation for it)? Is it considered rude, or condescending? Or is it considered a mark of being knowledgeable?  One of the things to consider are place names. When referring to Paris, France, should it be pronounced with the silent 'S' as the French would say it, or with the 'S'? Should Hiroshima be pronounced as a Japanese speaker would pronounce it slightly more emphasis on RO, rather than the SHI? Should Mount Pinatubo be pronounced as a Filipino speaker would pronounce it with shorter stronger vowels, or the longer vowels? (e.g. Pi as is 'pick' rather than 'pea'.) Should Wichita be pronounces as the original "shi" rather than the modern "chi"?  Pronounce pesos or sombrero as a Spanish speaker 'eh' or the English 'ay'?  (I can't think of other common words right now that aren't words taken from other languages like hurricane, boondocks, tornado, etc. which I think have (correctly) changed to English pronunciations.) I have been reading Men With Brooms for a few days. I read a sentence which did not make any sense to me, so I landed here to get some help on it. I have pasted the sentence from the novel. Please go through it and let me know your valuable feedback on it.  An excerpt from the novel (Men With Brooms: A Sweeping Epic, paperback 2002, a novelization by Diane Baker Mason):     Gordon was about to walk away from the Impala when he saw it stop and his son get out. So it was real the boy had come.   As per my opinion it should have been:     Gordon was about to walk away from the Impala when he saw it stopped and his son got out. So it was real the boy had come.   I have two questions here:   Why did the writer use stop and get instead of stopped and got? As she is telling us a story which happened in the past. Let's say the writer is telling us her mind's situation and she used the present tense for it but why did she use stop and get instead of stops and gets? How many $m\times n$ tables are there if they are subject to the following two conditions?    In each cell of the table we have $1$ or $-1$. The product of all the cells in any given row and the product of all cells in any given column equals $-1$.    I already know if Parity of $m$ and $n$ are different there is no such a table. what is the answer in general? Often times I am not interested in the question so much as I am the answer. Unfortunately, the only site option for bookmarking an optimal answer is to bookmark the question. This leaves you having to remember that no the question wasn't that great, but there's an answer in here worth a million. It'd be nice to be able to favorite an answer allowing you to jump immediately to the answer that was much more valuable than the question. I understand that I can bookmark the answers link using my browser but it seems being able to favorite the answer within the site's ecosystem is useful. Regular Expression Uppercase Replacement in C# I have the following C# which simply replaces parts of the input string that look like EQUIP:19d005 into URLs, like this:  input = Regex.Replace(input, @"(EQUIP:)(\S+)", @"&lt;a title=""View equipment item $2"" href=""/EquipmentDisplay.asp?eqnum=$2""&gt;$1$2&lt;/a&gt;", RegexOptions.IgnoreCase);   The HTML ends up looking like this.  &lt;a title="View equipment item 19d005" href="/EquipmentDisplay.asp?eqnum=19d005"&gt;EQUIP:19d005&lt;/a&gt;   The only trouble is that the destination page expects the eqnum querystring to be all UPPERCASE so it returns the correct equipment when eqnum=19D005 but fails if it receives eqnum=19d005.  I guess I can modify and correct EquipmentDisplay.asp's errant requirement of uppercase values however, if possible I'd like to make the C# code comply with the existing classic ASP page by uppercasing the $2 in the Regex.Replace statement above.  Ideally, I'd like the HTML returned to look like this:  &lt;a title="View equipment item 19d005" href="/EquipmentDisplay.asp?eqnum=19D005"&gt;EQUIP:19d005&lt;/a&gt;   Notice although the original string was EQUIP:19d005 (lowercase), only the eqnum= value is uppercased.  Can it be done and if so, what's the tidiest way to do it? Entire function $f$ such that $|f(z)| ≤ K |z|^3$ for $|z|\ge1$ and $f(z) = f(iz)$ for every $z∈C$ Let $f(z)$ be an entire function such that for some constant $K$ , $|f(z)| ≤ K$ $|z|^3&lt;$ for $|z|\ge1$ and $f(z) = f(iz)$ $∀z∈C$ , then which of the following are correct ?  (A)$ f(z) = Kz^3$ $∀ z∈\Bbb{C}$ (B) $f(z)$ is a constant function (C) $f(z)$ is quadratic function (D) No such $f$ exists   My attempt : I can deduce that since $f(z) = f(iz)$ and since $i^4=1 \implies f$ must involve only fourth powers of $z$ . Hence $f(z)=a_1z^4+b_1z^8+c_1z^{16}+ \cdots $-------(1)  Also for $|z| \ge 1 $: let C.   Then $|g(z)| = |f(z)|/|z^3| \le k$ => By the Cauchy estimate theorem we can prove that :  $f(z)/z^3$ is a constant function $= c $   $\implies f(z)=cz^3$ for $|z| &gt; 1$     ...........(2)  From (1) and (2) ; we get  there does not exist a function like this ?  My textbook answer says it's a constant function ? Examples of a non-commutative division ring How to draw arrows between circled coefficients of polynomials using tikz Should operator<< be implemented as a friend or as a member function? That's basically the question, is there a "right" way to implement operator&lt;&lt; ? Reading  I can see that something like:  friend bool operator&lt;&lt;(obj const&amp; lhs, obj const&amp; rhs);   is preferred to something like   ostream&amp; operator&lt;&lt;(obj const&amp; rhs);   But I can't quite see why should I use one or the other.  My personal case is:  friend ostream &amp; operator&lt;&lt;(ostream &amp;os, const Paragraph&amp; p) {     return os &lt;&lt; p.to_str(); }   But I could probably do:  ostream &amp; operator&lt;&lt;(ostream &amp;os) {     return os &lt;&lt; paragraph; }   What rationale should I base this decision on?  Note:   Paragraph::to_str = (return paragraph)    where paragraph's a string. I am under DDoS. What can I do?