We know how to typeset the commonly used diagonal dots, i.e., \ddots, whose slope is -1.  Now I want to produce dots with slope 1, which is perpendicular to \ddots.   Does anyone know how to do it?
122851
I was wondering if anyone knew how to draw 3 dots in a matrix that go in a diagonal way usually to represent that element flow similarily in the slots. I know there's one that goes in the bottom right corner. But is there one that goes in the bottom left?
58126
I'm about to use this diagram to make computing 3x3 determinants clear and easy for my students. I tried to add this picture to my lecture but it's blurred due to the bad quality of the picture. So i want to create exactly the same figure using latex.  I've searched and read a few topics about highlighting elements and diagonals in matrix but it does not work in this case. I want to add plus signs and minus signs as in the picture, and highlight elements in the matrice with elements outside the matrice.  I appreciate all your help and your advice. Thank you in advance !
54642
Above is an example produced by S\overline{n}=\overline{n+1}. The overlines are not aligned vertically, which is highly undesirable. Is there any way to produce overlines of the same height throughout a whole formula?
112761
I would like to know is there any way to express mathematically (using matrices multiplication, addition, etc.) a vector which is formed from the diagonal elements of a matrix. For example, I have a matrix called $\mathbf{M}$ and I want to create a vector  $\mathbf{v}$ such that $\mathbf{v}= \text{diagonal elements of } \mathbf{M} $  Any matrix algebra or operation do that ? Thank you.
88633
I did this using trial and error, but I was just wandering if there is an algebraic way of solving this?  I thought about double angle formula but that doesn't work properly does it?  I then tried writing it in the form $x \cdot 2 \cot(x) = 1$ but even then, I can't solve it properly by re-writing $\cot(x)$ because of the $x$ outside the brackets.  Anybody know how I would do this algebraically?
103305
How can I typeset the following symbol for direct current (i.e. a = symbol where the lower line is dashed)?     The image above is an enlarged version, the size I need is like =, \approx, \simeq etc.
91168
What would the graph of $(-1)^{x}$ look like?   I know that the value of the function alternates between $1$ and $-1$ when it is defined so I think it would just be points spread over the lines $x=1$ and $x=-1$.   Is this correct? Also, will there be any definite pattern of the points.
77118
I will use binary.  I claim to have a bijection $f \colon \mathcal{N} \to \left[ 0, 1 \right)$ where $\mathcal{N}$ is the set of natural numbers $\left\{ 0, 1, 10, 11, 100, \dotsc \right\}$ as follows:  $$ \begin{array}{l|l} x \in \mathcal{N} &amp; f(x) \in \left[ 0, 1 \right)\\\hline 0 &amp; 0.0 \\ 1 &amp; 0.1 \\ 10 &amp; 0.01 \\ 11 &amp; 0.11 \\ 100 &amp; 0.001 \\ 101 &amp; 0.011 \\ 110 &amp; 0.101 \\ 111 &amp; 0.111 \\ 1000 &amp; 0.0001 \\ \hspace{.6em}\vdots &amp; \hspace{.9em}\vdots \end{array} $$  It's basically the .  I claim to have everything on this list.  Diagonalization?  Go ahead!  Constructing a number that differs from $f(i + 1)$ in its $i$-th fractional digit yields $0.11111\!\ldots = 0.\overline{1}$ which is $1$, and that should not be on the list anyway!  Why does this not work? I think it has something to do with integers supposedly not having infinite digits, while reals supposedly do.
5905
can anybody help me to draw this minimal example of an xy axis? I couldn't find anything similar on the forum.  The axis should be black but the grid behind it a shade o gray, dashed. It would be great if the axis were a little bit thicker than the grid also.
11126
I want to take the inverse of a dot product, where both vectors have complex components.  In other words, if $\textbf{A} \cdot \textbf{B} = d$, and I know $\textbf{A}$ and $d$, I want to find a $\textbf{B}$.  I know that I cannot do so uniquely, which is fine; I have a procedure for creating a set of vectors that will satisfy $\textbf{A} \cdot \textbf{B} = d$.  But, it relies on finding vectors that are orthogonal to $\textbf{A}$ and each other.  Normally this would not be a problem; just take $\textbf{A}$, zero all but two of its components, switch the last two, and negate one of them.  Make sure it's a different pair every time.  For more orthogonal vectors, you can take cross products.  The difficulties arise when I consider vectors with complex components.  I want to normalize each of the orthogonal vectors, which means divide by their magnitudes.  I have read that you need to divide by $\sqrt{|v_x|^2 + |v_y|^2 + |v_z|^2 ...}$.  What is the justification for this?  Since I am just normalizing vectors, do I absolutely have to take the magnitudes of the components?  Also, My technique of finding the inverse dot product relies on the identity $\textbf{A} \cdot \textbf{B} = d = |\textbf{A}||\textbf{B}|$.  What modifications might I need to make?  I can post more details if people want.  Also, if anyone has links to stuff to read about, especially with regards to the reasons why, I would be most appreciative.
39446
I would like to find a way to cross out a term in a mathematical expression with a diagonal arrow pointing from lower left to upper right and with a little "zero" next to the pointy end of the arrow.
7924
From , I have tried the using \hspace{0pt} and   \makeatletter \renewcommand{\dddot}[1]{%   {\mathop{\kern\z@#1}\limits^{\vbox to-1.4\ex@{\kern-\tw@\ex@    \hbox{\normalfont ...}\vss}}}} \renewcommand{\ddddot}[1]{%   {\mathop{\kern\z@#1}\limits^{\vbox to-1.4\ex@{\kern-\tw@\ex@    \hbox{\normalfont....}\vss}}}} \makeatother   after loading mathtools but the problem still persists.    \documentclass{article} \usepackage{mathtools} \begin{document} \[ \dot{\mathbf{r}}\times\ddot{\mathbf{r}}\times\dddot{\mathbf{r}} \] \end{document}   In the case the font scheme I am using has any affect as well, the fonts are:  \usepackage[utf8]{inputenx} \renewcommand{\rmdefault}{ppl}                  \linespread{1.05}                   \usepackage[scaled]{helvet} \usepackage{courier}               \usepackage{eulervm} \normalfont       \usepackage[T1]{fontenc}            \usepackage{textcomp}
94166
Can someone show me step-by-step how to diagonalize this matrix? I'm trying to teach myself differential equations + linear algebra, but I'm stumped on how to do this. I'd really appreciate if someone would take the time to do this with me!   $\begin{bmatrix} 1 &amp; 2 &amp; 0 \\ 2 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; -3 \\ \end{bmatrix}$
104754
I'm trying, but something is wrong with my code in LaTeX: \begin{asy} pair dynkinMark(real x, real y, string s = &quot;&quot;, pen p = defaultpen, pair B = dir(45)) { if (s != &quot;&quot;) { dot(&quot;$&quot;+s+&quot;$&quot;, (x,y), B, p); }     else { dot((x,y), p); }     return (x,y); }  dynkinMark(0,1, &quot;1&quot;, dir(135)); dynkinMark(0,2, &quot;2&quot;, dir(135)); dynkinMark(0,3, &quot;3&quot;, dir(135)); dynkinMark(0,4, &quot;4&quot;, dir(135)); dynkinMark(7,1, &quot;1'&quot;); dynkinMark(7,2, &quot;2'&quot;); dynkinMark(7,3, &quot;3'&quot;); dynkinMark(7,4, &quot;4'&quot;); path p1 = (0,1)..(2,2.0)..(5,3.5)..(7,4); path p2 = (0,2)..(2,1.2)..(5,2.5)..(7,3); path p3 = (0,3)..(2,2.8)..(5,1.5)..(7,1); path p4 = (0,4)..(2,3.8)..(5,3.5)..(7,2); draw(p1, blue); draw(p2, heavycyan); draw(p3, mediumcyan); draw(p4, lightblue); dot(&quot;$12$&quot;, IP(p1, p2), dir(90), heavygreen); dot(&quot;$13$&quot;, IP(p1, p3), dir(90), heavygreen); dot(&quot;$23$&quot;, IP(p2, p3), dir(90), heavygreen); dot(&quot;$14$&quot;, IP(p1, p4), dir(90), heavygreen); dot(&quot;$24$&quot;, IP(p2, p4), dir(90), heavygreen); real y = 0.8; label(&quot;$s_1$&quot;, (IP(p1,p2).x, y), heavygreen); label(&quot;$s_2$&quot;, (IP(p1,p3).x, y), heavygreen); label(&quot;$s_1$&quot;, (IP(p2,p3).x, y), heavygreen); label(&quot;$s_3$&quot;, (IP(p1,p4).x, y), heavygreen); label(&quot;$s_2$&quot;, (IP(p2,p4).x, y), heavygreen); \end{asy}
112360
I have the following code:  \draw[step=1.0,black,thin] (0.5,0.5) grid (5.5,4.5);   which I would like to have generate grid lines in the Y direction at y = {0.5, 1.5, 2.5, 3.5, 4.5} and grid lines in the X direction at x = {0.5, 1.5, 2.5, 3.5, 4.5, 5.5}.   However, I get grid lines at y = {1.0, 2.0, 3.0, 4.0} and x = {1.0, 2.0, 3.0, 4.0, 5.0} with half cells around the edges of the grid. So instead of having a closed uniform grid, I get one with open edges and two half cells at the borders.   Does anybody know why it doesn't honor the starting point and stride?
32739
I am using the algorithmicx package to write an algorithm. However I am using the noend option:  \usepackage[noend]{algpseudocode}   Is there anyway to get vertical lines for each block of code? This question has been answered here:    However, it does not work when using the noend option.  EDIT: I'm looking for something like the algorithms here:
91574
I don't know exactly how to describe it, but in a programmatic way I want to spiral outward from coordinates 0,0 to infinity (for practical purposes, though really I only need to go to about +/-100,000:+/-100,000)  So if this is my grid:  [-1, 1][ 0, 1][ 1, 1][ 2, 1] [-1, 0][ 0, 0][ 1, 0][ 2, 0] [-1,-1][ 0,-1][ 1,-1][ 2,-1]   I want to spiral in an order sort of like:  [7][8][9][10] [6][1][2][11] [5][4][3][12] etc...   Is there a formula or method of doing this?