[Tex/LaTex] Straight math arrows

arrowssymbolstikz-pgf

Is there any way to redefine all arrows in math mode (\rightarrow, \leftarrow, \Rightarrow, \Leftarrow, \to, etc.) to another arrows that finish with angle 90 TikZ option?

I have done some tries and I've got good results with normal math size:
enter image description here

Using the following code:

\documentclass{minimal}

\usepackage{tikz}
\usetikzlibrary{arrows}

%Redefine \rightarrow
\renewcommand{\rightarrow}{\mathbin{\tikz[baseline]\draw[arrows={-angle 90},yshift=0.75ex] (0,0) -- (.95em,0);}}

\begin{document}

%\mathchar"3221 is just the original code of \rightarrow
\noindent $x^2 - 2x + 1 = 0 \mathchar"3221 (x-1)^2 = 0 \mathchar"3221 x = 1$\\
$x^2 - 2x + 1 = 0 \rightarrow (x-1)^2 = 0 \rightarrow x = 1$

\end{document}

But I have problems when I use that arrow in special cases like limits, and I would like that all those arrows work in that situation (if possible):

enter image description here

I know that in this question is also explained how to get symbols that are like mine, but the package doesn't include the arrows that I would like to have. Also, I would like that that new arrows behave like normal characters, so if I put $2x \textcolor{red}{\rightarrow} 0$ , I would obtain a red arrow.

Any idea or suggestion?

Best Answer

The problem is that the tikz definition you gave doesn't scale with the smaller math styles. To overcome that, we use the scalerel package to scale (on the fly) the good normal-sized arrow to the same vertical extent as the original arrow that you are replacing, in the current math style.

\documentclass{minimal}
\usepackage{scalerel}
\usepackage{tikz}
\usetikzlibrary{arrows}
\let\svrightarrow\rightarrow
\newcommand{\TSrightarrow}{\mathbin{\tikz[baseline]\draw[arrows={-angle 90},yshift=0.75ex] (0,0) -- (.95em,0);}}
\renewcommand\rightarrow{\mathrel{\scalerel*{\TSrightarrow}{\svrightarrow}}}
\parindent 0pt
\begin{document}
$x^2 - 2x + 1 = 0 \svrightarrow (x-1)^2 = 0 \svrightarrow x = 1$\\~\\
$x^2 - 2x + 1 = 0 \rightarrow (x-1)^2 = 0 \rightarrow x = 1$\\~\\
\( \lim_{h\svrightarrow0}\frac{f(x+h)-f(x)}{h} \)\\~\\
\( \lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h} \)
\end{document}

enter image description here

Related Solutions

TikZ Graphics – Draw Curved Arrow Parallel to Curved Line

The curve drawn by a to[bend left] is probably a bezier curve, and thus is not an arc nor a parabola. Unfortunately this syntas hides the control points used.

Even if the control points were known, they won't be useful for your problem, since basically you want to get another bezier curve parallel to a "segment" or portion of another given bezier curve. I have no idea of the equations that should be solved to find those control points.

But not all hope is lost. I had a crazy idea. Once a bezier path is computed by tikz, any of its intermediate points are available via [pos] key, so it is possible to find a number of those points, between the desired start and end point, and connect all of them via straight lines. If the points are close enough, the straight segments will not be visible and they will look like a curve.

So this is the proof of concept:

\documentclass{article}
\usepackage{tikz}
\begin{document}
\thispagestyle{empty}
\usetikzlibrary{calc}

\begin{tikzpicture}

  % First, define nodes
  \draw (0,0) node[circle, inner sep=0.8pt, fill=black, label={below:{$E$}}] (E) {};  
  \draw (5,0) node[circle, inner sep=0.8pt, fill=black, label={below:{$F$}}] (F) {}; 

  % Draw the curved line. No to[bend] is allowed, only explicit control points
  \draw  (E) .. controls +(1.9,1) and +(-1.9,1).. (F);
  % Now, repeat the same curve, shifted up, and define 20 inner points named
  % p1, p2, p3, etc.. for positions 0.15, 0.16, 0.17, etc up to 0.35 inside that curve
  \path  ($(E)+(0,0.2)$) .. controls +(1.9,1) and +(-1.9,1) ..  ($(F)+(0, 0.2)$)
     {\foreach \t [count=\i] in {0.15,0.16,...,0.35} {  coordinate[pos=\t] (p\i) } };

  % Finally draw the "curve" (polygonal line indeed) through those 20 inner points
  \draw[->>] (p1) { \foreach \i in {1,...,20} {-- (p\i) } };

  % A second example, with the same ideas but different path
  \draw[red]  (E) .. controls +(2,-2) and +(-3,-1).. (F);
  \path  ($(E)+(0,0.2)$) .. controls +(2,-2) and +(-3,-1)..  ($(F)+(0, 0.2)$) 
     {\foreach \t [count=\i] in {0.15,0.16,...,0.55} {  coordinate[pos=\t] (p\i) } };

  \draw[red, ->>] (p1) { \foreach \i in {1,...,40} {-- (p\i) } };
\end{tikzpicture}
\end{document}

This is the result:

Result

Update

A little refinement can be done in above code. Instead of specifying the foreach loop as from 0.15 to 0.55 in steps of 0.1, it looks more natural to simply specify the desired number of inner points (i.e: the "resolution" of the parallel curve), and let to some other expression to compute the pos for each of those points. This is done like this:

\path  ..curve specification..
     {\foreach \i in {1,...,40} {  coordinate[pos=0.15+0.55*\i/40] (p\i) } };

So, in this example, 40 intermediate points are computed, and the formula for the pos key specifies 0.15 and 0.55 as the extremes of the parallel curve.

However, this approach does not use a "true" parallel line. The curve used to compute the points is simply the same orginal curve, only shifted in the Y axis. This could work for smooth arcs which are mostly horizontal, but it will fail for parts of the curve where the tangent is not so horizontal. Consider for example:

\begin{tikzpicture}
  % First, define nodes
  \draw (0,0) node[circle, inner sep=0.8pt, fill=black, label={below:{$E$}}] (E) {};  
  \draw (5,2) node[circle, inner sep=0.8pt, fill=black, label={below:{$F$}}] (F) {}; 

  \draw[red]  (E) .. controls +(5,-3) and +(-4,1).. (F);
  \path  ($(E)+(0,0.2)$) .. controls +(5,-3) and +(-4,1)..  ($(F)+(0,0.2)$) 
     {\foreach \i in {1,...,40} {  coordinate[pos=0.15+0.75*\i/40] (p\i) } };

  \draw[blue, ->>] (p1) { \foreach \i in {1,...,40} {-- (p\i) } };
\end{tikzpicture}

This produces:

Bad Result

It can be seen that the distance between the red and blue lines is not constant. How can this be solved?

The following idea works: instead of shifting the curve, compute the sequence of points (p1), (p2) etc. on the original curve. But then, when drawing the parallel line, do not use those points, but points which are at a fixed distance (eg: 0.2cm) from each of these points in the direction perpendicular to the curve. This direction can be computed as the vector from current point (p_i) to next point (p_j), rotated 90 degrees.

Fortunately, tikz has the interpolated coordinates expression which allows for this with a simple formula: ($(p\i)!0.2cm!90:(p\j)$).

So, using this idea:

\begin{tikzpicture}
  % First, define nodes
  \draw (0,0) node[circle, inner sep=0.8pt, fill=black, label={below:{$E$}}] (E) {};  
  \draw (5,2) node[circle, inner sep=0.8pt, fill=black, label={below:{$F$}}] (F) {}; 

  % Draw curved path
  \draw[red]  (E) .. controls +(5,-3) and +(-4,1).. (F);
  % Compute points on the same curve
  \path  (E) .. controls +(5,-3) and +(-4,1)..  (F) 
     {\foreach \i in {1,...,40} {  coordinate[pos=0.15+0.75*\i/40] (p\i) } };
  % Draw parallel curve
  % (note that first and last points are specified out of the loop)
  \draw[blue, ->>] ($(p1)!.2cm!90:(p2)$) 
     { \foreach \i [count=\j from 3] in {2,...,39} {-- ($(p\i)!.2cm!90:(p\j)$) } }
     -- ($(p40)!.2cm!-90:(p39)$);
\end{tikzpicture}

And this gives a perfect result!

Result2

TikZ-PGF – Drawing Angled Lines for Calligraphy Templates

You can use polar coordinates such as (55:100cm). In the code below, I compute the precise number of oblique lines to draw in function of the chosen angle \myAngle and the distance \myDist between two consecutive oblique lines. My code also computes the largest length needed for these lines.

\documentclass[a4paper]{article}
\usepackage{tikz}
\usetikzlibrary{calc}

\pagestyle{empty}

\newcommand*{\myDist}{1cm} % distance between consecutive oblique lines
\newcommand*{\myAngle}{55} % angle of said lines wrt horizontal, in degrees
% Distance between consecutive oblique lines, projected on the horizontal axis
\pgfmathsetlengthmacro{\horizIntervWidth}{\myDist/sin(\myAngle)}

% Length of the longest oblique lines we'll need. I add 10pt to be 100% safe
% with respect to rounding errors (the lines will be clipped anyway).
\pgfmathsetlengthmacro{\maxLength}{10pt + \paperheight/sin(\myAngle)}

% Number of oblique lines to draw
\pgfmathtruncatemacro{\maxIndex}{
  round((\paperheight/tan(\myAngle) + \paperwidth)/\horizIntervWidth)}

\begin{document}

\begin{tikzpicture}[remember picture, overlay]
% Just to be sure we don't paint outside the page. :-)
\clip (current page.south west) rectangle (current page.north east);

\foreach \i in {1,2,...,30} {
  \draw[black] ($(current page.north west)+(0,-\i)$) --
               ($(current page.north east)+(0,-\i)$);
}

\foreach \i in {0.5,1.5,...,60} {
  \draw[lightgray] ($(current page.north west)+(0,-\i)$) --
                   ($(current page.north east)+(0,-\i)$);
}

\foreach \i in {1,2,...,\maxIndex} {
  \draw[red] ([xshift=-\i*\horizIntervWidth]current page.south east) --
            +(\myAngle:\maxLength);
}
\end{tikzpicture}

\end{document}

Best Answer

Related Solutions

TikZ Graphics – Draw Curved Arrow Parallel to Curved Line

Update

TikZ-PGF – Drawing Angled Lines for Calligraphy Templates

Related Question