Is there a way, using TikZ, to plot a smooth curve satisfying
f( 0) = -1, f'( 0) = 0
f( 2) = 2, f'( 2) = -1
f(-2) = -1, f'(-2) = 1
Of course, I could find a polynomial satisfying the given conditions, but I'd rather just feed the conditions to TikZ and let it do the work.
This solution is similar to that by @percusse and @Altermundus in that control points are computed using the slope at the given points. One small benefit of the method proposed here is that it is extensible to more than just three points, and at each point a unique value of <delta x> can be specified so that one can have more control the behavior at that point
To start the curve use
\ExtrapolateStart{<delta x>}{<x>}{<y>}{<y'>}
All the points following (except the last) are specified with
\Extrapolate{<delta x>}{<x>}{<y>}{<y'>}
and the last point is specified with
\ExtrapolateEnd{<delta x>}{<x>}{<y>}{<y'>}
Here is the output with various settings:
The control points are shown in blue for debugging purposes, but that part of the code can be commented out if that is not needed.
Here is an example where the x, y, and y' values are specified for 5 points with:
\ExtrapolateStart{\DeltaXStart}{-2}{0}{3.0}% delta x, x, y, y'
\Extrapolate{\DeltaXMiddle}{-1}{2}{1}
\Extrapolate{\DeltaXMiddle}{0}{1}{-2}
\Extrapolate{\DeltaXMiddle}{1}{-1}{-0.5}
\ExtrapolateEnd{\DeltaXEnd}{2}{1}{2.0}
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc}
\newcommand{\ExtrapolateStart}[4]{%
($(#2,#3)-(#1, #1#4)$) --%
(#2,#3) .. controls%
($(#2,#3)+(#1, #1#4)$)%
}%
\newcommand{\Extrapolate}[4]{%
and%
($(#2,#3)-(#1, #1#4)$) ..%
(#2,#3) .. controls%
($(#2,#3)+(#1, #1#4)$)%
}%
\newcommand{\ExtrapolateEnd}[4]{%
and%
($(#2,#3)-(#1, #1#4)$) ..%
(#2,#3) --%
($(#2,#3)+(#1, #1#4)$)%
}%
%--------- For debugging use only
\newcommand{\ShowPoints}[4]{%
\filldraw [blue, opacity=0.5]%
(#2,#3) circle (2pt)%
($(#2,#3)-(#1, #1#4)$) circle (1pt) --%
($(#2,#3)+(#1, #1#4)$) circle (1pt)%
}%
\newcommand*{\AddLabel}[2][]{%
\node [draw,shape=rectangle,fill=white,#1] at (-1,2.5) {#2};
}
%
\begin{document}
\noindent
Example using 3 points where $x$, $f(x)$ and $f^\prime(x)$ are given for various $\Delta x$:
\bigskip\par\noindent
\begin{tikzpicture}
\newcommand{\DeltaXStart}{0.5}%
\newcommand{\DeltaXMiddle}{0.5}%
\newcommand*{\DeltaXEnd}{0.5}%
%
\draw [thin, gray, opacity=0.5] (-3,-3) grid (3,3);
\edef\MyPath{
\ExtrapolateStart{\DeltaXStart}{-2}{-1}{1}
\Extrapolate{\DeltaXMiddle}{0}{-1}{0}
\ExtrapolateEnd{\DeltaXEnd}{2}{2}{-1}
}
\draw [ultra thick, red, smooth] \MyPath;
% Following for debugging use only:
\ShowPoints{\DeltaXStart}{-2}{-1}{1};
\ShowPoints{\DeltaXMiddle}{0}{-1}{0};
\ShowPoints{\DeltaXEnd}{2}{2}{-1};
%
\AddLabel{$\Delta x = \DeltaXMiddle$};
\end{tikzpicture}
%
\begin{tikzpicture}
\newcommand{\DeltaXStart}{1.0}%
\newcommand{\DeltaXMiddle}{1.0}%
\newcommand*{\DeltaXEnd}{1.0}%
%
\draw [thin, gray, opacity=0.5] (-3,-3) grid (3,3);
\edef\MyPath{
\ExtrapolateStart{\DeltaXStart}{-2}{-1}{1}
\Extrapolate{\DeltaXMiddle}{0}{-1}{0}
\ExtrapolateEnd{\DeltaXEnd}{2}{2}{-1}
}
\draw [ultra thick, red, smooth] \MyPath;
% Following for debugging use only:
\ShowPoints{\DeltaXStart}{-2}{-1}{1};
\ShowPoints{\DeltaXMiddle}{0}{-1}{0};
\ShowPoints{\DeltaXEnd}{2}{2}{-1};
%
\AddLabel{$\Delta x = \DeltaXMiddle$};
\end{tikzpicture}
%
\begin{tikzpicture}
\newcommand{\DeltaXStart}{0.4}%
\newcommand{\DeltaXMiddle}{1.0}%
\newcommand*{\DeltaXEnd}{0.6}%
%
\draw [thin, gray, opacity=0.5] (-3,-3) grid (3,3);
\edef\MyPath{
\ExtrapolateStart{\DeltaXStart}{-2}{-1}{1}
\Extrapolate{\DeltaXMiddle}{0}{-1}{0}
\ExtrapolateEnd{\DeltaXEnd}{2}{2}{-1}
}
\draw [ultra thick, red, smooth] \MyPath;
% Following for debugging use only:
\ShowPoints{\DeltaXStart}{-2}{-1}{1};
\ShowPoints{\DeltaXMiddle}{0}{-1}{0};
\ShowPoints{\DeltaXEnd}{2}{2}{-1};
%
\AddLabel{$\Delta x = \DeltaXStart,\DeltaXMiddle,\DeltaXEnd$};
\end{tikzpicture}
\newpage\noindent
Example using 5 points where $x$, $f(x)$ and $f^\prime(x)$ are given:
\bigskip
\par\noindent
\begin{tikzpicture}
\newcommand{\DeltaXStart}{0.5}%
\newcommand{\DeltaXMiddle}{0.3}%
\newcommand*{\DeltaXEnd}{0.5}%
%
\draw [thin, gray, opacity=0.5] (-3,-3) grid (3,3);
\edef\MyPath{
\ExtrapolateStart{\DeltaXStart}{-2}{0}{3.0}
\Extrapolate{\DeltaXMiddle}{-1}{2}{1}
\Extrapolate{\DeltaXMiddle}{0}{1}{-2}
\Extrapolate{\DeltaXMiddle}{1}{-1}{-0.5}
\ExtrapolateEnd{\DeltaXEnd}{2}{1}{2.0}
}
\draw [ultra thick, red, smooth] \MyPath;
% Following for debugging use only:
\ShowPoints{\DeltaXStart}{-2}{0}{3.0};
\ShowPoints{\DeltaXMiddle}{-1}{2}{1};
\ShowPoints{\DeltaXMiddle}{0}{1}{-2};
\ShowPoints{\DeltaXMiddle}{1}{-1}{-0.5};
\ShowPoints{\DeltaXEnd}{2}{1}{2.0};
%
\AddLabel[xshift=2cm]{$\Delta x = \DeltaXStart,\DeltaXMiddle,\DeltaXEnd$};
\end{tikzpicture}
%
\begin{tikzpicture}
\newcommand{\DeltaXStart}{0.5}%
\newcommand{\DeltaXMiddle}{0.3}%
\newcommand*{\DeltaXEnd}{0.5}%
%
\draw [thin, gray, opacity=0.5] (-3,-3) grid (3,3);
\edef\MyPath{
\ExtrapolateStart{\DeltaXStart}{-2}{-1}{1}
\Extrapolate{\DeltaXMiddle}{-1}{-2}{0}
\Extrapolate{\DeltaXMiddle}{0}{1.5}{0}
\Extrapolate{\DeltaXMiddle}{1}{-2}{0}
\ExtrapolateEnd{\DeltaXEnd}{2}{2}{-2}
}
\draw [ultra thick, red, smooth] \MyPath;
% Following for debugging use only:
\ShowPoints{\DeltaXStart}{-2}{-1}{1};
\ShowPoints{\DeltaXMiddle}{-1}{-2}{0};
\ShowPoints{\DeltaXMiddle}{0}{1.5}{0};
\ShowPoints{\DeltaXMiddle}{1}{-2}{0};
\ShowPoints{\DeltaXEnd}{2}{2}{-2};
%
\AddLabel{$\Delta x = \DeltaXStart,\DeltaXMiddle,\DeltaXEnd$};
\end{tikzpicture}
\end{document}
I think it is much better to use MetaPost or Asymptote to draw the curve. They provide direction curve specifier ({dir}) to obtain proper splines. TikZ is not good at interpolation and curve fitting. The algorithm used by MetaPost and Asymptote is too complex for pure TeX drawing packages.
An Asymptote example,
unitsize(2cm);
draw ( (-2,-1) {(1,1)} .. (0,-1) {(1,0)} .. (2,2) {(1,-1)} );
dot((0,-1) ^^ (2,2) ^^ (-2,1));
Here, specifier {(1,1)} after (-2,1) means the direction of tangent is (1,1), i.e. (1,f'(x)).
And it is easy to define a function to generalize the method:
unitsize(2cm);
// points should be sorted by x
guide fit_curve(pair[] points, real[] diffs)
{
guide g;
for (int i = 0; i < points.length; ++i)
g = g .. points[i] {(1,diffs[i])};
return g;
}
pair[] pts = { (-2,-1), (0,-1), (2,2) };
real[] diffs = { 1, 0, -1 };
draw (fit_curve(pts, diffs));
dot(pts);
As Peter commented the curve that you define is not unique. If this is not a problem then you can use the control points for the curve generation by giving tangent specifications at points that you would like. I tried to sketch a little bit how the mechanism works in this answer.
\documentclass{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
\draw[style=help lines] (-3,-3) grid[step=1cm] (3,3);
\node (O) at (0,0) {0};
\draw[thick] (-3,0) -- (3,0) (0,-3) -- (0,3);
\draw (-2,-1) node[below] {$f(-2)$}.. controls (-1,0) and (-1,-1) .. (0,-1) %
node[below right] {$f(0)$}.. controls ++(1,0) and (1,3) .. (2,2)node[below] {$f(2)$};
\end{tikzpicture}
\end{document}
You can further play around with the control points' distance (by keeping the direction constant) to the defined points to increase the contribution of that control point.
You could use the (a,b) to[in=<degrees>,out=<degrees>] (c,d) options. Without any looseness specified, it is set to one. The second picture illustrates it's influence in the range from zero (blue) to two (red). If you don't want to compute the <degrees> yourself, you could do it via the atan function of pgfmath, it should be atan(incline)+180 for in and atan(incline) for the out value.
\documentclass[parskip]{scrartcl}
\usepackage[margin=15mm]{geometry}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}[scale=2]
\draw (-2,-1) to[out=45,in=180] (0,-1) to[out=0,in=135] (2,2);
\draw (-2,-1) circle (0.1) (0,-1) circle (0.1) (2,2) circle (0.1);
\end{tikzpicture}
\begin{tikzpicture}[scale=2]
\foreach \x in {0,10,...,100}
{ \pgfmathsetmacro{\loose}{\x/50}
\draw[looseness=\loose,color=red!\x!blue] (-2,-1) to[out=45,in=180] (0,-1) to[out=0,in=135] (2,2);
}
\draw (-2,-1) circle (0.1) (0,-1) circle (0.1) (2,2) circle (0.1);
\end{tikzpicture}
\end{document}
Update I added a macro to draw curve defined by three points and the tangent specifications at these points. The macro is named \expression
The polynomial here is f(x)= -(9/128)*x^5-(5/32)*x^4+(15/32)*x^3+x^2-1
I draw three functions
1) in orange the first try :
(a).. controls ((a)+(1,f'(a)) and ((b)+(-1,f'(b)) .. (b) ..
controls ((b)+(1,f'(b)) and ((c)+(-1,f'(c)) .. (c)
2)in blue :
(a).. controls ((a)+(1,f'(a)) and ((b)+(-1/2,f'(b)/2) .. (b) ..
controls ((b)+(1,f'(b)) and ((c)+(-1/2,f'(c)/2) .. (c)
3) In red the polynomial, I used my package tkz-fct to draw this last function, because I can draw easily the tangents.
As you can see on the next picture, the results are not very bad. It's possible to make a macro to get the control points automatically.
Picture
The code
\documentclass[11pt]{scrartcl}
\usepackage{tkz-fct}
\usetikzlibrary{calc}
\newcommand\expression[9]{
(#1,#2) .. controls ($(#1,#2)+(1,#3)$) and ($(#4,#5)+(-1,-#6)$) ..(#4,#5)
..controls ($(#4,#5)+(1,#6)$) and ($(#7,#8)+(-1,-#9)$)..(#7,#8)
}
\begin{document}
\begin{tikzpicture}
\tkzInit[xmin=-3,xmax=3,ymax=3,ymin=-3];
\tkzAxeXY
\draw[help lines](-2,-2) grid (3,3);
\draw[orange,line width=1pt] \expression {-2} {-1} {1}%
{0} {-1} {0}%
{2} {2} {-1};
\draw[blue,line width=1pt] (-2,-1) .. controls (-1,0) and (-0.5,-1) ..(0,-1)..controls(1,-1) and (1.5,2.5)..(2,2) ;
\tkzFct[color = red,domain =-2:2,line width=3pt,opacity=.5]{(-9./128)*x**5-(5./32)*x**4+(15./32)*x**3+x**2-1}
\begin{scope}[line width=.5pt]
\tkzDrawTangentLine[draw,color=green](-2)
\tkzDrawTangentLine[draw,color=green](0)
\tkzDrawTangentLine[draw,color=green](2)
\end{scope}
\end{tikzpicture}
\end{document}
