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A curve pass via points at TiKz
How to use the siunitx package within Python/matplotlib?How to include a graph from python in latex textRotate a node but not its content: the case of the ellipse decorationHow to define the default vertical distance between nodes?TikZ: Drawing a curve using controlsTikZ wrong node placement on curveTikZ: Drawing an arc from an intersection to an intersectionDrawing rectilinear curves in Tikz, aka an Etch-a-Sketch drawingConcentric arc arrows in TikZTikzpicture and draw function producing an uneven lineDrawing a Cayley treeTikZ: fill text color different than fill
Look at this image:
This is what I get from this:
begintikzpicture
draw[style=help lines] (-5,-5) grid (5,5);
draw (-4,0)--(4,0);
draw (0,-4)--(0,4);
foreach y in -4,-3,...,4
draw (0 - 0.1,y) -- (0+0.1,y);
draw (y,0 - 0.1) -- (y,0+0.1);
%Nodes:
node (a0) at (-4,-4) ;
draw[fill] (a0) circle [radius=1.5pt];
node (a1) at (-2,4) ;
draw[fill] (a1) circle [radius=1.5pt];
node (a2) at (2,-2) ;
draw[fill] (a2) circle [radius=1.5pt];
node (a3) at (4,2) ;
draw[fill] (a3) circle [radius=1.5pt];
draw (-4,-4) to (-2,4) to (2,-2) to (4,2); % to (a2) to (a3);
endtikzpicture
I'm trying to to get a line between them (the dots) that will be like a function (not a straight line - a curve like a polynomial).
Is this possible?
Thank you!
tikz-pgf
asked 1 hour ago
heblyxheblyx
1,0511020
add a comment |
Look at this image:
This is what I get from this:
begintikzpicture
draw[style=help lines] (-5,-5) grid (5,5);
draw (-4,0)--(4,0);
draw (0,-4)--(0,4);
foreach y in -4,-3,...,4
draw (0 - 0.1,y) -- (0+0.1,y);
draw (y,0 - 0.1) -- (y,0+0.1);
%Nodes:
node (a0) at (-4,-4) ;
draw[fill] (a0) circle [radius=1.5pt];
node (a1) at (-2,4) ;
draw[fill] (a1) circle [radius=1.5pt];
node (a2) at (2,-2) ;
draw[fill] (a2) circle [radius=1.5pt];
node (a3) at (4,2) ;
draw[fill] (a3) circle [radius=1.5pt];
draw (-4,-4) to (-2,4) to (2,-2) to (4,2); % to (a2) to (a3);
endtikzpicture
I'm trying to to get a line between them (the dots) that will be like a function (not a straight line - a curve like a polynomial).
Is this possible?
Thank you!
tikz-pgf
asked 1 hour ago
heblyxheblyx
1,0511020
1
yes, mathematically it's possible: a cubic interpolation polynomial.
– Bernard
1 hour ago
add a comment |
Look at this image:
This is what I get from this:
begintikzpicture
draw[style=help lines] (-5,-5) grid (5,5);
draw (-4,0)--(4,0);
draw (0,-4)--(0,4);
foreach y in -4,-3,...,4
draw (0 - 0.1,y) -- (0+0.1,y);
draw (y,0 - 0.1) -- (y,0+0.1);
%Nodes:
node (a0) at (-4,-4) ;
draw[fill] (a0) circle [radius=1.5pt];
node (a1) at (-2,4) ;
draw[fill] (a1) circle [radius=1.5pt];
node (a2) at (2,-2) ;
draw[fill] (a2) circle [radius=1.5pt];
node (a3) at (4,2) ;
draw[fill] (a3) circle [radius=1.5pt];
draw (-4,-4) to (-2,4) to (2,-2) to (4,2); % to (a2) to (a3);
endtikzpicture
I'm trying to to get a line between them (the dots) that will be like a function (not a straight line - a curve like a polynomial).
Is this possible?
Thank you!
tikz-pgf
asked 1 hour ago
heblyxheblyx
1,0511020
Look at this image:
This is what I get from this:
begintikzpicture
draw[style=help lines] (-5,-5) grid (5,5);
draw (-4,0)--(4,0);
draw (0,-4)--(0,4);
foreach y in -4,-3,...,4
draw (0 - 0.1,y) -- (0+0.1,y);
draw (y,0 - 0.1) -- (y,0+0.1);
%Nodes:
node (a0) at (-4,-4) ;
draw[fill] (a0) circle [radius=1.5pt];
node (a1) at (-2,4) ;
draw[fill] (a1) circle [radius=1.5pt];
node (a2) at (2,-2) ;
draw[fill] (a2) circle [radius=1.5pt];
node (a3) at (4,2) ;
draw[fill] (a3) circle [radius=1.5pt];
draw (-4,-4) to (-2,4) to (2,-2) to (4,2); % to (a2) to (a3);
endtikzpicture
I'm trying to to get a line between them (the dots) that will be like a function (not a straight line - a curve like a polynomial).
Is this possible?
Thank you!
tikz-pgf
tikz-pgf
asked 1 hour ago
heblyxheblyx
1,0511020
asked 1 hour ago
heblyxheblyx
1,0511020
asked 1 hour ago
heblyxheblyx
1,0511020
asked 1 hour ago
heblyxheblyx
1,0511020
asked 1 hour ago
heblyxheblyx
1,0511020
1,0511020
1
yes, mathematically it's possible: a cubic interpolation polynomial.
– Bernard
1 hour ago
add a comment |
1
yes, mathematically it's possible: a cubic interpolation polynomial.
– Bernard
1 hour ago
1
1
yes, mathematically it's possible: a cubic interpolation polynomial.
– Bernard
1 hour ago
yes, mathematically it's possible: a cubic interpolation polynomial.
– Bernard
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
You can use plot [smooth] coordinates (which is not a single polynom but a spline):
documentclass[tikz]standalone
begindocument
begintikzpicture
draw[style=help lines] (-5,-5) grid (5,5);
draw (-4,0)--(4,0);
draw (0,-4)--(0,4);
foreach y in -4,-3,...,4
draw (0 - 0.1,y) -- (0+0.1,y);
draw (y,0 - 0.1) -- (y,0+0.1);
%Nodes:
node (a0) at (-4,-4) ;
draw[fill] (a0) circle [radius=1.5pt];
node (a1) at (-2,4) ;
draw[fill] (a1) circle [radius=1.5pt];
node (a2) at (2,-2) ;
draw[fill] (a2) circle [radius=1.5pt];
node (a3) at (4,2) ;
draw[fill] (a3) circle [radius=1.5pt];
draw plot [smooth] coordinates (-4,-4) (-2,4) (2,-2) (4,2); % to (a2) to (a3);
endtikzpicture
enddocument
Solution which forces the middle points to have a horizontal tangent:
documentclass[tikz,border=3.14]standalone
begindocument
begintikzpicture
draw[style=help lines] (-5,-5) grid (5,5);
draw (-4,0)--(4,0);
draw (0,-4)--(0,4);
foreach y in -4,-3,...,4
draw (0 - 0.1,y) -- (0+0.1,y);
draw (y,0 - 0.1) -- (y,0+0.1);
%Nodes:
node (a0) at (-4,-4) ;
draw[fill] (a0) circle [radius=1.5pt];
node (a1) at (-2,4) ;
draw[fill] (a1) circle [radius=1.5pt];
node (a2) at (2,-2) ;
draw[fill] (a2) circle [radius=1.5pt];
node (a3) at (4,2) ;
draw[fill] (a3) circle [radius=1.5pt];
draw (-4,-4) to[out=90,in=180] (-2,4) to[out=0,in=180] (2,-2) to[out=0,in=-95] (4,2); % to (a2) to (a3);
endtikzpicture
enddocument
I don't know how to compute this in LaTeX easily, so I fitted a plot using Python's numpy.polyfit and used the result to plot the fit in TikZ:
documentclass[tikz,border=3.14]standalone
%% polynomial coefficients found with Python (numpy.polyfit)
%% $f(x) = 0.1875 x^3 - 1/6 x^2 - 2.25 x^1 + 10/6 x^0$
begindocument
begintikzpicture
draw[style=help lines] (-5,-5) grid (5,5);
draw (-4,0)--(4,0);
draw (0,-4)--(0,4);
foreach y in -4,-3,...,4
draw (0 - 0.1,y) -- (0+0.1,y);
draw (y,0 - 0.1) -- (y,0+0.1);
%Nodes:
node (a0) at (-4,-4) ;
draw[fill] (a0) circle [radius=1.5pt];
node (a1) at (-2,4) ;
draw[fill] (a1) circle [radius=1.5pt];
node (a2) at (2,-2) ;
draw[fill] (a2) circle [radius=1.5pt];
node (a3) at (4,2) ;
draw[fill] (a3) circle [radius=1.5pt];
draw plot[domain=-4:4,samples=100] (x, .1875*x*x*x - x*x/6 - 2.25*x + 10/6);
endtikzpicture
enddocument
Just for your information. You can calculate and plot the interpolation polynomial with Python and the two libraries Matplotlib and NumPy:
import numpy as np
import matplotlib.pyplot as plt
x = (-4, -2, 2, 4)
y = (-4, 4, -2, 2)
p = np.polyfit(x,y,3)
t = np.linspace(min(x),max(x),num=100)
f = np.polyval(p,t)
plt.plot(t,f)
Matplotlib supports export to TikZ code (actually it exports to PGF) and to save the plots directly as PDF created with TikZ and LaTeX (see for example https://tex.stackexchange.com/a/426071/117050 and https://tex.stackexchange.com/a/391078/117050 for some code that might get you started).
answered 1 hour ago
SkillmonSkillmon
24.7k12250
1
It's a piece wise defined function built from multiple polynomial functions (between each two points there is one cubic polynomial, the differentiates of two neighbouring polynomials must be equal in the points).
– Skillmon
1 hour ago
1
@heblyx it does exactly pass them, but the middle points aren't the local extrema of the function. Else the function wouldn't be smooth.
– Skillmon
1 hour ago
1
Yes, but afterwards it's no longer a mathematical function, but a parametric curve.
– Skillmon
1 hour ago
1
@heblyx a piece wise defined function which maps a single $y$ value to each $x$. The second one doesn't have a single $y$ to each $x$ and therefore is no longer a function, but a parametric curve.
– Skillmon
1 hour ago
2
@heblyx There is also thehobbylibrary (which is not documented in the pgfmanual) which allows you to draw all sorts of smooth curves through a set of points, and you can fix the slopes and so on.
– marmot
1 hour ago
|
show 17 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can use plot [smooth] coordinates (which is not a single polynom but a spline):
documentclass[tikz]standalone
begindocument
begintikzpicture
draw[style=help lines] (-5,-5) grid (5,5);
draw (-4,0)--(4,0);
draw (0,-4)--(0,4);
foreach y in -4,-3,...,4
draw (0 - 0.1,y) -- (0+0.1,y);
draw (y,0 - 0.1) -- (y,0+0.1);
%Nodes:
node (a0) at (-4,-4) ;
draw[fill] (a0) circle [radius=1.5pt];
node (a1) at (-2,4) ;
draw[fill] (a1) circle [radius=1.5pt];
node (a2) at (2,-2) ;
draw[fill] (a2) circle [radius=1.5pt];
node (a3) at (4,2) ;
draw[fill] (a3) circle [radius=1.5pt];
draw plot [smooth] coordinates (-4,-4) (-2,4) (2,-2) (4,2); % to (a2) to (a3);
endtikzpicture
enddocument
Solution which forces the middle points to have a horizontal tangent:
documentclass[tikz,border=3.14]standalone
begindocument
begintikzpicture
draw[style=help lines] (-5,-5) grid (5,5);
draw (-4,0)--(4,0);
draw (0,-4)--(0,4);
foreach y in -4,-3,...,4
draw (0 - 0.1,y) -- (0+0.1,y);
draw (y,0 - 0.1) -- (y,0+0.1);
%Nodes:
node (a0) at (-4,-4) ;
draw[fill] (a0) circle [radius=1.5pt];
node (a1) at (-2,4) ;
draw[fill] (a1) circle [radius=1.5pt];
node (a2) at (2,-2) ;
draw[fill] (a2) circle [radius=1.5pt];
node (a3) at (4,2) ;
draw[fill] (a3) circle [radius=1.5pt];
draw (-4,-4) to[out=90,in=180] (-2,4) to[out=0,in=180] (2,-2) to[out=0,in=-95] (4,2); % to (a2) to (a3);
endtikzpicture
enddocument
I don't know how to compute this in LaTeX easily, so I fitted a plot using Python's numpy.polyfit and used the result to plot the fit in TikZ:
documentclass[tikz,border=3.14]standalone
%% polynomial coefficients found with Python (numpy.polyfit)
%% $f(x) = 0.1875 x^3 - 1/6 x^2 - 2.25 x^1 + 10/6 x^0$
begindocument
begintikzpicture
draw[style=help lines] (-5,-5) grid (5,5);
draw (-4,0)--(4,0);
draw (0,-4)--(0,4);
foreach y in -4,-3,...,4
draw (0 - 0.1,y) -- (0+0.1,y);
draw (y,0 - 0.1) -- (y,0+0.1);
%Nodes:
node (a0) at (-4,-4) ;
draw[fill] (a0) circle [radius=1.5pt];
node (a1) at (-2,4) ;
draw[fill] (a1) circle [radius=1.5pt];
node (a2) at (2,-2) ;
draw[fill] (a2) circle [radius=1.5pt];
node (a3) at (4,2) ;
draw[fill] (a3) circle [radius=1.5pt];
draw plot[domain=-4:4,samples=100] (x, .1875*x*x*x - x*x/6 - 2.25*x + 10/6);
endtikzpicture
enddocument
Just for your information. You can calculate and plot the interpolation polynomial with Python and the two libraries Matplotlib and NumPy:
import numpy as np
import matplotlib.pyplot as plt
x = (-4, -2, 2, 4)
y = (-4, 4, -2, 2)
p = np.polyfit(x,y,3)
t = np.linspace(min(x),max(x),num=100)
f = np.polyval(p,t)
plt.plot(t,f)
Matplotlib supports export to TikZ code (actually it exports to PGF) and to save the plots directly as PDF created with TikZ and LaTeX (see for example https://tex.stackexchange.com/a/426071/117050 and https://tex.stackexchange.com/a/391078/117050 for some code that might get you started).
answered 1 hour ago
SkillmonSkillmon
24.7k12250
1
It's a piece wise defined function built from multiple polynomial functions (between each two points there is one cubic polynomial, the differentiates of two neighbouring polynomials must be equal in the points).
– Skillmon
1 hour ago
1
@heblyx it does exactly pass them, but the middle points aren't the local extrema of the function. Else the function wouldn't be smooth.
– Skillmon
1 hour ago
1
Yes, but afterwards it's no longer a mathematical function, but a parametric curve.
– Skillmon
1 hour ago
1
@heblyx a piece wise defined function which maps a single $y$ value to each $x$. The second one doesn't have a single $y$ to each $x$ and therefore is no longer a function, but a parametric curve.
– Skillmon
1 hour ago
2
@heblyx There is also thehobbylibrary (which is not documented in the pgfmanual) which allows you to draw all sorts of smooth curves through a set of points, and you can fix the slopes and so on.
– marmot
1 hour ago
|
show 17 more comments
You can use plot [smooth] coordinates (which is not a single polynom but a spline):
documentclass[tikz]standalone
begindocument
begintikzpicture
draw[style=help lines] (-5,-5) grid (5,5);
draw (-4,0)--(4,0);
draw (0,-4)--(0,4);
foreach y in -4,-3,...,4
draw (0 - 0.1,y) -- (0+0.1,y);
draw (y,0 - 0.1) -- (y,0+0.1);
%Nodes:
node (a0) at (-4,-4) ;
draw[fill] (a0) circle [radius=1.5pt];
node (a1) at (-2,4) ;
draw[fill] (a1) circle [radius=1.5pt];
node (a2) at (2,-2) ;
draw[fill] (a2) circle [radius=1.5pt];
node (a3) at (4,2) ;
draw[fill] (a3) circle [radius=1.5pt];
draw plot [smooth] coordinates (-4,-4) (-2,4) (2,-2) (4,2); % to (a2) to (a3);
endtikzpicture
enddocument
Solution which forces the middle points to have a horizontal tangent:
documentclass[tikz,border=3.14]standalone
begindocument
begintikzpicture
draw[style=help lines] (-5,-5) grid (5,5);
draw (-4,0)--(4,0);
draw (0,-4)--(0,4);
foreach y in -4,-3,...,4
draw (0 - 0.1,y) -- (0+0.1,y);
draw (y,0 - 0.1) -- (y,0+0.1);
%Nodes:
node (a0) at (-4,-4) ;
draw[fill] (a0) circle [radius=1.5pt];
node (a1) at (-2,4) ;
draw[fill] (a1) circle [radius=1.5pt];
node (a2) at (2,-2) ;
draw[fill] (a2) circle [radius=1.5pt];
node (a3) at (4,2) ;
draw[fill] (a3) circle [radius=1.5pt];
draw (-4,-4) to[out=90,in=180] (-2,4) to[out=0,in=180] (2,-2) to[out=0,in=-95] (4,2); % to (a2) to (a3);
endtikzpicture
enddocument
I don't know how to compute this in LaTeX easily, so I fitted a plot using Python's numpy.polyfit and used the result to plot the fit in TikZ:
documentclass[tikz,border=3.14]standalone
%% polynomial coefficients found with Python (numpy.polyfit)
%% $f(x) = 0.1875 x^3 - 1/6 x^2 - 2.25 x^1 + 10/6 x^0$
begindocument
begintikzpicture
draw[style=help lines] (-5,-5) grid (5,5);
draw (-4,0)--(4,0);
draw (0,-4)--(0,4);
foreach y in -4,-3,...,4
draw (0 - 0.1,y) -- (0+0.1,y);
draw (y,0 - 0.1) -- (y,0+0.1);
%Nodes:
node (a0) at (-4,-4) ;
draw[fill] (a0) circle [radius=1.5pt];
node (a1) at (-2,4) ;
draw[fill] (a1) circle [radius=1.5pt];
node (a2) at (2,-2) ;
draw[fill] (a2) circle [radius=1.5pt];
node (a3) at (4,2) ;
draw[fill] (a3) circle [radius=1.5pt];
draw plot[domain=-4:4,samples=100] (x, .1875*x*x*x - x*x/6 - 2.25*x + 10/6);
endtikzpicture
enddocument
Just for your information. You can calculate and plot the interpolation polynomial with Python and the two libraries Matplotlib and NumPy:
import numpy as np
import matplotlib.pyplot as plt
x = (-4, -2, 2, 4)
y = (-4, 4, -2, 2)
p = np.polyfit(x,y,3)
t = np.linspace(min(x),max(x),num=100)
f = np.polyval(p,t)
plt.plot(t,f)
Matplotlib supports export to TikZ code (actually it exports to PGF) and to save the plots directly as PDF created with TikZ and LaTeX (see for example https://tex.stackexchange.com/a/426071/117050 and https://tex.stackexchange.com/a/391078/117050 for some code that might get you started).
answered 1 hour ago
SkillmonSkillmon
24.7k12250
1
It's a piece wise defined function built from multiple polynomial functions (between each two points there is one cubic polynomial, the differentiates of two neighbouring polynomials must be equal in the points).
– Skillmon
1 hour ago
1
@heblyx it does exactly pass them, but the middle points aren't the local extrema of the function. Else the function wouldn't be smooth.
– Skillmon
1 hour ago
1
Yes, but afterwards it's no longer a mathematical function, but a parametric curve.
– Skillmon
1 hour ago
1
@heblyx a piece wise defined function which maps a single $y$ value to each $x$. The second one doesn't have a single $y$ to each $x$ and therefore is no longer a function, but a parametric curve.
– Skillmon
1 hour ago
2
@heblyx There is also thehobbylibrary (which is not documented in the pgfmanual) which allows you to draw all sorts of smooth curves through a set of points, and you can fix the slopes and so on.
– marmot
1 hour ago
|
show 17 more comments
You can use plot [smooth] coordinates (which is not a single polynom but a spline):
documentclass[tikz]standalone
begindocument
begintikzpicture
draw[style=help lines] (-5,-5) grid (5,5);
draw (-4,0)--(4,0);
draw (0,-4)--(0,4);
foreach y in -4,-3,...,4
draw (0 - 0.1,y) -- (0+0.1,y);
draw (y,0 - 0.1) -- (y,0+0.1);
%Nodes:
node (a0) at (-4,-4) ;
draw[fill] (a0) circle [radius=1.5pt];
node (a1) at (-2,4) ;
draw[fill] (a1) circle [radius=1.5pt];
node (a2) at (2,-2) ;
draw[fill] (a2) circle [radius=1.5pt];
node (a3) at (4,2) ;
draw[fill] (a3) circle [radius=1.5pt];
draw plot [smooth] coordinates (-4,-4) (-2,4) (2,-2) (4,2); % to (a2) to (a3);
endtikzpicture
enddocument
Solution which forces the middle points to have a horizontal tangent:
documentclass[tikz,border=3.14]standalone
begindocument
begintikzpicture
draw[style=help lines] (-5,-5) grid (5,5);
draw (-4,0)--(4,0);
draw (0,-4)--(0,4);
foreach y in -4,-3,...,4
draw (0 - 0.1,y) -- (0+0.1,y);
draw (y,0 - 0.1) -- (y,0+0.1);
%Nodes:
node (a0) at (-4,-4) ;
draw[fill] (a0) circle [radius=1.5pt];
node (a1) at (-2,4) ;
draw[fill] (a1) circle [radius=1.5pt];
node (a2) at (2,-2) ;
draw[fill] (a2) circle [radius=1.5pt];
node (a3) at (4,2) ;
draw[fill] (a3) circle [radius=1.5pt];
draw (-4,-4) to[out=90,in=180] (-2,4) to[out=0,in=180] (2,-2) to[out=0,in=-95] (4,2); % to (a2) to (a3);
endtikzpicture
enddocument
I don't know how to compute this in LaTeX easily, so I fitted a plot using Python's numpy.polyfit and used the result to plot the fit in TikZ:
documentclass[tikz,border=3.14]standalone
%% polynomial coefficients found with Python (numpy.polyfit)
%% $f(x) = 0.1875 x^3 - 1/6 x^2 - 2.25 x^1 + 10/6 x^0$
begindocument
begintikzpicture
draw[style=help lines] (-5,-5) grid (5,5);
draw (-4,0)--(4,0);
draw (0,-4)--(0,4);
foreach y in -4,-3,...,4
draw (0 - 0.1,y) -- (0+0.1,y);
draw (y,0 - 0.1) -- (y,0+0.1);
%Nodes:
node (a0) at (-4,-4) ;
draw[fill] (a0) circle [radius=1.5pt];
node (a1) at (-2,4) ;
draw[fill] (a1) circle [radius=1.5pt];
node (a2) at (2,-2) ;
draw[fill] (a2) circle [radius=1.5pt];
node (a3) at (4,2) ;
draw[fill] (a3) circle [radius=1.5pt];
draw plot[domain=-4:4,samples=100] (x, .1875*x*x*x - x*x/6 - 2.25*x + 10/6);
endtikzpicture
enddocument
Just for your information. You can calculate and plot the interpolation polynomial with Python and the two libraries Matplotlib and NumPy:
import numpy as np
import matplotlib.pyplot as plt
x = (-4, -2, 2, 4)
y = (-4, 4, -2, 2)
p = np.polyfit(x,y,3)
t = np.linspace(min(x),max(x),num=100)
f = np.polyval(p,t)
plt.plot(t,f)
Matplotlib supports export to TikZ code (actually it exports to PGF) and to save the plots directly as PDF created with TikZ and LaTeX (see for example https://tex.stackexchange.com/a/426071/117050 and https://tex.stackexchange.com/a/391078/117050 for some code that might get you started).
answered 1 hour ago
SkillmonSkillmon
24.7k12250
You can use plot [smooth] coordinates (which is not a single polynom but a spline):
documentclass[tikz]standalone
begindocument
begintikzpicture
draw[style=help lines] (-5,-5) grid (5,5);
draw (-4,0)--(4,0);
draw (0,-4)--(0,4);
foreach y in -4,-3,...,4
draw (0 - 0.1,y) -- (0+0.1,y);
draw (y,0 - 0.1) -- (y,0+0.1);
%Nodes:
node (a0) at (-4,-4) ;
draw[fill] (a0) circle [radius=1.5pt];
node (a1) at (-2,4) ;
draw[fill] (a1) circle [radius=1.5pt];
node (a2) at (2,-2) ;
draw[fill] (a2) circle [radius=1.5pt];
node (a3) at (4,2) ;
draw[fill] (a3) circle [radius=1.5pt];
draw plot [smooth] coordinates (-4,-4) (-2,4) (2,-2) (4,2); % to (a2) to (a3);
endtikzpicture
enddocument
Solution which forces the middle points to have a horizontal tangent:
documentclass[tikz,border=3.14]standalone
begindocument
begintikzpicture
draw[style=help lines] (-5,-5) grid (5,5);
draw (-4,0)--(4,0);
draw (0,-4)--(0,4);
foreach y in -4,-3,...,4
draw (0 - 0.1,y) -- (0+0.1,y);
draw (y,0 - 0.1) -- (y,0+0.1);
%Nodes:
node (a0) at (-4,-4) ;
draw[fill] (a0) circle [radius=1.5pt];
node (a1) at (-2,4) ;
draw[fill] (a1) circle [radius=1.5pt];
node (a2) at (2,-2) ;
draw[fill] (a2) circle [radius=1.5pt];
node (a3) at (4,2) ;
draw[fill] (a3) circle [radius=1.5pt];
draw (-4,-4) to[out=90,in=180] (-2,4) to[out=0,in=180] (2,-2) to[out=0,in=-95] (4,2); % to (a2) to (a3);
endtikzpicture
enddocument
I don't know how to compute this in LaTeX easily, so I fitted a plot using Python's numpy.polyfit and used the result to plot the fit in TikZ:
documentclass[tikz,border=3.14]standalone
%% polynomial coefficients found with Python (numpy.polyfit)
%% $f(x) = 0.1875 x^3 - 1/6 x^2 - 2.25 x^1 + 10/6 x^0$
begindocument
begintikzpicture
draw[style=help lines] (-5,-5) grid (5,5);
draw (-4,0)--(4,0);
draw (0,-4)--(0,4);
foreach y in -4,-3,...,4
draw (0 - 0.1,y) -- (0+0.1,y);
draw (y,0 - 0.1) -- (y,0+0.1);
%Nodes:
node (a0) at (-4,-4) ;
draw[fill] (a0) circle [radius=1.5pt];
node (a1) at (-2,4) ;
draw[fill] (a1) circle [radius=1.5pt];
node (a2) at (2,-2) ;
draw[fill] (a2) circle [radius=1.5pt];
node (a3) at (4,2) ;
draw[fill] (a3) circle [radius=1.5pt];
draw plot[domain=-4:4,samples=100] (x, .1875*x*x*x - x*x/6 - 2.25*x + 10/6);
endtikzpicture
enddocument
Just for your information. You can calculate and plot the interpolation polynomial with Python and the two libraries Matplotlib and NumPy:
import numpy as np
import matplotlib.pyplot as plt
x = (-4, -2, 2, 4)
y = (-4, 4, -2, 2)
p = np.polyfit(x,y,3)
t = np.linspace(min(x),max(x),num=100)
f = np.polyval(p,t)
plt.plot(t,f)
Matplotlib supports export to TikZ code (actually it exports to PGF) and to save the plots directly as PDF created with TikZ and LaTeX (see for example https://tex.stackexchange.com/a/426071/117050 and https://tex.stackexchange.com/a/391078/117050 for some code that might get you started).
answered 1 hour ago
SkillmonSkillmon
24.7k12250
edited 31 mins ago
answered 1 hour ago
SkillmonSkillmon
24.7k12250
answered 1 hour ago
SkillmonSkillmon
24.7k12250
answered 1 hour ago
SkillmonSkillmon
24.7k12250
24.7k12250
1
It's a piece wise defined function built from multiple polynomial functions (between each two points there is one cubic polynomial, the differentiates of two neighbouring polynomials must be equal in the points).
– Skillmon
1 hour ago
1
@heblyx it does exactly pass them, but the middle points aren't the local extrema of the function. Else the function wouldn't be smooth.
– Skillmon
1 hour ago
1
Yes, but afterwards it's no longer a mathematical function, but a parametric curve.
– Skillmon
1 hour ago
1
@heblyx a piece wise defined function which maps a single $y$ value to each $x$. The second one doesn't have a single $y$ to each $x$ and therefore is no longer a function, but a parametric curve.
– Skillmon
1 hour ago
2
@heblyx There is also thehobbylibrary (which is not documented in the pgfmanual) which allows you to draw all sorts of smooth curves through a set of points, and you can fix the slopes and so on.
– marmot
1 hour ago
|
show 17 more comments
1
It's a piece wise defined function built from multiple polynomial functions (between each two points there is one cubic polynomial, the differentiates of two neighbouring polynomials must be equal in the points).
– Skillmon
1 hour ago
1
@heblyx it does exactly pass them, but the middle points aren't the local extrema of the function. Else the function wouldn't be smooth.
– Skillmon
1 hour ago
1
Yes, but afterwards it's no longer a mathematical function, but a parametric curve.
– Skillmon
1 hour ago
1
@heblyx a piece wise defined function which maps a single $y$ value to each $x$. The second one doesn't have a single $y$ to each $x$ and therefore is no longer a function, but a parametric curve.
– Skillmon
1 hour ago
2
@heblyx There is also thehobbylibrary (which is not documented in the pgfmanual) which allows you to draw all sorts of smooth curves through a set of points, and you can fix the slopes and so on.
– marmot
1 hour ago
1
1
It's a piece wise defined function built from multiple polynomial functions (between each two points there is one cubic polynomial, the differentiates of two neighbouring polynomials must be equal in the points).
– Skillmon
1 hour ago
It's a piece wise defined function built from multiple polynomial functions (between each two points there is one cubic polynomial, the differentiates of two neighbouring polynomials must be equal in the points).
– Skillmon
1 hour ago
1
1
@heblyx it does exactly pass them, but the middle points aren't the local extrema of the function. Else the function wouldn't be smooth.
– Skillmon
1 hour ago
@heblyx it does exactly pass them, but the middle points aren't the local extrema of the function. Else the function wouldn't be smooth.
– Skillmon
1 hour ago
1
1
Yes, but afterwards it's no longer a mathematical function, but a parametric curve.
– Skillmon
1 hour ago
Yes, but afterwards it's no longer a mathematical function, but a parametric curve.
– Skillmon
1 hour ago
1
1
@heblyx a piece wise defined function which maps a single $y$ value to each $x$. The second one doesn't have a single $y$ to each $x$ and therefore is no longer a function, but a parametric curve.
– Skillmon
1 hour ago
@heblyx a piece wise defined function which maps a single $y$ value to each $x$. The second one doesn't have a single $y$ to each $x$ and therefore is no longer a function, but a parametric curve.
– Skillmon
1 hour ago
2
2
@heblyx There is also the
hobby library (which is not documented in the pgfmanual) which allows you to draw all sorts of smooth curves through a set of points, and you can fix the slopes and so on.– marmot
1 hour ago
@heblyx There is also the
hobby library (which is not documented in the pgfmanual) which allows you to draw all sorts of smooth curves through a set of points, and you can fix the slopes and so on.– marmot
1 hour ago
|
show 17 more comments
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yes, mathematically it's possible: a cubic interpolation polynomial.
– Bernard
1 hour ago