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I wanted to plot a user-defined Piecewise function (pagoda function) in Mathematica 10.2.
It seems straightforward to me unfortunately the easy command leads to a bad result.
My first approach was:
f[x_] := Piecewise[{{0, x <= -1}, {-Abs[x] + 1, -1 < x < 1}, {0,
x >= 1}}]
Plot3D[ 5*f[x]*f[y], {x, -1.5, 1.5}, {y, -1.5, 1.5}]
I also tried to set MaxRecursion which lead to more terrible results in a few cases (e.g. 2,3).
Can anybody tell me how to plot this function in a smooth nice way?
Thanks,
Felix
As far as I can remember, making visible gaps was introduced as a feature. Before that, piecewise or discontinuous functions were plotted like this:
Plot[Piecewise[{{x, x <= 1}, {3, x > 1}}], {x, 0, 3}, Exclusions -> None]
That behavior gives the wrong impression. I would have to check when this was default or if I'm completely off here. Anyway, as already noted in the comments, you can use the Exclusions option to get connected graphs.
You don't need to increase PlotPoints because Mathematica will (hopefully always) recognize the boundaries of the pieces as places where it needs to recursively increase points. Therefore, the MaxRecursion option is far more important to give a smooth plot. This example was rendered with only 10 points, but a recursion value of 5:
Therefore, your function renders extremely well even with 10 plot-points when the recursion is high enough. Look how many subdivisions you get on the cracks
Plot3D[5*f[x]*f[y], {x, -1.5, 1.5}, {y, -1.5, 1.5}, PlotRange -> All,
Exclusions -> None, PlotPoints -> 10, MaxRecursion -> 6, Mesh -> All]
Finally, note that the gaps are not restricted to Piecewise functions. As you can verify yourself, UnitStep will also show gaps. You can try it with your example by using an undocumented function to turn everything to UnitStep:
Simplify`PWToUnitStep[5*f[x]*f[y]]
(*
5 (1 - Abs[x]) (1 - Abs[y]) (1 - UnitStep[-1 - x]) (1 -
UnitStep[-1 + x]) (1 - UnitStep[-1 - y]) (1 - UnitStep[-1 + y])
*)
With all due respect to #halirutan, by itself MaxRecursion set to 6 was not enough in the following plot to adjust the peak values of a piecewise function to be a monotonic increasing sequence:
This improved, when, in addition, I set PlotPoints to 240, as follows.
However, that does not completely solve all the display problems. For example, note in the plots above, that the initial value y = 0, is not correctly plotted as a blue vertical line despite use of Exclusions->None. Moreover, the grid lines, which are Dotted, do not display as dots, but as dashes which run off below the x-axis. All of these problems can be solved, probably more efficiently, by generating the points or dots as list data and using ListPlot or, as in this case ListLogPlot and using Joined->True when appropriate. That is low level solution, but is needed in more complicated plots to obtain a plot with shorter run time and more accessible control over the display features.
I'm having some troubles with the Dynamic command in Mathematica, the next code shows an interactive graphic of the function f(x) = 1 - x^2. The graphic's title also shows the current area under the curve (definite integral) which is modified using the slider.
Manipulate[Show[Plot[1 - x^2, {x, 0, 1}, PlotLabel -> Integrate[1 - x^2, {x, 0, Limite - 0.000000000001}]],
Plot[-x^2 + 1, {x, 0, Limite}, PlotRange -> {0, 1}, Filling -> Axis] ], {Limite, 0.000000000001, 1}, LocalizeVariables -> False]
I would like to show the current area using this command:
Integrate[1 - x^2, {x, 0, Dynamic[Limite]}]
but the result is not what i expected. Mathematica evaluates this like
0.529 - (0.529)^3 / 3
which is correct but i don't understand why it displays an expression instead of a single number. The //FullSimplify and//N commands just don't solve the problem.
Is there a better way to obtain the result?
Am I using the Dynamic command correctly?
Thanks!
With your example the Integrate command is performed once with a symbolic upper limit. When the value of that upper limit changes the integral is not recomputed. You will get your desired result if you move the Dynamic[] wrapper from the iterator specification and wrap it around the Integrate command, which will cause the integral to be recomputed whenever Limite changes.
Dynamic[Integrate[1 - x^2, {x, 0, Limite}]]
I need to solve this equation in Mathematica:
d/dx v(x) = A . v(x)
here v is the column vector {v1(x),v2(x),v3(x),v4(x)} and
A is a 4 x 4 matrix.
I want to solve for the functions v1, v2, v3, v4 with any initial conditions.
The range of x is from 0 to 1000.
How do I write Mathematica code for this type of differential equation using NDSolve?
So, if you have some horrible matrix
A = RandomReal[0.1, {4, 4}]; (* A horrible matrix *)
which we make anti-symmetric (so the solution is oscillatory)
A = A - Transpose#A;
Define the vector of functions and their initial conditions
v[x_] := {v1[x], v2[x], v3[x], v4[x]};
init = v[0] == RandomReal[1, 4]
Then the NDSolve command looks like
sol = NDSolve[LogicalExpand[v'[x] == A.v[x] && init],
{v1, v2, v3, v4}, {x, 0, 1000}]
And the solutions can be plotted with
Plot[Evaluate[v[x] /. sol], {x, 0, 1000}]
Note that that the above differential equation is a linear, first order equation with constant coefficients, so is simply solved using a matrix exponential.
However, if the matrix A was a function of x, then analytic solutions become hard, but the numerical code stays the same.
For example, try:
A = RandomReal[1/10, {4, 4}] - Exp[-RandomReal[1/100, {4, 4}] x^2];
A = A - Transpose#A;
Which can produce solutions like
I wanted to do the same with a matrix instead of a vector v. As long as equation for it can be read correctly without knowing that this symbol represents a vector or a matrix, NDSolve deduced its character from initial condition, however in case when dimensionality of variable is explicit:
M'[t]==a[t]*IdentityMatrix[2]+M[t]
it fails.
An "ordinary" solution is to define matrix explicitly and flatten it when giving as a list of variables.
However I omitted this issue (and many relatex syntax problems) just introducing a reduntant variable which only role is to be the identity matrix but without introducing a list (matrix is 2d list, so Mathematica acts as while adding lists to each other, generating the error):
eqn = {w'[t] == a[t]*identity[t] + w[t], a'[t] == 2, identity'[t] == {{0, 0}, {0, 0}}}
init={ w[0] == {{1, 2}, {2, 1}}, a[0] == 1, identity[0] == {{1, 0}, {0, 1}}}
sol = NDSolve[eqn&&init, {w, a, identity}, {t, 0, 1}]
Some evidence of work:
Plot[{Evaluate[w[t] /. sol][[1, 1, 1]], Evaluate[w[t] /. sol][[1, 1, 2]]}, {t, 0, 1}]
Try something like this (I do not have Mathematica on my home notebook :))
NDSolve[Transpose[{v1[x],v2[x],v3[x],v4[x]}']=={{a11,a12,a13,a14},{a21,a22,a23,a24},{a31,a32,a33,a34},{a41,a42,a43,a44}}.Transpose[{v1[x],v2[x],v3[x],v4[x]}], {v1,v2,v3,v4},{x,0,1000}]
ps: you can rewrite it in a different way, replacing your record as a set of equations
{v1'[x]==a11*v1[x]+a12*v2[x]+a13*v3[x]+a14*v4[x],v2'[x]==a21*v1[x]+a22*v2[x]+a23*v3[x]+a24*v4[x], and so on..} if you want )
I've got a list of three dimensional points, ordered by time. Is there a way to plot the points so that I can get a visual representation that also includes information on where in the list the point occurred? My initial thought is to find a way to color the points by the order in which they were plotted.
ListPlot3D drapes a sheet over the points, with no regard to the order which they were plotted.
ListPointPlot just shows the points, but gives no indication as to the order in which they were plotted. It's here that I am thinking of coloring the points according to the order in which they appear in the list.
ListLinePlot doesn't seem to have a 3D cousin, unlike a lot of the other plotting functions.
You could also do something like
lst = RandomReal[{0, 3}, {20, 3}];
Graphics3D[{Thickness[0.005],
Line[lst,
VertexColors ->
Table[ColorData["BlueGreenYellow"][i], {i,
Rescale[Range[Length[lst]]]}]]}]
As you did not provide examples, I made up some by creating a 3d self-avoiding random walk:
Clear[saRW3d]
saRW3d[steps_]:=
Module[{visited},
visited[_]=False;
NestList[
(Function[{randMove},
If[
visited[#+randMove]==False,
visited[#+randMove]=True;
#+randMove,
#
]
][RandomChoice[{{1,0,0},{-1,0,0},{0,1,0},{0,-1,0},{0,0,1},{0,0,-1}}]])&,
{0,0,0},
steps
]//DeleteDuplicates
]
(this is sort of buggy but does the job; it produces a random walk in 3d which avoids itself, ie, avoids revisiting the same place in subsequent steps).
Then we produce 100000 steps like this
dat = saRW3d[100000];
this is like I understood your data points to be. We then make these change color depepnding on which step it is:
datpairs = Partition[dat, 2, 1];
len = Length#datpairs;
dressPoints[pts_, lspec_] := {RGBColor[(N#First#lspec)/len, 0, 0],
Line#pts};
datplt = MapIndexed[dressPoints, datpairs];
This can also be done all at once like the other answers
datplt=MapIndexed[
{RGBColor[(N#First##2)/Length#dat, 0, 0], Line##1} &,
Partition[dat, 2, 1]
]
but I tend to avoid this sort of constructions because I find them harder to read and modify.
Finally plot the result:
Graphics3D[datplt]
The path gets redder as time advances.
If this is the sort of thing you're after, I can elaborate.
EDIT: There might well be easier ways to do this...
EDIT2: Show a large set of points to demonstrate that this is very useful to see the qualitative trend in time in cases where arrows won't scale easily.
EDIT3: Added the one-liner version.
I think Heike's method is best, but she made it overly complex, IMHO. I would use:
Graphics3D[{
Thickness[0.005],
Line[lst,
VertexColors ->
ColorData["SolarColors"] /# Rescale#Range#Length#lst ]
}]
(acl's data)
Graphics3D#(Arrow /# Partition[RandomInteger[{0, 10}, {10, 3}], 2, 1])
As to your last question: If you want to have a kind of ListLinePlot3D instead of a ListPointPlot you could simply do the following:
pointList =
Table[{t, Sin[t] + 5 Sin[t/10], Cos[t] + 5 Cos[t/10],
t + Cos[t/10]}, {t, 0, 100, .5}];
ListPointPlot3D[pointList[[All, {2, 3, 4}]]] /. Point -> Line
Of course, in this way you can't set line properties so you have to change the rule a bit if you want that:
ListPointPlot3D[pointList[[All, {2, 3, 4}]]] /.
Point[a___] :> {Red, Thickness[0.02], Line[a]}
or with
ListPointPlot3D[pointList[[All, {2, 3, 4}]]] /.
Point[a___] :> {Red, Thickness[0.002], Line[a], Black, Point[a]}
But then, why don't you use just Graphics3D and a few graphics primitives?
I have a few differential equations that I'd like to draw solutions for, for a variety of start values N_0
Here are the equations:
dN\dt= bN^2 - aN
dN\dt = bN^2 (1 - N\K) - aN
How would I go about it?
I don't really care about the language is used. In terms of dedicated math I have mathematica and matlab on my computer. I've got access to maple. I have to do more of this stuff, and I'd like to have examples from any language, as it'll help me figure out which one I want to use and learn it.
I'll pretend the first one cannot be solved analytically so as to show how one would go about playing with a general ODE in mathematica.
Define
p1[n0_, a_, b_, uplim_: 10] :=(n /. First#NDSolve[
{n'[t] == b*n[t]^2 - a*n[t], n[0] == n0},n, {t, 0, uplim}]
which returns the solution of the ODE, i.e., a = p1[.1, 2., 3.] and then e.g. a[.3] tells you n(.3). One can then do something like
Show[Table[ans = p1[n0, 1, 1];
Plot[ans[t], {t, 0, 10}, PlotRange \[Rule] Full],
{n0, 0, 1, .05}], PlotRange \[Rule] {{0, 5}, {0, 1}}]
which plots a few solutions with different initial values:
or, to gain some insight into the solutions, one can interactively manipulate the values of a, b and n0:
Manipulate[
ans = p1[n0, a, b];
Plot[ans[t], {t, 0, 10},PlotRange -> {0, 1}],
{{n0, .1}, 0, 1},
{{a, 1}, 0, 2},
{{b, 1}, 0, 2}]
which gives something like
with the controls active (i.e. you move them and the plot changes; try it live to see what I mean; note that you can set parameters for which the initial conditions gives diverging solutions).
Of course this can be made arbitrarily more complicated. Also in this particular case this ODE is easy enough to integrate analytically, but this numerical approach can be applied to generic ODEs (and many PDEs, too).
Adding to the several good answers, if you just want a quick sketch of an ODE's solutions for many starting values, for guidance, you can always do a one-line StreamPlot. Suppose a==1 and b==1, and dy/dx == x^2 - x.
StreamPlot[{1, x^2 - x}, {x, -3, 3}, {y, -3, 3}]
StreamStyle -> "Line" will give you just lines, no arrows.
In Mathematica you use NDSolve (unless it can be solved analytically, in which case you use DSolve. So for your first equation I tried:
b = 1.1; a = 2;
s = NDSolve[{n'[t] == b n[t]^2 - a n[t], n[0] == 1}, n, {t, 0, 10}];
Plot[Evaluate[n[t] /. s], {t, 1, 10}, PlotRange -> All]
I didn't know what to use for a, b or N0, but I got this result:
If you're happy to solve the equations numerically, MATLAB has a set of ODE solvers that might be useful. Check out the documentation for the ode45 function here.
The general approach is to define an "ode function" that describes the right-hand-side of the differential equations. You then pass this function, along with initial conditions and an integration range to the ode solvers.
One attractive feature of this type of approach is that it extends in a straight-forward way to complex systems of coupled ODE's.
Hope this helps.