14
votes
Accepted
How to divide the contour in three parts with the same arclength?
1. Specify Mesh as a list
...
7
votes
Accepted
7
votes
How to divide the contour in three parts with the same arclength?
Using arrows and arrowheads.
...
6
votes
Accepted
Approximation of the Fabius function with a quotient of exponentials
This question is very interesting since the Fabius function is a little mathematical treasure and many of its properties can be successfully demonstrated with Mathematica. It has a potential of ...
5
votes
Accepted
Plotting the z(x,y) = 0 plane with Plot3D. How can I make the Mesh in polar coordinates?
Try
Plot3D[0, {x, -10, 10}, {y, -10, 10},MeshFunctions -> {(Sqrt[#1^2 + #2^2]&), ArcTan[#1, #2] & } , MaxRecursion -> 4, PlotPoints -> 100]
5
votes
How to define step size of y axis in mathematica plot
It's not the "stepsize", try PlotRange
...
5
votes
Accepted
How to show a contourplot within a region?
RegionFunction -> Function[{x, y, z}, {x, y, z} ∈ R]
Or
RegionFunction -> Function[{x, y, z}, RegionMember[R, {x, y, z}]]
...
5
votes
Accepted
Plotting two variables from multiple lists
If your lists are together in one larger list like this,
lists = {l1, l2, l3, l4, <... potentially many more lists ...>};
then you can use ...
4
votes
Accepted
How to modify this code so that the first derivative image of the function displays different colors above and below the x-axis?
Replace {functions} in the first argument of Plot with
...
4
votes
Accepted
How to plot properly the disconnected regions if using the option Joined->True?
We can remove Flatten and export the two groups of points to xls with two sheets for further use.
...
4
votes
Plotting two variables from multiple lists
If I am understanding correctly, you have a large number of lists that look like {x1 -> val1, x2 -> val2, x3 -> val3, x4 -> val4}. But if you gave them all variable names by hand (l1, l2 ,...
4
votes
How to draw a line on an existing region using RegionPlot with manipulate?
Next time please paste the code used. Since it is small, I retyped it now.
The problem is that you have global z and control variable also called z. These should not be mixed. Here is one workaround. ...
4
votes
Approximation of the Fabius function with a quotient of exponentials
The Fabius function is approximated well by this piecewise polynomial function:
...
3
votes
How to get DFT from Sound/Audio/Image?
If you already, have Audio or lists of samples, then the answer is simple.
For Sound objects, you should convert them to ...
3
votes
Approximation of the Fabius function with a quotient of exponentials
Extended comment:
Optimal exponent a=? for the approximation
If we take the general ansatz ...
2
votes
Plotting the z(x,y) = 0 plane with Plot3D. How can I make the Mesh in polar coordinates?
Here some approaches:
Using MeshFunctions
...
2
votes
Accepted
2
votes
Accepted
Vectorizing in 3D environments
replace ViewPoint -> {Infinity, 0, 0} to ViewPoint -> {1, 0, 0}, ViewProjection -> "Orthographic"
...
2
votes
Accepted
Solving 3D heat equation with an off-center boundary condition
Note, that periodic boundary conditions and Dirichlet type condition could not be apply in one border simultaneously as it shown in a picture above. Therefore we should replace of center b.c. from the ...
1
vote
Plane embedded in Non-Euclidean spacetime
Here's the simple approach in case it helps (not a full answer), noting that we can phrase the constraint $u=x$ in terms of $t$ by substituting $x$ in for $u$ in $t = u + \int \frac{dz}{f(u,z)}$:
<...
1
vote
Problem during the analysis and plotting of a large data
When I try to plot only 1 single contour line with a reduced data set (you have over 4 millions points) I get:
...
1
vote
Accepted
1
vote
Accepted
Solving an implicit equation involving Elliptic integral
Clear["Global`*"]
eq1 = EllipticE[1/(1 + 16*(ja*me + jb*mo)^2)] +
16*(ja*me + jb*mo)^2*EllipticK[1/(1 + 16*(ja*me + jb*mo)^2)] == 0;
As recommended ...
1
vote
Solving an implicit equation involving Elliptic integral
The equation can be reduced to a (complex) quadratic equation in four variables. One may obtain one variable in terms of the other three, but one may not obtain two of the variables in terms of the ...
1
vote
Accepted
How to adjust PlotRange automatically depending on the value of the function?
You could make a function to calculate the vertical bounds of the plot that takes the function as input and calculates the min and max of the function over the range {0.01, 100}. If the min is less ...
1
vote
when we solve the simple ODE then
Being new, you should be aware built-in Mathematica functions and commands begin with upper-case letters including the single-letter commands: C,D, E, I, K,N O. If you were to begin a user-defined ...
1
vote
LOG transformation
As an alternative to the already given approach, try this:
data/.{x_,y_}->{x,Log(y+1)}
Have fun!
1
vote
Vectorizing in 3D environments
Not really an answer, but a solution to my particular problem:
...
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