evaluation - General::ivar is not a valid variable when plotting - what actually causes this and how to avoid it?

I was just evaluating a couple of expressions and started to get errors like this:

General::ivar: -1.49994 is not a valid variable. >>
General::ivar: -1.43871 is not a valid variable. >>
General::ivar: -1.37749 is not a valid variable. >>
General::stop: Further output of General::ivar will be 
               suppressed during this calculation. >>

I'm doing nothing complicated - currently, I simply did this:

f[x_]:=x^2 + 2x + 1

Plot[f[x], {x, -4, 4}]
Solve[f[x] == 4]
g[x_]:=D[f[x], x]
Plot[g[x], {x, -2, 2}]
// ^ errors caused by this

Actually, this isn't the exact quadratic I am investigating, but it is a quadratic and I expected this to work. I googled, as you'd expect, and found this Stack Overflow question which suggested:

Plot[Evaluate[g[x]], {x, -2, 2}]

As a workaround.

It works - my question is, why doesn't that set of instructions generate that error (I can see it is something to do with replacing, but why is one plot different from the other?) and how can I avoid it? Is there something I should specifically have done in forming g?

Answer

The problem lies in g[x_] := D[f[x], x]; remember that what SetDelayed (that is, :=) does is to replace stuff on the right corresponding to patterns on the left before evaluating. Thus, when one does something like g[2] (and something like this happens within Plot[]), you are in fact evaluating D[f[2], 2], and since one cannot differentiate with respect to a constant ;), you get the General::ivar error message.

If you use Set instead (that is, g[x_] = D[f[x], x]), f[x] is differentiated first before the result of D[] is assigned to g[x_]. Since what's on the right of g[x_] is now an actual function, Plot[] no longer has a reason to complain.

Comments

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.

Blog

Search This Blog