Skip to main content

evaluation - What happens when you divide by ##?



I've been playing around with sequences a bit. In particular with using ## with unary and binary operators.


Let's start simple, the following all make some kind of sense:


  + ## & [a,b] (* a + b *)
x + ## & [a,b] (* x + a + b *)
x * ## & [a,b] (* x * a * b *)
x ^ ## & [a,b] (* x ^ a ^ b *)

Now here is a slightly weird case:


  - ## & [a,b] (* -a*b *)
x - ## & [a,b] (* x - a*b *)


I guess, this sort of makes sense if - is actually interpreted as something like +(-1)*. But it also means that +##-## is generally non-zero.


But now here's the real puzzle:


x / ## & [a,b]   (* x a^(1/b) *)
x / ## & [a,b,c] (* x a^b^(1/c) *)

Wh... what? Can anyone explain what's happening here or at least give some justification like the one for subtraction? Answers which correct my explanation for subtraction are also welcome!


(No, I would never use this stuff in production code. But knowing what exactly is going on under the hood could come in handy some time.)


Bonus Question: Are there any other operators that yield unexpected and potentially "useful" results? (I mean, !## will yield Not[a,b] but that's neither very unexpected nor useful.)



Answer




The documentation for Minus states that



-x is converted to Times[-1,x] on input.



So -Sequence[a,b] == Times[-1,Sequence[a,b]] == Times[-1,a,b] by this definition. Similarly the documentation for Divide states that



x/y is converted to x y^-1 on input.



and therefore x / Sequence[a,b] == x Sequence[a,b]^-1. Sequence[a,b]^c == Power[a, Power[b,c]]. When c == -1 you get Power[b, -1] == 1/b.


Comments

Popular posts from this blog

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 - 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 - 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....