Skip to main content

function construction - Can SetDelayed (:=) be compiled?


Is it possible to put an If statement as below within a compile (see below)? I received a warning about SetDelayed.


  eef = Compile[{{μ, _Real}, {NNN, _Integer}}, 
Module[{NN = NNN},
kd[m_, n_] := If[m == n, 1, 0];
cosfun[m_, n_] :=
If[m == n,
0,
(1 - Cos[(m + n) π])/( (m + n) π)
+ ((Cos[(m - n) π] - 1)/( (m - n) π))

];

mm = Table[kd[m, n] (MM - B2 k k - B1 (m Pi/L)^2),
{m, 1, NN}, {n, 1, NN}];
kz = Table[cosfun[m, n] (-I A2 m Pi/L), {m, 1, NN}, {n, 1, NN}];
kxM = Table[kd[m, n] A1 k, {m, 1, NN}, {n, 1, NN}];
μM = Table[kd[m, n] μ, {m, 1, NN}, {n, 1, NN}];

HH = ArrayFlatten[{
{μM + mm, 0 mm, kz, kxM},

{0 mm, μM + mm, kxM, -kz},
{kz, kxM, μM - mm, 0 mm},
{kxM, -kz, 0 mm, μM - mm}
}];

ees = Table[Eigenvalues[HH], {k, -.1, .1, .01}]

] (* End Module *)
] (* End Compile *)


where A1, A2, B1, B2, and MM are global variables.



Answer



Compile can't handle SetDelayed.


In your specific case, you might be able to avoid the need for SetDelayed altogether through the use of Boole.


kd[m_, n_] :=If[m == n, 1, 0] can be replaced by Boole[n==m]. (or as FJRA pointed out in comments, by KroneckerDelta.)


cosfun[m_, n_] := 
If[m == n, 0, (1 - Cos[(m + n) π])/((m + n) π) +
((Cos[(m - n) π] - 1)/((m - n) π))]

can be replaced by



Boole[m == n] (1 - Cos[(m + n) π])/((m +  n) π) + 
((Cos[(m - n) π] - 1)/((m - n) π)) (-I A2 m Pi/L)

Giving


eenew = Compile[{{μ, _Real}, {NNN, _Integer}}, 
Module[{NN = NNN}, (* do you even need this placeholder for the input? *)
mm = Table[Boole[m == n] (MM - B2 k k - B1 (m Pi/L)^2), {m, 1, NN}, {n, 1, NN}];
kz = Table[Boole[m != n] (1 - Cos[(m + n) π])/((m + n) π) +
((Cos[(m - n) π] - 1)/((m - n) π)) (-I A2 m Pi/L),
{m, 1, NN}, {n, 1, NN}];

kxM = Table[Boole[m == n] A1 k, {m, 1, NN}, {n, 1, NN}];
μM = Table[Boole[m == n] μ, {m, 1, NN}, {n, 1, NN}];
HH = ArrayFlatten[{{μM + mm, 0 mm, kz, kxM},
{0 mm, μM + mm, kxM, -kz}, {kz, kxM, μM - mm,0 mm},
{kxM, -kz, 0 mm, μM - mm}}];
ees = Table[Eigenvalues[HH], {k, -.1, .1, .01}]]]

What is k for in this last line? There is no way for the iterator to matter for HH.


This doesn't give the error involving SetDelayed anymore.


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...

calculus and analysis - Integral of the Sinc product

Let us consider the following integral $$ B_n = \int_0^\infty \prod_{k=1,3,5,\dots}^n\frac{\sin (x/k)}{x/k}dx $$ By definition, Sinc[x] == Sin[x]/x , therefore B[n_?OddQ] := Integrate[Product[Sinc[x/k], {k, 1, n, 2}], {x, 0, Infinity}] B /@ Range[1, 13, 2] {Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2} OK, everything is fine. But... B[15] 467807924713440738696537864469 Pi/935615849440640907310521750000 What's going on? Answer As Eckhard wrote in comments B[n] is the n-th Borwein integral . (The letter B was not accidental :) ) This funny properties of Borwein integrals is related to the Fourier transform of Sinc function FourierTransform[Sinc[x], x, k] 1/2 Sqrt[Pi/2] (Sign[1 - k] + Sign[1 + k]) Plot[%, {k, -2, 2}, Filling -> 0] which is the box function. The result is $\pi/2$ while the sum $$ 1/3+1/5+\dots+1/n $$ B_n = \frac{\pi}{2} - \pi \bigg(\sum_{k=3,5,\ldots}^n\frac{1}{k} -1\biggr)^\frac{n-1}{2}\prod_{k=3,5,\dots}^n\frac{k}{k-1}. $$ For $n=15$ it is equal to $$ \frac{46780...