Skip to main content

calculus and analysis - How to speed up the calculation of a multi-dimension matrix involving symbolic integral?


The following program succeeds in getting matrix CC, but it takes time badly, especially in the case varNumber becomes larger just as the following varNumber = 35. Who can speed up the process of calculation? Thanks!


ClearAll["Global`*"];
varNumber = 35; end = Infinity;
s1 = 112*^9; s2 = 25*^9; s3 = 15.1; s4 = 5.5*10^-9;


a[m_] := Exp[-x/2]*LaguerreL[m, x];
b[m_, i_, j_, l_] := Integrate[a[m]*x^i*D[a[l], {x, j}], {x, 0, end}];
d[m_, i_, j_, l_] :=
Integrate[
a[m]*x^i*D[
a[l], {x, j}]*(DiracDelta[x] -
DiracDelta[x - end]), {x, -Infinity, Infinity}];

c[1, 1][m_, l_] := s2*d[m, 0, 1, l] + s2*b[m, 0, 2, l];
c[1, 2][m_, l_] := 0;

c[1, 3][m_, l_] := 0;
c[2, 1][m_, l_] := 0;
c[2, 2][m_, l_] := s1*d[m, 0, 1, l] + s1*b[m, 0, 2, l];
c[2, 3][m_, l_] := s3*d[m, 0, 1, l] + s3*b[m, 0, 2, l];
c[3, 1][m_, l_] := 0;
c[3, 2][m_, l_] := s3*d[m, 0, 1, l] + s3*b[m, 0, 2, l];
c[3, 3][m_, l_] := -s4*d[m, 0, 1, l] - s4*b[m, 0, 2, l];

CC = ArrayFlatten@
Table[c[m, n][i, j], {m, 3}, {n, 3}, {i, 0, varNumber - 1}, {j, 0,

varNumber - 1}]; // AbsoluteTiming
{2283.69, Null}

Answer



Try this instead of your definitions of b and d:


b[m_, 0, 2, l_] /; l == m = 1/4;
b[m_, 0, 2, l_] /; l > m = l - m;
b[m_, 0, 2, l_] /; l < m = 0;
d[m_, 0, 1, l_] = -l - 1/2;

With these I can assemble CC in 0.047589 seconds (varNumber = 35).



For different values of $i$ and $j$, I find the fast definition


d[m_, i_, j_, l_] := If[i == 0, a[m]*D[a[l], {x, j}] /. x -> 0, 0];

following directly from the integral over Dirac $\delta$-functions. As for b[m,i,j,l] I would recommend asking at math.SE to see if anyone knows a closed formula for these integrals.


More generally, in the absence of such formulas you can use classical memoization, which gains you a big factor because you don't keep re-calculating the same values. Alternatively, you can use persistent memoization that will remember values forever, even when the kernel is restarted:


cacheloc = PersistenceLocation["Local", 
FileNameJoin[{$UserBaseDirectory, "caches", "bintegrals"}]];
end = Infinity;
a[m_] = Exp[-x/2]*LaguerreL[m, x];
b[m_Integer, i_Integer, j_Integer, l_Integer] := b[m, i, j, l] =

Once[Integrate[a[m]*x^i*D[a[l], {x, j}], {x, 0, end}], cacheloc];

Here I've used a combination of classical memoization and persistent storage, which complement each other: the former is very fast but impermanent, while the latter is a bit sluggish but permanent. Together, we get both advantages: the first lookup from permanent storage is still sluggish, but afterwards we get very fast lookup.


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[ ...

equation solving - Invert and fit implicitly defined curve

I need to fit an implicitly defined curve. I thought I could get some data out of Solve , and then using FindFit . Therefore, I would like to find the relation the parametric curve defined by $F(x,y)=0$: Solve[-(1/2) + 1/2 (0.41202 BesselK[0, 0.1 Sqrt[x^2 + y^2]] + (0.101483 x BesselK[1, 0.1 Sqrt[x^2 + y^2]])/Sqrt[x^2 + y^2]) == 0, y] But I can't get an output: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve require exact input, providing Solve with an exact version of the system may help. >> Edit: In particular, I would like to fit the data coming from the curve with the expression of another curve, and not with a function $f(x)$. In particular, since this clearly looks like a cardioid , I would like it to fit to something like it. What other strategies could I try?

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...