Skip to main content

programming - Calculating an integral by the Romberg Algorithm



In my "Numerical Analysis" course, I learned the Romberg Algorithm to numerically calculate the integral.


The Romberg Algorithm as shown below:



tableau



$$T_{2n}(f)+\frac{1}{4^1-1}[T_{2n}(f)-T_{n}(f)]=S_n(f) \\ S_{2n}(f)+\frac{1}{4^2-1}[S_{2n}(f)-S_{n}(f)]=C_n(f) \\ C_{2n}(f)+\frac{1}{4^3-1}[C_{2n}(f)-C_{n}(f)]=R_n(f)$$


where $$T_n=\frac{h}{2}\left[ f(a)+ 2\sum_{i=1}^{n-1}f(x_i) +f(b)\right]$$ and $h=\frac{b-a}{n}$.


My code:


 trapezium[func_, n_, {a_, b_}] :=
  With[{h = (b - a)/n},

   1/2 h (func@a + 2 Sum[func[a + i h], {i, 1, n - 1}] + func@b)
 ]

rombergCalc[func_,iter_, {a_, b_}] := 
  Module[{m = 1},
   Nest[
    MovingAverage[#, {-1,4^(m++)}] &, 
    Table[trapezium[func, 2^i, {a, b}], {i, 0., iter}], 3]
  ]


The calculation process comes from my textbook



enter image description here



Fixed one bug 1



enter image description here



Update


Fixed bug 2




enter image description here



Test


 rombergCalc[Exp, 5, {0, 1}]//InputForm


 {1.7182818287945305, 1.7182818284603885, 1.7182818284590502}

My Question update




  • In function rombergCalc, I utilized the usage m++ that I believe is not suitable in Mathematica Programming. Is there any other method to replace m++ or implement Romberg Algorithm?


- Why Block[{$MinPrecision = precision, $MaxPrecision = precision}..] cannot give the result that contain significance digit that I gave(seeing the graphic of textbook)?


(Thanks for @xzczd's solution for dealing with precision problem)


N[{a, b}, precision]

and replace


trapezium[func, 2^i, {a, b}], {i, 0., iter}]


with


trapezium[func, 2^i, {a, b}], {i, 0, iter}]

- Except for SetOptions[SelectedNotebook[], PrintPrecision -> 16] or InputForm, is there other solutions to set precision conveniently?



Answer



Nice to meet you, Mr. Shu.


Bug fix first. Your function doesn't work under desired precision because:



Table[trapezium[func, 2^i, {a, b}], {i, **0.**, iter}]


Changing it to


Table[trapezium[func, 2^i, {a, b}], {i, 0, iter}]

still doesn't fix the problem, because all the numbers taking part in the calculation have infinite precision. Adding an


N[…, precision]

somewhere still doesn't fix the problem, because your utilization of m++ is not only unsuitable, but also wrong. Try changing your m++ into (Print[i = m++]; i) and see what will happen.


Fixed code:


trapezium[func_, n_, {a_, b_}] := 
 With[{h = (b - a)/n}, 

  1/2 h (func@a + 2 Sum[func[a + i h], {i, 1, n - 1}] + func@b)]

rombergCalc[func_, iter_, {a_, b_}, precision_] := 
 Block[{$MinPrecision = precision, $MaxPrecision = precision}, 
  Module[{m = 1}, 
   NestList[
    With[{n = 4^(m++)}, 
      Flatten@(MovingAverage[#, {-1, n}] & /@ Partition[#, 2, 1])] &, 
    Table[trapezium[func, 2^(i - 1), N[{a, b}, precision]], {i, 
      iter}], 3]]]


It's so ugly now that I'd like to turn to:


trapezium[func_, n_, {a_, b_}] := 
 With[{h = (b - a)/n}, 
  Module[{f = Function[x, func@x, Listable]}, h (Total@f@Range[a, b, h] - 1/2 (f@a + f@b))]]

rombergCalc[func_, iter_, {a_, b_}, precision_] := 
 FoldList[Rest@# + 1/(4^#2 - 1) Differences@# &, 
  trapezium[func, 2^(# - 1), N[{a, b}, precision]] & /@ Range@iter, Range@3]


Finally "visualize" the result:


TableForm[Flatten[rombergCalc[Exp, 6, {0, 1}, 13], {{2}, {1}}], 
   TableHeadings -> {2^Range[0, 5], 
   {"\!\(\*SubscriptBox[\(T\), \(n\)]\)", 
    "\!\(\*SubscriptBox[\(S\), \(n\)]\)", 
    "\!\(\*SubscriptBox[\(C\), \(n\)]\)", 
    "\!\(\*SubscriptBox[\(R\), \(n\)]\)"}}, TableAlignments -> Center]

enter image description here




For completeness, here's a compiled version of the function above. Notice that it only speeds up Listable compilable internal functions or pure functions formed by Listable compilable internal function.


trapezium[f_, n_, {a_, b_}] := 
 With[{h = (b - a)/n}, 
  h (Total[f@Range[a, b, h]] - 1/2 (f@a + f@b))]

compiledrombergCalc[f_, {a_, b_}] := 
 ReleaseHold[
  Hold@Compile[{{i, _Integer}}, 
     Fold[Rest@# + 1/(4^#2 - 1) Differences@# &, 
      trapezium[f, 2^(# - 1), {a, b}] & /@ Range@i, Range@3]] /. 

   DownValues@trapezium]

g = Exp[#] &
NumberForm[compiledrombergCalc[g, {0, 1}][18], 13] // AbsoluteTiming

enter image description here


It's a 23X speedup compared to the uncompiled version.


Also notice that inside the compiled function, the calculation is under MachinePrecision so the Precision of the result is also MachinePrecision, though its appearance is changed by NumberForm. However, I think this treatment may be closer to what your text book has done: as said in the comment below, significance arithmetic is the secret recipe of Mathematica anyway.


Comments

Post a Comment

Popular posts from this blog

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}]
Post a Comment
Read more

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}]
Post a Comment
Read more

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.
Post a Comment
Read more