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[✓] Perform an approximation with radial basis functions?

GROUPS:

Hi!

I want to approximate some functions with basisfunctions, which can be easy Fourier-transformed. So I got the idea to approximate my function with Gaussian normal-distribution curves. This led me to the idea of approximating my function with radial basis functions (RBF), such as the Gaussian curves are.

Due to this I found the "NeuralNetworks`" Mathematica package, which seems to be exactly what I was looking for. I even found this very helpful Example. Poorly this package doesn't seem to be part of Mathematica 11 anymore. The newer built-in Version doesn't seem to include functions, which do similar work.

Because I am quite new at Mathematica I have no idea if and how I can find another way to solve my problem.

Does anybody have an idea how I can approximate my function (maybe similar to the example?) with Gaussian normal-Distribution curves?

I would be very grateful, if anyone could help me!

Thanks you very much!

POSTED BY: Peter a
Answer
24 days ago

1) Are you looking to find

$$f(x) \approx \sum_j a_j \phi(x - x_j)$$

where $\phi(x) = \exp(-(b/h^2)\,x^2)$, where $b > 0$ is a parameter and $h$ is the distance between the evenly spaced centers $x_j$? (A linear problem.)

If RBF interpolation is the sort of thing you want, the approach of Boyd & Wang (2009) seems easy to program (eq. (38)), assuming it's easy to determine when to truncate the series and the parameters $b$ ( $=\alpha^2$ in Boyd & Wang) and $h$. It seems to approximate an RBF interpolation with extremely good error properties. The approximation (eq. (12)) is obtained by approximating the Fourier Transform (eq. (9)) of the "cardinal function" for RBF interpolation, taking the inverse transform and approximating it as well.

2) Or will $b = b_j$ be different at each center and optimized? (A nonlinear problem.)

3) Or will the centers be irregularly spaced?

4) How many centers do you think you might need? (This might determine whether a dumb method can be used on a small number of centers, or an efficient method is needed.)


Here's what is surely a "dumb method" for problem 2) above, or at least a naive one. But the result is not too bad on this example.

(* data from Peter's linked example *)
Ndata = 20;
x = Table[10 N[{i/Ndata}], {i, 0, Ndata - 1}];
y = Sin[0.1 x^2];

model = Sum[a[i] Exp[-b[i] (t - x[[i, 1]])^2], {i, 1, Length@x, 2}];  (* one Gaussian per pair *)
nlm = NonlinearModelFit[
  Transpose[Join[Transpose[x], Transpose[y]]],
  model,
  Transpose[{  (* parameters + initial values *)
    Join[Array[a, Length@x][[;; ;; 2]], Array[b, Length@x][[;; ;; 2]]],
    Join[Flatten@y[[;; ;; 2]], ConstantArray[0.5, Length@x][[;; ;; 2]]]
    }],
  t, MaxIterations -> 1000]

nlm["BestFitParameters"]
(*
{a[1] -> 23.1599, a[3] -> -13.0745, a[5] -> 0.73477, 
 a[7] -> 0.0989986, a[9] -> 0.155762, a[11] -> 0.0914005, 
 a[13] -> -9.50923, a[15] -> -2.4148, a[17] -> 0.392815, 
 a[19] -> 1.76915, b[1] -> 0.008367, b[3] -> 0.028266, 
 b[5] -> 0.123698, b[7] -> 0.544691, b[9] -> 0.595277, 
 b[11] -> 0.599547, b[13] -> -0.00378785, b[15] -> 0.289993, 
 b[17] -> 0.797485, b[19] -> 0.685441}
*)

For inspecting accuracy:

Show[
 Plot[nlm[t], {t, 0, 9.5}],
 ListPlot[Transpose[Join[Transpose[x], Transpose[y]]]],
 PlotRange -> All
 ]

Plot[nlm[t] - Sin[0.1 t^2], {t, 0, 9.5}, PlotRange -> All]

enter image description here

POSTED BY: Michael Rogers
Answer
22 days ago

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