Skip to main content

equation solving - FindFit with a sophisticated function (integral)


I am trying to find a fit to the distribution function (empiricial data) in terms of a function which is itself an integral of a product of two simpler functions (two polynomials), that is the model. In particular, I observe T(x) and the model is that T(x)≈∫x0F(x−y1−y)g(y)dy


My assumption is that F(.) and g(.) are polynomial functions. And so my problem would be to find the best fit polynomial functions, i.e. to assume that F(.)=k1+a1(x−y1−y)+b1(x−y1−y)2 and similarly g(.)=k2+a2y+b2y2 and let FindFit run.


Here is the data (just a small sample)


data={{0.002, 4}, {0.01, 5}, {0.02, 1}, {0.025, 1}, {0.0333, 1}, {0.05, 

1}, {0.0905, 1}, {0.09995, 1}, {0.105, 1}, {0.114, 1}, {0.2,
5}, {0.222, 2}, {0.25, 1}, {0.3, 1}, {0.35, 1}, {0.4, 7}, {0.5,
29}, {0.501, 2}, {0.505, 2}, {0.51, 1}, {0.52, 1}, {0.55,
1}, {0.55555, 1}, {0.6, 12}, {0.64, 2}, {0.65, 5}, {0.666, 1}, {0.7,
18}, {0.73, 1}, {0.74, 1}, {0.75, 30}, {0.76, 3}, {0.77266,
2}, {0.775, 1}, {0.8, 57}, {0.801, 2}, {0.8018, 1}, {0.802,
1}, {0.81, 1}, {0.81554, 1}, {0.82, 3}, {0.825, 1}, {0.82888,
1}, {0.83, 1}, {0.84, 4}, {0.85, 30}, {0.859, 1}, {0.86, 3}, {0.861,
1}, {0.862, 1}, {0.875, 7}, {0.88, 8}, {0.888, 2}, {0.9,
46}, {0.901, 3}, {0.9018, 1}, {0.902, 1}, {0.9022, 1}, {0.9026,

1}, {0.9027, 1}, {0.904, 1}, {0.9094, 1}, {0.91, 2}, {0.9202,
1}, {0.925, 1}, {0.926, 1}, {0.93, 1}, {0.94, 2}, {0.95, 5}, {0.96,
3}, {0.976, 1}, {0.98, 1}, {0.995, 1}, {1., 11}};

I have been trying to solve this pretty naively as follows:


D=SmoothKernelDistribution[data, 0.02];    
FindFit[CDF[D, x],
Integrate[(a1 ((x - y)/(1 - y))^2 + b1 ((x - y)/(1 - y)) +
k1) (a2 y^2 + b2 y + k2), {y, 0, x}], {a1, b1, k1, a2, b2,
k2}, {(x - y)/(1 - y), y}]


However I got the error: "(-y+x)/(1-y) is not a valid variable". I am sure that I need more sophisticated approach, but which steps would be there I am pretty lost...


(I did not specify in my code the domains (both x and y belong to [0,1]), but I doubt that this is of any relevance at this stage.)


I would appreciate any hints how to proceed!



Answer



Your datatest has one too many parenthesis.


Dont use func, replace it with the evaluated integral.


Integrate[(a ((x - y)/(1 - y))^2 + b ((x - y)/(1 - y)) + 
c) (d y^2 + e y + f), {y, 0, x}, Assumptions -> {0 < x < 1}]
(*

1/6 x (6 (c f + b (d + e + f) + a (4 d + 3 e + 2 f)) -
3 (-c e + b (d + e) + a (8 d + 5 e + 2 f)) x + (2 a - b +
2 c) d x^2) + (-b (d + e + f) +
a (-2 f + e (-3 + x) + 2 d (-2 + x))) (-1 + x) Log[1 - x] *)

Now run FindFit


sol = FindFit[datatest,
1/6 x (6 (c f + b (d + e + f) + a (4 d + 3 e + 2 f)) -
3 (-c e + b (d + e) + a (8 d + 5 e + 2 f)) x + (2 a - b +
2 c) d x^2) + (-b (d + e + f) +

a (-2 f + e (-3 + x) + 2 d (-2 + x))) (-1 + x) Log[1 - x], {a,
b, c, d, e, f}, x, Method -> NMinimize,
NormFunction -> (Norm[#, Infinity] &)]

(* {a -> -2.70862, b -> 0.555548, c -> -0.243013,
d -> -0.124391, e -> -0.310328, f -> -0.516147} *)

Plot the results


Show[
ListPlot[datatest],

Plot[Evaluate[func[a, b, c, d, e, f, x] /. sol], {x, 0, 1},
PlotStyle -> Red]
]

Mathematica graphics


Comments

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

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}]