(* test data*)
plot1=Plot[PDF[LogNormalDistribution[0,1],x],{x,0,5}]
mypts=plot1[[1,1,-1,2,1]];
ListLinePlot[mypts]
(* models and parameters *)
myrule=b[coef_]:>ToExpression["b"<>ToString[coef]];
mymodel=Accumulate[b[#]*x^#&/@Range[0,10,1]/.myrule][[2;;-1]]
myvar=b[#]&/@Range[0,10]/.myrule
(* fitting models *)
myfit=FindFit[mypts,mymodel[[#]],myvar[[1;;#+1]],x]&/@
Range[Length@mymodel]
(* plot data and model *)
Plot[mymodel[[#]]/.myfit[[#]],{x,0,5},Epilog->Point[mypts],
PlotRange->{{0,5},{0,0.8}}]&/@Range[Length@mymodel]
Mathematica 教學:Interpolation 的函數要如何輸出
由 戴忠淵 於 2015年1月6日星期二
上午11:02 發表

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