How to fit data to only half of a model

Question

pIDmm = {{1, 4.}, {2, 4.}, {3, 0.}, {4, 5.}, {5, 4.}, {6, 8.}, {7, 
  12.}, {8, 11.}, {9, 27.}, {10, 28.}, {11, 41.}, {12, 49.}, {13, 
  36.}, {14, 133.}, {15, 97.}, {16, 168.}, {17, 196.}, {18, 
  189.}, {19, 250.}, {20, 175.}, {21, 368.}, {22, 349.}, {23, 
  345.}, {24, 475.}, {25, 427.}};

sic = 11.6186 + 1.72833 x - 0.105655 x^2 + 0.0588155 x^3 - 
   0.00346728 x^4 + 0.0000699641 x^5 - 4.71206*10^-7 x^6;

sic is the model for the data pCD (blue dots). I would like to adapt or shape the model to fit the data in red dots (pIDmm) by phase-shifting the model and increasing its amplitude and range. But the red dots represent only the first half of the model. Is it possible to use FindFit using the red dots data on only the first half of the model so as to predict the likely subsequence of data points?

Answer 1

It's what Jim said- what parameters you can change. If, say, all the coefficients are related to phase you might have to put all of them as parameters in Findfit and your model will be like:

    model= a + b x + c x^2 + c x^3 + e x^4 + f x^5 + g x^6;

For the first set of data you provided (pIDmm) it will fit as:

    fit = FindFit[pIDmm,  a + b x + c x^2 + c x^3 + e x^4 + f x^5 + g 
          x^6, {a, b, c, d , e, f, g}, x];
    Show[ListPlot[pIDmm],  Plot[model /. fit, {x, 0, 25}, PlotStyle -> Red]]

However, If you want to fit combined data points of pIDmm and sic, first combine them in a list and obtain a fit which will depend on relative strength and number of points. For ex:

      data = Table[{x, sic + RandomVariate[NormalDistribution[0, 3]]}, {x, 0, 56.5, 56.5/25}];
      fit2 = FindFit[Sort[Join[pIDmm, data]], 
             a + b x + c x^2 + c x^3 + e x^4 + f x^5 + g x^6, {a, b, c, 
             d , e, f, g}, x];
      Show[ListPlot[Sort[Join[pIDmm, data]]], Plot[model /. fit2, {x, 0, 55}, PlotStyle -> Red], PlotRange -> All]

Answer 2

China model adapted to Italy, black dots are new data, the model was developed from the first 60 days and then not changed.

China model adapted to US

China model adapted to Iran

Answer 3

It seems that the best fit to pCD data is a gaussian bell as can be depicted. Hopelessly we don't have pCD data...

pIDmm = {{1, 4.}, {2, 4.}, {3, 0.}, {4, 5.}, {5, 4.}, {6, 8.}, {7, 12.}, {8, 11.}, {9, 27.}, {10, 28.}, {11, 41.}, {12, 49.}, {13, 36.}, {14, 133.}, {15, 97.}, {16, 168.}, {17, 196.}, {18, 189.}, {19, 250.}, {20, 175.}, {21, 368.}, {22, 349.}, {23, 345.}, {24, 475.}, {25, 427.}}

f[x_] := 11.6186 + 1.72833 x - 0.105655 x^2 + 0.0588155 x^3 - 0.00346728 x^4 + 0.0000699641 x^5 - 4.71206*10^-7 x^6
fa[x_, a_, b_, c_] := a Exp[-(x - b)^2/c]
data = Sum[(f[x] - fa[x, a, b, c])^2, {x, 0, 60}];
sol = NMinimize[data, {a, b, c}, Method -> "DifferentialEvolution"]
fa0 = fa[x, a, b, c] /. sol[[2]]
gr2 = Plot[{fa0, f[x]}, {x, 0, 60}, PlotStyle -> {{Thick, Blue}, {Dashed, Red}}]

so we follow adjusting to pIDmm a gauss bell also

data2 = Sum[(fa[pIDmm[[k, 1]], a, b, c] - pIDmm[[k, 2]])^2, {k, 1, Length[pIDmm]}];
sol2 = NMinimize[data2, {a, b, c}, Method -> "DifferentialEvolution"]
fa02 = fa[x, a, b, c] /. sol2[[2]]
gr0 = ListPlot[pIDmm];
gr1 = Plot[{fa02, fa0}, {x, 0, 60}, PlotStyle -> {{Thick, Black}, {Thick, Blue}}];
Show[gr1, gr0]

Answer 4

a new model based on trig -( a Sin[b x + c] + d ) on the completed trend, results

   In[1246]:= trigfitCD =  a Sin[b x + c] + d /. {a -> -56.11407429814903`, 
   b -> 0.12234789054358615`, c -> 8.054346767593303`, 
   d -> 61.16172326460761`}

   Out[1246]= 61.1617 - 56.1141 Sin[8.05435 + 0.122348 x]

I multiplied trigfitCD * Max[ItalyDeaths]/Max[ChinaDeaths] and plotted the results

It is close but the predictive line is just the trendline amplified. I would like to fit the old model to the new data. The new data is only half of the story, so I need to fit half the old model to the new data - to get a predictive trend, is that possible with Fit or FindFit or NonlinearModelFit or something else???