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I have the following test data, and I need to interpolate over these data. I attempted to use Interpolation. However, the interpolated data are not smooth enough. Is there any better way to interpolate my data?

a = Table[DATA[[i, 1]], {i, 1, 35}];
b = Table[DATA[[i, 2]], {i, 1, 35}]; 
xx = Interpolation[Transpose[{a, b}], InterpolationOrder -> 1];

{{1., 0.}, {1.3432, 57.4713}, {1.39448, 344.828}, {1.41026, 
  459.77}, {1.43195, 574.713}, {1.45168, 632.184}, {1.4714, 
  747.126}, {1.49901, 747.126}, {1.53057, 862.069}, {1.54832, 
  977.011}, {1.57791, 919.54}, {1.59763, 1034.48}, {1.61933, 
  1091.95}, {1.63708, 1149.43}, {1.6568, 1379.31}, {1.67258, 
  1551.72}, {1.69428, 1839.08}, {1.71992, 2126.44}, {1.73964, 
  2471.26}, {1.78107, 3103.45}, {1.7929, 3850.57}, {1.80868, 
  4540.23}, {1.81854, 5344.83}, {1.83235, 6206.9}, {1.84221, 
  7011.49}, {1.86193, 8045.98}, {1.87179, 10057.5}, {1.87771, 
  10747.1}, {1.88363, 11781.6}, {1.89546, 12988.5}, {1.89941, 
  14712.6}, {1.90335, 15517.2}, {1.9073, 16034.5}, {1.90927, 
  16781.6}, {1.91124, 17528.7}}
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  • $\begingroup$ How are you using Interplation? $\endgroup$
    – user21
    Commented Jun 7, 2017 at 14:36
  • $\begingroup$ a = Table[DATA[[i, 1]], {i, 1, 35}]; b = Table[DATA[[i, 2]], {i, 1, 35}]; xx = Interpolation[Transpose[{a, b}], InterpolationOrder -> 1]; $\endgroup$
    – user49047
    Commented Jun 7, 2017 at 14:40
  • $\begingroup$ Please add this information to your post. Also add a comment about what it is that you got as a result and what you would have expected. $\endgroup$
    – user21
    Commented Jun 7, 2017 at 14:41
  • $\begingroup$ Interpolation will go through all the points, is that what you want? Or do you want a regression, i.e. a best fit to a model function? $\endgroup$
    – Marius Ladegård Meyer
    Commented Jun 7, 2017 at 14:49
  • $\begingroup$ yes i need it to go thorough all the points as smooth as possible. $\endgroup$
    – user49047
    Commented Jun 7, 2017 at 14:56

1 Answer 1

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this is fit (not interpolation) using a piecewise cubic function:

data = {{1.3432, 57.4713}, {1.39448, 344.828}, {1.41026, 
   459.77}, {1.43195, 574.713}, {1.45168, 632.184}, {1.4714, 
   747.126}, {1.49901, 747.126}, {1.53057, 862.069}, {1.54832, 
   977.011}, {1.57791, 919.54}, {1.59763, 1034.48}, {1.61933, 
   1091.95}, {1.63708, 1149.43}, {1.6568, 1379.31}, {1.67258, 
   1551.72}, {1.69428, 1839.08}, {1.71992, 2126.44}, {1.73964, 
   2471.26}, {1.78107, 3103.45}, {1.7929, 3850.57}, {1.80868, 
   4540.23}, {1.81854, 5344.83}, {1.83235, 6206.9}, {1.84221, 
   7011.49}, {1.86193, 8045.98}, {1.87179, 10057.5}, {1.87771, 
   10747.1}, {1.88363, 11781.6}, {1.89546, 12988.5}, {1.89941, 
   14712.6}, {1.90335, 15517.2}, {1.9073, 16034.5}, {1.90927, 
   16781.6}, {1.91124, 17528.7}}
np = 20
unk = Transpose[{Subdivide[Sequence @@ data[[{1, -1}, 1]], np - 1], 
    Table[Symbol["yi" <> ToString[i]], {i, np}]}];
NMinimize[(Interpolation[unk][data[[All, 1]]] - data[[All, 2]])^2 // 
   Total, unk[[All, 2]]];
fit = Interpolation[unk /. %[[2]]];

Show[{
  ListPlot[data],
  Plot[fit[x], {x, data[[1, 1]], data[[-1, 1]]}, PlotRange -> All]}]

enter image description here

As you see it aims to be a smooth approximation without attempting to hit every one of the scattered points.

Note I dropped the very first point from the data as it was too far outlying and messed up the fit.

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2
  • $\begingroup$ it looks promising after x=1.34 but still i am missing good chunk of the data from 1:1.34. lets say n=100;a= Table[1 + (1.91124 - 1)/n*i, {i, 0, n}]; b = Table[fit[x], {x, a}]; then you will see it needs to extrapolate between 1:1.34 which are some ridiculous values. $\endgroup$
    – user49047
    Commented Jun 8, 2017 at 9:24
  • $\begingroup$ so collect more data.. seriously just do a linear interploation between the first two points. $\endgroup$
    – george2079
    Commented Jun 8, 2017 at 12:51

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