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I am trying to get Mathematica to interpolate a list of points so that I can find precise x values x(f) that won't have a corresponding data point. Here's what I've tried:

 data={{19.668, 0.161158}, {18.5235, 0.17188}, {17.581, 0.183499}, {16.7985,
   0.196057}, {16.1445, 0.209601}, {15.595, 0.224179}, {15.1315, 
  0.239842}, {14.7395, 0.256644}, {14.4075, 0.274639}, {14.1265, 
  0.293882}, {13.8885, 0.314433}, {13.6875, 0.336349}, {13.518, 
  0.359688}, {13.3765, 0.384507}, {13.2585, 0.410862}, {13.162, 
  0.43881}, {13.083, 0.468401}, {13.0205, 0.499684}, {12.9715, 
  0.532704}, {12.935, 0.567499}, {12.909, 0.604103}, {12.892, 
  0.642541}, {12.8835, 0.682831}, {12.8815, 0.724979}, {12.885, 
  0.768983}, {12.8935, 0.814829}, {12.906, 0.862489}, {12.9215, 
  0.911922}, {12.9395, 0.963073}, {12.9595, 1.01587}, {12.98, 
  1.07023}, {13.001, 1.12604}, {13.022, 1.1832}, {13.0425, 
  1.24156}, {13.0615, 1.30096}, {13.0785, 1.36126}, {13.094, 
  1.42226}, {13.1065, 1.48377}, {13.1165, 1.54559}, {13.1235, 
  1.6075}, {13.1265, 1.66927}, {13.126, 1.73069}, {13.122, 
  1.79152}, {13.1135, 1.85153}, {13.1005, 1.91048}, {13.0835, 
  1.96817}, {13.062, 2.02436}, {13.0355, 2.07887}, {13.005, 
  2.13148}, {12.97, 2.18203}, {12.9305, 2.23035}, {12.8865, 
  2.27629}, {12.838, 2.31974}, {12.786, 2.36058}, {12.7295, 
  2.39873}, {12.6695, 2.43413}, {12.606, 2.46673}, {12.539, 
  2.49651}, {12.469, 2.52347}, {12.396, 2.54761}, {12.32, 
  2.56897}, {12.242, 2.58759}, {12.1615, 2.60354}, {12.079, 
  2.61689}, {11.995, 2.62772}, {11.9095, 2.63613}, {11.823, 
  2.64222}, {11.7355, 2.64609}, {11.647, 2.64786}, {11.558, 
  2.64765}, {11.469, 2.64558}, {11.38, 2.64177}, {11.291, 
  2.63633}, {11.2025, 2.6294}, {11.1145, 2.62109}, {11.0275, 
  2.61152}, {10.9415, 2.6008}, {10.8565, 2.58906}, {10.7725, 
  2.57639}, {10.69, 2.5629}, {10.609, 2.5487}, {10.53, 
  2.53389}, {10.453, 2.51855}, {10.3775, 2.50278}, {10.304, 
  2.48667}, {10.233, 2.47028}, {10.164, 2.45371}};

Show[ListPlotdata], Plot[Interpolation[data,][x], {x, 10.2, 16}]], InterpolationOrder -> 2

And it gives me the following plot:

Interpolated function

Which obviously doesn't look very good in the middle where the function becomes very steep! Unfortunately, this is right where I need to find x values, where f is between .8 and 1.8. I have also tried other InterpolationOrders and InterpolationMethods, but they all look just as bad.

Is there anyway to get a better fit? Or any other method for finding values of f? I tried using FindClosest but that wasn't precise enough for my means.

Thanks in advance!

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  • 1
    $\begingroup$ The problem is that your function "f" is not a one-variable function, as for one value of x you may have more than one value of f(x) This may be of your interest: mathematica.stackexchange.com/questions/47332/… $\endgroup$
    – Fraccalo
    Commented Jun 6, 2018 at 16:06
  • $\begingroup$ Do you have some way to decide which of the two possible y values you want when x is near the middle? $\endgroup$
    – Rudy Potter
    Commented Jun 6, 2018 at 16:15
  • $\begingroup$ This could be a duplicate of this post. $\endgroup$
    – march
    Commented Jun 6, 2018 at 16:20

1 Answer 1

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Here's how I would do it. Interpolate both x and y data separately, and then use ParametricPlot:

Show[
 ListPlot[data],
 ParametricPlot[{Interpolation[data[[All, 1]]][x], Interpolation[data[[All, 2]]][x]},
  Evaluate@{x, Sequence @@ First@Interpolation[data[[All, 1]]]["Domain"]}, 
  PlotStyle -> Black
 ]
]

enter image description here

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