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I have some data that I'd like to find a periodic, continuous function that will approximately fit to:

data={{0.0936962,0.0270862},{0.115233,0.145407},{0.144849,0.262185},{0.18072,0.377282},{0.221018,0.49056},{0.257301,0.627628},{0.295673,0.757493},{0.337199,0.868766},{0.382255,0.952742},{0.43682,0.981143},{0.503706,0.936757},{0.569108,0.860408},{0.628899,0.759121},{0.678952,0.639921},{0.690301,0.522011},{0.685416,0.399376},{0.674975,0.271783},{0.663962,0.141791},{0.663175,0.014363},{0.685254,-0.104778},{0.713799,-0.221913},{0.747581,-0.337052},{0.785366,-0.450208},{0.822,-0.582662},{0.859839,-0.714971},{0.900764,-0.830273},{0.94515,-0.919859},{0.99656,-0.965099},{1.06515,-0.926806},{1.13231,-0.855449},{1.19332,-0.758203},{1.24344,-0.642242},{1.25456,-0.525299},{1.23874,-0.406979},{1.2157,-0.283425},{1.19162,-0.156898},{1.17516,-0.0303219},{1.19331,0.0885252},{1.2202,0.20588},{1.25401,0.321605},{1.2929,0.43556},{1.33099,0.563469},{1.36831,0.697935},{1.40865,0.816976},{1.45238,0.911886},{1.49987,0.973959},{1.56822,0.942449},{1.63599,0.87657},{1.69847,0.783497},{1.75092,0.670404},{1.77092,0.551412},{1.7555,0.433008},{1.73251,0.309592},{1.70868,0.183855},{1.69069,0.0584907},{1.71575,-0.0522204},{1.75403,-0.15947},{1.79892,-0.265626},{1.847,-0.371852},{1.88809,-0.493204},{1.91582,-0.636881},{1.94382,-0.768884},{1.97407,-0.879205},{2.00856,-0.957838},{2.0674,-0.947164},{2.13315,-0.888161},{2.19701,-0.799206},{2.25431,-0.687758},{2.29135,-0.56586},{2.2916,-0.447008},{2.28357,-0.321803},{2.27225,-0.192802},{2.26262,-0.0625656},{2.27484,0.0600786},{2.30001,0.178306},{2.33109,0.294533},{2.36684,0.408769},{2.4053,0.526898},{2.44296,0.666954},{2.483,0.793954},{2.52537,0.898515},{2.57002,0.971256},{2.62383,0.964648},{2.68179,0.896665},{2.73659,0.797775},{2.78497,0.676355},{2.82119,0.54281},{2.8216,0.422406},{2.81429,0.297515},{2.80384,0.170022},{2.79479,0.0418083}};

I'd like it to continue with the same period indefinitely, with period as close to average of 2.5 periods presented here.

I have tried parametrize and then interpolating it with

parametrizeCurve[pts_ /; MatrixQ[pts, NumericQ], 
a : (_?NumericQ) : 1/2] := FoldList[Plus, 0, Normalize[(Norm /@ Differences[pts])^a, Total]]

and then trying to find a periodic fit for x and y separately (as sums of periodic curves, for example), but I am coming up short. Is there a better way to do this?

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    $\begingroup$ Your x values are not monotonously increasing. Is this correct? $\endgroup$
    – Daniel Huber
    May 20, 2021 at 9:39
  • $\begingroup$ @DanielHuber yes, it is I'm afraid! $\endgroup$
    – martin
    May 20, 2021 at 9:45
  • $\begingroup$ Please can you edit your question to include that you want the fitted function to be periodic. $\endgroup$
    – flinty
    May 20, 2021 at 14:09
  • $\begingroup$ @flinty have done. thank you $\endgroup$
    – martin
    May 20, 2021 at 16:45

1 Answer 1

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Here I present a one-dimensional interpolation ip[t] with curveparameter 0<t<1:

ti = Subdivide[0, 1, Length[data] - 1];(* curve parameter*)
ip = Interpolation[MapThread[{#1, #2} &, {ti, data}] ]
Show[{ParametricPlot[ip[t], {t, 0, 1}], ListPlot[data]}]

enter image description here

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