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Maybe the following meets your desiderata: if2[x_ /; 1 <= x <= 2] = if[x]; if2[0.9] (* if2[0.9] *) if2[1.1] (* 1.21 *)

Show[ ListPlot[data, PlotStyle -> Black], Plot[Evaluate@Interpolation[data][x], {x, 0, 120}, PlotRange -> All] ] If those wiggles bother you, use linear interpolation instead: Show[ ListPlot[data, … try this approach to regularised Interpolation. …

Most (if not all) of my actual InterpolatingFunctions use InterpolationMethod Hermite, so I tried tweaking @Mr.Wizard's and @MichaelE2's answers to deal with them, while still addressing discontinuous …

NumericQ}] := Interpolation[Join[ {{xmin, if[xmin]}}, Select[Transpose[{if["Coordinates"][[1]], if["ValuesOnGrid"]}], xmin < #[[1]] < xmax &], {{xmax, if[xmax]}} ]]; It generally works OK, but … Simply deleting one of the duplicate points is no good because it completely messes up the interpolation. This very issue was explored in depth in this answer by Michael E2. …

With help from a few clues in @MichaelE2's comments, I think I've solved Problem 2 (discontinuous InterpolatingFunctions). First, define discontInterpolation from this answer. Then define discontIn …

Interpolation can make an InterpolatingFunction periodic with the option PeriodicInterpolation. Can we make an InterpolatingFunction that already exists periodic, perhaps the output of NDSolve? …

Yes, if we're willing to risk modifying the internals of the InterpolatingFunction. Using info from this answer from MichaelE2 and some guesswork, it looks like we need to change three things in the …

options *) interpolationopts,interpolationpoints, (* other variables *) xmin,xmax,ifs,grid,tmp}, (* handle options *) interpolationopts=FilterRules[Flatten[{opts,Options[Reinterpolation]}],Options[Interpolation … If[interpolationpoints===Automatic, grid=Union[Flatten[Through[ifs["Grid"]],1]], {xmin,xmax}=ifs[[1,1,1]]; grid=Table[x,{x,xmin,xmax,(xmax-xmin)/(interpolationpoints-1)}]; ]; Quiet[ tmp=Interpolation …

Just replace the list of variables being solved for with an empty list. E.g. NDSolve[{n'[t] == n[t] (1 - n[t]), n[0] == 0.1, WhenEvent[n[t] == 0.5, Print[t]]}, n, {t, 0, 10}] (* 2.19722 *) NDSol …

The following works as expected: if = Interpolation[Table[{x, Sin[x]}, {x, 10^-3, 2 \[Pi], 0.1}], "ExtrapolationHandler" -> {Indeterminate &, "WarningMessage" -> False}] (* InterpolatingFunction[<0.001 … , 6.2>] *) if[0] (* Indeterminate *) However if the extrapolated point is just outside the range, ExtrapolationHandler lets me down: if = Interpolation[Table[{x, Sin[x]}, {x, 10^-6, 2 \[Pi], 0.1}], …

I'm numerically solving a PDE (time+1D space) with NDSolve and want to plot some spatial profiles. Plot works, but is very slow compared to ListPlot. Here's an example (Fisher-KPP equation): l = 10 …

Encapsulating the right-hand side in a NumericQ-protected function and setting the MaxStepSize to 1 seems to work for me: df[t_?NumericQ] := Piecewise[{{7, datafunc[t] > 0.}, {-5, datafunc[t] <= 0.}} …

NumericQ] := Module[{if, grid}, if = InterpolatingFunction[stuff]; grid = (InterpolatingFunction[stuff])["Grid"]; Return[(Interpolation@Transpose@{ grid[[All, 1, {1}]], (* extract x-grid *) … NumericQ, var_Symbol] := Module[{if, grid}, if = InterpolatingFunction[stuff]; grid = (InterpolatingFunction[stuff])["Grid"]; Return[(Interpolation@Transpose@{ grid[[1, All, {2}]], (* extract …

sol5IFN = Interpolation[Transpose[{ln["Grid"], Exp[ln["ValuesOnGrid"]]} /.First@sol4]]; Plot[sol5IFN[t], {t, 0, 100}] Looking at the underlying grid makes it obvious what the problem is: ListPlot[ … [sol6[t], {t, 0, 100}] UPDATE 2: Higher InterpolationOrder sol5IFN = Interpolation[Transpose[{ln["Grid"], Exp[ln["ValuesOnGrid"]]} /. …