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# Tag Info

## Hot answers tagged interpolation

15

You might be interested in the approach with optimal transport. Let $F(x)=\int_{-\infty}^x f(y)\,dy$ and $G(x)=\int_{-\infty}^x g(y)\,dy$ be the repartition functions. Then $T(x)=G^{-1}\bigl(F(x)\bigr)$ is the optimal transport map from $f$ to $g$. The map $T_t(x)=(1-t)x+tT(x)$ is the displacement geodesic, so that the intermediate densities are given by \$(...

13

One good way to interpolate a function of this nature is to take its Log, interpolate, and take Exp of the result. lntrp = Interpolation[data4[100] /. {x_, y_} -> {x, Log[y]}] Plot[Exp[lntrp[x]], {x, 99.98, 100.01}, PlotRange -> All]

7

Clear["Global`*"] f[x_] := Exp[-(x + 3)^2] g[x_] := 1/2 Exp[-(x - 3)^2/4] Treating f and g as unnormalized distributions distf = ProbabilityDistribution[f[x], {x, -Infinity, Infinity}, Method -> "Normalize"]; distg = ProbabilityDistribution[g[x], {x, -Infinity, Infinity}, Method -> "Normalize"]; disth = TransformedDistribution[(x + y)/2, {...

3

if0 = Interpolation[data0, "ExtrapolationHandler" -> {Automatic, "WarningMessage" -> False}]; difference = {#, #2 - if0 @ #} & @@@ data9k; ListLinePlot[{data0, data9k, difference}, ImageSize -> Large, PlotLegends -> {"data0", "data9k", "difference"}] An alternative approach is to use TemporalData: td0 = TemporalData[#2, {#}, ...

2

This is not complete answer, I only show how to overcome your issues with Plot. Your first check Do[Print[Plot[ Evaluate[Lh[hfuns[[i]], x] - hvals[[i]]*hfuns[[i]][x]], {x, 0, π}]], {i, 6}] can be done in the following way: GraphicsGrid@ Partition[ Table[Plot[ Evaluate[(Lh[hfuns[[i]], x]) - hvals[[i]]*hfuns[[i]][x]], {x, ...

2

This is an extended comment versus an answer, but you can improve your interpolation by performing it on log-log transformed. If you ListPlot mydata, you will see that much of the data is up against the x=0 axis. ListPlot[mydata] Since your data looks much nicer and spread out on a ListLogLogPlot, you can perform the interpolation on Log-Log transformed ...

2

Too long for a comment to John Doty's answer: While the answer might suit your needs (i.e. a smooth curve that passes through your points and remains positive), I would not say that it is a good way of interpolating. In the example below, I illustrate two bad properties of taking the exponential of log-interpolated values: sensitivity to vertical ...

2

BezierCurve only is a Graphics3D-object. Use BezierFunction to describe a function. Examplary here I give the calculation of the curvature kappa of the BezierFunction pts = Table[{2 Cos[t], 2 Sin[t], 3*t}, {t, 0, 6, 0.01}]; x = BezierFunction[pts] kappa = Sqrt[#.#] &[ D[x'[u]/Sqrt[x'[u].x'[u]], u]/Sqrt[x'[u].x'[u]]]; Plot[kappa, {u, 0, 1}, PlotRange -&...

1

Linear interpolation doesn't show this over/under-swinging behavior of polynomial interpolation of higher order. Clear[Interpolate] Interpolate[100] = Interpolation[data4[100], InterpolationOrder -> 1]; Plot[Interpolate[100][x], {x, 100.0033, 100.0034}, PlotRange -> {-2*10^4, +10*10^4}] Alternatively, if you don't want to use linear interpolation ...

1

I could get some results by defining before the my data block: diffeq = {T'[t]/T[t] == -1/(1 + (1/3)*T[t]/f[T[t]]*f'[T[t]])} boundary = {T[-40] == 10^10} odeqs = diffeq~Join~boundary Where I took freedom to change 0.3 to (1/3), since they're not equal and your equation seems to use the latter. And then using after defining f[x_]: sol = NDSolve[odeqs, T, {...

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