Interpolation[] gives negative values when all the initial data is positive

Question

I have this data:

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I can interpolate the data with

Clear[Interpolate]
Interpolate[100] = Interpolation[data4[100], InterpolationOrder -> 2];

However, I noticed that Interpolation[] gives negative values in a small domain of my interpolation:

Plot[Interpolate[100][x], {x, 99.98, 100.01}, PlotRange -> All]

Plot[Interpolate[100][x], {x, 100.0033, 100.0034}, 
 PlotRange -> {-2*10^4, 0}]

Why is Interpolation[] giving a negative value when none of the data gives negative values? I had looked at this question, which suggested adding InterpolationOrder, but unfortunately this solution had no effect.

Answer 1

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]

Answer 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 translation (the interpolation is sensitive to a vertical translation)
• sensitivity to small absolute variations near 0

Naturally, these two "properties" are related.

 data1 = {{0, 1}, {1, 10^-3}, {2, 1}};
 data2 = {{0, 1}, {1, 0}, {2, 1}} /. {x_, y_} -> {x, y + 1};
 data3 = {{0, 1}, {1, 10^-5}, {2, 1}};

One would expect interpolation of these three datasets to have quasi-exact same shapes. It is the case with "classic" interpolation:

int1 = Interpolation[data1, InterpolationOrder -> 2];
int2 = Interpolation[data2, InterpolationOrder -> 2];
int3 = Interpolation[data3, InterpolationOrder -> 2];
Plot[{int1[t], int2[t] - 1, int3[t]}, {t, 0, 2}]

But not at all with the "Log-Exp trick":

int4 = Interpolation[data1 /. {x_, y_} -> {x, Log[y]}, InterpolationOrder -> 2];
int5 = Interpolation[data2 /. {x_, y_} -> {x, Log[y]}, InterpolationOrder -> 2];
int6 = Interpolation[data3 /. {x_, y_} -> {x, Log[y]}, InterpolationOrder -> 2];
Plot[{Exp@int4[t], -1 + Exp@int5[t], Exp@int6[t]}, {t, 0, 2}]

So maybe it is good enough for you in this case, but I would not recommend it generally speaking (and I would interrogate myself on the grounds to create an interpolation method that works only on a limited domain - positive values).

Answer 3

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 you can use Akima Interpolation which gives very smooth results with little over/undershooting.

f = ResourceFunction["AkimaInterpolation"][data4[100]];
Plot[f[x], {x, 99.98, 100.01}, PlotRange -> All]

Plot[Interpolate[100][x], {x, 100.0033, 100.0034},  PlotRange -> {-2*10^4, +10*10^4}]