# Tag Info

5

The OP linked to an answer of mine for interpolating over general point sets; for constructing a single interpolating function, a slight modification of my procedure is needed. (In particular, you don't need centripetal or chord-length parametrization in this case.) Let's start with some data: data = {{0, 0}, {1/10, 3/10}, {1/2, 3/5}, {1, -1/5}, {2, 3}, ...

4

Based on the information you gave, you definitely can NOT. Especially since this is a highly non-linear device (as you mentioned it is a power MOSFET). You need first to have a mathematical "model" that describes the Id, Vds, and gate charge relation in your device. In fact, this model need to be a predictive model, i.e., one that can be extrapolated to ...

0

you can try FindRoot f = Interpolation[{1, 2, 3, 5, 8, 5}]; sol = FindRoot[f[x] == 2.5, {x, 3}] (*{x -> 2.5641}*) Plot[{f[x], 2.5}, {x, 1, 6}, GridLines -> {{x /. sol}, None}]

3

Method 1 You can reverse the value of {x,y} to {y,x}, and then interpolate them. Note:In this case, the value of $y$ cannot be duplicate lst= {{3.61648, 5.64818}, {7.53428, 4.52803}, {4.21088, 2.35117}, {4.48224,1.08325}, {4.63735, 5.5877}, {2.24299, 3.10376}} x = Interpolation[Reverse /@ data]; x[3.] 2.44086 Method 2 If the the values of $y$ ...

4

I guess what you are looking for can be easily achieved (and adjusted) with the RegionFunction option. Let me give an example using Plot3D. It should work similarly with ListPlot3D: Manipulate[ Plot3D[Sin[10 x + 0.1 y]^2*Exp[-(x^2 + y^2)], {x, -2, 2}, {y, -2, 2}, PlotRange -> {Automatic, Automatic, {0, 1}}, PlotPoints -> 25, RegionFunction ...

1

Here is my second stab at the problem. The first time I tried this, I was using Interpolation to find the expression of the parabolas, but the Manipulate wrapper was quite sluggish. @GuessWhoItIs. pointed out that InterpolatingPolynomials might be a snappier choice in this case. As I understand it, this function constructs a Newton divided difference ...

0

This is a possible implementation of what I was proposing, to get you started: Plot[ Interpolation[#, InterpolationOrder -> 1][x] & /@ {r1a, r2a}, {x, 0, 9.1}, Evaluated -> True, Epilog -> {{Red, Point[r1a]}, {Blue, Point[r2a]}} ]

1

Yes, for this you can use Map (as operator /@) which is one of the most often used functions when doing the same job for several inputs. For instance creating 3 data sets {g1, g2, g3} = Range /@ {10, 20, 30} and then creating your interpolations {i1, i2, i3} = Interpolation[#, InterpolationOrder -> 1] & /@ {g1, g2, g3} If you don't know what the ...

0

Recently, after trying several methods, I found that the following easy trick works the best and the fastest (in case of large number of function calls): Suppose that f is a deliberate interpolated function or spline (or B-spline), (* note that 'points' cover the range [xmin, xmax]; see the original question above *) splineintpF = Interpolation[points, ...

5

I like the answer of @Michael E2 but I'm still searching, if available, a simpler way. Experimenting I found this way. My way probably can be applied also for a 3D mesh/region. The data to interpolate: << NDSolveFEM f[t_, x_, y_] := t (1 - x^2 - y^2); mesh = ToElementMesh[Disk[]]; coords = mesh["Coordinates"]; tl = Range[0, 1, .05]; values = ...

6

Somewhat manual construction: << NDSolveFEM mesh = ToElementMesh[Disk[], MaxCellMeasure -> Infinity]; tl = Range[0, 1, 0.2]; mesh3 = MeshOrderAlteration[ With[{gc = ElementMeshToGraphicsComplex@MeshOrderAlteration[mesh, 1]}, gc /. GraphicsComplex[pts_, g_, o___] :> ToElementMesh[ "Coordinates" -> Flatten[ ...

3

If you plot f, f', and f'' over a much smaller range, the answer will be obvious: the interpolation function f is piecewise linear, f' is piecewise constant (i.e., a step function), and f'' is identically zero within each interval.

1

Chaining extrapolation handlers We can chain together the extrapolation handlers. It will overwrite any existing extrapolation handler except in the last interpolating function; however, that seems consistent with the goal of splicing together interpolating functions. We can find the position of the extrapolation handler this way (see also What's ...

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