Mathematica如何创建一个InterpolatingFunction对象？
问题描述：
Mathematica如何创建一个InterpolatingFunction对象？例如：Mathematica如何创建一个InterpolatingFunction对象？
test1 = FunctionInterpolation[Sin[x],{x,0,2*Pi}]
TEST1的FullForm很长，但是与 对应的y值主要x值。然而，插值是非线性 （因为我没有设置InterpolationOrder - > 1）
我知道Mathematica使用三次样条，部分原因是默认 InterpolationOrder是3，也因为：
Plot[D[test1[t],t,t,t,t] /. t->x, {x,0,2*Pi}]
显示四阶导数是一致的0.
那么，Mathematica如何计算这个三次样条？
我的目标是在Perl中使用FunctionInterpolation对象。
编辑：谢谢你，萨沙！这正是我想要的，有一个小小的 小故障。以下是我尝试以易于转换为Perl的 方式重新实现Hermite插值（也可在 https://github.com/barrycarter/bcapps/blob/master/bc-approx-sun-ra-dec.m#L234处获得）。
问题：最后3个图有很小但非零的值。我不能 告诉我是否实施了Hermite错误，或者这只是一个数字 小故障。
(* the Hermite <h>(not Hermione)</h> polynomials *)
h00[t_] = (1+2*t)*(1-t)^2
h10[t_] = t*(1-t)^2
h01[t_] = t^2*(3-2*t)
h11[t_] = t^2*(t-1)
(*
This confirms my understanding of InterpolatingFunction by calculating
the value in a different, Perl-friendly, way; this probably does NOT
work for all InterpolatingFunction's, just the ones I'm using here.
f = interpolating function, t = value to evaluate at
*)
altintfuncalc[f_, t_] := Module[
{xvals, yvals, xint, tisin, tpos, m0, m1, p0, p1},
(* figure out x values *)
xvals = Flatten[f[[3]]];
(* and corresponding y values *)
yvals = Flatten[f[[4,3]]];
(* and size of each x interval; there are many other ways to do this *)
(* <h>almost all of which are better than this?</h> *)
xint = (xvals[[-1]]-xvals[[1]])/(Length[xvals]-1);
(* for efficiency, all vars above this point should be cached *)
(* which interval is t in?; interval i = x[[i]],x[[i+1]] *)
tisin = Min[Max[Ceiling[(t-xvals[[1]])/xint],1],Length[xvals]-1];
(* and the y values for this interval, using Hermite convention *)
p0 = yvals[[tisin]];
p1 = yvals[[tisin+1]];
(* what is t's position in this interval? *)
tpos = (t-xvals[[tisin]])/xint;
(* what are the slopes for the intervals immediately before/after this one? *)
(* we are assuming interval length of 1, so we do NOT divide by int *)
m0 = p0-yvals[[tisin-1]];
m1 = yvals[[tisin+2]]-p1;
(* return the Hermite approximation *)
(* <h>Whoever wrote the wp article was thinking of w00t</h> *)
h00[tpos]*p0 + h10[tpos]*m0 + h01[tpos]*p1 + h11[tpos]*m1
]
(* test cases *)
f1 = FunctionInterpolation[Sin[x],{x,0,2*Pi}]
f2 = FunctionInterpolation[x^2,{x,0,10}]
f3 = FunctionInterpolation[Exp[x],{x,0,10}]
Plot[{altintfuncalc[f1,t] - f1[t]},{t,0,2*Pi}]
Plot[{altintfuncalc[f2,t] - f2[t]},{t,0,10}]
Plot[{altintfuncalc[f3,t] - f3[t]},{t,0,10}]
答
通常它使用分段Hermite cubic interpolation。不过，我不确定节点的选择。看起来他们是在整个区间内统一选择的。我相信在假设函数平滑的情况下，对于所请求的精度的间隔的下界有结果，但我没有具体的细节。
它可能使用Hermite插值作为默认值（而不是Splines），至少这是Interpolation []的作用 – 2011-04-16 14:02:26