math - Mathematica 如何创建 InterpolatingFunction 对象？

Question

Mathematica 如何创建 InterpolatingFunction 对象？示例:

test1 = FunctionInterpolation[Sin[x],{x,0,2*Pi}]

test1 的 FullForm 很长，但主要是 x 值相应的 y 值。但是，插值不是线性的(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 为 3，也是因为:

Plot[D[test1[t],t,t,t,t] /. t->x, {x,0,2*Pi}]

表明四阶导数一致为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，Sasha!这正是我想要的，有一个未成年人毛刺。下面是我尝试在 a 中重新实现 Hermite 插值很容易转换为 Perl 的方式(也可在 https://github.com/barrycarter/bcapps/blob/master/bc-approx-sun-ra-dec.m#L234 ).

问题:最后 3 个图的值很小，但非零。我不能告诉我是否错误地实现了 Hermite，或者这只是一个数字毛刺。

(* 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}] 

Answer 1

一般使用分段Hermite cubic interpolation .不过，我不确定节点的选择。似乎它们是在整个区间内统一选择的。我确信在假定平滑函数的情况下，对于所请求精度的间隔下限会有结果，但我没有详细信息。