Question

Mathematica如何创建InterpolatingFunction对象？例如：

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 - ＆gt; 1）

我知道Mathematica使用三次样条，部分原因是默认 InterpolationOrder是3，但也是因为：

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

表示4阶导数均为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Sasha！这完全是我想要的，一个未成年人毛刺。下面是我尝试重新实现Hermite插值这种方式很容易转换为Perl（也可以在 https://github.com/barrycarter/bcapps/blob/master/bc-approx-sun-ra-dec.m#L234）。

问题：最后3个图有小但非零的值。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

通常它使用分段Hermite cubic interpolation。不过，我不确定节点的选择。看来它们是在整个区间内均匀选择的。我确信假设平滑函数的请求精度的间隔下限有结果，但我没有详细信息。

Mathematica如何创建InterpolatingFunction对象？

1 个答案: