与Mathematica中的Lagrange插值多项式有关的Mathematica中的列表处理

问题描述

我正在尝试使用以可以的方式传递给函数的列表

I am trying to use a list which is passed to a function in such a way that I can

• 获取该列表的长度
• 获取单独的x和y值进行操作

我要操作的列表如下所示:

The list I am trying to manipulate can be seen below:

dataTan = Table[{x, Tan[x]}, {x, -1.5, 1.5, .75}];

此问题是对在此处看到的问题的一种后续措施.我最终想要在mathematica中编写自己的函数，该函数针对给定的点集生成拉格朗日插值多项式

this question is a sort of a follow-up to the question seen here. I eventually want to write my own function in mathematica that generates the Lagrange Interpolation polynomial for a given set of points

 {{x0, y0}, ... , {xn, yn}}

我需要某种方式来访问上述要点，以便可以利用以下代码:

I need some way to access the points above so that I may utilize the code below:

Sum[Subscript[y, j]*Product[If[j != m, (x - Subscript[x, m])/
       (Subscript[x, j] - Subscript[x, m]), 1], {m, 0, k}], {j, 0, k}]

推荐答案

给出您的最后一个问题，我猜表示拉格朗日多项式，因此

LagrangePoly[pts_?MatrixQ, var_: x] /; MatchQ[Dimensions[pts], {_, 2}] := 
   With[{k = Length[pts]}, Sum[pts[[j, 2]] Product[
     If[j != m, (var - pts[[m, 1]])/(pts[[j, 1]] - pts[[m, 1]]), 1], 
     {m, 1, k}], {j, 1, k}]]

我们可以针对切线函数

In[2]:= points = Table[{x, Tan[x]}, {x, -1.2, 1.2, .2}]
Out[2]= {{-1.2, -2.57215}, {-1., -1.55741}, {-0.8, -1.02964}, 
         {-0.6, -0.684137}, {-0.4, -0.422793}, {-0.2, -0.20271}, 
         {0., 0.}, {0.2, 0.20271}, {0.4, 0.422793}, 
         {0.6, 0.684137}, {0.8, 1.02964}, {1., 1.55741}, {1.2, 2.57215}}

In[3]:= Plot[Evaluate[Expand[LagrangePoly[points, x]]], {x, -1.2, 1.2}, 
     Epilog -> Point[points]]

在这种情况下，插值良好，与原始函数的最大偏差为

In this case, the interpolation is good, the maximum deviation from the original function is

In[4]:= FindMaximum[{Abs[Tan[x] - LagrangePoly[points, x]], -1.2<x<1.2}, x]   
Out[4]= {0.000184412, {x -> 0.936711}}

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还要注意，内插多项式实际上是内置在Mathematica中的: /p>

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Also note, that interpolating polynomials are actually built into Mathematica:

In[5]:= InterpolatingPolynomial[points, x]-LagrangePoly[points, x]//Expand//Chop
Out[5]= 0

在比较之前，我必须将它们都展开，因为InterpolatingPolynomial在高效的 HornerForm ，而我的LagrangePoly以非常低效的形式返回.

I had to expand them both before comparison, since InterpolatingPolynomial returns the result in the efficient HornerForm, while my LagrangePoly is returned in a very inefficient form.

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