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需要切比雪夫多项式的高阶导数

[英]High order derivatives of chebyshev polynomials needed

I need to evaluate high order (up to 4) derivatives of Chebyshev polynomials at points of the so called Chebyshev grid,我需要在所谓的 Chebyshev 网格的点处评估 Chebyshev 多项式的高阶(最多 4 个)导数,

x(j)=cos(πj/N), j=0,...,N x(j)=cos(πj/N), j=0,...,N

Anyone know how to do that?有谁知道该怎么做? I tried iterative methods but they are too clumsy.我尝试了迭代方法,但它们太笨拙了。 I remember seeing something like that in an old paper but now it's nowhere to be found.我记得在旧报纸上看到过类似的东西,但现在找不到了。

Any help appreciated.任何帮助表示赞赏。

One way to do this (though this might be the iterative method you reject) is to use the recurrence:一种方法(尽管这可能是您拒绝的迭代方法)是使用递归:

T[n+1]'/(n+1) - T[n-1]'/(n-1) = 2T[n]   n>=2

This requires one to be able to compute the derivatives of the first 3 polynomials by hand, but since这需要一个人能够手动计算前 3 个多项式的导数,但是因为

T[0](x) = 1
T[1](x) = x
T[2](x) = 2*x*x-1

this is straightforward.这很简单。

The coefficients in the recurrence are independent of x, so that if T[j,k] is the k'th derivative of the j'th Chebyshev poly, we can easily differentiate it, getting递归中的系数与 x 无关,因此如果 T[j,k] 是第 j 个 Chebyshev 多边形的第 k 个导数,我们可以很容易地对其进行微分,得到

T[n+1, k]/(n+1) - T[n-1,k]/(n-1) = 2T[n,k-1]    n>=2

So the code could be:所以代码可能是:

compute the T[n,0] (ie the polynomials) at the point, for n=0..deg
initialise T[j,d] for j=0,1,2 and the required degrees
for j=1..deg
   use the recurrence to compute the remaining polynomials

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