Minimal Lebesgue Constants
In the case of equidistant nodes, the Lebesgue constant grows exponentially. More precisely, we have the following asymptotic estimate
On the other hand, the Lebesgue constant grows only logarithmically if Chebyshev nodes are used, since we have
where a = 0.9625….
We conclude again that Chebyshev nodes are a very good choice for polynomial interpolation. However, there is an easy (linear) transformation of Chebyshev nodes that gives a better Lebesgue constant. Let ti denote the ith Chebyshev node. Then, define si = ti⁄cos(π⁄2(n+1)). For such nodes:
where b = 0.7219….
Those nodes are, however, not optimal (i.e. they do not minimize the Lebesgue constants) and the search for an optimal set of nodes (which has already been proved to be unique under some assumptions) is still one of the most intriguing topics in mathematics today. Using a computer, one can approximate the values of the minimal constants, here for the canonical interval :
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n 1 2 3 4 5 6 7 8 9 Λn(T) 1.0000 1.2500 1.4229 1.5595 1.6722 1.7681 1.8516 1.9255 1.9917
There are several sets of nodes that minimize, for fixed n, the Lebesgue constant. Though if we assume that we always take −1 and 1 as nodes for interpolation, then such a set is unique. To illustrate this property, we shall see what happens when n = 2 (i.e. we consider 3 interpolation nodes in which case the property is not trivial). One can check that each set of nodes of type (−a,0,a) is optimal when √8⁄3 ≤ a ≤ 1 (we consider only nodes in ). If we force the set of nodes to be of the type (−1,b,1), then b must equal 0 (look at the Lebesgue function, whose maximum is the Lebesgue constant).
If we assume that we take −1 and 1 as nodes for interpolation, then as shown by H.-J. Rack, for the case n = 3, the explicit values of the optimal nodes and the explicit value of the minimal Lebesgue constant are known.
Read more about this topic: Lebesgue Constant (interpolation)
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