1549301742-The_Theory_of_Difference_Schemes__Samarskii

Two-layer iteration sche1nes 673

does follow ( 44) in conformity with the values m = nj ,2nj, ... ,

2r1-^1 nj. Since 1lj+J = 2"1nj + 1, the set en1+1 IS described by

formula. ( 46) under the condition 2in = 2"1 nj;.

(2) if rj = 1, that is, nj = (nj+l - 1)/2, then the set e 2 n 1 is given by

formulas ( 45) and the set en;+i = e 2111 + 1 is in line with formulas

(46). After that, we a.re moving from en .J+l to en 1+2 , etc.

Starting from j = 1 with n 1 = 1 and en J = e 1 = { 1}, we proceed to

calculations by a nun1ber of different formulas: if kt = 0, then n = nt is

an odd number and the con1puta.tions must terminate for j = t - l upon

receipt of forn1ulas (46); if kt > 0, then n is an even nm11ber such that

n = 2k' nt and nt+l = 2n + 1. Since kt+l = -1, we employ r, > 2 and

formulas ( 44) in passing from en, to e,, for the values

We will elaborate on this later for several particular cases. In con-

clusion it should be noted that formulas ( 44) must be used 2r^1 -^1 tin1es as

opposed to a single application of formulas ( 45) with the accompanymg

formula ( 46). Thus, the algorithn1 is completely described.

Example 1 Let n = 90, n = 26 + 24 + 23 + 21 , that is, k 1 = 6, k 2 = 4,

k 3 = 3, k 4 = l; n 1 = 1, n 2 = 5, n 3 = 11, n 4 = 45; r 1 = 2, r 2 = 1, r 3 = 2.

The chain of describing sets is designated by the symbols

leading to en = e 90. Here e 2 m for rn = 5 refers to the set specified by
formulas ( 45) in contrast to the others e 2 m. The transition to the next
member of that chain can be done using formulas ( 44), ( 45) or ( 46) as
suggested before.

Exan1ple 2 Let n = 25, n = 24 + 23 + 2°, that is, k: 1 = 4, k 2 = :3, k: 3 = O;

n 1 = 1, n 2 = 3, n 3 = 25; -r 1 = 1, r 2 = 3. The sets

constitute a perfect chain for the purposes of the present section.
Sumn1arizing, the preceding algorithm is showing a way of obtaining
the ordered set of n zeroes of Chebyshev's polynomial T,, ( t) and a stable