Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 1 / 39B / i

Table of contents
- Solution i

Solution

Assume the diagonal entries of

A

are nonzero. The relaxed Jacobi method simultaneously updates every component using the preceding iterate:

x_{i}^{(ν + 1)} = (1 - ω) x_{i}^{(ν)} + \frac{ω}{a _{ii}} (b_{i} - \sum_{j \neq = i} a_{ij} x_{j}^{(ν)}) .

In matrix form, with

D = diag (A)

, this is

x^{(ν + 1)} = x^{(ν)} + ω D^{- 1} (b - A x^{(ν)})

. The ordinary Jacobi method has

ω = 1

.

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