Convergence Of The Iteration Methods

Expand the terms in Taylor’s series. Using the fact that f(α) = 0, and canceling f ′(α), we obtain

Neglecting the terms containing ε_k³ and higher powers of ε_k, we get

and

| ε_{k 1} | = | C | | ε_k |². ........................1.1

Therefore, Newton’s method is of order 2 or has quadratic rate of convergence.

Note :

What is the importance of defining the order or rate of convergence of a method?

Ans : Suppose that we are using Newton’s method for computing a root of f(x) = 0. Let us assume that at a particular stage of iteration, the error in magnitude in computing the root is 10^-1 = 0.1. We observe from (1), that in the next iteration, the error behaves like C(0.1)² = C(10^-2). That is, we may possibly get an accuracy of two decimal places. Because of the quadratic convergence of the method, we may possibly get an accuracy of four decimal places in the next
iteration. However, it also depends on the value of C. From this discussion, we conclude that both fixed point iteration and regula-falsi methods converge slowly as they have only linear rate of convergence. Further, Newton’s method converges at least twice as fast as the fixed point iteration and regula-falsi methods.

2. When does the Newton-Raphson method fail?

Ans: (a) The method may fail when the initial approximation x₀ is far away from the exact root α . However, if the root lies in a small interval (a, b) and x₀ ∈ (a, b), then the method converges.

(b) From Eq.(1), we note that if f ′(α) ≈ 0, and f″ (x) is finite then C → ∞ and the method may fail. That is, in this case, the graph of y = f(x) is almost parallel to x-axis at the root α.

3. Let us have a re-look at the error equation. We have defined the error of approximation at the k_th iterate as ε_k = x_k – α, k = 0, 1, 2,...

From x_{k 1}= φ(x_k), k = 0, 1, 2,... and α = φ(α),

we obtain

or ε_{k 1} = a₁ε_k a₂ε_k² ..

where a1 = φ′(α), a2 = (1/2)φ″(α), etc.