Solutions of equations in one variable

Solutions of equations in one variable

October 21, 2012

Problem: Solve the following nonlinear equation f (x) ≡ π +1

2sinx 2

− x = 0, x ∈ [0, 2π]. (1)

• Fixed-point iteration or functional iteration: Given a continuous function g, choose an initial point x0and generate {xk}^∞_k=0 by

xk+1= g(xk), k ≥ 0.

Take g(x) = π +¹₂sin ^x₂
.

Given x0, tolerance T OL, maximum number of iteration M . Set i = 1 and x = g(x0).

While i ≤ M and ^|x−x_|x|^0| ≥ T OL Set i = i + 1, x₀= x and x = g(x₀).

End While

Algorithm 1: Fixed point iteration

• Newton’s method: Starts with an initial approximation x0and generates the sequence {xn}^∞_n=0 defined by

xn+1= xn− f (xn) f⁰(x_n).

Given x0, tolerance T OL, maximum number of iteration M . Set i = 1 and x = x0− f (x0)/f⁰(x0).

While i ≤ M and ^|x−x_|x|^0| ≥ T OL

Set i = i + 1, x₀= x and x = x₀− f (x₀)/f⁰(x₀).

End While

Algorithm 2: Newton’s method

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Figure 1: Newton’s method

• Secant method: Using the approximation

f⁰(x_n−1) ≈ f (xn−1) − f (xn−2) xn−1− xn−2

.

for f⁰(xn−1) in Newton’s formula gives

xn = xn−1−f (xn−1)(xn−1− xn−2) f (x_n−1) − f (x_n−2) .

Given x0, x1, tolerance T OL, maximum number of iteration M . Set i = 2; y0= f (x0); y1= f (x1); x = x1− y1(x1− x0)/(y1− y0).

While i ≤ M and ^|x−x_|x|^1|≥ T OL

Set i = i + 1; x0= x1; y0= y1; x1= x; y1= f (x);

x = x₁− y1(x₁− x0)/(y₁− y0).

End While

Algorithm 3: Secant method

Home works

1. Plot the figure of the function f (x) on [0, 2π].

2. Use fixed point iteration, Newton’s method and Secant method to solve (1). In each iteration, please output the approximation x₁and the relative error ^|x1_|x^−x0|

1| .

(a) Fixed point iteration: 40040112S, 498711079, 40040126S

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Figure 2: Secant method

(b) Newton’s method: 40040114S, 498402331, 40040316S, 40140229S (c) Secant method: 40040118S, 498401052, 40040326S

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