Solutions of equations in one variable
December 23, 2013
Problem: Solve the following nonlinear equation f (x) ≡ π +1
2sinx 2
− x = 0, x ∈ [0, 2π]. (1)
• Bisection method: Iff (x) ∈ C[a, b]and f (a)f (b) < 0, then ∃ c ∈ (a, b)such thatf (c) = 0.
Figure 1: Bisection method
Given f (x) defined on (a, b), the maximal number of iterations M , and stop criteria δ and ε, this algorithm tries to locate one root of f (x).
Compute u = f (a), v = f (b), and e = b − a If sign(u) = sign(v), then stop
For k = 1, 2, . . . , M
e = e/2, c = a + e, w = f (c) If |e| < δ or |w| < ε, then stop If sign(w) 6= sign(u)
b = c, v = w Else
a = c, u = w End If
End For
Algorithm 1: Bisection method
• Fixed-point iteration or functional iteration: Given a continuous function g, choose an initial point x0 and generate {xk}∞k=0 by
xk+1= g(xk), k ≥ 0.
Take g(x) = π +12sin x2.
1
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 2: Fixed point iteration
• Newton’s method: Starts with an initial approximation x0 and generates the sequence {xn}∞n=0 defined by
xn+1= xn− f (xn) f0(xn).
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 3: Newton’s method
Figure 2: Newton’s method
Consider the nonlinear eigenvalue problem
A(v)v ≡
A0+ sin v>Bv v>v
A1
v = λv, where
A0= 1 10
10 21 13 16
21 −26 24 2
13 24 −26 37
16 2 37 −4
, A1= β 10
20 28 12 32
28 4 14 6
12 14 32 34 32 6 34 16
,
B = 1 10
−14 16 −4 15
16 10 15 −9
−4 15 16 6
15 −9 6 −6
.
2
The Jacobian matrix of A(v)v is
J (v) = ∂
∂v(A(v)v) = A(v) + 2 cos
v>Bv v>v
(v>v)2 A1v v>v v>B − v>Bv v>
Apply Newton method to solve the nonlinear eigenvalue problem and rewrite the nonlinear eigen- value problem as
F
v λ
=
A(v)v − λv
`Tv − 1
= 0, (2)
where ` ∈ Rn is a suitable fixed nonzero vector. The Jacobian matrix of (2) is J F
vk
λk
=J(vk) − λkI −vk
`T 0
. (3)
The Newton iteration is of the form
vk+1
λk+1
= vk
λk
−
J F
vk
λk
−1A(vk)vk− λkvk
`Tvk− 1
. (4)
• Secant method: Using the approximation
f0(xn−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 (xn−1) − f (xn−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 4: Secant method
Figure 3: Secant method
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Home works
1. Plot the figure of the function f (x) on [0, 2π].
2. Use Bisection method, fixed point iteration and Newton’s method to solve (1). In each iteration, please output the approximation x1 and the relative error |x1|x−x0|
1| . Use MATLAB command function:
1. Build two functions, said “fun f” and “fun df”, to compute the values f (x) and f0(x), respectively.
2. Rewrite your previous MATLAB code of Bisection method, fixed point iteration and Newton’s method with using the functions “fun f” and “fun df”.
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