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BISECTION METHOD
Root-Finding Problem
- Given computable f (x) ∈ C[a, b], problem is to find for x ∈ [a, b] a solution to f (x) = 0.
- Solution r with f (r) = 0 is root or zero of f.
- Maybe more than one solution; rearrangement some- times needed: x^2 = sin(x) + 0.5.
Bisection Algorithm
- Input: computable f (x) and [a, b], accuracy level .
- Initialization: find [a 1 , b 1 ] ⊂ [a, b], with f (a 1 )f (b 1 ) < 0, set i = 1.
- Basic Bisection Algorithm:
- Set ri = (ai + bi)/2;
- If f (ai)f (ri) < 0, set bi+1 = ri, ai+1 = ai; otherwise, set ai+1 = ri, bi+1 = bi;
- If (bi+1 − ai+1)/ 2 > , set i = i + 1 and go to step 1
- Stop with r ≈ (bi+1 + ai+1)/2. Notes: careful algorithm checks for f (ri) = 0, limits i; other stopping conditions possible, e.g. |f (ri)| < .
−1 −0.5 0 0.5 1 1.5 2
−0.
0
1
2
x
x^2 −sin(x)−.
BISECTION METHOD CONTINUED
Bisection Method Analysis
- Convergence:
- Steps needed for absolute error ?
- Summary: simple method with slow (linear) but guarenteed convergence; size of f (rn) is not used.
FIXED POINT-ITERATION METHODS
Background
- Terminology: given g ∈ C[a, b] a fixed point r for g(x) is a point where r = g(r).
- Relation to rootfinding: given g(x), a fixed point for g(x) is root of f (x) = x − g(x); given f (x), a root of f (x) is a fixed point for g(x) = x − f (x).
- Fixed points from graphs: a fixed point exists if the graph y = x crosses the graph y = g(x). Example: g(x) = cos(x):
−3−3 −2 −1 0 1 2 3
−
−
0
1
2
3 y = x and y = cos(x)
(^11) 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3
2
3
4
5
6
7 y=x and y=2/x+5/x
2
(^11) 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3
2
3 y=x and y=(2*x+5)
1/
FIXED-POINT METHODS CONTINUED
- Another example: x^2 −sin(x) = 0.5, with root in [0, 2]. Some possible rearrangments: a) x = (sin(x) + .5)/x has [g(0), g(2)] → [. 7 , ∞); b) x =
√ sin(x) + .5 has [g(0), g(2)] ⊂ [. 70 , 1 .19]. Fixed-Point Iterations: i a) b) 0 2 2 1 0.70465 1. 2 1.6288 1. 3 0.91986 1. 4 1.4084 1. 5 1.0557 1. 6 1.2979 1. 7 1.1272 1. Another solution in [− 1 , 0]? [ga(−.5), ga(0)] ⊂ [−. 05 , ∞), [gb(−.5), gb(0)] ⊂ [0, .71]. c) x = sin−^1 (x^2 − .5), has [g(−.5), g(0)] ⊂ [−. 53 , .53]. Fixed-Point Iterations: i a) b) c) 0 -0.5 -0.5 -0. 1 -0.04115 0.14344 -0. 2 -11.151 0.80184 -0. 3 -0.13343 1.1039 -0. 4 -2.7503 1.1802 -0. 5 -0.04312 1.1936 -0. 6 -10.596 1.1957 -0. 7 -0.13413 1.1960 -0.
FP METHODS ANALYSIS CONTINUED
Fixed Point Theorem: if g ∈ C[a, b], g(x) ∈ [a, b] and |g′(x)| ≤ k < 1 for x ∈ (a, b), then for any r 0 ∈ [a, b], rn = g(rn− 1 ) converges to a unique fixed point r ∈ [a, b]. Proof: consider |rn − r| = |g(rn− 1 ) − g(r)|
Corollary: given same conditions on g(x), |rn − r| ≤ kn^ max{r 0 − a, b − r 0 }, |rn − r| ≤ kn^ |r 11 −−kr^0 |. Proof: consider k|r − r 0 | = |(r 1 − r 0 ) + (r − r 1 )|
Notes: a) Linear convergence: if ei is the error at each step i, and
ilim→∞ |
ei+ ei
| = S < 1 ,
the method has linear convergence with rate S. b) FP convergence is usually linear, with |rn − r| ≈ Ckn; if k > 1 /2, the bisection method is better. c) Estimating k? Let en = rn − r. Then for signed k, rn− 2 = r + en− 2 , rn− 1 ≈ r + ken− 2 , rn ≈ r + k^2 en− 2 , so signed k is k ≈ (rn − rn− 1 )/(rn− 1 − rn− 2 ). d) Taylor series analysis: from rn+1 = g(rn), near r r + en+1 = g(r + en) = g(r) + eng′(r) + e^2 ng′′(ξn)/ 2 , so en+1 ≈ eng′(r), linear convergence with S = |g′(r)|.
