Lecture note 2 Numerical Analysis 1.1 Solution of nonlinear equations in one variable Chapter 2 in the textbook. ·Give a function f(z),iffp)=O,then we say that p is a"zero”ora"root” of f(x). ·Solve f(x)=0in[a,. -f(r)=22-4x+4=0,we can find the exact solution by hand. -f(r)=3x-et=0,need an algorithm to approximate the solution. An example of Bisection Method. Intermediate Value Theorem (special form):If f(x)is continuous on [a,b]and f(a)f(b)<0,then there exist a p[a,b]such that f(p)=0. f(b) y=f(x) f(a) Figure 1.1:Intermediate Value theorem Example of the Bisection method:Find a zero of f(z)=3x-e on the interval [1,2]. Step1f(1)=3-e≈0.282 f(2)=6-e2≈-1.389 There is a root in [a,b]. Step 2 Divide [1,2]into two equal sections(by the middle point of it):[1,1.5] and[1.5,2]. Where is the zero? Step3f(1.5)=3.1.5-e1.5≈0.018. f(1.5)f(2)<0,so the root is in [1.5,2]. Step 4 Divide [1.5,2]into equal sections(by the middle point of it):[1.5,1.75] and[1.75,2]. f(1.75)≈-0.505 f(1.5)f(1.75)<0 So the root is in [1.5,1.75] 2Lecture note 2 Numerical Analysis 1.1 Solution of nonlinear equations in one variable • Chapter 2 in the textbook. • Give a function f(x), if f(p) = 0, then we say that p is a ”zero” or a ”root” of f(x). • Solve f(x) = 0 in [a, b]. – f(x) = x 2 − 4x + 4 = 0, we can find the exact solution by hand. – f(x) = 3x − e x = 0, need an algorithm to approximate the solution. An example of Bisection Method. • Intermediate Value Theorem (special form): If f(x) is continuous on [a, b] and f(a)f(b) < 0, then there exist a p ∈ [a, b] such that f(p) = 0. a p b y = f(x) x y f(b) f(a) Figure 1.1: Intermediate Value theorem • Example of the Bisection method: Find a zero of f(x) = 3x − e x on the interval [1, 2]. Step 1 f(1) = 3 − e ≈ 0.282 f(2) = 6 − e 2 ≈ −1.389 There is a root in [a, b]. Step 2 Divide [1, 2] into two equal sections (by the middle point of it): [1, 1.5] and [1.5, 2]. Where is the zero? Step 3 f(1.5) = 3 · 1.5 − e 1.5 ≈ 0.018. f(1.5)f(2) < 0 , so the root is in [1.5, 2]. Step 4 Divide [1.5, 2] into equal sections (by the middle point of it): [1.5, 1.75] and [1.75, 2]. f(1.75) ≈ −0.505 f(1.5)f(1.75) < 0 So the root is in [1.5, 1.75] 2