3 Fixed point iteration methodI Fixed point iteration methodEquivalentf (x) = 0x=g (x)transformationthe fixed point g(x)the rooot of f (x) =0Starting from an initial value xo,calculate xi = g(xo), x2 =Thinkingg(x),.., k+ = g(xx),...if (, J- is convergence, existx* to make lim x, = x*, and g is continuous, fromthe lim Xk+1 = lim g(xk) can get x* = g(x* ), that x* isαa fixed point of g , which is the root of f =0
f (x) = 0 x = g (x) Equivalent transformation the rooot of f (x) ＝ 0 the fixed point g(x) Thinking Starting from an initial value x0 ,calculate x1 = g(x0 ), x2 = g(x1 ), ., xk+1 = g(xk ), . if is convergence，exist x* to make , and g is continuous，from the can get x* = g(x* )，that x* is a fixed point of g ，which is the root of f ＝0。 k k=0 x lim x x * k k = → ( ) k k k k x g x → + → lim 1 = lim §3 Fixed point iteration method Ⅰ、Fixed point iteration method
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x y y = x x y y = x x y y = x x y y = x x* x* x* x* y=g(x) y=g(x) y=g(x) y=g(x) x0 p0 x1 p1 ✓ x0 p0 x1 p1 ✓ x0 p0 x1 p1 x0 p0 x1 p1
Example 1.Solving equations 2x3 -x-1= 0 byiterative methodSlove: (1) Take the original equations into the equivalentequationx=2x3 -1If we take the initial value xo = O,According to the iterative method(3),we can getX = 2x -1=-1X2 =2xi -1=-3x = 2x -1 = -55Obviously iterative method is divergence
Example 1. 2 1 0 3 x − x − = Slove: 2 1 3 x = x − (1) Take the original equations into the equivalent equation 0, x0 = 2 1 3 x1 = x0 − = −1 2 1 3 x2 = x1 − = −3 2 1 3 x3 = x2 − = −55 Obviously iterative method is divergence Solving equations by iterative method If we take the initial value According to the iterative method(3),we can get
(2) If the original equation into an equivalent equationx+1x=32Taking the initial valueX = 0Xo +11xi=~0.79373122X +11.79373X2=～ 0.9644322
3 2 +1 = x x (2) If the original equation into an equivalent equation x0 = 0 Taking the initial value 0.7937 3 1 2 2 +1 = x x 3 2 1.7937 = 0.9644 3 0 1 2 +1 = x x 3 2 1 =
And so on, tox2 = 0.9644x3 = 0.9940x4 = 0.9990x5 =0.9998x6 = 1.0000x7 = 1.0000Have convergence, therefore, the solution of the equation isx = 1.0000There are different results with different Iterativeschemes in the sameequation,relate to Iterative function structureWhat forms of iteration to converge?
x = 1.0000 There are different results with different Iterative schemes in the same equation， What forms of iteration to converge? relate to Iterative function structure x2 = 0.9644 x3 = 0.9940 x4 = 0.9990 x5 = 0.9998 x6 = 1.0000 x7 = 1.0000 And so on, to Have convergence, therefore, the solution of the equation is