NumericalMethodsChaper05

NumericalMethodsChaper05 - Numerical Methods Chapter 5...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Numerical Methods - Chapter 5 – Solution of Non-linear equations 1. Function of One Variable Given a continuous function ) ( x f in a certain interval [ x = a , x = b ], i.e. dx x df ) ( is finite in all points of the interval [ a , b ]. We want to find the solution to the equation: 0 ) ( = x f Example: () 0 5 . 0 3 sin 8 ) cos( 4 5 ) ( 2 4 = + + = x x x x x x f We want to find the value(s) of x for which 0 ) ( = x f . Those values are called roots of ) ( x f or solutions of the equation 0 ) ( = x f Existence of a solution Suppose we evaluate a continuous function ) ( x f at 2 points 1 x and 2 x . If ) ( 1 x f and ) ( 2 x f have different signs, then there exist at least one point x * in the interval [ 1 x , 2 x ] such that ( ) 0 * = x f x * is a root of ) ( x f . Conditioning Condition number = () * 1 x f if ( ) * x f is zero or close to zero then finding the root * x is difficult. Convergence rate Convergence rate definition: C x x x x r k k k = + * * 1 lim k is the iteration number. C is a constant to be chosen (C > 0). Typically C is chosen as ( ) * x f * x : is the root of () x f r : is the convergence rate. r = 1: convergence rate is linear. r > 1: convergence rate is super-linear. r = 2: convergence rate is quadratic. r = 3: convergence rate is cubic.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2 1. Interval Bisection Method The algorithm starts by finding 1 x and 2 x such that 0 ) ( ) ( 2 1 < × x f x f i.e. ) ( 1 x f and ) ( 2 x f have different signs (This can be done by trial and error). double IntervalBisection( double x1, double x2, double Tolerance) { double m, F1 ,F2, Fm; F1=f(x1); F2=f(x2); while (1){ if(fabs(F2)<Tolerance) return (x2); if(fabs(F1)<Tolerance) return (x1); m=0.5*(x2+x1); Fm=f(m); if (sgn(F1)==sgn(Fm){x1=m;F1=Fm;} else {x2=m;F2=Fm;} } } x f ( x ) x 2 x 1 m Convergence rate is typically linear r = 1 2. Fixed Point Iteration We rewrite the equation: 0 ) ( = x f as x x g = ) ( Example: 0 2 ) ( 2 = = x x x f Can be rewritten as: 1. x x x g = = 2 ) ( 2 2. x x x g = + = 2 ) ( 3. x x x x g = + = 2 ) ( 4. x x x x g = + = 1 2 2 ) ( 2 Algorithm steps: 1. Start from an initial value 0 x 2. 1 0 ) ( x x g = 3. 1 0 x x = 4. Go to step 2. The algorithm converges to the solution of ) ( x f if 1 ) ( < x g around the solution point.
Background image of page 2
3 3. Newton-Raphson Method L + Δ + Δ + Δ + = Δ + 3 3 1 3 2 2 1 2 1 1 1 ) ( ! 3 1 ) ( ! 2 1 ) ( ) ( ) ( x x x f x x x f x x x f x f x x f Taking the first term in the Taylor series:
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 9

NumericalMethodsChaper05 - Numerical Methods Chapter 5...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online