Ee1033aS10

Ee1033aS10 - Finding a Solution of a Nonlinear Equation...

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Finding a Solution of a Nonlinear Equation f(x)=0 x f(x) f min ( ) F x x e(x) = v f(x) = e(x) - v 0 ' () def F x f x  0 T h a t is fin d a n x so th a t fx , EE103 (SEJ) SLIDES 3A 1 Fixed Point Approach (Method of Successive Approximations): xg x f xx g x  5 21 0 Ex x x  : 5 1 2 x x EE103 (SEJ) SLIDES 3A 2
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Fixed Point Approach (Method of Successive Approximations): xg x () 10 x 21 x  1 kk x EE103 (SEJ) SLIDES 3A 3 5 () ( 1 ) /2 gx x  5 0 xx  EE103 (SEJ) SLIDES 3A 4
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Fixed Point Rate of Convergence Assume as k x k x  1 lim ( ) (lim ) ( ) kk k k xx g x g x g x  Assume k x x and assume ' g is continuous. Let exx  denote the error at the th k th iteration. Then, by Taylor’s 0 order theorem , the mean value theorem, we have 1, () () ' ( ) ( ) k x x k x gx g x x  () x '( )( ) k kx x k x xg x x  1 || | ' k g , |( ) | k k x x 1 lim | '( ) | k k  |' |1  EE103 (SEJ) SLIDES 3A 5 k x x Fixed Point Rate of Convergence Therefore, if |0 , we see that the convergence rate is 1 , a linear (or geometric) rate of convergence. On the other hand, suppose it is the case that | ' | 0 g x ? Since the convergence rate is “super linear” , it may be the case that the convergence rate is actually greater than 1 . Assume and assume '' g the second derivative exists and is continuous Let Assume k and assume , the second derivative, exists and is continuous. Let denote the error at the th k iteration. 2 1 1 2 , () () ' ( ) ( ) k k k x g x g xx x g '( ) 0 x 2 1 , ''( ) k g x x 1 1 2 2 lim | ''( ) | k k k  EE103 (SEJ) SLIDES 3A 6
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The basic idea of Newton's method for finding a root of f x 0 is quite simple Given Newton’s Method The basic idea of Newton s method for finding a root of ( ) 0 is quite simple. Given an initial point, x 0 , one computes the linear approximation for f
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Ee1033aS10 - Finding a Solution of a Nonlinear Equation...

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