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justifier que xn+1 = xn − [f (xn)/ f ' (xn)]   puis que, pour la fonction f considérée ici, xn+1 = 1/2 ( xn + 5/xn  ) pour f(x)=x*x-5

Guest Jul 14, 2017
edited by Guest  Jul 14, 2017
 #1
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Have a look at https://en.wikipedia.org/wiki/Newton%27s_method for a derivation of the Newton-Raphson method.

 

Applying the technique tp f(x) = x2 - 5 we have:

 

f(x) = x2 - 5

 

f`(x) = 2x

 

xn+1 = xn - (xn2 - 5)/(2xn)

 

xn+1 = (2xn2 - (xn2 - 5))/(2xn

 

xn+1 = (2xn2 - xn2 + 5))/(2xn

 

xn+1 = (xn2 + 5))/(2xn

 

xn+1 = (x+ 5/xn)/2 

 

.

Alan  Jul 14, 2017

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