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
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 = (xn + 5/xn)/2
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