Dear Boubakeur, in Newton's method try to substitute F'(x0) with finite difference (F(x) - F(x0))/(x - x0) for some suitable x; note that you can not use x0 = xn and x = xn+1, else Newton's formula is only an identity and so not useful; try using x = xn-1, so your finite difference (backward) uses a value of x already computed in previous step. Gianluca

Hi Gianluca. It is possible if we work in IR. But in IRn it is not possible .