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Numerical Mathematics - A Collection of Solved Problems

Numerical Mathematics - A Collection of Solved Problems

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NELINEARNE JEDNAČINE 127pa, dalje, sledujef ′ (x k ) = f ′ (a) + f ′′ (a)e k + 1 2 f ′′′ (a)e 2 k + O(e 3 k),(6)f ′ (x k ) − f ′ (x k−1 )e k − e k−1= f ′′ (a) + 1 2 f ′′′ (a)(e k + e k−1 ) + O(e 2 k−1) .pa jeNa osnovu formule (5) imam<strong>of</strong>(x k )f ′ (x k ) = e k − f ′′ (a)2f ′ (a) e2 k + O(e 3 k)(7)„ « 21 f(xk )2f ′ (x k ) f ′ = 1(x k ) 2f ′ (a) e2 k + O(e 3 k) .Na osnovu (4), a korišćenjem relacuje (3), (7) i (6) dobijamoe k+1 = − 1 4f ′′′ (a)f ′ (a) e2 k e k−1 + O(e 3 k),ili, u dovoljno bliskoj okolini tačke x = a, možemo pisati1 f ′′ (a)(8) |e k+1 | ∼˛4f ′ (a) ˛ |e k| 2 |e k−1 | .Ako iterativni proces (1) ima red konvergencije r, tada je(9) |e k+1 | ∼ A |e k | r (A > 0).Na osnovu (8) i (9) sledujeA |e k | r 1∼˛4f ′′′ (a)f ′ (a)˛ |e k| 2 |e k−1 |,odakle, rešavanjem po |e k | dobijamo1(10) |e k | ∼˛4Af ′′′ (a)f ′ (a)˛1/(r−2)Pored¯enjem (9) i (10) zaključujemo da mora bitir = 1r − 2 ,˛˛˛˛˛A = f ′′′ (a)4f ′ (a) ˛|e k−1 | 1/(r−2) .1/(r−1).

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