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Culegere de probleme de Analiz˘a numeric˘a

Culegere de probleme de Analiz˘a numeric˘a

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10.1. Ecuat¸ii în R 159<br />

Observat¸ia 10.1.7 Numărul <strong>de</strong> zecimale corecte se dublează la fiecare pas, comparativ<br />

cu numărul original <strong>de</strong> zecimale corecte.<br />

x0 = √ a(1+δ)<br />

x1 = 1<br />

<br />

x0 +<br />

2<br />

a<br />

<br />

=<br />

x0<br />

1 √ √ −1<br />

a(1+δ)+ a(1+δ)<br />

2<br />

=<br />

= 1<br />

<br />

√ √ 2<br />

a(1+δ +1−δ +δ ) = x 1+<br />

2<br />

δ2<br />

<br />

2<br />

b) Folosim metoda secantei<br />

xn+1 = xn − (xn −xn−1)f(xn)<br />

f(xn)−f(xn−1) =<br />

= xn − (xn −xn−1)(x 2 n −a)<br />

x 2 n −x2 n−1<br />

= xn − x2 n −a<br />

xn +xn−1<br />

=<br />

= x2 n +xnxn−1 −x 2 n +a<br />

xn +xn−1<br />

x0 > 0<br />

Problema 10.1.8 La fel pentru rădăcina cubică 3√ x.<br />

yn+1 = 1<br />

<br />

2yn +<br />

3<br />

x<br />

y2 <br />

n<br />

y0 > 0<br />

Problema 10.1.9 Strict aplicabilitatea meto<strong>de</strong>i lui Newton pentru rădăcini multiple.<br />

Solut¸ie. Fiex ∗ o rădăcină multiplă <strong>de</strong> ordinulm.<br />

Dorim convergent¸ă <strong>de</strong> ordinul 2.<br />

g(x) = x−m(f ′ (x)) −1 f(x)<br />

g(x ∗ ) = x ∗<br />

Presupunem căf(x ∗ ) = f ′ (x ∗ ) = ··· = f (m−1) (x ∗ ) = 0<br />

f (m) (x ∗ ) = 0

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