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Culegere de probleme de Analiz˘a numeric˘a

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94 Interpolare<br />

=<br />

∞<br />

aix i<br />

i=0<br />

m<br />

qix i −<br />

i=0<br />

q(x)<br />

n<br />

pix i<br />

f−r are o rădăcină multiplă <strong>de</strong> ordinN. Pentru coeficientul luixk <strong>de</strong> la numărător<br />

avem<br />

k<br />

aiqk−1 −pk = 0, k = 0,N<br />

i=0<br />

Luămq0 = 1 s¸i pn+1 = pn+2 = ··· = pN = 0 s¸i qm+1 = qm+2 = ··· = qN = 0<br />

i=0<br />

x 5 : 1<br />

2 q3 + 1<br />

6 q2 + 1<br />

24 q1 = − 1<br />

120<br />

x 4 : q3 + 1<br />

2 q2 + 1<br />

6 q1 + 1<br />

= 0<br />

24<br />

x 3 : q3 +q2 + 1<br />

2 q1 + 1<br />

= 0<br />

6<br />

x 2 : q2 +q1 −p2 + 1<br />

= 0<br />

2<br />

x 1 : q1 −p1 +1 = 0<br />

x 0 : p+0 = 1<br />

p0 = 1, p1 = 3<br />

5 , p2 = 2<br />

20 , q1 = − 2<br />

5 q2 = 3<br />

20 , q3 = − 1<br />

60<br />

r(x) =<br />

1− 2<br />

5<br />

1+ 3 1 x+ 5 20x2 x+ 3<br />

20 x2 − 1<br />

60 x3<br />

Problema 6.5.2 Determinat¸i aproximarea Padé <strong>de</strong> grad 6 pentruf(x) = sinx s¸i<br />

n = m = 3.<br />

Solut¸ie.<br />

k<br />

akqk−i −pk = 0, k = 0,6<br />

i=0<br />

p4 = p5 = p6 = 0 q0 = 1<br />

qn = q5 = q6 = 0 a0 = 0, a1 = 1, a2 = 0<br />

sinx = x− x3<br />

3!<br />

+ x5<br />

5!<br />

− x7<br />

7! +...

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