Culegere de probleme de Analiz˘a numeric˘a
Culegere de probleme de Analiz˘a numeric˘a
Culegere de probleme de Analiz˘a numeric˘a
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m−1 <br />
+ (xm −x0)[x0,...,xk;f][xk,...,xm;g]+<br />
k=1<br />
+(xm −x0)[x0,...,xm;f][xm;g]<br />
=<br />
<br />
m<br />
[x0,...,k;f][xk,...,xm;g]<br />
k=0<br />
Observat¸ia 5.0.22 Diferent¸a divizată se poate introduce ca s¸i coeficient dominant<br />
în PIL.<br />
Problema 5.0.23 (Aplicat¸ie) O modalitate rapidă <strong>de</strong> a calcula valorile unui polinom<br />
<strong>de</strong> grad 3 în puncte echidistante folosind diferent¸e divizate.<br />
∆ 3 P(0)<br />
P(x) = ax 3 +bx 2 +cx+d<br />
∆P(x) = P(x+h)−P(x) ⇒ P(x+h) = P(x)+∆P(x)<br />
∆ 2 P(x) = ∆P(x+h)−∆P(x)<br />
∆P(x+h) = ∆P(x)+∆ 2 P(x)<br />
∆ 3 P(x) = ∆ 2 P(x+h)−∆ 2 P(x)<br />
∆ 2 P(x+h) = ∆ 2 P(x)+∆ 3 P(x)<br />
∆ 3 P(x) = 6ah 3<br />
∆P(0) = ah 3 +bh 2 +ch = h(h(ah+b)+c)<br />
∆ 2 P(0) = P(2h)−2P(h)+P(0) =<br />
= 8ah 3 +4bh 2 +2ch+d−2ah 3 −2bh 2 −2ch−2d+d =<br />
= 6ah 3 +2bh 2 = 2h 2 (3ah+b)<br />
∆k,i+1 = ∆k−1,i +∆k−1,i+1<br />
Problema 5.0.24 Dacă f,g : M → R are loc<br />
(∆ m h<br />
fg)(a) =<br />
m<br />
i=0<br />
=<br />
<br />
m<br />
(∆<br />
i<br />
i hf)(a)(∆m−i h g)(a+ih)<br />
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