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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6.6. Interpolare spline 97<br />
Demonstrat¸ie. f ∈ C m [a,b] ⇒ f (m) ∈ [a,b] ⇒ f (m) poate fi aproximată<br />
uniform pe [a,b] printr-o funct¸ie în scară, continuă la dreapta s¸i discontinuă în<br />
x1,x2,...,xn ∈ [a,b], notată cu hm.<br />
Fie problema diferent¸ială<br />
s (m) (x) = hm(x), x ∈ [a,b]<br />
s (r) (a) = f (r) (a), r = 0,m−1<br />
Solut¸ia acestei <strong>probleme</strong> pe[a,b] este<br />
s(x) = f(a)+(x−a)f ′ (a)+···+ (x−a)m−1<br />
(m−1)! f(m−1) x<br />
(x−t)<br />
(a)+<br />
a<br />
m−1<br />
(m−1)! hm(t)dt<br />
(6.2)<br />
s este o funct¸ie spline <strong>de</strong> grad m căci<br />
s|(xi,xi+1) ∈ Pm−1, s ∈ C m−1 [a,b]<br />
f ∈ C m [a,b] ⇒<br />
f(x) = f(a)+(x−a)f ′ (a)+···+ (x−a)m−1<br />
(m−1)! f(m−1) x<br />
(a)+<br />
a<br />
(6.2), (6.3) ⇒ f (r) (x) − s (r) (x) = x<br />
a<br />
0,m−1<br />
f (r) −s (r) ∞ ≤ (b−a)m−r<br />
(x−t) m−1<br />
(m−1)! f(m) (t)dt<br />
(6.3)<br />
(x−t) m−r−1<br />
(m−r−1)! [f(m) (t) − hm(t)]dt, r =<br />
(m−r)! f(m) −hm ∞, r = 0,m−1<br />
<br />
Problema 6.6.2 Fiea,b ∈ R, a < 0, b > 1, f : [a,b] → R s¸tiind căf ∈ C 1 [a,b]<br />
s¸i cunoscând f(0),f<br />
1<br />
2<br />