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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92 Interpolare<br />
Rezolvând sistemul se obt¸ine<br />
(B2f)(x) = (2x−h)(3h−2x)<br />
f<br />
8h<br />
′ (0)+f<br />
(B2f)(x) = b01(x)f ′ (0)+b10(x)f<br />
<br />
h<br />
2<br />
+ 4x2 −h2 f<br />
8h<br />
′ (h)<br />
<br />
h<br />
+b21(x)f<br />
2<br />
′ (h)<br />
b01(x) = (2x−h)(3h−2x)<br />
8h2 , b10(x) = 1, b21(x) = 4x2 −h2 8h<br />
h<br />
(R2f)(x) =<br />
0<br />
ϕ2(x;s)f ′′′ (s)ds<br />
ϕ2(x;s) = 1<br />
2 {(x−s)2 −b01(x)[(0−s) 2 + ]+b10(x)<br />
<br />
h<br />
2 −s<br />
= 1<br />
<br />
(x−s)<br />
2<br />
2 <br />
h<br />
+ −<br />
2 −s<br />
2 − 4x2 −h2 <br />
(h−s) .<br />
4h<br />
+<br />
<br />
ϕ2(x;s) ≥ 0 dacă x ∈ 0, h<br />
<br />
, s ∈ [0,h]<br />
2<br />
<br />
h<br />
ϕ2(x;s] ≤ 0 pentru x ∈<br />
2 ,h<br />
<br />
, s ∈ [0,h]<br />
Pentru x ∈ [0,h], ϕ2(x,·) are semn constant pe[0,h]<br />
(R2f)(x) = f ′′′ (ξ)<br />
b<br />
a<br />
2 −S21[(h−s)<br />
+<br />
2 + ]′<br />
(x;s)ds = (2x−h)(2x2 −2hx−h 2 )<br />
f<br />
24<br />
′′′ (ξ), 0 ≤ ξ ≤ h<br />
Problema 6.4.3 Să se <strong>de</strong>termine un polinom <strong>de</strong> grad minim care verifică<br />
P(0) = f(0), P ′ (h) = f ′ (h), P ′′ (2h) = f ′′ (2h),<br />
un<strong>de</strong> f ∈ C 3 [0,2h] (Problema Abel-Goncearov cu două noduri). Dat¸i expresia<br />
restului.<br />
Solut¸ie. Din condit¸iile <strong>de</strong> interpolare se obt¸ine<br />
P(x) = f′′ (2h)<br />
x<br />
2<br />
2 +[f ′ (h)−hf ′′ (2h)]x+f(0)