Culegere de probleme de Analiz˘a numeric˘a

76 Calculul cu diferent¸e
căci
Demonstrat¸ie. Induct¸ie dupăm
m = 1
(∆hfg)(a) = f(a)(∆hg)(a)+g(a+h)(∆hf)a (5.3)
(∆hfg)(a) = f(a+h)g(a+h)−f(a)g(a)|±f(a)g(c+h)
(∆hfg)(a) = f(a)[g(a+h)−g(a)]+g(a+h)[f(a+h)−f(a)]
Presupunem relat¸ia adevărată pentru m−1
(∆ m−1
m−1
h fg)(a) =
(5.3) ⇒ (∆ m h
(∆ m h
fg)(a) =
+
i=0
fg)(a) =
m−1
i=0
i
(∆ o hf)(a)(∆ m−i−1
h g)(a+ih)
m
m−1
[(∆
i
i hf)(a)(∆m−i h g(a+ih)+
+(∆ i+1
h f)(a)(∆m−i−1
h g(a+(i+1)h)]
m
k=1
m−1
i=0
m−1
= f(a)(∆ m h g)(a)+ m−1
i
i
(∆ i h f)(a)(∆m−i
h g)(a+ih)+
m−1
(∆
k −1
k hf)(a)(∆m−k h g)(a+kh) =
+
+(∆ m h f)(a)g(a+mh).
Problema 5.0.25 (Formula lui Vandermonde)
(a+b) [m,h] =
m
j=0
m−1
(∆
i−1
m−i
h g)(a+ih)+
m
a
j
[m−j,h] b [j,h] .
Demonstrat¸ie. Induct¸ie dupăm
m = 1
(a+b) [1,h] = a+b
1
a
0
[1,h] b [0,h]
1
+
1
a [0,h] b [1,h] = a+b