Se considera \( f:[a, b]\to\mathbb{R} \) o functie de \( n+1 \)-ori derivabila pe \( [a, b] \) astfel incat \( f^{(i)}(a)=f^{(i)}(b)=0 \) pentru \( i=0, 1, \ldots, n \). Aratati ca exista \( \xi\in (a, b) \) astfel incat \( f^{(n+1)}(\xi)=f(\xi) \).
(Problema E 3214, 1987, American Mathematical Monthly)
The n-th mean value theorem
Moderators: Bogdan Posa, Laurian Filip, Beniamin Bogosel, Radu Titiu, Marius Dragoi
-
Cezar Lupu
- Site Admin
- Posts: 612
- Joined: Wed Sep 26, 2007 2:04 pm
- Location: Bucuresti sau Constanta
- Contact:
The n-th mean value theorem
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
-
Bogdan Cebere
- Thales
- Posts: 145
- Joined: Sun Nov 04, 2007 1:04 pm
Fie \( g(x)=e^{-x}(f^{(n)}(x)+f^{(n-1)}(x)+...+f^{\prime}(x)+ f(x)) \). Din ipoteza avem ca \( g(a)=g(b)=0 \). Rezulta ca exista \( \xi\in (a, b) \) astfel incat \( g^{\prime}(x)=0 \). Dar
\( g^{\prime}(x)=e^{-x}((f^{(n+1)}(x)-f^{(n)}(x))+(f^{(n)}(x)-f^{(n-1)}(x))+...+(f^{\prime}(x)-f(x)))=e^{-x}(f^{(n+1)}(x)-f(x)) \).
Deci exista \( \xi\in (a, b) \) a.i. \( f^{(n+1)}(\xi )-f(\xi )=0 \).
\( g^{\prime}(x)=e^{-x}((f^{(n+1)}(x)-f^{(n)}(x))+(f^{(n)}(x)-f^{(n-1)}(x))+...+(f^{\prime}(x)-f(x)))=e^{-x}(f^{(n+1)}(x)-f(x)) \).
Deci exista \( \xi\in (a, b) \) a.i. \( f^{(n+1)}(\xi )-f(\xi )=0 \).