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sir recurent care e polinomial (clasic?)

Posted: Wed Feb 17, 2010 8:16 pm
by Dragos Fratila
Definim urmatorul sir
\( t_0=1,t_{n+1} = \frac{t_n^2-1}{t_{n-1}} \)
Sa se demonstreze ca \( t_n \) sunt polinoame cu coeficienti intregi in \( t_1 \).

Pentru a fi mai clar scriu cativa termeni ai sirului:
\( t_2 = t_1^2-1 \)
\( t_3 = t_1^3-2t_1 \)
\( t_4 = t_1^4 - 3t_1^2 + 1 \)

Posted: Thu Sep 16, 2010 9:48 pm
by lasamasatelas
Relatia de recurente se scrie si \( t_{n+1}t_{n-1}-t_n^2=1 \). Daca scadem din \( t_{n+2}t_n-t_{n+1}^2=1 \) obtinem \( t_{n+2}t_n+t_n^2-t_{n+1}t_{n-1}-t_{n+1}^2=0 \) deci \( t_n(t_n+t_{n+2})=t_{n+1}(t_{n-1}+t_{n+1}) \). Rezulta ca \( \frac{t_n+t_{n+2}}{t_{n-1}+t_{n+1}}=\frac{t_{n+1}}{t_n} \). Facem produs de la \( n=1 \) la \( N \) si obtinem \( \frac{t_N+t_{N+2}}{t_0+t_2}=\frac{t_{N+1}}{t_1} \), i.e. \( \frac{t_N+t_{N+2}}{t_1^2}=\frac{t_{N+1}}{t_1} \) deci \( t_N+t_{N+2}=t_1t_{N+1} \). Rezulta ca \( t_{N+2}=t_1t_{N+1}-t_N \) deci prin inductie \( t_N=P_N(t_1) \), unde \( P_0=1 \), \( P_1=X \) si \( P_{N+2}=XP_{N+1}-P_N \). Evident \( P_N \) e un polinom cu coeficienti intregi pt. orice \( N \).