Fie \( (x_n)_n \) si \( (y_n)_n \) si \( 0<x_0<y_0 \) definite astfel \(
x_{n+1}=\frac{x_n+y_n}{2},\ y_{n+1}=sqrt{x_{n+1}y_n} \). Sa se determine limitele celor doua siruri.
Limita a doua siruri date prin recurenta
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Re: Limita a doua siruri date prin recurenta
\(
x_1 < y_1 ,si{\rm prin inductie demonstram ca x}_{\rm n} < y_n ; \)\\
\( y_1 = \sqrt {x_1 y_0 } = \sqrt {\frac{{x_0 + y_0 }}{2}*y_0 } < y_0 {\rm si prin inductie si cateva calcule y}_{\rm n} < y_{n - 1} ; \\ \)
\( {\rm prin inductie demonstram ca x}_{\rm n} > x_{n - 1} ; \\ \)
\( Obtinem\;{\rm lantul 0 < x}_{\rm 0} < x_1 < ... < x_n < y_n < .... < y_{0{\rm }} deci{\rm sirurile sunt convergente } \\ \)
\( {\rm teorema lui Weierstrass;x = {{\rm lim}}\limits_{{\rm n} \to \infty } x_n ;y = {{\rm lim}}\limits_{{\rm n} \to \infty } y_n {\rm si trecand la limita in relatia de } \\ \)
\( {\rm recurenta obtinem x = y}{\rm .} \\ \)
x_1 < y_1 ,si{\rm prin inductie demonstram ca x}_{\rm n} < y_n ; \)\\
\( y_1 = \sqrt {x_1 y_0 } = \sqrt {\frac{{x_0 + y_0 }}{2}*y_0 } < y_0 {\rm si prin inductie si cateva calcule y}_{\rm n} < y_{n - 1} ; \\ \)
\( {\rm prin inductie demonstram ca x}_{\rm n} > x_{n - 1} ; \\ \)
\( Obtinem\;{\rm lantul 0 < x}_{\rm 0} < x_1 < ... < x_n < y_n < .... < y_{0{\rm }} deci{\rm sirurile sunt convergente } \\ \)
\( {\rm teorema lui Weierstrass;x = {{\rm lim}}\limits_{{\rm n} \to \infty } x_n ;y = {{\rm lim}}\limits_{{\rm n} \to \infty } y_n {\rm si trecand la limita in relatia de } \\ \)
\( {\rm recurenta obtinem x = y}{\rm .} \\ \)
La inceput a fost numarul. El este stapanul universului.