Question

Fie functia f:R-R ; f(x)= 2x-1. Calculati (fofo...of)(2) , unde fof= f compus cu f, de n ori. Multumesc!

Answer 1

VARIIANTA 1, mai grea...tereceti la varianta 2


f(x) =2x-1
f°f(x)= f(f(x))= 2(2x-1)-1=4x-3
 f°f°f (x)=f(4x-3) = 2(4x-3)-1=8x-7
 f°f°f°f(x)=f(7x-7)=2(8x-7)-1=16x-15


presupunem ca f°f°....°f (x)-=2^n * x-(2^n-1)
unde fs0-a comopusd cu f de n ori
verificare pt n=1
 f(x) =2^1  *x- (2-1)=2x-1 adevarat

presupunem adevarata relatia Pn
f ^ n (x)=f°f°....°f (x)-=2^n * x-(2^n-1) unde f s-a compus cu f de n ori
 atuncio f^ (n+1) (x) = f(f^n (x))= f(2^n * x-(2^n-1))=

=2(2^n * x-(2^n-1)) -1=
=2 ^(n+1) * x -2^(n+1)+2-1=

=2^(n+1)*x-2^(n+1)+1=
=2^(n+1)*x- (2^(n+1)-1)

Pn⇒Pn+1
 Relatia este adevarta , demonmstrata prin inductie matematica completa
 
 f^n (x)=2^n * x-(2^n-1)
atunci
  f^n(2)= 2^n * 2 - (2^n-1)
                   = 2^(n+1)- (2^n-1)
                   =2^ (n+1)-2^n+1
                   =2^n(2-1) +1=
                     = 2^n+1


ALTFEL, VARIANTA 2
 cred ca era mai simplu
f(2) =2*2-1=3=2+1
 f°f (2) =2*3-1=5=4+1=2²+1
f°f°f (2)= f(5) =2*5-1=9=8+1=2³+1
f°f°f°f(2)= f(9)=2*9-1=17=16+1=2^4+1


presupunem f^n (2)=2^n+1
verificare pt n=1
f(2) =2^1+1=2+1=3 adevarat
 presupenem adevarat Pn
f^n (2)=2^n+1
 f^(n+1) (2)= f(2^n+1)= 2(2^n+1) -1= 2^ (n+1)+2-1=2^(n+1) +1
Pn⇒Pn+1
formula este demomnstratatb prin inductie matematica completa
deci
f^n (2)=2^n+1