a doua
f(x) =2x+1
f(f(x))=2(2x+1)+1= 4x+2+1=4x+3
f°f°f (x)=f(4x+3)=2(4x+3) +1= 8x+7
presupunem f°f°...°f= (2^n)*x + (2^n-1)
verificare pt n=1
f(x) = 2^1 *x+ (2-1) =2x+1 adevarat
presupunem adevarta relatia pt n Propozitia Pn
f°f°...°f de n ori = (2^n)*x + (2^n-1)
af tunci
f°f°...°fde (n+1) ori = f° ( f°f°...°f)de n ori = f( (2^n)*x + (2^n-1))
2((2^n)*x + (2^n-1))+1= 2^(n+1) *x + 2*2^n-2+1= 2^ (n+1)*x +2^(n+1)-1
Pn⇒Pn+1 relatia este demonstrata prin inductie completa
deci
f°f°...°f de n ori = (2^n)*x + (2^n-1)
problema 1
f(x)=x+[x]= [x]+{x}+[x]=2[x]+{x}
f°f(x) = f(2[x]+{x})=2[x]+{x}+ [2[x]+{x}]=2[x]+2[x]+{x}=4[x]+{x}
f°f°f(x)=4[x]+[x}+[4[x]+{x}]= 4[x]+{x}+4[x]=8[x]+{x}
presupunem ca
f°f°....°f (x)de n ori = 2^n*[x]+{x}
verificare pt n=1
f(x) = 2^1*[x]+{x}=2[x]+{x}
presupunem adevarata Pn
adica
f°f°....°f (x)de n ori = 2^n*[x]+{x}
atunci
f°f°....°f (x)de n+1 ori= f°(f°f°...°f)de n ori= f(2^n*[x]+{x})=
[
2^N*[x]+{x} + [2^n*[x]+{x}]= 2^n*[x]+{x}+2^n*[x]= 2*2n[x]+{x}= 2*2^n*[x]+{x}
=2^(n+1)*[x]+{x}
Pn⇒Pn+1 , relatia este demonstrata prin inductie completa
deci
f°f°....°f (x)de n ori = 2^n*[x]+{x}
atunci
f°f°....°f (x)de 2017 ori=2^2017[x]+{x}
