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Question Number 193247 by mokys last updated on 08/Jun/23
if f(x) = ((ax+1)/(x+b)) find f^( 100) (x) and f^(101) (x) ?
iff(x)=ax+1x+bfindf100(x)andf101(x)?
Answered by aba last updated on 08/Jun/23
f(x)=((ax+1)/(x+b))=a+((1−ab)/(x+b)) f^2 (x)=fof(x)=f(a+((1−ab)/(x+b)))=a+((1−ab)/(f(x)+b)) f^3 (x)=f^2 of(x)=f^2 (a+((1−ab)/(x+b)))=a+((1−ab)/(f(a+((1−ab)/(x+b)))+b))=a+((1−ab)/(a+((1−ab)/(f(x)+b))+b))=a+((1−ab)/(f^2 (x)+b)) f^k (x)=a+((1−ab)/(f^(k−1) (x)+b)), ∀ k≥2 ⇓ f^(100) (x)=a+((1−ab)/(f^(99) (x)+b)) f^(101) (x)=a+((1−ab)/(f^(100) (x)+b))
f(x)=ax+1x+b=a+1−abx+bf2(x)=fof(x)=f(a+1−abx+b)=a+1−abf(x)+bf3(x)=f2of(x)=f2(a+1−abx+b)=a+1−abf(a+1−abx+b)+b=a+1−aba+1−abf(x)+b+b=a+1−abf2(x)+bfk(x)=a+1−abfk−1(x)+b,∀k⩾2⇓f100(x)=a+1−abf99(x)+bf101(x)=a+1−abf100(x)+b
Answered by aba last updated on 09/Jun/23
f^k (x)=a+((1−ab)/(f^(k−1) (x)+b)), ∀k≥2 it is true for k=2. then assume true for k and prove that it is true for k+1 f^(k+1) (x)=f^k of(x)=f^k (f(x))=a+((1−ab)/(f^(k−1) (f(x))+b))=a+((1−ab)/(f^k (x)+b)) ∀k≥2 , f^k (x)=a+((1−ab)/(f^(k−1) (x)+b))
fk(x)=a+1−abfk−1(x)+b,∀k⩾2itistruefork=2.thenassumetrueforkandprovethatitistruefork+1fk+1(x)=fkof(x)=fk(f(x))=a+1−abfk−1(f(x))+b=a+1−abfk(x)+b∀k⩾2,fk(x)=a+1−abfk−1(x)+b

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