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Question Number 71147 by mr W last updated on 12/Oct/19

Commented by mr W last updated on 12/Oct/19

find f(5)=?

findf(5)=?

Answered by MJS last updated on 12/Oct/19

let f(f(f..._(k times) )) =f_k f_1 (5)=f_(200) (1000)=f_(199) (997)=f_(200) (1002)= =f_(199) (999)=f_(200) (1004)=f_(199) (1001)=f_(198) (998)= =f_(199) (1003)=f_(198) (1000)=...=f_(196) (1000)=...= =f_2 (1000)=f_1 (997)=f_2 (1002)=f_1 (999)= =f_2 (1004)=f_1 (1001)=998

letf(f(f...ktimes))=fkf1(5)=f200(1000)=f199(997)=f200(1002)==f199(999)=f200(1004)=f199(1001)=f198(998)==f199(1003)=f198(1000)=...=f196(1000)=...==f2(1000)=f1(997)=f2(1002)=f1(999)==f2(1004)=f1(1001)=998

Commented by mr W last updated on 12/Oct/19

thanks!

thanks!

Commented by mr W last updated on 12/Oct/19

it seems that f(2k)=997 and f(2k+1)=998, right?

itseemsthatf(2k)=997andf(2k+1)=998,right?

Answered by mind is power last updated on 12/Oct/19

f(5)=ff(10),f(10)=fff(15) ⇒f(5)=f........f(1000) ,1000=5×200 f(5)=(f^((200)) (1000))= f(1000)=997 ff(1000)=f(997)=ff(1002)=f(999)=ff(1004)=f(1001)=998 fff(1000)=f(998)=ff(1003)=f(1000)=997 ⇒f^(2n+1) (1000)=f(1000)=997 f^(2n+2) =f(f^(2n+1) (1000))=f(f(1000))=f^2 (1000)=998 by induction f^n (1000)= { ((998 if n=2k ,k≥1)),((997 if n=2k+1 ,k≥0)) :} f(5)=f^(200) (1000)=998

f(5)=ff(10),f(10)=fff(15)⇒f(5)=f........f(1000),1000=5×200f(5)=(f(200)(1000))=f(1000)=997ff(1000)=f(997)=ff(1002)=f(999)=ff(1004)=f(1001)=998fff(1000)=f(998)=ff(1003)=f(1000)=997⇒f2n+1(1000)=f(1000)=997f2n+2=f(f2n+1(1000))=f(f(1000))=f2(1000)=998byinductionfn(1000)={998ifn=2k,k⩾1997ifn=2k+1,k⩾0f(5)=f200(1000)=998

Commented by mr W last updated on 12/Oct/19

thanks!

thanks!

Commented by mind is power last updated on 12/Oct/19

y′re welcom

y′rewelcom

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