Question Number 1850 by 123456 last updated on 14/Oct/15
![f_(n+1) (z+1)=[z−f_n (0)]f_n (z) f_1 (z)=z+1 f_3 (z)=???](https://www.tinkutara.com/question/Q1850.png)
$${f}_{{n}+\mathrm{1}} \left({z}+\mathrm{1}\right)=\left[{z}−{f}_{{n}} \left(\mathrm{0}\right)\right]{f}_{{n}} \left({z}\right) \\ $$$${f}_{\mathrm{1}} \left({z}\right)={z}+\mathrm{1} \\ $$$${f}_{\mathrm{3}} \left({z}\right)=??? \\ $$
Answered by Rasheed Soomro last updated on 22/Oct/15
![f_(n+1) (z+1)=[z−f_n (0)]f_n (z).................(I) f_1 (z)=z+1⇒f_1 (z+1)=z+1_(−) +1=z+2 f_1 (0) =0+1=1_(−) •Substituting n=1 in (I) f_2 (z+1)=[z−f_1 (0)]f_1 (z) =[z−1](z+2) f_2 (z−1+1)=(z−1−1)(z−1+2) [Replace z by z−1] f_2 (z)=(z−2)(z+1)_(−) f_2 (0)=−2_(−) •Substituting n=2 in (I) f_3 (z+1)=[z−f_2 (0)]f_2 (z) =[z−(−2)][(z−2)(z+1)] =(z+2)(z−2)(z+1) f_3 (z−1+1)=(z−1+2)(z−1−2)(z−1+1) [Replace z by z−1] f_3 (z)=z(z+1)(z−3)](https://www.tinkutara.com/question/Q1852.png)
$${f}_{{n}+\mathrm{1}} \left({z}+\mathrm{1}\right)=\left[{z}−{f}_{{n}} \left(\mathrm{0}\right)\right]{f}_{{n}} \left({z}\right)……………..\left({I}\right) \\ $$$${f}_{\mathrm{1}} \left({z}\right)={z}+\mathrm{1}\Rightarrow\underset{−} {{f}_{\mathrm{1}} \left({z}+\mathrm{1}\right)={z}+\mathrm{1}}+\mathrm{1}={z}+\mathrm{2}\:\: \\ $$$$\underset{−} {{f}_{\mathrm{1}} \left(\mathrm{0}\right)\:=\mathrm{0}+\mathrm{1}=\mathrm{1}}\:\:\:\:\:\:\:\: \\ $$$$\bullet{Substituting}\:{n}=\mathrm{1}\:{in}\:\left({I}\right) \\ $$$${f}_{\mathrm{2}} \left({z}+\mathrm{1}\right)=\left[{z}−{f}_{\mathrm{1}} \left(\mathrm{0}\right)\right]{f}_{\mathrm{1}} \left({z}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[{z}−\mathrm{1}\right]\left({z}+\mathrm{2}\right) \\ $$$${f}_{\mathrm{2}} \left({z}−\mathrm{1}+\mathrm{1}\right)=\left({z}−\mathrm{1}−\mathrm{1}\right)\left({z}−\mathrm{1}+\mathrm{2}\right)\:\:\left[{Replace}\:{z}\:{by}\:{z}−\mathrm{1}\right] \\ $$$$\underset{−} {{f}_{\mathrm{2}} \left({z}\right)=\left({z}−\mathrm{2}\right)\left({z}+\mathrm{1}\right)} \\ $$$$\underset{−} {{f}_{\mathrm{2}} \left(\mathrm{0}\right)=−\mathrm{2}} \\ $$$$\bullet{Substituting}\:{n}=\mathrm{2}\:{in}\:\left({I}\right) \\ $$$${f}_{\mathrm{3}} \left({z}+\mathrm{1}\right)=\left[{z}−{f}_{\mathrm{2}} \left(\mathrm{0}\right)\right]{f}_{\mathrm{2}} \left({z}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[{z}−\left(−\mathrm{2}\right)\right]\left[\left({z}−\mathrm{2}\right)\left({z}+\mathrm{1}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({z}+\mathrm{2}\right)\left({z}−\mathrm{2}\right)\left({z}+\mathrm{1}\right) \\ $$$${f}_{\mathrm{3}} \left({z}−\mathrm{1}+\mathrm{1}\right)=\left({z}−\mathrm{1}+\mathrm{2}\right)\left({z}−\mathrm{1}−\mathrm{2}\right)\left({z}−\mathrm{1}+\mathrm{1}\right)\:\:\:\:\:\left[{Replace}\:{z}\:{by}\:{z}−\mathrm{1}\right] \\ $$$${f}_{\mathrm{3}} \left({z}\right)={z}\left({z}+\mathrm{1}\right)\left({z}−\mathrm{3}\right) \\ $$