Question Number 1848 by 123456 last updated on 14/Oct/15
![f_(n+1) (z+1)=[z−f_(n−1) (0)]f_n (z) f_0 (z)=0 f_1 (z)=z+1 f_3 (z)=????](Q1848.png)
$${f}_{{n}+\mathrm{1}} \left({z}+\mathrm{1}\right)=\left[{z}−{f}_{{n}−\mathrm{1}} \left(\mathrm{0}\right)\right]{f}_{{n}} \left({z}\right) \\ $$$${f}_{\mathrm{0}} \left({z}\right)=\mathrm{0} \\ $$$${f}_{\mathrm{1}} \left({z}\right)={z}+\mathrm{1} \\ $$$${f}_{\mathrm{3}} \left({z}\right)=???? \\ $$
Answered by Rasheed Soomro last updated on 21/Oct/15
![Note: Question has been chaged now and this answer belongs to its first version.First three lines of the following make the statement of original Question. f_(n+1) (z+1)=[z−f_(n−1) (0)]f_n (z)............(I) f_1 (z)=z+1⇒f(z+1)=z+2 f_3 (z)=???? −−−−−−−−− Substituting n=0 in (I) f_1 (z+1)=[z−f_(−1) (0)]f_0 (z) ⇒z+2=[z−f_(−1) (0)]f_0 (z).................(A) Subsgituting n=2 in (I) f_3 (z+1)=[z−f_1 (0)]f_2 (z) ⇒f_3 (z+1)=[z−1]f_2 (z).....................(B) Substituting n=1 in (I) f_2 (z+1)=[z−f_0 (0)]f_1 (z) ⇒f_2 (z+1)=[z−f_0 (0)](z+1)..............(C) (A) , (B) and (C) contain f_(−1) (0) , f_0 (z) ; f_3 (z+1) , f_2 (z) ; f_2 (z+1) , f_0 (0) Eleminating all other than f_3 (z+1) leads to f_3 (z).For this more such equations containing same function definitions are required. I think the definition of f_3 (z), indepndant of any other function definition is impossible.](Q1849.png)
$${Note}:\:{Question}\:{has}\:{been}\:{chaged}\:{now}\:{and}\:{this}\:{answer} \\ $$$${belongs}\:{to}\:{its}\:{first}\:{version}.{First}\:{three}\:{lines}\:{of}\:{the} \\ $$$${following}\:{make}\:{the}\:{statement}\:{of}\:{original}\:{Question}. \\ $$$${f}_{{n}+\mathrm{1}} \left({z}+\mathrm{1}\right)=\left[{z}−{f}_{{n}−\mathrm{1}} \left(\mathrm{0}\right)\right]{f}_{{n}} \left({z}\right)............\left(\mathrm{I}\right) \\ $$$${f}_{\mathrm{1}} \left({z}\right)={z}+\mathrm{1}\Rightarrow{f}\left({z}+\mathrm{1}\right)={z}+\mathrm{2} \\ $$$${f}_{\mathrm{3}} \left({z}\right)=???? \\ $$$$−−−−−−−−− \\ $$$${Substituting}\:{n}=\mathrm{0}\:{in}\:\left(\mathrm{I}\right) \\ $$$${f}_{\mathrm{1}} \left({z}+\mathrm{1}\right)=\left[{z}−{f}_{−\mathrm{1}} \left(\mathrm{0}\right)\right]{f}_{\mathrm{0}} \left({z}\right) \\ $$$$\:\:\:\:\:\Rightarrow{z}+\mathrm{2}=\left[{z}−{f}_{−\mathrm{1}} \left(\mathrm{0}\right)\right]{f}_{\mathrm{0}} \left({z}\right).................\left({A}\right) \\ $$$${Subsgituting}\:{n}=\mathrm{2}\:{in}\:\left(\mathrm{I}\right)\: \\ $$$${f}_{\mathrm{3}} \left({z}+\mathrm{1}\right)=\left[{z}−{f}_{\mathrm{1}} \left(\mathrm{0}\right)\right]{f}_{\mathrm{2}} \left({z}\right) \\ $$$$\:\:\:\:\:\Rightarrow{f}_{\mathrm{3}} \left({z}+\mathrm{1}\right)=\left[{z}−\mathrm{1}\right]{f}_{\mathrm{2}} \left({z}\right).....................\left({B}\right) \\ $$$${Substituting}\:{n}=\mathrm{1}\:{in}\:\left(\mathrm{I}\right) \\ $$$${f}_{\mathrm{2}} \left({z}+\mathrm{1}\right)=\left[{z}−{f}_{\mathrm{0}} \left(\mathrm{0}\right)\right]{f}_{\mathrm{1}} \left({z}\right) \\ $$$$\:\:\:\:\:\:\:\Rightarrow{f}_{\mathrm{2}} \left({z}+\mathrm{1}\right)=\left[{z}−{f}_{\mathrm{0}} \left(\mathrm{0}\right)\right]\left({z}+\mathrm{1}\right)..............\left({C}\right) \\ $$$$\left({A}\right)\:,\:\left({B}\right)\:{and}\:\left({C}\right)\:{contain}\:{f}_{−\mathrm{1}} \left(\mathrm{0}\right)\:,\:{f}_{\mathrm{0}} \left({z}\right)\:; \\ $$$${f}_{\mathrm{3}} \left({z}+\mathrm{1}\right)\:,\:{f}_{\mathrm{2}} \left({z}\right)\:;\:{f}_{\mathrm{2}} \left({z}+\mathrm{1}\right)\:,\:{f}_{\mathrm{0}} \left(\mathrm{0}\right) \\ $$$${Eleminating}\:{all}\:{other}\:{than}\:{f}_{\mathrm{3}} \left({z}+\mathrm{1}\right)\:{leads}\:{to} \\ $$$${f}_{\mathrm{3}} \left({z}\right).{For}\:{this}\:{more}\:{such}\:{equations} \\ $$$${containing}\:{same}\:{function}\:{definitions} \\ $$$${are}\:{required}. \\ $$$${I}\:{think}\:{the}\:{definition}\:{of}\:{f}_{\mathrm{3}} \left({z}\right),\: \\ $$$${indepndant}\:{of}\:{any}\:{other}\:{function}\:{definition} \\ $$$${is}\:{impossible}. \\ $$$$ \\ $$
Answered by Rasheed Soomro last updated on 20/Oct/15
![f_(n+1) (z+1)=[z−f_(n−1) (0)]f_n (z).....................I f_0 (z)=0⇒f_0 (0)=0 f_1 (z)=z+1⇒f_1 (0)=0+1=1 f_3 (z)=???? −−−−−−−−−−−−−−−−−−− •Let n=1,substituting in I f_2 (z+1)=[z−f_0 (0)]f_1 (z) =(z−0)(z+1)=z(z+1) ∗Let z=z−1 f_2 (z−1+1)=(z−1)(z−1+1) f_2 (z)=z(z−1) •Let n=2 , substituting in I f_3 (z+1)=[z−f_1 (0)]f_2 (z) =[z−1][z(z−1)] =z(z−1)^2 ∗Let z=z−1 f_3 (z−1+1)=(z−1)(z−1−1)^2 f_3 (z)=(z−1)(z−2)^2](Q1877.png)
$${f}_{{n}+\mathrm{1}} \left({z}+\mathrm{1}\right)=\left[{z}−{f}_{{n}−\mathrm{1}} \left(\mathrm{0}\right)\right]{f}_{{n}} \left({z}\right).....................{I} \\ $$$${f}_{\mathrm{0}} \left({z}\right)=\mathrm{0}\Rightarrow{f}_{\mathrm{0}} \left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}_{\mathrm{1}} \left({z}\right)={z}+\mathrm{1}\Rightarrow{f}_{\mathrm{1}} \left(\mathrm{0}\right)=\mathrm{0}+\mathrm{1}=\mathrm{1} \\ $$$${f}_{\mathrm{3}} \left({z}\right)=???? \\ $$$$−−−−−−−−−−−−−−−−−−− \\ $$$$\bullet{Let}\:{n}=\mathrm{1},{substituting}\:{in}\:\:{I} \\ $$$${f}_{\mathrm{2}} \left({z}+\mathrm{1}\right)=\left[{z}−{f}_{\mathrm{0}} \left(\mathrm{0}\right)\right]{f}_{\mathrm{1}} \left({z}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({z}−\mathrm{0}\right)\left({z}+\mathrm{1}\right)={z}\left({z}+\mathrm{1}\right) \\ $$$$\ast{Let}\:{z}={z}−\mathrm{1} \\ $$$${f}_{\mathrm{2}} \left({z}−\mathrm{1}+\mathrm{1}\right)=\left({z}−\mathrm{1}\right)\left({z}−\mathrm{1}+\mathrm{1}\right) \\ $$$${f}_{\mathrm{2}} \left({z}\right)={z}\left({z}−\mathrm{1}\right) \\ $$$$\bullet{Let}\:{n}=\mathrm{2}\:,\:{substituting}\:{in}\:\:{I} \\ $$$${f}_{\mathrm{3}} \left({z}+\mathrm{1}\right)=\left[{z}−{f}_{\mathrm{1}} \left(\mathrm{0}\right)\right]{f}_{\mathrm{2}} \left({z}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[{z}−\mathrm{1}\right]\left[{z}\left({z}−\mathrm{1}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={z}\left({z}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\ast{Let}\:{z}={z}−\mathrm{1} \\ $$$${f}_{\mathrm{3}} \left({z}−\mathrm{1}+\mathrm{1}\right)=\left({z}−\mathrm{1}\right)\left({z}−\mathrm{1}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$${f}_{\mathrm{3}} \left({z}\right)=\left({z}−\mathrm{1}\right)\left({z}−\mathrm{2}\right)^{\mathrm{2}} \\ $$