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