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Question Number 120049 by bramlexs22 last updated on 29/Oct/20

f(x+2)+f(x−2)=f(x) f(1)=1 ,f(2)=2,f(3)=3,f(4)=4 then f(100)=?

$$\:{f}\left({x}+\mathrm{2}\right)+{f}\left({x}−\mathrm{2}\right)={f}\left({x}\right) \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1}\:,{f}\left(\mathrm{2}\right)=\mathrm{2},{f}\left(\mathrm{3}\right)=\mathrm{3},{f}\left(\mathrm{4}\right)=\mathrm{4} \\ $$$${then}\:{f}\left(\mathrm{100}\right)=? \\ $$

Answered by bemath last updated on 29/Oct/20

⇒f(x−2)−f(x)+f(x+2)=0 replace x by x+2 ⇒f(x)−f(x+2)+f(x+4)=0 (i) replace again x by x+2⇒f(x+2)−f(x+4)+f(x+6)=0 (ii) adding(i) and (ii) ⇒f(x)+f(x+6) = 0 (iii) replace x by x+6 ⇒f(x+6)+f(x+12)=0 (iv) substract (iii) by (iv) ⇒ f(x+12)=f(x) it does meant f(x) is periodic function with period 12. ⇒f(100)=f(8.12+4)=f(4)=4

$$\Rightarrow{f}\left({x}−\mathrm{2}\right)−{f}\left({x}\right)+{f}\left({x}+\mathrm{2}\right)=\mathrm{0} \\ $$$${replace}\:{x}\:{by}\:{x}+\mathrm{2}\:\Rightarrow{f}\left({x}\right)−{f}\left({x}+\mathrm{2}\right)+{f}\left({x}+\mathrm{4}\right)=\mathrm{0}\:\left({i}\right) \\ $$$${replace}\:{again}\:{x}\:{by}\:{x}+\mathrm{2}\Rightarrow{f}\left({x}+\mathrm{2}\right)−{f}\left({x}+\mathrm{4}\right)+{f}\left({x}+\mathrm{6}\right)=\mathrm{0}\:\left({ii}\right) \\ $$$${adding}\left({i}\right)\:{and}\:\left({ii}\right) \\ $$$$\Rightarrow{f}\left({x}\right)+{f}\left({x}+\mathrm{6}\right)\:=\:\mathrm{0}\:\left({iii}\right) \\ $$$${replace}\:{x}\:{by}\:{x}+\mathrm{6}\:\Rightarrow{f}\left({x}+\mathrm{6}\right)+{f}\left({x}+\mathrm{12}\right)=\mathrm{0}\:\left({iv}\right) \\ $$$${substract}\:\left({iii}\right)\:{by}\:\left({iv}\right)\:\Rightarrow\:{f}\left({x}+\mathrm{12}\right)={f}\left({x}\right) \\ $$$${it}\:{does}\:{meant}\:{f}\left({x}\right)\:{is}\:{periodic}\:{function} \\ $$$${with}\:{period}\:\mathrm{12}. \\ $$$$\Rightarrow{f}\left(\mathrm{100}\right)={f}\left(\mathrm{8}.\mathrm{12}+\mathrm{4}\right)={f}\left(\mathrm{4}\right)=\mathrm{4} \\ $$$$ \\ $$$$ \\ $$

Answered by Ar Brandon last updated on 29/Oct/20

f(x+2)+f(x−2)=f(x)...eqn(1) Replacing x with x−2 we have f(x)+f(x−4)=f(x−2)...eqn(2) eqn(1)+eqn(2) ⇒f(x+2)=−f(x−4) ⇒f(x)=−f(x−6){replacing x=x−2} ⇒f(x+6)=−f(x)=f(x−6) ⇒f(x)=f(x−12) Period is 12⇒f(100)=f[4+12(8)] ⇒f(100)=f(4)=4

$${f}\left(\mathrm{x}+\mathrm{2}\right)+{f}\left(\mathrm{x}−\mathrm{2}\right)={f}\left(\mathrm{x}\right)...\mathrm{eqn}\left(\mathrm{1}\right) \\ $$$$\mathrm{Replacing}\:\mathrm{x}\:\mathrm{with}\:\mathrm{x}−\mathrm{2}\:\mathrm{we}\:\mathrm{have} \\ $$$${f}\left(\mathrm{x}\right)+{f}\left(\mathrm{x}−\mathrm{4}\right)={f}\left(\mathrm{x}−\mathrm{2}\right)...\mathrm{eqn}\left(\mathrm{2}\right) \\ $$$$\mathrm{eqn}\left(\mathrm{1}\right)+\mathrm{eqn}\left(\mathrm{2}\right) \\ $$$$\Rightarrow{f}\left(\mathrm{x}+\mathrm{2}\right)=−{f}\left(\mathrm{x}−\mathrm{4}\right) \\ $$$$\Rightarrow{f}\left(\mathrm{x}\right)=−{f}\left(\mathrm{x}−\mathrm{6}\right)\left\{\mathrm{replacing}\:\mathrm{x}=\mathrm{x}−\mathrm{2}\right\} \\ $$$$\Rightarrow{f}\left(\mathrm{x}+\mathrm{6}\right)=−{f}\left(\mathrm{x}\right)={f}\left(\mathrm{x}−\mathrm{6}\right) \\ $$$$\Rightarrow{f}\left(\mathrm{x}\right)={f}\left(\mathrm{x}−\mathrm{12}\right) \\ $$$$\mathrm{Period}\:\mathrm{is}\:\mathrm{12}\Rightarrow{f}\left(\mathrm{100}\right)={f}\left[\mathrm{4}+\mathrm{12}\left(\mathrm{8}\right)\right] \\ $$$$\Rightarrow{f}\left(\mathrm{100}\right)={f}\left(\mathrm{4}\right)=\mathrm{4} \\ $$

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