Question Number 1988 by Rasheed Soomro last updated on 28/Oct/15
$${x}^{\mathrm{2}} =\:\frac{{f}\left({x}\right)+{f}\left(−{x}\right)}{\mathrm{2}} \\ $$$${f}\left({x}\right)=? \\ $$
Answered by 123456 last updated on 28/Oct/15
![supossing that f is poly f(x)=Σ_(i=0) ^n a_i x^i f(−x)=Σ_(i=0) ^n a_i (−1)^i x^i x^2 =((f(x)+f(−x))/2) f(x)+f(−x)=2x^2 Σ_(i=0) ^n a_i x^i +Σ_(i=0) ^n a_i (−1)^i x^i =2x^2 (n≥2) Σ_(i=0) ^n a_i [1+(−1)^i ]x^i =2x^2 i≠2⇒a_i [1+(−1)^i ]=0⇒a_i =0∨i≡1(mod 2) i=2⇒2a_2 =2⇒a_2 =1 so f(x)=Σ_(i=0) ^n a_i x^i ,a_i = { ((∀a_i ,i≡1(mod 2))),((a_i =0,i≡0(mod 2)∨i≠2)),((a_2 =1)) :} −−−−−−−−− f(x)=a_1 x+x^2 +a_3 x^3 +a_5 x^5 +∙∙∙ f(−x)=−a_1 x+x^2 −a_3 x^3 −a_5 x^5 −∙∙∙ f(x)+f(−x)=2x^2 ((f(x)+f(−x))/2)=x^2](https://www.tinkutara.com/question/Q1989.png)
$$\mathrm{supossing}\:\mathrm{that}\:{f}\:\mathrm{is}\:\mathrm{poly} \\ $$$${f}\left({x}\right)=\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{i}} {x}^{{i}} \\ $$$${f}\left(−{x}\right)=\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{i}} \left(−\mathrm{1}\right)^{{i}} {x}^{{i}} \\ $$$${x}^{\mathrm{2}} =\frac{{f}\left({x}\right)+{f}\left(−{x}\right)}{\mathrm{2}} \\ $$$${f}\left({x}\right)+{f}\left(−{x}\right)=\mathrm{2}{x}^{\mathrm{2}} \\ $$$$\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{i}} {x}^{{i}} +\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{i}} \left(−\mathrm{1}\right)^{{i}} {x}^{{i}} =\mathrm{2}{x}^{\mathrm{2}} \:\:\:\left({n}\geqslant\mathrm{2}\right) \\ $$$$\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{i}} \left[\mathrm{1}+\left(−\mathrm{1}\right)^{{i}} \right]{x}^{{i}} =\mathrm{2}{x}^{\mathrm{2}} \\ $$$${i}\neq\mathrm{2}\Rightarrow{a}_{{i}} \left[\mathrm{1}+\left(−\mathrm{1}\right)^{{i}} \right]=\mathrm{0}\Rightarrow{a}_{{i}} =\mathrm{0}\vee{i}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{2}\right) \\ $$$${i}=\mathrm{2}\Rightarrow\mathrm{2}{a}_{\mathrm{2}} =\mathrm{2}\Rightarrow{a}_{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{so} \\ $$$${f}\left({x}\right)=\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{i}} {x}^{{i}} ,{a}_{{i}} =\begin{cases}{\forall{a}_{{i}} ,{i}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{2}\right)}\\{{a}_{{i}} =\mathrm{0},{i}\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{2}\right)\vee{i}\neq\mathrm{2}}\\{{a}_{\mathrm{2}} =\mathrm{1}}\end{cases} \\ $$$$−−−−−−−−− \\ $$$${f}\left({x}\right)={a}_{\mathrm{1}} {x}+{x}^{\mathrm{2}} +{a}_{\mathrm{3}} {x}^{\mathrm{3}} +{a}_{\mathrm{5}} {x}^{\mathrm{5}} +\centerdot\centerdot\centerdot \\ $$$${f}\left(−{x}\right)=−{a}_{\mathrm{1}} {x}+{x}^{\mathrm{2}} −{a}_{\mathrm{3}} {x}^{\mathrm{3}} −{a}_{\mathrm{5}} {x}^{\mathrm{5}} −\centerdot\centerdot\centerdot \\ $$$${f}\left({x}\right)+{f}\left(−{x}\right)=\mathrm{2}{x}^{\mathrm{2}} \\ $$$$\frac{{f}\left({x}\right)+{f}\left(−{x}\right)}{\mathrm{2}}={x}^{\mathrm{2}} \\ $$
Answered by prakash jain last updated on 28/Oct/15
$$\mathrm{Let}\:{g}\left({x}\right)\:\mathrm{be}\:\mathrm{any}\:\mathrm{odd}\:\mathrm{function}. \\ $$$${g}\left({x}\right)=−{g}\left(−{x}\right) \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} +{g}\left({x}\right) \\ $$$${f}\left(−{x}\right)={x}^{\mathrm{2}} +{g}\left(−{x}\right)={x}^{\mathrm{2}} −{g}\left({x}\right) \\ $$$$\frac{{f}\left({x}\right)+{f}\left(−{x}\right)}{\mathrm{2}}={x}^{\mathrm{2}} \\ $$$${examples} \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{sin}\:{x} \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} +\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{i}} {x}^{\mathrm{2}{i}+\mathrm{1}} \\ $$