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If-f-R-R-is-an-odd-function-such-that-a-f-1-x-1-f-x-b-x-2-f-1-x-f-x-x-0-Then-find-f-x-

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Question Number 33375 by rahul 19 last updated on 15/Apr/18
If f:R → R is an odd function such that : a) f(1+x) = 1+f(x) . b) x^2 f((1/x)) = f(x) , x≠0. Then find f(x) ?
$${If}\:{f}:{R}\:\rightarrow\:{R}\:{is}\:{an}\:\boldsymbol{{odd}}\:{function}\:{such} \\ $$$${that}\:: \\ $$$$\left.{a}\right)\:{f}\left(\mathrm{1}+{x}\right)\:=\:\mathrm{1}+{f}\left({x}\right)\:. \\ $$$$\left.{b}\right)\:{x}^{\mathrm{2}} \:{f}\left(\frac{\mathrm{1}}{{x}}\right)\:=\:{f}\left({x}\right)\:,\:{x}\neq\mathrm{0}. \\ $$$${Then}\:{find}\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\:? \\ $$
Commented by prof Abdo imad last updated on 15/Apr/18
f is odd⇒f(0)=0 ⇒f(1+0)=1+f(0)⇒f(1)=1 f(2)=f(1+1)=1+f(1) =2 let suppose f(n)=n ⇒f(n+1)=1+f(n)=1+n⇒ ∀n∈N f(n)=n if n∈Z^− n=−m f(n)=f(−m)=−f(m)=−m =n ⇒ ∀n∈Z f(n)=n f(1)=f((n/n))=nf((1/n))=1⇒f((1/n))=(1/n) f((p/n))=f((1/n) +..+(1/n))(p×) =pf((1/n))=(p/n) ⇒ ∀x∈Q f(x)=x but Q is dense inside R ⇒ ∀x∈R f(x)=x we have for x≠o x^2 f((1/x))=x^2 .(1/x) =x =f(x) cond.b is verified.
$${f}\:{is}\:{odd}\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow{f}\left(\mathrm{1}+\mathrm{0}\right)=\mathrm{1}+{f}\left(\mathrm{0}\right)\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${f}\left(\mathrm{2}\right)={f}\left(\mathrm{1}+\mathrm{1}\right)=\mathrm{1}+{f}\left(\mathrm{1}\right)\:=\mathrm{2} \\ $$$${let}\:{suppose}\:{f}\left({n}\right)={n}\:\Rightarrow{f}\left({n}+\mathrm{1}\right)=\mathrm{1}+{f}\left({n}\right)=\mathrm{1}+{n}\Rightarrow \\ $$$$\forall{n}\in{N}\:{f}\left({n}\right)={n}\:{if}\:{n}\in{Z}^{−} \:{n}=−{m} \\ $$$${f}\left({n}\right)={f}\left(−{m}\right)=−{f}\left({m}\right)=−{m}\:={n}\:\Rightarrow \\ $$$$\forall{n}\in{Z}\:{f}\left({n}\right)={n}\: \\ $$$${f}\left(\mathrm{1}\right)={f}\left(\frac{{n}}{{n}}\right)={nf}\left(\frac{\mathrm{1}}{{n}}\right)=\mathrm{1}\Rightarrow{f}\left(\frac{\mathrm{1}}{{n}}\right)=\frac{\mathrm{1}}{{n}} \\ $$$${f}\left(\frac{{p}}{{n}}\right)={f}\left(\frac{\mathrm{1}}{{n}}\:+..+\frac{\mathrm{1}}{{n}}\right)\left({p}×\right)\:={pf}\left(\frac{\mathrm{1}}{{n}}\right)=\frac{{p}}{{n}} \\ $$$$\Rightarrow\:\forall{x}\in{Q}\:{f}\left({x}\right)={x}\:\:\:{but}\:{Q}\:{is}\:{dense}\:{inside}\:{R}\:\Rightarrow \\ $$$$\forall{x}\in{R}\:{f}\left({x}\right)={x}\:\:{we}\:{have}\:{for}\:{x}\neq{o} \\ $$$${x}^{\mathrm{2}} {f}\left(\frac{\mathrm{1}}{{x}}\right)={x}^{\mathrm{2}} .\frac{\mathrm{1}}{{x}}\:={x}\:={f}\left({x}\right)\:{cond}.{b}\:{is}\:{verified}. \\ $$$$ \\ $$
Commented by rahul 19 last updated on 15/Apr/18
thank you sir !
$${thank}\:{you}\:{sir}\:! \\ $$

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