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Question Number 112681 by bemath last updated on 09/Sep/20

solve the equation of function   (f(3x))^2  = (f(2x))^2 +(f(x))^2

$$\mathrm{solve}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{function}\: \\ $$$$\left(\mathrm{f}\left(\mathrm{3x}\right)\right)^{\mathrm{2}} \:=\:\left(\mathrm{f}\left(\mathrm{2x}\right)\right)^{\mathrm{2}} +\left(\mathrm{f}\left(\mathrm{x}\right)\right)^{\mathrm{2}} \\ $$

Answered by bobhans last updated on 09/Sep/20

 (⧫) (f(3x))^2  = (f(2x))^2 +(f(x))^2   by letting h(x)=(f(x))^2   now we get h(3x)=h(2x) + h(x)  plugging in x = 0  ⇒ h(0) = 2h(0), ⇒h(0)=0  since the equation is linear , let h(x)=Cx  therefore f(x)=± (√(Cx)) or f(x) = A(√x)  where A can be any constant

$$\:\left(\blacklozenge\right)\:\left(\mathrm{f}\left(\mathrm{3x}\right)\right)^{\mathrm{2}} \:=\:\left(\mathrm{f}\left(\mathrm{2x}\right)\right)^{\mathrm{2}} +\left(\mathrm{f}\left(\mathrm{x}\right)\right)^{\mathrm{2}} \\ $$$$\mathrm{by}\:\mathrm{letting}\:\mathrm{h}\left(\mathrm{x}\right)=\left(\mathrm{f}\left(\mathrm{x}\right)\right)^{\mathrm{2}} \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{get}\:\mathrm{h}\left(\mathrm{3x}\right)=\mathrm{h}\left(\mathrm{2x}\right)\:+\:\mathrm{h}\left(\mathrm{x}\right) \\ $$$$\mathrm{plugging}\:\mathrm{in}\:\mathrm{x}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{h}\left(\mathrm{0}\right)\:=\:\mathrm{2h}\left(\mathrm{0}\right),\:\Rightarrow\mathrm{h}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\mathrm{since}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{linear}\:,\:\mathrm{let}\:\mathrm{h}\left(\mathrm{x}\right)=\mathrm{Cx} \\ $$$$\mathrm{therefore}\:\mathrm{f}\left(\mathrm{x}\right)=\pm\:\sqrt{\mathrm{Cx}}\:\mathrm{or}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{A}\sqrt{\mathrm{x}} \\ $$$$\mathrm{where}\:\mathrm{A}\:\mathrm{can}\:\mathrm{be}\:\mathrm{any}\:\mathrm{constant} \\ $$

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