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Question Number 106023 by ZiYangLee last updated on 02/Aug/20

Given that f(0)≠0 for all x,y∈R; 2f(x)f(y)=f(x+y)+f(x−y), express f(2x) in term of f(x).

$$\mathrm{Given}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{0}\right)\neq\mathrm{0}\:\mathrm{for}\:\mathrm{all}\:\mathrm{x},\mathrm{y}\in\mathbb{R}; \\ $$$$\mathrm{2f}\left(\mathrm{x}\right)\mathrm{f}\left(\mathrm{y}\right)=\mathrm{f}\left(\mathrm{x}+\mathrm{y}\right)+\mathrm{f}\left(\mathrm{x}−\mathrm{y}\right),\: \\ $$$$\mathrm{express}\:\mathrm{f}\left(\mathrm{2x}\right)\:\mathrm{in}\:\mathrm{term}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right). \\ $$

Commented by mr W last updated on 02/Aug/20

y=0: 2f(x)f(0)=f(x)+f(x) ⇒f(0)=1 y=x: 2f(x)f(x)=f(2x)+f(0) ⇒f(2x)=2(f(x))^2 −1

$${y}=\mathrm{0}: \\ $$$$\mathrm{2}{f}\left({x}\right){f}\left(\mathrm{0}\right)={f}\left({x}\right)+{f}\left({x}\right) \\ $$$$\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${y}={x}: \\ $$$$\mathrm{2}{f}\left({x}\right){f}\left({x}\right)={f}\left(\mathrm{2}{x}\right)+{f}\left(\mathrm{0}\right) \\ $$$$\Rightarrow{f}\left(\mathrm{2}{x}\right)=\mathrm{2}\left({f}\left({x}\right)\right)^{\mathrm{2}} −\mathrm{1} \\ $$

Commented by ZiYangLee last updated on 02/Aug/20

Wow Nice Thanks for your solutions

$$ \\ $$$$\mathrm{Wow}\:\mathrm{Nice}\:\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{solutions} \\ $$

Answered by Aziztisffola last updated on 02/Aug/20

2f(x)f(x)=f(2x)+f(0) f(2x)=2f^2 (x)−f(0)

$$\mathrm{2f}\left(\mathrm{x}\right)\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{2x}\right)+\mathrm{f}\left(\mathrm{0}\right) \\ $$$$\mathrm{f}\left(\mathrm{2x}\right)=\mathrm{2f}^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{f}\left(\mathrm{0}\right) \\ $$

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