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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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