f-2x-2-f-x-2-1-0-f-x-

Question Number 163942 by Rasheed.Sindhi last updated on 12/Jan/22

$${f}\left(\mathrm{2}{x}\right)−\mathrm{2}\left[\:{f}\left({x}\right)\:\right]^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$${f}\left({x}\right)=? \\ $$

Answered by mr W last updated on 12/Jan/22

x ∈ R so it′s also valid for x∈N, say x=n, f(2n)=2(f(n))^2 −1 a_(2n) =2a_n ^2 −1 ???? similar to cos (2θ)=2 (cos θ)^2 −1 ???? let a_n =cos θ_n cos θ_(2n) =2 cos^2 θ_n −1=cos (2θ_n ) ⇒θ_(2n) =2θ_n θ_2 =2θ_1 θ_4 =2^2 θ_1 ... θ_2^n =2^n θ_1 generally θ_n =nθ_1 with θ_1 =constant=c f(n)=a_n =cos θ_n =cos (nθ_1 )=cos (nc) or ⇒f(x)=cos (cx) example: f(1)=0 ⇒c=(π/2) ⇒f(x)=cos (((xπ)/2))

$${x}\:\in\:{R} \\ $$$${so}\:{it}'{s}\:{also}\:{valid}\:{for}\:{x}\in{N},\:{say}\:{x}={n}, \\ $$$${f}\left(\mathrm{2}{n}\right)=\mathrm{2}\left({f}\left({n}\right)\right)^{\mathrm{2}} −\mathrm{1} \\ $$$${a}_{\mathrm{2}{n}} =\mathrm{2}{a}_{{n}} ^{\mathrm{2}} −\mathrm{1} \\ $$$$????\:{similar}\:{to}\:\mathrm{cos}\:\left(\mathrm{2}\theta\right)=\mathrm{2}\:\left(\mathrm{cos}\:\theta\right)^{\mathrm{2}} −\mathrm{1}\:\:???? \\ $$$${let}\:{a}_{{n}} =\mathrm{cos}\:\theta_{{n}} \\ $$$$\mathrm{cos}\:\theta_{\mathrm{2}{n}} =\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\theta_{{n}} −\mathrm{1}=\mathrm{cos}\:\left(\mathrm{2}\theta_{{n}} \right) \\ $$$$\Rightarrow\theta_{\mathrm{2}{n}} =\mathrm{2}\theta_{{n}} \\ $$$$\theta_{\mathrm{2}} =\mathrm{2}\theta_{\mathrm{1}} \\ $$$$\theta_{\mathrm{4}} =\mathrm{2}^{\mathrm{2}} \theta_{\mathrm{1}} \\ $$$$… \\ $$$$\theta_{\mathrm{2}^{{n}} } =\mathrm{2}^{{n}} \theta_{\mathrm{1}} \\ $$$${generally} \\ $$$$\theta_{{n}} ={n}\theta_{\mathrm{1}} \: \\ $$$${with}\:\theta_{\mathrm{1}} ={constant}={c} \\ $$$${f}\left({n}\right)={a}_{{n}} =\mathrm{cos}\:\theta_{{n}} =\mathrm{cos}\:\left({n}\theta_{\mathrm{1}} \right)=\mathrm{cos}\:\left({nc}\right) \\ $$$${or} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{cos}\:\left({cx}\right) \\ $$$$ \\ $$$${example}:\:{f}\left(\mathrm{1}\right)=\mathrm{0}\:\Rightarrow{c}=\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{cos}\:\left(\frac{{x}\pi}{\mathrm{2}}\right) \\ $$

Commented by Rasheed.Sindhi last updated on 13/Jan/22

$$\mathcal{V}\:\mathcal{F}{ine}\:\mathcal{S}{ir}! \\ $$

Commented by Rasheed.Sindhi last updated on 13/Jan/22

Sir this is the reposted question(Q#1307)

$$\mathcal{S}{ir}\:{this}\:{is}\:{the}\:{reposted}\:{question}\left({Q}#\mathrm{1307}\right) \\ $$

Commented by mr W last updated on 13/Jan/22

thanks sir! since cos 2x=2 cos^2 x−1 and cosh 2x=2 cosh^2 x−1, we have generally two possiblities: f(x)=cos (cx) or f(x)=cosh (cx). but depending on initial condition, there is only one solution suitable.

$${thanks}\:{sir}! \\ $$$${since}\:\mathrm{cos}\:\mathrm{2}{x}=\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:{x}−\mathrm{1}\:{and} \\ $$$$\mathrm{cosh}\:\mathrm{2}{x}=\mathrm{2}\:\mathrm{cosh}^{\mathrm{2}} \:{x}−\mathrm{1},\:{we}\:{have}\: \\ $$$${generally}\:{two}\:{possiblities}: \\ $$$${f}\left({x}\right)=\mathrm{cos}\:\left({cx}\right)\:{or}\:{f}\left({x}\right)=\mathrm{cosh}\:\left({cx}\right). \\ $$$${but}\:{depending}\:{on}\:{initial}\:{condition}, \\ $$$${there}\:{is}\:{only}\:{one}\:{solution}\:{suitable}. \\ $$

Commented by Rasheed.Sindhi last updated on 13/Jan/22

$$\mathcal{T}{hanks}\:{a}\:{lot}\:\boldsymbol{{sir}}! \\ $$

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