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

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

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

$$x\in R$$$$so{it}^{\prime }salsovalidforx\in N,sayx=n,$$$$f\left(2n\right)=2{\left(f\left(n\right)\right)}^2-1$$$$a_{2n}=2a_n^2-1$$$$????similarto\cos \left(2\theta \right)=2{\left(\cos \theta \right)}^2-1????$$$$let{a}_n=\cos {\theta }_n$$$$\cos {\theta }_{2n}=2{\cos }^2{\theta }_n-1=\cos \left(2{\theta }_n\right)$$$$\Rightarrow {\theta }_{2n}=2{\theta }_n$$$${\theta }_2=2{\theta }_1$$$${\theta }_4=2^2{\theta }_1$$$$...$$$${\theta }_{2^n}=2^n{\theta }_1$$$$generally$$$${\theta }_n=n{\theta }_1$$$$with{\theta }_1=constant=c$$$$f\left(n\right)=a_n=\cos {\theta }_n=\cos \left(n{\theta }_1\right)=\cos \left(nc\right)$$$$or$$$$\Rightarrow f\left(x\right)=\cos \left(cx\right)$$$$$$$$example:f\left(1\right)=0\Rightarrow c=\frac{\pi }{2}$$$$\Rightarrow f\left(x\right)=\cos \left(\frac{x\pi }{2}\right)$$

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

$$\text{You can't use 'macro parameter character \#' in math mode}$$

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

$$thankssir!$$$$since\cos 2x=2{\cos }^{2}x-1and$$$$\cosh 2x=2{\cosh }^{2}x-1,wehave$$$$generallytwopossiblities:$$$$f\left(x\right)=\cos \left(cx\right)orf\left(x\right)=\cosh \left(cx\right).$$$$butdependingoninitialcondition,$$$$thereisonlyonesolutionsuitable.$$

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

$$\mathcal{T}hanksalot\mathit{s}\mathit{i}\mathit{r}!$$

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