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Question-164027

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Question Number 164027 by HongKing last updated on 13/Jan/22
Commented by HongKing last updated on 13/Jan/22
Yes my dear Sir
$$\mathrm{Yes}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir} \\ $$
Commented by mr W last updated on 13/Jan/22
now i got it. A is the number of roots of f(f(f(....)))=x. A=2^(2020−1) .
$${now}\:{i}\:{got}\:{it}. \\ $$$${A}\:{is}\:{the}\:{number}\:{of}\:{roots}\:{of} \\ $$$${f}\left({f}\left({f}\left(….\right)\right)\right)={x}. \\ $$$${A}=\mathrm{2}^{\mathrm{2020}−\mathrm{1}} . \\ $$
Answered by mr W last updated on 13/Jan/22
due to symmetry we just consider x≥0. f(x)=2x(√(1−x^2 ))≥0 let x=sin θ with 0≤θ≤(π/2) f(x)=2 sin θ (√(1−sin^2 θ))=2 sin θ cos θ=sin 2θ f(f(x))=sin (2×2θ)=sin (2^2 θ) f(f(f(...f(x)))_(n times) =sin (2^n θ) f(f(f(...f(x)))_(n times) =x≥0 ⇒sin (2^n θ)=sin θ ⇒2^n θ=2kπ+θ ⇒θ=((2kπ)/(2^n −1)) or ⇒2^n θ=(2m+1)π−θ ⇒θ=(((2m+1)π)/(2^n +1)) θ=((2kπ)/(2^n −1))≤(π/2) ⇒0≤k≤((2^n −1)/4) θ=(((2m+1)π)/(2^n +1))≤(π/2) ⇒0≤m≤((2^n +1)/4)−(1/2) x=sin ((2kπ)/(2^n −1)) or x=sin (((2k+1)π)/(2^n +1)) with 0≤k≤2^(n−2) −1 and n=2020 totally we have A=2^(n−1) =2^(2019) roots. A (mod 1000)=288
$${due}\:{to}\:{symmetry}\:{we}\:{just}\:{consider}\:{x}\geqslant\mathrm{0}. \\ $$$${f}\left({x}\right)=\mathrm{2}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\geqslant\mathrm{0} \\ $$$${let}\:{x}=\mathrm{sin}\:\theta\:{with}\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}} \\ $$$${f}\left({x}\right)=\mathrm{2}\:\mathrm{sin}\:\theta\:\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\theta}=\mathrm{2}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta=\mathrm{sin}\:\mathrm{2}\theta \\ $$$${f}\left({f}\left({x}\right)\right)=\mathrm{sin}\:\left(\mathrm{2}×\mathrm{2}\theta\right)=\mathrm{sin}\:\left(\mathrm{2}^{\mathrm{2}} \theta\right) \\ $$$$\underset{{n}\:{times}} {{f}\left({f}\left({f}\left(…{f}\left({x}\right)\right)\right)}=\mathrm{sin}\:\left(\mathrm{2}^{{n}} \theta\right)\right. \\ $$$$\underset{{n}\:{times}} {{f}\left({f}\left({f}\left(…{f}\left({x}\right)\right)\right)}={x}\geqslant\mathrm{0}\right. \\ $$$$\Rightarrow\mathrm{sin}\:\left(\mathrm{2}^{{n}} \theta\right)=\mathrm{sin}\:\theta \\ $$$$\Rightarrow\mathrm{2}^{{n}} \theta=\mathrm{2}{k}\pi+\theta\:\Rightarrow\theta=\frac{\mathrm{2}{k}\pi}{\mathrm{2}^{{n}} −\mathrm{1}}\:{or} \\ $$$$\Rightarrow\mathrm{2}^{{n}} \theta=\left(\mathrm{2}{m}+\mathrm{1}\right)\pi−\theta\:\Rightarrow\theta=\frac{\left(\mathrm{2}{m}+\mathrm{1}\right)\pi}{\mathrm{2}^{{n}} +\mathrm{1}} \\ $$$$\theta=\frac{\mathrm{2}{k}\pi}{\mathrm{2}^{{n}} −\mathrm{1}}\leqslant\frac{\pi}{\mathrm{2}}\:\Rightarrow\mathrm{0}\leqslant{k}\leqslant\frac{\mathrm{2}^{{n}} −\mathrm{1}}{\mathrm{4}} \\ $$$$\theta=\frac{\left(\mathrm{2}{m}+\mathrm{1}\right)\pi}{\mathrm{2}^{{n}} +\mathrm{1}}\leqslant\frac{\pi}{\mathrm{2}}\:\Rightarrow\mathrm{0}\leqslant{m}\leqslant\frac{\mathrm{2}^{{n}} +\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}=\mathrm{sin}\:\frac{\mathrm{2}{k}\pi}{\mathrm{2}^{{n}} −\mathrm{1}}\:{or} \\ $$$${x}=\mathrm{sin}\:\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}^{{n}} +\mathrm{1}}\: \\ $$$${with}\:\mathrm{0}\leqslant{k}\leqslant\mathrm{2}^{{n}−\mathrm{2}} −\mathrm{1}\:{and}\:\:{n}=\mathrm{2020} \\ $$$$ \\ $$$${totally}\:{we}\:{have}\:{A}=\mathrm{2}^{{n}−\mathrm{1}} =\mathrm{2}^{\mathrm{2019}} \:{roots}. \\ $$$$ \\ $$$${A}\:\left({mod}\:\mathrm{1000}\right)=\mathrm{288} \\ $$
Commented by HongKing last updated on 13/Jan/22
perfect my dear Sir thank you so much
$$\mathrm{perfect}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$
Commented by Rasheed.Sindhi last updated on 13/Jan/22
G_( S I_(!) R ) ^( R^( E ) A) T
$$\mathbb{G}_{\:\:\mathbb{S}\:\:\underset{!} {\mathbb{I}}\:\:\mathbb{R}\:} ^{\:\:\mathbb{R}^{\:\mathbb{E}\:} \mathbb{A}} \mathbb{T} \\ $$

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