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Question Number 154675 by mathdanisur last updated on 20/Sep/21
if we let  f(n) = (n/2) (n + 1)  then solve the equation for real numbers  f(f(f(f(n)))) = (1/(32))
$$\mathrm{if}\:\mathrm{we}\:\mathrm{let} \\ $$$$\mathrm{f}\left(\mathrm{n}\right)\:=\:\frac{\mathrm{n}}{\mathrm{2}}\:\left(\mathrm{n}\:+\:\mathrm{1}\right) \\ $$$$\mathrm{then}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers} \\ $$$$\mathrm{f}\left(\mathrm{f}\left(\mathrm{f}\left(\mathrm{f}\left(\mathrm{n}\right)\right)\right)\right)\:=\:\frac{\mathrm{1}}{\mathrm{32}} \\ $$
Answered by MJS_new last updated on 20/Sep/21
n^2 +n−2k=0 ∧ n>0 ⇒ n=(((√(8k+1))−1)/2)  g(k)=(((√(8k+1))−1)/2)  g(g(k))=(((√(4(√(8k+1))−3))−1)/2)  g(g(g(k)))=(((√(4(√(4(√(8k+1))−3))−3))−1)/2)  g(g(g(g(k))))=(((√(4(√(4(√(4(√(8k+1))−3))−3))−3))−1)/2)  g(g(g(g((1/(32))))))=(((√(4(√(4(√(2(√5)−3))−3))−3))−1)/2)≈.281886524469
$${n}^{\mathrm{2}} +{n}−\mathrm{2}{k}=\mathrm{0}\:\wedge\:{n}>\mathrm{0}\:\Rightarrow\:{n}=\frac{\sqrt{\mathrm{8}{k}+\mathrm{1}}−\mathrm{1}}{\mathrm{2}} \\ $$$${g}\left({k}\right)=\frac{\sqrt{\mathrm{8}{k}+\mathrm{1}}−\mathrm{1}}{\mathrm{2}} \\ $$$${g}\left({g}\left({k}\right)\right)=\frac{\sqrt{\mathrm{4}\sqrt{\mathrm{8}{k}+\mathrm{1}}−\mathrm{3}}−\mathrm{1}}{\mathrm{2}} \\ $$$${g}\left({g}\left({g}\left({k}\right)\right)\right)=\frac{\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{8}{k}+\mathrm{1}}−\mathrm{3}}−\mathrm{3}}−\mathrm{1}}{\mathrm{2}} \\ $$$${g}\left({g}\left({g}\left({g}\left({k}\right)\right)\right)\right)=\frac{\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{8}{k}+\mathrm{1}}−\mathrm{3}}−\mathrm{3}}−\mathrm{3}}−\mathrm{1}}{\mathrm{2}} \\ $$$${g}\left({g}\left({g}\left({g}\left(\frac{\mathrm{1}}{\mathrm{32}}\right)\right)\right)\right)=\frac{\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{3}}−\mathrm{3}}−\mathrm{3}}−\mathrm{1}}{\mathrm{2}}\approx.\mathrm{281886524469} \\ $$
Commented by mathdanisur last updated on 20/Sep/21
awesome solution thanks Ser
$$\mathrm{awesome}\:\mathrm{solution}\:\mathrm{thanks}\:\mathrm{Ser} \\ $$

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