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Question Number 154675 by mathdanisur last updated on 20/Sep/21

$$\mathrm{if}\mathrm{we}\mathrm{let}$$$$\mathrm{f}\left(\mathrm{n}\right)=\frac{\mathrm{n}}{2}(\mathrm{n}+1)$$$$\mathrm{then}\mathrm{solve}\mathrm{the}\mathrm{equation}\mathrm{for}\mathrm{real}\mathrm{numbers}$$$$\mathrm{f}\left(\mathrm{f}\left(\mathrm{f}\left(\mathrm{f}\left(\mathrm{n}\right)\right)\right)\right)=\frac{1}{32}$$

Answered by MJS_new last updated on 20/Sep/21

$$n^2+n-2k=0\wedge n>0\Rightarrow n=\frac{\sqrt{8k+1}-1}{2}$$$$g\left(k\right)=\frac{\sqrt{8k+1}-1}{2}$$$$g\left(g\left(k\right)\right)=\frac{\sqrt{4\sqrt{8k+1}-3}-1}{2}$$$$g\left(g\left(g\left(k\right)\right)\right)=\frac{\sqrt{4\sqrt{4\sqrt{8k+1}-3}-3}-1}{2}$$$$g\left(g\left(g\left(g\left(k\right)\right)\right)\right)=\frac{\sqrt{4\sqrt{4\sqrt{4\sqrt{8k+1}-3}-3}-3}-1}{2}$$$$g\left(g\left(g\left(g\left(\frac{1}{32}\right)\right)\right)\right)=\frac{\sqrt{4\sqrt{4\sqrt{2\sqrt{5}-3}-3}-3}-1}{2}\approx .281886524469$$

Commented by mathdanisur last updated on 20/Sep/21

$$\mathrm{awesome}\mathrm{solution}\mathrm{thanks}\mathrm{Ser}$$

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