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Question Number 217579 by Rasheed.Sindhi last updated on 16/Mar/25

$$\mathrm{f}:\mathbb{R}\to \mathbb{R}$$$$\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)={\mathrm{x}}^2-\mathrm{x}+1$$$$\mathrm{f}\left(\mathrm{x}\right)=?$$$$\text{You can't use 'macro parameter character \#' in math mode}$$

Commented by Ghisom last updated on 16/Mar/25

$$\mathrm{I}\mathrm{thought}\mathrm{of}\mathrm{this},\mathrm{there}\mathrm{might}\mathrm{be}\mathrm{no}\mathrm{closed}$$$$\mathrm{term}\mathrm{for}f\left(x\right)$$$$\mathrm{because}:$$$$g\left(g\left(x\right)\right)=x^2\Rightarrow g\left(x\right)=x^{\sqrt{2}}$$$$[\mathrm{not}\mathrm{possible}\mathrm{for}x<0]$$$$\mathrm{but}\mathrm{if}\mathrm{we}\mathrm{try}\mathrm{to}\mathrm{add}\mathrm{terms}:$$$$g\left(x\right)=x^{\sqrt{2}}+c\Rightarrow g\left(g\left(x\right)\right)={(x^{\sqrt{2}}+c)}^{\sqrt{2}}+c$$$$g\left(x\right)=x^{\sqrt{2}}+h\left(x\right)\Rightarrow \mathrm{weirdness}$$

Commented by mr W last updated on 16/Mar/25

$$withgivenconditionwecanfind$$$$f\left(0\right),f\left(1\right),butwecanevennotfind$$$$f\left(2\right)etc.,nottomentionf\left(x\right)$$$$generally.ithink.$$

Commented by Rasheed.Sindhi last updated on 16/Mar/25

$$\mathbb{T}\mathbf{han}\mathbb{k}\mathbf{s}\mathbf{sirs}!$$

Commented by Ghisom last updated on 17/Mar/25

$$\mathrm{just}\mathrm{to}\mathrm{explore}\mathrm{some}\mathrm{problems}:$$$$g\left(x\right)={ax}^2+bx+c$$$$G\left(x\right)={Ax}^4+{Bx}^3+{Cx}^2+Dx+E=g\left(g\left(x\right)\right)=$$$$A=a^3$$$$B=2a^{2}b$$$$C=a(2ac+b(b+1))$$$$D=(2ac+b)b$$$$E=(ac+b+1)c$$$$\mathrm{this}\mathrm{cannot}\mathrm{be}\mathrm{solved}\mathrm{for}\mathrm{every}\mathrm{given}\mathrm{quintuplet}$$$$A,B,C,D,E$$$$$$$$\mathrm{even}\mathrm{if}\mathrm{we}\mathrm{want}G\left(x\right)=x^4+{Cx}^2+E$$$$\Rightarrow a=1\wedge b=0$$$$g\left(x\right)=x^2+c$$$$G\left(x\right)=x^4+2{cx}^2+c(c+1)$$$$\mathrm{only}\mathrm{very}\mathrm{few}\mathrm{are}\mathrm{possible}...$$

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