f-R-R-f-f-x-x-2-x-1-f-x-Altered-Question-217541-

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

$$\mathrm{f}\::\:\mathbb{R}\:\rightarrow\:\mathbb{R} \\ $$$$\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\:=\:\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{x}\:+\:\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\:? \\ $$$$\mathrm{Altered}\:\mathrm{Question}#\:\mathrm{217541} \\ $$

Commented by Ghisom last updated on 16/Mar/25

I thought of this, there might be no closed term for f(x) because: g(g(x))=x^2 ⇒ g(x)=x^(√2) [not possible for x<0] but if we try to add terms: g(x)=x^(√2) +c ⇒ g(g(x))=(x^(√2) +c)^(√2) +c g(x)=x^(√2) +h(x) ⇒ weirdness

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

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

with given condition we can find f(0), f(1), but we can even not find f(2) etc., not to mention f(x) generally. i think.

$${with}\:{given}\:{condition}\:{we}\:{can}\:{find}\: \\ $$$${f}\left(\mathrm{0}\right),\:{f}\left(\mathrm{1}\right),\:\:{but}\:{we}\:{can}\:{even}\:{not}\:{find}\: \\ $$$${f}\left(\mathrm{2}\right)\:{etc}.,\:{not}\:{to}\:{mention}\:{f}\left({x}\right) \\ $$$${generally}.\:{i}\:{think}. \\ $$

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

$$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{sirs}}! \\ $$

Commented by Ghisom last updated on 17/Mar/25

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

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

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