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Question-197034

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Question Number 197034 by Mingma last updated on 06/Sep/23
Answered by TheHoneyCat last updated on 08/Sep/23
let f: { (R,→,R_+ ),(α, ,((4α^2 )/(1+4α^2 ))) :} if (x,y,z) is a “good” triplet then f(x)=y f(y)=z f(z)=x we notice that therefore, using f each triplet can be unicly defined by it′s first elelment x, the others beeing f(x) and f^2 (x):=f(f(x)) Thus there are as many triples as there are values x such that f^3 (x)=x f(x)=((4x^2 )/(1+4x^2 )) f^2 (x)=((4(((4x^2 )/(1+4x^2 )))^2 )/(1+4(((4x^2 )/(1+4x^2 )))^2 )) =((4(4x^2 )^2 )/((1+4x^2 )^2 +4(4x^2 )^2 )) =((2^6 x^4 )/(1+2^3 x^2 +2^6 x^4 )) f^3 (x)=((2^6 (((4x^2 )/(1+4x^2 )))^4 )/(1+2^3 (((4x^2 )/(1+4x^2 )))^2 +2^6 (((4x^2 )/(1+4x^2 )))^4 )) =((2^(14) x^8 )/((1+4x^2 )^4 +2^7 (1+4x^2 )^2 x^4 +2^(14) x^8 )) =((2^(14) x^8 )/(1+2^4 x^2 +224x^4 +1280x^6 +18688x^8 )) f^3 (x)=x has 3 solutions... x=(1/2) (I let you check it) x=0 (I let you check it) and one last point that I was not able to compute algebraicaly. (actualy I was, but we shall see that later) you can verify that there are only 3 because: f^3 (x)≥0 ⇒x≥0 f^3 (x) is strictly monotonous over R_+ between 0 and 1/2 ∃y∣f^3 (y)<y between 1/2 and 0.56 ∃y∣f^3 (y)>y beetween 0.57 and ∞ ∃y∣f^3 (y)<y this gives us exactly 3 values for x and therefore exactly 3 triples.
$$\mathrm{let} \\ $$$${f}:\begin{cases}{\mathbb{R}}&{\rightarrow}&{\mathbb{R}_{+} }\\{\alpha}&{ }&{\frac{\mathrm{4}\alpha^{\mathrm{2}} }{\mathrm{1}+\mathrm{4}\alpha^{\mathrm{2}} }}\end{cases} \\ $$$$ \\ $$$$\mathrm{if}\:\left({x},{y},{z}\right)\:\mathrm{is}\:\mathrm{a}\:“\mathrm{good}''\:\mathrm{triplet}\:\mathrm{then} \\ $$$${f}\left({x}\right)={y} \\ $$$${f}\left({y}\right)={z} \\ $$$${f}\left({z}\right)={x} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{notice}\:\mathrm{that}\:\mathrm{therefore},\:\mathrm{using}\:{f} \\ $$$$\mathrm{each}\:\mathrm{triplet}\:\mathrm{can}\:\mathrm{be}\:\mathrm{unicly}\:\mathrm{defined}\:\mathrm{by}\:\mathrm{it}'\mathrm{s} \\ $$$$\mathrm{first}\:\mathrm{elelment}\:{x},\:\mathrm{the}\:\mathrm{others}\:\mathrm{beeing}\:{f}\left({x}\right)\:\mathrm{and} \\ $$$${f}^{\mathrm{2}} \left({x}\right):={f}\left({f}\left({x}\right)\right)\: \\ $$$$ \\ $$$$\mathrm{Thus}\:\mathrm{there}\:\mathrm{are}\:\mathrm{as}\:\mathrm{many}\:\mathrm{triples}\:\mathrm{as}\:\mathrm{there}\:\mathrm{are} \\ $$$$\mathrm{values}\:{x}\:\mathrm{such}\:\mathrm{that}\:{f}^{\mathrm{3}} \left({x}\right)={x} \\ $$$$ \\ $$$${f}\left({x}\right)=\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} } \\ $$$${f}^{\mathrm{2}} \left({x}\right)=\frac{\mathrm{4}\left(\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\right)^{\mathrm{2}} }{\mathrm{1}+\mathrm{4}\left(\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}\left(\mathrm{4}{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}\left(\mathrm{4}{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}^{\mathrm{6}} {x}^{\mathrm{4}} }{\mathrm{1}+\mathrm{2}^{\mathrm{3}} {x}^{\mathrm{2}} +\mathrm{2}^{\mathrm{6}} {x}^{\mathrm{4}} } \\ $$$${f}^{\mathrm{3}} \left({x}\right)=\frac{\mathrm{2}^{\mathrm{6}} \left(\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\right)^{\mathrm{4}} }{\mathrm{1}+\mathrm{2}^{\mathrm{3}} \left(\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{6}} \left(\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\right)^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{2}^{\mathrm{14}} {x}^{\mathrm{8}} }{\left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)^{\mathrm{4}} +\mathrm{2}^{\mathrm{7}} \left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)^{\mathrm{2}} {x}^{\mathrm{4}} +\mathrm{2}^{\mathrm{14}} {x}^{\mathrm{8}} } \\ $$$$=\frac{\mathrm{2}^{\mathrm{14}} {x}^{\mathrm{8}} }{\mathrm{1}+\mathrm{2}^{\mathrm{4}} {x}^{\mathrm{2}} +\mathrm{224}{x}^{\mathrm{4}} +\mathrm{1280}{x}^{\mathrm{6}} +\mathrm{18688}{x}^{\mathrm{8}} } \\ $$$$ \\ $$$${f}^{\mathrm{3}} \left({x}\right)={x}\:\mathrm{has}\:\mathrm{3}\:\mathrm{solutions}… \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}\:\left(\mathrm{I}\:\mathrm{let}\:\mathrm{you}\:\mathrm{check}\:\mathrm{it}\right) \\ $$$${x}=\mathrm{0}\:\left(\mathrm{I}\:\mathrm{let}\:\mathrm{you}\:\mathrm{check}\:\mathrm{it}\right) \\ $$$$\mathrm{and}\:\mathrm{one}\:\mathrm{last}\:\mathrm{point}\:\mathrm{that}\:\mathrm{I}\:\mathrm{was}\:\mathrm{not}\:\mathrm{able}\:\mathrm{to} \\ $$$$\mathrm{compute}\:\mathrm{algebraicaly}.\:\left(\mathrm{actualy}\right. \\ $$$$\left.\mathrm{I}\:\mathrm{was},\:\mathrm{but}\:\mathrm{we}\:\mathrm{shall}\:\mathrm{see}\:\mathrm{that}\:\mathrm{later}\right) \\ $$$$ \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{verify}\:\mathrm{that}\:\mathrm{there}\:\mathrm{are}\:\mathrm{only}\:\mathrm{3}\:\mathrm{because}: \\ $$$${f}^{\mathrm{3}} \left({x}\right)\geqslant\mathrm{0}\:\Rightarrow{x}\geqslant\mathrm{0} \\ $$$${f}^{\mathrm{3}} \left({x}\right)\:\mathrm{is}\:\mathrm{strictly}\:\mathrm{monotonous}\:\mathrm{over}\:\mathbb{R}_{+} \\ $$$$\mathrm{between}\:\mathrm{0}\:\mathrm{and}\:\mathrm{1}/\mathrm{2}\:\exists{y}\mid{f}^{\mathrm{3}} \left({y}\right)<{y} \\ $$$$\mathrm{between}\:\mathrm{1}/\mathrm{2}\:\mathrm{and}\:\mathrm{0}.\mathrm{56}\:\exists{y}\mid{f}^{\mathrm{3}} \left({y}\right)>{y} \\ $$$$\mathrm{beetween}\:\mathrm{0}.\mathrm{57}\:\mathrm{and}\:\infty\:\exists{y}\mid{f}^{\mathrm{3}} \left({y}\right)<{y} \\ $$$$\mathrm{this}\:\mathrm{gives}\:\mathrm{us}\:\mathrm{exactly}\:\mathrm{3}\:\mathrm{values}\:\mathrm{for}\:{x} \\ $$$$\mathrm{and}\:\mathrm{therefore}\:\mathrm{exactly}\:\mathrm{3}\:\mathrm{triples}. \\ $$

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