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Question Number 122128 by bemath last updated on 14/Nov/20
For which numbers a,b,c,d will the function ψ(x)= ((ax+b)/(cx+d)) satisfy ψ(ψ(x))= x for all x.
$$\:{For}\:{which}\:{numbers}\:{a},{b},{c},{d}\: \\ $$$${will}\:{the}\:{function}\: \\ $$$$\:\:\:\:\psi\left({x}\right)=\:\frac{{ax}+{b}}{{cx}+{d}}\:{satisfy}\:\psi\left(\psi\left({x}\right)\right)=\:{x}\: \\ $$$${for}\:{all}\:{x}. \\ $$
Commented by liberty last updated on 14/Nov/20
from ψ(ψ(x)) = x , we get ψ(x)=ψ^(−1) (x) first, we find ψ^(−1) (x). ⇒ψ^(−1) (x) = ((−dx+b)/(cx−a)) = ((ax+b)/(cx+d)) ⇒(−dx+b)(cx+d)=(ax+b)(cx−a) ⇒ −cd x^2 +(bc−d^2 )x+bd = ac x^2 +(bc−a^2 )x−ba → { ((−cd = ac , a=−d)),((bc−d^2 =bc−a^2 ; a^2 = d^2 )),((bd = −ba ; a = −d)) :} therefore we get a = −d and { ((b is arbitary constant)),((c is arbitary constant)) :}
$$\:\mathrm{from}\:\psi\left(\psi\left(\mathrm{x}\right)\right)\:=\:\mathrm{x}\:,\:\mathrm{we}\:\mathrm{get}\:\psi\left(\mathrm{x}\right)=\psi^{−\mathrm{1}} \left(\mathrm{x}\right) \\ $$$$\mathrm{first},\:\mathrm{we}\:\mathrm{find}\:\psi^{−\mathrm{1}} \left(\mathrm{x}\right). \\ $$$$\Rightarrow\psi^{−\mathrm{1}} \left(\mathrm{x}\right)\:=\:\frac{−\mathrm{dx}+\mathrm{b}}{\mathrm{cx}−\mathrm{a}}\:=\:\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{cx}+\mathrm{d}} \\ $$$$\Rightarrow\left(−\mathrm{dx}+\mathrm{b}\right)\left(\mathrm{cx}+\mathrm{d}\right)=\left(\mathrm{ax}+\mathrm{b}\right)\left(\mathrm{cx}−\mathrm{a}\right) \\ $$$$\Rightarrow\:−\mathrm{cd}\:\mathrm{x}^{\mathrm{2}} +\left(\mathrm{bc}−\mathrm{d}^{\mathrm{2}} \right)\mathrm{x}+\mathrm{bd}\:=\:\mathrm{ac}\:\mathrm{x}^{\mathrm{2}} +\left(\mathrm{bc}−\mathrm{a}^{\mathrm{2}} \right)\mathrm{x}−\mathrm{ba} \\ $$$$\:\rightarrow\begin{cases}{−\mathrm{cd}\:=\:\mathrm{ac}\:,\:\mathrm{a}=−\mathrm{d}}\\{\mathrm{bc}−\mathrm{d}^{\mathrm{2}} =\mathrm{bc}−\mathrm{a}^{\mathrm{2}} ;\:\mathrm{a}^{\mathrm{2}} \:=\:\mathrm{d}^{\mathrm{2}} \:}\\{\mathrm{bd}\:=\:−\mathrm{ba}\:;\:\mathrm{a}\:=\:−\mathrm{d}}\end{cases} \\ $$$$\mathrm{therefore}\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:=\:−\mathrm{d}\:\mathrm{and}\:\begin{cases}{\mathrm{b}\:\mathrm{is}\:\mathrm{arbitary}\:\mathrm{constant}}\\{\mathrm{c}\:\mathrm{is}\:\mathrm{arbitary}\:\mathrm{constant}}\end{cases} \\ $$$$ \\ $$
Answered by TANMAY PANACEA last updated on 14/Nov/20
ψ(ψ(x))=((a(((ax+b)/(cx+d)))+b)/(c(((ax+b)/(cx+d)))+d))=x ((a^2 x+ab+bcx+bd)/(acx+bc+cdx+d^2 ))=x ((x(a^2 +bc)+ab+bd)/(x(ac+cd)+bc+d^2 ))= a^2 +bc=1 ab+bd=0→b(a+d)=0 ac+cd=0→c(a+d)=0 bc+d^2 =1 now if a+d≠0 b=0 and c=0 a^2 =1 a=1 and d=1 a=1 b=0 c=0 d=1
$$\psi\left(\psi\left({x}\right)\right)=\frac{{a}\left(\frac{{ax}+{b}}{{cx}+{d}}\right)+{b}}{{c}\left(\frac{{ax}+{b}}{{cx}+{d}}\right)+{d}}={x} \\ $$$$\frac{{a}^{\mathrm{2}} {x}+{ab}+{bcx}+{bd}}{{acx}+{bc}+{cdx}+{d}^{\mathrm{2}} }={x} \\ $$$$\frac{{x}\left({a}^{\mathrm{2}} +{bc}\right)+{ab}+{bd}}{{x}\left({ac}+{cd}\right)+{bc}+{d}^{\mathrm{2}} }= \\ $$$${a}^{\mathrm{2}} +{bc}=\mathrm{1} \\ $$$${ab}+{bd}=\mathrm{0}\rightarrow{b}\left({a}+{d}\right)=\mathrm{0} \\ $$$${ac}+{cd}=\mathrm{0}\rightarrow{c}\left({a}+{d}\right)=\mathrm{0} \\ $$$${bc}+{d}^{\mathrm{2}} =\mathrm{1} \\ $$$${now}\:\:{if}\:{a}+{d}\neq\mathrm{0}\:\:{b}=\mathrm{0}\:\:{and}\:{c}=\mathrm{0} \\ $$$${a}^{\mathrm{2}} =\mathrm{1}\:\:{a}=\mathrm{1}\:\:{and}\:{d}=\mathrm{1} \\ $$$$\boldsymbol{{a}}=\mathrm{1}\:\:\:\boldsymbol{{b}}=\mathrm{0}\:\:\:\boldsymbol{{c}}=\mathrm{0}\:\:\:\boldsymbol{{d}}=\mathrm{1} \\ $$$$ \\ $$

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