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Question Number 151080 by malwan last updated on 18/Aug/21

$$if(f\circ f\circ f\circ f)\left(x\right)=16x+15$$$$findf\left(x\right)$$

Answered by Mokmokhi last updated on 18/Aug/21

$$\mathrm{It}\mathrm{is}\mathrm{reasonable}\mathrm{to}\mathrm{say}f\left(x\right)\mathrm{is}\mathrm{linear}\mathrm{by}\mathrm{rejecting}\mathrm{other}\mathrm{possibilities}.$$$$\mathrm{Then}f\left(x\right)=ax+b\mathrm{for}\mathrm{some}\mathrm{unknowns}a\mathrm{and}b.$$$$\mathrm{The}\mathrm{value}\mathrm{of}a\mathrm{should}\mathrm{be}2\mathrm{or}-2.$$$$\mathrm{If}a=2.b=1.$$$$\mathrm{If}a=-2.b=20.$$

Commented by malwan last updated on 18/Aug/21

$$thankyousir$$

Answered by qaz last updated on 18/Aug/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x},\mathrm{we}\mathrm{get}\mathrm{fixed}\mathrm{point}\mathrm{x}=-1$$$$\mathrm{so}\mathrm{f}\left(\mathrm{x}\right)=16(\mathrm{x}+1)-1$$$$\Rightarrow (\mathrm{f}\circ \mathrm{f})\left(\mathrm{x}\right)={16}^2(\mathrm{x}+1)-1$$$$(\mathrm{f}\circ \mathrm{f}\circ \mathrm{f})\left(\mathrm{x}\right)={16}^3(\mathrm{x}+1)-1$$$$(\mathrm{f}\circ \mathrm{f}\circ \mathrm{f}\circ \mathrm{f})\left(\mathrm{x}\right)={16}^4(\mathrm{x}+1)-1$$$$\left(\underset{\mathrm{n}}{\mathrm{f}\circ \mathrm{f}\circ ...\circ \mathrm{f}}\right)\left(\mathrm{x}\right)={16}^{\mathrm{n}}(\mathrm{x}+1)-1$$

Commented by malwan last updated on 18/Aug/21

$$thankyousir$$

Commented by qaz last updated on 19/Aug/21

$$\mathrm{sorry}.\mathrm{there}\mathrm{is}\mathrm{a}\mathrm{error}.$$$$\mathrm{we}\mathrm{need}\mathrm{make}\mathrm{a}\mathrm{reverse}\mathrm{iteration}.$$$$(\mathrm{f}\circ \mathrm{f}\circ \mathrm{f}\circ \mathrm{f})\left(\mathrm{x}\right)=2^4(\mathrm{x}+1)-1$$$$....$$$$\mathrm{f}\left(\mathrm{x}\right)=2(\mathrm{x}+1)-1=2\mathrm{x}+1$$

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