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Question Number 64404 by aliesam last updated on 17/Jul/19

Commented by Prithwish sen last updated on 17/Jul/19

$$\mathrm{x}+1=\sqrt{{(\mathrm{x}+1)}^2}=\sqrt{1+\mathrm{x}(\mathrm{x}+2)}=\sqrt{1+\mathrm{x}\sqrt{{(\mathrm{x}+2)}^2}}$$$$=\sqrt{1+\mathrm{x}\sqrt{1+(\mathrm{x}+1)(\mathrm{x}+3)}}=\sqrt{1+\mathrm{x}\sqrt{1+(\mathrm{x}+1)\sqrt{{(\mathrm{x}+3)}^2}}}$$$$=\sqrt{1+\mathrm{x}\sqrt{1+(\mathrm{x}+1)\sqrt{1+(\mathrm{x}+2)(\mathrm{x}+4)}}}$$$$=\sqrt{1+\mathrm{x}\sqrt{1+(\mathrm{x}+1)\sqrt{1+(\mathrm{x}+2)\sqrt{1+....}}}}$$$$\therefore \mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+1\mathrm{please}\mathrm{check}.$$

Commented by aliesam last updated on 17/Jul/19

$$thankyousirthatistherightresult.$$$$butihaveonethingtosay,aboutthepassagetotheconclusion.$$$$youassumedthatiscorrect$$$$basedonobservation,howcanweprovethatf\left(x\right)$$$$iswrittenlikethatingeneralcase.$$$$youcannotuseinductionprincipletoproveit$$$$becausethevariableisnotaninteger,its$$$$realpositivenumber.butitsgoodjopgodblessyou$$

Commented by Prithwish sen last updated on 17/Jul/19

$$\mathrm{Yes}\mathrm{sir}\mathrm{you}\mathrm{are}\mathrm{right}.\mathrm{There}\mathrm{is}\mathrm{no}\mathrm{such}\mathrm{process}$$$$\mathrm{atleast}\mathrm{not}\mathrm{known}\mathrm{to}\mathrm{me}\mathrm{right}\mathrm{now}.$$

Commented by mr W last updated on 17/Jul/19

$$generally$$$$x+n=$$$$=\sqrt{{(x+n)}^2}$$$$=\sqrt{1+{(x+n)}^2-1}$$$$=\sqrt{1+(x+n-1)(x+n+1)}$$$$=\sqrt{1+(x+n-1)\sqrt{1+(x+n)(x+n+2)}}$$$$=\sqrt{1+(x+n-1)\sqrt{1+(x+n)\sqrt{1+(x+n+1)(x+n+3)}}}$$$$....$$$$withx=2,n=1$$$$3=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}}$$

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