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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

x+1 = (√((x+1)^2 )) = (√(1+x(x+2))) = (√(1+x(√((x+2)^2 )))) = (√(1+x(√(1+(x+1)(x+3))) )) = (√(1+x(√(1+(x+1)(√((x+3)^2 )))))) = (√(1+x(√(1+(x+1)(√(1+(x+2)(x+4))))) )) =(√(1+x(√(1+(x+1)(√(1+(x+2)(√(1+....)))))))) ∴ f(x) = x+1 please check.

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

Commented by aliesam last updated on 17/Jul/19

thank you sir that is the right result. but i have one thing to say,about the passage to the conclusion. you assumed that is correct based on observation,how can we prove that f(x) is written like that in general case. you cannot use induction principle to prove it because the variable is not an integer,its real positive number.but its good jop god bless you

$${thank}\:{you}\:{sir}\:{that}\:{is}\:{the}\:{right}\:{result}. \\ $$$${but}\:{i}\:{have}\:{one}\:{thing}\:{to}\:{say},{about}\:{the}\:{passage}\:{to}\:{the}\:{conclusion}. \\ $$$${you}\:{assumed}\:{that}\:{is}\:{correct} \\ $$$${based}\:{on}\:{observation},{how}\:{can}\:{we}\:{prove}\:{that}\:{f}\left({x}\right) \\ $$$${is}\:{written}\:{like}\:{that}\:{in}\:{general}\:{case}. \\ $$$${you}\:{cannot}\:{use}\:{induction}\:{principle}\:{to}\:{prove}\:{it} \\ $$$${because}\:{the}\:{variable}\:{is}\:{not}\:{an}\:{integer},{its} \\ $$$$\:{real}\:{positive}\:{number}.{but}\:{its}\:{good}\:{jop}\:{god}\:{bless}\:{you} \\ $$

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

Yes sir you are right. There is no such process atleast not known to me right now.

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

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

generally x+n= =(√((x+n)^2 )) =(√(1+(x+n)^2 −1)) =(√(1+(x+n−1)(x+n+1))) =(√(1+(x+n−1)(√(1+(x+n)(x+n+2))))) =(√(1+(x+n−1)(√(1+(x+n)(√(1+(x+n+1)(x+n+3))))))) .... with x=2, n=1 3=(√(1+2(√(1+3(√(1+4(√(1+5(√(1+...))))))))))

$${generally} \\ $$$${x}+{n}= \\ $$$$=\sqrt{\left({x}+{n}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{\mathrm{1}+\left({x}+{n}\right)^{\mathrm{2}} −\mathrm{1}} \\ $$$$=\sqrt{\mathrm{1}+\left({x}+{n}−\mathrm{1}\right)\left({x}+{n}+\mathrm{1}\right)} \\ $$$$=\sqrt{\mathrm{1}+\left({x}+{n}−\mathrm{1}\right)\sqrt{\mathrm{1}+\left({x}+{n}\right)\left({x}+{n}+\mathrm{2}\right)}} \\ $$$$=\sqrt{\mathrm{1}+\left({x}+{n}−\mathrm{1}\right)\sqrt{\mathrm{1}+\left({x}+{n}\right)\sqrt{\mathrm{1}+\left({x}+{n}+\mathrm{1}\right)\left({x}+{n}+\mathrm{3}\right)}}} \\ $$$$.... \\ $$$${with}\:{x}=\mathrm{2},\:{n}=\mathrm{1} \\ $$$$\mathrm{3}=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+\mathrm{5}\sqrt{\mathrm{1}+...}}}}} \\ $$

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