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Question Number 2275 by Filup last updated on 13/Nov/15

Is there a way to evaluate the following: S=2+(√(2+(√(2+(√(2+(√(2+...))))))))

$$\mathrm{Is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{way}\:\mathrm{to}\:\mathrm{evaluate}\:\mathrm{the}\:\mathrm{following}: \\ $$$$ \\ $$$${S}=\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+...}}}} \\ $$

Answered by Rasheed Soomro last updated on 13/Nov/15

S=2+(√(2+(√(2+(√(2+(√(2+...)))))))) (S−2)^2 =((√(2+(√(2+(√(2+(√(2+...)))))))) )^2 S^2 −4S+4 =2+(√(2+(√(2+(√(2+(√(2+...)))))))) =S S^2 −5S+4=0 (S−1)(S−4)=0 S=1 ∣ S=4 Since we have squared both sides the resulting equation is not completely equivalent to the original and there is possibility of extraneous roots. Clearly S>2 because (√(2+(√(2+(√(2+(√(2+...)))))))) is positve. Hence S=1 is extraneous root and S=4 is required sum.

$${S}=\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+...}}}} \\ $$$$\left({S}−\mathrm{2}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+...}}}}\:\right)^{\mathrm{2}} \\ $$$${S}^{\mathrm{2}} −\mathrm{4}{S}+\mathrm{4}\:=\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+...}}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={S} \\ $$$${S}^{\mathrm{2}} −\mathrm{5}{S}+\mathrm{4}=\mathrm{0} \\ $$$$\left({S}−\mathrm{1}\right)\left({S}−\mathrm{4}\right)=\mathrm{0} \\ $$$${S}=\mathrm{1}\:\mid\:\:{S}=\mathrm{4} \\ $$$${Since}\:{we}\:{have}\:{squared}\:{both}\:{sides}\:{the}\:{resulting}\:{equation} \\ $$$${is}\:{not}\:{completely}\:{equivalent}\:{to}\:{the}\:{original}\:{and}\:{there} \\ $$$${is}\:{possibility}\:{of}\:{extraneous}\:{roots}. \\ $$$${Clearly}\:{S}>\mathrm{2}\:{because}\:\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+...}}}}\:{is}\:{positve}. \\ $$$${Hence}\:{S}=\mathrm{1}\:{is}\:{extraneous}\:{root}\:{and} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{S}=\mathrm{4}\:{is}\:{required}\:{sum}. \\ $$

Commented by Filup last updated on 13/Nov/15

AMAZING! I am honestly fascinated!

$${AMAZING}!\:\mathrm{I}\:\mathrm{am}\:\mathrm{honestly}\:\mathrm{fascinated}! \\ $$

Commented by Filup last updated on 13/Nov/15

So how can we be sure S>2 I can′t see how, because (√x)≤x So, if: 2+(√(2+(√(...)))) is small, (√(2+(√(2+(√(...)))))) is smaller

$$\mathrm{So}\:\mathrm{how}\:\mathrm{can}\:\mathrm{we}\:\mathrm{be}\:\mathrm{sure}\:{S}>\mathrm{2} \\ $$$$\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{see}\:\mathrm{how},\:\mathrm{because}\:\sqrt{{x}}\leqslant{x} \\ $$$$\mathrm{So},\:\mathrm{if}:\:\:\:\:\:\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{...}}\:\mathrm{is}\:\mathrm{small}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{...}}}\:\mathrm{is}\:\mathrm{smaller} \\ $$

Commented by RasheedAhmad last updated on 13/Nov/15

S=2+(√(2+(√(2+(√(2+(√(2+...))))))))_(−−−−−−−−−−−−−−−) [Given] Underlined expression>0 ∴ S=2+positive number ∴ S>2

$${S}=\mathrm{2}+\underset{−−−−−−−−−−−−−−−} {\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+...}}}}}\:\:\:\left[{Given}\right] \\ $$$${Underlined}\:{expression}>\mathrm{0} \\ $$$$\:\therefore\:{S}=\mathrm{2}+{positive}\:{number} \\ $$$$\therefore\:{S}>\mathrm{2} \\ $$

Commented by Filup last updated on 14/Nov/15

ahh i must of had a massive brain fart

$$\mathrm{ahh}\:\mathrm{i}\:\mathrm{must}\:\mathrm{of}\:\mathrm{had}\:\mathrm{a}\:\mathrm{massive}\:\mathrm{brain}\:\mathrm{fart} \\ $$

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