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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+...))))))))
Isthereawaytoevaluatethefollowing:S=2+2+2+2+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=2+2+2+2+2+(S2)2=(2+2+2+2+)2S24S+4=2+2+2+2+2+=SS25S+4=0(S1)(S4)=0S=1S=4Sincewehavesquaredbothsidestheresultingequationisnotcompletelyequivalenttotheoriginalandthereispossibilityofextraneousroots.ClearlyS>2because2+2+2+2+ispositve.HenceS=1isextraneousrootandS=4isrequiredsum.
Commented by Filup last updated on 13/Nov/15
AMAZING! I am honestly fascinated!
AMAZING!Iamhonestlyfascinated!
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
SohowcanwebesureS>2Icantseehow,becausexxSo,if:2+2+issmall,2+2+issmaller
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=2+2+2+2+2+[Given]Underlinedexpression>0S=2+positivenumberS>2
Commented by Filup last updated on 14/Nov/15
ahh i must of had a massive brain fart
ahhimustofhadamassivebrainfart

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