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Find the value of the infinite continued fraction

\(\cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \dotsb}}}\)

 Dec 7, 2019
 #1
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+1

This short computer code gives the value of:

 

c=1E100; listforeach(b,reverse(1, 2, 1, 2, 1, 2), c=b + 1/c)

 

Sqrt(3) =1.7320508075688772935274463415059......

 

P.S. One of the moderators will solve it  algebraically.

 Dec 7, 2019
 #2
avatar+108629 
+1

 

 

\(x=1+\dfrac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+...}}}}\\ x-1=\dfrac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+...}}}}\\ x-1=\dfrac{1}{1+\frac{1}{2+(x-1)}}\\ x-1=\dfrac{1}{\frac{2+x-1+1}{2+x-1}}\\ x-1=\dfrac{1}{\frac{2+x}{1+x}}\\ x-1=\dfrac{1+x}{x+2}\\ (x-1)(x+2)=x+1\\ x^2+x-2=x+1\\ x^2=3\\ x=\pm\sqrt3\)

 

The answer must be positive so the only valid andwer is     \(x=\sqrt3\)

.
 Dec 7, 2019

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