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Evaluate the following "continued fraction" : {1; 1, 2, 1, 2, 1, 2, 1, 2.........etc.}. Thanks for help.

 Sep 21, 2017
 #1
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Can you please rephrase that? I'm not sure what {1; 1, 2, 1, 2, 1, 2, 1, 2.........etc.}. Also, I think the mathy term you are trying to say is "repeating decimal", or "continued decimal".

 Sep 21, 2017
 #2
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No, it is called "continued fraction" of a well-known number: It means:

1 + 1/(1 + 1/(2 + 1/(1 + 1/2 +.......etc.)))

 Sep 21, 2017
 #3
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That is right! This is called "continued fraction" as the questioner rightly says so:

 

We can represent it as follows: x = 1; 1, 2, 1, 2, 1, 2.......etc., which,in turn, can be written as:

1+ 2 / {1+x} = x, solve for x


2/(x + 1) + 1 = x

Bring 2/(x + 1) + 1 together using the common denominator x + 1:
(x + 3)/(x + 1) = x

Multiply both sides by x + 1:
x + 3 = x (x + 1)

Expand out terms of the right hand side:
x + 3 = x^2 + x

Subtract x^2 + x from both sides:
3 - x^2 = 0

Subtract 3 from both sides:
-x^2 = -3

Multiply both sides by -1:
x^2 = 3

Take the square root of both sides:
 x = sqrt(3) or x = -sqrt(3). Since the continued fraction has all positive signs, then the correct is: sqrt(3).

 Sep 21, 2017
 #4
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\(x=1+\dfrac{1}{2+\frac{1}{1+...}}\\ 1+\dfrac{1}{x}=1+\dfrac{1}{1+\frac{1}{2+...}}\\ \)

First we let x = {1;2,1,2,1,...}, then 1+1/x = {1;1,2,1,2,1,2,...}

Then we find the value of x.

\(x=1+\dfrac{1}{2+\frac{1}{1+...}}\\ x=1+\dfrac{1}{2+x}\\ (x-1)(2+x)=1\\ x^2+x-3=0\\ (x+\dfrac{1}{2})^2=\dfrac{13}{4}\\ x+\dfrac{1}{2}=\pm\dfrac{\sqrt{13}}{2}\\ x=\dfrac{-1+\sqrt{13}}{2}\text{ or }x=\dfrac{-1-\sqrt{13}}{2}(rej.)\\ \)

The value of the original expression is 1+1/x which is:

\(1+\dfrac{2}{\sqrt{13}-1}\\ =\dfrac{\sqrt{13}+1}{\sqrt{13}-1}\\ =\dfrac{14+2\sqrt{13}}{12}\\ =\dfrac{7+\sqrt{13}}{6}\)

:D

~The smartest cookie in the world.

 Sep 21, 2017
edited by MaxWong  Sep 21, 2017
 #5
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x=1+\dfrac{1}{2+\frac{1}{1+...}}\\ 1+\dfrac{1}{x}=1+\dfrac{1}{1+\frac{1}{2+...}}\\

 

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

 

As our guest said, all the signs are positive so the answer is    \(\sqrt3\)

.
 Sep 21, 2017

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