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Find the value of \(x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}.\)

 May 7, 2019
 #1
avatar+100519 
+1

Add 1 to both sides     and we have that

 

                                   1                                

x + 1  =    2  +       __________

                             2 +  1

                                  ________

                                    2 +  .......

 

So we have that

 

x  =   1 +    1

                ______          multiply  through by   x + 1

                 x + 1

 

x(x + 1)  = (x + 1)  + 1

 

x^2 + 1x =  1x + 2         subtract 1x from both sides

 

x^2    =  2          take the positive square root.....since the original right side is positive

 

x = √2

 

 

cool cool cool

 May 7, 2019
edited by CPhill  May 7, 2019
 #2
avatar+5074 
+1

\(\text{let }a = \dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\ddots}}}\)

 

\(a = \dfrac{1}{2+a}\\ a^2+2a-1=0\\ a = \dfrac{-2\pm \sqrt{4+4}}{2} = -1 \pm \sqrt{2}\\ \text{we can rule out the negative solution since everything is positive}\\ a = -1 + \sqrt{2}\)

 

\(x= 1+a = \sqrt{2}\)

.
 May 7, 2019
 #3
avatar+22180 
+2

Find the value of 

\(x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}.\)

 

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}} } \\\\ x-1 &=& \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}} \\\\ \dfrac{1}{x-1} &=& 2+ \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}} \\\\ \dfrac{1}{x-1} &=& 2+ (x-1) \\\\ 1 &=& (x-1)(2+x-1) \\ 1 &=& (x-1)(x+1) \\ 1 &=& x^2-1 \\ x^2 &=& 2 \\ \mathbf{x} &=& \mathbf{\sqrt{2}} \\ \hline \end{array} \)

 

 

laugh

 May 7, 2019

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