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What is the value of $\sqrt{x + \sqrt{ x + \sqrt{ x + \cdots } } }\,$? I know the basic trick to calculate this using $f = \sqrt{ x + f }$. But, I want more accurate answer which is I am not getting with this formula.

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Squaring we get $f^2=x+f\iff f^2-f-x=0\implies f=\dfrac{1\pm\sqrt{1+4x}}2$

Now as $f>0,$ discard the negative root assuming $x>0$

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    $\begingroup$ Reaching 100k! Congrats! $\endgroup$ – Jack D'Aurizio Sep 20 '14 at 4:46
  • $\begingroup$ @JackD'Aurizio, Thanks. How to discard $$\frac{1-\sqrt{1+4x}}2$$ if $x<0$? $\endgroup$ – lab bhattacharjee Sep 20 '14 at 4:50
  • $\begingroup$ Hey lab ! Congratulations $\endgroup$ – Claude Leibovici Sep 20 '14 at 4:58
  • $\begingroup$ @ClaudeLeibovici, Thanks. Could you please have another look into math.stackexchange.com/questions/938585/… ? $\endgroup$ – lab bhattacharjee Sep 20 '14 at 5:01
  • $\begingroup$ I did but may I confess that I do not see how you finish ! $\endgroup$ – Claude Leibovici Sep 20 '14 at 5:09
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If $f=\sqrt{x+f}$ then $f^2-f-x=0$, hence $$f=\frac{1+\sqrt{1+4x}}{2}.$$

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