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April 28th, 2016, 11:10 AM  #1 
Senior Member Joined: Feb 2014 From: Louisiana Posts: 156 Thanks: 6 Math Focus: algebra and the calculus  Value of an infinite surd
What is the value of $\displaystyle \sqrt{3 + 2\sqrt{3 + 2\sqrt{3 + 2\sqrt{3 + \cdots}}}}$? What is a general way of finding the value of infinite surds like this?

April 28th, 2016, 11:21 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,600 Thanks: 2588 Math Focus: Mainly analysis and algebra 
If is has a value, $s$, then we can take your expression, multiply by 2, add 3 and take the square root and be left with the same value. That is, $s = \sqrt{3+2s}$ Alternatively, if you square it, subtract 3 and divide by 2, you are left with the same value: $s=\frac12(s^23)$. (Note that this gives, in addition to the positive solution, a negative solution which in an obvious  but not necessarily correct  way cannot be valid). I get $s \in \{1,3\}$. In other words, the general way of solving these is to see what operations are required to add or remove a "term". Last edited by v8archie; April 28th, 2016 at 11:25 AM. 
April 28th, 2016, 12:29 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,600 Thanks: 2588 Math Focus: Mainly analysis and algebra 
I think I would go so far as to say that the negative solution is not correct because the square root function is never negative for finite input. The argument can be made more technical and less certain, but in the end it boils down to the above.

April 28th, 2016, 02:14 PM  #4 
Senior Member Joined: Dec 2015 From: iPhone Posts: 389 Thanks: 62 
first we should claim if the infinite surd converges before we can equal it as a real number 
April 28th, 2016, 02:29 PM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,600 Thanks: 2588 Math Focus: Mainly analysis and algebra 
That was the very first thing that I said. Except that, strictly speaking, this is not a limit problem so there is no "convergence". My phraseology reflects that. If the question talked about limits or convergence, you would form a series based on the recurrence formula $s_{n+1}=\sqrt{3+2s_n}$ and show that it converges for all positive $s_1$. Actually, I think it converges to 3 for all $s_1 \ge \frac32$. For other $s_1$, $s_2$ is not real. Last edited by v8archie; April 28th, 2016 at 02:36 PM. 
October 19th, 2016, 09:51 AM  #6 
Newbie Joined: Oct 2016 From: Minna Posts: 8 Thanks: 0 
Express 148√60÷6√2010√12 in the form a√b+6√3; where a & b are irrational numbers.

October 19th, 2016, 09:59 AM  #7 
Math Team Joined: Jul 2011 From: Texas Posts: 2,820 Thanks: 1464  

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