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April 28th, 2016, 10:10 AM   #1
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Question 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?
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April 28th, 2016, 10:21 AM   #2
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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: $s=\frac12(s^2-3)$. (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".
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Last edited by v8archie; April 28th, 2016 at 10:25 AM.
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April 28th, 2016, 11:29 AM   #3
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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.
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April 28th, 2016, 01:14 PM   #4
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first we should claim if the infinite surd converges before
we can equal it as a real number
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April 28th, 2016, 01:29 PM   #5
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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 01:36 PM.
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October 19th, 2016, 08:51 AM   #6
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Express 14-8√60÷6√20-10√12 in the form a√b+6√3; where a & b are irrational numbers.
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October 19th, 2016, 08:59 AM   #7
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Quote:
Originally Posted by Hezekiah View Post
Express 14-8√60÷6√20-10√12 in the form a√b+6√3; where a & b are irrational numbers.
do not piggy back onto an existing thread from a different problem, start a new thread ... you've also been asked to show some effort.
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