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Mr Davis 97 April 28th, 2016 10:10 AM

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?

v8archie April 28th, 2016 10:21 AM

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".

v8archie April 28th, 2016 11:29 AM

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.

idontknow April 28th, 2016 01:14 PM

first we should claim if the infinite surd converges before
we can equal it as a real number

v8archie April 28th, 2016 01:29 PM

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.

Hezekiah October 19th, 2016 08:51 AM

Express 14-8√60÷6√20-10√12 in the form a√b+6√3; where a & b are irrational numbers.

skeeter October 19th, 2016 08:59 AM

Quote:

Originally Posted by Hezekiah (Post 551009)
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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