Albert G.

asked • 01/29/18

Evaluate an infinite staircase of square roots of square roots plus square roots .

This is an expression involving a concatenation of overlapping square roots of square roots,
 
x =  √16¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
              + √16¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
                        + √16¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
                                  + √16¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
                                            + etc., whereby all radicals encompass ALL subsequent radicands (which are themselves radicals) for any number of iterations N. I want to be evaluate the limit as N approaches ∞.

Albert G.

Thanks for the approach, an insight that did not occur to me. I have subsequently applied your approach to a similar function with the value unity (1) instead of 16. Its solution is (1/2)[1 ± √5], which I realized was numerically equal to the Golden Ratio (1.618...)! A limit does indeed exist.
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01/30/18

Bobosharif S.

You are welcome.
Can you prove the existence? I think it is easy to see that it is bounded. 
There are other ways to find similar limits as well, but they are harder sometimes. In some cases trigonometric functions might help but not always.
With 1's: why it doesn't go to infinity when n→∞?. Is it indeed bounded!?
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01/30/18

1 Expert Answer

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Bobosharif S. answered • 01/29/18

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