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| Rooting For The Limit II (Posted on 2009-11-26) |
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Given that:
Fm(x) = √(2 - √(2 + √(2 + √(2 + .....√(2 + √x))))) (m square roots)
Evaluate this limit:
Limit Fm(2)/Fm(3)
m → ∞
Solution
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Comment 4 of 4 | 
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I’ve returned several times to this problem in an attempt to establish an analytical basis for the previously posted limit and, at last, having drawn a graph of F(m,x) for large m, I noticed that it resembles an inverse cosine whose wavelength is halved by every unit increase in m. So this suggests the following trig substitution which collapses the series of radicals. Much fun.
Let x = 2(1 + cos(2mA)) (1)
which gives x = 4 cos2(2m-1A)
Therefore sqrt(x) = 2 cos(2m-1A)
which gives 2 + sqrt(x) = 2(1 + cos(2m-1A)) = 4 cos2(2m-2A)
and sqrt(2 + sqrt(x)) = 2 cos(2m-2A)
Then, after a further m - 3 inductive steps, the LHS will have m - 1 sqrts ...
sqrt(2 + sqrt(2 + sqrt(2 + ..... +sqrt(2 + sqrt(x))..))) = 2 cos(2A)
So the formula given in the problem can be written as:
F(m,x) = sqrt(2 - 2 cos(2A)) = 2 sin(A)
But, from (1) A = 2-m arccos(x/2 - 1)
therefore F(m,x) = 2 sin(2-m arccos(x/2 - 1))
Now, as m becomes large, 2-m causes the bracket to become small so that we
can use the small-angle approximation (sin B = B) to give a limiting formula:
Lim F(m,x) as m tends to infinity = 21-m arcos(x/2 - 1)
Thus F(m,2) tends to 21-m arccos(0) = 21-m pi/2
and F(m,3) tends to 21-m arccos(1/2) = 21-m pi/3
Therefore the ratio F(m,2)/F(m,3) tends to 3/2
- which agrees with the computed results posted much earlier.
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Posted by Harry
on 2010-04-16 00:05:22 |
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