Consider the infinite power tower a a a ⋅ ⋅ ⋅
with a = 0 . 0 5 , representing the sequence x 1 = a and x n + 1 = a x n . Find L = n → ∞ lim x 2 n .
Feel free to use a computer for some of your work.
Bonus Question: What about n → ∞ lim x 2 n + 1 ?
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The two limits of this tower are 0.66 and 0.14 (for limit with odd index) asked in the Bonus.
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Jawohl, genau! For x < c 1 we have f ( x ) > x , so that the sequence x 2 n + 1 is increasing and will approach c 1 .
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Since we are interested in the even terms only (Herr @Andreas Wendler will take care of the odd terms) , we compute x 2 = a a = 0 . 0 5 0 . 0 5 ≈ 0 . 8 6 . Now we have the recursive formula x n + 2 = a a x n , and we define the iteration function f ( x ) = a a x . It turns out that f ( x ) has three fixed points, where f ( x ) = x , namely, c 1 ≈ 0 . 1 4 , c 2 ≈ 0 . 3 5 , c 3 ≈ 0 . 6 6 . The function f ( x ) is increasing and f ( x ) < x for x > c 3 ; see the attached graph. Thus the sequence x 2 n is decreasing and bounded by c 3 below; draw a cob web! We have lim n → ∞ x 2 n = c 3 ≈ 0 . 6 6