One more Power Tower

Calculus Level 5

Consider the infinite power tower a a a \huge a^{a^{a^{\cdot^{\cdot^\cdot}}}}

with a = 0.05 a=0.05 , representing the sequence x 1 = a x_1=a and x n + 1 = a x n x_{n+1}=a^{x_n} . Find L = lim n x 2 n \displaystyle L=\lim_{n\to\infty}x_{2n} .

Feel free to use a computer for some of your work.

Bonus Question: What about lim n x 2 n + 1 \displaystyle \lim_{n\to\infty}x_{2n+1} ?

L L fails to exist L 0.14 L\approx 0.14 L 0.66 L\approx 0.66 L = 0 L=0 L 0.35 L\approx 0.35 L = 1 L=1

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1 solution

Otto Bretscher
Feb 22, 2016

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 0.05 0.86 x_2=a^a= 0.05^{0.05}\approx0.86 . Now we have the recursive formula x n + 2 = a a x n x_{n+2}=a^{a^{x_n}} , and we define the iteration function f ( x ) = a a x f(x)=a^{a^x} . It turns out that f ( x ) f(x) has three fixed points, where f ( x ) = x f(x)=x , namely, c 1 0.14 , c 2 0.35 , c 3 0.66 c_1\approx 0.14,c_2\approx 0.35, c_3\approx 0.66 . The function f ( x ) f(x) is increasing and f ( x ) < x f(x)<x for x > c 3 x>c_3 ; see the attached graph. Thus the sequence x 2 n x_{2n} is decreasing and bounded by c 3 c_3 below; draw a cob web! We have lim n x 2 n = c 3 0.66 \lim_{n\to\infty}x_{2n}=c_3\approx \boxed{0.66}

The two limits of this tower are 0.66 and 0.14 (for limit with odd index) asked in the Bonus.

Andreas Wendler - 5 years, 3 months ago

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Jawohl, genau! For x < c 1 x<c_1 we have f ( x ) > x f(x)>x , so that the sequence x 2 n + 1 x_{2n+1} is increasing and will approach c 1 c_1 .

Otto Bretscher - 5 years, 3 months ago

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