a 0 = a and a n + 1 = a a n
What is the smallest a > 0 such that a n is convergent?
Note: This is a follow up question from Limit of The Tower
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Can you provide a bit more details?
At the very least, explain your previous misconception of why simply graphing x x 1 and seeing that it tends to 0 at x = 0 , doesn't give the right answer.
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well, i use non-rigorous method, its kinda weird, mixing some intuition, and use computer program to check my final answer is valid.
im waiting others to give a more rigorous solution, or prove my answer is wrong.
edit: solution added
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@Brian Lie – thanks for your link Brian, but it seems like a dead link to me.
"This site can’t be reached"
Can we state that a > 0 here.
a = − 1 is a perfectly valid solution as is.
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e − e
let f ( x ) = a x , if f n ( x ) converges to L , then f ( L ) = a L = L , f ′ ( L ) = a L ln a = ln L .
when x ≈ L we approximate f ( x ) using its tangent at x = L , f ( x ) f ( L + ε ) f 2 ( L + ε ) f 3 ( L + ε ) ≈ L + f ′ ( L ) ( x − L ) = L + ( ln L ) ( x − L ) ≈ L + ( ln L ) ε ≈ L + ( ln L ) 2 ε ≈ L + ( ln L ) 3 ε for f n ( L + ε ) to converge, we must have − 1 e − 1 ( e − 1 ) e − 1 1 e − e < ln L < 1 < L < e < a < e e 1 ( ∵ a = L L 1 increases monotonically between e − 1 < L < e ) < a < e e 1
p/s: im not sure if this solution is valid, so the result is double checked using computer program, which seems to support my answer.