Just another tetration.

Calculus Level 4

x = e ( e − 1 ) e ( e − 1 ) . . . \displaystyle {x=e^ {(e^{-1})^{e^{(e^{-1})^{...}}}}}

Find x . \text{Find } x \text{.}

Details and assumptions

If you can’t read the expression, it’s an alternating \text{If you can't read the expression, it's an alternating} infinite tetration, with the terms being e and e − 1 . \text{infinite tetration, with the terms being } e \text{ and } {e}^{-1} \text{.}


The answer is 1.3098.

Mikael Marcondes by Mikael Marcondes , 24, Brazil

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2 solutions

I used the same method as Caleb Townsend . Just to attach the graph and Newton's method calculation, I have done with an Excel spreadsheet.

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Caleb Townsend
Mar 1, 2015

x = e e − x x = e^{e^{-x}} ln ⁡ ( x ) = 1 e x \ln(x) = \frac{1}{e^x} I don't think there's a closed form for the solution with standard mathematical functions, but correct me if I'm wrong. I will solve it with Newton's method. Define x n + 1 = x n − f ( x n ) f ′ ( x n ) x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)} where f ( x ) = ln ⁡ ( x ) − 1 e x f(x) = \ln(x) - \frac{1}{e^x} and x 1 = 1.3 x_1 = 1.3 since the root is near there by looking at the graph.

After several terms, x n ≈ 1.31 x_n \approx \boxed{1.31} and it's not changing by much.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

It's a completely fair method to solve it. I was pretty lazy to write a solution, but yeah, this is the "easiest" way I know to solve a tetration, which probably hasn't a closed form. You've nailed it!

Mikael Marcondes - 6 years, 3 months ago

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