Weird definition (2)

Algebra Level 3

Let f f be a function of x x defined such that f 2 x = x 2 f − 1 \dfrac{f^2}{x} = x^{2f-1} . If x n x_n are solutions to f ( x ) = 1 n f(x) = \dfrac{1}{n} for 1 ≤ n ≤ 14 1 \le n \le 14 find, to three decimal places, ∑ n = 1 14 x n \displaystyle \sum_{n=1}^{14} {x_n} .


The answer is 1.291.

Akeel Howell by Akeel Howell , 22, USA

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1 solution

Tom Engelsman
Sep 19, 2020

If f ( x ) = 1 n f(x) = \frac{1}{n} , then we have 1 n 2 x = x 2 / n − 1 ⇒ 1 n 2 = x 2 / n ⇒ x n = 1 n n . \frac{1}{n^{2}x} = x^{2/n - 1} \Rightarrow \frac{1}{n^2} = x^{2/n} \Rightarrow x_{n} = \frac{1}{n^{n}}. We can now compute:

Σ n = 1 14 1 n n ≈ 1.291 . \Sigma_{n=1}^{14} \frac{1}{n^n} \approx \boxed{1.291}.

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