Don't mess with tetrations.

Calculus Level 4

$\displaystyle {x=\pi^ {(\pi^{-1})^{\pi^{(\pi^{-1})^{...}}}}}$

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

Details and assumptions

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

The answer is 1.296.

1 solution

Chew-Seong Cheong
Apr 8, 2015

$x = \pi^{(\pi^{-1})^{\pi^{(\pi^{-1})^{...}}}} = \pi^{(\frac{1}{\pi})^{\pi^{(\frac{1}{\pi} )^{...}}}} = \pi^{(\frac{1}{\pi})^x} \quad \Rightarrow \ln{x} = \left( \dfrac {1}{\pi} \right)^x \ln{\pi}$

Let $\Rightarrow f(x) = \ln{x} - \left( \dfrac {1}{\pi} \right)^x \ln{\pi}$ .

I can only solve it using numerical method and discover that Newton's method cannot solve it because it does not converge nicely. I solved it by plotting increasing accuracy ( $\Delta x$ getting smaller) graph of $f(x)$ and checking for its root. I show hear the first and last graphs in the process.

The answer was found to be $\boxed{1.296349}$ in six decimal places.

Sorry, what is wrong with the convergence? A little number of iterations of Newton-Raphson's method weren't working out?

Mikael Marcondes - 6 years, 2 months ago

I solved it out graphically first, analytically after, but approximation was quite good...

Mikael Marcondes - 6 years, 2 months ago

Don't mess with tetrations.

The answer is 1.296.

1 solution

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