To solve equations involving power towers (infinite tetration) we usually do something like this:

$$x^{x^{x^{x^{\dots}}}} =k$$

$$x^{(x^{x^{x^{\dots}}})} =k$$



But what if I do something like this:

$$x^{x^{(x^{x^{\dots}})}} =k$$


$$x^k\ln x=k$$

$$e^{k\ln x}\ln x=k$$

$$e^{k\ln x}k\ln x=k^2$$

$$x=e^{\frac {W(k^2)} k}$$

Where $W(z)$ is the Lambert W function.

This two expression must be equal so how can I prove this ? If they are not equal am I missing something whit the definition of infinite tetration ? And what about if I try to solve the equation doing this (and so on) ?:

$$x^{x^{x^{(x^{\dots})}}} =k$$


  • $\begingroup$ The mistake: In the second calculations you took the logarithm of LHS and you did not take the logarithm of the RHS. (see my answer for the correct expression) $\endgroup$ – Elaqqad Apr 26 '15 at 13:05
  • $\begingroup$ A power tower is not an "infinite" tetration it's just an ordinary one. $\endgroup$ – mathreadler Apr 26 '15 at 13:09

Actually you fixed $x$ and you're trying to find the limit so in the first case, if $k$ is the limit then $\sqrt[k]k=x$ and in the second case if $k$ is the limit then $x=e^{\frac{W(k\ln(k))}{k}}$ ( you made a mistake) but actually the two equations are equivalents : $$x=\sqrt[k]k\iff x=e^{\frac{\ln(k)}{k}}\iff x=e^{\frac{W(k\ln(k))}{k}}$$ using the fact that $W(k\ln(k))=\ln(k)$.

For higher order you will have the same expression, the limit is unique and can be proved rigorously under some conditions.

  • $\begingroup$ Note also the limit in the two cases is $k=e^{-W(-\ln(x))}=\frac{W(-\ln(x)}{-\ln(x)}$ the conditions are $e^{−e }≤ x ≤ e^{1/e}$ $\endgroup$ – Elaqqad Apr 26 '15 at 13:08
  • $\begingroup$ Thanks! Now everything makes sense, I've made a stupid error. $\endgroup$ – Renato Faraone Apr 26 '15 at 13:15

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