given $a,x \in (1,\infty]$then $x$ and $\sqrt[x]{a}$ are different numbers, except for a single value of $x$ which satisfies: $$ x^x = a $$ to solve this equation, therefore, it might help to look at the sequence defined by: $$ x_{n+1} = \frac12\bigg(x_n+e^{\frac{\log a}{x_n}}\bigg) $$ a few trials suggest that this sequence does converge to the required result.

if this suggestion is correct, how does one prove the fact?

  • $\begingroup$ This answer is partially correct, there are actually two solutions for $a$ in finite interval $a\in \left(e^{-1/e},1\right)$. There are none for smaller $a$. $\endgroup$ – Machinato Feb 17 '18 at 21:59
  • $\begingroup$ @Machinato, the OP clearly states that a>1 $\endgroup$ – Yuriy S Feb 17 '18 at 22:05
  • $\begingroup$ @Machinato It's a question, not an answer. $\endgroup$ – Professor Vector Feb 17 '18 at 22:06

Taking logarithm:

$$x \log x= \log a$$

$$x= \exp \left( \frac{\log a}{x} \right)$$

$$2x=x+ \exp \left( \frac{\log a}{x} \right)$$

$$x=\frac{1}{2} \left(x+ \exp \left( \frac{\log a}{x} \right) \right)$$

We have obtained the equation for the fixed point iterations method.

Now we need to consider the convergence conditions.

$$f(x)=\frac{1}{2} \left(x+ \exp \left( \frac{\log a}{x} \right) \right)$$

$$f'(x)=\frac{1}{2} \left(1- \frac{\log a}{x^2} \exp \left( \frac{\log a}{x} \right) \right)$$

For the iterations to converge we need to have:

$$\left|\frac{1}{2} \left(1- \frac{\log a}{x^2} \exp \left( \frac{\log a}{x} \right) \right) \right|<1$$


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.