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?

  • 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 $a$. – Machinato Feb 17 at 21:59
  • @Machinato, the OP clearly states that a>1 – Yuriy S Feb 17 at 22:05
  • @Machinato It's a question, not an answer. – Professor Vector Feb 17 at 22:06
up vote 3 down vote accepted

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:

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

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