Is there a formal way of solving the equation $$x^x = \frac{1}{\sqrt{2}}\ ?$$The solutions are $x = \frac{1}{4},\frac{1}{2}$. This can be easily obtained by plotting the function or just by guessing the solutions. However, is there a general approach for handling this kind of equations?
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When you write "this kind of equations", what do you mean exactly? Equations of the form "$x^x = k$ for constants $k$? – Ulrik Dec 04 '14 at 00:57
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@Svinepels Yes, that's exactly what I mean. – Dec 04 '14 at 00:59
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Probably not. If $a$ is algebraic and $x^x=a$, then $x$ is rational or trascendental. See Gelfond-Schneier theorem.
That means that is impossible to express the solution of, say, $x^x=5$, with radicals.
ajotatxe
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Taking logs of both sides gets $x \log x=\log\frac{1}{\sqrt2}$, whose solution is $\frac{\log \frac{1}{\sqrt2}}{W(\frac{1}{\sqrt2})}$. W(x) is the productlog function. http://mathworld.wolfram.com/LambertW-Function.html
More generally, for $x^x=k$, the solution would be $x=\frac{log(k)}{W(k)}$
Teoc
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As answered by MathNoob, the solution of $x^x=k$ is given by Lambert function and the solution is $$x=\frac{\log (k)}{W(\log (k))}$$ So, if you have access to Lambert function, the problem is simple.
If not, the best is to solve for $x$ the equation $$f(x)=x \log(x)-\log(k)=0$$ and Newton method is one of the simplest root finder; starting from a reasonable guess $x_0$, the process will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ which will write $$ x_{n+1}=\frac{\log (k)+x_n}{\log (x_n)+1}$$
Claude Leibovici
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