$x^x-x+5=\frac{29}{4}$

Question

A friend of mine is claiming to have a closed form solution to $x^x-x+5=\frac{29}{4}$, plotting it into wolfram alpha gives an approximation, and the equation doesn't seem very easy to solve. Can any of you help me finding a solution?

Answer 1

A friend of mine is claiming to have a closed form solution to $x^x-x+5=\frac{29}{4}$

I assume he means the rather obvious $x=-2$.

Plotting it into Wolfram Alpha gives an approximation.

I assume you are referring to the positive transcendental root, somewhere in the vicinity of $2$.

Can any of you help me finding a solution?

I believe we already did. :-)

Answer 2

The equation reduces to $$f(x)=x^x-x-\frac{9}{4}$$ and, by inspection $f(2)=-\frac{1}{4}$ while $f(3)=\frac{87}{4}$. So, as the plot shows it, there is a solution slightly above $x=2$.

A easy procedure to solve nonlinear equation is based on Newton method which, starting from a "reasonable" guess $x_0$, produces the following iteration scheme $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ Using $x_0=2$, the following iterates are generated :$2.04331$, $2.04127$, $2.04126$ which is the solution for six significant figures.

The other solution is "quite" obvious, corresponding to $x=-2$.