Solution of the equation $\left(x+\frac{1}{x}\right)^{\frac{1}{x}}=A$

Question

Is it possible to solve analytically the following equation? $$\left(x+\frac{1}{x}\right)^{\frac{1}{x}}=A$$ with $A\gt 1$? I tried to transform it in the following: $\frac{1}{x}\ln\left(x+\frac{1}{x}\right)=B$ with $B=\ln(A)$, but it seems to be still unsolvable. Is there some trick to solve it? Thanks.

Answer 1

It can, of course, but not analytically, only numerically. It can't even be expressed in terms of the Lambert W function, since it is equivalent to $xA^x=x^2+1$, where the right hand side is variable, and not constant, as it would be needed.

Answer 2

Your equation is very interesting but, as said before, there is no analytical solution and everything must be done numerically. Using Newton iterative scheme is probably the simplest way providing a reasonable estimate of the solution. Another point to remember is that, more "linear" will be the function, better conditioned will be the path to solution.

I think that writing the equation as
$ f(x) = \ln\left(x+\frac{1}{x}\right) - x \ln(A)$
is probably the most convenient (you almost wrote it).

Starting with a guess (let us call it x_old), the Newton iterates will write
x_new = x_old - f(x_old) / f'(x_old).

For illustration purposes, I shall consider two cases :

In the first case, A=10^6. The solution is close to 0.2. Then let us start with x_old = 0.2 and start the Newton process. The successive iterates will be 0.139534, 0.142479, 0.142491; at this last point, f(x) = -9.1623*10^-6.

In the second case, A = 1.25. The solution is close to 10. Then let us start with x_old = 10 and start the Newton process. The successive iterates will be 10.6482, 10.6336; at this last point, f(x) = 4.46362*10^-6.

For sure, you can continue iterating until you reach the desired level of accuracy.