Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))

[bo198214]

It is well known that the (lower real) solution of \( x=y^{1/y} \) is
\( h(x)=y=\frac{W(-\ln(x))}{-\ln(x)} \) where \( W \) is the Lambert W function which is the solution of \( ye^y=x \) where \( W(x)=y \).
Short derivation:
\( x^y=y \)
\( e^{\ln(x)y}=y=y\ln(x)/\ln(x) \)
\( \ln(x)e^{\ln(x)y}=y\ln(x) \)
\( \ln(x)=y\ln(x)e^{-\ln(x)y} \)
\( -\ln(x)=-y\ln(x)e^{-\ln(x)y} \)
\( W(-\ln(x))=-y\ln(x) \)
\( y=\frac{W(-\ln(x))}{-\ln(x)} \)

Now the series development of the Lambert W function is well known to be:
\( W(z)=\sum_{n=1}^\infty \frac{(-n)^{n-1}}{n!} z^n \).
Then
\( h(z)=\frac{W(-\ln(z))}{-\ln(z)}=\sum_{n=1}^\infty \frac{(-n)^{n-1}}{n!} (-\ln(z))^{n-1}=\sum_{n=0}^\infty \frac{(n+1)^n}{(n+1)!} \log(z)^n \)
with \( \frac{(n+1)^n}{(n+1)!}=\frac{(n+1)^{n-1}}{n!} \) your formula is confirmed

However the radius of convergence of the \( W \) series is \( 1/e \) and hence the radius of convergence of \( h(z) \) is \( e^{1/e} \) but Gottfried was just interested in a solution for \( z>e^{1/e} \).

The radius of convergence is limited because the LambertW function has a singularity at \( -1/e \). It can however analytically continued past that singularity to values below \( -1/e \) which are then complex.

Such an analytic continuation is somewhat cumbersome.
You start with the development point \( a_0=0 \).
Then you compute the development at a point \( a_1 \) with \( |a_1-(-1/e)|<1/e \) then you compute the development at a point \( a_2 \) with \( |a_2-a_1|<|a_1-(-1/e)| \) and so on until you can reach an \( a_n<-1/e \).
See the attachment for a depiction.

A development \( \sum_{k=0}^\infty f_{n,k} (x-a_n)^k \) at point \( a_n \) turns into the development \( \sum_{k=0}^\infty f_{n+1,k} (x-a_{n+1})^k \) at point \( a_{n+1} \) via
\( f_{n+1,k} = \sum_{j=k}^\infty \left(j\\k\right) f_{n,j} (a_{n+1}-a_n)^{j-k} \).

For our task it should suffice to have 4 steps to reach a development at \( a_4=-\frac{2}{e} \).
\( a_n=-\frac{1}{e} + \frac{1}{e}e^{i\frac{\pi}{4}n}, n=0..4 \)

Taking 3 steps (each 60 degrees) would result in \( |a_{n+1}-a_n|=1/e \) which is not allowed.

So the four step solution would result in
\( f_{4,k} =
\sum_{j_4=k }^\infty\sum_{j_3=j_4}^\infty
\sum_{j_2=j_3}^\infty\sum_{j_1=j_2}^\infty
\left(j_4\\k\right)\left(j_3\\j_4\right)\left(j_2\\j_3\right) \left( j_1 \\ j_2 \right)
f_{0,j_4}a_1^{j_1-j_2}(a_2-a_1)^{j_2-j_3} (a_3-a_2)^{j_3-j_4} (a_4-a_3)^{j_4-k}
\)

But nobody wants to use this *lol*