02/01/2008, 05:57 PM
I think I have to intervene and to clarify the basics.
1. There is the concept of a function in mathematics: We assign to every argument of the domain exactly one value of the codomain of the function. An example is \( f(x)=x^2 \) with the domain of the real numbers and also (!) \( f(x)=\sqrt{x} \) with the domain (and codomain!) being the positive real numbers. In the same way is \( f(x)=\ln(x) \) a function defined on the positive real numbers and especially \( f(x)={^\infty}x \) is a function defined on \( e^{-e}<x<e^{1/e} \).
Those above functions can be (uniquely) developed into power series (at points of their domain) and hence being continued somewhat outside the real line on the complex plane. This continuation is performed along a path and usually the continuation along two different paths gives the same value at the common end point, however in certain cases - if you start at point \( z \) wind the path around a singularity and finish again at \( z \) - the continuation value is different from the starting value, and these different values are to be said the branches of the function \( f \).
For example if you take the function (single valued!) \( f(x)=\sqrt{x} \), you can develop it at 1 by
\( \sqrt{z+1}=\sum_{n=0}^\infty \left(0.5\\n\right) z^n \).
Surely there is a (/the only) singularity at 0.
If you continue the function to -1 along a path above 0 you get \( +i \) and if you continue the function to -1 along a path below 0 you get \( -i \).
To avoid this scenario you slice the complex plane at the negative real axis. This means no path for continuation is allowed to cross the negative real axis, more precisely the argument \( \alpha(t) \) of such a path must be continuous and completely lie in \( -\pi < \alpha(t) \le \pi \).
With this sliced complex plane by continuation you get exactly one value of \( \sqrt{z} \) for each point on the complex plane, particularely you get \( \sqrt{-1}=i \).
From there you can get to the branches. You can continue the function around 0 and get the values \( e^{2\pi I k/2} \sqrt{z} = (-1)^k \sqrt{z} \) where \( k \) is the winding number of the path, i.e. how often it does anticlockwise cross the negative real axis (minus the clockwise crossings). This is the same as the number of discontinuities in \( \alpha(t) \) going from \( \pi \) to \( -\pi \) (minus the discontinuities going from \( -\pi \) to \( \pi \)).
For the logarithm you use also this sliced complex plane (which guaranties a unique value everywhere except 0) and for the branches you get \( \ln(z)+2\pi I k \) where \( k \) is again the winding number.
Now back to \( h(x)={^\infty}x \) this is a function (single valued!) on \( e^{-e}<x<e^{1/e} \). It can be developed there into a power series (did Ioannis derive a powerseries?) and hence continued to the complex plane.
It has (the only?) singularity at \( e^{1/e} \) and hence it would make sense to define it uniquely on the at \( [e^{1/e},\infty) \) sliced complex plane, more precisely a path for continuation \( p(t) \) must satisfy \( 0\le \text{arg}\left(p(t)-e^{1/e}\right)<2\pi \).
Starting fromt this definition \( h_k(z) \) would be the continuation for a path with winding number \( k \) in the above sliced plane.
However I am not completely sure if that defintion of sliced plane is compatible with the definition of the branches of the lambert function, such that we could write:
\( h_k(z)=\frac{W_k(-\ln(z))}{-\ln(z)} \)
or if we even should take into account the branches of the logarithm.
Perhaps someone takes pity on that somewhat tedious task.
1. There is the concept of a function in mathematics: We assign to every argument of the domain exactly one value of the codomain of the function. An example is \( f(x)=x^2 \) with the domain of the real numbers and also (!) \( f(x)=\sqrt{x} \) with the domain (and codomain!) being the positive real numbers. In the same way is \( f(x)=\ln(x) \) a function defined on the positive real numbers and especially \( f(x)={^\infty}x \) is a function defined on \( e^{-e}<x<e^{1/e} \).
Those above functions can be (uniquely) developed into power series (at points of their domain) and hence being continued somewhat outside the real line on the complex plane. This continuation is performed along a path and usually the continuation along two different paths gives the same value at the common end point, however in certain cases - if you start at point \( z \) wind the path around a singularity and finish again at \( z \) - the continuation value is different from the starting value, and these different values are to be said the branches of the function \( f \).
For example if you take the function (single valued!) \( f(x)=\sqrt{x} \), you can develop it at 1 by
\( \sqrt{z+1}=\sum_{n=0}^\infty \left(0.5\\n\right) z^n \).
Surely there is a (/the only) singularity at 0.
If you continue the function to -1 along a path above 0 you get \( +i \) and if you continue the function to -1 along a path below 0 you get \( -i \).
To avoid this scenario you slice the complex plane at the negative real axis. This means no path for continuation is allowed to cross the negative real axis, more precisely the argument \( \alpha(t) \) of such a path must be continuous and completely lie in \( -\pi < \alpha(t) \le \pi \).
With this sliced complex plane by continuation you get exactly one value of \( \sqrt{z} \) for each point on the complex plane, particularely you get \( \sqrt{-1}=i \).
From there you can get to the branches. You can continue the function around 0 and get the values \( e^{2\pi I k/2} \sqrt{z} = (-1)^k \sqrt{z} \) where \( k \) is the winding number of the path, i.e. how often it does anticlockwise cross the negative real axis (minus the clockwise crossings). This is the same as the number of discontinuities in \( \alpha(t) \) going from \( \pi \) to \( -\pi \) (minus the discontinuities going from \( -\pi \) to \( \pi \)).
For the logarithm you use also this sliced complex plane (which guaranties a unique value everywhere except 0) and for the branches you get \( \ln(z)+2\pi I k \) where \( k \) is again the winding number.
Now back to \( h(x)={^\infty}x \) this is a function (single valued!) on \( e^{-e}<x<e^{1/e} \). It can be developed there into a power series (did Ioannis derive a powerseries?) and hence continued to the complex plane.
It has (the only?) singularity at \( e^{1/e} \) and hence it would make sense to define it uniquely on the at \( [e^{1/e},\infty) \) sliced complex plane, more precisely a path for continuation \( p(t) \) must satisfy \( 0\le \text{arg}\left(p(t)-e^{1/e}\right)<2\pi \).
Starting fromt this definition \( h_k(z) \) would be the continuation for a path with winding number \( k \) in the above sliced plane.
However I am not completely sure if that defintion of sliced plane is compatible with the definition of the branches of the lambert function, such that we could write:
\( h_k(z)=\frac{W_k(-\ln(z))}{-\ln(z)} \)
or if we even should take into account the branches of the logarithm.
Perhaps someone takes pity on that somewhat tedious task.