# Omega constant

The omega constant is a mathematical constant defined by

${\displaystyle \Omega \,e^{\Omega }=1.}$

It is the value of W(1) where W is Lambert's W function. The name is derived from the alternate name for Lambert's W function, the omega function. The value of Ω is approximately

${\displaystyle 0.5671432904097838729999686622...}$ (sequence A030178 in the OEIS).

## Properties

### Fixed point representation

The defining identity can be expressed, for example, as

${\displaystyle \ln({\tfrac {1}{\Omega }})=\Omega .}$

or

${\displaystyle -\ln(\Omega )=\Omega }$

or

${\displaystyle e^{-\Omega }=\Omega .}$

Plot[{x, Exp[-x], -Log[x]}, {x, 0, 1}]

### Computation

One can calculate Ω iteratively, by starting with an initial guess Ω0, and considering the sequence

${\displaystyle \Omega _{n+1}=e^{-\Omega _{n}}.}$

init = 0.5;

FixedPoint[Exp[-#] &, init]

This sequence will converge towards Ω as n→∞. This convergence is because Ω is an attractive fixed point of the function ex.

It is much more efficient to use the iteration

${\displaystyle \Omega _{n+1}={\frac {1+\Omega _{n}}{1+e^{\Omega _{n}}}},}$

because the function

${\displaystyle f(x)={\frac {1+x}{1+e^{x}}}={\frac {1+x}{1+e^{-x}}}\cdot e^{-x}}$

has the same fixed point but features a zero derivative at this fixed point, therefore the convergence is quadratic (the number of correct digits is roughly doubled with each iteration).

### Integral representation

A beautiful identity due to Victor Adamchik[1][1] is given by the relationship

${\displaystyle \displaystyle \int \limits _{-\infty }^{\infty }{\frac {\,dt}{(e^{t}-t)^{2}+\pi ^{2}}}={\dfrac {1}{1+\Omega }}}$

or

${\displaystyle \Omega ={\frac {1}{\displaystyle \int \limits _{-\infty }^{\infty }{\tfrac {\,dt}{(e^{t}-t)^{2}+\pi ^{2}}}}}-1.}$

## Irrationality and transcendence

Ω can be proven irrational from the fact that e is transcendental; if Ω were rational, then there would exist integers p and q such that

${\displaystyle {\frac {p}{q}}=\Omega }$

so that

${\displaystyle 1={\frac {pe^{\left({\frac {p}{q}}\right)}}{q}}}$

and

${\displaystyle e=\left({\frac {q}{p}}\right)^{\left({\frac {q}{p}}\right)}={\sqrt[{p}]{\frac {q^{q}}{p^{q}}}}}$

The number e would therefore be algebraic of degree p. However e is transcendental, so Ω must be irrational.

Ω is in fact transcendental as the direct consequence of Lindemann–Weierstrass theorem. If Ω were algebraic, e−Ω would be transcendental; but Ω=exp(-Ω), so these cannot both be true.

1. ^ Adamchik, Victor. http://www-bcf.usc.edu/~adamchik/. Missing or empty |title= (help)