Algebra> value of one omega...

value of one omega

Abhay Shivansh Bhardwaj , 10 Years ago

Grade 11

2 Answers

Gayatri Jayesh Bondriya

Last Activity: 10 Years ago

The omega constant is a mathematical constant defined by

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

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 0.5671432904097838729999686622... It has properties that

$e^{-\Omega}=\Omega,\,$

or equivalently,

$\ln \Omega = - \Omega.\,$

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

$\Omega_{n+1}=e^{-\Omega_n}.\,$

This sequence will converge towards Ω as n→∞. This convergence is due to the fact that Ω is an attractive fixed point of the function e^−x.

It is much more efficient to use the iteration

$\Omega_{n+1} = \frac{1+\Omega_n}{1+e^{\Omega_n}},$

because the function

$f(x) = \frac{1+x}{1+e^x},$

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).

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

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

«»^[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

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

so that

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

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

and 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.

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