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value of one omega

value of one omega

Grade:11

2 Answers

Gayatri Jayesh Bondriya
521 Points
8 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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Gayatri Jayesh Bondriya
521 Points
8 years ago
 if  you   satisfy  with   my  answe  so please   just click on approve  button  ...and  if  you  have  any  other  question   or  dought     keep asking

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