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Grade: 11
value of one omega
4 years ago

Answers : (2)

Gayatri Jayesh Bondriya
521 Points

The omega constant is a mathematical constant defined by


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


or equivalently,

 \ln \Omega = - \Omega.\,

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


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 ex.

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 would be transcendental; but Ω=exp(-Ω), so these cannot both be true.

….….….….…..if  you  like  my  advise just click on approve  button  ...and  please type  for any other  quary  if  you  have,,,,,,,,,,,,,,,,,,,

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