Expanding Present Time Universe at Omega 0 = 1: Help Needed

[knightq]

the present time of matter dominated universe for [tex]\Omega_{0}>1[/tex] is :
[tex]t_{0}=H_{0}^{-1}\frac{\Omega_{0}} {2(\Omega_{0}-1)^{3/2}}[cos^{-1}(2\Omega_{0}^{-1}-1)-\frac{2}{\Omega_{0}}(\Omega_{0}-1)^{1/2}][/tex]
how to expand this at[tex]\Omega_{o}=1[/tex]??help me

 

[Chalnoth]

This seems like a much more complicated formula than it should be for a matter-only universe. How did you come by it?

 

[knightq]

you can find this fomula in many books,say Kolb and Turner P53

 

[Chalnoth]

you can find this fomula in many books,say Kolb and Turner P53

Hmm, okay. I'll trust that you have it correct, then.

A first trick is just to make a simple substitution:

[tex]x = \Omega_0 - 1[/tex]

This doesn't change the problem, just makes it easier to work with.

Now, the derivatives are obviously going to get a little bit messy, but whenever you find yourself running into the problem where you have a fraction with both numerator and denominator equal to zero, you merely make use of l'Hôpital's Rule to find the result.

 

[knightq]

yes,let x=[tex]\((Omega_{0}-1)^{1/2}[/tex]will be more simple.the difficulty is we can't expand cos^-1(x) at x=1,the caculation through l'hospital is too difficult ,we may differentiate 3 times

 

[Dickfore]

Hint:

Make:

[tex]
2 \, \Omega^{-1}_{0} - 1 = \cos{2 \, p} \Leftrightarrow \Omega_{0} = \frac{2}{1 + \cos{2 \, p}} = \frac{1}{\cos^{2}{p}}
[/tex]

and

[tex]
\Omega_{0} - 1 = \frac{1}{\cos^{2}{p}} - 1 = \tan^{2}{p}
[/tex]

The limit [itex]\Omega_{0} \rightarrow 1 + 0[/itex] is equivalent to the limit [itex]p \rightarrow 0[/itex] (from any direction).

 

[mathman]

Be careful: arccos(x) is not analytic around x=1. To first order it behaves like √(1-x), so that it is real for x < 1 and imaginary for x > 1.

 

[knightq]

I can get the right answer through Dickfore's method ,thank you!

 

[Dickfore]

cool. post it here for future generations to learn from your example.