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

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the present time of matter dominated universe for \Omega_{0}>1 is :
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}]
how to expand this at\Omega_{o}=1??help me
 
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This seems like a much more complicated formula than it should be for a matter-only universe. How did you come by it?
 
you can find this fomula in many books,say Kolb and Turner P53
 
knightq said:
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:

x = \Omega_0 - 1

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.
 
yes,let x=\((Omega_{0}-1)^{1/2}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
 
Hint:

Make:

<br /> 2 \, \Omega^{-1}_{0} - 1 = \cos{2 \, p} \Leftrightarrow \Omega_{0} = \frac{2}{1 + \cos{2 \, p}} = \frac{1}{\cos^{2}{p}}<br />

and

<br /> \Omega_{0} - 1 = \frac{1}{\cos^{2}{p}} - 1 = \tan^{2}{p}<br />

The limit \Omega_{0} \rightarrow 1 + 0 is equivalent to the limit p \rightarrow 0 (from any direction).
 
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.
 
I can get the right answer through Dickfore's method ,thank you!
 
cool. post it here for future generations to learn from your example.
 

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