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I was reading about the Taylor expansion of the scale factor from visser,2004

He writes:

$$ \frac{a(t)}{a_0} = 1 +H_0 (t -t_0) - \frac{q_0}{2} {H_0}^{2}(t-t_0)^2 + \frac{j_0}{3 !} {H_0}^{3}(t-t_0)^3 + \\ + \frac{s_0}{4 !} {H_0}^{4}(t-t_0)^4 + \frac{l_0}{5!} {H_0}^{5}(t-t_0)^5 + \mathcal{O}(t- t_0)^6 $$

where $q,j$ and $s$ are the deceleration, the jerk and the snak parameters. But I would like to see it expanded in terms of the redshift $z$. I suppose I should use the fact that:

$$a(t) = \frac{1}{1+z} \ ,$$

but how do I transform the dependence from $t$ to $z$?

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The answer is right there.

$$ \frac{a(t)}{a(t_0)} = (1+z)^{-1} = 1-z+z^2-z^3 + \cdots $$

which converges for $z<1$.

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    $\begingroup$ $(1+z)^{-1} $ will converge for any positive value of $z$ $\endgroup$
    – KP99
    Commented Apr 29, 2023 at 15:12
  • $\begingroup$ No, the radius of convergence is $1$. $\endgroup$
    – Avantgarde
    Commented Apr 30, 2023 at 4:03

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