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The spatial part of the FRW metric can be written as $$d\Sigma^2=d\rho^2+f^2(\rho)(d\theta^2+{sin}^2\theta d\phi^2)$$ where $f(\rho)$ satisfies $$\frac{df}{d\rho}=\frac{f(2\rho)}{2f(\rho)}.$$ I am trying to derive the form of $f(\rho)$ by using a power series expansion $f(\rho)=\sum a_n \rho^n$ and show that $f(\rho)$ can be $\rho$, $R\sin(\rho/R)$ or $R\sinh(\rho/R)$. I am getting stuck.

What should be further steps?

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Should be:

$f(\rho) = \sum a_n \rho^n$
$df/d\rho = \sum n a_n \rho^{n-1}$
$f(2 \rho) = \sum a_n (2 \rho)^n = \sum 2^n a_n \rho^n$

You plug what above in the equation, you multiply L.H.S. and R.H.S. times the expression for $f(\rho)$ and move all the terms to the L.H.S. Then order the terms as per powers of $\rho$, e.g. $\rho^0$, $\rho^1$, $\rho^2$, etc. For each power you equate the coefficient to zero. You should get a recurring formula relating $a_n$ to $a_{n-1}$.

DERIVATION
$df/d\rho = f(2 \rho) / (2 f(\rho))$
$(df/d\rho) (2 f(\rho)) = f(2 \rho)$
$(\sum_n n a_n \rho^{n-1}) 2 (\sum_m a_m \rho^m) - \sum_l 2^l a_l \rho^l = 0$
$\sum_n \sum_m 2 n a_n a_m \rho^{n-1+m} - \sum_l 2^l a_l \rho^l = 0$
Equating the coefficients of power $l$ requires $n - 1 + m = l$, that is $m = l + 1 - n$, hence
$a_l = 2^{-(l - 1)} \sum_{n = 0}^{l + 1} n a_n a_{l + 1 - n}$

Solution I
$l = 0, n = 0, 1$
$a_0 = 2 a_1 a_0$
$a_0 (2 a_1 - 1) = 0$
Let us consider $a_0 = 0$ and $a_1$ undefined

$l = 1, n = 0, 1, 2$
$a_1 = a_1^2$
$a_1 (a_1 - 1) = 0$
$a_1 = 1$ defined

$l = 2, n = 0, 1, 2, 3$
$a_2 = 2^{-1} (a_2 + 2 a_2)$
$a_2 = 0$

$l = 3, n = 0, 1, 2, 3, 4$
$a_3 = 2^{-2} (a_3 + 3 a_3)$
$a_3$ undefined

If we choose $a_3 = 0$, then the equation for $a_l$ with $l \ge 4$ provides a zero value if the previous coefficients up to $a_{l - 1} $ are zero, except for $a_1$.

To summarize we got the following solution:
$a_1 = 1$
$a_l = 0$ for $l \ne 1$
That is $f (\rho) = \rho$
It is the zero curvature geometry of FRW metric, i.e. a flat universe.

Note: To look for the positive/negative curvature solutions you have to exploit the fact that $a_3$ is undefined. You assume it is non zero and then proceed. However it is more laborious.

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  • $\begingroup$ I used the same method. I am getting stuck. Could you please include a small part of the derivation? $\endgroup$ – Tejas P Nov 15 '18 at 11:15
  • $\begingroup$ I edited my answer adding the section DERIVATION. $\endgroup$ – Michele Grosso Nov 17 '18 at 17:05

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