math bonus question: can you derive the Taylor expansion of the logarithm using the geometric series of $1/(1+x)$?

## Wednesday, May 29, 2013

### dark energy takes over...

what possibilities are there for defining the moment where the Universe switches from matter domination to dark energy domination? the two most obvious choices would be the instant $a$ where the density parameters are equal, $\Omega_m(a) = \Omega_\Lambda(a)$, or perhaps when the deceleration parameter changes sign, $q(a)=0$... how would you define it and what numerical values for $a$ would you get with the two above definitions?

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the first definition $\Omega_m(a)=\Omega_\Lambda(a)$ can be used together with the adiabatic relation $\Omega(a)/\Omega=H_0^2/a^{3(1+w)}/H(a)^2$ to give $a^3=\Omega_m/\Omega_\Lambda$. with $\Omega_m=1/4$ and $\Omega_\Lambda=3/4$ one arrives at $z\simeq0.4$.

ReplyDeleteAs for the math question (I will do this in some kind of latex as I don't know how to embed math here):

ReplyDeleteLet f(x) = (1+x)^{-1} which has geometric series (as confirmed by Taylor Expansion around x=0) of

f(x) = 1-x+x^2-x^3...

Then noting that d[ln(1+x)]/dx = (1+x)^{-1} we see that ln(1+x) = int[ (1+x)^{-1} ]

Integrating, we get

ln(1+x) = x-(x^2)/2+(x^3)/3-(x^4)/4+...

Which is the Taylor Expansion of ln(1+x) which can be easily verified. Note, ln(1+x) around x=0 as ln is undefined at x=0.

Thank you very much, Christopher, for providing the answer!

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