Solving Cosmology Problem on Page 12 of Damtp.cam.ac.uk

[latentcorpse]

I'm looking through the solution to a problem and am stuck with some of the maths.
The solution can be found here:
http://www.damtp.cam.ac.uk/user/r.ribeiro/supervisions_files/Cosmo1sol_M2010.pdf

The pages are numbered in the bottom right corner. I am on p12.

We have an expression $1+z \simeq \frac{1}{\dots}$ half way down the page. I can get to here fine.

Then she solves for z. How has she done this? Is it binomial? If so how has that worked? We have two terms here not just the usual $(1+x)^n$

And there is a new term that appears when we solve for z: $H_0^2(t-t_0)^2$. Where's that from?

Then under that we "invert" to find $t-t_0$ Iliterally have no idea what has happened here!

Thanks for any help!

 

[fzero]

The first part is just a Taylor expansion in powers of $$t-t_0$$. You can use the binomial expansion to do this, but if that's giving you problems just compute the necessary derivatives.

The "inversion" is just solving a quadratic equation. Did you even try working this part out for yourself before posting?

 

[latentcorpse]

The first part is just a Taylor expansion in powers of $$t-t_0$$. You can use the binomial expansion to do this, but if that's giving you problems just compute the necessary derivatives.

The "inversion" is just solving a quadratic equation. Did you even try working this part out for yourself before posting?

Hmm. well say you have $(1+x)^{-1}=1-x$ where for us $x=H_0(t-t_0) - \frac{1}{2} q_0 H_0^2(t-t_0)^2$

then $(1+x)^{-1}= 1-H_0(t-t_0) - \frac{1}{2} q_0 H_0^2(t-t_0)^2$ but I'm missing a term?And then for the next bit i think i was confused by the word "inverting". Nonetheless, it's a quadratic so we should be able to use $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

In the form $(1+\frac{1}{2}q_0 ) H_0^2 x^2 - H_0 x - z = 0$ with $x=t-t_0$

we get $x = \frac{H_0 \pm \sqrt{H_0^2 + 4z(1+\frac{1}{2}q_0)}}{(2+q_0)H_0^2}$
which doesn't look very promising

 

[fzero]

Hmm. well say you have $(1+x)^{-1}=1-x$ where for us $x=H_0(t-t_0) - \frac{1}{2} q_0 H_0^2(t-t_0)^2$

then $(1+x)^{-1}= 1-H_0(t-t_0) - \frac{1}{2} q_0 H_0^2(t-t_0)^2$ but I'm missing a term?

You have to include terms of order $$x^2$$ to obtain all of the terms of order $$(t-t_0)^2$$.

And then for the next bit i think i was confused by the word "inverting". Nonetheless, it's a quadratic so we should be able to use $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

In the form $(1+\frac{1}{2}q_0 ) H_0^2 x^2 - H_0 x - z = 0$ with $x=t-t_0$

we get $x = \frac{H_0 \pm \sqrt{H_0^2 + 4z(1+\frac{1}{2}q_0)}}{(2+q_0)H_0^2}$
which doesn't look very promising

They've expanded the square root in powers of $$z$$.