Chapter 44 - Quarks, Leptons, and the Big Bang - Problems - Page 1367: 51b

$\frac{\Delta \lambda}{\lambda} = \frac{r~\alpha}{c-r~\alpha}$

The detected wavelength is $~~\lambda' = \lambda+\lambda~\alpha~\Delta t$ Then: $\lambda' = \lambda+\lambda~\alpha~\Delta t$ $\lambda' - \lambda = \lambda~\alpha~\Delta t$ $\lambda' - \lambda = (\lambda~\alpha)~(\frac{r}{c-r~\alpha})$ $\frac{\lambda' - \lambda}{\lambda} = \frac{r~\alpha}{c-r~\alpha}$ $\frac{\Delta \lambda}{\lambda} = \frac{r~\alpha}{c-r~\alpha}$