I know that the group velocity of a light pulse is defined as
$$\begin{split}v_g&=v_p\left(1+\frac{\lambda}{n}\frac{dn}{d\lambda}\right)\\
&=\frac{c}{n}\left(1+\frac{\lambda}{n}\frac{dn}{d\lambda}\right).\end{split} \tag{1}$$
On the other hand, it is also defined as
$$\begin{split}v_g &= \frac{c}{\left(n-\lambda\frac{dn}{d\lambda}\right)}\\
&=\frac{c}{n}\left(1-\frac{\lambda}{n}\frac{dn}{d\lambda}\right)^{-1}.\end{split}\tag{2}$$
That leads to the equation
$$0=\frac{\lambda^2}{n^2}\left(\frac{dn}{d\lambda}\right)^2.\tag{3}$$
After neither $\lambda$ nor $n$ is $0$, does that mean that $\left(\frac{dn}{d\lambda}\right)^2=0$? And how can I justify that?
Source: http://en.wikipedia.org/wiki/Group_velocity#Other_expressions
Alternative explanation: Wiki is crap?
Best Answer
The first equation correctly states that $$ v_g = \frac{c}{n}\left(1 + \frac{\lambda}{n}\frac{dn}{d\lambda}\right). $$ But if you look at the wikipedia page that you linked to, you'll see that the second equation should read $$ v_g = \frac{c}{n}\left(1 - \frac{\lambda_0}{n}\frac{dn}{d\lambda_0}\right)^{-1}, $$ where $\lambda_0$ is the wavelength in vacuum, and $$ n = \frac{\lambda_0}{\lambda}, $$ where $\lambda$ is the wavelength in the medium. The second equation can be derived from the first as follows: $$ \begin{align} v_g &= \frac{c}{n}\left(1 + \frac{\lambda}{n}\frac{dn}{d\lambda}\right)\\ &= \frac{c}{n}\left[1 + \frac{\lambda}{n}\left(\frac{d\lambda}{dn}\right)^{-1}\right]\\ &= \frac{c}{n}\left[1 + \frac{\lambda_0}{n^2}\left(\frac{d(\lambda_0/n)}{dn}\right)^{-1}\right]\\ &= \frac{c}{n}\left[1 + \frac{\lambda_0}{n^2}\left(-\frac{\lambda_0}{n^2} + \frac{1}{n}\frac{d\lambda_0}{dn}\right)^{-1}\right]\\ &= \frac{c}{n}\left[1 + \left(-1 + \frac{n}{\lambda_0}\frac{d\lambda_0}{dn}\right)^{-1}\right]\\ &= \frac{c}{n}\left[\frac{n}{\lambda_0}\frac{d\lambda_0}{dn}\left(-1 + \frac{n}{\lambda_0}\frac{d\lambda_0}{dn}\right)^{-1}\right]\\ &= \frac{c}{n}\left[\left(-\frac{\lambda_0}{n}\left(\frac{d\lambda_0}{dn}\right)^{-1} + 1\right)^{-1}\right]\\ &= \frac{c}{n}\left(1 - \frac{\lambda_0}{n}\frac{dn}{d\lambda_0}\right)^{-1}. \end{align} $$