How does the zeroth component of four-momentum reduce to kinetic energy and mass at low velocities?

Question

I'm currently reading Sean Carroll's book on general relativity. In the first chapter, he states something like:

In a moving frame, we can find the components of $p^\mu$ by performing a Lorentz transformation; for a particle moving with three-velocity $v = dx / dt$ along the $x$ axis, we have $$p^\mu = (\gamma m, v \gamma m, 0, 0)$$ where $\gamma = 1 / \sqrt{1 - v^2}$. For small $v$, this gives $p^0 = m + \frac{1}{2}mv^2$ (what we usually think of as rest energy plus kinetic energy)...

... How does having small $v$ lead to that, though? It looks to me that having a small $v$ means $\gamma = 1$, so $p^0 = m$.

Answer 1

A Taylor expansion of order 1:

$$\gamma = (1-v^2)^{-\frac{1}{2}} = 1 + \frac{v^2}{2} + O(v^4)$$

where $O(v^4)$ means terms at least of order $v^4$.