I need to expand the equation:
$$E^2 = p^2c^2+m_0^2c^4$$
to the fourth term according to the momentum (p)
My question is, usually when expanding something in a Taylor series, I am given that something is much smaller relative to something else, allowing me to use the series. Here I am just told to expand, and would appreciate some direction
On
It is helpful to think in terms of dimensional analysis: The only independent physical quantity in this problem which has units of momentum besides $p$ is $m_0c$. Hence for this expansion to make sense we should really be expanding in powers of $p/m_0c$. To make this more familiar, note that in special relativity we have $$\frac{p}{m_0 c}=\frac{\gamma m_0 v}{m_0c}=\frac{v/c}{\sqrt{1-v^2/c^2}}.$$ For this to be small, we need $v/c$ to be small as well i.e. the system is non-relativistic for speeds much smaller than the speed of light.
For a particle of mass $m_0$, this equation can be rewritten as
$$ E = m_0c^2 \sqrt{1 + \left(\frac{p}{m_0 c} \right)^2} $$
The expansion you're looking for is in the variable $x = p/m_0c$, which is basically the speed of the particle in units of $c$, $x = v/c$, so the only thing you need is to know how to expand $\sqrt{1 + x^2}$:
$$ \sqrt{1 + x^2} \approx 1 + \frac{x^2}{2} - \frac{x^4}{8} $$
So that
\begin{eqnarray} E &\approx& m_0c^2\left[1 + \frac{1}{2}\left(\frac{p}{m_0 c} \right)^2 - \frac{1}{8}\left(\frac{p}{m_0 c} \right)^4\right] \\ &=& m_0c^2 + \frac{1}{2}m_0v^2 - \frac{1}{8c^4}m_0v^4 \end{eqnarray}