Hey jules027,

Captain has correctly specified what you have to do, and Mathstud28 continued with the definition of McLaurin series to help you.

$\displaystyle f(x) = \frac1{\sqrt{1-x^2}}$

Clearly f(0) = 1,

Then do a little computation to obtain f''(0) = 1.

Now from Mathstud28's formula

$\displaystyle f(x)=f(0)+f'(0)x+\frac{f''(0)x^2}{2!}+.....$

So $\displaystyle \frac1{\sqrt{1-x^2}} \approx 1 + \frac{x^2}{2!}$

Now $\displaystyle mc^2 = \dfrac{m_0 c^2}{\sqrt{1 - \frac{v^2}{c^2}}}\\\\\\\\\\ (*)$

But from the previous approximation if we let $\displaystyle x = \frac{v}{c}$, we get:

$\displaystyle \dfrac1{\sqrt{1- \frac{v^2}{c^2}}} \approx 1 + \frac{v^2}{2c^2}$

Substituting this in $\displaystyle (*)$ we get,

$\displaystyle mc^2 = m_0 c^2 \left(1 + \frac{v^2}{2c^2} \right)$

$\displaystyle mc^2 = m_0 c^2 + \frac{mv^2}{2}$

Done