From this article: http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/releng.html#c6
It shows how to derive the classical formula for Kinetic energy from Relativistic Kinetic energy. I've started at the solution for probably more than an hour, however my maths isn't strong enough to jump from this line to the next one here:
$$ \left(1-\frac{v^2}{c^2}\right)^{-1/2} = 1 + \frac{1}{2}\frac{v^2}{c^2}+ \frac{\frac{-1}{2}\frac{-3}{2}}{2!}\frac{v^4}{c^4}+\cdots $$ $$ KE = \frac{1}{2} m_0 v^2 + \frac{3}{8} m_0 \frac{v^4}{c^2} + \frac{5}{16}m_0 \frac{v^6}{c^4}+\cdots $$
If someone could show this mathematically, the denominator become too hard to rearrange for me. Thanks
From the top line, we have $$ KE = m_0 c^2 \left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1\right) $$ Using the binomial expansion, as Alexey Burdin noted with $a=1, x =-v^2/c^2,$ and $n=-1/2$, we have $$ \left(1-\frac{v^2}{c^2}\right)^{-1/2} = 1 + \frac{1}{2}\frac{v^2}{c^2}+ \frac{\frac{-1}{2}\frac{-3}{2}}{2!}\frac{v^4}{c^4}+\cdots $$ $$ \left(1-\frac{v^2}{c^2}\right)^{-1/2} -1= \frac{1}{2}\frac{v^2}{c^2}+ \frac{3}{8}\frac{v^4}{c^4}+\frac{5}{16}\frac{v^6}{c^6}+\cdots $$Multiply this through by $m_0c^2$: $$ KE = m_0 c^2\left(\left(1-\frac{v^2}{c^2}\right)^{-1/2} -1\right)= m_0 c^2\left(\frac{1}{2}\frac{v^2}{c^2}+ \frac{3}{8}\frac{v^4}{c^4}+\frac{5}{16}\frac{v^6}{c^6}+\cdots\right) $$ $$ \approx \frac{1}{2} m_0 v^2, $$for $v<<c$.