# High precision calculation in Mathematica

by brian0918
Tags: calculation, mathematica, precision
 Sci Advisor P: 905 It's not numerically efficient to deal directly with v/c when energy is very high. You should get into the habit of using the relevant expansions. So for example, $$\gamma={1\over\sqrt{1-\beta^2}}$$ where $\gamma=\mbox{Energy}/(mc^2)$, and $\beta=v/c$. Then $$\beta=\sqrt{1-1/\gamma^2}\approx 1-{1\over 2\gamma^2}$$ This gives you easily sufficient accuracy, and is very fast.