(3-12)  We have the relativistic definition of kinetic energy:

①  Deduce the value of Ek when v=0.  This, at least, should be reassuring.

②  Because we are interested in what happens when v is small (compared to the speed of light), we get v2/c2 << 1. This means that it is useful to approximate the term that includes this as a Maclaurin series:  the value of a function, f(x), can be determined approximately – provided x is fairly close to zero - from the polynomial expansion:

Here f’ indicates the first derivative, f’’ is the second derivative etc.  In our case, we’re going to let x = v2/c2, and use the expansion to find an expression for 1/√(1 – v2/c2) in the limit where v is small. If v2/c2 is much less than 1, these terms rapidly become vanishingly small as the power of x increases, and hence we can get a good approximation to f(x) just from the first few of them.

(If you haven’t come across Maclaurin series before, they’re worth looking up. You can see the basic idea from the form of the expression: the first term gives the value at x=0, the second term adds a linear correction based on the gradient of the function at x=0, the third term makes a further correction by bringing in the curvature, the next term brings in the curvature of the curvature and so on).

To prepare for using this expansion, you’re going to find the first two derivatives of 1/√(1 – x)  (you could go on to higher derivatives, but there’s no need). Use the substitution u = (1 – x) to help you do this with the chain rule.

③  Now use these expressions to write down the first three terms in the Maclaurin series for 1/√(1 – v2/c2).   You should be able to see that, since v2/c2 << 1,  as it is raised to increasing powers the resulting terms rapidly becomes negligibly small. Therefore we can make a reasonable approximation to the full function by just taking the first two terms of the series. So take this truncated form and build it into the full equation for the kinetic energy (equation III-6): see what you get now.