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I must be a little bit confused on the Maclaurin series, because I can't write out the first few terms of the gamma factor $\gamma(\beta)=\frac{1}{\sqrt{1-\beta^2}}$, which should be $1+\frac{1}{2}\beta^2-\frac{3}{8}\beta^4\dots$

The Maclaurin series for $\gamma$, if I'm not mistaken, goes as follows:

$\gamma(\beta)=\gamma(o)+\gamma'(o)\beta+\frac{1}{2}\gamma''(o)\beta^2+\frac{1}{6}\gamma^{(3)}(o)\beta^3 + \dots$

So what do I get?

$\gamma(o)=1\\ \gamma'(o)\beta=\frac{1}{2}\cdot0(1-\beta^2)^{-1\frac{1}{2}}\beta=0\\ \frac{1}{2}\gamma''(o)\beta^2=1\frac{1}{2}\cdot0(1-\beta^2)^{-2\frac{1}{2}}\beta^2=0$

But wait - everything after the first term is getting zero? What am I doing wrong?

1 Answers 1

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If you are so interested in your error, it is that

$$\gamma''(\beta)=\frac{\sqrt{1-\beta^2}+3\beta^2}{(1-\beta^2)^{5/2}}\to1$$

You should instead use the following:

$$(a+b)^n=\binom n0a^n+\binom n1a^{n-1}b+\binom n2a^{n-2}b^2+\dots$$

where

$$\binom nk=\frac{n(n-1)(n-2)\dots(n-k)}{k!}$$

$$\binom n0=1$$

and substitute $a=1$, $b=-\beta^2$, and $n=-1/2$.

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    Thank you a lot for pointing out the error! I don't see why you brought up the binomial expansion, but my question is fully answered by your first remark. – 2017-01-14
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    @ShaVuklia ah, wonderful! I brought up the binomial expansion since its so much faster than taking derivatives directly. – 2017-01-14