# $4$- vector Taylor expansion, sign confusion

I've been presented with a function expansion which I'm told is correct but I can't figure out where the sign in the second term might be coming from.

$$e^{i\alpha(x_\mu + \epsilon \, n_\mu)} = e^{i\alpha(x_\mu)} ( 1 - i\,\epsilon \,n^\nu\, \partial_\nu \alpha(x_\mu) )$$

$x_\mu$, $n_\mu$ are four vectors, the metric signature is + - - - and $\epsilon$ is infinitesimal.

Taylor expanding this myself about $x_\mu$ I get

$$e^{i\alpha(x_\mu + \epsilon \, n_\mu)} = e^{i\alpha(x_\mu)} ( 1 + i\,\epsilon \,n^\nu \,\partial_\nu \alpha(x_\mu) )$$

Am I missing something or is the presented equation wrong? Perhaps to do with the metric, or maybe they are doing something other than a Taylor expansion ? :-/

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I would trust your instincts. By what authority (book, important person, ...) you know that the formula is correct? – Fabian Mar 3 '11 at 18:21
Thanks Fabian, perhaps you are right. I really just wanted to check that I wasn't missing something 'obvious'. – fluXor Mar 4 '11 at 16:55