# Question about Ryder text (Generating functional)

The second equality in (6.88) he says was obtained by expanding the denomitator by the binomial theorem. It is probably very dumb but I'm not following. I see how the 1 and the vacuum term in the numerator cancel with the denominator and give a 1. But I don't follow how he got the rest.

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The expansion is $\frac{1}{1-x}=1+x+x^2+x^3+\cdots$ and you only get the linear term to order $g^1$. –  Michael Brown Feb 21 '13 at 12:53
I see, that answers my question. Thanks. You should have posted it as an answer tho. –  Barefeg Feb 21 '13 at 13:07
Done, with a little expansion. –  Michael Brown Feb 21 '13 at 13:12

## 1 Answer

The expansion is

$$\frac{1}{1-x} = 1 + x + x^2 + \cdots$$

where $x$ is the vacuum diagram. You only get the linear term to first order in $g$, which cancels the vacuum diagram in the numerator. There is a combinatorial proof that the cancellation of vacuum diagrams holds to all orders - Ryder should have it.

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Just completing with a reference, in case it's not in the Ryder, Peskin & Schroeder has the demonstration in the fourth chapter (the one developing the perturbation tools). –  Learning is a mess Feb 21 '13 at 13:36