Retain the specified leading poles and write

A = a_{1} / ϵ

B = b_{1} / ϵ^{2}

. Solving

g = g_{0} + A g_{0}^{2} + B g_{0}^{3}

recursively gives

g_{0} = g - A g^{2} + (2 A^{2} - B) g^{3} + O (g^{4}) .

The scale derivative must again be reexpressed in the renormalized coupling:

β (g) = - ϵ g_{0} (1 + 2 A g_{0} + 3 B g_{0}^{2}) = - ϵ g - ϵ A g^{2} + 2 ϵ (A^{2} - B) g^{3} + O (g^{4}) = - ϵ g - a_{1} g^{2} + \frac{2 ( a _{1}^{2} - b _{1} )}{ϵ} g^{3} + O (g^{4}) .

b_{1} = a_{1}^{2} .

The leading two-loop pole is fixed by the one-loop simple pole; it is not an independent parameter. This is an instance of the simple-pole formulas for minimal-subtraction renormalization group functions and their higher-pole consistency relations.

The source's leading-pole truncation does not fix the finite two-loop coefficient of the beta function. For example, adding the omitted term

c_{1} g_{0}^{3} / ϵ

leaves the double-pole condition unchanged but contributes

- 2 c_{1} g^{3}

to the finite beta function. Thus cancelling the displayed

1/ ϵ

term is not a claim that the physical two-loop running vanishes.

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