Typical form of the beta function of the renormalization group Why in "typical" cases (according to some non-English text I read), does the $\beta$-function have the form
$$
\beta (g) = ag^{2} + bg^{3} + O(g^{4})\ ?
$$
I.e., why are there no linear or logarithmic terms?
 A: Typically, when you calculate quantum effects, you will put some cut-off $\Lambda$, and typical integrals, say for a $ \lambda \phi^4$ theory, are, at first order,  going in $\log \Lambda$ (ex : $\int \frac{d^4k}{k^2 (p-k)^2}$)
While defining renormalized quantities at some energy scale $p_0$, you may remove the cut-off, and get typically equations like (at first order) $: 
$$\lambda_r(p) = \lambda_r(p_0) + C \lambda_r(p)^2 \log (\frac{p}{p_0}) \tag{1}$$ 
where $C$ is some constant, and $p$ represents a energy scale.
So finally, at first order, typical beta functions go like $\beta(\lambda_r) = \frac{\partial \lambda_r}{\partial \log p} = C \lambda_r^2$.
Going back, we see that a $\log \lambda$ factor would seem strange, because the factor $\lambda_r^2$ in $(1)$ comes from the fact that one has $2$ vertex in the Feynman diagram. If we would have $4$ vertex, we would have a $\lambda_r^4$ factor, and so on. So, unless you are doing a fine-tuning infinite sum, you would not obtain a factor $\log(\lambda)$ or more generally a factor  $\lambda^l \log(\lambda)^m$.
You do not obtain a linear factor, because simply, the quantum corrections implies more vertex, so necessarily the first order quantum correction gives at least a $\lambda_r^2$ factor.
