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This post is related to the answer given in Beta function in $\lambda_0\phi^4$ theory

The beta function calculus for that theory provides you of $$ \beta(\lambda_p) = - \frac{\epsilon \lambda_p + z\lambda_p^2}{1+2z\lambda_p/\epsilon} =z \lambda_p^2 +O(\epsilon)= \frac{3\lambda_p^2}{(4\pi)^2} \tag1 $$

But right hand side of Eq. (1) is obtained by Taylor expansion of the denominator making use of the already known formula:

$$ (1 + x)^r = \sum_{n = 0}^\infty {{r}\choose {n}}x^n \Leftrightarrow |x| \ll 1, \quad x, r \in \mathbb{R},\quad n\in \mathbb{N} \tag2 $$

But this is only true under the condition $|x| \ll 1$. In Eq. (1), $2z\lambda_p/\epsilon$ plays the role of $x$ in Eq. (2) but due to $\epsilon \rightarrow 0$ this quantity is infinite, not less that 1.

I understand that in principle you can make use of Eq. (2) into Eq. (1) because the $\epsilon$ limit is something that you apply only at the end of calculus, but that final result (still $\epsilon$ dependent) will be correct if only if the restriction coming out from Eq. (2) is fullfilled by the $\epsilon$ limit you take because Eq. (2) is only valid in that case. Otherwise, you are mathematically inconsistent and shouldn't use Eq. (2).

Therefore, why can we use (2) in (1)?

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  • $\begingroup$ The linked post just gives a squirrelly way to calculated $\beta$. You can do it perfectly well, using renormalized perturbation theory, without getting that kind of denominator. $\endgroup$ – Buzz Feb 27 at 2:45
  • $\begingroup$ I've seen this kind of computation of the beta function more than once and gives the correct result so there must be a reason. That's what I want to know $\endgroup$ – Vicky Mar 5 at 3:56

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