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I am reading Schwartz's chapter on renormalizing the $\phi^4$ theory and I have two questions. We define the renormalized coupling to be the matrix element of all contributing diagrams at a given energy scale

$$\lambda_R=\lambda+\frac{\lambda^2}{32 \pi^2} \ln\frac{s_0}{\Lambda^2}+...\tag{15.65}$$

Now we want to get an expression $\lambda=\lambda(\lambda_R)$ to substitute this in the matrix element for general energies. To this end, Schwartz now writes a series expansion

$$\lambda=\lambda_R+a \lambda^2_R+...\tag{15.66}$$

Why can we do this?

Second, the resulting expression for $\mathcal{M}(s)$ reads

$$\mathcal{M}(s)=-\lambda_R-\frac{\lambda^2_R}{32 \pi^2}\ln\frac{s}{s_0}+...\tag{15.69}$$

He then writes

This equation gives us an expression for $\mathcal{M}(s)$ for any $s$ that is finite order-by-order in perturbation theory.

I dont really understand this claim. He showed that its finite at one-loop-order. Why is it clear from the preceding calculation that there won't be divergences at some higher order?

Edit: It can be found on page 298 in Schwartz "QFT and the SM"

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    $\begingroup$ Feynman diagrams can be expanded as a Taylor series in the external momentum. The polynomial representing the diagram will have divergent coefficients, reflecting the divergence of the integral, and we set external momentum to p = 0. QFT for the Gifted Amateur (p 289) covers the same ground as in your post. $\endgroup$ – StudyStudy Jun 21 '20 at 15:06
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The expression for $\lambda$ in terms of $\lambda_R$ is just perturbation theory. Given an expression for $\lambda_R$ in terms of $\lambda$ $$ \lambda_R = \lambda + \frac{\lambda^2}{32\pi^2}\ln\frac{s_0}{\Lambda^2}+... $$ we solve perturbatively, assuming that $\lambda$ can be written as a power expansion in $\lambda_R$. That is, we guess a solution of the form $$ \lambda = \sum_{i} a_i\lambda^i $$ plug it into our expression for $\lambda_R$ $$ \lambda_R = \left(\sum_{i} a_i\lambda^i\right) + \frac{\left(\sum_{i} a_i\lambda^i\right)^2}{32\pi^2}\ln\frac{s_0}{\Lambda^2}+... $$ then solve for the $a_i$ at each order in $\lambda_R$. At order 1, we get $$ \lambda_R = \lambda_R $$ indicating that $a_1 = 1$. Plugging this in, we go to order 2, getting $$ 0 = a_2\lambda_R^2+ \frac{\lambda_R^2}{32\pi^2}\ln\frac{s_0}{\Lambda^2}\longrightarrow a_2 = -\frac{1}{32\pi^2}\ln\frac{s_0}{\Lambda^2} $$ and so forth for the other orders.

For the loop divergences: Schwartz doesn't actually prove that higher orders don't diverge in this section. For that, check out 21.1.3 in Schwartz, where he extends the argument from one loop to multiple loops.

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