I'm having trouble in understeanding a mathematical feature of RG, namely how it provides a way to resum the perturbation series and how that's defined mathematically.

From a conceptual point of view one applies the RG flow to a 'theory' from a scale $\Lambda$ down to an energy scale $\mu$ and then reparametrizes it with a finite set of parameters more suitable for describing the physics at that energy scale (renormalization) $\{g^i_R(\mu)\}_{i=1........N}$, the physics must be indipendent of the parametrization we choose or equivalently it must be indipendent of the energy scale $\mu<<\Lambda$ which defines the parametrization, so given a set of observables $\{O_i\}_{i=1......M}$ it must hold (near a fixed point where the RG can be diagonalized): $$\frac{d\ O_i(g,\mu) }{d\log(\mu)}=(\frac{\partial}{\partial log(\mu)}-\beta(g)\frac{\partial}{\partial g}\ )\ O_i=0 \\ \beta(g)=-\ \frac{d \ g}{d \ log(\mu)}$$

($log\ \mu$ derivative so we don't introduce any new energy scale) Now if we calculate perturbatevly $\beta=\sum_n g^n\alpha_n\approx -\alpha g^2$ we get : $$log(\mu'/\mu)=\int \frac{dg}{ \alpha g^2}$$ we get the well known leading logarithm resummation: $$g(\mu')=\frac{g(\mu)}{1-\alpha \ g(\mu) \ log(\mu'/\mu)}$$ So if now we expand in a perturbative series $O_i$ at every order in $g(\mu')$ we have a resummation of the leading logs.

Moreover, consider $O_i(p/\mu,g(\mu))=\sum_n g^n(\mu)\Omega_n(p/\mu)$ in order to obtain a resummation the perturbative series must be redefined in terms of a new expansion parameter $g(\mu')=g_p=f(g(p),p/\mu)$ where $f$ is an exact soultion of the RG equation, such that (using RG flow invariance of the physical quantities):$$O_i(p/\mu,g(\mu))=O_i(1,g_p)=\sum_nf(g(p),p/\mu)^n\Omega_n(1)=\sum_n g_p^n \ \Omega_n(1)$$ Where the coefficents $\Omega_n(1)$ are free from large logarithms problems too. Now $g_p=g'=f(g,\mu'/\mu)$ comes from the implicit equation: $$log(\mu'/\mu)=\int_g^{g'} \frac{d\tilde{g}}{\beta(\tilde{g})}$$ such that $f$ obeys a group like law following from the reparametrization invariance.

My problem now is (provided what i said before is correct and i apologize for the lenghty premise) how this redefinition of the perturbative series works? i.e. Why if i calculate $\beta(g)$ for a few terms i get a complete resummation of lelading,sub leading, sub sub leading terms and so on? Is there a way to define those concept of resummation and redefinition of the perturbative series in a clear and precise way which then can be applied to this particular case?

EDIT: Thank to octonion for the nice answer which cleared half of my question. i.e. how given the $\beta$ function one is able to partially resum $f$.

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    $\begingroup$ I think you are not getting answers because it is not clear enough from your question what exactly you don't understand. For instance if you keep only the order $g^2$ term in $\beta$ you seem to understand that it gives you the leading logarithms, so what is the question exactly? $\endgroup$ – octonion Jan 11 '16 at 0:44
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    $\begingroup$ Indeed, the (perturbative) RG does not sum all higher order terms, it sums classes of higher order terms such as large logs $[g^2\log(Q^2)]^n$. $\endgroup$ – Thomas Jan 11 '16 at 0:49

Here's an attempt at clarification. We can expand the exact solution $g'=f(g,\mu'/\mu)$ as a power series in $g$

$$f(g,\mu'/\mu)=g+f_1(\mu'/\mu)\,g^2 + f_2(\mu'/\mu)\,g^3+\dots$$

Now the idea is we can shift our renormalization scale from $\mu$ to $\mu''$

$$g'=f(g,\mu'/\mu)=f(g'',\mu'/\mu'').\quad (1)$$

$g''$ itself can be expanded as a power series in $g$

$$g''=g+f_1(\mu''/\mu)\,g^2 + f_2(\mu''/\mu)\,g^3+\dots$$

Expand both sides of equation (1) as a power series in $g$ and equate the coefficients. At order $g^2$: $$f_1(\mu'/\mu)=f_1(\mu''/\mu)+f_1(\mu'/\mu'')$$ This implies $f_1(x) = \alpha_1 \log x$ for some parameter $\alpha_1$.

At order $g^3$ in (1): $$f_2(\mu'/\mu)=f_2(\mu''/\mu)+2 f_1(\mu''/\mu)f_1(\mu'/\mu'')+f_2(\mu'/\mu''),$$ which implies $f_2(x) = (\alpha_1 \log x)^2+\alpha_2\log x$ for some new parameter $\alpha_2$. You can continue the procedure and figure out each $f_n(x)$ as a polynomial in $\log x$ getting one new free parameter at each order. If you work it out you can convince yourself:

$$f_3(x)=(\alpha_1 \log x)^3 +\frac{5}{2}\alpha_1\alpha_2(\log x)^2 + \alpha_3 \log x,$$ and in general $$f_n(x)=(\alpha_1 \log x)^n +\dots + \alpha_n \log x$$ where the coefficients of the intermediate powers of $\log x$ in the "$\dots$" are a bit more difficult to figure out.

Now what's the connection with the $\beta$ function? $$\beta(g) = -\left(\frac{\partial f}{\partial \log \mu'}\right)_{\mu'=\mu}\quad (2)$$ The only terms in the $f_n$ that survive when we take $\mu'=\mu$ are the first order $\log x$ terms. So the parameters we introduced above are just the coefficients in the power series for $\beta$ $$\beta(g) = -\alpha_1 g^2 -\alpha_2 g^3 + \dots - \alpha_n g^{n+1} + \dots$$

So if we terminate the power series for $\beta$ at some finite order and integrate (2) to find $f$ we will get a power series to all orders in $g$ since for instance $\alpha_1$ appears in every $f_n$.

But even so this is not the full power series and may not even be a good approximation of the coefficients $f_n$ for large powers $n$. If $\log x$ is very small (if $\mu'$ is close to $\mu$) then it is the $\alpha_n \log x$ term which is dominant.


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