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When we do $1/N$ expansions in, say, 2+1$D$ $O(N)$ models and try to extract all kinds of critical exponents from it, we get the following results for the scaling dimensions of various operators up to order $1/N^2$ (Here $\phi$ is the vector field, $S$ is the scalar operator and $T$ is the symmetric traceless operator) $$\Delta_\phi = 0.5+\frac{0.135}{N}-\frac{0.097}{N^2}$$ $$\Delta_S = 2-\frac{1.081}{N}-\frac{3.048}{N^2}$$ $$\Delta_T = 1+\frac{1.081}{N}-\frac{0.195}{N^2}$$ We see that especially for $S$ the expansion is not very good since the coefficient for $1/N^2$ is very big and it will not give us a sensible scaling dimensions. Similar problems will be encountered in epsilon expansions. How do we make sense of these results beyond leading orders?

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  • $\begingroup$ One can use suitable resummation methods. E.g. the simple fact that for small $N$ an expansion in positive powers of $N$ also exists (even if these expansion coefficients are not known), means that you can re-expand the series by putting $\frac{1}{N} = \lambda \frac{u}{1-u}$ where $\lambda$ is an arbitrary parameter that can be optimized by choosing it such that the last known term of the expansion becomes zero. $\endgroup$ – Count Iblis Feb 6 at 20:25
  • $\begingroup$ As shown by Zinn-Justin, this method yields rapidly converging approximations in the sense that with more and more terms and summing till the last term set to zero using $\lambda$, makes the error in the approximation smaller and smaller. $\endgroup$ – Count Iblis Feb 6 at 20:26
  • $\begingroup$ Thank you very much! Do you have a reference for that? $\endgroup$ – Weicheng Ye Feb 6 at 20:28
  • $\begingroup$ You can check out arxiv.org/abs/1001.0675 and references in there. $\endgroup$ – Count Iblis Feb 6 at 20:31

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