I'm trying to calculating the four point function for the 3D Ising model. To do so I need to calculate the 3D Conformal Blocks . I found a paperpaper which has a recursive relation for calculating the Conformal Blocks https://link.springer.com/content/pdf/10.1007%2FJHEP06%282014%29091.pdf: equation (3.10) but. The authors are Filip Kos, David Poland and David Simmons-Duffinb, and the title is Bootstrapping the O(N) vector models. $$h_{\Delta,\ell}(r,\eta)=h_\ell^{(\infty)}(r,\eta)+\sum_i\frac{c_i r^{n_i}}{\Delta-\Delta_i}h_{\Delta_i+n_i,\ell_i}(r,\eta)\tag{3.10}$$
However, I am unsure how to actually implement it. I want to compute 3(3.10) up to r^12$r^{12}$. I imagine my expression is going to look like a sum of the h^infinity$h^{(\infty)}$ times coefficients with powers of r$r$. However I'm not so sure that is the correct interpretation of this recursion relation. If anyone can point me in the right direction that would be much appreciated.Also I do not know how many times I need to plug the right hand side into the left hand side. If I want to order 12 I imagine I do it 12 times.
The authors are Filip Kos, David Poland and David Simmons-Duffinb
and the title is Bootstrapping the O(N) vector models of the referenced paper.
As an example of what I am doing. I'll sum up to k=1$k=1$ and pretend I'm looking for order up to r^2$r^2$ only
$h(V,L)=\sum _{k=1}^{\frac{L}{2}} \frac{c(k)_3 r^{2 k} h(L+2,L-2 k)}{2 k-L+V-2}+\frac{c(1)_1 r^2 h(1-L,L+2)}{L+V+1}+\frac{c(1)_2 r^2 h\left(\frac{5}{2},L\right)}{V-\frac{1}{2}}+H(\infty ,L)$$$h(V,L)=\sum _{k=1}^{\frac{L}{2}} \frac{c(k)_3 r^{2 k} h(L+2,L-2 k)}{2 k-L+V-2}+\frac{c(1)_1 r^2 h(1-L,L+2)}{L+V+1}+\frac{c(1)_2 r^2 h\left(\frac{5}{2},L\right)}{V-\frac{1}{2}}+H(\infty ,L)$$
after iterating this once so I only get r^2$r^2$ terms I get
$h(1.4,0)=1.11111 c(1)_2 r^2 \left(\frac{2}{7} c(1)_1 r^2 h(1,2)+\frac{1}{2} c(1)_2 r^2 h\left(\frac{5}{2},0\right)+H(\infty ,0)\right)+0.416667 c(1)_1 r^2 \left(\frac{1}{4} c(1)_1 r^2 h(-1,4)+2 c(1)_2 r^2 h\left(\frac{5}{2},2\right)-c(1)_3 r^2 h(4,0)+H(\infty ,2)\right)+H(\infty ,0)$
and now expanding up to r^2$r^2$ yields
$h(1.4,0)=r^2 \left(0.416667 c(1)_1 H(\infty ,2)+1.11111 c(1)_2 H(\infty ,0)\right)+H(\infty ,0)$
all of the h's$h$'s from the rhs are gone and I have something in terms of $H(∞,2)$ which I know how to evaluate. Is this the correct approach, I'm asking because even though I'm doing it this way I'm getting utter nonsense for my answer. To get the Conformal Block I then multiply this by r^V$r^V$ which gives .08$.08$ instead of .6707$.6707$ I should be getting.