angular dependence of $f_{1}$ is given by $f_{1}=cos(4\theta)g(r,\rho).$ The function $g$ satisfies

This equation admits a power series solution. To set up the solution, notice that both $r$ and $\rho$ have the dimensions of length ($L$). It is not difficult to see that $g$ has dimension $L^{-8}$. Thus, by dimensional analysis, it must have an expansion of the form

For consisteny, we require that $g\to 0$ at least as $r^{-4}$ as $ r\to\infty$. If this is not the case, $f_{1}$ is not a small correction to $f_{0}$. Thus, we restrict $m\ge 4$ in . With this restriction, after inserting  into , one finds for the first few $c_{m}$:

$${\eqalign{c_{m}&=0,\qquad m<8,\qquad
c_{8}=\alpha m_{1}m_{2}m_{3}m_{4}N,\cr
c_{9}&=0\qquad
c_{10}=-6m_{1}m_{2}m_{3}m_{4}N\Big[
{1+\alpha\over (\tau_{2cl})^{3}}
\Big].}}$$

The full solution is not needed, since only $f_{1}$ at $y=0$ enters the probe action. Notice that this solution for $f_{1}$ is labeled by an arbitrary parameter $\alpha$ which cannot be fixed by the above iterative calculation. It is an interesting open question to see if $\alpha$ can be fixed by a more sophisticated analysis . The correction to the leading term in $f$

$$f(y=0,a,\bar{a})={4N\pi\over (\tau_{2cl}a\bar{a})^{2}}+
\alpha m_{1}m_{2}m_{3}m_{4}N{1\over 2(\bar{a}a)^{2}}
\Big({q\over a^{4}}+{\bar{q}\over\bar{a}^{4}}\Big)
+O\Big({1\over |a|^{12}}\Big)$$

represents a correction to the AdS$_{5}$ geometry. This correction is 