Expansion of a potential in QFT 
Consider the $O(N)$ model in QFT. This is a collection of $N$ scalar fields $\boldsymbol{\phi} = ( \phi_{1}, \ldots, \phi_{N} )$ which have $O(N)$ symmetry. In $D$ dimensions the (Euclidean) action is:
$$S_{E}[\phi] = \int~\mathrm d^{D}x \left[ \frac{1}{2} ( \nabla \boldsymbol{\phi} )^{2}  + U \left( || \boldsymbol{\phi} ||^{2} \right) \right] $$
The potential $U(z)$ with $z = || \boldsymbol{\phi} ||^{2} $ is a well-behaved function.
We split apart the field such that $$\boldsymbol{\phi} = \boldsymbol{\phi}_{\mathrm{s}} + \boldsymbol{\phi}_{\mathrm{f}} ,$$ where $\boldsymbol{\phi}_{\mathrm{s}}$ is a "slow" component and $\boldsymbol{\phi}_{\mathrm{f}}$ is a "fast" component.
I've been told to fix $\boldsymbol{\phi}_{\mathrm{s}}$ and expand the potential to second order in $\boldsymbol{\phi}_{\mathrm{f}}$ in terms of derivatives $U'$ and $U''$.

I don't know how to do this. It would seem that I can write:
$$z = || \boldsymbol{\phi} ||^{2} = || \boldsymbol{\phi}_{\mathrm{s}} ||^{2} + 2 \boldsymbol{\phi}_{\mathrm{s}} \cdot \boldsymbol{\phi}_{\mathrm{f}} + || \boldsymbol{\phi}_{\mathrm{f}} ||^{2}$$
So that we have a function:
$$U \left( || \boldsymbol{\phi}_{\mathrm{s}} ||^{2} + 2 \boldsymbol{\phi}_{\mathrm{s}} \cdot \boldsymbol{\phi}_{\mathrm{f}} + || \boldsymbol{\phi}_{\mathrm{f}} ||^{2} \right)$$
My initial thought was to taylor expand this, but I have no idea how to deal with the $ 2 \boldsymbol{\phi}_{\mathrm{s}} \cdot \boldsymbol{\phi}_{\mathrm{f}}$ term. Can somebody help me out? I've got tunnel vision.
P.S. Does anyone know of a reference that examines the renormalization group flow for this model?
 A: Here is a suggestion. Use an auxiliary parameter for the "smallness" of $\boldsymbol{\phi}_{\mathrm{f}}$. Let's say $d$. Then you have
$$U \left( || \boldsymbol{\phi}_{\mathrm{s}} ||^{2} + 2 d \boldsymbol{\phi}_{\mathrm{s}} \cdot \boldsymbol{\phi}_{\mathrm{f}} + d^2 || \boldsymbol{\phi}_{\mathrm{f}} ||^{2} \right) . $$
Now just go ahead and do a Taylor series expansion in $d$. Afterwards, set $d=1$.
The Taylor series expansion in $d$ would be
$$ U \approx U|_{d=0} + d [\partial_d U]_{d=0} + \frac{1}{2} d^2 [\partial_d^2 U]_{d=0} . $$
Applying this to the expression of $U$, we get
$$ U \approx U \left(||\boldsymbol{\phi}_{\mathrm{s}} ||^{2}\right) 
+ 2 d \boldsymbol{\phi}_{\mathrm{s}} \cdot \boldsymbol{\phi}_{\mathrm{f}} U^{\prime} \left( || \boldsymbol{\phi}_{\mathrm{s}} ||^{2} \right) 
+ d^2 \left[ \left( \boldsymbol{\phi}_{\mathrm{s}} \cdot \boldsymbol{\phi}_{\mathrm{f}} \right)^2 U^{\prime\prime}\left(||\boldsymbol{\phi}_{\mathrm{s}} ||^{2}\right) + || \boldsymbol{\phi}_{\mathrm{f}} ||^{2} U^{\prime} \left( || \boldsymbol{\phi}_{\mathrm{s}} ||^{2} \right)  \right] . $$
Finally we can now set $d=1$. So the result is
$$ U \approx U \left(||\boldsymbol{\phi}_{\mathrm{s}} ||^{2}\right) 
+ \left( 2 \boldsymbol{\phi}_{\mathrm{s}} \cdot \boldsymbol{\phi}_{\mathrm{f}} + || \boldsymbol{\phi}_{\mathrm{f}} ||^{2} \right) U^{\prime} \left( || \boldsymbol{\phi}_{\mathrm{s}} ||^{2} \right) 
+ \left( \boldsymbol{\phi}_{\mathrm{s}} \cdot \boldsymbol{\phi}_{\mathrm{f}} \right)^2 U^{\prime\prime}\left(||\boldsymbol{\phi}_{\mathrm{s}} ||^{2}\right) . $$
