\section{The model neuron}
Central to the comparison of the accuracy of the traditional and
generalised compartmental models is the construction of a typical
branched neuron for which the mathematical model has a closed form
expression for the membrane potential in response to input. This
solution then stands as a reference against which the performance
of the traditional and generalised compartmental models may be
measured. The most effective way to construct a branched model
neuron with a closed form solution for the membrane potential is
to choose the radii and lengths of its sections such that the Rall
conditions for an equivalent cylinder are satisfied (Rall,
\cite{Rall64}). The Rall conditions require that at any branch
point the sum of the three-halves power of the diameters of the
child limbs is equal to the three-halves power of the parent limb,
and that the total electrotonic length from a branch point to
dendritic tip is independent of path. In particular, the
electrotonic distance from soma-to-tip is independent of path. The
model neuron illustrated in Figure \ref{TestNeuron} satisfies
these conditions. When the Rall conditions are satisfied, the
effect at the soma of any configuration of input on the branched
model of the neuron is identical to the effect at the soma of the
unbranched equivalent cylinder with biophysical properties and
configuration of input determined uniquely from those of the
original branched neuron (Lindsay \emph{et al.},
\cite{Lindsay03}).

\begin{figure}[!h]
\[
\begin{array}{c}
$\begin{mfpic}[1][1]{0}{220}{-20}{220}
\pen{2pt}
\dotsize=1pt
\dotspace=3pt
\lines{(-5,100),(5,110),(15,100),(5,90),(-5 ,100)}
% Upper dendrite
% Root branch
\dotted\lines{(5,115),(15,170),(20,170)}
\lines{(20.0,160),(36.7,160)}
\tlabel[tc](28.4,150){\textsf{(a)}}
% Level 1
\lines{(50.0,190),(88.3,190)}
\tlabel[bc](75,200){\textsf{(c)}}
\lines{(50.0,130),(91.0,130)}
\tlabel[tc](75,120){\textsf{(d)}}
\dotted\lines{(36.7,160),(45,200),(55,200)}
\dotted\lines{(36.7,160),(45,120),(55,120)}
% Level 2
\lines{(100.0,210),(153.2,210)}
\lines{(100.0,190),(153.2,190)}
\lines{(100.0,170),(153.2,170)}
\tlabel[cl](160,210){\textsf{(g)}}
\tlabel[cl](160,190){\textsf{(g)}}
\tlabel[cl](160,170){\textsf{(g)}}
\dotted\lines{(88.3,190),(95,220),(105,220)}
\dotted\lines{(88.3,190),(95,160),(105,160)}
\lines{(100.0,140),(165.1,140)}
\lines{(100.0,120),(165.1,120)}
\dotted\lines{(91.0,130),(95,150),(105,150)}
\dotted\lines{(91.0,130),(95,110),(105,110)}
\tlabel[cl](175,140){\textsf{(h)}}
\tlabel[cl](175,120){\textsf{(h)}}
%
% Lower dendrite
% Root branch
\lines{(20.0,40),(58.0,40)}
\dotted\lines{(5,85),(15,30),(25,30)}
\tlabel[bc](39,50){\textsf{(b)}}
% Level 1
\lines{(70.0,70),(133.1,70)}
\lines{(70.0,10),(127.1,10)}
\dotted\lines{(58,40),(66.5,80),(76.5,80)}
\dotted\lines{(58,40),(66.5,0),(76.5,0)}
\tlabel[bc](105,80){\textsf{(e)}}
\tlabel[tc](105,0){\textsf{(f)}}
% Level 2
\lines{(145,80),(195.1,80)}
\lines{(145,60),(195.1,60)}
\dotted\lines{(133.1,70),(140,90),(150,90)}
\dotted\lines{(133.1,70),(140,50),(150,50)}
\tlabel[cl](205,80){\textsf{(i)}}
\tlabel[cl](205,60){\textsf{(i)}}
\lines{(140,30),(179.6,30)}
\lines{(140,10),(179.6,10)}
\lines{(140,-10),(179.6,-10)}
\dotted\lines{(127.1,10),(134,40),(144,40)}
\dotted\lines{(127.1,10),(134,-20),(144,-20)}
\tlabel[cl](190,30){\textsf{(j)}}
\tlabel[cl](190,10){\textsf{(j)}}
\tlabel[cl](190,-10){\textsf{(j)}}
\end{mfpic}$
\end{array}\qquad
\begin{array}{ccc}
\hline
\mbox{Section} & \mbox{Length }\mu\mbox{m} & \mbox{Diameter }\mu\mbox{m}\\[2pt]
\hline
 (a) & 166.809245 & 7.089751 \\
 (b) & 379.828386 & 9.189790 \\
 (c) & 383.337494 & 4.160168 \\
 (d) & 410.137845 & 4.762203 \\
 (e) & 631.448520 & 6.345604 \\
 (f) & 571.445800 & 5.200210 \\
 (g) & 531.582750 & 2.000000 \\
 (h) & 651.053246 & 3.000000 \\
 (i) & 501.181023 & 4.000000 \\
 (j) & 396.218388 & 2.500000 \\
\hline
\end{array}
\]
\centering
\parbox{5in}{\caption{\label{TestNeuron} A branched neuron
satisfying the Rall conditions. The radii and lengths of the
dendritic section are given in the right hand panel of the figure.
A each branch point, $l/\sqrt{r}$ is fixed for all children of the
branch point.}}
\end{figure}

To guarantee that any apparent errors between the closed form
solution and the numerical solution are not due to the lack of
precision with which the branched dendrite is represented as an
equivalent cylinder, a high degree of accuracy is used in the
specification of dendritic radii and section lengths in the model
neuron. The model neuron illustrated in Figure \ref{TestNeuron}
will be assumed to have a dendritic membrane of specific
conductance $g_\mathrm{M}=0.091$ mS/cm$^2$ and specific
capacitance $1.0\mu$F/cm$^2$, and an intracellular medium of
conductance $g_\mathrm{A}=14.286$ mS/cm. With these biophysical
properties, the equivalent cylinder has length one electrotonic
unit. The soma of the test dendrite is assumed to have a membrane
of area $A_\mathrm{S}$, specific conductance $g_\mathrm{S}=g_\mathrm{M}$ and
specific capacitance $c_\mathrm{S}=c_\mathrm{M}$.

\subsection{Closed form solution for the equivalent cylinder}
The first step in the assessment of the performance of both
compartmental models requires the construction of the closed form
solution for the Rall equivalent cylinder under the action of
input on the cylinder and at its soma.

Consider a uniform cylindrical dendrite of radius $r$ and length
$l$ attached to a soma of area $A_\mathrm{S}$ at its left hand end
which is taken to be the point $x=0$ of a coordinate system with
axis oriented along the length of the dendrite. If $V(x,t)$ is the
deviation of the transmembrane potential from its resting value at
point $x$ and time $t>0$, then $V(x,t)$ satisfies the cable equation
\begin{equation}\label{es1}
\ds 2\pi r\Big(c_\mathrm{M}\frac{\partial V}{\partial t}
+g_\mathrm{M}V\Big)=\pi r^2 g_\mathrm{M}\frac{\partial^2V}
{\partial x^2}-I(x,t)\,,\quad (x,t)\in(0,l)\times(0,\infty)
\end{equation}
where $c_\mathrm{M}$, $g_\mathrm{M}$ and $g_\mathrm{A}$ have their
usual meanings and $I(x,t)$ is the linear density of exogenous
current along the dendrite. A solution of equation (\ref{es1}) is
sought satisfying the initial condition $V(x,0)=0$ and the
boundary conditions
\begin{equation}\label{es2}
A_\mathrm{S}\Big(c_\mathrm{M}\,\frac{\partial V(0,t)}{\partial t}
+g_\mathrm{M}V(0,t)\Big)=\pi r^2g_\mathrm{A}\,
\frac{\partial V(0,t)}{\partial x}-I_\mathrm{S}(t)\,,
\qquad\frac{\partial V(l,t)}{\partial x}=0\,,
\end{equation}
in which it has been recognised that the somal and dendritic
membranes have identical specific capacitances and conductances.
Prior to describing the critical steps in the construction of the
exact mathematical solution to the initial boundary value problem
posed by equations (\ref{es1}) and (\ref{es2}), it is convenient
to introduce the well-known non-dimensional electrotonic length
\begin{equation}\label{es3}
L=l\,\sqrt{\ds\frac{2 g_\mathrm{M}}{r g_\mathrm{A}}}\,.
\end{equation}
The required solution is now constructed by observing that the
series
\begin{equation}\label{es4}
V(x,t)=e^{-t/\tau}\,\Big[\,\phi_0(t)+\sum_\beta\;\phi_\beta(t)
e^{-\beta^2 t/L^2\tau}\,\cos{\beta(1-x/l)}\,\Big]\,,
\qquad\tau=\frac{c_\mathrm{M}}{g_\mathrm{M}}\,,
\end{equation}
satisfies the gradient boundary condition at $x=l$ for all values
of $\beta$ and functions $\phi_0(t)$ and $\phi_\beta(t)$, and will
also satisfy the initial condition $V(x,0)=0$ provided
$\phi_0(0)=\phi_\beta(0)=0$. This series solution for $V(x,t)$
also satisfies the partial differential equation (\ref{es1})
provided
\begin{equation}\label{es4}
\ds \frac{d\phi_0}{dt}+\sum_\beta\,\frac{d\phi_\beta}{dt}
\,e^{-\beta^2\,t/L^2\tau}\,\cos{\beta(1-x/l)}
=-\frac{I(x,t)e^{t/\tau}}{2\pi r c_\mathrm{M}}
\end{equation}
and the boundary condition at the soma provided
\begin{equation}\label{es5}
\frac{d\phi_0}{dt}+\sum_\beta\,\Big[\,\frac{d\phi_\beta}{dt}\,
\cos\beta-\frac{\beta\cos\beta}{\gamma\tau\,L^2}\,\Big(\,
\gamma\beta+\tan\beta\,\Big)\,\phi_\beta\,\Big]
\,e^{-\beta^2\,t/L^2\tau}= -\frac{I_\mathrm{S}(t)}{A_\mathrm{S}
c_\mathrm{M}}\; e^{t/\tau}
\end{equation}
where $\gamma=A_\mathrm{S}/2\pi r l$, that is, $\gamma$ is the
ratio of the membrane surface area of the soma to the membrane
surface area of the dendrite. Equation (\ref{es5}) suggests that
the values of $\beta$ in expression (\ref{es4}) should be chosen
to be the zeros of the transcendental equation
\begin{equation}\label{es6}
\tan\beta+\gamma\beta = 0\,.
\end{equation}
With this choice for the values of $\beta$, equations (\ref{es4})
and (\ref{es5}) take the simplified form
\begin{equation}\label{es7}
\begin{array}{rcl}
\ds \frac{d\phi_0}{dt}+\sum_\beta\,\frac{d\phi_\beta}{dt}
\,e^{-\beta^2\,t/L^2\tau}\,\cos{\beta(1-x/l)}
& = & -\ds\frac{I(x,t)e^{t/\tau}}{2\pi r c_\mathrm{M}}\,, \\[12pt]
\ds\frac{d\phi_0}{dt}+\sum_\beta\,\frac{d\phi_\beta}{dt}\,
\cos\beta\,e^{-\beta^2\,t/L^2\tau} & = & -
\ds\frac{I_\mathrm{S}(t)}{A_\mathrm{S}c_\mathrm{M}}
\; e^{t/\tau}\,.
\end{array}
\end{equation}
The coefficients $\phi_0(t)$ and $\phi_\beta(t)$ are determined
from equations (\ref{es7}) by two different procedures. To find
$\phi_0(t)$, the first of equations (\ref{es7}) is integrated over
$(0,l)$ to obtain
\begin{equation}\label{es8}
l\,\frac{d\phi_0}{dt}-\gamma\,l\,\sum_\beta\,\frac{d\phi_\beta}{dt}
\,e^{-\beta^2\,t/L^2\tau}\,\cos\beta
= -\frac{e^{t/\tau}}{2\pi r c_\mathrm{M}}\,\int_0^l\,I(x,t)\,dx\,.
\end{equation}
The summation in this expression is now eliminated using the
second of equations (\ref{es7}) to get
\begin{equation}\label{es9}
\frac{d\phi_0}{dt}= -\frac{e^{t/\tau}}
{(2\pi r l+A_\mathrm{S})c_\mathrm{M}}\,\Big[\,
I_\mathrm{S}(t)+\int_0^l\,I(x,t)\,dx\,\Big]\,.
\end{equation}
Note that $(2\pi r l+A_\mathrm{S})c_\mathrm{M}$ is simply the
total membrane capacitance of the dendrite and soma. The
coefficient $\phi_0(t)$ is obtained by integrating equation
(\ref{es9}) with respect to time with the initial condition
$\phi_0(0)=0$ in the simulations to carried out here.

The procedure to find $\phi_\beta(t)$ begins by subtracting
equations (\ref{es7}) to get
\begin{equation}\label{es10}
\sum_\beta\,\frac{d\phi_\beta}{dt}
\,e^{-\beta^2\,t/L^2\tau}\,\Big(\,\cos{\beta(1-x/l)}
-\cos\beta\,\Big) = \frac{e^{t/\tau}}{2\pi r c_\mathrm{M}}
\,\Big[\,\frac{I_\mathrm{S}(t)}{\gamma\,l}-I(x,t)\,\Big]\,.
\end{equation}
Further progress is based on the observation that if $\alpha$ and
$\beta$ are solutions of equation (\ref{es6}) then
\begin{equation}\label{es11}
\int_0^l\,\cos\alpha\big(1-x/l\big)\big(\cos\beta\big(1-x/l\big)
-\cos\beta\,\big)\,dx=
\left[\begin{array}{cc}
0 & \alpha\ne\beta\,, \\[10pt]
\ds\frac{l(1+\gamma\cos^2\alpha)}{2} & \alpha=\beta\,.
\end{array}\right.
\end{equation}
Equation (\ref{es10}) is multiplied by $\cos\alpha\big(1-x/l\big)$
and integrated over $[0,l]$ with respect to $x$ to obtain
\begin{equation}\label{es12}
\frac{d\phi_\alpha}{dt}=-\frac{2e^{(1+\alpha^2/L^2)/\tau}}
{(2\pi r l+A_\mathrm{S}\cos^2\alpha)c_\mathrm{M}}\,\Big[\,
\int_0^1\,I(x,t)\cos\alpha\big(1-x/l\big)\,dx
+\cos\alpha\,I_\mathrm{S}(t)\,\Big]\,.
\end{equation}
The coefficient $\phi_\alpha(t)$ is obtained by integrating
equation (\ref{es12}) with respect to time with the initial
condition $\phi_\alpha(0)=0$ in the simulations to be carried out
here. Once $\phi_0(t)$ and $\phi_\alpha(t)$ are known, the
potential is determined everywhere from expression (\ref{es4}).

\subsubsection{Computation of current on equivalent cylinder}
The special case in which the current $I(x,t)$ is constructed from
a series of point currents $I_1, \cdots,I_n$ at locations
$x_1,\cdots x_n$ from the soma of the equivalent cable is
particularly useful for the investigation of large scale synaptic
activity. In this case
\begin{equation}\label{ec1}
I(x,t)=\sum_{k=1}^n\;I_k(t)\delta(x-x_k)
\end{equation}
and the corresponding coefficient functions $\phi_0$ and
$\phi_\alpha$ satisfy
\begin{equation}\label{ec2}
\begin{array}{rcl}
\ds\frac{d\phi_0}{dt} & = &\ds -\frac{e^{t/\tau}}
{(2\pi r l+A_\mathrm{S})c_\mathrm{M}}\,\Big[\,
I_\mathrm{S}(t)+\sum_{k=1}^n\;I_k(t)\,\Big]\,,\\[10pt]
\ds\frac{d\phi_\alpha}{dt} & = & \ds-\frac{2e^{(1+\alpha^2/L^2)/\tau}}
{(2\pi r l+A_\mathrm{S}\cos^2\alpha)c_\mathrm{M}}\,\Big[\,
\sum_{k=1}^n \;I_k(t)\cos\alpha\big(1-x_k/l\big)
+\cos\alpha\,I_\mathrm{S}(t)\,\Big]\,.
\end{array}
\end{equation}
If $X_k$ is the electrotonic distance of the input $I_k(t)$ at
distance $x_k$ from the soma of the equivalent cylinder, then
$I_k(t)$ is the sum of the exogenous current inputs to the
branched neuron taken across all dendritic sections at
electrotonic distance $X_k$ from the soma.