# 8.7 Mathematical modeling of hippocampal spatial memory with place  (Page 5/6)

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## Computational method

To solve for $v\left(t\right)$ computationally, we first look at the times with no input spikes ( $kI ). Integrating both sides of equation [link] from $t-dt$ to $t$ and using the trapezoid rule, we find

$\begin{array}{cc}\hfill \tau \left(v\left(t\right)-v\left(t-dt\right)\right)& ={v}_{r}dt-\frac{v\left(t\right)+v\left(t-dt\right)}{2},\phantom{\rule{1.em}{0ex}}\text{which}\phantom{\rule{4.pt}{0ex}}\text{can}\phantom{\rule{4.pt}{0ex}}\text{be}\phantom{\rule{4.pt}{0ex}}\text{rearranged}\phantom{\rule{4.pt}{0ex}}\text{as}\hfill \\ \hfill v\left(t\right)& =\frac{2dt}{2\tau +1}·{v}_{r}+\frac{2\tau -1}{2\tau +1}·v\left(t-dt\right).\hfill \end{array}$

When there is an input spike, we add ${w}_{inp}$ to $v\left(t\right)$ , which is shown in

${v}_{inp}\left(t\right)=v\left(t\right)+{w}_{inp}.$

## Analytic method

To solve for $v\left(t\right)$ analytically, we first look at $v\left(t\right)$ between input spikes. From equation [link] , we get

$\tau {v}^{\text{'}}\left(t\right)=\left({v}_{r}-v\left(t\right)\right).$

Solving this ordinary differential equation gives us

$v\left(t\right)={v}_{r}+c{e}^{-t/\tau },$

where $c$ is the constant of integration. We know we want $v\left(0\right)={v}_{r}+{w}_{inp}$ , so $c$ must equal ${w}_{inp}$ . Thus, we have

$v\left(t\right)={v}_{r}+{w}_{inp}{e}^{-t/\tau },\phantom{\rule{4pt}{0ex}}\text{where}\phantom{\rule{4pt}{0ex}}0\le t

which simply tells us that after one input spike at $t=0$ , ${w}_{inp}$ decays so that $v\left(t\right)$ approaches ${v}_{r}$ . Consider the following calculations of $v\left(t\right)$ for up to three input spikes.

At $t=I$ , we have a second input spike, and at $I , we decay the input to find

$\begin{array}{cc}\hfill v\left(I\le t<2I\right)& ={v}_{r}+{w}_{inp}{e}^{-t/\tau }+{w}_{inp}{e}^{-\left(t-I\right)/\tau }.\hfill \end{array}$

Finally, at $t=2I$ , we have a third input spike and see

$v\left(2I\right)={v}_{r}+{w}_{inp}{e}^{-2I/\tau }+{w}_{inp}{e}^{-I/\tau }+{w}_{inp}.$

To determine when the voltage reaches threshold and the cell spikes, we need only examine the peak values of $v$ , which are when $t=kI,\phantom{\rule{4pt}{0ex}}0\le k\le n-1$ . Thus, we use the following generalized formula to calculate $v\left(\left(n-1\right)I\right)$ when there are $n$ total input spikes:

$\forall n,\phantom{\rule{1.em}{0ex}}v\left(\left(n-1\right)I\right)={v}_{r}+{w}_{inp}\sum _{k=0}^{n-1}{\left({e}^{-I/\tau }\right)}^{k}.$

[link] shows that in the absence of spikes, the peak voltages approach an asymptote. This asymptote can be calculated by

$\begin{array}{cc}\hfill {v}_{\infty }=\underset{n\to \infty }{lim}v\left(\left(n-1\right)I\right)& ={v}_{r}+{w}_{inp}\sum _{k=0}^{\infty }{\left({e}^{-I/\tau }\right)}^{k}\hfill \\ & ={v}_{r}+{w}_{inp}\left(\frac{1}{1-{e}^{-I/\tau }}\right).\hfill \end{array}$

If ${v}_{\infty }<{v}_{th}$ , then the cell will never spike.

## Computational vs. analytic method

We found the minimum input weight ${w}_{inp}$ necessary for the cell to spike at least once as a function of the input time interval $I$ when given a sufficiently long simulation.

Let the interspike interval $I$ and input weights ${w}_{inp}$ satisfy $2\le I\le 30$ and $2\le {w}_{inp}\le 20$ .

In the computational method, the Matlab program compW.m calculates $v\left(t\right)$ according to equations [link] and [link] . In AnalysisW.m , the minimum ${w}_{inp}$ is calculated by

${w}_{inp}=\left({v}_{th}-{v}_{r}\right)\left(1-{e}^{-I/\tau }\right),$

which was obtained by setting ${v}_{\infty }$ of equation [link] to ${v}_{\infty }\ge {v}_{th}$ where

${v}_{th}\le {v}_{r}+{w}_{inp}\left(\frac{1}{1-{e}^{-I/\tau }}\right).$

[link] shows that as the input time interval increases, greater input weight is necessary for the cell to spike at least once ( AnalysisW.m ). We note on the graph the value of ${w}_{inp}=10.11$ at $I=20$ because these two values will be put to use in the next section.

## Computational vs. analytic method

We determine the minimum number of input spikes necessary for the cell to spike as a function of input weight.

We use $I=20$ and consider only the weights that produce at least one spike with sufficient simulation, starting with ${w}_{inp}=10.2$ as shown in [link] . Let ${n}_{1}$ denote the minimum number of input spikes of weight ${w}_{inp}$ necessary for $v\left(t\right)$ to reach ${v}_{th}$ . We see that

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