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To solve for $v\left(t\right)$ computationally, we first look at the times with no input spikes ( $kI<t<(k+1)I$ ). Integrating both sides of equation [link] from $t-dt$ to $t$ and using the trapezoid rule, we find
When there is an input spike, we add ${w}_{inp}$ to $v\left(t\right)$ , which is shown in
To solve for $v\left(t\right)$ analytically, we first look at $v\left(t\right)$ between input spikes. From equation [link] , we get
Solving this ordinary differential equation gives us
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
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 $I<t<2I$ , we decay the input to find
Finally, at $t=2I$ , we have a third input spike and see
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 $v\left(\right(n-1\left)I\right)$ when there are $n$ total input spikes:
[link] shows that in the absence of spikes, the peak voltages approach an asymptote. This asymptote can be calculated by
If ${v}_{\infty}<{v}_{th}$ , then the cell will never spike.
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
which was obtained by setting ${v}_{\infty}$ of equation [link] to ${v}_{\infty}\ge {v}_{th}$ where
[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.
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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