-1
$\begingroup$

$\ X - X_{y/z}$

What does this notation mean? Does it have a defined meaning?

The context is from this paper, second paragraph of methods section. It is used in a formula as the parameter of 'the Dirac delta function': $\ \delta(t-t_{\text{pre}/\text{post}})$

$\ \delta(t)$ is the Dirac delta function that step-increases the variable $c$. Firings of pre- and post-synaptic neurons, occurring at times $\ \ t_{\text{pre}/\text{post}}$, respectively, [influence synaptic plasticity].

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    @Brian Seems you're right.2011-07-28

2 Answers 2

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From the piece of text you quoted, I'd guess it simply means that the firing of pre-synaptic neurons occurs at time $t_\mathrm{pre}$, and firing of post-synaptic neurons occurs at time $t_\mathrm{post}$.

Yes, I'd consider it bad notation, given that such a use of the slash could be confused with division.

Edit: As for equation (1) in the paper, $\dot c = -c/\tau_c + \operatorname{STDP}(\tau)\delta(t-t_\mathrm{pre/post})$, my best guess so far is that it should be interpreted as something like $t_\mathrm{pre/post} = \frac 1 2 (t_\mathrm{pre}+t_\mathrm{post})$.

Or rather, I think the point of the notation is that, apparently, $\tau = t_\mathrm{post}-t_\mathrm{pre}$ is on the order of milliseconds, while the decay timescale $\tau_c$ is measured in seconds, so that, on the longer timescale, the difference between $\delta(t-t_\mathrm{pre})$ and $\delta(t-t_\mathrm{post})$ doesn't really matter.

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    In other words, if $t-t_{pre/post} = 0$ then $\delta(t-t_{pre/post}) = 1$, else $\delta(t-t_{pre/post}) = 0$. Or, in plain english, if a pre or postsynaptic impulse just occured, update$c$according to STDP and also calculate decay, otherwise, just calculate decay.2011-07-28
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I haven't looked at the paper, but it seems to me that the word "respectively" indicates that the formula holds in two cases, with $\delta(t-t_{pre/post})$ meaning either $\delta(t-t_{pre})$ or $\delta(t-t_{post})$ depending on which case you're looking at.

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    Yeah, you were right.2011-07-29