Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$\ X - X_{y/z}$

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].

share|cite|improve this question
The answer will likely depend on the context. Where did you see this notation? – JavaMan Jul 28 '11 at 7:22
@DJC I added the context to the question – Matt Munson Jul 28 '11 at 7:35
Does your book derive your formula in some way? If so, during what time of the derivation do you first see the notation? – Fredrik Meyer Jul 28 '11 at 7:57
I think the set of people who could answer your question given more substantive mathematical context is much larger than the set of people who are specifically familiar with artificial neural synaptic plasticity. – anon Jul 28 '11 at 8:18
I found an actual instance, $\dot{c} = -c/\tau_c + \rm{STDP}(\tau)\delta(t-t_{pre/post})$, here; I’m not positive, but it appears to me that this is simply two equations in one, one with $\delta(t-t_{pre})$, giving a boost to $c$ at time $t_{pre}$, and one with $\delta(t-t_{post})$, giving a boost to $c$ at time $t_{post}$. – Brian M. Scott Jul 28 '11 at 8:41
up vote 2 down vote accepted

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.

share|cite|improve this answer
So then $\ t-t_{pre/post}$ is equivalent to...? Thanks for the heads up, I fixed the link. – Matt Munson Jul 28 '11 at 8:41
Ok, I'm more out of my depth here than I thought. Firstly, what does $\dot c$ mean as compared to $ c$? And secondly, is the $\delta (t)$ supposed to evaluate to zero for all $t$ other than 0, which evaluates to 1? Or how does $\delta (t)$ work? – Matt Munson Jul 28 '11 at 10:10
Also, why is it $\dot c = -c$, rather than $\dot c = c$? It seems like the former will just reverse the sign of c at each recursive computation. – Matt Munson Jul 28 '11 at 10:40
$\dot c$ is (almost certainly) Newton dot notation for the time derivative of $c$. So $\dot c = -c/\tau_c$ is a differential equation whose solution $c(t) = c(0) \exp(-t/\tau_c)$ describes exponential decay; the $\delta$ peak term in eqn (1) just provides an initial impulse from which the decay can start. – Ilmari Karonen Jul 28 '11 at 11:16
Alright, I think I understand whats going on now. This equation describes the change in $c$ at timestep $t$. Since timesteps are on the order of ms, $-c/\tau_c = -c/1000$, which gives exponentially decelerating decreases in $c$, in the same way that repeatedly dividing a number by 2 decreases that number by increasingly smaller amounts, but never reaches 0. Meanwhile, $\delta(t-t_{pre/post})$ is like a true/false function that evaluates to true only when the timestep of the most recent pre or postsynaptic AP (whichever is more recent) is equal to the current timestep. – Matt Munson Jul 28 '11 at 17:10

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.

share|cite|improve this answer
Yeah, you were right. – Matt Munson Jul 29 '11 at 5:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.