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This may be a silly question but I was wondering how to interpret f(0) for something like: $$ f(t) = \sum_{t=1}^T \log(g(t)) $$

or if this doesn't make any sense altogether?

The motivation for this is that I'm working with the following: for the operator: $\Delta A(t) = A(t+1)-A(t)$, let :

$$ \Delta C = \sum^{T-1}_{t=0} [~~H(\pi(t+1)~|~\mu(t+1))~~ - ~~H(\pi(t)~|~\mu(t+1))~~] $$ Where $H (\pi(t+1)|\mu(t+1)) = \sum^n_{i=1}\pi_i(t) \log \frac{\pi_i(t)}{\mu_i(t+1}$ is the relative entropy.

I believe this implies that: $$ C = \sum_{t=1}^{T-1}H(\pi(t)~|~\mu(t+1)) $$ and I want to find $C(0), C'(0) $

WeakLearner
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    The expression on the right is independent of $t$. It is a function of $T$. Did you mean $f(T)$ on the left? – 5xum Jul 15 '15 at 08:20
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    Assuming you mean $f(T)$, the convention is that $f(0)=0$, see Wikipedia. – Cerberus Jul 15 '15 at 08:22
  • @5xum I updated the question to include why I was asking, the way that the $\Delta$ operator is specified implies that it should be $f(t)$ so I am confused about what $C$ is a function of.. – WeakLearner Jul 15 '15 at 08:50
  • @5xum or is it that C is a function of $\pi(t)$? – WeakLearner Jul 15 '15 at 09:02
  • @dimebucker91 $C$, as you wrote it, is clearly not a function of $t$. It is a funcion of $T$. The sum $$\sum_{i=1}^n f(i)$$ is never a function of $i$, it is a function of $n$! – 5xum Jul 15 '15 at 09:28
  • @5xum so then $C(T) = \sum^{T-1}_{t=0} H(\pi(t) |\mu(t+1) )$, which makes it incompatible with $\Delta C = C(T+1) - C(T)$ doesn't it? I guess my question boils down to, if you were given $\Delta C$ as I have written it, what is $C(T)$ ? – WeakLearner Jul 15 '15 at 09:34

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