I was going through an article on counting principles and came across exponential generating functions.
is defined recursively as -
, 
This is the exp. generating function -
When differentiated, I was confused as to how this -
Changes to this -
Here are the complete steps before the differential equation part (for your reference) -
Could anyone help me here? I know how to differentiate, but this specifically wasn't clear to me.
Note: My markdown editor is acting up at the moment, so I'll update the post using MathJax.
Line (2) is application of the recurrence for $I$.
In line (3), write down the first few terms of the first series to help you see what is going on: $$ \frac{I(0)t^0}{0!} + \frac{I(1)t^1}{1!} + \frac{I(2)t^2}{2!} + \cdots \text{,} $$ which is $\theta(t)$
To get to line (4), pull out one of the factors to $t$ so the argument to $I$, the power of $t$, and the argument to the factorial are all the same, to try to do the same thing we did on the previous line. ... which works.
Another way to see what is going on is to change the index. For $$ \sum_{k \geq 1} \frac{I(k-1)t^{k-1}}{(k-1)!} $$ set $j = k-1$. Since $k$ ranges over $[1,\infty)$, $j$ ranges over $[0, \infty)$. We also have $k = j+1$, so \begin{align*} \sum_{k \geq 1} \frac{I(k-1)t^{k-1}}{(k-1)!} &= \sum_{j \geq 0} \frac{I((j+1)-1)t^{(j+1)-1}}{((j+1)-1)!} \\ &= \sum_{j \geq 0} \frac{I(j)t^{j}}{j!} \\ &= \theta(t) \text{,} \end{align*} by definition. Pulling out a power of $t$ from the second sum and making the substitution $j = k-2$ does the same thing with the second sum.