2

When we are working with a generating function of a given sequence, when we take the derivative, we normally multiply by $x$ to shift the series back due to the derivative causing a shift in the opposite direction. Now, say we are trying to find the $n^{th}$ derivative of a generating function. Do we multiply by $x$ each time? I am trying to find a pattern.

H5159
  • 969

1 Answers1

2

To get the derivative, you differentiate. The $z \mathrm{D}$ trick you describe is for something different. Say you have: $$ A(z) = \sum_{n \ge 0} a_n z^n $$ If you compute the derivative: $$ A'(z) = \sum_{n \ge 1} n a_n z^{n - 1} = \sum_{n \ge 0} (n + 1) a_{n + 1} z^n $$ and again: $$ A''(z) = \sum_{n \ge 2} n (n - 1) a_n z^{n - 2} = \sum_{n \ge 0} (n + 2) (n + 1) a_{n + 2} z^n $$ To put back the $z$ "lost" by differentiation, you multiply by $z$, i.e., $$ z A'(z) = z \mathrm{D} A(z) = \sum_{n \ge 0} n a_n z^n $$ and if you are interested in the sequence $n^r a_n$, you get it by: $$ (z \mathrm{D})^r A(z) = \sum_{n \ge 0} n^r a_n z^n $$

vonbrand
  • 27,812
  • For your third line, where we have $a_{n+2}$, doesn't this "lose" the first two numbers of the sequence? I am studying the generating function of a sequence and I would not want to lose a few terms. Or maybe I am mistaken? – H5159 Mar 27 '14 at 21:35
  • 1
    @Frumpy: When you differentiate $a_0 + a_1x + a_2x^2 + \dots$, you get $a_1 + 2a_2x + 3a_3x^2 + \dots$, i.e. the first (constant) term disapears. In terms of the sequence, multiplying the $n$th term by $n$ means that the $0$th term gets multiplied by $0$, and hence disappears. This is what is reflected in the differentiation. – ShreevatsaR Mar 28 '14 at 02:17