# zeroing out lower-order terms in generating function

Given a generating function $F(x)=a+bx+cx^2+dx^3+\dotsb$, how do I truncate the $n$ lower order terms to get, for example if $n=2$: $cx^2+dx^3+\dotsb$?

For example, if I wanted to find $0a+1b+2c+\dotsb$, I would evaluate $$\left.\frac{dF(x)}{dx}\right|_{x=1}$$

This procedure can be used to find the expected value of a probability distribution given its generating function.

I want something similar for truncation of lower-order terms. This would give a cdf for a probability distribution. Since the cdf has a nice form for a binomial generating function, this suggests that there might be a nice way to arrive at it using generating function operators.

I vaguely remember learning this once, but flipping through the book generatingfunctionology didn't yield it.

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By what available means? Obviously subtracting $a+bx$ from $F(x)$ works for your example, and is easily generalized... – anon May 31 '12 at 4:09
@anon: I've added to my question to make it more clear. – Neil G May 31 '12 at 4:12
If differentiation and evaluation are all you have at hand, you can truncate by subtracting a partial Taylor expansion - however, without convergence (e.g. $\sum n!x^n$) I'm unsure if truncation is obtainable with these two operations alone. – anon May 31 '12 at 4:16
@anon: There are other operations available, but I don't know what they are. – Neil G May 31 '12 at 4:23

If $f(x) = \sum\limits_{n=0}^\infty a_n x^n$, then $f''(x) = \sum\limits_{n=2}^\infty a_n n (n-1) x^{n-2}$, and $$\int_0^x (x-t) f''(t)\ dt = \sum_{n=2}^\infty a_n x^n.$$ EDIT: More generally, $$\int_0^x \dfrac{(x-t)^{k-1}}{(k-1)!} f^{(k)}(t)\ dt = \sum_{n=k}^\infty a_n x^n.$$
$\int_0^t f''(s)\ ds = \sum_{n=2}^\infty a_n \int_0^t n(n-1) s^{n-2}\ ds = \sum_{n=2}^\infty a_n n t^{n-1}$, and so $\int_0^x \int_0^t f''(s)\ ds\ dt = \sum_{n=2}^\infty a_n x^n$. Now interchange the order of integration. – Robert Israel May 31 '12 at 4:45