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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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    @anon: There are other operations available, but I don't know what they are.2012-05-31

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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. $

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    @RobertIsrael How does the interchange work. I cannot see why $\int_{0}^{x} \int_{0}^{t} f''(s)\ ds\ dt = \int_{0}^{x}\int_{s}^{x} f''(s)\ dt \ ds $ I would think that, upon interchanging what we have, we have $\int_{0}^{t} \int_{0}^{x} f''(s)\ dt \ ds = \int_{0}^{t} f''(s)x \ ds$2017-05-12