Integral of Generating Function
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Theorem
Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Then:
\(\ds \int_0^z \map G t \rd t\) | \(=\) | \(\ds \sum_{k \mathop \ge 1} \dfrac {a_{k - 1} z^k} k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a_0 z + \dfrac {a_1 z^2} 2 + \dfrac {a_2 z^3} 3 + \dfrac {a_3 z^4} 4 + \cdots\) |
Proof
\(\ds \int_0^z \map G t \rd t\) | \(=\) | \(\ds \int_0^z \paren {\sum_{k \mathop \ge 0} a_k t^k} \rd t\) | Definition of Generating Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop \ge 0} \paren {\int_0^z a_k t^k \rd t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop \ge 0} \paren {\intlimits {a_k \dfrac {t^{k + 1} } {k + 1} } 0 z}\) | Primitive of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop \ge 0} \paren {a_k \dfrac {z^{k + 1} } {k + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop \ge 1} \dfrac {a_{k - 1} z^k} k\) | Translation of Index Variable of Summation |
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.9$: Generating Functions: $(15)$