TAOCP 1.2.9 Exercise 13

Exercise 13. [HM22] The Laplace transform of a function $f(x)$ is the function

$$ Lf(s) = \int_0^\infty e^{-st}f(t),dt. $$

Given that $a_0,a_1,a_2,\ldots$ is an infinite sequence having a convergent generating function, let $f(x)$ be the step function $\sum_k a_k [0 \le k \le x]$. Express the Laplace transform of $f(x)$ in terms of the generating function $G$ for this sequence.

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Solution

Let $\langle a_k \rangle = a_0,a_1,a_2,\ldots$ be a sequence with generating function

$$ G(z) = \sum_{k \ge 0} a_k z^k. $$

Define the step function

$$ f(x) = \sum_{k \ge 0} a_k [0 \le k \le x], $$

so that $f(x) = a_0 + a_1 + \cdots + a_{\lfloor x \rfloor}$ for nonnegative $x$. The Laplace transform of $f$ is

$$ Lf(s) = \int_0^\infty e^{-sx} f(x), dx. $$

We partition the integral over intervals $[k,k+1)$ for $k=0,1,2,\ldots$. On each interval $[k,k+1)$, the function $f(x)$ is constant and equal to

$$ f(x) = \sum_{j=0}^{k} a_j. $$

Hence

$$ \begin{aligned} Lf(s) &= \sum_{k \ge 0} \int_k^{k+1} e^{-sx} \sum_{j=0}^k a_j , dx \ &= \sum_{k \ge 0} \left( \sum_{j=0}^k a_j \right) \int_k^{k+1} e^{-sx} , dx \ &= \sum_{k \ge 0} \left( \sum_{j=0}^k a_j \right) \frac{e^{-sk}-e^{-s(k+1)}}{s} \ &= \frac{1}{s} \sum_{k \ge 0} \left( \sum_{j=0}^k a_j \right) e^{-sk} \left(1 - e^{-s}\right). \end{aligned} $$

Factor $1-e^{-s}$:

$$ Lf(s) = \frac{1-e^{-s}}{s} \sum_{k \ge 0} \left( \sum_{j=0}^k a_j \right) e^{-sk}. $$

Observe that the sum $\sum_{j=0}^k a_j$ is the $k$th partial sum of the sequence $\langle a_j \rangle$. By Eq. (7), the generating function for the partial sums is

$$ \sum_{k \ge 0} \left( \sum_{j=0}^k a_j \right) z^k = \frac{G(z)}{1-z}. $$

Substitute $z = e^{-s}$:

$$ \sum_{k \ge 0} \left( \sum_{j=0}^k a_j \right) e^{-sk} = \frac{G(e^{-s})}{1 - e^{-s}}. $$

Hence

$$ Lf(s) = \frac{1-e^{-s}}{s} \cdot \frac{G(e^{-s})}{1-e^{-s}} = \frac{G(e^{-s})}{s}. $$

This expresses the Laplace transform of the step function $f(x)$ in terms of the generating function $G$ of the sequence $\langle a_k \rangle$.

$$ \boxed{Lf(s) = \frac{G(e^{-s})}{s}} $$

This completes the proof.

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