Exponential generating function of the sequence $1,0,1,0,\dots$

Question

I'm currently learning exponential generating functions and am working on a problem that needs me to find the corresponding closed form expression to a given sequence.

I've been given the sequence $1,0,1,0,...$ and decided to use the EGF $e^x$ to find my solution by substituting $0$ in every odd term in the expression:

$e^x = \sum_{r=1}^\infty \frac{x^r}{r!} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$

Doing so, I've arrived at this expression:

$1 + \frac{x^2}{2!} + \frac{x^4}{4!} + ...$

But I don't know where to go from here. Am I on the right track or is there a better way to find the closed form expression?

Answer 1

HINT: Have you considered $\frac{e^x+e^{-x}}{2}$.

ETA: And that is basically all there is. Let $(a_i)_{i=0}^{\infty}$ be a sequence. Then the exponential generating function $f$ for this sequence is by definition the function whose Taylor series is prescribed by $f(x) =$ $a_0 + \sum_{i=1}^{\infty} \frac{a_ix^i}{i!}$. If the sequence is $1,0,1,0,\ldots$ where every other term is $1$ and the remaining terms are $0$, or equivalently, if the generating sequence is $(a_i)_{i=0}^{\infty}$, where $a_i$ is $1$ iff $i$ is even and is $0$ otherwise, then by definition of exponential generating function is $f(x)=\frac{e^x+e^{-x}}{2}$, as this function has the Taylor series expansion $f(x)=1+\sum_{i=1}^{\infty} \frac{a_ix^i}{i!}$, where $a_i$ is $1$ iff $i$ is even, and $a_i =0$ otherwise.

Answer 2

A few different ways to approach this one. For your sequence: \begin{equation} S = 1, 0, 1,0\dots \end{equation} If we let $T_n$ be the $n$-th value of $S$ with $T_0 = 1$, then we can define $S$ via the iterative sequence: \begin{equation} T_{n+ 1} = T_{n} + \left(-1\right)^{n + 1} \end{equation} Let us know define our Exponential Generating Function: \begin{equation} T\left(x\right) = \sum_{n = 0}^\infty T_n \frac{x^n}{n!} \end{equation} We now return to our iterative sequence and multiply both sides by $\frac{x^n}{n!}$: \begin{equation} T_{n+1} \frac{x^n}{n!} = T_{n}\frac{x^n}{n!} + \left(-1\right)^{n +1}\frac{x^n}{n!} \end{equation} We now take the summation for $n \geq 0$:
\begin{equation} \sum_{n = 0}^\infty T_{n+1} \frac{x^n}{n!} = \sum_{n = 0}^\infty T_{n}\frac{x^n}{n!} - \sum_{n = 0}^\infty\left(-1\right)^n\frac{x^n}{n!} \end{equation} Starting with the RHS: \begin{equation} \sum_{n = 0}^\infty T_{n}\frac{x^n}{n!} = T\left(x\right); \qquad \qquad \sum_{n = 0}^\infty\left(-1\right)^n\frac{x^n}{n!} = e^{-x} \end{equation} For the LHS, we evaluate by taking the derivative of $T(x)$ w.r.t. $x$: \begin{equation} T'(x) = \frac{d}{dx} \sum_{n = 0}^\infty T_{n}\frac{x^n}{n!}= \sum_{n = 0}^\infty T_{n}\frac{\frac{d}{dx}\left[x^n\right]}{n!} = \sum_{n = 0}^\infty T_{n}\frac{nx^{n-1}}{n!} = \sum_{n = 1}^\infty T_{n}\frac{nx^{n-1}}{n!} = \sum_{n = 1}^\infty T_{n}\frac{x^{n-1}}{\left(n-1\right)!} \end{equation} Shifting the index to $ m = n - 1$ we yield: \begin{equation} T'(x) =\sum_{m=0}^\infty T_{m + 1}\frac{x^{m}}{m!} \end{equation} And so, we arrive at: \begin{equation} T'(x) = T(x) - e^{-x} \longrightarrow T'(x) - T(x) = -e^{-x} \end{equation} Here we solve via the Integrating Factor: \begin{equation} \frac{d}{dx}\left[e^{-x}T(x)\right] = -e^{-2x} \longrightarrow e^{-x}T(x) = \frac{1}{2}e^{-2x} + C \longrightarrow T(x) = \frac{1}{2}e^{-x} + Ce^{x} \end{equation} Where $C = $ Constant of Integration. To resolve $C$ evaluate: \begin{equation} T(0) = \sum_{n = 0}^\infty T_n \frac{0^n}{n!} = T_0 = 1 \end{equation} And so, \begin{equation} T(0) = 1 = \frac{1}{2}e^{-0} + Ce^{0} = \frac{1}{2} + C \longrightarrow C =\frac{1}{2} \end{equation} Thus, we arrive at our Exponential Generating Function: \begin{equation} T(x) = \frac{1}{2}e^{-x} + \frac{1}{2}e^{x} = \cosh(x) \end{equation}