Looking at an earlier post Finding the power series of a rational function, I am trying to get a closed formula for the n'th coefficient in the Taylor series of the rational function (1-x)/(1-2x-x^3). Is it possible to use any of the tricks in that post to not only obtain specific coefficients, but an expression for the n'th coefficient ?If T(x) is the Taylor polynomial I am looking at the equality (1-x) = (1-2x-x^3)*T(x) and differentiating, but I am not able to see a pattern giving me an explicit formula for the coefficients.
[Math] Coefficients for Taylor series of given rational function
taylor expansion
Related Solutions
$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$
$$ \expo{z} = \sum_{k = 0}^{\infty}{z^{k} \over k!} \quad \left\vert% \begin{array}{rcl} \qquad\sec\pars{z} & = & \sum_{k = 0}^{\infty}{\verts{E_{2k}} \over \pars{2k}!}z^{2k} \\ \verts{z} & < & {\pi \over 2} \\[2mm] &&E_{\nu}\ \mbox{is an}\ {\it\mbox{Euler Number}}. \end{array}\right. $$
\begin{align} &\vphantom{\Huge A^{A}} \\ {\expo{z} \over \cos\pars{z}} &= \expo{z}\sec\pars{z} = \sum_{k = 0}^{\infty}\sum_{k' = 0}^{\infty}{z^{k} \over k!}\, {\verts{E_{2k'}} \over \pars{2k'}!}z^{2k'} = \sum_{k = 0}^{\infty}\sum_{k' = 0}^{\infty}{z^{k} \over k!}\, {\verts{E_{2k'}} \over \pars{2k'}!} \sum_{n = 0}^{\infty}z^{n - k}\,\delta_{n,k + 2k'} \\[3mm]&= \sum_{n = 0}^{\infty}z^{n} \sum_{k' = 0 \atop {\vphantom{\LARGE A}n - 2k' \geq 0}}^{\infty} {1 \over \pars{n - 2k'}!}\, {\verts{E_{2k'}} \over \pars{2k'}!} \end{align}
$${\large% {\expo{z} \over \cos\pars{z}} = \sum_{n = 0}^{\infty}A_{n}\,z^{n}\,, \qquad\qquad A_{n} \equiv \sum_{k = 0}^{\floor{n/2} \atop } {\verts{E_{2k}} \over \pars{2k}!\pars{n - 2k}!}} $$
HINT:
Let $f(x)=\frac{1/5}{1-5x}$ so that $f'(x)=\frac{1}{(1-5x)^2}$. Now, differentiate the geometric series for $f(x)$ term by term.
Best Answer
Denoting the coefficients of $T(x)$ as $t_i$ (so $T(x) = t_0 + t_1 x + ...$), consider the coefficient of $x^a$ in $(1 - 2x - x^3) T(x)$. It shouldn't be hard to show that it's $t_a - 2t_{a-1} - t_{a-3}$.
So we have $t_0 - 2t_{-1} - t_{-3} = 1$, $t_1 - 2t_0 - t_{-2} = -1$, and $a > 1 \Rightarrow t_a - 2t_{a-1} - t_{a-3} = 0$. You can get $t_0$, $t_1$. Do you know how to solve the discrete recurrence $t_{a+3} = 2t_{a+2} + t_{a}$?
(If not, here's a hint: suppose you had a number $\alpha$ such that $t_a = \alpha^a$ satisfied the recurrence. What constraint on $\alpha$ can you prove? Now consider a linear combination of the possible such solutions, $t_a = b_0 \alpha_0^a + ... + b_n \alpha_n^a$, and solve $n$ simultaneous equations from your base cases $t_0$, $t_1$, and use $0 = t_{-1} = t_{-2} = ...$ if necessary).