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I am struggling to find $f^{(2013)}(0)$ for $$f(x) = \frac{1}{1 + x + x^3 + x^4}$$ I know that I should use power series, and following a hint I rewrote the equation as the following:

$$1 = (1 + x + x^3 + x^4)(\sum_{n = 0}^{\infty} a_n x^n) = \sum_{n = 0}^{\infty} a_n x^n + \sum_{n = 0}^{\infty} a_n x^{n + 1} + \sum_{n = 0}^{\infty} a_n x^{n + 3} + \sum_{n = 0}^{\infty} a_n x^{n + 4}$$

I can work out the value of $a_0$ easily, but I'm not really sure how to solve for the remaining coefficients and hence find the derivatives... I know that the pattern is supposed to be, but I'm completely at a loss. Any hints as to how to proceed?

lpsolver
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3 Answers3

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Following the hint and letting $f(x)=\sum_{n}a_n x^n$, you have $$ \begin{eqnarray} 1&=&\left(1+x+x^3+x^4\right) f(x) \\ &=& \left(1+x+x^3+x^4\right)\sum_{n}a_n x^n \\ &=& \sum_na_nx^n+\sum_na_nx^{n+1}+\sum_na_nx^{n+3}+\sum_na_nx^{n+4} \\ &=&\sum_na_nx^n+\sum_na_{n-1}x^n+\sum_na_{n-3}x^{n}+\sum_na_{n-4}x^n \\ &=&\sum_n\left(a_n+a_{n-1}+a_{n-3}+a_{n-4}\right)x^{n}. \end{eqnarray} $$ Equating identical powers of $x$ tells you that $a_0=1$ and $a_{n}=-a_{n-1}-a_{n-3}-a_{n-4}$ for $n>0$. Direct calculation of this recurrence relation shows that $a_{2013}=-672$. (There is also a closed-form solution, but that doesn't seem necessary here.) Since $f^{(2013)}(0)=2013!\times a_{2013}$, you find $$ f^{(2013)}(0)=-672\cdot2013! \approx -1.91\times 10^{5781}. $$

mjqxxxx
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Does it help that $1+x+x^2 +x^3 + x^4 = \frac{1-x^5}{1-x}$? You can then separate the fraction and apply the extended binomial theorem.

Brian Burns
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  • unfortunately the extended binomial theorem is explicitly not in the list of results we may use. but this does make sense – lpsolver Nov 15 '13 at 05:50
  • lpsolver, if you're allowed to use power series then you're very probably allowed to use it. The theorem just says that for $|x|<1$ we have $(1+x)^{\alpha} = \sum_{k=0}^{\infty}\binom{\alpha}{k}x^k$ – Brian Burns Nov 15 '13 at 06:02
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    His denominator is $1+x+x^3+x^4$, though. – mjqxxxx Nov 15 '13 at 06:05
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\ds{\fermi\pars{x} = \frac{1}{1 + x + x^3 + x^4} = {1 \over \pars{1 + x^{3}}\pars{1 + x}}\,,\quad \fermi^{\pars{2013}}\pars{0}:\ ?}$

\begin{align} \fermi\pars{x} &= \sum_{\ell = 0}^{\infty}\pars{-1}^{3\ell}x^{3\ell} \sum_{\ell' = 0}^{\infty}\pars{-1}^{\ell'}x^{\ell'} = \sum_{\ell = 0}^{\infty}\sum_{\ell' = 0}^{\infty}\pars{-1}^{\ell + \ell'} \sum_{n = 0}^{\infty}x^{n}\delta_{n,3\ell + \ell'} \\[3mm]&= \sum_{n = 0}^{\infty}x^{n}\sum_{\ell = 0}^{\infty}\pars{-1}^{\ell} \sum_{\ell' = 0}^{\infty} \pars{-1}^{\ell'}\delta_{\ell', n - 3\ell} = \sum_{n = 0}^{\infty}x^{n}\bracks{\sum_{\ell = 0}^{\infty} \pars{-1}^{\ell}\pars{-1}^{n - 3\ell}}_{n - 3\ell \geq 0} \\[3mm]&= \sum_{n = 0}^{\infty}{\fermi^{\pars{n}}\pars{0} \over n!}\,x^{n} \quad\mbox{where}\quad \fermi^{\pars{n}}\pars{0} = \pars{-1}^{n}\,n!\!\!\!\!\!\!\!\sum_{\ell = 0 \atop{\vphantom{\LARGE A}n - 3\ell\ \geq\ 0}}^{\infty}1 \end{align}

\begin{align} \fermi^{\pars{2013}}\pars{0} &= \pars{-1}^{2013}\,2013!\!\!\!\!\! \sum_{\ell = 0 \atop {\vphantom{\LARGE A}2013 - 3\ell\ \geq\ 0}}^{\infty} \!\!\!\!\!\!\!\!\!\!\!1 = -\pars{2013!}\sum_{\ell = 0}^{671}1 \end{align} $$ \color{#0000ff}{\large\fermi^{\pars{2013}}\pars{0} = -672 \times \pars{2013!}} $$

Felix Marin
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