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How can I find the coefficient of $x^{80}$ in the power series $$(1+x+x^{2}+x^{3}+x^{4}+\cdots)(x^{2}+x^{4}+x^{6}+x^{8}+\cdots)(1+x^{3}+x^{5})\,?$$

Is there a general method to this?

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We can use the fact that \begin{align} 1+x+x^2+\dotsb&=\frac{1}{1-x}\\ x^2+x^4+x^6+\dotsb&=\frac{x^2}{1-x^2} \end{align} so the expression is $$ \frac{x^2(1+x^3+x^5)}{(1-x)^2(1+x)} $$ and we can try doing partial fraction decomposition: $$ \frac{x^2(1+x^3+x^5)}{(1-x)^2(1+x)}= x^4+x^3+3x^2+3x+5+\frac{6x^2+2x-5}{(1-x)^2(1+x)} $$ so we want to decompose the fraction into $$ \frac{A}{(1-x)^2}+\frac{B}{1-x}+\frac{C}{1+x} $$ which gives $A=3/2$, $B=-25/4$, $C=-1/4$. Therefore your product can be written as $$ x^4+x^3+3x^2+3x+5 +\frac{3}{2}\sum_{k\ge0}(k+1)x^k -\frac{25}{4}\sum_{k\ge0}x^k -\frac{1}{4}\sum_{k\ge0}(-1)^kx^k $$ and the coefficient of $x^{80}$ can be read directly as $$ \frac{3}{2}\cdot 81-\frac{25}{4}-\frac{1}{4}=115 $$

Note that this method gives all coefficients.

I've used the formula $$ \frac{1}{(1-x)^2}=\sum_{k\ge0}(k+1)x^k $$ that can be deduced by differentiating $$ \frac{1}{1-x}=\sum_{k\ge0}x^k $$

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We have $(1+x+x^2+\cdots)(x^2+x^4+\cdots)(1+x^3+x^5)$ Then we want The coeficient of $x^{80}$, note that $x^nx^m=x^{n+m}$ then searching The ways of write $80$ using the sets $\{0,1,\cdots\},\{2,4,\cdots\},\{0,3,5\}$. $$\begin{align} 80&=0+80+0\\ &=2+78+0\\ &=4+76+0\\ &\vdots\\ &=78+2+0\\ &=1+76+3\\ &=3+74+3\\ &=5+72+3\\ &\vdots\\ &=75+2+3\\ &=1+74+5\\ &=3+72+5\\ &\vdots\\ &=73+2+5 \end{align}$$ Then The coeficiente will be $$\begin{align} C_{80}&=\stackrel{2n}{(40-1+1)}+\stackrel{2n+1}{(37-0+1)}+\stackrel{2n+1}{(36-0+1)}\\ &=40+38+37\\ &=115 \end{align}$$ Then if nothing as wrong The answer must be $115$

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  • $\begingroup$ Unfortunately when I inputted it in, the answer was wrong. I did however find out that the "power series representation" of all three products is x^5+x^3+1 / (1-x)(1-x^2). $\endgroup$ – Mathy Person Dec 9 '14 at 3:41
  • $\begingroup$ sure you have not forgotten any data from the given problem? i will revise and fix the answer later, thx. $\endgroup$ – cand Dec 9 '14 at 3:58
  • $\begingroup$ There was a typo in my original problem statement....it should have been *(1+x^2+x^5), rather than (1+x^3+x^5). Sorry for the confusion!! $\endgroup$ – Mathy Person Dec 9 '14 at 4:35
  • $\begingroup$ You sure that The middle product is $(x^2+x^4+\cdots)$ instead of $(1+x^2+x^4+\cdots)$? $\endgroup$ – cand Dec 9 '14 at 13:10
  • $\begingroup$ Yes, I'm sure. [[15 characters in length left]] $\endgroup$ – Mathy Person Dec 9 '14 at 21:08
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\bracks{x^{80}}\bracks{\pars{1 + x + x^{2} + x^{3} + x^{4} +\cdots} \pars{x^{2} + x^{4} + x^{6} + x^{8} + \cdots} \pars{1 + x^{3} + x^{5}}}:\ {\large ?}}$

$$ {\rm Lets}\quad \dsc{\fermi\pars{x}}\equiv \dsc{\pars{1 + x + x^{2} + x^{3} + x^{4} +\cdots} \pars{x^{2} + x^{4} + x^{6} + x^{8} + \cdots}} $$

Then, $$ \bracks{x^{80}}\bracks{\fermi\pars{x}\pars{1 + x^{3} + x^{5}}} =\bracks{x^{80}}\fermi\pars{x} + \bracks{x^{77}}\fermi\pars{x} + \bracks{x^{75}}\fermi\pars{x} $$

and \begin{align} \fermi\pars{x}&=\sum_{i\ =\ 0}^{\infty}x^{i}\sum_{j\ =\ 0}^{\infty}x^{2j + 2} =\sum_{i\ =\ 0}^{\infty}\sum_{j\ =\ 0}^{\infty}x^{i + 2j + 2} =\sum_{i\ =\ 0}^{\infty}\sum_{j\ =\ 0}^{\infty} \sum_{n\ =\ 2}^{\infty}x^{n}\delta_{n,i + 2j + 2} \\[5mm]&=\sum_{n\ =\ 2}^{\infty}x^{n}\pars{% \sum_{j\ =\ 0}^{\infty}\sum_{i\ =\ 0}^{\infty}\delta_{i,n - 2j - 2}} =\sum_{n\ =\ 2}^{\infty}x^{n} \pars{\left.\sum_{j\ =\ 0}^{\infty}1\right\vert_{n - 2j - 2\ \geq\ 0}} \\[5mm]&=\sum_{n\ =\ 2}^{\infty}x^{n} \pars{\left.\sum_{j\ =\ 0}^{\infty}1\right\vert_{j\ \leq\ n/2 - 1}} =\sum_{n\ =\ 2}^{\infty}x^{n}\pars{\sum_{j\ =\ 0}^{\floor{n/2 - 1}}1} =\sum_{n\ =\ 2}^{\infty}\pars{\floor{{n \over 2} - 1} + 1}x^{n} \end{align}

$$ \bracks{x^{n}}\fermi\pars{x} = \floor{n - 2 \over 2} + 1\,,\qquad n \geq 2 $$

The $\color{#66f}{\large\mbox{final result}}$ is given by: $$ \pars{\floor{78 \over 2} + 1} + \pars{\floor{75 \over 2} + 1} +\pars{\floor{73 \over 2} + 1}=40 + 38 + 37=\color{#66f}{\Large 115} $$

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