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I have a homework question which requires me to find the coefficient of $x^{14}$ in the expression: $\dfrac{5x^2-x^4}{(1-x)^3}$

I have not figured out a way to do this (I believe this is because my algebra is weak).

This is a question in a combinatorics class and has to do with Generating Functions if that is any help.

Help is greatly appreciated. Thanks, Jason

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Hint: $\frac{1}{1-x}=1+x+x^2+\ldots$. So $\frac{1}{(1-x)^3}=?$ When you multiply this by $5x^2-x^4$, where do the terms in $x^{14}$ come from?

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    $\begingroup$ Note also that the best way to get $\frac{1}{(1-x)^3}$ from $\frac{1}{1-x}$ in this context may involve differentiation rather than multiplication. $\endgroup$ – Robert Israel Sep 4 '11 at 19:30
  • $\begingroup$ hmm Interesting but I got $5*91 - 4*66 =264$ which I think is incorrect what did I do wrong? (EDIT: Actuall I got it right just made a mistake its $-1*66$ and then you get $389$ Thanks ALOT :) ) $\endgroup$ – Jason Sep 4 '11 at 19:37
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    $\begingroup$ Or one could use the binomial series. The series for $(1-x)^{-n}$ is particularly easy to remember (perhaps hard to calculate). $\endgroup$ – Srivatsan Sep 4 '11 at 19:38
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Worth mention is the use of partial fractions to derive differential operator representations, viz. $$\rm\displaystyle \frac{5-x^2}{(1-x)^3}\ =\ \frac{4}{(1-x)^3}- \frac{2}{(1-x)^2} - \frac{1}{1-x}\ =\ (2\ D^2 + 2\ D - 1)\ \bigg(\frac{1}{1-x}\bigg),\quad D\ =\ \frac{d}{dx} $$

Consider a general $2\:$nd-order constant coefficient differential operator applied to $\rm\:1/(1-x)\:.$

$\rm\qquad\displaystyle \frac{2\:a\:+b+c-(b+2\:c)\ x + c\ x^2}{(1-x)^3}\ =\ (a\ D^2 + b\ D + c)\ \bigg(\!\!\frac{1}{1-x}\bigg)$ $\rm\displaystyle\phantom{\qquad \frac{2\:a\:+b+c-(b+2\:c)\ x + c\ x^2}{(1-x)^3}}\ =\ (a\ D^2 + b\ D + c)\ \sum_{k\:=\:0}^{\infty}\ x^k $ $\rm\displaystyle\phantom{\qquad \frac{2\:a\:+b+c-(b+2\:c)\ x + c\ x^2}{(1-x)^3}}\ =\ \sum_{k\:=\:0}^{\infty}\ ((k+2)\:(k+1)\:a+(k+1)\:b+c)\ x^k$
Thus $\rm\ a = 2 = b,\ c = -1\ $ yields

$$\rm \frac{5-x^2}{(1-x)^3}\: =\ \sum_{k\:=\:0}^\infty\ (2\:(k+3)\:(k+1)-1)\ x^k $$

Hence we find that the coefficient of $\rm\:x^{12}\:$ equals $\ 2\cdot 15\cdot 13 -1\ =\ 389\:.$

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  • $\begingroup$ Nice. An economical calculation :) $\endgroup$ – Srivatsan Sep 4 '11 at 21:39
  • $\begingroup$ @Bill: I was looking over your tex code (which is very lovely), and I noticed a few things. Firstly, your phantoming is wonderful. I noticed you started with \\rm and that it changed your variables to non-italics. What is it, and is there a way to toggle it on/off? $\endgroup$ – davidlowryduda Sep 4 '11 at 21:42
  • $\begingroup$ @mix I use roman fonts (\rm) for math because I find the MathJax rendering of italic too ugly. The use of phantom is a kludge. It should be eqnarray but that is sometimes buggy in MathJax. $\endgroup$ – Bill Dubuque Sep 4 '11 at 21:53
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To complement @Bill's answer, I will explain one way to quickly obtain a partial fraction expansion in this example.

Use substitution. Set $x = 1 -y$, so that the given expression becomes: $$ \frac{5(1-y)^2 - (1-y)^4}{y^3} = \frac{4 - 6y - y^2 + 4y^3 - y^4}{y^3} = \frac{4}{y^3} - \frac{6}{y^2} - \frac{1}{y} + 4 - y. $$ Substituting back $1-x$ for $y$, we get: $$ \frac{4}{(1-x)^3} - \frac{6}{(1-x)^2} - \frac{1}{(1-x)} + 3 + x. $$

Using the binomial series1, we can calculate the coefficient of $x^{14}$ to be: $$ 4 \binom{14+3-1}{14} - 6 \binom{14+2-1}{14} - \binom{14+1-1}{14} + 0 + 0 = 4 \binom{16}{2} - 6 \times 15 - 1 = 389. $$

1As I remarked in a comment above, the binomial series for $(1-x)^{\beta+1}$ has a particularly simple form.

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  • $\begingroup$ +1 Very cool. How did you come up with that? The substitution part I mean. $\endgroup$ – Robert S. Barnes Jan 19 '12 at 18:30
  • $\begingroup$ @Robert: It is inspired by the special form of the denominator $(1-x)^k$. Admittedly, the scope of this method is quite narrow: the denominator should be a power of a degree-1 polynomial. But the method is usually straightforward when it applies. (Since the other answers covered the usual approaches, I also wanted to point out another way to tackle the problem. :-)) $\endgroup$ – Srivatsan Jan 19 '12 at 19:20
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First pull a $5$ and an $x^2$ from the rational function. This gives you $$ \frac{5x^2\left(1 - \left(\frac{1}{5}x^2\right) \right)}{(1-x)^3}. $$ Then, use the expansions $$ (1+x^m)^n = 1 - \binom{n}{1}x + \binom{n}{2}x^2 + \dots + (-1)^k\binom{n}{k}x^{km} + \dots + (-1)^n \binom{n}{n}x^{nm} $$ and $$ \frac{1}{(1-x)^n} = 1 + \binom{1 + n -1}{1}x + \binom{2 + n -1}{2}x^2 + \dots + \binom{r + n -1}{r}x^r + \dots. $$ To find the coefficient of $x^{12}$ (Why $x^{12}$?)

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You are after: \begin{align} [x^{14}] \frac{5 x^2 - x^4}{(1 - x)^3} &= 5 [x^{12}] (1 - x)^{-3} - [x^{10}] (1 - x)^{-3} \\ &= 5 (-1)^{12} \binom{-3}{12} - (-1)^{10} \binom{-3}{10} \\ &= 5 \binom{12 + 3 - 1}{3 - 1} - \binom{10 + 3 - 1}{3 - 1} \\ &= 389 \end{align} Here I used: $$ \binom{-n}{k} = (-1)^k \binom{k + n - 1}{n - 1} $$

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