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Let $\displaystyle f(x) = x^4 (\frac{1-x^6}{1-x})^4$.

Now I want to find the series representation of this function.

Note that $\displaystyle \frac{d^3}{dx^3} (\frac{1}{1-x})= \frac{6}{(1-x)^4}$.

Then $\frac{1}{(1-x)^4} = \frac{1}{6} \sum n (n-1) (n-2) x^{n-3}$, what gives

$x^4 \frac{1}{(1-x)^4} = \frac{1}{6} \sum n (n-1) (n-2) x^{n+1} = \frac{1}{6} \sum (n-1) (n-2) (n-3) x^{n}$

But how further with the $(1-x^6)^4$ in the numerator?

clubkli
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1 Answers1

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Hint: $1-x^6=(1-x) (x+1) (x^2-x+1) (x^2+x+1)$

In particular $1-x^6=(1-x)p(x)$ for a polynomial $p$.

Reveillark
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  • I do not see how this can help me further. – clubkli Apr 22 '15 at 22:50
  • @clubkli By simplifying the factor $(1-x)$ the fraction disappears and you're left with a polynomial. The series representation of a polynomial is itself. – Reveillark Apr 22 '15 at 22:52
  • In that case: $\displaystyle f(x) = x^4 (\frac{1-x^6}{1-x})^4 = x^4 (x+1)^4 (x^2 -x+1)^4 (x^2 +x+1)^4 = x^4 (1+x+x^2 + x^3 + x^4 + x^5)^4 = x^4 \frac{1}{6} \sum_{n=1} ^{n=5} n (n-1) (n-2) x^{n-3} = \frac{1}{6} \sum_{n=4} ^{n=9} (n-1) (n-2) (n-3) x^{n}$. But now I need to find the value of a coefficient for $n>9$. That can not be computed from this expression. Or is the expression incorrect? – clubkli Apr 23 '15 at 00:05
  • Wolfram Alpha gives the outcomes for different n, but I want to compute this from the power series. – clubkli Apr 23 '15 at 00:12