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I was given the generating function $f(x)=\frac{1+x^3}{(1+x)^3}$ and I was asked to find $a_9$.

I attempted to break it down into two parts:

$$f(x)=\frac{1}{(1+x)^3}+\frac{x^3}{(1+x)^3}$$

For the first part, I utilized a known formula: $$\frac{1}{(1+ax)^k}=\sum_{n=0}^{\infty}(-1)^n\binom{n+k-1}{n}a^nx^n.$$

However, I encountered difficulty in handling the second term.

RobPratt
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Ariana
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4 Answers4

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Hint $$\frac{(1+x^3)}{(1+x)^3}=\frac{(1+x)(x^2-x+1)}{(1+x)^3}=\frac{x^2-x+1}{(1+x)^2}=\frac{-3}{(1+x)}+\frac{3}{(1+x)^2}+1$$

So $f(x)=1+\sum_{n=1}^{\infty}3x^nn(-1)^n$

DARK
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  • I can't understand why $f(x)=1+\sum_{n=1}^{\infty}3x^nn(-1)^n$? also there is +1 and n starts from 1 not 0. I tried $a_n=3n(-1)^n$ so $a_9=-27$ which is correct! – Ariana Feb 17 '24 at 19:22
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    @Ariana since there is an $n$ term in the sum, it doesn't matter if you start the sum from 0 or 1 – D S Feb 17 '24 at 19:24
  • @Ariana D S is correct. you can write from 0 to it doesnt mtter – DARK Feb 17 '24 at 19:27
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HINT

Alternatively, you can proceed as follows: \begin{align*} \frac{1 + x^{3}}{(1 + x)^{3}} & = \frac{(1 + x)(1 - x + x^{2})}{(1 + x)^{3}} = \frac{1 - x + x^{2}}{(1 + x)^{2}} = -\left(1 - x + x^{2}\right)\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{1 + x}\right) \end{align*} where the derivative's argument corresponds to a geometric series that converges when $|x| < 1$.

Can you take it from here?

Átila Correia
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A more direct approach: $$\begin{align}(1+x^3)\cdot \frac 1{(1+x)^3} &= (1+x^3)\sum_{r=0}^\infty (r+2)(r+1)(-x)^r/2 \\ &= \sum_{r=0}^\infty (r+2)(r+1)(-x)^r/2 - \sum_{r=3}^\infty (r-2)(r-1)(-x)^r/2 \\ &= \sum_{r=0}^\infty (r+2)(r+1)(-x)^r/2 - \left(\sum_{r=0}^\infty (r-2)(r-1)(-x)^r/2-1\right) \\ &= \sum_{r=0}^\infty 3r(-x)^r + 1 \end{align}$$ The same as the other answer.

D S
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Hint: It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of a series. This way we can write for instance for a series $f(x)=\sum_{n=0}^{\infty}a_nx^n$ \begin{align*} [x^k]f(x)=[x^k]\sum_{n=0}^{\infty}a_nx^n=a_k \end{align*}

Here we consider \begin{align*} a_9&=[x^9]f(x)=[x^9]\left(1+x^3\right)\frac{1}{(1+x)^3}\\ &=\left([x^9]+[x^9]x^3\right)\frac{1}{(1+x)^3}\\ &=\left([x^9]+[x^6]\right)\frac{1}{(1+x)^3} \end{align*} noting that $[x^p]x^qf(x)=[x^{p-q}]f(x)$ and the series expansion of $\frac{1}{(1+x)^3}$ is already known to you.

Markus Scheuer
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