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I’m stuck on something in generating functionology. The first problem asks: Find the ordinary power series generating functions of the sequence in simple closed form for the sequence $a_n = n$. The sequence is defined as $n ≥ 0$.

I figured out how to get to $A(x) = x/((1-x)^2)$. That’s not an issue.

However, the book lists the answer as $(xD)(1/(1-x)) = x/((1-x)^2)$

Where did the D come from? How can I get my answer in terms of D?

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  • $\begingroup$ The $D$ is a differential operator. $\endgroup$ – John Wayland Bales Aug 8 '18 at 21:17
  • $\begingroup$ $D$ is $\frac d{dx}$. $\endgroup$ – Lord Shark the Unknown Aug 8 '18 at 21:34
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You should think of "$xD$" as an operator. "$xD$" read right to left means differentiate and then multiply by $x$. So $$ (xD)\frac{1}{1-x}=x\left(\frac{1}{1-x}\right)'=x\times\frac{1}{1-x^2}=\frac{x}{1-x^2}. $$ Note that if $$ B(x)=\sum_{n\geq 0} b_n z^n $$ then $(xD)B$ is the ogf which corresponds to the sequence $(nb_n)_0^\infty$.

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\begin{eqnarray} A(x)&=&0+1\cdot x+2\cdot x^2+3\cdot x^3+\cdots\\ &=&x(1+2x+3x^2+\cdots)\\ &=&x\frac{d}{dx}(1+x+x^2+x^3+\cdots)\\ &=&x\frac{d}{dx}\left(\frac{1}{1-x}\right)\\ &=&\frac{x}{(1-x)^2} \end{eqnarray}

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