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Artist 52 [7]
3 years ago
8

Find a compact form for generating functions of the sequence 1, 8,27,... , k^3

Mathematics
1 answer:
pantera1 [17]3 years ago
5 0

This sequence has generating function

F(x)=\displaystyle\sum_{k\ge0}k^3x^k

(if we include k=0 for a moment)

Recall that for |x|, we have

\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k

Take the derivative to get

\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k

\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k

Take the derivative again:

\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k

\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k

Take the derivative one more time:

\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k

\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k

so we have

\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}

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