1 Answer

Rayshon · Answer 1 · 2020-02-28T11:08:44+0000

The generating function is
f(x) where

$f(x)=\displaystyle\sum_(k=0)^\infty a_kx^k$

with
a_k=k^3 for
$k\ge0$ .

Recall that for
|x|<1 , we have

$g(x)=\frac1{1-x}=\displaystyle\sum_(k=0)^\infty x^k$

Taking the derivative gives

$g'(x)=\frac1{(1-x)^2}=\displaystyle\sum_(k=1)^\infty kx^(k-1)=\sum_(k=0)^\infty(k+1)x^k$

$\implies g'(x)-g(x)=\frac x{(1-x)^2}=\displaystyle\sum_(k=0)^\infty kx^k$

Taking the derivative again, we get

$g''(x)=\frac2{(1-x)^3}=\displaystyle\sum_(k=2)^\infty k(k-1)x^(k-2)=\sum_(k=0)^\infty(k^2+3k+2)x^k$

$\implies g''(x)-3g'(x)+g(x)=(x^2+x)/((1-x)^3)=\displaystyle\sum_(k=0)^\infty k^2x^k$

Take the derivative one last time to get

$g'''(x)=\frac6{(1-x)^4}=\displaystyle\sum_(k=3)^\infty k(k-1)(k-2)x^(k-3)=\sum_(k=0)^\infty(k^3+6k^2+11k+6)x^k$

$\implies g'''(x)-6g''(x)+7g'(x)-g(x)=(x^3+4x^2+x)/((1-x)^4)=\displaystyle\sum_(k=0)^\infty k^3x^k$

So the generating function is

$\boxed{f(x)=(x^3+4x^2+x)/((1-x)^4)}$

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