Question

Find a generating function for the sequence of squares: g(x) = P∞ n=0 n 2x n . Then for fun, as above evaluate your expression g(1/100) or g(1/1000) to get a fraction that contains the squares in its decimal expansion.

Answer 1

The sequence of squares,
`\{n^2\}_(n\ge0)`, has generating function

`g(x)=\displaystyle\sum_(n=0)^\infty n^2x^n`

Recall that for
`|x|<1`,

`f(x)=\frac1{1-x}=\displaystyle\sum_(n=0)^\infty x^n`

Taking the derivative, we have

`f'(x)=\frac1{(1-x)^2}=\displaystyle\sum_(n=0)^\infty nx^(n-1)=\frac1x\sum_(n=0)^\infty nx^n`

and taking the derivative again, we have

`f''(x)=\frac2{(1-x)^3}=\displaystyle\frac1x\sum_(n=0)^\infty n^2x^(n-1)-\frac1{x^2}\sum_(n=0)^\infty nx^n`

`f''(x)=\frac2{(1-x)^3}=\displaystyle\frac1{x^2}\left(\sum_(n=0)^\infty n^2x^n-\sum_(n=0)^\infty nx^n\right)`

From this we can get an expression for
`g(x)` in terms of the derivatives of
`f(x)`:

`f''(x)=(g(x)-xf'(x))/(x^2)`

`\implies g(x)=x^2f''(x)+xf'(x)`

`\implies g(x)=(2x^2)/((1-x)^3)+\frac x{(1-x)^2}`

`\implies\boxed{g(x)=(x^2+x)/((1-x)^3)}`

Then

`g\left(\frac1{100}\right)=(10,100)/(970,299)\approx0.\underline{01}\,\underline{04}\,\underline{09}\,\underline{16}\ldots`

`g\left(\frac1{1000}\right)=(1,001,000)/(997,002,999)\approx0.\underline{001}\,\underline{004}\,\underline{009}\,\underline{016}\ldots`