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 $|x|<1$,

$f(x)=\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n$

Taking the derivative, we have

$f'(x)=\dfrac1{(1-x)^2}=\displaystyle\sum_{n=0}^\infty nx^{n-1}=\frac1x\sum_{n=0}^\infty nx^n$

and taking the derivative again, we have

$f''(x)=\dfrac2{(1-x)^3}=\displaystyle\frac1x\sum_{n=0}^\infty n^2x^{n-1}-\frac1{x^2}\sum_{n=0}^\infty nx^n$

$f''(x)=\dfrac2{(1-x)^3}=\displaystyle\frac1{x^2}\left(\sum_{n=0}^\infty n^2x^n-\sum_{n=0}^\infty nx^n\right)$

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

$f''(x)=\dfrac{g(x)-xf'(x)}{x^2}$

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

$\implies g(x)=\dfrac{2x^2}{(1-x)^3}+\dfrac x{(1-x)^2}$

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

Then

$g\left(\dfrac1{100}\right)=\dfrac{10,100}{970,299}\approx0.\underline{01}\,\underline{04}\,\underline{09}\,\underline{16}\ldots$

$g\left(\dfrac1{1000}\right)=\dfrac{1,001,000}{997,002,999}\approx0.\underline{001}\,\underline{004}\,\underline{009}\,\underline{016}\ldots$