# A little bit of generating-function-ology

Posted by: Gary Ernest Davis on: April 2, 2011

James Tanton (@jamestanton) raised the interesting question of what numbers we get if we from the infinite decimals $0.123456789\vert 10\vert 11\vert 12\vert 13\ldots$ and $0.\vert 1^2\vert 2^2\vert 3^2\vert 4^2 \ldots$

Here we are placing the numbers $n$ and $n^2$ in the $n^{th}$ digit place in the infinite decimals.

This means that we will have to perform a lot of carries to figure out what these  numbers are.

Or does it?

What James Tanton is asking us to calculate are the numbers $\frac{1}{10}+\frac{2}{10^2}+\frac{3}{10^3}+\frac{4}{10^4}+\ldots$ and $\frac{1}{10}+\frac{2^2}{10^2}+\frac{3^2}{10^3}+\frac{4^2}{10^4}+\ldots$

By combining two ideas of Isaac newton – calculus and power series – we can, relatively easily, calculate these numbers, and more.

They are in fact rational numbers – fractions.

Curiously, Isaac Newton thought of power series as generalizations of decimal numbers where we expand a number using the digits $0,1,\ldots,9$ and powers of $\frac{1}{10}$.

We will do just the reverse.

Let’s denote the infinite sum $x+2^kx^2+3^kx^3+4^kx^4+\ldots$, where $k$ is a positive integer, by $S(k;x)$.

We will treat $S(k;x)$ as a formal power series to be manipulated term-by-term, but will interpret it as a number when $x$ is chosen appropriately – mainly as $x=\frac{1}{10}$.

The formal derivative, $S'(k;x)$ of $S(k;x)$, is the infinite sum

$1+2\times 2^kx+3\times 3^kx^2+4\times 4^kx^3+\ldots= 1+2^{k+1}x+3^{k+1}x^2+4^{k+1}x^3+\ldots$

x

When we multiply $S'(k;x)$ by $x$ we get the infinite sum $x+2^{k+1}x^2+3^{k+1}x^3+4^{k+1}x^4+\ldots=S(k+1;x)$

That is, $S(k+1;x)=xS'(k;x)$.

Now we can fairly easily figure out $S(1;x)=x+2x^2+3x^3+4x^4+\ldots=x(1+2x+3x^2+4x^3+\ldots)$

That’s because if $\Sigma=1+2x+3x^2+4x^3+\ldots$ then $\Sigma-x\Sigma=1+x+x^2+x^3+x^4+\ldots = \frac{1}{1-x}$

so $\Sigma=\frac{1}{(1-x)^2}$ and $S(1:x)=\frac{x}{(1-x)^2}$.

In particular, $S(1:\frac{1}{10})=\frac{1}{10}+\frac{2}{10^2}+\frac{3}{10^3}+\frac{4}{10^4}+\ldots$

$=\frac{1/10}{(1-10)^2}=\frac{1}{10}\frac{100}{81}=\frac{10}{81}$

This allows us also to calculate $S(2:x)=xS'(1;x)=x\frac{dx/(1-x)^2}{dx}= x\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\frac{x(1+x)}{(1-x)^3}$

In particular, $S(2:\frac{1}{10})=\frac{1}{10}+\frac{2^2}{10^2}+\frac{3^2}{10^2}+\ldots$

$=\frac{1+1/10}{(1-1/10)^3} =\frac{11}{10}\frac{100}{9^3} =\frac{110}{729}$

We can continue this way to calculate $\frac{1}{10}+\frac{2^k}{10^2}+\frac{3^k}{10^3}+\ldots$ for higher values of $k$.

For example $S(3:x)=xS'(2:x)=\frac{x(1+4x+x^2)}{(1-x)^4}$ so $S(3:\frac{1}{10})=\frac{470}{2187}$

One thing we notice is that we always get a power of 9 in the denominator.

Of course a 3 in the numerator might cancel a 3 in the denominator, so we can say that in reduced form $\frac{1}{10}+\frac{2^k}{10^2}+\frac{3^k}{10^3}+\ldots$ will have a denominator that is a power of 3.