# Another repertoire method example from Concrete Mathematics

This is my third post on the repertoire method. This method seems quite magical (or perhaps that means I don’t fully understand it 😅). In this case I am referring, as is often the case, to the book Concrete Mathematics by Graham, Patishnick, and Knuth - in particular the example 3, equation 2.40 under “General Methods” for solving a summation. There they present many ways to find the sum of squares $\unicode{9744}_n$ and what I am attempting to cement my knowledge in is the repertoire method approach in the aforementioned example. It’s frightfully clever.

To reproduce it here:

\[\begin{align} R_0 &= \alpha \\ R_n &= R_{n-1} + \beta + \gamma n + \delta n^2 \ \ \ \text{for} \ n > 0 \\ \end{align}\]This will have a solution of the general form

\[R_n = A(n)\alpha + B(n)\beta + C(n)\gamma + D(n)\delta\]Conveniently we have already determined $A(n)$, $B(n)$, and $C(n)$ which I wrote about in this blog post. More rigorously we can say that because if we set $\delta$ to $0$ then we end up with what we determined before.

The book then says, rather tersely for me anyway,

If we now plug in $R_n = n^3$, we find that $n^3$ is the solution when $\alpha = 0$, $\beta = 1$, $\gamma = –3$, $\delta = 3$. Hence

\[3D(n) - 3C(n) + B(n) = n^3\]

The process to arrive at that is pretty much the same as the $n^2$ as discussed here. Which goes something like this:

1 - We start with our recurrence form for which we want to find a closed form \(R_n = R_{n-1} + \beta + \gamma n + \delta n^2 \ \ \ \text{for} \ n > 0\)

As above the closed form will have the below general structure; we need to find $D(n)$ - we already know the rest from before: \(R_n = A(n)\alpha + B(n)\beta + C(n)\gamma + D(n)\delta\)

2a - This is the magic bit: we set the recurrence to $n^3$ and see what $\alpha$, $\beta$, $\gamma$, and $\delta$ need to be in order to make the RHS of the recurrence equal to $n^3$

\[\begin{align} n^3 &= (n-1)^3 + \alpha + \beta + \gamma n + \delta n^2 \\ &= n^3 + -3n^2 -1 + \alpha + \beta + \gamma n + \delta n^2 \\ \end{align}\]2b - Re-arrange things to make the process clearer

\[\begin{align} n^3 &= n^3 + \delta n^2 - 3n^2 + \gamma n + 3n + \beta -1 + \alpha \\ &= n^3 + n^2(\delta - 3) + n(\gamma + 3) + \beta -1 + \alpha \end{align}\]2c - Now we can see what we need to set $\alpha$, $\beta$, $\gamma$, and $\delta$ to in order to make that equation true i.e. $n^3 = n^3$. Specifically $\alpha = 0$, $\beta = 1$, $\gamma = -3$, and $\delta = 3$

3 - Then we substitute (linearly apply?) the found values for $\alpha$, $\beta$, $\gamma$, and $\delta$ into the general form set to n^3.

\[\begin{align} n^3 = A(n)\cdot0 + B(n)\cdot1 + C(n)\cdot-3 + D(n)\cdot3 \\ n^3 = B(n) - 3C(n) + 3D(n) \\ 3D(n) - 3C(n) + B(n) = n^3 \\ \end{align}\]Last re-ordering to match what is in the book at the top of page 45. This will tell us what $D(n)$ needs to be linearly combined with - namely $3$ - for that general form to be $n^3$.

3 - Express in terms of our target - sum of squares $\unicode{9744}_n$. Now that we have built a repertoire let’s see what things have to be set to in order to find $\unicode{9744}_n$. So $\unicode{9744}_n$ equals $\unicode{9744} _{n-1} + n^2$ thus we’ll get $\unicode{9744}_n = R_n$ by setting $\alpha = \beta = \gamma = 0$ and $\delta = 1$. In other words $R_n = A(n)\cdot0 + B(n)\cdot0 + C(n)\cdot0 + D(n)\cdot1$.

So this shows us that $D(n)$ is what we are after - namely - $\unicode{9744}_n$.

4 - Finally substitute in the already known values for $A(n)$, $B(n)$, $C(n)$ and *finally* solving for $D(n)$.

As mentioned above, surprisingly, solving for $D(n)$ gives us our closed form! Let’s take a look:

\[\begin{align} 3D(n) &= n^3 + 3C(n) - B(n) \\ 3D(n) &= n^3 - \frac{3(n+1)n}{2} - n \\ 3D(n) &= n(n + 1/2)(n + 1) \\ D(n) &= \frac{n(n + 1/2)(n + 1)}{3} \\ \end{align}\]So sometimes, in the process of building a repertoire, one of the items actually ends up being The Solution!