# The Unapologetic Mathematician

## Faulhaber’s Fabulous Formula

Tonight I’ll be telling the undergrad math club here about a nifty little thing I picked up from Scott Carter last week, who in turn picked it up from John Conway shortly before that: Faulhaber’s Fabulous Formula. The formula itself expresses the sum of the first $n$ $p$-th powers as (sort of) an integral: $\displaystyle\sum\limits_{k=1}^nk^p=\int\limits_b^{b+n}x^p\,dx$

The question here is what this mysterious $b$ is, and the answer to that takes us into the shadowy netherworld of 19th-century analysis: the umbral calculus.

Our mysterious character $b$ is defined by a simple equation: $b^n=(b-1)^n$. For $n=0$ this is trivially true, as both sides are $1$. For $n=1$ this is clearly nonsense, as it says $b^1=b^1-1$, or $0=-1$, so let’s just not apply the relation for this value of $n$. For higher $n$, it’s still nonsense, but of a subtler sort. So let’s dive in.

When we consider $n=2$ we get $b^2=b^2-2b^1+1$. The $b^2$ terms cancel, and what’s left tells us that $b^1=\frac{1}{2}$. So $b=\frac{1}{2}$? Not quite, because that would make sense!

Let’s look at $n=3$. Now the relation reads $b^3=b^3-3b^2+3b^1-1$. Again the $b^3$ terms cancel, and now we know that $b^1=\frac{1}{2}$. Thus we can solve to find $b^2=1/6$. Huh?

Moving on to $n=4$, we find $b^4=b^4-4b^3+6b^2-4b^1+1$. The $b^4$ terms cancel and we substitute in the values we know for $b^2$ and $b^1$ to find $b^3=0$. As I told Scott, “you do realize that this is f___ing nuts, right?”

We can keep going on like this, spinning out values of $b^n$ for all $n\geq1$ $b^4=-\frac{1}{30}$, $b^5=0$, $b^6=\frac{1}{42}$, $b^7=0$, $b^8=-\frac{1}{30}$, $b^9=0$ — which have absolutely nothing to do with powers of any base $b$. But we use them as powers in the relation. Remember: this is prima facie nonsense.

But onwards, mathematical soldiers, integrating as to war: we take the integral and break it up into $n$ pieces: $\displaystyle\int\limits_b^{b+n}x^p\,dx=\sum\limits_{k=1}^n\int\limits_{b+k-1}^{b+k}x^p\,dx$

Even (most of) my calculus students can take the antiderivatives on the right to get $\displaystyle\frac{x^{p+1}}{p+1}\Biggl\vert_{b-1+k}^{b+k}=\frac{(b+k)^{p+1}-((b-1)+k)^{p+1}}{p+1}$

Now we use the binomial formula to expand each of the terms in the numerator. $\displaystyle\frac{1}{p+1}\left(\sum\limits_{l=0}^{p+1}\binom{p+1}{l}b^lk^{p+1-l}-\sum\limits_{l=0}^{p+1}\binom{p+1}{l}(b-1)^lk^{p+1-l}\right)$

and combine to find $\displaystyle\frac{1}{p+1}\sum\limits_{l=0}^{p+1}\binom{p+1}{l}k^{p+1-l}\left(b^l-(b-1)^l\right)$

But we said that $b^n=(b-1)^n$ unless $n=1$! That is, all these terms cancel except for one, which is just $k^p$. And so the integral we started with is the sum of the first $n$ $p$-th powers, as we wanted to show.

So how do we apply this? Well, we can evaluate the integral more directly rather than breaking it into a big sum. $\displaystyle\int_b^{b+n}x^p\,dx=\frac{(b+n)^{p+1}-b^{p+1}}{p+1}=\sum\limits_{l=0}^{p}\binom{p+1}{l}\frac{n^{p+1-l}}{p+1}b^l$

And now we use the values we cranked out before. Let’s write it out for $p=1$, which many of you already know the answer for. $\displaystyle\binom{2}{0}\frac{n^2}{2}b^0+\binom{2}{1}\frac{n}{2}b^1=\frac{n^2}{2}+\frac{n}{2}=\frac{n(n+1)}{2}$

What about $p=2$? It’s not as well-known, but some of you might recognize it when we write it out. $\displaystyle\binom{3}{0}\frac{n^3}{3}b^0+\binom{3}{1}\frac{n^2}{3}b^1+\binom{3}{2}\frac{n}{3}b^2=\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}=\frac{n(n+1)(2n+1)}{6}$

In fact, given the numbers we cranked out back at the start of the post, we can use this formula to easily write out the sum of the first $n$ 9th powers: $\displaystyle\frac{n^2(n+1)^2(n^2+n-1)(2n^4+4n^3-n^2-3n+3)}{20}$

all thanks to Faulhaber’s Fabulous Formula.

November 14, 2007 Posted by | Uncategorized | 10 Comments