The Unapologetic Mathematician

Mathematics for the interested outsider

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