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:


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\geq1b^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:


Even (most of) my calculus students can take the antiderivatives on the right to get


Now we use the binomial formula to expand each of the terms in the numerator.


and combine to find


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.


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.


What about p=2? It’s not as well-known, but some of you might recognize it when we write it out.


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:


all thanks to Faulhaber’s Fabulous Formula.

November 14, 2007 Posted by | Uncategorized | 10 Comments