# Faulhaber's formula

In mathematics, Faulhaber's formula, named after Johann Faulhaber, expresses the sum

$\sum_{k=1}^n k^p = 1^p + 2^p + 3^p + \cdots + n^p$

as a (p + 1)th-degree polynomial function of n, the coefficients involving Bernoulli numbers Bj.

The formula says

$\sum_{k=1}^n k^p = {1 \over p+1} \sum_{j=0}^p (-1)^j{p+1 \choose j} B_j n^{p+1-j},\qquad \mbox{where}~B_1 = -\frac{1}{2}.$

Faulhaber himself did not know the formula in this form, but only computed the first seventeen polynomials; the general form was established with the discovery of the Bernoulli numbers (see History section below). The derivation of Faulhaber's formula is available in The Book of Numbers by John Horton Conway and Richard K. Guy.[1]

There is also a similar (but somehow simpler) expression: using the idea of telescoping, one gets:

$(n+1)^{k+1} - 1 = \sum_{m = 1}^n (m+1)^{k+1} - m^{k+1} = \sum_{p = 0}^k \binom{k+1}{p} (1^p+2^p+ \dots + n^p)$.

This in particular yields the examples below.

## Examples

$1 + 2 + 3 + \cdots + n = {n(n+1) \over 2} = {n^2 + n \over 2}$ (the triangular numbers)
$1^2 + 2^2 + 3^2 + \cdots + n^2 = {n(n+1)(2n+1) \over 6} = {2n^3 + 3n^2 + n \over 6}$ (the square pyramidal numbers)
$1^3 + 2^3 + 3^3 + \cdots + n^3 = \left({n(n+1) \over 2}\right)^2 = {n^4 + 2n^3 + n^2 \over 4}$ (the squared triangular numbers)
$1^4 + 2^4 + 3^4 + \cdots + n^4 = {n(n+1)(2n+1)(3n^2+3n-1) \over 30} = {6n^5 + 15n^4 + 10n^3 - n \over 30}$
$1^5 + 2^5 + 3^5 + \cdots + n^5 = {n^2(n+1)^2(2n^2+2n-1)\over 12} = {2n^6 + 6n^5 + 5n^4 - n^2 \over 12}$
$1^6 + 2^6 + 3^6 + \cdots + n^6 = {n(n+1)(2n+1)(3n^4+6n^3-3n+1)\over 42} = {6n^7 + 21n^6 + 21n^5 -7n^3 + n \over 42}$

## Proof

Let the sums Sp(n) be defined as

$S_p(0)=0$
$S_p(n+1)=S_p(n)+(n+1)^p.$

If the closed forms for Sp(n) can be expressed as polynomials they will not have constant terms, and will be of the form

$\sum_{k=0}^\infty a_{p,k}n^{k+1}$

for some matrix of coefficients ap,k. Substituting this into the inductive part of the definition above, applying the binomial theorem, and rearranging, we have

$\sum_{k=0}^\infty a_{p,k}(n+1)^{k+1}=\sum_{k=0}^\infty a_{p,k}n^{k+1}+(n+1)^p$
$\sum_{k=0}^\infty a_{p,k}\left[\left(\sum_{j=0}^{k+1}\binom{k+1}{j}n^j\right)-n^{k+1}\right]=(n+1)^p$
$\sum_{k=0}^\infty a_{p,k}\sum_{j=0}^k\binom{k+1}{j}n^j=\sum_{j=0}^\infty\binom{p}{j}n^j.$

The double sum on the left hand side can be rearranged, noting that jk

$\sum_{j=0}^\infty n^j\sum_{k=j}^\infty\binom{k+1}{j}a_{p,k}=\sum_{j=0}^\infty\binom{p}{j}n^j.$

Aligning coefficients then gives the relation

$\sum_{k=j}^\infty\binom{k+1}{j}a_{p,k}=\binom{p}{j}.$

If p is a natural number (the reasoning so far has not assumed it is), then the right hand side is 0 when j>p, and we can assert that ap,k=0 for all k>p. Note that this is not the only possible solution; an infinite series that satisfied this property for its coefficients would also be valid, and would correspond to a closed form with a periodic component that was 0 at every natural number, such as sin(πn). The multiple solutions are an immediate result of the closed forms being defined only at the integers.

Asserting now that the matrix of coefficients is triangular, and multiplying both sides by j!, we have the recurrence relation giving all coefficients for any p (the first at j=p, the next at j=p-1, etc.).

$\sum_{k=j}^p(k+1)_j a_{p,k}=(p)_j$

which makes use of the Pochhammer symbol for the falling factorial.

Just as there is a diagonal relationship between binomial coefficients and between falling factorials, there is such a relationship between these coefficients, which can be demonstrated by 'breaking' the falling factorials at an arbitrary number tj.

$\sum_{k=j}^p(k+1)_t (k+1-t)_{j-t}a_{p,k}=(p)_t (p-t)_{j-t}$

Substituting k=k'+t and rearranging gives

$\sum_{k'=j-t}^{p-t}(k'+1)_{j-t}\left[\frac{(k'+t+1)_t}{(p)_t}a_{p,k'+t}\right]=(p-t)_{j-t}.$

This is of the same form as the original recurrence relation, which shows that

$\frac{(k'+t+1)_t}{(p)_t}a_{p,k'+t}=a_{p-t,k'}.$

In particular, for k'=0, we have

$a_{p,t}=\frac{(p)_t}{(t+1)_t}a_{p-t,0}=\binom{p}{t}\frac{a_{p-t,0}}{t+1}.$

The linear coefficients ap,0 are thus fundamental. These are the Bernoulli numbers, and so we have the final form

$\sum_{k=1}^nk^p=\sum_{j=0}^p\binom{p}{j}\frac{B_{p-j}}{j+1}n^{j+1}$

with each Bernoulli number derivable from this formula itself at n=1

$\sum_{j=0}^p\binom{p}{j}\frac{B_{p-j}}{j+1}=1$

and so B1=1/2 here.

## Alternate expression

Due to the relationship between the Riemann zeta function and Bernoulli numbers, and the observations below, an arguably more natural expression in terms of B1=1/2 is

$\sum_{k=1}^nk^p=\sum_{j=0}^p\binom{p}{j}\frac{B_{p-j}}{j+1}n^{j+1}.$

The older expression given earlier follows from this one as a result of the (somewhat coincidental) fact that when the sum Sp(n) is extended to negative integers using the property Sp(n−1)=Sp(n)−np, we have

\begin{align} S_p(-n) & = S_p(-1)-\sum_{k=1}^{n-1}(-k)^p=S_p(-1)-\left(\sum_{k=1}^n(-k)^p-(-n)^p\right) \\ & = S_p(-1)-(-1)^p\left(\sum_{k=1}^nk^p-n^p\right). \end{align}

Since Sp(−1)=Sp(0)−0p, it is 0 for p>0, but is indeterminate when p=0. S0(n)=n, however, so the convention is that Sp(−1) is the negated Kronecker delta, giving

$S_p(-n)=(-1)^{p+1}\left(S_p(n)-n^p\right)-\delta_p.$

The subtraction of np (when it is not cancelled by the delta for p=0) changes the sign of B1 (it is the coefficient of np), giving

$S_p(n)=(-1)^{p+1}\sum_{j=0}^p\binom{p}{j}\frac{B_{p-j}}{j+1}(-n)^{j+1}$

in terms of B1=−1/2, which is essentially the same as the conventional expression given earlier.

The use of this expression at n=−1 also gives

$\sum_{j=0}^p\binom{p}{j}\frac{B_{p-j}}{j+1}=\delta_p$

for a recurrence relation giving the B1=−1/2 sequence.

## Relation to Bernoulli polynomials

One may also write

$\sum_{k=1}^{n} k^p = \frac{B_{p+1}(n+1)-B_{p+1}(0)}{p+1},$

where Bj (t) is the jth Bernoulli polynomial.

## Umbral form

In the classic umbral calculus one formally treats the indices j in a sequence Bj as if they were exponents, so that, in this case we can apply the binomial theorem and say

$\sum_{k=1}^n k^p = {1 \over p+1} \sum_{j=0}^p {p+1 \choose j} B_j n^{p+1-j} = {1 \over p+1} \sum_{j=0}^p {p+1 \choose j} B^j n^{p+1-j}$

$= {(B+n)^{p+1} - B^{p+1} \over p+1}.$

In the modern umbral calculus, one considers the linear functional T on the vector space of polynomials in a variable b given by

$T(b^j) = B_j.\,$

Then one can say

$\sum_{k=1}^n k^p = {1 \over p+1} \sum_{j=0}^p {p+1 \choose j} B_j n^{p+1-j} = {1 \over p+1} \sum_{j=0}^p {p+1 \choose j} T(b^j) n^{p+1-j}$

$= {1 \over p+1} T\left(\sum_{j=0}^p {p+1 \choose j} b^j n^{p+1-j} \right) = T\left({(b+n)^{p+1} - b^{p+1} \over p+1}\right).$

## Faulhaber polynomials

The term Faulhaber polynomials is used by some authors to refer to something other than the polynomial sequence given above. Faulhaber observed that if p is odd, then

$1^p + 2^p + 3^p + \cdots + n^p$

is a polynomial function of

$a=1+2+3+\cdots+n= \frac{n(n+1)}{2}.$

In particular:

$1^3 + 2^3 + 3^3 + \cdots + n^3 = a^2;$

$1^5 + 2^5 + 3^5 + \cdots + n^5 = {4a^3 - a^2 \over 3};$

$1^7 + 2^7 + 3^7 + \cdots + n^7 = {12a^4 -8a^3 + 2a^2 \over 6};$

$1^9 + 2^9 + 3^9 + \cdots + n^9 = {16a^5 - 20a^4 +12a^3 - 3a^2 \over 5};$

$1^{11} + 2^{11} + 3^{11} + \cdots + n^{11} = {32a^6 - 64a^5 + 68a^4 - 40a^3 + 10a^2 \over 6}.$

More generally,

\begin{align} 1^{2p+1} + 2^{2p+1} &+ 3^{2p+1} + \cdots + n^{2p+1}\\ &= \frac{1}{2^{2p+2}(2p+2)} \sum_{q=0}^p \binom{2p+2}{2q} (2-2^{2q})~ B_{2q} ~\left[(8a+1)^{p+1-q}-1\right]. \end{align}

The first of these identities, for the case p = 3, is known as Nicomachus's theorem. Some authors call the polynomials on the right hand sides of these identities "Faulhaber polynomials in a". The polynomials in the right-hand sides are divisible by a 2 because for j > 1 odd the Bernoulli number Bj is 0.

Faulhaber also knew that if a sum for an odd power is given by

$\sum_{k=1}^n k^{2m+1} = c_1 a^2 + c_2 a^3 + \cdots + c_m a^{m+1}$

then the sum for the even power just below is given by

$\sum_{k=1}^n k^{2m} = \frac{n+1/2}{2m+1}(2 c_1 a + 3 c_2 a^2+\cdots + (m+1) c_m a^m).$

Note that the polynomial in parentheses is the derivative of the polynomial above with respect to a.

Since a = n(n + 1)/2, these formulae show that for an odd power (greater than 1), the sum is a polynomial in n having factors n2 and (n + 1)2, while for an even power the polynomial has factors n, n + ½ and n + 1. As an application, the atomic numbers of every other alkaline earth metal (Be, Ca, Ba) are given by (4/3)n(n + 1/2)(n + 1)[citation needed].

## History

Faulhaber's formula is also called Bernoulli's formula. Faulhaber did not know the properties of the coefficients discovered by Bernoulli. Rather, he knew at least the first 17 cases, as well as the existence of the Faulhaber polynomials for odd powers described above.[2]

A rigorous proof of these formulas and his assertion that such formulas would exist for all odd powers took until Carl Jacobi (1834).