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Faulhaber's formula

## Summary

In mathematics, Faulhaber's formula, named after the early 17th century mathematician Johann Faulhaber, expresses the sum of the p-th powers of the first n positive integers

${\displaystyle \sum _{k=1}^{n}k^{p}=1^{p}+2^{p}+3^{p}+\cdots +n^{p}}$

as a (p + 1)th-degree polynomial function of n, the coefficients involving Bernoulli numbers Bj, in the form submitted by Jacob Bernoulli and published in 1713:

${\displaystyle \sum _{k=1}^{n}k^{p}={\frac {n^{p+1}}{p+1}}+{\frac {1}{2}}n^{p}+\sum _{k=2}^{p}{\frac {B_{k}}{k!}}p^{\underline {k-1}}n^{p-k+1},}$

where ${\displaystyle p^{\underline {k-1}}=(p)_{k-1}={\dfrac {p!}{(p-k+1)!}}}$ is a falling factorial.

## History

Faulhaber's formula is also called Bernoulli's formula. Faulhaber did not know the properties of the coefficients later 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 below.[1]

A rigorous proof of these formulas and Faulhaber's assertion that such formulas would exist for all odd powers took until Carl Jacobi (1834), two centuries later.

## 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

${\displaystyle 1^{p}+2^{p}+3^{p}+\cdots +n^{p}}$

is a polynomial function of

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

In particular:

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

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

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

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

${\displaystyle 1^{11}+2^{11}+3^{11}+\cdots +n^{11}={16a^{6}-32a^{5}+34a^{4}-20a^{3}+5a^{2} \over 3}.}$

The first of these identities (the case p = 3) is known as Nicomachus's theorem.

More generally,[citation needed]

{\displaystyle {\begin{aligned}1^{2m+1}+2^{2m+1}&+3^{2m+1}+\cdots +n^{2m+1}\\&={\frac {1}{2^{2m+2}(2m+2)}}\sum _{q=0}^{m}{\binom {2m+2}{2q}}(2-2^{2q})~B_{2q}~\left[(8a+1)^{m+1-q}-1\right].\end{aligned}}}

Some authors call the polynomials in a on the right-hand sides of these identities Faulhaber polynomials. These polynomials are divisible by a2 because the Bernoulli number Bj is 0 for odd j > 1.

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

${\displaystyle \sum _{k=1}^{n}k^{2m+1}=c_{1}a^{2}+c_{2}a^{3}+\cdots +c_{m}a^{m+1}}$

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

${\displaystyle \sum _{k=1}^{n}k^{2m}={\frac {n+1/2}{2m+1}}(2c_{1}a+3c_{2}a^{2}+\cdots +(m+1)c_{m}a^{m}).}$

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.

## Summae Potestatum

Jakob Bernoulli's Summae Potestatum, Ars Conjectandi, 1713

In 1713, Jacob Bernoulli published under the title Summae Potestatum an expression of the sum of the p powers of the n first integers as a (p + 1)th-degree polynomial function of n, with coefficients involving numbers Bj, now called Bernoulli numbers:

${\displaystyle \sum _{k=1}^{n}k^{p}={\frac {n^{p+1}}{p+1}}+{\frac {1}{2}}n^{p}+{1 \over p+1}\sum _{j=2}^{p}{p+1 \choose j}B_{j}n^{p+1-j}.}$

Introducing also the first two Bernoulli numbers (which Bernoulli did not), the previous formula becomes

${\displaystyle \sum _{k=1}^{n}k^{p}={1 \over p+1}\sum _{j=0}^{p}{p+1 \choose j}B_{j}n^{p+1-j},}$

using the Bernoulli number of the second kind for which ${\displaystyle B_{1}={\frac {1}{2}}}$, or

${\displaystyle \sum _{k=1}^{n}k^{p}={1 \over p+1}\sum _{j=0}^{p}(-1)^{j}{p+1 \choose j}B_{j}n^{p+1-j},}$

using the Bernoulli number of the first kind for which ${\displaystyle B_{1}=-{\frac {1}{2}}.}$

For example, as

${\displaystyle B_{0}=1,~B_{1}=1/2,~B_{2}=1/6,~B_{3}=0,~B_{4}=-1/30,}$

one has for p = 4,

{\displaystyle {\begin{aligned}1^{4}+2^{4}+3^{4}+\cdots +n^{4}&={1 \over 5}\sum _{j=0}^{4}{5 \choose j}B_{j}n^{5-j}\\&={1 \over 5}\left(B_{0}n^{5}+5B_{1}n^{4}+10B_{2}n^{3}+10B_{3}n^{2}+5B_{4}n\right)\\&={\frac {1}{5}}n^{5}+{\frac {1}{2}}n^{4}+{\frac {1}{3}}n^{3}-{\frac {1}{30}}n.\end{aligned}}}

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). The derivation of Faulhaber's formula is available in The Book of Numbers by John Horton Conway and Richard K. Guy.[2]

There is also a similar (but somehow simpler) expression: using the idea of telescoping and the binomial theorem, one gets Pascal's identity:[3]

{\displaystyle {\begin{aligned}(n+1)^{k+1}-1&=\sum _{m=1}^{n}\left((m+1)^{k+1}-m^{k+1}\right)\\&=\sum _{p=0}^{k}{\binom {k+1}{p}}(1^{p}+2^{p}+\dots +n^{p}).\end{aligned}}}

This in particular yields the examples below – e.g., take k = 1 to get the first example. In a similar fashion we also find

{\displaystyle {\begin{aligned}n^{k+1}=\sum _{m=1}^{n}\left(m^{k+1}-(m-1)^{k+1}\right)=\sum _{p=0}^{k}(-1)^{k+p}{\binom {k+1}{p}}(1^{p}+2^{p}+\dots +n^{p}).\end{aligned}}}

## Examples

${\displaystyle 1+2+3+\cdots +n={\frac {n(n+1)}{2}}={\frac {n^{2}+n}{2}}}$ (the triangular numbers)
${\displaystyle 1^{2}+2^{2}+3^{2}+\cdots +n^{2}={\frac {n(n+1)(2n+1)}{6}}={\frac {2n^{3}+3n^{2}+n}{6}}}$ (the square pyramidal numbers)
${\displaystyle 1^{3}+2^{3}+3^{3}+\cdots +n^{3}=\left[{\frac {n(n+1)}{2}}\right]^{2}={\frac {n^{4}+2n^{3}+n^{2}}{4}}}$ (the triangular numbers squared)
{\displaystyle {\begin{aligned}1^{4}+2^{4}+3^{4}+\cdots +n^{4}&={\frac {n(n+1)(2n+1)(3n^{2}+3n-1)}{30}}\\&={\frac {6n^{5}+15n^{4}+10n^{3}-n}{30}}\end{aligned}}}
{\displaystyle {\begin{aligned}1^{5}+2^{5}+3^{5}+\cdots +n^{5}&={\frac {[n(n+1)]^{2}(2n^{2}+2n-1)}{12}}\\&={\frac {2n^{6}+6n^{5}+5n^{4}-n^{2}}{12}}\end{aligned}}}
{\displaystyle {\begin{aligned}1^{6}+2^{6}+3^{6}+\cdots +n^{6}&={\frac {n(n+1)(2n+1)(3n^{4}+6n^{3}-3n+1)}{42}}\\&={\frac {6n^{7}+21n^{6}+21n^{5}-7n^{3}+n}{42}}\end{aligned}}}

## From examples to matrix theorem

From the previous examples we get:

${\displaystyle \sum _{i=1}^{n}i^{0}=n}$
${\displaystyle \sum _{i=1}^{n}i^{1}={1 \over 2}n+{1 \over 2}n^{2}}$
${\displaystyle \sum _{i=1}^{n}i^{2}={1 \over 6}n+{1 \over 2}n^{2}+{1 \over 3}n^{3}}$
${\displaystyle \sum _{i=1}^{n}i^{3}={1 \over 4}n^{2}+{1 \over 2}n^{3}+{1 \over 4}n^{4}}$
${\displaystyle \sum _{i=1}^{n}i^{4}=-{1 \over 30}n+{1 \over 3}n^{3}+{1 \over 2}n^{4}+{1 \over 5}n^{5}}$
${\displaystyle \sum _{i=1}^{n}i^{5}=-{1 \over 12}n^{2}+{5 \over 12}n^{4}+{1 \over 2}n^{5}+{1 \over 6}n^{6}}$
${\displaystyle \sum _{i=1}^{n}i^{6}={1 \over 42}n-{1 \over 6}n^{3}+{1 \over 2}n^{5}+{1 \over 2}n^{6}+{1 \over 7}n^{7}}$
Writing these polynomials as a product between matrices gives
${\displaystyle {\begin{pmatrix}\sum _{i=1}^{n}i^{0}\\\sum _{i=1}^{n}i^{1}\\\sum _{i=1}^{n}i^{2}\\\sum _{i=1}^{n}i^{3}\\\sum _{i=1}^{n}i^{4}\\\sum _{i=1}^{n}i^{5}\\\sum _{i=1}^{n}i^{6}\\\end{pmatrix}}=G_{7}\cdot {\begin{pmatrix}n\\n^{2}\\n^{3}\\n^{4}\\n^{5}\\n^{6}\\n^{7}\\\end{pmatrix}}\qquad {\text{where}}\qquad G_{7}={\begin{pmatrix}1&0&0&0&0&0&0\\{1 \over 2}&{1 \over 2}&0&0&0&0&0\\{1 \over 6}&{1 \over 2}&{1 \over 3}&0&0&0&0\\0&{1 \over 4}&{1 \over 2}&{1 \over 4}&0&0&0\\-{1 \over 30}&0&{1 \over 3}&{1 \over 2}&{1 \over 5}&0&0\\0&-{1 \over 12}&0&{5 \over 12}&{1 \over 2}&{1 \over 6}&0\\{1 \over 42}&0&-{1 \over 6}&0&{1 \over 2}&{1 \over 2}&{1 \over 7}\\\end{pmatrix}}}$

Surprisingly, inverting the matrix of polynomial coefficients yields something more familiar:

${\displaystyle G_{7}^{-1}={\begin{pmatrix}1&0&0&0&0&0&0\\-1&2&0&0&0&0&0\\1&-3&3&0&0&0&0\\-1&4&-6&4&0&0&0\\1&-5&10&-10&5&0&0\\-1&6&-15&20&-15&6&0\\1&-7&21&-35&35&-21&7\\\end{pmatrix}}}$

In the inverted matrix, Pascal's triangle can be recognized, without the last element of each row, and with alternating signs. More precisely, let ${\displaystyle A_{7}}$ be the matrix obtained from Pascal's triangle by removing the last element of each row, and filling the rows by zeros on the right, that is the matrix obtained from the lower triangular Pascal matrix, filling the main diagonal by zeros and shifting up all the elements one place:

${\displaystyle A_{7}={\begin{pmatrix}1&0&0&0&0&0&0\\1&2&0&0&0&0&0\\1&3&3&0&0&0&0\\1&4&6&4&0&0&0\\1&5&10&10&5&0&0\\1&6&15&20&15&6&0\\1&7&21&35&35&21&7\\\end{pmatrix}}}$

Let ${\displaystyle {\overline {A}}_{7}}$ be the matrix obtained from ${\displaystyle A_{7}}$ by changing the signs of the entries in odd diagonals, that is by replacing ${\displaystyle a_{i,j}}$ by ${\displaystyle (-1)^{i+j}a_{i,j}}$. Then

${\displaystyle G_{7}^{-1}={\overline {A}}_{7}.}$

This is true for every order,[4] that is, for each positive integer m, one has ${\displaystyle G_{m}^{-1}={\overline {A}}_{m}.}$ Thus, it is possible to obtain the coefficients of the polynomials of the sums of powers of successive integers without resorting to the numbers of Bernoulli but by inverting the matrix easily obtained from the triangle of Pascal.

One has also[5]

${\displaystyle A_{7}^{-1}={\overline {G}}_{7}={\begin{pmatrix}1&0&0&0&0&0&0\\{1 \over 2}&{1 \over 2}&0&0&0&0&0\\{1 \over 6}&{1 \over 2}&{1 \over 3}&0&0&0&0\\0&{1 \over 4}&{1 \over 2}&{1 \over 4}&0&0&0\\{1 \over 30}&0&{1 \over 3}&{1 \over 2}&{1 \over 5}&0&0\\0&{1 \over 12}&0&{5 \over 12}&{1 \over 2}&{1 \over 6}&0\\{1 \over 42}&0&{1 \over 6}&0&{1 \over 2}&{1 \over 2}&{1 \over 7}\end{pmatrix}},}$

where ${\displaystyle {\overline {G}}_{7}}$ is obtained from ${\displaystyle G_{7}}$ by removing the minus signs.

## Proof with exponential generating function

Let

${\displaystyle S_{p}(n)=\sum _{k=1}^{n}k^{p},}$

denote the sum under consideration for integer ${\displaystyle p\geq 0.}$

Define the following exponential generating function with (initially) indeterminate ${\displaystyle z}$

${\displaystyle G(z,n)=\sum _{p=0}^{\infty }S_{p}(n){\frac {1}{p!}}z^{p}.}$

We find

{\displaystyle {\begin{aligned}G(z,n)=&\sum _{p=0}^{\infty }\sum _{k=1}^{n}{\frac {1}{p!}}(kz)^{p}=\sum _{k=1}^{n}e^{kz}=e^{z}.{\frac {1-e^{nz}}{1-e^{z}}},\\=&{\frac {1-e^{nz}}{e^{-z}-1}}.\end{aligned}}}

This is an entire function in ${\displaystyle z}$ so that ${\displaystyle z}$ can be taken to be any complex number.

We next recall the exponential generating function for the Bernoulli polynomials ${\displaystyle B_{j}(x)}$

${\displaystyle {\frac {ze^{zx}}{e^{z}-1}}=\sum _{j=0}^{\infty }B_{j}(x){\frac {z^{j}}{j!}},}$

where ${\displaystyle B_{j}=B_{j}(0)}$ denotes the Bernoulli number with the convention ${\displaystyle B_{1}=-{\frac {1}{2}}}$. This may be converted to a generating function with the convention ${\displaystyle B_{1}^{+}={\frac {1}{2}}}$ by the addition of ${\displaystyle j}$ to the coefficient of ${\displaystyle x^{j-1}}$ in each ${\displaystyle B_{j}(x)}$ (${\displaystyle B_{0}}$ does not need to be changed):

{\displaystyle {\begin{aligned}\sum _{j=0}^{\infty }B_{j}^{+}(x){\frac {z^{j}}{j!}}=&{\frac {ze^{zx}}{e^{z}-1}}+\sum _{j=1}^{\infty }jx^{j-1}{\frac {z^{j}}{j!}}\\=&{\frac {ze^{zx}}{e^{z}-1}}+\sum _{j=1}^{\infty }x^{j-1}{\frac {z^{j}}{(j-1)!}}\\=&{\frac {ze^{zx}}{e^{z}-1}}+ze^{zx}\\=&{\frac {ze^{zx}+ze^{z+zx}-ze^{zx}}{e^{z}-1}}\\=&{\frac {ze^{zx}}{1-e^{-z}}}\end{aligned}}}

It follows immediately that

${\displaystyle S_{p}(n)={\frac {B_{p+1}^{+}(n)-B_{p+1}^{+}(0)}{p+1}}}$

for all ${\displaystyle p}$.

## Alternate expressions

• By relabelling we find the alternative expression
${\displaystyle \sum _{k=1}^{n}k^{p}=\sum _{k=0}^{p}{(-1)^{p-k} \over k+1}{p \choose k}B_{p-k}n^{k+1}.}$
• We may also expand ${\displaystyle G(z,n)}$ in terms of the Bernoulli polynomials to find
{\displaystyle {\begin{aligned}G(z,n)=&{\frac {e^{(n+1)z}}{e^{z}-1}}-{\frac {e^{z}}{e^{z}-1}}\\=&\sum _{j=0}^{\infty }\left(B_{j}(n+1)-(-1)^{j}B_{j}\right){\frac {z^{j-1}}{j!}},\end{aligned}}}
which implies
${\displaystyle \sum _{k=1}^{n}k^{p}={\frac {1}{p+1}}\left(B_{p+1}(n+1)-(-1)^{p+1}B_{p+1}\right)={\frac {1}{p+1}}\left(B_{p+1}(n+1)-B_{p+1}(1)\right).}$
Since ${\displaystyle B_{n}=0}$ whenever ${\displaystyle n>1}$ is odd, the factor ${\displaystyle (-1)^{p+1}}$ may be removed when ${\displaystyle p>0}$.
${\displaystyle \sum _{k=0}^{n}k^{p}=\sum _{k=0}^{p}\left\{{p \atop k}\right\}{\frac {(n+1)_{k+1}}{k+1}},}$
${\displaystyle \sum _{k=1}^{n}k^{p}=\sum _{k=1}^{p+1}\left\{{p+1 \atop k}\right\}{\frac {(n)_{k}}{k}}.}$
This is due to the definition of the Stirling numbers of the second kind as mononomials in terms of falling factorials, and the behaviour of falling factorials under the indefinite sum.

## Relationship to Riemann zeta function

Using ${\displaystyle B_{k}=-k\zeta (1-k)}$, one can write

${\displaystyle \sum \limits _{k=1}^{n}k^{p}={\frac {n^{p+1}}{p+1}}-\sum \limits _{j=0}^{p-1}{p \choose j}\zeta (-j)n^{p-j}.}$

If we consider the generating function ${\displaystyle G(z,n)}$ in the large ${\displaystyle n}$ limit for ${\displaystyle \Re (z)<0}$, then we find

${\displaystyle \lim _{n\rightarrow \infty }G(z,n)={\frac {1}{e^{-z}-1}}=\sum _{j=0}^{\infty }(-1)^{j-1}B_{j}{\frac {z^{j-1}}{j!}}}$

Heuristically, this suggests that

${\displaystyle \sum _{k=1}^{\infty }k^{p}={\frac {(-1)^{p}B_{p+1}}{p+1}}.}$

This result agrees with the value of the Riemann zeta function ${\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}}$ for negative integers ${\displaystyle s=-p<0}$ on appropriately analytically continuing ${\displaystyle \zeta (s)}$.

## Umbral form

In the classical 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

${\displaystyle \sum _{k=1}^{n}k^{p}={1 \over p+1}\sum _{j=0}^{p}{p+1 \choose j}B_{j}n^{p+1-j}={1 \over p+1}\sum _{j=0}^{p}{p+1 \choose j}B^{j}n^{p+1-j}}$

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

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

${\displaystyle T(b^{j})=B_{j}.\,}$

Then one can say

${\displaystyle \sum _{k=1}^{n}k^{p}={1 \over p+1}\sum _{j=0}^{p}{p+1 \choose j}B_{j}n^{p+1-j}={1 \over p+1}\sum _{j=0}^{p}{p+1 \choose j}T(b^{j})n^{p+1-j}}$

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

## Notes

1. ^ Donald E. Knuth (1993). "Johann Faulhaber and sums of powers". Mathematics of Computation. 61 (203): 277–294. arXiv:math.CA/9207222. doi:10.2307/2152953. JSTOR 2152953. The arxiv.org paper has a misprint in the formula for the sum of 11th powers, which was corrected in the printed version. Correct version.
2. ^ John H. Conway, Richard Guy (1996). The Book of Numbers. Springer. p. 107. ISBN 0-387-97993-X.
3. ^ Kieren MacMillan, Jonathan Sondow (2011). "Proofs of power sum and binomial coefficient congruences via Pascal's identity". American Mathematical Monthly. 118 (6): 549–551. arXiv:1011.0076. doi:10.4169/amer.math.monthly.118.06.549.
4. ^ Pietrocola, Giorgio (2017), On polynomials for the calculation of sums of powers of successive integers and Bernoulli numbers deduced from the Pascal's triangle (PDF).
5. ^ Derby, Nigel (2015), "A search for sums of powers", The Mathematical Gazette, 99 (546): 416–421, doi:10.1017/mag.2015.77.
6. ^ Concrete Mathematics, 1st ed. (1989), p. 275.