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
$\sum _{k=1}^{n}k^{p}=1^{p}+2^{p}+3^{p}+\cdots +n^{p}$
as a polynomial in n. In modern notation, Faulhaber's formula is
$\sum _{k=1}^{n}k^{p}={\frac {1}{p+1}}\sum _{r=0}^{p}{\binom {p+1}{r}}B_{r}n^{p-r+1}.$
Here, ${\textstyle {\binom {p+1}{r}}}$ is the binomial coefficient "p + 1 choose r", and the B_{j} are the Bernoulli numbers with the convention that ${\textstyle B_{1}=+{\frac {1}{2}}}$.
The result: Faulhaber's formula
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Faulhaber's formula concerns expressing the sum of the p-th powers of the first n positive integers
$\sum _{k=1}^{n}k^{p}=1^{p}+2^{p}+3^{p}+\cdots +n^{p}$
as a (p + 1)th-degree polynomial function of n.
The first few examples are well known. For p = 0, we have
$\sum _{k=1}^{n}k^{0}=\sum _{k=1}^{n}1=n.$
For p = 1, we have the triangular numbers$\sum _{k=1}^{n}k^{1}=\sum _{k=1}^{n}k={\frac {n(n+1)}{2}}={\frac {1}{2}}(n^{2}+n).$
For p = 2, we have the square pyramidal numbers$\sum _{k=1}^{n}k^{2}={\frac {n(n+1)(2n+1)}{6}}={\frac {1}{3}}(n^{3}+{\tfrac {3}{2}}n^{2}+{\tfrac {1}{2}}n).$
The coefficients of Faulhaber's formula in its general form involve the Bernoulli numbersB_{j}. The Bernoulli numbers begin
${\begin{aligned}B_{0}&=1&B_{1}&={\tfrac {1}{2}}&B_{2}&={\tfrac {1}{6}}&B_{3}&=0\\B_{4}&=-{\tfrac {1}{30}}&B_{5}&=0&B_{6}&={\tfrac {1}{42}}&B_{7}&=0,\end{aligned}}$
where here we use the convention that ${\textstyle B_{1}=+{\frac {1}{2}}}$. The Bernoulli numbers have various definitions (see Bernoulli number#Definitions), such as that they are the coefficients of the exponential generating function
${\frac {t}{1-\mathrm {e} ^{-t}}}={\frac {t}{2}}\left(\operatorname {coth} {\frac {t}{2}}+1\right)=\sum _{k=0}^{\infty }B_{k}{\frac {t^{k}}{k!}}.$
Then Faulhaber's formula is that
$\sum _{k=1}^{n}k^{p}={\frac {1}{p+1}}\sum _{k=0}^{p}{\binom {p+1}{k}}B_{k}n^{p-k+1}.$
Here, the B_{j} are the Bernoulli numbers as above, and
${\binom {p+1}{k}}={\frac {(p+1)!}{(p+1-k)!\,k!}}={\frac {(p+1)p(p-1)\cdots (p-k+3)(p-k+2)}{k(k-1)(k-2)\cdots 2\cdot 1}}$
is the binomial coefficient "p + 1 choose k".
Examples
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So, for example, one has for p = 4,
${\begin{aligned}1^{4}+2^{4}+3^{4}+\cdots +n^{4}&={\frac {1}{5}}\sum _{j=0}^{4}{5 \choose j}B_{j}n^{5-j}\\&={\frac {1}{5}}\left(B_{0}n^{5}+5B_{1}n^{4}+10B_{2}n^{3}+10B_{3}n^{2}+5B_{4}n\right)\\&={\frac {1}{5}}\left(n^{5}+{\tfrac {5}{2}}n^{4}+{\tfrac {5}{3}}n^{3}-{\tfrac {1}{6}}n\right).\end{aligned}}$
The history of the problem begins in antiquity and coincides with that of some of its special cases. The case $p=1$ coincides with that of the calculation of the arithmetic series, the sum of the first $n$ values of an arithmetic progression. This problem is quite simple but the case already known by the Pythagorean school for its connection with triangular numbers is historically interesting:
$1+2+\dots +n={\frac {1}{2}}n^{2}+{\frac {1}{2}}n,$ polynomial $S_{1,1}^{1}(n)$ calculating the sum of the first $n$ natural numbers.
For $m>1,$ the first cases encountered in the history of mathematics are:
$1+3+\dots +2n-1=n^{2},$ polynomial $S_{1,2}^{1}(n)$ calculating the sum of the first $n$ successive odds forming a square. A property probably well known by the Pythagoreans themselves who, in constructing their figured numbers, had to add each time a gnomon consisting of an odd number of points to obtain the next perfect square.
$1^{2}+2^{2}+\ldots +n^{2}={\frac {1}{3}}n^{3}+{\frac {1}{2}}n^{2}+{\frac {1}{6}}n,$ polynomial $S_{1,1}^{2}(n)$ calculating the sum of the squares of the successive integers. Property that is demonstrated in Spirals, a work of Archimedes.^{[1]}
$1^{3}+2^{3}+\ldots +n^{3}={\frac {1}{4}}n^{4}+{\frac {1}{2}}n^{3}+{\frac {1}{4}}n^{2},$ polynomial $S_{1,1}^{3}(n)$ calculating the sum of the cubes of the successive integers. Corollary of a theorem of Nicomachus of Gerasa.^{[1]}
L'insieme $S_{1,1}^{m}(n)$ of the cases, to which the two preceding polynomials belong, constitutes the classical problem of powers of successive integers.
Middle period
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Over time, many other mathematicians became interested in the problem and made various contributions to its solution. These include Aryabhata, Al-Karaji, Ibn al-Haytham, Thomas Harriot, Johann Faulhaber, Pierre de Fermat and Blaise Pascal who recursively solved the problem of the sum of powers of successive integers by considering an identity that allowed to obtain a polynomial of degree $m+1$ already knowing the previous ones.^{[1]}
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.^{[2]}
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 B_{j}, now called Bernoulli numbers:
Introducing also the first two Bernoulli numbers (which Bernoulli did not), the previous formula becomes
$\sum _{k=1}^{n}k^{p}={1 \over p+1}\sum _{j=0}^{p}{p+1 \choose j}B_{j}n^{p+1-j},$
using the Bernoulli number of the second kind for which ${\textstyle B_{1}={\frac {1}{2}}}$, or
$\sum _{k=1}^{n}k^{p}={1 \over p+1}\sum _{j=0}^{p}(-1)^{j}{p+1 \choose j}B_{j}^{-}n^{p+1-j},$
using the Bernoulli number of the first kind for which ${\textstyle B_{1}^{-}=-{\frac {1}{2}}.}$
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. Jacobi benefited from the progress of mathematical analysis using the development in infinite series of an exponential function generating Bernoulli numbers.
Modern period
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In 1982 A.W.F. Edwards publishes an article ^{[3]} in which he shows that Pascal's identity can be expressed by means of triangular matrices containing the Pascal's triangle deprived of 'last element of each line:
The example is limited by the choice of a fifth order matrix but is easily extendable to higher orders. The equation can be written as: ${\vec {N}}=A{\vec {S}}$ and multiplying the two sides of the equation to the left by $A^{-1}$ , inverse of the matrix A, we obtain $A^{-1}{\vec {N}}={\vec {S}}$ which allows to arrive directly at the polynomial coefficients without directly using the Bernoulli numbers. Other authors after Edwards dealing with various aspects of the power sum problem take the matrix path ^{[6]} and studying aspects of the problem in their articles useful tools such as the Vandermonde vector.^{[7]} Other researchers continue to explore through the traditional analytic route ^{[8]} and generalize the problem of the sum of successive integers to any geometric progression^{[9]}^{[10]}
Proof with exponential generating function
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Let
$S_{p}(n)=\sum _{k=1}^{n}k^{p},$
denote the sum under consideration for integer $p\geq 0.$
Define the following exponential generating function with (initially) indeterminate $z$$G(z,n)=\sum _{p=0}^{\infty }S_{p}(n){\frac {1}{p!}}z^{p}.$
We find
${\begin{aligned}G(z,n)=&\sum _{p=0}^{\infty }\sum _{k=1}^{n}{\frac {1}{p!}}(kz)^{p}=\sum _{k=1}^{n}e^{kz}=e^{z}\cdot {\frac {1-e^{nz}}{1-e^{z}}},\\=&{\frac {1-e^{nz}}{e^{-z}-1}}.\end{aligned}}$
This is an entire function in $z$ so that $z$ can be taken to be any complex number.
We next recall the exponential generating function for the Bernoulli polynomials$B_{j}(x)$${\frac {ze^{zx}}{e^{z}-1}}=\sum _{j=0}^{\infty }B_{j}(x){\frac {z^{j}}{j!}},$
where $B_{j}=B_{j}(0)$ denotes the Bernoulli number with the convention $B_{1}=-{\frac {1}{2}}$. This may be converted to a generating function with the convention $B_{1}^{+}={\frac {1}{2}}$ by the addition of $j$ to the coefficient of $x^{j-1}$ in each $B_{j}(x)$ ($B_{0}$ does not need to be changed):
${\begin{aligned}\sum _{j=0}^{\infty }B_{j}^{+}(x){\frac {z^{j}}{j!}}=&{\frac {ze^{zx}}{e^{z}-1}}+\sum _{j=1}^{\infty }jx^{j-1}{\frac {z^{j}}{j!}}\\=&{\frac {ze^{zx}}{e^{z}-1}}+\sum _{j=1}^{\infty }x^{j-1}{\frac {z^{j}}{(j-1)!}}\\=&{\frac {ze^{zx}}{e^{z}-1}}+ze^{zx}\\=&{\frac {ze^{zx}+ze^{z}e^{zx}-ze^{zx}}{e^{z}-1}}\\=&{\frac {ze^{zx}}{1-e^{-z}}}\end{aligned}}$
It follows immediately that
$S_{p}(n)={\frac {B_{p+1}^{+}(n)-B_{p+1}^{+}(0)}{p+1}}$
for all $p$.
Faulhaber polynomials
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The term Faulhaber polynomials is used by some authors to refer to another polynomial sequence related to that given above.
Write
$a=\sum _{k=1}^{n}k={\frac {n(n+1)}{2}}.$
Faulhaber observed that if p is odd then ${\textstyle \sum _{k=1}^{n}k^{p}}$ is a polynomial function of a.
For p = 1, it is clear that
$\sum _{k=1}^{n}k^{1}=\sum _{k=1}^{n}k={\frac {n(n+1)}{2}}=a.$
For p = 3, the result that
$\sum _{k=1}^{n}k^{3}={\frac {n^{2}(n+1)^{2}}{4}}=a^{2}$
is known as Nicomachus's theorem.
Further, we have
${\begin{aligned}\sum _{k=1}^{n}k^{5}&={\frac {4a^{3}-a^{2}}{3}}\\\sum _{k=1}^{n}k^{7}&={\frac {6a^{4}-4a^{3}+a^{2}}{3}}\\\sum _{k=1}^{n}k^{9}&={\frac {16a^{5}-20a^{4}+12a^{3}-3a^{2}}{5}}\\\sum _{k=1}^{n}k^{11}&={\frac {16a^{6}-32a^{5}+34a^{4}-20a^{3}+5a^{2}}{3}}\end{aligned}}$
(see OEIS: A000537, OEIS: A000539, OEIS: A000541, OEIS: A007487, OEIS: A123095).
More generally, ^{[citation needed]}$\sum _{k=1}^{n}k^{2m+1}={\frac {1}{2^{2m+2}(2m+2)}}\sum _{q=0}^{m}{\binom {2m+2}{2q}}(2-2^{2q})~B_{2q}~\left[(8a+1)^{m+1-q}-1\right].$
Some authors call the polynomials in a on the right-hand sides of these identities Faulhaber polynomials. These polynomials are divisible by a^{2} because the Bernoulli numberB_{j} is 0 for odd j > 1.
Inversely, writing for simplicity $s_{j}:=\sum _{k=1}^{n}k^{j}$, we have
${\begin{aligned}4a^{3}&=3s_{5}+s_{3}\\8a^{4}&=4s_{7}+4s_{5}\\16a^{5}&=5s_{9}+10s_{7}+s_{5}\end{aligned}}$
and generally
$2^{m-1}a^{m}=\sum _{j>0}{\binom {m}{2j-1}}s_{2m-2j+1}.$
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
$\sum _{k=1}^{n}k^{2m}={\frac {n+{\frac {1}{2}}}{2m+1}}(2c_{1}a+3c_{2}a^{2}+\cdots +(m+1)c_{m}a^{m}).$
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 n^{2} and (n + 1)^{2}, while for an even power the polynomial has factors n, n + 1/2 and n + 1.
Expressing products of power sums as linear combinations of power sums
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Products of two (and thus by iteration, several) power sums $s_{j_{r}}:=\sum _{k=1}^{n}k^{j_{r}}$ can be written as linear combinations of power sums with either all degrees even or all degrees odd, depending on the total degree of the product as a polynomial in $n$, e.g. $30s_{2}s_{4}=-s_{3}+15s_{5}+16s_{7}$.
Note that the sums of coefficients must be equal on both sides, as can be seen by putting $n=1$, which makes all the $s_{j}$ equal to 1. Some general formulae include:
${\begin{aligned}(m+1)s_{m}^{2}&=2\sum _{j=0}^{\lfloor {\frac {m}{2}}\rfloor }{\binom {m+1}{2j}}(2m+1-2j)B_{2j}s_{2m+1-2j}.\\m(m+1)s_{m}s_{m-1}&=m(m+1)B_{m}s_{m}+\sum _{j=0}^{\lfloor {\frac {m-1}{2}}\rfloor }{\binom {m+1}{2j}}(2m+1-2j)B_{2j}s_{2m-2j}.\\2^{m-1}s_{1}^{m}&=\sum _{j=1}^{\lfloor {\frac {m+1}{2}}\rfloor }{\binom {m}{2j-1}}s_{2m+1-2j}.\end{aligned}}$
Note that in the second formula, for even $m$ the term corresponding to $j={\dfrac {m}{2}}$ is different from the other terms in the sum, while for odd $m$, this additional term vanishes because of $B_{m}=0$.
Surprisingly, inverting the matrix of polynomial coefficients yields something more familiar:
$G_{7}^{-1}={\begin{pmatrix}1&0&0&0&0&0&0\\-1&2&0&0&0&0&0\\1&-3&3&0&0&0&0\\-1&4&-6&4&0&0&0\\1&-5&10&-10&5&0&0\\-1&6&-15&20&-15&6&0\\1&-7&21&-35&35&-21&7\\\end{pmatrix}}={\overline {A}}_{7}$
In the inverted matrix, Pascal's triangle can be recognized, without the last element of each row, and with alternating signs.
Let $A_{7}$ be the matrix obtained from ${\overline {A}}_{7}$ by changing the signs of the entries in odd diagonals, that is by replacing $a_{i,j}$ by $(-1)^{i+j}a_{i,j}$, let ${\overline {G}}_{7}$ be the matrix obtained from $G_{7}$ with a similar transformation, then
$A_{7}={\begin{pmatrix}1&0&0&0&0&0&0\\1&2&0&0&0&0&0\\1&3&3&0&0&0&0\\1&4&6&4&0&0&0\\1&5&10&10&5&0&0\\1&6&15&20&15&6&0\\1&7&21&35&35&21&7\\\end{pmatrix}}$
and
$A_{7}^{-1}={\begin{pmatrix}1&0&0&0&0&0&0\\-{1 \over 2}&{1 \over 2}&0&0&0&0&0\\{1 \over 6}&-{1 \over 2}&{1 \over 3}&0&0&0&0\\0&{1 \over 4}&-{1 \over 2}&{1 \over 4}&0&0&0\\-{1 \over 30}&0&{1 \over 3}&-{1 \over 2}&{1 \over 5}&0&0\\0&-{1 \over 12}&0&{5 \over 12}&-{1 \over 2}&{1 \over 6}&0\\{1 \over 42}&0&-{1 \over 6}&0&{1 \over 2}&-{1 \over 2}&{1 \over 7}\end{pmatrix}}={\overline {G}}_{7}.$
Also
${\begin{pmatrix}\sum _{k=0}^{n-1}k^{0}\\\sum _{k=0}^{n-1}k^{1}\\\sum _{k=0}^{n-1}k^{2}\\\sum _{k=0}^{n-1}k^{3}\\\sum _{k=0}^{n-1}k^{4}\\\sum _{k=0}^{n-1}k^{5}\\\sum _{k=0}^{n-1}k^{6}\\\end{pmatrix}}={\overline {G}}_{7}{\begin{pmatrix}n\\n^{2}\\n^{3}\\n^{4}\\n^{5}\\n^{6}\\n^{7}\\\end{pmatrix}}$
This is because it is evident that
${\textstyle \sum _{k=1}^{n}k^{m}-\sum _{k=0}^{n-1}k^{m}=n^{m}}$ and that therefore polynomials of degree $m+1$ of the form ${\textstyle {\frac {1}{m+1}}n^{m+1}+{\frac {1}{2}}n^{m}+\cdots }$ subtracted the monomial difference $n^{m}$ they become ${\textstyle {\frac {1}{m+1}}n^{m+1}-{\frac {1}{2}}n^{m}+\cdots }$.
This is true for every order, that is, for each positive integer m, one has $G_{m}^{-1}={\overline {A}}_{m}$ and ${\overline {G}}_{m}^{-1}=A_{m}.$
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.^{[12]}^{[13]}
Variations
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Replacing $k$ with $p-k$, we find the alternative expression: $\sum _{k=1}^{n}k^{p}=\sum _{k=0}^{p}{\frac {1}{k+1}}{p \choose k}B_{p-k}n^{k+1}.$
Subtracting $n^{p}$ from both sides of the original formula and incrementing $n$ by $1$, we get ${\begin{aligned}\sum _{k=1}^{n}k^{p}&={\frac {1}{p+1}}\sum _{k=0}^{p}{\binom {p+1}{k}}(-1)^{k}B_{k}(n+1)^{p-k+1}\\&=\sum _{k=0}^{p}{\frac {1}{k+1}}{\binom {p}{k}}(-1)^{p-k}B_{p-k}(n+1)^{k+1},\end{aligned}}$
where $(-1)^{k}B_{k}=B_{k}^{-}$ can be interpreted as "negative" Bernoulli numbers with $B_{1}^{-}=-{\tfrac {1}{2}}$.
We may also expand $G(z,n)$ in terms of the Bernoulli polynomials to find ${\begin{aligned}G(z,n)&={\frac {e^{(n+1)z}}{e^{z}-1}}-{\frac {e^{z}}{e^{z}-1}}\\&=\sum _{j=0}^{\infty }\left(B_{j}(n+1)-(-1)^{j}B_{j}\right){\frac {z^{j-1}}{j!}},\end{aligned}}$ which implies $\sum _{k=1}^{n}k^{p}={\frac {1}{p+1}}\left(B_{p+1}(n+1)-(-1)^{p+1}B_{p+1}\right)={\frac {1}{p+1}}\left(B_{p+1}(n+1)-B_{p+1}(1)\right).$ Since $B_{n}=0$ whenever $n>1$ is odd, the factor $(-1)^{p+1}$ may be removed when $p>0$.
It can also be expressed in terms of Stirling numbers of the second kind and falling factorials as^{[14]}$\sum _{k=0}^{n}k^{p}=\sum _{k=0}^{p}\left\{{p \atop k}\right\}{\frac {(n+1)_{k+1}}{k+1}},$$\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.
Interpreting the Stirling numbers of the second kind, $\left\{{p+1 \atop k}\right\}$, as the number of set partitions of $\lbrack p+1\rbrack$ into $k$ parts, the identity has a direct combinatorial proof since both sides count the number of functions $f:\lbrack p+1\rbrack \to \lbrack n\rbrack$ with $f(1)$ maximal. The index of summation on the left hand side represents $k=f(1)$, while the index on the right hand side is represents the number of elements in the image of f.
There is also a similar (but somehow simpler) expression: using the idea of telescoping and the binomial theorem, one gets Pascal's identity:^{[15]}
Faulhaber's formula was generalized by Guo and Zeng to a q-analog.^{[16]}
Relationship to Riemann zeta function
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Using $B_{k}=-k\zeta (1-k)$, one can write
$\sum \limits _{k=1}^{n}k^{p}={\frac {n^{p+1}}{p+1}}-\sum \limits _{j=0}^{p-1}{p \choose j}\zeta (-j)n^{p-j}.$
If we consider the generating function $G(z,n)$ in the large $n$ limit for $\Re (z)<0$, then we find
$\lim _{n\rightarrow \infty }G(z,n)={\frac {1}{e^{-z}-1}}=\sum _{j=0}^{\infty }(-1)^{j-1}B_{j}{\frac {z^{j-1}}{j!}}$
Heuristically, this suggests that
$\sum _{k=1}^{\infty }k^{p}={\frac {(-1)^{p}B_{p+1}}{p+1}}.$
This result agrees with the value of the Riemann zeta function${\textstyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}}$ for negative integers $s=-p<0$ on appropriately analytically continuing $\zeta (s)$.
In the umbral calculus, one treats the Bernoulli numbers ${\textstyle B^{0}=1}$, ${\textstyle B^{1}={\frac {1}{2}}}$, ${\textstyle B^{2}={\frac {1}{6}}}$, ... as if the index j in ${\textstyle B^{j}}$ were actually an exponent, and so as if the Bernoulli numbers were powers of some object B.
Using this notation, Faulhaber's formula can be written as
$\sum _{k=1}^{n}k^{p}={\frac {1}{p+1}}{\big (}(B+n)^{p+1}-B^{p+1}{\big )}.$
Here, the expression on the right must be understood by expanding out to get terms ${\textstyle B^{j}}$ that can then be interpreted as the Bernoulli numbers. Specifically, using the binomial theorem, we get
${\begin{aligned}{\frac {1}{p+1}}{\big (}(B+n)^{p+1}-B^{p+1}{\big )}&={1 \over p+1}\left(\sum _{k=0}^{p+1}{\binom {p+1}{k}}B^{k}n^{p+1-k}-B^{p+1}\right)\\&={1 \over p+1}\sum _{k=0}^{p}{\binom {p+1}{j}}B^{k}n^{p+1-k}.\end{aligned}}$
A derivation of Faulhaber's formula using the umbral form is available in The Book of Numbers by John Horton Conway and Richard K. Guy.^{[17]}
Classically, this umbral form was considered as a notational convenience. In the modern umbral calculus, on the other hand, this is given a formal mathematical underpinning. One considers the linear functionalT on the vector space of polynomials in a variable b given by ${\textstyle T(b^{j})=B_{j}.}$ Then one can say
${\begin{aligned}\sum _{k=1}^{n}k^{p}&={1 \over p+1}\sum _{j=0}^{p}{p+1 \choose j}B_{j}n^{p+1-j}\\&={1 \over p+1}\sum _{j=0}^{p}{p+1 \choose j}T(b^{j})n^{p+1-j}\\&={1 \over p+1}T\left(\sum _{j=0}^{p}{p+1 \choose j}b^{j}n^{p+1-j}\right)\\&=T\left({(b+n)^{p+1}-b^{p+1} \over p+1}\right).\end{aligned}}$
A general formula
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The series $1^{m}+2^{m}+3^{m}+..n^{m}$ as a function of m is often abbreviated as $S_{m}$. Beardon (see External Links) have published formulas for powers of $S_{m}$. For example, Beardon 1996 stated this general formula for powers of $S_{1}:\;\;\;S_{1}^{\;N}={\frac {1}{2^{N}}}\sum _{r=0}^{N}{N \choose r}S_{N+r}(1-(-1)^{N-r})$, which shows that $S_{1}$ raised to a power N can be written as a linear sum of terms $S_{3},\;\;S_{5},\;\;S_{7}$... For example, by taking N to be 2, then 3, then 4 in Beardon's formula we get the identities $S_{1}^{\;2}=S_{3},\;\;S_{1}^{\;3}={\frac {1}{4}}S_{3}+{\frac {3}{4}}S_{5},\;\;S_{1}^{\;4}={\frac {1}{2}}S_{5}+{\frac {1}{2}}S_{7}$.
Other formulae, such as $S_{2}^{\;2}={\frac {1}{3}}S_{4}+{\frac {2}{3}}S_{5}$ and $S_{2}^{\;3}={\frac {1}{12}}S_{4}+{\frac {7}{12}}S_{6}+{\frac {1}{3}}S_{8}$ are known but no general formula for $S_{m}^{\;N}$, where m, N are positive integers, has been published to date. In an unpublished paper by Derby (2019) ^{[18]} the following formula was stated and proved:
This can be calculated in matrix form, as described above. In the case when m = 1 it replicates Beardon's formula for $S_{1}^{\;N}$. When m = 2 and N = 2 or 3 it generates the given formulas for $S_{2}^{\;\;2}$ and $S_{2}^{\;3}$ . Examples of calculations for higher indices are
$S_{2}^{\;4}={\frac {1}{54}}S_{5}+{\frac {5}{18}}S_{7}+{\frac {5}{9}}S_{9}+{\frac {4}{27}}S_{11}$ and $S_{6}^{\;3}={\frac {1}{588}}S_{8}-{\frac {1}{42}}S_{10}+{\frac {13}{84}}S_{12}-{\frac {47}{98}}S_{14}+{\frac {17}{28}}S_{16}+{\frac {19}{28}}S_{18}+{\frac {3}{49}}S_{20}$.
Notes
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^ ^{a}^{b}^{c}Beery, Janet (2009). "Sum of powers of positive integers". MMA Mathematical Association of America. doi:10.4169/loci003284 (inactive 1 November 2024).{{cite news}}: CS1 maint: DOI inactive as of November 2024 (link)
^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. Archived 2010-12-01 at the Wayback Machine
^Edwards, Anthony William Fairbank (1982). "Sums of powers of integers: A little of the History". The Mathematical Gazette. 66 (435): 22–28. doi:10.2307/3617302. JSTOR 3617302. S2CID 125682077.
^The first element of the vector of the sums is $n$ and not $\sum _{k=0}^{n-1}k^{0}$ because of the first addend, the indeterminate form $0^{0}$, which should otherwise be assigned a value of 1
^Edwards, A.W.F. (1987). Pascal's Arithmetical Triangle: The Story of a Mathematical Idea. Charles Griffin & C. p. 84. ISBN 0-8018-6946-3.
^Kalman, Dan (1988). "Sums of Powers by matrix method". Semantic scholar. S2CID 2656552.
^Helmes, Gottfried (2006). "Accessing Bernoulli-Numbers by Matrix-Operations" (PDF). Uni-Kassel.de.
^Howard, F.T (1994). "Sums of powers of integers via generating functions" (PDF). CiteSeerX10.1.1.376.4044.
^Lang, Wolfdieter (2017). "On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli numbers". arXiv:1707.04451 [math.NT].
^Tan Si, Do (2017). "Obtaining Easily Sums of Powers on Arithmetic Progressions and Properties of Bernoulli Polynomials by Operator Calculus". Applied Physics Research. 9. Canadian Center of Science and Education. ISSN 1916-9639.
^Gulley, Ned (March 4, 2010), Shure, Loren (ed.), Nicomachus's Theorem, Matlab Central
^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).
^Derby, Nigel (2015), "A search for sums of powers", The Mathematical Gazette, 99 (546): 416–421, doi:10.1017/mag.2015.77, S2CID 124607378.
^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. S2CID 207521003.
^Guo, Victor J. W.; Zeng, Jiang (30 August 2005). "A q-Analogue of Faulhaber's Formula for Sums of Powers". The Electronic Journal of Combinatorics. 11 (2). arXiv:math/0501441. Bibcode:2005math......1441G. doi:10.37236/1876. S2CID 10467873.
Johann Faulhaber (1631). Academia Algebrae - Darinnen die miraculosische Inventiones zu den höchsten Cossen weiters continuirt und profitiert werden. A very rare book, but Knuth has placed a photocopy in the Stanford library, call number QA154.8 F3 1631a f MATH. (online copy at Google Books)
Beardon, A. F. (1996). "Sums of Powers of Integers" (PDF). American Mathematical Monthly. 103 (3): 201–213. doi:10.1080/00029890.1996.12004725. Retrieved 2011-10-23. (Winner of a Lester R. Ford Award)
Schumacher, Raphael (2016). "An Extended Version of Faulhaber's Formula". Journal of Integer Sequences. Vol. 19, no. 16.4.2.
Orosi, Greg (2018). "A Simple Derivation Of Faulhaber's Formula" (PDF). Applied Mathematics E-Notes. Vol. 18. pp. 124–126.