PlanetMath: proof of Faulhaber's formula

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proof of Faulhaber's formula

(Theorem)

Theorem 0.1 If $k\in\Nats, 2\leq n\in\Ints$ , then$$\sum_{m=1}^{n-1}m^k = \frac{1}{k+1}\sum_{i=0}^k \binom{k+1}{i}B_in^{k+1-i}=\int_1^n b_k(x)d$$ where the $B_i$ are the Bernoulli numbers and $b_i$ the Bernoulli polynomials.

The exponential generating function for the Bernoulli numbers is$$\sum_{n=0}^{\infty}B_n\frac{x^n}{n!}=\frac{x}{e^x-1$$ We develop an equation involving sums of Bernoulli numbers on one side, and a simple generating involving powers of $e$ that gives us the appropriate sum of powers on the other side. Equating coefficients of powers of $x$ then gives the result.

To get a generating function where the coefficient of $x^n/n!$ is $\sum_{m=1}^{n-1}m^k$ , we can use

$\displaystyle \sum_{m=0}^{n-1} e^{mx}$	$\displaystyle = \sum_{m=0}^{n-1} \sum_{k=0}^{\infty} \frac{m^k x^k}{k!}$
	$\displaystyle = \sum_{k=0}^{\infty}\left(\sum_{m=1}^{n-1} m^k\right)\frac{x^k}{k!}$

But this is also a geometric series, so

$\displaystyle \sum_{k=0}^{n-1}e^{kx}$	$\displaystyle = \frac{1-e^{nx}}{1-e^x}$
	$\displaystyle = \frac{e^{nx}-1}{x}\cdot\frac{x}{e^x-1}$
	$\displaystyle = \frac{e^{nx}-1}{x}\sum_{l=0}^{\infty}B_l\frac{x^l}{l!}$
	$\displaystyle = \left(\sum_{k=0}^{\infty}\frac{n^{k+1}}{k+1}\cdot\frac{x^k}{k!}\right) \left(\sum_{l=0}^{\infty}B_l\frac{x^l}{l!}\right)$
	$\displaystyle =\sum_{k=0}^{\infty} \left(\sum_{i=0}^j \frac{1}{k-i+1}\binom{k}{i}B_in^{k+1-i}\right)\frac{x^k}{k!}$

Equating coefficients of $x^k/k!$ we get

$\displaystyle \sum_{m=1}^{n-1}m^k$	$\displaystyle =\sum_{i=0}^k \frac{1}{k-i+1}\binom{k}{i}B_i n^{k+1-i}$
	$\displaystyle =\sum_{i=0}^k\frac{k!}{(k-i+1)i!(k-i)!}B_in^{k+1-i}=\frac{1}{k+1}\sum_{i=0}^k\binom{k+1}{i}B_in^{k+1-i}$

which proves the first equality.

If $f(x)$ is a polynomial, write $[x^r]f(x)$ for the coefficient of $x^r$ in $f(x)$ . Then$$[x^r]b_k(x)=\frac{1}{r}[x^{r-1}]b_k'(x)=\frac{k}{r}[x^{r-1}]b_{k-1}(x$$ and thus if $r\leq k$ , iterating, we get$$[x^r]b_k(x)=\binom{k}{r}[x^0]b_{k-r}(x)=\binom{k}{r}B_{k-r$$ Then using the fact that $b_k'=kb_{k-1}$ , we have

$\displaystyle \int_1^n b_k(x)$	$\displaystyle =\frac{1}{k+1}(b_{k+1}(n)-b_{k+1}(1)) = \frac{1}{k+1}\sum_{r=0}^{k+1}[x^r]b_{k+1}(x)(n^r-1)$
	$\displaystyle = \frac{1}{k+1}\sum_{r=0}^{k+1}\binom{k+1}{r}B_{k+1-r}(n^r-1) = \frac{1}{k+1}\sum_{r=1}^{k+1}\binom{k+1}{r}B_{k+1-r}n^r$

Now reverse the order of summation (i.e. replace $r$ by $k+1-r$ ) to get$$ \int_1^n b_k(x)=\frac{1}{k+1}\sum_{r=0}^k\binom{k+1}{k+1-r}B_rn^{k+1-r}=\frac{1}{k+1}\sum_{r=0}^k\binom{k+1}{r}B_r n^{k+1-r$$

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Cross-references: order, polynomial, equality, geometric series, coefficients, sum of powers, generating, simple, side, sums, equation, generating function, exponential, Bernoulli polynomials, Bernoulli numbers

This is version 1 of proof of Faulhaber's formula, born on 2009-01-13.
Object id is 11499, canonical name is ProofOfFaulhabersFormula.
Accessed 625 times total.

Classification:

AMS MSC:

11B68 (Number theory :: Sequences and sets :: Bernoulli and Euler numbers and polynomials)

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