PlanetMath: Bernoulli polynomials and numbers

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Bernoulli polynomials and numbers

(Definition)

For $n = 0,\,1,\,2,\,\ldots$ , the Bernoulli polynomial may be defined as the uniquely determined polynomial satisfying

$\displaystyle \int_x^{x+1}b_n(t)\,dt = x^n.$

(1)

The constant term of is the $n^{\mathrm{th}}$ Bernoulli number .

The Bernoulli polynomial is often denoted also .

The uniqueness of the solution in (1) is justificated by the

Lemma. For any polynomial , there exists a unique polynomial with the same degree satisfying

$\displaystyle \int_x^{x+1}g(t)\,dt = f(x).$

(2)

Proof. For every $n = 0,\,1,\,2,\,\ldots$ , the polynomial

$\displaystyle g_n(x) := \int_x^{x+1}t^n\,dt = \frac{(x+1)^{n+1}-x^{n+1}}{n+1}$

is monic and its degree is

. If the coefficient of

, then the difference $f(x)\!-\!a_0g_n(x)$ is a polynomial of degree $n\!-\!1$ . Correspondingly we obtain $f(x)-a_0g_n(x)-a_1g_{n-1}(x)$ having the degree $n\!-\!2$ and so on. Finally we see that

$\displaystyle f(x)-a_0g_n(x)-a_1g_{n-1}(x)-\ldots-a_ng_0(x)$

must be the zero polynomial. Therefore

$\displaystyle f(x)$	$\displaystyle = a_0g_n(x)+a_1g_{n-1}(x)+\ldots+a_ng_0(x)$
	$\displaystyle = \sum_{i=0}^na_ig_{n-i}(x)$
	$\displaystyle = \sum_{i=0}^na_i\int_x^{x+1}t^{n-i}\,dt$
	$\displaystyle = \int_x^{x+1}\sum_{i=0}^na_it^{n-i}\,dt$

whence we have $\displaystyle g(x) = \sum_{i=0}^na_ix^{n-i}$ .

The proof implies also that the coefficients of are rational, if the coefficients of are such. So we know that all Bernoulli polynomials have only rational coefficients.