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For $n = 0,\,1,\,2,\,\ldots$ , the Bernoulli polynomial may be defined as the uniquely determined polynomial $b_n(x)$ satisfying
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(1) |
The constant term of $b_n(x)$ is the $n^{\mathrm{th}}$ Bernoulli number $B_n$ .
The Bernoulli polynomial is often denoted also $B_n(x)$ .
The uniqueness of the solution $b_n(x)$ in (1) is justificated by the
Lemma. For any polynomial $f(x)$ , there exists a unique polynomial $g(x)$ with the same degree satisfying
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(2) |
Proof. For every $n = 0,\,1,\,2,\,\ldots$ , the polynomial $$g_n(x) := \int_x^{x+1}t^n\,dt = \frac{(x+1)^{n+1}-x^{n+1}}{n+1}$$ is monic and its degree is $n$ . If the coefficient of $x^n$ in $f(x)$ is $a_0$ , 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 $$f(x)-a_0g_n(x)-a_1g_{n-1}(x)-\ldots-a_ng_0(x)$$ must be the zero polynomial. Therefore
whence we have $\displaystyle g(x) = \sum_{i=0}^na_ix^{n-i}$ .
The proof implies also that the coefficients of $g(x)$ are rational, if the coefficients of $f(x)$ are such. So we know that all Bernoulli polynomials have only rational coefficients.
- 1
- . . :. ``''. (1982).
English translation:
M. M. Postnikov: Introduction to algebraic number theory. Science Publs (``Nauka''). Moscow (1982).
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