# coefficients of Bernoulli polynomials

The coefficient of ${x}^{k}$ in ${b}_{r}(x)$ for $k=1,2,\mathrm{\dots},r$ is $\left(\genfrac{}{}{0pt}{}{r}{k}\right){B}_{r-k}$.

The proof is by induction^{} on $r$. For $r=1$, note that ${b}_{1}(x)=x-\frac{1}{2}$, so that $[x]{b}_{1}(x)=1=\left(\genfrac{}{}{0pt}{}{1}{1}\right){B}_{0}$.

Writing $[{x}^{k}]f(x)$ for the coefficient of ${x}^{k}$ in a polynomial $f(x)$, note that for $k=1,2,\mathrm{\dots},r$,

$$[{x}^{k}]{b}_{r}(x)=\frac{1}{k}[{x}^{k-1}]{b}_{r}^{\prime}(x)=\frac{r}{k}[{x}^{k-1}]{b}_{r-1}(x)$$ |

since ${b}_{r}^{\prime}(x)=r{b}_{r-1}(x)$. By induction,

$$\frac{r}{k}[{x}^{k-1}]{b}_{r-1}(x)=\frac{r}{k}\left(\genfrac{}{}{0pt}{}{r-1}{k-1}\right){B}_{r-k}=\left(\genfrac{}{}{0pt}{}{r}{k}\right){B}_{r-k}$$ |

Thus the Bernoulli polynomials^{} can be written

$${b}_{r}(x)=\sum _{k=1}^{r}\left(\genfrac{}{}{0pt}{}{r}{k}\right){B}_{r-k}{x}^{k}+{B}_{r}$$ |

Title | coefficients of Bernoulli polynomials |
---|---|

Canonical name | CoefficientsOfBernoulliPolynomials |

Date of creation | 2013-03-22 17:46:08 |

Last modified on | 2013-03-22 17:46:08 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 4 |

Author | rm50 (10146) |

Entry type | Derivation |

Classification | msc 11B68 |

Related topic | BernoulliPolynomialsAndNumbers |