Symmetry of Bernoulli Polynomial

Theorem

Let $\map {B_n} x$ denote the nth Bernoulli polynomial.

Then:

$\map {B_n} {1 - x} = \paren {-1}^n \map {B_n} x$

$\map G {t, x} = \dfrac {t e^{t x} } {e^t - 1}$

Then:

$\ds \map G {t, 1 - x}$

$=$

$\ds \frac {t e^{t \paren {1 - x} } } {e^t - 1}$

$\ds $

$=$

$\ds \frac {t e^{t - t x} } {e^t - 1}$

$\ds $

$=$

$\ds \frac {t e^{-t x} } {1 - e^{-t} }$

$\ds $

$=$

$\ds \frac {-t e^{-t x} } {e^{-t} - 1}$

$\ds $

$=$

$\ds \map G {-t, x}$

Thus:

$\ds \map G {t, 1 - x}$

$=$

$\ds \map G {-t, x}$

$\ds \leadsto \ \ $

$\ds \sum_{k \mathop = 0}^\infty \frac {\map {B_k} {1 - x} } {k!} t^k$

$=$

$\ds \sum_{k \mathop = 0}^\infty \frac {\paren {-1}^k \map {B_k} x} {k!} t^k$

$\ds \leadsto \ \ $

$\ds \map {B_k} {1 - x}$

$=$

$\ds \paren {-1}^k \map {B_k} x$

$\blacksquare$