Symmetry of Bernoulli Polynomial

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

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

Proof
Let $\map G {t, x}$ denote the Generating Function of Bernoulli Polynomials:


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

Then:

Thus: