Value of Odd Bernoulli Polynomial at One Half

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Theorem

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


Then:

$\map {B_{2 n + 1} } {\dfrac 1 2} = 0$


Proof

\(\ds \map {B_{2 n + 1} } {1 - x}\) \(=\) \(\ds \paren {-1}^{2 n + 1} \map {B_{2 n + 1} } x\) Symmetry of Bernoulli Polynomial
\(\ds \) \(=\) \(\ds \paren {-1} \map {B_{2 n + 1} } x\)
\(\ds \leadsto \ \ \) \(\ds \map {B_{2 n + 1} } {\frac 1 2}\) \(=\) \(\ds \paren {-1} \map {B_{2 n + 1} } {\frac 1 2}\)
\(\ds \leadsto \ \ \) \(\ds 2 \map {B_{2 n + 1} } {\frac 1 2}\) \(=\) \(\ds 0\)
\(\ds \map {B_{2 n + 1} } {\frac 1 2}\) \(=\) \(\ds 0\)

$\blacksquare$