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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1} \map {B_{2 n + 1} } x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {B_{2 n + 1} } {\frac 1 2}\) | \(=\) | \(\ds \paren {-1} \map {B_{2 n + 1} } {\frac 1 2}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 \map {B_{2 n + 1} } {\frac 1 2}\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \map {B_{2 n + 1} } {\frac 1 2}\) | \(=\) | \(\ds 0\) |
$\blacksquare$