Equivalence of Definitions of Bernoulli Numbers

From ProofWiki
Jump to navigation Jump to search

Theorem

The following definitions of the concept of Bernoulli Numbers are equivalent:

Generating Function

$\displaystyle \frac x {e^x - 1} = \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!}$

Recurrence Relation

$B_n = \begin{cases} 1 & : n = 0 \\ \displaystyle - \sum_{k \mathop = 0}^{n - 1} \binom n k \frac {B_k} {n - k + 1} & : n > 0 \end{cases}$


Proof

From the generating function definition:

\(\displaystyle \frac x {e^x - 1}\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \frac{B_n x^n} {n!}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!} \left({ \sum_{k \mathop = 0}^\infty \frac {x^k} {k!} - 1}\right)\) Definition of Real Exponential Function
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \frac{B_n x^n} {n!} \sum_{k \mathop = 0}^\infty \frac {x^{k + 1} } {\left({k + 1}\right)!}\) $1 = \dfrac {x^0} {0!}$
\(\displaystyle \leadsto \ \ \) \(\displaystyle 1\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \frac{B_n x^n} {n!} \sum_{k \mathop = 0}^\infty \frac {x^k } {\left({k + 1}\right)!}\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!} \sum_{k \mathop = 0}^n \binom n k \frac {B_k} {n - k + 1}\) Cauchy Product


Equating coefficients:

For $n = 0$:

\(\displaystyle 1\) \(=\) \(\displaystyle \sum_{k \mathop = 0}^0 \binom 0 k \frac {B_k} {0 - k + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \binom 0 0 \frac {B_0} {0 - 0 + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle B_0\) Binomial Coefficient with Zero


For $n > 0$:

\(\displaystyle 0\) \(=\) \(\displaystyle \frac 1 {n!} \sum_{k \mathop = 0}^n \binom n k \frac {B_k} {n - k + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {n!} \left({\sum_{k \mathop = 0}^{n - 1} \binom n k \frac {B_k} {n - k + 1} + \binom n n \frac {B_n} {n - n + 1} }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 0}^{n - 1} \binom n k \frac {B_k} {n - k + 1} + B_n\) Binomial Coefficient with Self and simplifying
\(\displaystyle \leadsto \ \ \) \(\displaystyle B_n\) \(=\) \(\displaystyle -\sum_{k \mathop = 0}^{n - 1} \binom n k \frac {B_k} {n - k + 1}\)


Hence the result:

$B_n = \begin{cases} 1 & : n = 0 \\ \displaystyle - \sum_{k \mathop = 0}^{n-1} \binom n k \frac {B_k} {n - k + 1} & : n > 0 \end{cases}$

$\blacksquare$