Equivalence of Definitions of Bernoulli Numbers
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Theorem
The following definitions of the concept of Bernoulli Numbers are equivalent:
Generating Function
- $\ds \frac x {e^x - 1} = \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!}$
Recurrence Relation
- $B_n = \begin {cases} 1 & : n = 0 \\ \ds - \sum_{k \mathop = 0}^{n - 1} \binom n k \frac {B_k} {n + 1 - k} & : n > 0 \end {cases}$
or equivalently:
- $B_n = \begin {cases} 1 & : n = 0 \\ \ds - \frac 1 {n + 1} \sum_{k \mathop = 0}^{n - 1} \binom {n + 1} k B_k & : n > 0 \end {cases}$
Proof
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From the generating function definition:
\(\ds \frac x {e^x - 1}\) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!} \paren {\sum_{k \mathop = 0}^\infty \frac {x^k} {k!} - 1}\) | Definition of Real Exponential Function | ||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!} \sum_{k \mathop = 0}^\infty \frac {x^{k + 1} } {\paren {k + 1}!}\) | $1 = \dfrac {x^0} {0!}$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 1\) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!} \sum_{k \mathop = 0}^\infty \frac {x^k} {\paren {k + 1}!}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!} \sum_{k \mathop = 0}^n \binom n k \frac {B_k} {n - k + 1}\) | Cauchy Product, Product of Absolutely Convergent Series |
Equating coefficients:
For $n = 0$:
\(\ds 1\) | \(=\) | \(\ds \sum_{k \mathop = 0}^0 \binom 0 k \frac {B_k} {0 - k + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \binom 0 0 \frac {B_0} {0 - 0 + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds B_0\) | Binomial Coefficient with Zero |
For $n > 0$:
\(\ds 0\) | \(=\) | \(\ds \frac 1 {n!} \sum_{k \mathop = 0}^n \binom n k \frac {B_k} {n - k + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {n!} \paren {\sum_{k \mathop = 0}^{n - 1} \binom n k \frac {B_k} {n - k + 1} + \binom n n \frac {B_n} {n - n + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^{n - 1} \binom n k \frac {B_k} {n - k + 1} + B_n\) | Binomial Coefficient with Self and simplifying | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds B_n\) | \(=\) | \(\ds -\sum_{k \mathop = 0}^{n - 1} \binom n k \frac {B_k} {n - k + 1}\) |
Hence the result:
- $B_n = \begin{cases} 1 & : n = 0 \\
\ds - \sum_{k \mathop = 0}^{n-1} \binom n k \frac {B_k} {n - k + 1} & : n > 0 \end{cases}$
$\blacksquare$