Bernoulli Number in terms of Euler Numbers

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Then:

\(\ds B_{2 n}\) \(=\) \(\ds \frac {2n} {2^{2 n} \left({2^{2 n} - 1}\right)} \left({\sum_{k \mathop = 1}^n \dbinom {2 n - 1} {2 k - 1} E_{2 n - 2 k} }\right)\)
\(\ds \) \(=\) \(\ds \frac {2n} {2^{2 n} \left({2^{2 n} - 1}\right)} \left({\binom {2 n - 1} 1 E_{2 n - 2} + \binom {2 n - 1} 3 E_{2 n - 4} + \binom {2 n - 1} 5 E_{2 n - 6} + \cdots + 1}\right)\)

where:

$B_n$ denotes the $n$th Bernoulli number
$E_n$ denotes the $n$th Euler number.


Proof


This theorem requires a proof.
In particular: Exposition may be incorrect. Spiegel gives this in terms of the alternative form Euler and Bernoulli numbers.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.



Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Bernoulli_Number_in_terms_of_Euler_Numbers&oldid=517417"