Talk:Aufgaben:Problem 10
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Revision as of 16:20, 13 January 2015 by Trubo-Warrior (Talk | contribs) (Created page with "There's a small mistake in the proof of the last claim. In the induction step you want to use \( \dagger \) - but this is only defined for \( n \geq 2 \). So the base case sho...")
There's a small mistake in the proof of the last claim. In the induction step you want to use \( \dagger \) - but this is only defined for \( n \geq 2 \). So the base case should be:
\( n = 2 \): Using \( \dagger : \sum_{k=1}^n \frac{B_{n-k}}{(n-k)!k!} = 0 \) we get the following set of equations:
$$ \begin{align} \begin{cases} B_1 + \frac{1}{2} B_0 = 0\\ \frac{1}{2}B_2 + \frac{1}{2}B_1 + \frac{1}{3!}B_0 = 0 \end{cases} \\ \end{align} $$
which solving is a piece of cake, resulting in \( B_2 = \frac{1}{6} \) and \( B_1 = - \frac{1}{2} \) if you remember \( B_0 = 1 \).
So the base case \( B_2 (0) = \frac{\pi^2}{3} = -\frac{(2\pi i)^2 B_2}{2} \) is satisfied.
Best, A. aka Hawkeye-Bill
