I'm reading the Amann and Escher Analysis II. I'm try to work out Exercise VI.6.3.
Show that for $n \in \mathbb{N}^{+}$ \begin{align*} \text{(i)} \; & B_{2n+1}(X)\text{ has in $[0,1]$ only the zeros $0$, $1/2$, and $1$.} \\ \text{(ii)} \; & B_{2n} \text{ has in $[0,1]$ only two zeros } x_{2m} \text{ and } x_{2m}', \text{ with } x_{2m} +x_{2m}' = 1. \end{align*}
This is my attempt:
- (i) I used the proposition $B_{2n+1}(X)=(-1)B_{2n+1}(1-X)$,which implies $B_{2n+1}(\frac 1 2 )=0$. Additionally, since the odd Bernoulli numbes are zero, we have $B_{2n+1}(0)=0$.
- (ii) If $B_{2n}(X)=0$, then $B_{2n}(1-X)=0$ as well.
I attempted to use the induction to complete the proof. Assuming the proposition holds for $k \leqslant N$, I tried to show also for the $k+1$. However, I realized that for the induction to work, I need to prove that the Bernoulli numbers are alternating. Using the derivative $B'_{n+1}(X)=(n+1)B_n(X)$, I believe the proof can be completed. However, I am stuck on proving that the Bernoulli numbers are alternating.
Questions:
- Could someone help me finish the proof by showing that the Bernoulli numbers are alternating?
- Alternatively, is there a better approach to proving the proposition?
Any hints or suggestions would be greatly appreciated!
