 Solution

Set $z=\pi h/2$, so $0<z\le \pi /4$. The formulas in (a) and (b) giveFor $0\le t\le \pi /4$, $0\le \sin t\le t$ and $1-\cos t=2{\sin }^2(t/2)\le t^2/2$. ThereforeThese trigonometric inequalities implySubstitution yields the explicit eigenvalue error boundThis constant is asymptotically sharp: Taylor's theorem gives ${\sin }^{2}z=z^2-z^4/3+O(z^6)$, so $({\lambda }_{1,1}-{\Lambda }_{1,1})/h^2\to {\pi }^4/6$. Here $O$ denotes Big O notation.