= Solution
On the panel $[2j/n,2(j+1)/n]$, the maximum fourth <derivative> is at the left endpoint because it is decreasing. Thus
$$
E_n\le\frac1{90n^5}\sum_{j=0}^{n/2-1}
\left(e^{2j/n}-\frac{2ej}{n}\right).
$$
The <geometric series> and the sum of consecutive integers give
$$
\sum_{j=0}^{n/2-1}e^{2j/n}=\frac{e-1}{e^{2/n}-1},\qquad
\frac{2e}{n}\sum_{j=0}^{n/2-1}j=\frac{en}{4}-\frac e2.
$$
Substitution proves
$$
\boxed{E_n\le B(n):=\frac1{90n^5}
\left(\frac{e-1}{e^{2/n}-1}-\frac{en}{4}+\frac e2\right).}
$$
For $n>2$, at least one panel's maximum is smaller than the global maximum $1$, so this bound is strictly tighter than the global bound. For $n=2$, the two bounds coincide.
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