ϵ=log10(|f″computed−f″exactf″exact|), ϵtot=ϵapprox+ϵro.
For the computed second derivative we have f″0=fh−2f0+f−hh2−2∞∑j=1f(2j+2)0(2j+2)!h2j, and the truncation or approximation error goes like ϵapprox≈f(4)012h2.