Past exam of the numerical analysis course of the University of Oxford / 2015 / 2 / c / iii / Solution

Write ${\tilde{p}}_1$ and $\tilde{q}$ for the quantities computed from the perturbed second sample. The endpoint linear interpolation and its derivative areCompare them on their interpolation interval $0\le x\le h$, with fixed $a>0$. Put $x=th$, $0\le t\le 1$. The Taylor series of the sine gives, uniformly in $t$,Here the second equality compares with $f^{\prime }(0)=a$. More generally, the exact derivative is $f^{\prime }(th)=a\cos (ath)$ andThe input perturbation itself changes the value interpolant by exactly $aϵx/h$, and changes its derivative by exactly $aϵ/h$. ThusFor fixed nonzero $ϵ$, the value approximation has a bounded noise floor: its maximum error approaches $a|ϵ|$, and at $x=h$ that error is exactly $a|ϵ|$. It does not converge uniformly to the exact values as $h\to 0$. However, it does not amplify the absolute sample perturbation on $[0,h]$. The derivative error instead grows without bound like $a|ϵ|/h$.

With exact samples, $ϵ=0$, both approximations do converge on the shrinking interval: the value error is $O(h^3)$ for this particular sine, and the derivative error is $O(h^2)$. With step-dependent noise, value convergence requires $ϵ(h)\to 0$, whereas derivative convergence requires the stronger $ϵ(h)=o(h)$, where $o$ denotes little o notation.

This is noise amplification by differentiation of a linear interpolant. Value interpolation has an absolute perturbation bound independent of $h$ on the sampled interval. Numerical differentiation has a perturbation factor $1/h$, expressing the ill-posedness of numerical differentiation from values with fixed absolute uncertainty. The amplification follows from the mathematical data map even in exact arithmetic; it is distinct from additional roundoff error in a particular implementation. For fixed $x$ outside $[0,h]$, the operation becomes extrapolation, and the value perturbation $aϵx/h$ can also diverge.