Past exam of the numerical analysis course of the University of Oxford / 2017 / 1 / d

Read the differential order $dxdy$ literally: take quadrature nodes $X=(c,(c+d)/2,d)$ in $x$ and $Y=(a,(a+b)/2,b)$ in $y$. Applying Simpson's rule first in $x$ and then in $y$ multiplies the one-dimensional quadrature weights. With $s=(1,4,1)$, this tensor-product quadrature isEquivalently, the nine sampled function values have the quadrature weight arraywith rows indexed by the $x$ quadrature nodes and columns by the $y$ quadrature nodes. The quadrature weights sum to the rectangle's area. If the intended coordinate assignment is instead $x\in [a,b]$, $y\in [c,d]$, exchange $X$ and $Y$ in the formula. Both conventions give the same calculation on the unit square used below.

For the separable integrand, tensor-product quadrature factors into two one-dimensional applications of Simpson's rule:soThe exact double integral, evaluated by Fubini's theorem and the power rule for integration, isThe $x$ factor is always integrated exactly. All quadrature error therefore comes from applying Simpson's rule to the $y$ factor.

At $q=2$, the preceding tensor-product quadrature givesBoth factors of the separable integrand are quadratic polynomials, so exactness follows directly from the construction of Simpson's rule.

At $q=3$,The $y$ factor is now a cubic polynomial. Simpson's rule still integrates it exactly because its degree of precision of a quadrature rule is three. Exactness of tensor-product quadrature depends on the polynomial degree in each variable, rather than their sum.

At $q=4$,This is an overestimate, with signed quadrature errorThe relative error is $(1/720)/(1/30)=1/24$, approximately $4.17\%$. The quartic polynomial $y^4$ lies beyond the degree of precision of a quadrature rule of Simpson's rule, so the exactness of the previous cases no longer applies.