The L4 quadrature rule for (-1,1] uses four nodes: t1 = -1, t4 = 1 and tz and t3 chosen optimally,
to minimise the error.
(a) Write down the system of equations for the nodes and weights and solve this exactly.
Hint: aim for an equation involving 1 + tz but not wi or w2.
(b) Another way to find the nodes for the L4 rule is to observe that it should integrate polynomials
of degree five exactly, because it has six free parameters. Any such polynomial Qs(t) can be
expressed in the form
Q5 (t) = (f2 - 1).Jz(t) A,(t) + B3(t),
where J2(t) = t* + at + B and A,(t) and B3(t) are polynomials of degree one and three,
respectively. Here, A1 and Bz depend on Q5, but Ja is the same in all cases.
(0) Prove that there exist unique values for a and 3 such that
{(* - 1) /(t)° dt = 0, for r=0,1,
and verify that the roots of Jy are then the same as the nodes found in part (a).
(ii) Explain why these choices for a and ß enable the La rule to integrate Q, exactly.
Hint: this is very similar to the argument for Gaussian quadrature in the lecture notes.
(c) Calculate the leading-order error for the L4 rule on a subinterval of width Д.r.
(d) (i) In view of your answer to part (c), make a fair comparison of the L4 quadrature rule
and the three-point Gaussian rule.
(ii) Suppose that after performing a calculation with the La rule on N subintervals, a second
calculation us made, this time using 2N subintervals. How many new evaluations of the
integrand are required? What is the corresponding value for the three-point Gaussian
rule (i.e. how many new evaluations are required to apply the rule on 2N subintervals if
it has already been applied on N subintervals)? Explain vour answers.