The L. quadrature rule for (-1,1) uses four nodes: t1 = -1, ta = 1 and tą and t3 chosen optimaly, to minimise the error.

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The L Quadrature Rule For 1 1 Uses Four Nodes T1 1 Ta 1 And Ta And T3 Chosen Optimaly To Minimise The Error 1 (84.65 KiB) Viewed 47 times

The L. quadrature rule for (-1,1) uses four nodes: t1 = -1, ta = 1 and tą and t3 chosen optimaly, 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 + t but not wi or w2. (b) Another way to find the nodes for the La 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 Qs(t) = (t? – 1)J2(t)A(t) + B3(t), where Ju(t) = ť? + at + B and Ai(t) and B3(t) are polynomials of degree one and three, respectively. Here, A, and B3 depend on Qs, but Jy is the same in all cases. Provided J2 is such that L. the argument used to locate the nodes for Gaussian quadrature in the lecture notes will now work (you are not asked to write out the details of this argument). Prove that there are unique values for a and B such that (*) is satisfied, and verify that the roots of Jy are then the same as the nodes found in part (a). (c) Calculate the leading order error for the La rule on a subinterval of width Ar. (d) (i) In view of your answer to part (c), make a fair comparison of the La 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 your answers. (– 1),z(t)\ dt = 0, for r = 0, 1,