1.  In the chapter on polynomial interpolation, we investigated the construction of cubic splines to

interpolate the data set

(x0 , y0 ),  (x1 , y1 ),  . . . ,  (xn , yn )  .                                               (∗)

In this question, we use simpler quadratic splines of the form

Qj (x) = αj (x − xj )2 + βj (x − xj ) + γj ,    xj  ≤ x ≤ xj+1,    j = 0, 1, . . . , n − 1.

The function Q(x) is formed from the union of the individual splines, and the notation hj  = xj+1−xj is used throughout.

(a)   (i)  Given that Q(x) has a continuous first derivative, how many equations are available to

determine the coefficients αj , βj  and γj ? How many coefficients will be left undetermined when these have been applied? Justify your answers.

(ii)  Determine γj , and show that

2αj hj  + βj  = βj+1      and   βj+1  = 2y[xj , xj+1] − βj .

(iii)  In view of the above results, what is the main advantage of quadratic splines over cubic

splines?

(b)  Calculate the magnitude of the discontinuity in the curvature of Q(x) at x = xj . Simplify your answer as far as possible.

(c) To use quadratic splines, we must choose a value for the coefficient β0 .  Here we try to determine a good choice by using the Newton polynomial through the three points (x0 , y0 ) , (x1 , y1 ) and (x2 , y2 ), which we denote by P (x) .

(i) Show that setting Q0(′)(x) = P′ (x0 ) yields β0  = y[x0 , x1] − y[x1 , x2] + y[x0 , x2] .

(ii)  By considering the roots of the difference d(x) = P (x) − Q0 (x), prove that, with this choice for β0 , P (x) and Q0 (x) are representations of the same function.

(iii) Verify algebraically that d(x) = 0 for all x (still with β0  defined as in part (i)). Hint:  write α0  as a second divided difference.

(d)   (i) Write a Maple procedure that takes three arguments:  arrays containing the x and y values from the data set (∗) and a boolean newton. If newton is true, then the value for β0  obtained in part (c) should be used. Otherwise, set β0  = 0 . The procedure should return an array containing the coefficients for interpolating quadratic splines as its result. This array should have the constant, linear and quadratic coefficients for the splines (i.e. γj , βj  and αj ) in columns 0, 1 and 2, respectively.

(ii)  Define the function

f (x) = cos(x2 )ex .

Generate a set of quadratic splines with β0  = 0 , fitting to the function f (x) at eleven equally spaced data points, with x0  = 0 and x10  = 2. Plot the splines along with f (x) on the same graph.

(iii)  Repeat part (ii), now setting newton to true, so that the value for β0  obtained in part

(c) is used. What do you notice?

You can plot the splines by loading the NumericalMethods package and using

plot(  ’eval_spline’( t  , X  ,  s  )  , t =  0  . .  2  )  ;

Here, X is the array containing the nodes xj  and s is the array returned by your procedure. Make sure the latter has the correct structure, as set out in part (i).

2. The Cn  quadrature rule for the interval [ −1, 1] uses the points at which Tn −1 (t) = ±1 as its nodes (here Tn −1   is the Chebyshev polynomial of degree n − 1).  The C3   rule is just Simpson’s rule because T2 (t) = 2t2 − 1 .

(a)   (i)  Find the nodes and weights for the C5  quadrature rule.

(ii)  Determine the first nonzero coefficient Sj  for the C5  rule.

(iii)  If the C5  rule and the five-point Newton– Cotes rule are applied on the same number of

subintervals, what approximate relationship do you expect the two errors to satisfy?     (iv) Suppose that the C5  rule has been applied on N subintervals, and that all of the function

evaluations have been stored. How many new function evaluations are required to apply the C9  rule on the same set of subintervals? Justify your answer.

(b)  Consider the approximation

−11 g(t) dt ≈  wq g(tq ).

where tq  and wq  are the nodes and weights for the Cn  quadrature rule. Assume that n is odd, and let k = (n − 1)/2 .

(i) Set g(t) = T2j(t), and then make the substitution t = cos θ to evaluate the integral. Hence show that

 −  = wq cos  !,    j = 0, 1, . . . , k .

(ii)  Consider the operator

Sk [cj ] =  −  +  cj .

j=0

It can be shown (proof: exercise for fun) that

Sk "cos  !# = 0,    for   q = 2, . . . , n − 1.

Use this fact to show that

w1  =