Solutions of Homework 6: CS321, Fall 2010 Please show all steps in your work. Please be reminded that you should do your homework independently. 1. (10 points) Let S be a cubic spline that has knots t0 < t1 < · · · < tn . Suppose that on the two intervals [t0 , t1 ] and [t2 , t3 ], S reduces to linear polynomials. What does the polynomial S look like on the interval [t1 , t2 ] (linear, quadratic, or cubic)? Solution. Let S0 (x) = c0 x + d0 be the linear polynomial defined on [t0 , t1 ], and S2 (x) = c2 x + d2 the linear polynomial defined on [t2 , t3 ]. We have S0′ (x) = c0 , S2′ (x) = c2 , S0′′ (x) = 0, S2′′ (x) = 0. and Assume the cubic spline polynomial defined on [t1 , t2 ] to be S1 (x) = a1 x3 + b1 x2 + c1 x + d1 . It follows that S1′ (x) = 3a1 x2 + 2b1 x + c1 , and S1′′ (x) = 6a1 x + 2b1 . Since S, S ′ and S ′′ have to be continuous on [t0 , t3 ], we must have S0′′ (t1 ) = S1′′ (t1 ), S1′′ (t2 ) = S2′′ (t2 ). 6a1 t1 + 2b1 = 0, 6a1 t2 + 2b1 = 0. Hence (1) These two equations lead to 6a1 (t1 − t2 ) = 0. Since t1 6= t2 , we have a1 = 0. From Eq. (1), we have b1 = 0. We can now conclude that S1 (x) = c1 x + d1 , so S is also a linear polynomial on [t1 , t2 ]. 2. (10 points) Find a quadratic spline interpolant for these data x −1 0 1/2 1 2 5/2 y 2 1 0 1 2 3 Solution. Suppose the quadratic spline interpolant has the following form Q0 (x), Q1 (x), x ∈ [−1, 0] x ∈ [0, 1/2] Q(x) = Q2 (x), x ∈ [1/2, 1] Q 3 (x), x ∈ [1, 2] Q4 (x), x ∈ [2, 5/2] ′ Define zi = Qi (ti ), the following is the formula for Qi zi+1 − zi (x − ti )2 + zi (x − ti ) + yi . Qi (x) = 2(ti+1 − ti ) It follows that zi+1 = −zi + 2 yi+1 − yi , ti+1 − ti (0 ≤ i ≤ n − 1). By setting z0 = 0, we can compute recursively, 1−2 = −2, 0 − (−1) 0−1 = −2, = −(−2) + 2 1/2 − 0 1−0 = −(−2) + 2 = 6, 1 − 1/2 2−1 = −4, = −6 + 2 2−1 3−2 = −(−4) + 2 = 8. 5/2 − 2 z1 = −0 + 2 z2 z3 z4 z5 Hence, the quadratic spline interpolant is Q(x) = Q0 (x) = −(x + 1)2 + 2, Q1 (x) = −2x + 1, x ∈ [−1, 0] x ∈ [0, 1/2] Q2 (x) = 8(x − 1/2)2 − 2(x − 1/2), x ∈ [1/2, 1] Q3 (x) = −5(x − 1)2 + 6(x − 1) + 1, x ∈ [1, 2] Q4 (x) = 12(x − 2)2 − 4(x − 2) + 2, x ∈ [2, 5/2] 3. (10 points) Determine if this function is a quadratic spline? Explain why or why not. −∞ < x ≤ 1 x 2 Q(x) = x 1≤x≤2 4 2≤x<∞ Solution. Q(x) is not a quadratic spline. The domain of definition is (−∞, ∞) which is not finite. Furthermore, lim Q′ (x) = lim 1 = 1, x→1− x→1− ′ lim Q (x) = lim+ 2x = 2. x→1+ x→1 It follows that Q′ (x) is not continuous at x = 1, which violates the definition of a quadratic spline. 4. (10 points) Determine the parameters a, b, c, d and e so that S is a natural cubic spline ( a + b(x − 1) + c(x − 1)2 + d(x − 1)3 x ∈ [0, 1] S(x) = 3 2 (x − 1) + ex − 1 x ∈ [1, 2] Solution. We first compute the derivatives of the individual functions S0′ (x) = b + 2c(x − 1) + 3d(x − 1)2 S0′′ (x) = 2c + 6d(x − 1) S1′ (x) = 3(x − 1)2 + 2ex S1′′ (x) = 6(x − 1) + 2e We will make use of the continuity condition and the definition of the natural cubic spline. From S0 (1) = S1 (1), we have a = c − 1. From S0′ (1) = S1′ (1), we have b = 2e. From S0′′ (1) = S1′′ (1), we have 2c = 2e, with c = e. We also have S0′′ (0) = S ′′ (2) = 0 2c − 6d = 0, 2e + 6 = 0, we have e = −3, c = −3. Then a = e − 1 = −4, b = 2e = −6, and d = c/3 = −1. 5. (10 points) Determine the coefficients so that the function S(x) = ( x2 + x3 a + bx + cx2 + dx3 0≤x≤1 1≤x≤2 is a cubic spline and has the property S1′′′ (x) = 12. Solution. The derivatives are S0′ (x) = 2x + 3x2 S0′′ (x) = 2 + 6x S1′ (x) = b + 2cx + 3dx2 S1′′ (x) = 2c + 6dx S1′′′ (x) = 6d Given the condition S1′′′ (x) = 12, we have d = 2. By continuity, we have S0′ (1) = S1′ (1), which yields 5 = b + 2c + 3d. From S0′′ (1) = S1′′ (1), we have 8 = 2c + 6d. We can solve the above two equations and get c = −2 and b = 3. From S0 (1) = S1 (1), we have 2 = a + b + c + d, which gives us a = −1. 6. (10 points) List all the ways in which the following function fails to be a natural cubic spline: ( x3 + x − 1 (−1 ≤ x ≤ 0) f (x) = x3 − x − 1 (0 ≤ x ≤ 1) Solution. S ′ (x) is not continuous at x = 0. S ′′ (x) is not equal to zero at x = −1, 1.

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