HW1 Solution

Computational Math II Assignment 1Math578/Math4781. Sec. 6.1 of the textbook by Kincard and Cheney: 13, 15,16, 22.3. Sec. 6.2 of the textbook by Kincard and Cheney: 4, 10, 24.5. Sec. 6.1 of the textbook by Kincard and Cheney: 27.I. Section 6.113. Prove that if we take any set of 23 nodes in the interval [1, 1] and interpolate the functionf(x) = cosh x with a polynomial p of degree 22, then the relative error ‖p(x) − f(x)‖/‖f(x)‖ is nogreater than 5 × 10 6 on [−1, 1].Proof:It is obvious f(x) = cosh x = ex +e −x2∈ C 23 [−1, 1]. Let x 0 , x 1 , . . . , x 22 denote 23 nodes in theinterval [1, 1] and the interpolating polynomial is p(x), then we can have the error by Theorem 2.f(x) − p(x) = 123! f (23) (ξ x )22∏i=0(x − x i ), where ξ x , x ∈ [−1, 1].We know that f (23) (ξ x ) = sinh(ξ x ) = eξx −e −(ξx)2is a monotonous increasing function,|f (23) (ξ x )| ≤ sinh(1) < 3 2 , ξ x ∈ [−1, 1].∏22And we know∣ (x − x i )∣ < 223 as x ∈ [−1, 1].i=0Combine all these inequality, we have |f(x) − p(x)| < 123! · 32 · 223 . And |f(x)| ≥ 1, then|p(x) − f(x)||f(x)|

2In this algorithm, there are n − i + 1 times of additions and no subtractions.16. Write an efficient algorithm for evaluatingu =n∑ i∏d j .i=1 j=1Solution:The algorithmu ← d nfor i = n − 1 to 1 step -1 dou ← (u + 1)d ienddo22. Find the Lagrange and Newton forms of the interpolating polynomial for these data: Writex −2 0 1f(x) 0 1 −1both polynomials in the form a + bx + cx 2 in order to verify that they are identical as functions.Solution:Lagrange form:l 0 (x) =l 1 (x) =l 2 (x) =x(x − 1)(−2)(−2 − 1)(x + 2)(x − 1)2(−1)(x + 2)x(1 + 2)1x(x − 1)= ,6=(x + 2)x3p L (x) = 0l 0 (x) + l 1 (x) − l 2 (x)(x + 2)(x − 1)= −2(x + 2)(x − 1) (x + 2)x= − −23= 1 − 7 6 x − 5 6 x2 .

3Newton:p 0 (x) = c 0 ⇒ c 0 = f(x 0 ) = 0,p 1 (x) = c 0 + c 1 (x + 2) ⇒ c 1 = 1 2 ,p 2 (x) = c 0 + c 1 (x + 2) + c 2 (x + 2)x ⇒ c 2 = − 5 6 .p N (x) = c 0 + c 1 (x + 2) + c 2 (x + 2)x= 1 2 (x + 2) − 5 (x + 2)x6= 1 − 7 6 x − 5 6 x2 .We can find the Lagrange form and Newton form are the same.III. Section 6.24. Prove that if f is a polynomial of degree k, then for n > kf[x 0 , x 1 , . . . , x n ] = 0.Proof:Suppose p(x) is the interpolation polynomial of at most degree n for f, thenp(x i) = f(x i ). i = 0, 1, . . . , n.Let q(x) = p(x) − f(x) be a polynomial of at most degree n. From above, we know that q(x) atleast has n + 1 roots. Henceq(x) = 0. ⇒ p(x) − f(x) = 0 ⇔ p(x) = f(x).Which means p(x) is a polynomial of degree k. By divided difference, we havep(x) =n∑∏i−1f[x 0 , x 1 , . . . , x i ] · (x − x j ).i=0Then f[x 0 , x 1 , . . . , x n ] is the coefficient of x n . Hence, f[x 0 , x 1 , . . . , x n ] = 0 when n > k.j=010. Compare the efficiency of the divided difference algorithm to the procedure de- scribed inSection 6.1 for computing the coefficients in a Newton interpolating polynomial.Solution:Algorithm in Section 6.1:c 0 ← y 0for k = 1 to n dod ← x k − x k−1u ← c k−1for i = k − 2 to 0 step − 1 dou ← u(x k − x i ) + c i

4d ← d(x k − x i )end doc k ← (y k − u)/dend doWe just need to calculate the number of operations in inner loop. There are 1 addition, 2subtractions and 2 multiplies. Then the number of operations is:n∑0∑k=1 i=k−25 =n∑5(k − 1) = 5 n 2 − 5 2 2 n ∼ 5 2 n2 .k=1Algorithm in section 6.2:for i = 0 to n dod i ← f xiend dofor j = 1 to n dofor i = j to n dod i ← (d i − d i−1 )/(x i − x i−j )end doend doWe just need to calculate the number of operations in inner loop. There are 2 multiplies and 1dividings. Then the number of operations is:n∑j=1 i=jn∑3 =n∑3(n − j + 1) = 3 2 n2 + 3 2 n ∼ 3 2 n2 .j=1Hence, the algorithm in section 6.2 is more efficient than the algorithm in section 6.1.24. Write the Newton interpolating polynomial by divided differences for these data:x 4 2 0 3f(x) 63 11 7 28Solution:By divided differences, we have table as following:4 63 26 6 12 11 2 50 7 73 28Then the Newton interpolating polynomial is:p(x) = 63 + 26(x − 4) + 6(x − 4)(x − 2) + (x − 4)(x − 2)x.

5IV. Section 6.1:27. If we interpolate the function f(x) = e x−1 with a polynomial p of degree 12 using 13 nodesin [−1, 1], what is a good upper bound for |f(x) − p(x)| on [−1, 1]?Solution:As f(x) = e x−1 is smooth enough, we can choose Chebyshev points to interpolate f(x). Therefore,we can get the error formula1|f(x) − p(x)| ≤2 12 (12 + 1)! max |f 13 (t)||t|≤11=2 12 · 13! max|t|≤1 |et−1 |1=2 12 · 13! .1Hence, the good upper bound is2 12·13! ≈ 3.9207 × 10−14 .