EVALUATING A POLYNOMIAL

Transcription

1 EVALUATING A POLYNOMIAL Consider having a polynomial p(x) = a + a 1 x + a 2 x a n x n which you need to evaluate for many values of x. How do you evaluate it? This may seem a strange question, but the answer is not as obvious as you might think. The standard way, written in a loose algorithmic format: poly = a for j = 1 : n poly = poly + a j x j end

2 To compare the costs of different numerical methods, we do an operations count, and then we compare these for the competing methods. Above, the counts are as follows: additions : n multiplications : n = n(n + 1) 2 This assumes each term a j x j is computed independently of the remaining terms in the polynomial.

3 Next, do the terms x j recursively: x j = x x j 1 Then to compute { x 2, x 3,..., x n} will cost n 1 multiplications. Our algorithm becomes The total operations cost is poly = a + a 1 x power = x for j = 2 : n power = x power poly = poly + a j power end additions : n multiplications : n + n 1 = 2n 1 When n is not small, this is much less than for the first method of evaluating p(x). E.g., with n = 2, the first method has 21 multiplications, whereas the second has 39 multiplications.

4 We now considered nested multiplication. As examples of particular degrees, write n = 2 : p(x) = a + x(a 1 + a 2 x) n = 3 : p(x) = a + x (a 1 + x (a 2 + a 3 x)) n = 4 : p(x) = a + x (a 1 + x (a 2 + x (a 3 + a 4 x))) These contain, respectively, 2, 3, and 4 multiplications. This is less than the preceding method, which would have need 3, 5, and 7 multiplications, respectively. For the general case, write p(x) = a + x (a 1 + x (a x (a n 1 + a n x) )) This requires n multiplications, which is only about half that for the preceding method. For an algorithm, write poly = a n for j = n 1 : 1 : poly = a j + x poly end With all three methods, the number of additions is n; but the number of multiplications can be dramatically different for large n.

5 NESTED MULTIPLICATION Imagine we are evaluating the polynomial p(x) = a + a 1 x + a 2 x a n x n at a point x = z. Thus with nested multiplication p(z) = a + z (a 1 + z (a z (a n 1 + a n z) )) We can write this as the following sequence of operations: b n = a n b n 1 = a n 1 + zb n b n 2 = a n 2 + zb n 1. b = a + zb 1 The quantities b n 1,..., b are simply the quantities in parentheses, starting from the inner most and working outward.

6 Introduce q(x) = b 1 + b 2 x + b 3 x b n x n 1 Claim: p(x) = b + (x z)q(x) ( ) Proof: Simply expand b + (x z) ( b 1 + b 2 x + b 3 x b n x n 1) and use the fact that zb j = b j 1 a j 1, j = 1,..., n With this result (*), we have p(x) x z = b x z + q(x) Thus q(x) is the quotient when dividing p(x) by x z, and b is the remainder.

7 If z is a zero of p(x), then b = ; and then p(x) = (x z)q(x) For the remaining roots of p(x), we can concentrate on finding those of q(x). In rootfinding for polynomials, this process of reducing the size of the problem is called deflation. Another consequence of (*) is the following. Form the derivative of (*) with respect to x, obtaining p (x) = (x z)q (x) + q(x) p (z) = q(z) Thus to evaluate p(x) and p (x) simultaneously at x = z, we can use nested multiplication for p(z) and we can use the intermediate steps of this to also evaluate p (z). This is useful when doing rootfinding problems for polynomials by means of Newton s method.

8 APPROXIMATING SF (x) Define SF (x) = 1 x x sin t dt, x t We use Taylor polynomials to approximate this function, to obtain a way to compute it with accuracy and simplicity. 1. y x

9 As an example, begin with the degree 3 Taylor approximation to sin t, expanded about t = : with c t between and t. Then x 1 x x sin t = t 1 6 t t5 cos c t sin t = 1 1 t 6 t t4 cos c t sin t x [ dt = 1 1 t 6 t2 + 1 ] 12 t4 cos c t dt sin t t = x 1 18 x x dt = x 2 + R 2 (x) R 2 (x) = 1 1 x t 4 cos c t dt 12 x t 4 cos c t dt

10 How large is the error in the approximation SF (x) x 2 on the interval [ 1, 1]? Since cos c t 1, we have for x > that R 2 (x) x x t 4 dt = 1 6 x 4 and the same result can be shown for x <. Then for x 1, we have R 2 (x) 1 6 To obtain a more accurate approximation, we can proceed exactly as above, but simply use a higher degree approximation to sin t.

11 In the book we consider finding a Taylor polynomial approximation to SF (x) with its error satisfying R 8 (x) 5 1 9, x 1 A MATLAB program, plot sint.m, implementing this approximation is given in the text and is made available to the class.

12 Begin with a Taylor series for sin t, sin t = t t3 3! + t5 5! + ( 1)n 1 t 2n 1 (2n 1)! + ( 1) n t 2n+1 (2n + 1)! cos(c t) with c t between and t. Then write SF (x) = 1 x [1 t2 x 3! + t4 5! + t 2n 2 ] ( 1)n 1 dt (2n 1)! R 2n 2 (x) = 1 x + R 2n 2 (x) = 1 x 2 3!3 + x 4 5!5 + ( 1)n 1 x 2n 2 (2n 1)!(2n 1) + R 2n 2 (x) x t 2n ( 1) n (2n + 1)! cos(c t) dt

13 R 2n 2 (x) = 1 x x t 2n ( 1) n (2n + 1)! cos(c t) dt To simplify matters, let x >. Since cos(c t ) 1, R 2n 2 (x) 1 x x t 2n (2n + 1)! dt = x 2n (2n + 1)!(2n + 1) It is easy to see that this bound is also valid for x <. As required, choose the degree so that From the error bound, R 2n 2 (x) max R 2n 2(x) x 1 1 (2n + 1)!(2n + 1) Choose n so that this upper bound is itself bounded by This is true if 2n , i.e. n 5.

14 The polynomial is and p(x) = 1 x 2 3!3 + x 4 5!5 x 6 7!7 + x 8 9!9, 1 x 1 SF (x) p(x) 5 1 9, x 1 To evaluate it efficiently, we set u = x 2 and evaluate g(u) = 1 u 18 + u2 6 u u After the evaluation of the coefficients (done once), the total number of arithmetic evaluations is 4 additions and 5 multiplications to evaluate p(x) for each value of x.