安徽大学2009-2010(1)数值分析A卷

(a) Construct the quadratic Lagrange interpolation polynomial to approximate the function. (b) Use the interpolation polynomial in part (a) to approximate f (x) at x = 4. 2
L2 (x) =
Putting these values of the Lagrange coefficients in p2 (x), we have p2 (x) = 1 (37x2 − 55x + 168), 84 . . . . . . 10 marks (b) Now taking x = 4 in the polynomial above, we have p2 (4) = 1 (37(4)2 − 55(4)x + 168) = 6.4286, 84 . . . . . . 15 marks 4. (15 marks) Using the method of least squares, fit a straight line to the four points, (1, 1), (2, 2), (3, 2), and (4, 3). Solution. The sums required for normal equation are easily obtained using the values in table i xi yi x2 xi yi i 1 1.0000 1.0000 1.0000 1.0000 2 2.0000 2.0000 4.0000 4.0000 3 3.0000 2.0000 9.0000 6.0000 4 4.0000 3.0000 16.000 12.000 ∑ ∑ ∑ 2 ∑ n=4 xi = 10 yi = 8 xi = 30 xi yi = 23
∫
1 2
ln(x + 1)dx ≈ T6 (f ) = 0.9092.
. . . . . . 8 marks (b) The second derivative of the function can be obtained as f ′ ( x) = 1 (x + 1) and f ′′ (x) = 4 −1 . (x + 1)2
The bound f ′′ (x) on [1, 2] is M = max |f ′′ (x)| = max
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Hence, the function changes sign on [a1 , c1 ] = [1.5, 1.75]. To continue, we squeeze from right and set a2 = a1 and b2 = c1 . Then the midpoint is: a2 + b2 = 1.625 : f (c2 ) = 0.0410. 2 Then continuing in this manner we obtain c3 = 1.5625. c2 = . . . . . . 10 marks Here, a = 1.5, b = 2.0. Then using inequality (b − a)/2n < 10−2 , we get ln[102 (2 − 1.5)] ≈ 5.644. ln 2 Hence, the number of iterations needed to achieve an approximation within accuracy 10−2 is 6. . . . . . . 15 marks n≥ 2. (15 marks) In the following linear equation system
( 1/6 [ 7 8 9 ln(1 + 1) + 2 ln( + 1) + ln( + 1) + ln( + 1) 2 6 6 6
+ ln( T6 (f ) = Hence,
) ] 10 11 + 1) + ln( + 1) + ln(2 + 1) 6 6
1 [0.6932 + 2(4.5591) + 1.0986] = 0.9092. 12
∫
2
I (f ) =
1
ln(x + 1)dx,
where x ln(x + 1) 1 7/6 8/6 9/6 10/6 11/6 2 0.6931 0.7732 0.8473 0.9163 0.9808 1.0415 1.0986
(a) Compute the approximation of the integral using the composite Trapezoidal rule. (b) Compute the error bound for your approximation using formula. Solution. (a) Given f (x) = ln(x + 1), n = 6, and so h = (2 − 1)/6 = 1/6, then composite Trapezoidal rule for n = 6, can be written as T6 (f ) =
Solution. (a) Obviously, a quadratic polynomial can be determined so that it passes through the three points. Consider the quadratic Lagrange interpolating polynomial as follows: p2 (x) = L0 (x)f (x0 ) + L1 (x)f (x1 ) + L2 (x)f (x2 ) that is p2 (x) = 2L0 (x) + 4L1 (x) + 19L2 (x). . . . . . . 5 marks The Lagrange coefficients can be calculated as follows: L0 (x) = L1 (x) = (x − x1 )(x − x2 ) 1 = (x2 − 10x + 21), (x0 − x1 )(x0 − x2 ) 21 (x − x0 )(x − x2 ) 1 = − (x2 − 7x), (x1 − x0 )(x1 − x − x2 ) 12 (x − x0 )(x − x1 ) 1 = (x2 − 3x). (x2 − x0 )(x2 − x1 ) 28
Anhui University
Semester 1, 2009-2010 Final Examination Model Answer and Referee Criterion for Numerical Analysis (Paper A)
In the computation of problem 1-6 below, keep the forth decimal place with truncation error. 1. (15 marks) Use the Bisection method to find the approximation {cn , }(n = 1, 2, 3) of the root of the equation x3 = 2x + 1 that is located on the interval [1.5, 2.0], and determine the number of iterations needed to achieve an approximation within accuracy 10−2 . Solution. Since the given function f (x) = x3 − 2x − 1 is a polynomial function and so is continuous on [1.5, 2.0], starting with a1 = 1.5 and b1 = 2, we compute: a1 = 1.5 : f (a1 ) = −0.625 b1 = 2.0 : f (b1 ) = 3.0 and since f (1.5) · f (2.0) < 0, so that a root of f (x) = 0 lies in the interval [1.5, 2.0]. Using bisection formula, we get: c1 = a1 + b1 = 1.75; 2 f (c1 ) = 0.8593.
which is the required quadratic interpolating polynomial.
which is the required estimate value of f (4).
3
. . . . . . 5 marks The linear system involving A and B in the normal equation form 4A + 10B = 8 10A + 30B = 23 . . . . . . 10 marks The solution of the linear system is A = 0.5 and B = 0.6. Thus, the least squares line is p1 (x) = 0.5 + 0.6x. . . . . . . 15 marks 5. (15 marks) Consider the integral
5x − y + z
= 10 2x + 8y − z = 11 −x + y + 4z = 3 1
(a) Starting with P0 = (0, 0, 0), and use Gauss-iteration to find P1 , P2 , P3 . (b) Prove this Gauss-iteration is convergent. Solution. (a) The Gauss-Seidel iteration for the given system is 1 xk+1 = (10 + yk − zk ) 5 1 yk+1 = (11 − 2xk+1 + zk ) 8 1 zk+1 = (3 + xk+1 − yk+1 ) 4 . . . . . . 5 marks and starting with the initial approximation x0 = 0, y0 = 0, z0 = 0, then for k = 0, we obtain 1 1 x1 = (10 + y0 − z0 ) = (10 + 0 − 0) = 2 5 5 1 1 y1 = (11 − 2x1 + z0 ) = (11 − 4 + 0) = 0.875 8 8 1 1 z1 = (3 + x1 − y1 ) = (3 + 2 − 0.875) = 1.0312 4 4 The subsequent iterations are P2 = (x2 , y2 , z2 ) = (1.9687, 1.0117, 0.9892) P3 = (x3 , y3 , z3 ) = (2.0044, 0.9975, 1.0017) . . . . . . 10 marks (b) Since the coefficient matrix 5 −1 1 8 −1 A= 2 −1 1 4 is strictly diagonally dominant, by the theorem, the vector sequence Pn will converge to P for any choice of the starting vector P0 . . . . . . . 15 marks 3. (15 marks) Consider the following table: x f (x) 0 3 7 2 4 19