CS514 Fall '00 Numerical Analysis (Sketched) Solution of Homework 3 1. Questions from text in Chapter 3 Problem 1: 0 f 0 1 f 8 1 8(f 1 f 0 ) 1 f 4 2 8

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1 CS514 Fll '00 Numericl Anlysis (Sketched) Solution of Homework 3 1. Questions from text in Chpter 3 Problem 1: 0 f 0 1 f 8 1 8(f 1 f 0 ) 1 f 4 2 8(f 2 f 1 ) 32(f 2 2f 1 f 0 ) 1 32 f 2 3 4(f 3 f 2 ) (f f 2 + 2f 1 ) (f 3 3 6f 2 + 8f 1 3f 0 ) Therefore, f 0 (0) = 1 3 (2f 3 24f f 1 42f 0 ) + e, nd jej» M 4 4! = M Problem 6: () I(h) = 1 3 h h7=2 ; T (h) = 1 2 h h7=2 : (b) Then, E(h) = O(h 3 ) s h! 0: I(h) = 1 3 h h3=2 ; T (h) = 1 2 h h3=2 : Then, E(h) = O(h 3=2 ) s h! 0: Problem 7: The error is lrger by order compred to (). f in () is in C 2 [0; h]; but f in (b) is in C[0; h]. 1

2 () By Tylor's Theorem, letting x k + 1 h be written s 2 x0, f(x) = f(x 0 ) + (x x 0 )f 0 (x 0 ) (x x0 ) 2 f 00 (ο(x)); x k» x» x k + h; where ο 2 [x k ; x k + h]. Integrting gives, Z +h f(x)dx = hf(x 0 ) + f 0 (x 0 ) Z +h (x x 0 )dx Z +h (x x 0 ) 2 f 00 (ο(x))dx: The second term on the right hnd side is zero nd pply Men Vlue Theorem, yields, Z +h f(x)dx = hf(x 0 ) f 00 (ο(x)) Clculte Hence, Z +h (b) Sum ll subintervls, Problem 8: () tht is, Z b 1 2 f 00 (ο(x)) Z +h Z +h (x x 0 ) 2 dx = h3 12 : (x x 0 ) 2 dx: f(x)dx = hf(x 0 )f(x 0 ) + h3 24 f 00 (ο k ); x k» ο k» x k + h: Z b X n 1 f(x)dx = h f(x 0 ) + h2 n 1 b f 00 (ο k ); 24 n X k=0 X k=0 n 1 f(x)dx = h f(x 0 ) + h2 24 (b )f 00 (ο);» ο» b: k=0 1 y 1 0 y 0 y 0 y 1 0 y 0 y y 1 y 1 y 0 y 0 y y 1 y 1 y 0 y (y 2 1 y 1 2y0) 0 p 3 (y; t) = y 1 + (t + 1)(y 0 y 1 ) + (t + 1)t(y 0 0 y 0 + y 1 ) (t + 1)t2 (y 1 y 1 2y 0 0): 2

3 Then, Z 1 p 3 (y; t)dt 1 = 2y 1 + 2(y 0 y 1 ) (y0 0 y 0 + y 1 ) (y 1 y 1 2y 0 0) = 1 3 (y 1 + 4y 0 + y 1 ): (b) E S (y) = Z 1 1 (t + 1)t 2 (t 1) y(4) (fi(t)) dt 4! = 1 90 y(4) (fi); 1» fi» 1: (c) Let n be even, h = (b )=n, x k = + kh, nd f k = f(x k ). Then, Z +2 f(x)dx = h Z 1 Let y(t) = f(x k+1 + th) in () nd (b), we obtin 1 f(x k+1 + th)dt: Z +2 Sum ll even k from 0 to n-2, f(x)dx = hf 1 3 (f k + 4f k+1 + f k+2 ) 1 90 h4 f (4) (ο k )g; Z b ο k = x k+1 + fih; 1» fi» 1: f(x)dx = n 2 X k is even. After clculting, we will get k=0 Z +2 f(x)dx; Z b f(x)dx = h 3 (f 0 + 4f 1 + 2f 2 + 4f f n 2 + 4f n 1 + f n ) + E S n (f); where Problem 10: E S n (f) = b 180 h4 f (4) (ο);» ο» b: f = e x2 ; f 00 = 2(2x 2 1)e x2 ; f (4) = 4(4x 4 12x 2 + 3)e x2 : so, jf 00 j» 2; jf (4) j» 12 on [0; 1]: 3

4 () E T n = 1 12n 2 f 00 (ο); 0» ο» 1; we get, Tht is, je T n j» 1 6n 2» if n : n 103 p 3 ß 578: (b) we get, if Tht is, E S n = 1 180n 4 f (4) (ο); 0» ο» 1; je S n j» 1 180n 4 12 = 1 15n 4» n : n [ ]1=4 10 ß 20: Problem 35 The re of unit disk cn be computed s A = 2 Z 1 1 (1 t 2 ) 1=2 dt = 2 Z 1 1 (1 t 2 )(1 t 2 ) 1=2 dt; hence cn be evluted exctly by 2 point Guss-Chebyshev qudrture rule pplied to f(t) = 1 t 2 : A = 2 ß 2 (1 t t 2 2) = ß(2 t 2 1 t2 2): But, nd Thus, t 1 = cos ß 4 = 1 p 2 t 2 = cos 3ß 4 = 1 p 2 : A = ß(2 1=2 1=2) = ß: 4

5 2. Mchine Assignment 1 in Chpter 3. After computing, we get The progrm is s following: e 1 (h) = ß (f 1 f 0 ) ; e 2 (h) = 2ß 2 f 1 2f 0 + f 1 ; h h 2 e 3 (h) = 6ß 3 f 2 3f 1 + 3f 0 f 1 h 3 ; e 4 (h) = 24ß 4 f 2 4f 1 + 6f 0 4f 1 + f 2 h 4 : % % Mchine ssignment 1 in Chpter 3, the problem description is not cler % fixed nd solve it. % index for in f is % formt long; i=[ ]; PI = 4*tn(1); E1 = PI; E2 = 2*pi^2; E3 = 6*pi^3; E4 = 24*pi^4; k = 0:10; h=2.^(-k-2); f = 1./ (1 - PI * (i'*h)); P = [ ((f(4,:)-f(3,:))./ h) ; ((f(4,:)-2*f(3,:)+f(2,:))./ (h.^2)); ((f(5,:)-3*f(4,:)+3*f(3,:)-f(2,:))./ (h.^3)); ((f(5,:)-4*f(4,:)+6*f(3,:)-4*f(2,:)+f(1,:))./ (h.^4)) ]; E = [E1 - ((f(4,:)-f(3,:))./ h) ; E2 - ((f(4,:)-2*f(3,:)+f(2,:))./ (h.^2)); E3 - ((f(5,:)-3*f(4,:)+3*f(3,:)-f(2,:))./ (h.^3)); E4 - ((f(5,:)-4*f(4,:)+6*f(3,:)-4*f(2,:)+f(1,:))./ (h.^4)) ]; R=[]; 5

6 for j=2:11 R = [R bs(e(:,j)./ E(:,j-1))]; end fid1 = fopen('ma3_1','w'); fprintf(fid1,' exct derivtes n'); fprintf(fid1,' %8.3f %8.3f %8.3f %8.3f n', E1, E2, E3, E4); fprintf(fid1,' pproximte derivtes n'); fprintf(fid1,' %8.3f %8.3f %8.3f %8.3f n', P); fprintf(fid1,' pproximte errors n'); fprintf(fid1,' %8.3f %8.3f %8.3f %8.3f n', E); fprintf(fid1,' r1 r2 r3 r4 n'); fprintf(fid1,' %8.3f %8.3f %8.3f %8.3f n', R); fclose(fid1); The result is s exct derivtes pproximte derivtes pproximte errors

7 r1 r2 r3 r One cn relize tht the convergence order is O(h) for n = 1 nd n = 3, the rtion tends to 1. However, for n = 2 nd n = 4, the order is 2 O(h2 ) nd the rtio tends to be 1. 4 Rounding error occurs in the lst few vlues of n = 4 which corrupts the limit of the rtion. 3. Problem 3 on HW sheet. () trpezoid rule: Z 1 0 e x dx = 0:1( 1 2 e0 + 9X i=1 e i e 1 ) ß 0:63265: Then the error is Z 1 0 e x dx = 0:

8 (b) gussin qudrture: Let ß 3 (t) = t 3 +p 1 t 2 +p t 2+p 3 : Since it is orthogonl to 1, t, nd t 2, one cn solve the coefficients which p 1 = 3, p 2 2 = 3, nd p 5 3 = 1. Then, we 20 cn get 3 roots for ß 3. Then, one need to find the weights w 1, w 2, nd w 3. Consider f(t) 1, f(t) t, nd f(t) t 2, then solve the liner eqution system nd get w 1, w 2, nd w 3. The results re s t 1 = 0:887298; t 2 = 0:5; t 3 = 0:112702; w 1 = 0:277778; w 2 = 0:444444; w 3 = 0:277778: And, Z 1 0 e x dx ß 0: Z 1 0 e x dx = 0: Then the error is (more precision pplied here). Therefore, gussin qudrture preforms better for this problem 4. Questions from text in Chpter 4. Problem 7: () Plot the function. Since there re three intersections with the line x = 0, there (b) No. Problem 21: () re 3 rel root. Their pproximted loction is -2, 2, nd 10.5 respectively. x n+1 ff = 1 2 (x n + x n ) ff = x2 n + 2ffx n 2x n = x2 n 2ffx n + ff 2 2x n = (x n + ff) 2 2x n : ) 1 2x n (b) Let f(x) = x 3. Then f 0 (x) = 3x 2. = x n+1 ff x n ff : x n+1 ff lim n!1 x n ff = n!1 lim 1 = 1 2x n 2ff : x n+1 = x n f(x n) f 0 (x n ) = x n x3 n = 3x3 x 3 + n n 3x 2 3x 2 n n 8

9 = 1 3 (2x n + x 2 n ): Then, Problem 22: () Therefore, x n+1 ff = 1 3 (2x n + x 2 n = 2x3 n + ff 3 3ffx 2 n 2x 2 n = (ff3 x 3 n) 3x 2 n(ff x n ) 3x 2 n ) ff = 2x3 n + 3ffx 2 n 3x 2 n = ff3 x 3 n + 3(x 3 n ffx 2 n) 3x 2 n ) x n+1 ff (x n ff) = ff + 2x n : 2 3x 2 n x n+1 ff lim n!1 (x n ff) = lim ff + 2x n 2 n!1 3x 2 n = (ff x n) 2 (ff + 2x n ) : 3x 2 n = 1 ff : d n = x n+1 x n = 1 2 ( x n x n ): (b) ) 2d n x n = x 2 n : ) x n 2 + 2dx n = 0: ) x n = 2d n ± p 4dn = d n ± q d 2 n + : We ssume x 0 > 0, then x n > 0. Therefore, q x n = d n + d 2 + = + d2 p d 2 n n n d n + d 2 + : n ) x n = d n + p d 2 n + : ) x n = d n + p d 2 n + = 1 2 (x n 1 + = 1 2 ( d n 1 + qd 2n (d n 1 + x n 1 ) q d 2 n 1 + )): q p d n + d 2 + = 1 2 ((d n 1 d 2 + ) n 1 d 2 n n 1 d2 + d n 1 + n 1 = 1 2 ( qd 2 n 1 + d n 1 + d n 1 + q d 2 n 1 + ) qd 2n 1 + ) = q d 2 n 1 + : 9

10 () '(x) = 2 x, x 6= 0. Then, ' 0 (ff) = 2ff 1 6= 0: ) p q d n + d 2 + = d 2 + : n 1 n q ) d n + d 2 + = n qd 2n 1 + : ) q d 2 n + = ) d 2 n + = d 2 n ) q 2d n d 2 + = n 1 q d n : d 2 n 1 2d n 2 qd 2n d 2 + : n 1 d2 n 1 d 2 n 1 + 2: Problem 35: Therefore, d 2 n 1 ) d n = 2 qd 2n 1 + : d 2 n 1 jd n j = 2 qd 2n 1 + : Clim j 2 x 2 j < 1, then x 2 ( 1; p 2) [ ( p 2; 1). It will converge only when x 0 = ± p 2. Otherwise, it never converges. (b) '(x) = x 2 + x 2. Then, ' 0 (ff) = 2ff + 1 6= 0: Clim j2x + 1j < 1, then x 2 ( 1; 0). It will never converge to the root. (c) '(x) = x+2, x 6= 1. Then, x+1 ' 0 (ff) = 1 6= 0: (ff + 1) 2 1 Clim j j < 1, then, x 2 ( 1; 2) [ (0; 1). (x+1) 2 converge when x 0 > 0. Since p 2 > 2, it will 5. Mchine Assignment 1 in Chpter 4. The progrm is s following: FORTRAN(f90) prt, min progrm 10

11 !! This is for Mchine Assignment 1 in Chpter 4! PROGRAM m4_1 rel e rel from, to, h integer i, n n=200 e = Eps()! For prt (): open(20, file='m4_1') from = 0.93 to = 1.07 h = (to-from)/n write(20,2) from write(20,2) to do i=0, n x = from+i*h write(20,1)(p(x)/e) enddo close(20) 1 formt(e16.7) 2 formt(f16.7)! For prt (b): open(20, file='m4_1b') from = 21.7 to = 22.2 h = (to-from)/n write(20,2) from write(20,2) to do i=0, n 11

12 x = from+i*h write(20,1)(pb(x)/e) enddo close(20) end rel function Eps() rel, b, c =4./3. b=-1 c=b+b+b Eps=bs(c-1) return end rel function p(x) rel x p=x**5-5*x**4+10*x**3-10*x**2+5*x-1; return end rel function pb(x) rel x pb=x**5-100*x**4+3995*x** *x** *x ; return end MATLAB prt, for plot only % % for plot in Mchine ssignment 1 of Chpter 4, red file m4_1 nd m4_1b % fid1 = fopen('m4_1', 'r'); 12

13 p = fscnf(fid1, '%g', [203]); fclose(fid1); fid1 = fopen('m4_1b', 'r'); pb = fscnf(fid1, '%g', [203]); fclose(fid1); h = (p(2)-p(1))/200; hb = (pb(2)-pb(1))/200; i = p(1):h:p(2); ib = pb(1):hb:pb(2); figure; subplot(2,1,1); plot(i, p(3:203), '-'); title('polynomil /eps'); grid on; subplot(2,1,2); plot(ib, pb(3:203), '-'); title('polynomil b/eps') grid on; The plots re s below 13

14 20 polynomil /eps x 107 polynomil b/eps Figure 1: The plots for two functions 14