# Solution we have f 0 2 3 69600 f 0 7 2 48400 the next

This preview shows pages 4–6. Sign up to view the full content.

and true absolute errors at the second iteration? Solution We have f (0 . 2) = 3 . 69600 f (0 . 7) = - 2 . 48400 The next root estimate is x r = x u - f ( x u )( x l - x u ) / ( f ( x l ) - f ( x u )) = 0 . 7 - ( - 2 . 484)(0 . 2 - 0 . 7) / (3 . 696 - ( - 2 . 484)) = 0 . 499029 and f (0 . 499029) = 0 . 0121359. The next iteration is x r = 0 . 7 - ( - 2 . 484)(0 . 499029 - 0 . 7) / (0 . 0121359 - ( - 2 . 484)) = 0 . 500006 This has | E t | = 0 . 000006 and | E a | = | 0 . 499029 - 0 . 500006 | = 0 . 000977.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
NAME 5 5. One can construct a fixed point iteration for a root of f ( x ) as follows: 0 = 2 x 3 - 3 x 2 - 11 x + 6 11 x = 2 x 3 - 3 x 2 + 6 x = (2 x 3 - 3 x 2 + 6) / 11 Show that this will converge if the iterations are started suﬃciently close to x = 0 . 5. Perform two iterations of the algorithm starting at x = 1. Find the true and estimated absolute errors in each case. Solution This will converge when | g ( x ) | < 1, where g ( x ) = (2 x 3 - 3 x 2 + 6) / 11. We have g ( x ) = (6 x 2 - 6 x ) / 11 = 6( x 2 - x ) / 11. At x = 0 . 5, this is (6 / 11)( - 1 / 4) = - 3 / 22 = - 0 . 13636 < 1. All this, so long as the starting point is near enough to 0.5. 1 x 0 = 1 , | E t | = 0 . 5 x 1 = 5 / 11 = 0 . 45454 , | E t | = 0 . 04545 | E a | = . 54545 x 2 = 0 . 50618 | E t | = 0 . 00618 | E a | = . 05164 1 The derivative is less than 1 in absolute value so long as the starting value is between -0.943 and 1.943, but the fixed point iteration actually converges for a larger region. There are three roots to f ( x ), which are -2, 0.5, and 3. If the starting iterate is between -2 and 3, the iterations converge to the root x = 0 . 5. If the starting point is larger than 3 or smaller than -2, it diverges.
6. Perform two iterations of Newton-Raphson to find a root of f ( x ) starting at x = 1. Carry six significant figures.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern