CS205A hw5_solutions

# Scientific Computing

• Homework Help
• PresidentHackerCaribou10582
• 4
• 100% (1) 1 out of 1 people found this document helpful

This preview shows pages 1–2. Sign up to view the full content.

CS205 Homework #5 Solutions Problem 1 [Heath 5.5, p.248] 1. Show that the iterative method x k +1 = x k - 1 f ( x k ) - x k f ( x k - 1 ) f ( x k ) - f ( x k - 1 ) is mathematically equivalent to the secant method for solving a scalar nonlinear equa- tion f ( x ) = 0. 2. When implemented in finite-precision floating-point arithmetic, what advantages or disadvantages does the formula given in part (1) have compared with the formula for the secant method (given in the notes and in Heath, section 5.5.4)? Solutions 1. Starting with the secant method update formula is given as x k +1 = x k - f ( x k )( x k - x k - 1 ) f ( x k ) - f ( x k - 1 ) = x k ( f ( x k ) - f ( x k - 1 )) - f ( x k )( x k - x k - 1 ) f ( x k ) - f ( x k - 1 ) = x k f ( x k ) - x k f ( x k - 1 ) - x k f ( x k ) + x k - 1 f ( x k ) f ( x k ) - f ( x k - 1 ) = x k - 1 f ( x k ) - x k f ( x k - 1 ) f ( x k ) - f ( x k - 1 ) 2. When we’re near the solution x k - 1 and x k are close to each other and thus their difference is near zero. Similarly f ( x k - 1 ) and f ( x k ) are close to each other so their difference is near zero. Thus we have an indefinite form in secant computing ( x k - x k - 1 ) / ( f ( x k ) - f ( x k - 1 )). However, since we’re near the root we have f ( x k ) near zero so

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

View Full Document
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