{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw14_solutions

# hw14_solutions - Homework#4 Solutions for problems from...

This preview shows pages 1–3. Sign up to view the full content.

Homework #4 Solutions for problems from Section 2.5, Section 2.6 can be found in the solution to Homework #3. Section 3.1 1. ( ) Three iterations are required. We have ( ) ( ) The first iteration gives us: ( ) Therefore the new interval is [ ] [ ] The second iteration gives us: ( ) Therefore the new interval is [ ] [ ] The third iteratio n gives us: ( ) Therefore the new interval is [ ] [ ] 2a. ( ) Desired accuracy is We have ( ) ( ) The first iteration gives us: ( ) Therefore the new interval is [ ] [ ]

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

View Full Document
Length of the interval therefore we need to continue. The second iteration gives us: ( ) Therefore the new interval is [ ] [ ] Length of the interval therefore we need to continue. The third iteratio n gives us: ( ) Therefore the new interval is [ ] [ ] Length of the interval therefore we need to continue. The fourth iteratio n gives us: ( ) Therefore the new interval is [ ] [ ] Length of the interval desired accuracy is achieved. It took 4 iterations to get accuracy of . This result coincides with formula ( ) ( ) ( ) 3a, f. Here is the MATLAB code (2 functions): First m.file %Function funvalue returns value of function f at point x function [ f ] = funvalue(x) % function for problem 3a
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}