# Lecture 5 handout.pdf - Math 472 Lecture 5 Chapter 1...

• 18

This preview shows page 1 - 6 out of 18 pages.

Math 472 Lecture 5Chapter 1Solving EquationsRuoyu WuU Michigan1 / 18
1 Solving equationsProblem: solve the equationf(x) = 0forx.Definition 1.1The functionf(x)has arootatx=riff(r) = 0.Why we study it?Shows up in almost any practical application.Sometimes the functionfis explicit and easy to work with, e.g.,f(x) =ex-sinx. But sometimes not, e.g.,f(x)denotes the freezingtemperature of some mineral water underxatmospheres of pressure.We will learn several methods:trade-off between efficiency and the amount of input neededmore efficient methods may require extra regularity forf, the cost ofevaluatingf(x)is an important factor here
2 / 18
1.1 The bisection methodMain idea of thebisection method:BracketingThink of how you would look up a word ina dictionaryHow to make sure that a root exists?Figure:f(x) =x3+x-1Theorem 1.2 (by Intermediate Value Theorem)Letfbe acontinuousfunction on[a, b], satisfyingf(a)f(b)<0. Then,fhasa root in(a, b), i.e. there existsa < r < bsuch thatf(r) = 0.3 / 18
1.1 The bisection methodThe Bisection Method:Assume we are given an interval[a, b]wheref(a)f(b)<0.Bisection: Letc= (a+b)/2. Iff(c)6= 0, then either(i)f(c)f(a)<0in which case the root must be in[a, c]; or(ii)f(c)f(b)<0in which case the root is in[c, b].By repeating the bisection step, we can make the interval containing theroot arbitrarily small.4 / 18
1.1 The bisection methodExample 1.1Find the root ofx3+x-1 = 0