   Chapter 3.8, Problem 17E

Chapter
Section
Textbook Problem

# 17-22 Use Newton’s method to find all solutions of the equation correct to six decimal places. 3 cos x = x + 1

To determine

To use:

Newton’s method to find all solutions of the equation correct to six decimal places

Explanation

1) Concept:

Use Newton’s formula to find for nth approximation

2) Formula:

i. Newton’s formula for nth approximation is xn+1=xn-fxnf'xn for n=1,2,3,

ii. Power rule of differentiation ddxxn=nxn-1

3) Given:

3cosx=x+1

4) Calculations:

Given 3cosx=x+1

Subtract x+1 and simplify

3cosx-x-1=0

Solving 3cosx=x+1 is same as solving for zero of 3cosx-x-1

Let fx=3cosx-x-1

Differentiate f(x)=3cosx-x-1 with respect to x

f'x=3-sinx-1

So Newton’s formula for nth approximation becomes

xn+1=xn-3cosxn-xn-1-3sinxn-1

Looking at graph, it intersects the x-axis 3 times so we start with a guess of x1=-3.7

To find x2

Substitute x1=-3.7  in formula xn+1=xn-3cosxn-xn-1-3sinxn-1

x2=-3.7-3cos-3.7--3.7-1-3sin-3.7-1

-3.63987279

To find x3

Substitute x2=-3.63987279 in formula xn+1=xn-3cosxn-xn-1-3sinxn-1

x3=-3.63987279-3cos-3.63987279--3.63987279-1-3sin-3.63987279-1

-3.637959

To find x4

Substitute x3=-3.637959  in formula xn+1=xn-3cosxn-xn-1-3sinxn-1

x4=-3.637959-3cos-3.637959--3.637959-1-3sin-3.637959-1

-3.637958

To find x5

Substitute x4=-3.637958 in formula xn+1=xn-3cosxn-xn-1-3sinxn-1

x5=-3.637958-3cos-3.637958--3.637958-1-3sin-3.637958-1

-3.637958

x4 -3.637958  x5

Therefore, x-3.637958

Next guess x1=-1.9

To find x2

Substitute x1=-1.9 in formula xn+1=xn-3cosxn-xn-1-3sinxn-1

x2=-1

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