   Chapter 4.8, Problem 27E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Use Newton’s method to find all the solutions of the equation correct to eight decimal places. Start by drawing a graph to find initial approximations. 4 e − x 2 sin  x = x 2 − x + 1

To determine

To find: the solution of f(x)=4ex2sinxx2+x1=0 correct to eight decimal places.

Explanation

Formula used:

Newton’s method

xn+1=xnf(xn)f(xn)

Given:

f(x)=4ex2sinxx2+x1=0

Calculation:

f(x)=4ex2sinxx2+x1

Differentiate with respect to x,

f(x)=8xex2sinx2x+4ex2cosx+1

In Figure 1, the curves intersect at about 0.2

x1=0.2

Substitute n=1  in,xn+1=xn4exn2sinxnxn2+xn18xnexn2sinxn2xn+4exn2cosxn+1

x2=x14ex12sinx1x12+x118x1ex12sinx12x1+4ex12cosx1+1

Substitute x1=0.2 in x2=x14ex12sinx1x12+x118x1ex12sinx12x1+4ex12cosx1+1

x2=0.24e0.22sin0.20.22+0.218×0.2e0.22sin0.22×0.2+4e0.22cos0.2+1=0.21883273

Substitute n=2  in,xn+1=xn4exn2sinxnxn2+xn18xnexn2sinxn2xn+4exn2cosxn+1

x3=x24ex22sinx2x22+x218x2ex22sinx22x2+4ex22cosx2+1

Substitute x2=0.21883273 in x3=x24ex22sinx2x22+x218x2ex22sinx22x2+4ex22cosx2+1

x3=0.218832734e0.218832732sin0.218832730.218832732+0.2188327318×0.21883273e0.2188327306109192sin0.218832732×0.21883273+4e0.218832732cos0.21883273+1=0.21916357

Substitute  n=3  in,xn+1=xn4exn2sinxnxn2+xn18xnexn2sinxn2xn+4exn2cosxn+1

x4=x34ex32sinx3x32+x318x3ex32sinx32x3+4ex32cosx3+1

Substitute x3=0.21916357 in x4=x34ex32sinx3x32+x318x3ex32sinx32x3+4ex32cosx3+1

x4=0.219163574e0.219163572sin0.219163570.219163572+0.2191635718×0.21916357e0.219163572sin0.2191635720.21916357+4e0.219163572cos0.21916357+1=0.21916367

Substitute n=4  in,xn+1=xn4exn2sinxnxn2+xn18xnexn2sinxn2xn+4exn2cosxn+1

x5=x44ex42sinx4x42+x418x4ex42sinx42x4+4ex42cosx4+1

Substitute x4=0

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