BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 4.7, Problem 20E
To determine

To calculate: The roots of the equation,

  3sin(x2)=2x .

Expert Solution

Answer to Problem 20E

The roots are

  x2=0.69299995

  x3=0.69299996 .

Explanation of Solution

Given information:

The equation is given as:

  3sin(x2)=2x .

Formula used:

Newton’s Method:

We seek a solution of f(x)=0 , starting from an initial estimate x=x1 .

For x=xn , compute the next approximation xn+1 by

  xn+1=xnf(xn)f'(xn) and so on.

Calculation:

Consider the equation,

  3sin(x2)=2x

The initial approximation x1 can be guessed by sketching both the graphs

  y=3sin(x2) and y=2x .

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 4.7, Problem 20E

Both the graph intersect at two points whose x-coordinate can be seen in graph.

Let x1=0.693 where

  f(x)=3sin(x2)2xf'(x)=3[2xcos(x2)]2

  =6xcos(x2)2

Now, at n=1

  x2=x1f(x1)f'(x1)

  x2=x1{3sin(x12)2x16x1cos(x12)2}

  x2=(0.693)(3sin[(0.693)2]2(0.693)6(0.693)cos[(0.693)2]2)x2=(0.693)(1.3860000691.3861.687646675)x2=(0.693)(0.0000000691.687646675)

  x2=0.693(0.0000000408853352)x2=0.6930.0000000408853352x2=0.692999959

The second approximation is x2=0.69299995 .

Let n=2 ,

  x3=x2f(x2)f'(x2)

  x3=x2{3sin(x22)2x26x2cos(x22)2}

  x3=(0.69299995)(3sin[(0.69299995)2]2(0.69299995)6(0.69299995)cos[(0.69299995)2]2)x3=(0.69299995)(1.3859998831.38599991.687646544)x3=(0.69299995)(0.0000000171.687646544)

  x3=0.69299995(0.00000001007319931)x3=0.69299995+0.00000001007319931x3=0.69299996

The third approximation with six decimal places is x3=0.69299996 .

Therefore, the roots with eight decimal places are :-

  x2=0.69299995

  x3=0.69299996

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