
To calculate: The roots of the equation,
(x−2)2=lnx .

Answer to Problem 15E
The roots are
x2≈1.333333
x3≈0.816537
x4≈1.262929
Explanation of Solution
Given information:
The equation is given as:
(x−2)2=lnx .
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=xn−f(xn)f'(xn) and so on.
Calculation:
Consider the equation,
(x−2)2=lnx
Now,
(x−22)−lnx=0f(x)=(x−22)−lnxf'(x)=2(x−2)−1x
Let the initial approximation x1=1
Now, at n=1
x2=x1−f(x1)f'(x1)
x2=x1−(x1−2)2−lnx12(x1−2)−1x1
x2=(1)−((1−2)2−ln(1)2(1−2)−11)x2=(1)−((−1)2−02(−1)−1)x2=(1)−(1−2−1)
x2=1−(1−3)x2=1−(−0.333333333)x2=1.333333333
The second approximation is x2=1.333333 .
Let n=2 ,
x3=x2−f(x2)f'(x2)
x3=x2−(x2−2)2−lnx22(x2−2)−1x2
x3=(1.333333)−((1.333333−2)2−ln(1.333333)2(1.333333−2)−11.333333)x3=(1.333333)−(−0.444444888−0.287681822−0.666666−0.750000187)x3=1.333333−(−0.73212671−1.416666187)
x3=1.333333−0.5167955x3=0.816537499
The third approximation with six decimal places is x3=0.816537 .
Let n=3 ,
x4=x3−f(x3)f'(x3)
x4=x3−(x3−2)2−ln(x3)2(x3−2)−1x3
x4=(0.816537)−((0.816537−2)2−ln(0.816537)2(0.816537−2)−10.816537)x4=(0.816537)−(1.400584672−(−0.202683052)−2.366926−1.224684246)x4=0.816537−(1.603267724−3.591610246)
x4=0.816537+0.446392457x4=1.262929457
The fourth approximation with six decimal places is x4=1.262929 .
Therefore, the roots with six decimal places are :-
x2≈1.333333
x3≈0.816537
x4≈1.262929
Chapter 4 Solutions
Single Variable Calculus: Concepts and Contexts, Enhanced Edition
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