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Slav-nsk [51]
3 years ago
9

Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equ

ation. (Round your answer to four decimal places.) 2 x − x2 + 1 = 0, x1 = 2
Physics
1 answer:
sdas [7]3 years ago
8 0

Answer:

x_{3}=2.35

Explanation:

Given x^2-2x-1=0,x_1=2

From Newton's method

x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}

f(x)=x^2-2x-1

f'(x)=2x-2

Now

x_{2}=x_1-\dfrac{f(x_1)}{f'(x_1)}

f(x)=x^2-2x-1

f(2)=2^2-2\times 2-1

f(x)=-1

f'(2)=2x-2

f'(1)=2\times 2-2

f'(1)=2

x_{2}=2+\dfrac{1}{2}

x_{2}=2.5

x_{3}=x_2-\dfrac{f(x_2)}{f'(x_2)}

f(2.5)=2.5^2-2\times 2.5-1

f(2.5)=0.45

f'(2.5)=2\times 2.5-2

f'(2.5)=3

x_{3}=2.5-\dfrac{0.45}{3}

So

x_{3}=2.35

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