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Use Newton's method to approximate a root of the equation

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cos(x^2+3)=x^3 as follows . Let x_1=1 be the initial approximation. x_2?

asked Nov 13, 2014 in CALCULUS by anonymous

1 Answer

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Given equation :

x³ - cos(x²+3) = 0

Let f(x) = x³ - cos(x²+3)

x1 = 1

Find x2 = ?

f '(x) = 3x3-1 - (- sin(x²+3)) (2x)

f '(x) = 3x2 + 2xsin(x2+3)

According to Newton Method  image

x2 = x1 – [ f(x1) / f’(x1) ]

f(x1) = (x1)³ - cos(x1²+3)

f(x1) = (1)³ - cos(1²+3)

=  1 - cos4

=  1 - 0.99756

= 0.00244

f '(x1) =  3(x1)2 + 2x1sin(x12+3)

=  3(1)2 + 2*1*sin(12+3)

=  3 + 2sin(4)

=  3 + 0.1395

=  3.1395

x2 = 1 – [ 0.00244 / 3.1395 ]

x2 = 1 – 0.0007772

x2 = 0.999223

Solution :

x2 = 0.999223

 

answered Nov 13, 2014 by Shalom Scholar

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