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Newton's Method to find xn

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I am not very familiar with Newton's method, could you please show the steps so that I can learn how to solve this problem? thank you

asked Oct 2, 2014 in CALCULUS by anonymous

2 Answers

0 votes

(a)

The function is g(x) = x² - cosx

Given X1 = 1

According to newton Raphson Method  image

g(1) = 1² - cos 1

g(1) = 1 - 0.9998

g(1) = 0.0002

g'(x) = 2x + sin x

g'(1) = 2(1) + sin 1

g'(1) = 2 + 0.01745

g'(1) = 2.01745

image

image

 

According to newton Raphson Method X2 = 0.9999

 

answered Oct 2, 2014 by friend Mentor
0 votes

(b)

The function is g(x) = X7 - 4

Given X1 = 1

According to newton Raphson Method  image

g'(x) = 7X6

Now solve for  X2

g(1) = (1)7 - 4 = 1 - 4 = -3

g'(1) = 7(1)= 7

image

X2 = 1 - (-3/7) =1.428

 

Now solve for  X3

g(1.428) = (1.428)7 - 4 = 12.108 - 4 = 8.108

g'(1.428) = 7(1.428)= 59.35

image

X3 = 1.428 - (8.108/59.35) =1.291

 

Now solve for  X4

g(1.291) = (1.291)7 - 4 = 5.977 - 4 =1.977

g'(1.291) = 7(1.291)= 32.408

image

X4  = 1.291 - (1.977/32.408) =1.229

 

According to newton Raphson Method the value of X4 = 1.229

answered Oct 2, 2014 by friend Mentor

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