Newton's Roots (1 Viewer)

[kimmeh]

I dont know how to start on this question:

Using newton's approximation method, find, correct to 2 dp, then continue the process to obtain an approximation correct to 8 dp

a) sqrt.(13)

 

[Estel]

x = rt(13)
x^2 - 13 = 0

f(x) = x^2 - 13
so x = 4 - f(4)/f'(4)
= 3.5

Repeating
x = 3.5 - f(3.5)/f'(3.5)
= 3 - 0.107...
= 3.607142857...

Again,
x = 3.60 - f(3.60)/f'(3.60) [use calc saved]
= 3.605551627, repeating
= 3.605551275, repeating
= 3.605551275

Since the result is the same, to 8 dp
x = 3.60555128

[wow, didn't realise it would converge so fast]
As for whether you are allowed to leave it at that, I have no clue as I just learnt Newton's method [just for u ]

 

[Xayma]

Repeating
x = 3.5 - f(3.5)/f'(3.5)
= 3 - 0.107...
= 3.607142857...

Again,
x = 3.60 - f(3.60)/f'(3.60) [use calc saved]
= 3.605551627, repeating
= 3.605551275, repeating
= 3.605551275

x=3.61 not 3.60 for those estimations . You had to find it to 2 dp which was 3.61

 

[Estel]

ooooooops!
I was reading the example in the book which said 8 dp for rt 2.... ooh dear