Asked by Betty
Use Newtons Method to find 13^(1/4) correct to four decimal places. I know the formula X_(n+1)= X_n - [f(X_n)]/[f'(X_n)]. I am not sure how to go on from there. I made the equation into y=X^(1/4), but I can't seem to figure out how to go on. Please help?
Thanks in advance!
Thanks in advance!
Answers
Answered by
Reiny
let x = 13^(1.4)
x^4 = 13
x^4 - 13 = 0
let y f(x) = x^4 - 13
dy/dx = 4x^3
using your formula:
x<sub>n+1</sub> = x<sub>n</sub> - f(x<sub>n</sub>/f ' (x<sub>n</sub>)
= x - (x^4 - 13)/(4x^3)
= (4x^4 - x^4 + 13)/(4x^3)
= (3x^4 + 13)/(4x^3)
start with x = 1.5
oldx newx
1.5 --> 2.0879...
2.0879... ---> 1.923..
1.923.. ---> 1.89928...
1.89928.. ---> 1.898829...
1.898829 --> 1.898828922
well , how about that ?
my calaculator says 13^(1/4) = 1.898828922
x^4 = 13
x^4 - 13 = 0
let y f(x) = x^4 - 13
dy/dx = 4x^3
using your formula:
x<sub>n+1</sub> = x<sub>n</sub> - f(x<sub>n</sub>/f ' (x<sub>n</sub>)
= x - (x^4 - 13)/(4x^3)
= (4x^4 - x^4 + 13)/(4x^3)
= (3x^4 + 13)/(4x^3)
start with x = 1.5
oldx newx
1.5 --> 2.0879...
2.0879... ---> 1.923..
1.923.. ---> 1.89928...
1.89928.. ---> 1.898829...
1.898829 --> 1.898828922
well , how about that ?
my calaculator says 13^(1/4) = 1.898828922
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