Asked by Elizabeth
                Use newtons method and the function f(x)=x^2-a to derive the mechanics rule.
Mechanics Rule for approaching sqrt(a), a>0, is
Xn+1=1/2(Xn+a/Xn), n=1,2,3,...
            
        Mechanics Rule for approaching sqrt(a), a>0, is
Xn+1=1/2(Xn+a/Xn), n=1,2,3,...
Answers
                    Answered by
            Count Iblis
            
    Newton's method for the root of f(x) is:
x_{n+1} = x_{n} - f(x_{n})/f'(x_{n})
= x_{n} - (x_{n}^2 - a)/(2 x_{n}) =
1/2 (x_{n} + a/(x_{n}))
This rule is not so convenient, because you need to divide by x_{n} at each step (division is slower than multiplication). A rule without division can be obtained by writing the square root as the root of the function:
f(x) = x^(-2) - 1/a
Newton's method then yields:
x_{n+1} = x_{n} -
( x_{n}^(-2) - 1/a)/(-2x_{n}^(-3)) =
1/2 x_{n} (3 - x_{n}^2/a)
So, in each step you now only have a division by a, so you only have to compute 1/a once.
    
x_{n+1} = x_{n} - f(x_{n})/f'(x_{n})
= x_{n} - (x_{n}^2 - a)/(2 x_{n}) =
1/2 (x_{n} + a/(x_{n}))
This rule is not so convenient, because you need to divide by x_{n} at each step (division is slower than multiplication). A rule without division can be obtained by writing the square root as the root of the function:
f(x) = x^(-2) - 1/a
Newton's method then yields:
x_{n+1} = x_{n} -
( x_{n}^(-2) - 1/a)/(-2x_{n}^(-3)) =
1/2 x_{n} (3 - x_{n}^2/a)
So, in each step you now only have a division by a, so you only have to compute 1/a once.
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.