# Math Help - newton's method problem

1. ## newton's method problem

Use Newton's method to find the 100th root of 100.

so I got f(x)=x^100-100
used x=1 as my starting point

x2=1-(1^100-100)/(100*1^99)
=1-(-99/100)
=1.99

then I did it a couple more times and ended up with 1.97 which is not right. What did I do wrong?

eta: is their a tutorial or something for the notation on these forums?

2. Originally Posted by muffinman987
Use Newton's method to find the 100th root of 100.

so I got f(x)=x^100-100
used x=1 as my starting point

x2=1-(1^100-100)/(100*1^99)
=1-(-99/100)
=1.99

then I did it a couple more times and ended up with 1.97 which is not right. What did I do wrong?

eta: is their a tutorial or something for the notation on these forums?
hey mate,

Newtons method is a powerful tool used to iterate towards the solution of a non-linear equation, however it is not globally convergent and is based on certain attributes that the function itself must satisfy; in this case the strength of your derivative of the function is what is causing your solution to move away from the actual solution,

Thankfully there is a method or strategy to overcome and or dampen the divergence by attempting to force the newton step to be globally convergent which I will attempt to explain,

x^(n+1) = x^(n) + dx^(n)

where dx(n) = -f(x^(n))/f'(x^(n));

here we alter the current strategy so that a dampening factor is included so that
| dx^(n+1) | < | dx^(n) | --> the solution is converging

Thus,

x^(n+1) = x^(n) + g(damp^(n))*dx^(n)
where damp or dampening factor is bounded between 0 and 1.
The choice of the function g(damp^(n)) is dependant upon the problem, but the simplest is,
g(damp^(n)) = damp^(n)

in terms of its implementation, a general method is at each iteration set damp^(n) = 1 and thus compute the newton step normally, and if the new dx is greater than the older dx, set damp^(n) = (1/2)*damp^(n) and repeat until
|dx new | < |dx old|

Give this strategy a go, and if you have any trouble in its implementation and or what I've stated, please let me know and I will be more than happy to assist further,

Regards,

David