1. ## Newton's method

I need help on a Newton's Method problem. I have to "Use Newton's method to approximate the given number correct to eight decimal places".

The number is the hundredth root of 100 (100^(1/100)). When done on calculator, the result is 1.04712855. I've tried the approximation but am not able to get close.

The method is Xn+1=Xn-(f(n)/f'(n))

Thanks

2. So what are you using for f(x) and f'(x)? Can you show me at least one iteration of what you've been trying so I can see where you went wrong?

3. From the examples I've seen on the book, f(x) = x^(100)-100 and
f'(x) = 100x^99

I've tried using n=1, n=2, n=3, also tried big numbers and cannot seem to come up with the true approximation.

4. Originally Posted by djo201
From the examples I've seen on the book, f(x) = x^(100)-100 and
f'(x) = 100x^99

I've tried using n=1, n=2, n=3, also tried big numbers and cannot seem to come up with the true approximation.
Well you are using the correct f and f'. I started doing this with n=1 as the initial guess and you are right it will take forever to converge. I found a cool website that give you 9 iterations of Newton's method for any equation and an initial guess.

Newton's Method

You can see that a guess of 1 or 2 won't get close after 9 iterations. 1.5 is much closer and 1.1 gets very close. So I guess the point of this is you have to make an approximation without a calculator, so using 1.1 needs to come from somewhere. If you graph the equation you can see that it's just above 1. Did your teacher make restrictions on how close your initial guess could be?

5. We have no restrictions on how to look for n. The only thing the professor said was to try using the Intermediate Value Theorem to "have an idea" of where the root or zero is located.

I guess the best number on this case is 1.1 indeed.

Thanks for your help and for the very useful site.

6. Originally Posted by djo201
We have no restrictions on how to look for n. The only thing the professor said was to try using the Intermediate Value Theorem to "have an idea" of where the root or zero is located.

I guess the best number on this case is 1.1 indeed.

Thanks for your help and for the very useful site.
This particular example converges very slowly because of the large exponent, or small exponent depending on how you're looking at it. This causes slope approximations to be poorer than usual since the slope is changing so rapidly compared to say f = x^2.

The formula is easy to follow but like I said just make sure you get a good starting guess so you don't go crazy doing the iteration 100 times.

7. Thanks for all your help.