Newton's Method to find 100^(1/100) correct to 8 dec. places

[djo201]

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

 

[royhaas]

Re: Newton's Method

What are you using for the function \(\displaystyle f(x)\)?

 

[djo201]

Re: Newton's Method

I'm using f(x)=x^100-100

Am I incorrect?

 

[Deleted member 4993]

Re: Newton's Method

I'm using f(x)=x^100-100

Am I incorrect?

No - that is correct - but this is what I call "pathological" problem. This problem shows the "problem" with Newton's method.

However, it can be done.

Hint:

Instead of trying to find the hundreth root in one shot - break it up into factors of 100

 

[PAULK]

Re: Newton's Method

I'm using f(x)=x^100-100

Am I incorrect?

No - that is correct - but this is what I call "pathological" problem. This problem shows the "problem" with Newton's method.

However, it can be done.

Hint:

Instead of trying to find the hundreth root in one shot - break it up into factors of 100

It should work. But a lot depends on your choice of x[0]. Choose something bad and it explodes. Choose something good and it converges nicely.

P.S. Excel is very good for this. I tried it with x[0] = 1 and got nothing good after 25 iterations. But x[0] = 1.2 worked beautifully.

 

[royhaas]

Re: Newton's Method

If \(\displaystyle x_0=1, x_1=1.99\), so convergence will be real slow.

 

[Deleted member 4993]

Re: Newton's Method

the way I propose

find

\(\displaystyle x_1 \,= \,100^{\frac{1}{5}\)

Then

\(\displaystyle x_2 \,= \,(x_1)^{\frac{1}{5}\)

finally

\(\displaystyle x \,= \,(x_2)^{\frac{1}{4}\)

Convergence is pretty fast in each case - with beginning value of 1 (for the first one it would faster to start with 3)

 

[PAULK]

Re: Newton's Method

If \(\displaystyle x_0=1, x_1=1.99\), so convergence will be real slow.

Yes, exactly what I tried, with the same results.