Newton's Method General Formula

1. Jun 17, 2011

crybllrd

1. The problem statement, all variables and given/known data

The statement $\sqrt[4]{a}=x$ means that $x^{n}=a$.

Using this, we can approximate the radical $\sqrt[n]{a}$ by approximating the

solution to the equation $x^{n}-a=0$.

Consider the function $f(x)=x^{n}-a$.

We can use Newton's Method to approximate where f(x)=0 and thus approximate the radical

$\sqrt[n]{a}$.

a) Use Newton's Method with the function $f(x)=x^{n}-a$

to obtain a general formula approximating $\sqrt[4]{a}$.

b) Enter $\sqrt[4]{100}$ using your calculator and give the approximation to the

c) Use the formula found in (a) and make a table of values to approximate

$\sqrt[4]{100}$ to the same accuracy as your calculator. Use 3

d) How many iterations are required to obtain this same accuracy?

e)The fundamental theorem of algebra guarantees 4 solutions to x^4-100=0.

you just found one. Are there more real solutions? Use your tools of calculus to sustain

2. Relevant equations

$x_{2}=x_{1}-\frac{f(x_{1}}{f'(x_{1}}$

3. The attempt at a solution

a)$f(x)=x^{n}-a$

Am I supposed to assume a is constant here? If so, then:

$f '(x)=nx^{n-1}$

$x_{2}=x_{1}-\frac{x^{n}-a}{nx^{n-1}}$

b) Easy enough, plugged it in to the calc to get 3.16227766.

c) I want to make sure I have part a) right before making a chart.

d) This will be simple after part c)

e) I can tell I will be stuck on this final part. Any tips to help me get started?

2. Jun 17, 2011

I like Serena

Hi crybllrd!

You have (a) right!
Just fill in n=4 and a=100 and you're good to go!

Cheers!

3. Jun 17, 2011

crybllrd

I'm working on part c). She has a set up a chart like this:

I guess I'm confused as to where I use n=4 and n=iteration

4. Jun 17, 2011

I like Serena

Yes. It's a pity n is used for 2 different things.

Let's just say that the n in your formula needs to be replaced by 4, while all the occurrences of n in your table should have been named k or i.

5. Jun 17, 2011

crybllrd

Oh ok, not sure why my teacher would do that to us :P

However, my numbers are not converging.

My n=1 is 3.175925926

n=2 is 2.986439461

n=3 is 3.191986488

n=4 is 2.970703369

I stopped there because they should be converging to 3.16227766.

Here's what I am using:

f(x)=x^4-100

f '(x)=4x^3

so, $x-\frac{x^{4}-100}{4x^{3}}$, where x is the previous approximation (x1=3)

Am I missing something?

6. Jun 17, 2011

I like Serena

Not sure what you did, because starting from n=2 I get a different sequence, which is converging.

To make it explicit:

$x_{k+1} = x_k-\frac{{x_k}^{4}-100}{4{x_k}^{3}}$

and

$3.1759-\frac{3.1759^{4}-100}{4 \cdot 3.1759^{3}} = 3.162$

7. Jun 17, 2011

crybllrd

Not sure what I did either, but I redid it and got it in 3 iterations.

For the final part, I should have 4 solutions to x^4-100=0,

3.16227766 being one of them, and -3.16227766 would be another.

How would I go about getting the others? I'm assuming they are not real solutions.

8. Jun 17, 2011

I like Serena

Quite right.

The other 2 roots are:
3.16227766 i
and
-3.16227766 i

where i is the imaginary constant, or sqrt(-1).

What they would have intended (I think), is that you can find the negative root from the fact that looking at the graph of y = x^4 - 100, you'll see 2 real solutions. ;)