# Newton's Method General Formula

• crybllrd
In summary, the homework statement is that the equation x^{n}-a=0 can be approximated by approximating the solution to x^{n}-a=0 with Newton's Method. This can be done by solving for f(x) where f(x) is x^{n}-a. Once f(x) is found, a table can be made that approximates \sqrt[4]{100} with the same accuracy as a calculator. Finally, after n iterations, the approximation should be 3.16227766.
crybllrd

## Homework Statement

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

## Homework Equations

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

## 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 into 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?

Hi crybllrd!

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

Cheers!

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

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

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.

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?

crybllrd said:
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?

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

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

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.

crybllrd said:
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.

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. ;)

## 1. What is Newton's Method General Formula?

Newton's Method General Formula is a mathematical algorithm used to find the roots of a given function. It is also known as the Newton-Raphson Method and is named after Sir Isaac Newton and Joseph Raphson.

## 2. How does Newton's Method General Formula work?

The formula uses an iterative process to approximate the roots of a given function. It starts with an initial guess and then uses the tangent line at that point to find a better approximation. This process is repeated until the desired level of accuracy is achieved.

## 3. What is the significance of Newton's Method General Formula?

Newton's Method General Formula is an important tool in calculus and numerical analysis. It is widely used in various fields of science, engineering, and mathematics to approximate solutions to complex equations and systems of equations.

## 4. What are the advantages of using Newton's Method General Formula?

One of the major advantages of this formula is that it can converge to the root of a function quickly, often in just a few iterations. It is also relatively easy to implement and can handle a wide range of functions.

## 5. What are the limitations of Newton's Method General Formula?

One limitation of this formula is that it may fail to converge or may converge to a different root if the initial guess is not chosen carefully. It also requires knowledge of the derivative of the function, which may not always be readily available.

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