# Homework Help: Newtons Method - A little help

1. Aug 2, 2010

### Asphyxiated

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

Solve each equation using newtons method. The problem I am working on right now is:

$$x^{5}+x^{3}+x=1$$

which is the same as:

$$x^{5}+x^{3}+x-1=0$$

2. Relevant equations

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

with x1 being the first guess to what an f(x)=0 would be.

3. The attempt at a solution

So I put the equation into my calculator to graph it and It is very clear from the graph that f(x)=0 on the x>0 side so I choose x1=1 and continue like so:

$$x_{2} = 1 - \frac {f(x)}{f'(x)} = 1- \frac {2}{9} = \frac {7}{9}$$

and after 5 iterations I get:

$$x_{5} = .6368843716$$

which looks right to me but when I check the answer in the back of the book it says that it should be -.63688, so just the opposite of what I got. The graph of the equation clearly shows that f(x)=0 on x>0 so did I do something wrong or does the book have another typo?

thanks

2. Aug 2, 2010

### Char. Limit

Another typo.

3. Aug 3, 2010

### hunt_mat

I've just spent the last 5 months play with Newton's method... The general method is
$$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$$
$$f(x)=x^{5}+x^{3}+x-1$$
The derivative of this is:
$$f'(x)=5x^{4}+3x^{2}+1$$
And in your case Newton's method is
$$x_{n+1}=x_{n}-\frac{x_{n}^{5}+x_{n}^{3}+x_{n}-1}{5x_{n}^{4}+3x_{n}^{2}+1}$$
So all you have to do is plug numbers in. Newtons method converges very quickly in my experience, 2 or 3 iterations will usually do the trick.

4. Aug 3, 2010

### hunt_mat

As a check, plug your solution back into the equation and check.