Using Newtons Method to Solve Equations | Homework Help

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Homework Statement



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

[tex]x^{5}+x^{3}+x=1[/tex]

which is the same as:

[tex]x^{5}+x^{3}+x-1=0[/tex]


Homework Equations



[tex]x_{2} = x_{1}-\frac {f(x)}{f'(x)}[/tex]

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

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:

[tex]x_{2} = 1 - \frac {f(x)}{f'(x)} = 1- \frac {2}{9} = \frac {7}{9}[/tex]

and after 5 iterations I get:

[tex]x_{5} = .6368843716[/tex]

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
 
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Another typo.
 
I've just spent the last 5 months play with Newton's method... The general method is
[tex] x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}[/tex]
Where in your case:
[tex] f(x)=x^{5}+x^{3}+x-1[/tex]
The derivative of this is:
[tex] f'(x)=5x^{4}+3x^{2}+1[/tex]
And in your case Newton's method is
[tex] x_{n+1}=x_{n}-\frac{x_{n}^{5}+x_{n}^{3}+x_{n}-1}{5x_{n}^{4}+3x_{n}^{2}+1}[/tex]
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.